Math 602
1
1
Partial derivatives and Chain rule
Question 1: Let u be a solution to the PDE ∂t u(x, t) + 12 ∂x u2 (x, t) − ν∂xx u(x, t) = 0, x ∈
(−∞, +∞), t > 0.R
x
(a) Let ψ(x, t) = −∞ ∂t u(ξ, t)dξ + 21 u2 (x, t) − ν∂x u(x, t). Compute ∂x ψ(x, t).
Solution: The definition of ψ implies that
Z x
1
∂t u(ξ, t)dx + u2 (x, t) − ν∂x u(x, t))
∂x ψ(x, t) = ∂x (
2
−∞
1
= ∂t u(x, t)x + ∂x u2 (x, t) − ∂xx u(x, t) = 0
2
i.e., ∂x ψ(x, t) = 0. This means that ψ depends on t only.
1
(b) Let φ(x, t) := e− 2ν
Rx
−∞
u(ξ,t)dξ
. Compute ∂t φ, ∂x φ, and ∂xx φ.
Solution: The definition of φ, together with the chain rule, implies that
Z x
Rx
1
1
u(ξ, t)dξ e− 2ν −∞ u(ξ,t)dξ
∂t φ(x, t) = ∂t −
2ν −∞
Z x
Rx
1
1
= −
∂t u(ξ, t)dξ e− 2ν −∞ u(ξ,t)dξ
2ν −∞
and
Z x
Rx
1
1
u(ξ, t)dξ e− 2ν −∞ u(ξ,t)dξ
∂x φ(x, t) = ∂x −
2ν −∞
Rx
1
1
= − u(x, t) e− 2ν −∞ u(ξ,t)dξ
2ν
and
∂xx φ(x, t) =
2
Rx
Rx
1
1
1
1
u(ξ,t)dξ
− 2ν
−∞
− ∂x u(x, t) e
+ − u(x, t) e− 2ν −∞ u(ξ,t)dξ
2ν
2ν
(c) Compute ∂t φ − ν∂xx φ, assuming ψ(x, t) = 0.
Solution: The above computations give
Rx
1
1
1 2
−ν∂xx φ(x, t) = −
−ν∂x u(x, t) + u (x, t) e− 2ν −∞ u(ξ,t)dξ
2ν
2
In conclusion
Z x
Rx
1
1
1 2
∂t φ − ν∂xx φ = −
∂t u(ξ, t)dξ + u (x, t) − ν∂x u(x, t) e− 2ν −∞ u(ξ,t)dξ
2ν
2
−∞
Rx
1
1
− 2ν
u(ξ,t)dξ
−∞
= − ψ(x, t)e
.
2ν
This means ∂t φ − ν∂xx φ = 0.
Question 2: Let d be positive integer and let u : Rd ×R+ −→ R be the solution to the nonlinear
PDE ∂t u(x, t) − a∆u(x, t) + b(∇u(x, t))·(∇u(x, t)) = 0, with u(x, 0) = f (x), where a > 0 and
b
b 6= 0 are real numbers, and f (x) is a smooth function. Let v(x, t) = e− a u(x,t) . (i) Compute
Pd
∇v(x, t) and ∇·(∇v(x, t)) (Hint: compute ∂xi v(x, t) for i = 1, . . . , d, then i=1 ∂xi xi v(x, t). It
is the same question as question 1 in mdt1; it is worded differently.)
Solution: The chain rule gives
b b
∂xi v(x, t) = − e− a u(x,t) ∂xi u(x, t),
a
Math 602
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and
d
X
d
d
X
X
b b
b2 b
∂xi xi v(x, t) = − e− a u(x,t)
∂xi xi u(x, t) + 2 e− a u(x,t)
∂xi u(x, t)∂xi u(x, t).
a
a
i=1
i=1
i=1
This implies that
b b
∇v(x, t) = − e− a u(x,t) ∇u(x, t),
a
and
b2 b
b b
∆v(x, t) = − e− a u(x,t) ∆u(x, t) + 2 e− a u(x,t) (∇u(x, t))·(∇u(x, t)).
a
a
(ii) Compute ∂t v(x, t), ∂t v(x, t) − a∆v(x, t), and v(x, 0).
Solution: Using again the chain rule we obtain
b b
∂t v(x, t) = − e− a u(x,t) ∂t u(x, t).
a
Combining the above results gives
b b
∂t v(x, t) − a∆v(x, t) = − e− a u(x,t) (∂t u(x, t) − a∆u(x, t) + b(∇u(x, t))·(∇u(x, t))) = 0
a
b
i.e., v solves the heat equation ∂t v(x, t) − a∆v(x, t) = 0 with initial condition v(x, 0) = e− a f (x) .
(iii) Recall that the solution to the heat equation ∂t φ(x, t)−a∆φ(x, t) = 0 with initial condition
R
kx−yk2
φ(x, 0) = ψ0 (x) is given by ψ0 (x)(4πat)−d/2 Rd e− 4at dy. Give the integral representation
of u(x, t) in terms of a, b, and f .
b
Solution: Since ∂t v(x, t) − a∆v(x, t) = 0 with initial condition v(x, 0) = e− a f (x) , the integral
representation of the function u is then given by
Z
kx−yk2
b
a
e− 4at dy
u(x, t) = − log(v(x, t)), with, v(x, t) = e− a f (x) (4πat)−d/2
b
Rd
Question 3: Let u be a vector field in Rd , (d is the space dimension). Let p be a scalar field
in Rd .
(a) Using the product rule, express div(pu) in terms of ∇·u and ∇p. (You may use divu or ∇·u
to denote the divergence of u and ∇p or gradp to denote the gradient of p. The space dimension
is d.)
Solution: Using the product rule we have
∇·(pu) =
d
X
∂i (pui ) =
i=1
d
X
ui ∂i p + p∂i ui
i=1
= u·∇p + p∇·u
In conclusion ∇·(pu) = u·∇p + p∇·u.
(b) Let u be a smooth vector field in Ω ⊂ Rd with zero divergence and zero
R normal component
at
the
boundary
of
Ω.
Let
p
be
a
smooth
scalar
field
in
Ω.
Compute
p(x)∇·u(x)dx and
Ω
R
u(x)·∇p(x)dx.
Ω
Solution: (i) Using (a) and the fact that u is divergence free, we have
Z
Z
Z
u·∇pdx = (∇·(pu) − p∇·u)dx =
∇·(pu)dx.
Ω
Ω
Ω
The divergence theorem (fundamental theorem of calculus in d space dimension) implies that
Z
Z
u·∇pdx =
pu·nds,
Ω
∂Ω
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Rwhere ∂Ω denotes the boundary of Ω. Since u·n = 0 at the boundary, we finally infer that
u·∇pdx = 0.
Ω
R
(ii) Note finally that Ω p∇·udx = 0 since ∇·u = 0.
Question 4: Let φ be a smooth scalar field in Rd , (d is the space dimension). (a) Prove that
∇φ·∂i (∇φ) = k∇φk∂i k∇φk. Give all the details.
Solution: Using the product rule, we have
d
X
d
1X
1
∇φ·∂i (∇φ) =
∂j φ∂i (∂j φ) = ∂i (
(∂j φ)2 ) = ∂i ( k∇φk2 ) = k∇φk∂i k∇φk.
2
2
j=1
j=1
p
(b) Let > 0. Show that ∇·( k∇φk2 + ei ) = √
∇φ
·∂i (∇φ)
k∇φk2 +
where ei is the unit vector in
the direction i.
Solution: Using the chain rule, we have
p
p
1
1
∂i (k∇φk2 + )
∇·( k∇φk2 + ei ) = ∂i ( k∇φk2 + ) = p
2 k∇φk2 + k∇φk
=p
∂i k∇φk.
k∇φk2 + Then using (a) we infer that
p
∇φ
·∂i (∇φ).
∇·( k∇φk2 + ei ) = p
k∇φk2 + (c)
that φ(x) = 0 for all kxk ≥ R where R is a positive real number. Compute
R Assume
√ ∇φ 2 ·∂i (∇φ)dx, for all i = 1, . . . , d. Give all the details.
Rd
k∇φk +
Solution: Using (b) we have
Z
Z
p
∇φ
p
·∂i (∇φ)dx =
∇·( k∇φk2 + ei )dx,
k∇φk2 + Rd
Rd
where ei is the unit vector in the direction i. The divergence theorem together with the assumption
that φ(x) = 0 for all kxk ≥ R implies that
Z
∇φ
p
·∂i (∇φ)dx = 0.
k∇φk2 + Rd
Question 5: Let y(x, t) = x cos(2t + log(|x|)). Compute ∂tt y + x2 ∂xx y − x∂x y + 6y.
Solution: This exercise is meant to check whether you understand the notion of partial derivatives
and the chain rule
∂tt y(x, t) = −4x cos(2t + log(|x|)) = −4y(x, t),
∂x y(x, t) = cos(2t + log(|x|)) − x sin(2t + log(|x|))
1
= cos(2t + log(|x|)) − sin(2t + log(|x|)),
x
1
1
∂xx y(x, t) = − sin(2t + log(|x|)) − cos(2t + log(|x|))
x
x
In conclusion
∂tt y + x2 ∂xx y − x∂x y + 6y = −4x cos(2t + log(|x|)) − x sin(2t + log(|x|))
− x cos(2t + log(|x|)) − x cos(2t + log(|x|)) + x sin(2t + log(|x|)) + 6x cos(2t + log(|x|))
= 0,
that is to say, y(x, t) solve the PDE ∂tt y + x2 ∂y − x∂x y + 6y = 0.
Math 602
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Question 6: Let ∇× be the curl operator acting on vector fields: i.e., let A = (A1 , A2 , A3 ) :
R3 −→ R3 be a three-dimensional vector field over R3 , then ∇×A = (∂2 A3 − ∂3 A2 , ∂3 A1 −
∂1 A3 , ∂1 A2 − ∂2 A1 ). Accept as a fact that ∇·(A×B) = B·∇×A − A·∇×B for all smooth vector
3
fields A and B. Let
R Ω be a subset of R with a smooth boundary ∂Ω. Find an integration by
parts formula for Ω B·∇×Adx.
Solution: Using the divergence Theorem we infer that
Z
Z
Z
(B·∇×A − A·∇×B)dx =
∇·(A×B) =
Ω
Ω
(A×B)·nds.
∂Ω
which implies that
Z
Z
Z
B·∇×Adx =
Ω
A·∇×Bdx +
Ω
(A×B)·nds.
∂Ω
Question 7: Let u, f : R −→ R be two functions of class C 1 . (a) Compute ∂x f (u(x)).
Solution: Using the chain rule we obtain
∂x f (u(x)) = f 0 (u(x))∂x u.
where f 0 denotes the derive of f .
(b)
ψ : R −→ R be functions of class C 1 . Let F : R −→ R be defined by F (v) =
R v Let
0
f (t)ψ 0 (t)dt. Use (a) to compute ∂x (F (u(x)) − ∂x (f (u(x)))ψ 0 (u(x)).
0
Solution: Using the chain rule we obtain
∂x (F (u(x)) = F 0 (u(x))∂x u(x) = f 0 (u(x))ψ 0 (u(x))∂x u(x) = ∂x (f (u(x)))ψ 0 (u(x)).
This means that ∂x (F (u(x)) = ∂x (f (u(x)))ψ 0 (u(x)).
(c) Using the notation of (a) and (b), assume that u(±∞) = 0 and compute
R +∞
−∞
∂x (f (u(x)))ψ 0 (u(x))dx.
Solution: Using (b) and u(±∞) = 0 we have
Z
+∞
∂x (f (u(x)))ψ 0 (u(x))dx =
Z
−∞
+∞
∂x (F (u(x)))dx = F (u(x))|+∞
−∞ = F (0) − F (0) = 0.
−∞
Question 8: Let φ(x, y) = cos(x + sin(x − y)). Compute ∂x φ(x, y) and ∂y φ(x, y). (Do not try
to simplify the results).
Solution: We apply the chain rule repeatedly
∂x φ(x, y) = − sin(x + sin(x − y))(1 + cos(x − y)).
∂y φ(x, y) = − sin(x + sin(x − y))(− cos(x − y)).
Question 9: Let φ = x4 − y 4 (a) ComputeR∆φ(x, y). (b) Consider the square Ω = [0, 1]×[0, 1]
and let Γ be the boundary of Ω. Compute Γ ∂n φdΓ.
Solution: (a) The definition ∆φ = ∂xx φ + ∂yy φ implies that
∆φ = ∂xx φ + ∂yy φ = 12x2 − 12y 2 = 12(x2 − y 2 ).
(b)The definition ∆φ = div(∇φ) and the fundamental theorem of calculus (also known as the
divergence theorem) imply that
Z
Z
Z
Z
Z
∂n φdΓ =
n·∇φdΓ =
div(∇φ)dΩ =
∆φdΩ =
12(x2 − y 2 )dxdy = 0.
Γ
Γ
Ω
Ω
Ω
Math 602
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Question 10: Let φ = sin(x) − sin(y) (a) Compute ∆φ(x,
y). (b) Consider the square Ω =
R
[0, 1]×[0, 1] and let Γ be the boundary of Ω. Compute Γ ∂n φdΓ.
Solution: (a) The definition ∆φ = ∂xx φ + ∂yy φ implies that
∆φ = ∂xx φ + ∂yy φ = − sin(x) + sin(y).
(b)The definition ∆φ = div(∇φ) and the fundamental theorem of calculus (also known as the
divergence theorem) imply that
Z
Z
Z
Z
Z
Z
∂n φdΓ =
n·∇φdΓ =
div(∇φ)dΩ =
∆φdΩ = −
sin(x)dxdy +
sin(x)dxdy = 0.
Γ
Γ
Ω
Ω
Ω
Ω
Question 11: Let φ(x, y) = log(2 + sin(x − y)). Compute ∂x φ(x, y) and ∂y φ(x, y). (Do not try
to simplify the results).
Solution: We apply the chain rule repeatedly
∂x φ(x, y) =
1
cos(x − y)
2 + sin(x − y)
∂y φ(x, y) = −
1
cos(x − y).
2 + sin(x − y)
R sin(x)
Question 12: Let φ(x, y) = 0
log(1+y 2 +z 2 )dz. Given α, β, γ ∈ R, compute ∂x φ(α, α+β)
and ∂y φ(α − β, γ). (Do not try to simplify the results).
Solution: Recalling the fundamental theorem of calculus
Z t
∂t
f (z)dz = f (t),
0
we apply the chain rule repeatedly
∂x φ(x, y) = log(1 + y 2 + sin(x)2 ) cos(x)
This means that
∂x φ(α, α + β) = log(1 + (α + β)2 + sin(α)2 ) cos(α)
Recalling that
Z
v
∂t
Z
v
f (s, t)ds =
u
∂t f (s, t)ds,
u
we apply the chain rule repeatedly
sin(x)
Z
∂y φ(x, y) =
0
This means that
Z
∂y φ(α − β, γ) =
0
2y
dz
1 + y2 + z2
sin(α−β)
2γ
dz.
1 + γ2 + z2
Question 13: Let φ = x2 + 2y 2 (a) Compute ∆φ(x, y). (b)
R Consider the disk of radius 1
centered at (0, 0) and let Γ be the boundary of Ω. Compute Γ ∂n φdΓ.
Solution: (a) The definition ∆φ = ∂xx φ + ∂yy φ implies that
∆φ = ∂xx φ + ∂yy φ = 2 + 4 = 6.
(b)The definition ∆φ = div(∇φ) and the fundamental theorem of calculus (also known as the
divergence theorem) imply that
Z
Z
Z
Z
Z
∂n φdΓ =
n·∇φdΓ =
div(∇φ)dΩ =
∆φdΩ = 6
dΩ = 6π,
Γ
because the surface of Ω,
Γ
R
Ω
Ω
dΩ, is equal to π.
Ω
Ω
Math 602
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6
Heat equation
Question 14: Let u be the solution of ∂t u = ∂xx u + 2, x ∈ (0, L), with ∂x u(0, t) = α,
∂x u(L, t) = 3, u(x, 0) = f (x).
RL
(a) Let α = 1. Compute 0 u(x, t)dx as a function of t.
RL
RL
Solution: dt 0 u(x, t)dx = ∂x u|L
0 + 2L. That is dt 0 u(x, t)dx = 2 + 2L. This implies
RL
RL
u(x, t)dx = (2 + 2L)t + 0 f (x)dx.
0
RL
(b) Let α be an arbitrary number. For which value of α, 0 u(x, t)dx does not depend on t?
RL
RL
Solution: The above computation yields 0 u(x, t)dx = (3 − α + 2L)t + 0 f (x)dx. This is
independent of t if (3 − α + 2L) = 0, meaning α = 3 + 2L.
Question 15: Let u solve ∂t u − ∂xx u = 5, x ∈ (0, L), with ∂x u(0, t) = α, ∂x u(L, t) = −3,
u(x, 0) = f (x).
RL
(a) Let α = 1. Compute 0 u(x, t)dx as a function of t.
Solution: Integrate the equation over (0, L):
Z
dt
L
Z
u(x, t)dx =
0
RL
L
Z
∂t u(x, t)dx =
0
L
∂xx udx + 5L = ∂x u|L
0 + 5L.
0
RL
RL
u(x, t)dx = (−4 + 5L)t + 0 f (x)dx.
RL
(b) Let α be an arbitrary number. For which value of α, 0 u(x, t)dx does not depend on t?
RL
RL
Solution: The above computation yields 0 u(x, t)dx = (−3 − α + 5L)t + 0 f (x)dx. This is
independent of t if (−3 − α + 5L) = 0, meaning α = −3 + 5L.
That is dt
u(x, t)dx = −4 + 5L. This implies
0
0
Question 16: Let u solve ∂t u − ∂x ((2x + 1)∂x u) = 3, x ∈ (0, L), with ∂x u(0, t) = α, ∂x u(L, t) =
−1, u(x, 0) = f (x).
RL
(a) Let α = 1. Compute 0 u(x, t)dx as a function of t.
Solution: Integrate the equation over the domain (0, L):
L
Z
dt
L
Z
u(x, t)dx =
0
L
Z
∂t u(x, t)dx =
∂x ((2x + 1)∂x u)dx + 3L
0
0
= (2L + 1)∂x u(L, t) − ∂x u(0, t) + 3L = −(2L + 1) − α + 3L
=L−1−α
RL
u(x, t)dx = (L − 2)t + 0 f (x)dx.
RL
(b) Let α be an arbitrary number. For which value of α, 0 u(x, t)dx does not depend on t?
RL
RL
Solution: The above computation yields 0 u(x, t)dx = (L − 1 − α)t + 0 f (x)dx. This is
independent of t if L − 1 − α = 0, meaning α = L − 1.
That is dt
RL
0
u(x, t)dx = L − 2. This implies
RL
0
Question 17: Let u solve ∂t u−∂x ((3x+1)∂x u) = −3, x ∈ (0, L), with ∂x u(0, t) = 1, ∂x u(L, t) =
α, u(x, 0) = f (x).
RL
(a) Compute 0 u(x, t)dx as a function of t.
Solution: Integrate the equation over the domain (0, L):
Z
dt
L
Z
u(x, t)dx =
0
L
Z
0
L
∂x ((3x + 1)∂x u)dx − 3L
∂t u(x, t)dx =
0
= (3L + 1)∂x u(L, t) − ∂x u(0, t) − 3L = (3L + 1)α − 1 − 3L
= (3L + 1)(α − 1)
That is dt
RL
0
u(x, t)dx = (3L+1)(α−1). This implies
RL
0
u(x, t)dx = (3L+1)(α−1)t+
RL
0
f (x)dx.
Math 602
(b) For which value of α the quantity
7
RL
u(x, t)dx does not depend on t?
RL
RL
Solution: The above computation yields 0 u(x, t)dx = (3L + 1)(α − 1)t + 0 f (x)dx. This is
independent of t if (3L + 1)(α − 1) = 0, meaning α = 1.
0
2
Question 18: Consider the differential equation − ddtφ2 = λφ, t ∈ (0, π), supplemented with the
boundary conditions φ(0) = 0, 3φ(π) = −φ0 (π).
(a) Prove that it is necessary that λ be positive for a non-zero solution to exist.
Solution: (i) Let φ be a non-zero solution to the problem. Multiply the equation by φ and integrate
over the domain.
Z π
Z π
φ2 (t)dt.
(φ0 (t))2 dt − φ0 (π)φ(π) + φ0 (0)φ(0) = λ
0
0
Using the BCs, we infer
π
Z
(φ0 (t))2 dt + 3φ(π)2 = λ
Z
π
φ2 (t)dt,
0
0
which means that λ is non-negative since φ is non-zero.
Rπ
(ii) If λ = 0, then 0 (φ0 (t))2 dt = 0 and φ(π)2 = 0, which implies that φ0 (t) = 0 and φ(π) = 0.
Rt
The fundamental theorem of calculus implies φ(t) = φ(π) + π φ0 (τ )dτ = 0. Hence, φ is zero if
λ = 0. Since we want a nonzero solution, this implies that λ cannot be zero.
(iii) In conclusion, it is necessary that λ be positive for a nonzero solution to exist.
(b) Find the equation that λ must solve for the above problem to have a nonzero solution.
Solution: Since λ is positive, φ is of the following form
√
√
φ(t) = c1 cos( λt) + c2 sin( λt).
The boundary
condition
φ(0) = 0 implies
c1 = 0. The other boundary condition φ0 (π) = −3φ(π)
√
√
√
implies λc2 cos( λπ) = −3c2 sin( λπ). The constant c2 cannot be zero since we want φ to be
nonzero; as a result, λ must solve the following equation
√
√
√
λ cos( λπ) + 3 sin( λπ) = 0,
for a nonzero solution φ to exist.
Question 19: Consider the Laplace equation −∆u = 0 in the rectangle {(x, y); x ∈ [0, L], y ∈
y
), ∂y u(x, 0) = 0,
[0, H]} with the boundary conditions u(0, y) = 0, u(L, y) = 3 cos( 52 π H
u(x, H) = 0. Solve the equation using the method of separation of variables. (Give all the
details.)
Solution: Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies
φ”(x)
ψ”(y)
0
φ(x) = − ψ(y) = λ. Observe that ψ (0) = 0 and ψ(H) = 0. The energy technique applied to
−ψ”(y) = λψ(y) gives
Z
H
Z
−ψ”(y)ψ(y)dy =
0
H
0
2
0
0
Z
ψ (y) dy − ψ (H)ψ(H) + ψ (0)ψ(0) = λ
0
H
ψ(y)2 y,
0
RH
RH
which implies 0 ψ 0 (y)2 dy = λ 0 ψ(y)2 y since ψ 0 (0) = 0 and ψ(H) = 0. This in turn implies
that λ is nonnegative. Actually λ cannot be zero since it would mean that ψ = 0, which would
0
contradict the fact that the solution u is nonzero (λ = 0 ⇒ ψ
ψ(y) = ψ(H) = 0 for
√ (y) = 0 ⇒ √
all y ∈ [0, H]). As a result λ is positive and ψ(y) = a cos( λy) + b sin(
√ λy). The Neumann
condition
at
y
=
0
gives
b
=
0.
The
Dirichlet
condition
at
H
implies
cos(
λH) = 0, which implies
√
y
λH = (n + 21 )π, where n is any integer. This means that ψ(y) = a cos((n + 21 )π H
). The fact
√
√
that λ is positive implies φ(x) = c cosh( λx) + d sinh( λx). The boundary condition at x = 0
implies c = 0. Then
y
x
u(x, y) = A cos((n + 21 )π ) sinh((n + 12 )π ).
H
H
Math 602
8
The boundary condition at x = L gives
5 y
y
L
3 cos( π ) = A cos((n + 21 )π ) sinh((n + 12 )π ),
2 H
H
H
5 πL
which implies n = 2 and A = sinh−1 2H , i.e.,
sinh
5
sinh
5
u(x, y) = 3
2 πx
H
2 πL
H
5 y
cos( π ).
2 H
Question 20: Let k : [−1, +1] −→ R be such that k(x) = 1, if x ∈ [−1, 0] and k(x) = 2
if x ∈ (0, 1]. Solve the boundary value problem −∂x (k(x)∂x T (x)) = 0 with T (−1) = 0 and
∂x T (1) = 1.
(i) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 1 and k + (0) = 2.
(ii) Solve the problem, i.e., find T (x), x ∈ [−1, +1].
Solution: On [−1, 0] we have k − (x) = 1, which implies ∂xx T − (x) = 0. This in turn implies
T − (x) = a + bx. The Dirichlet boundary condition at x = −1 implies T − (−1) = 0 = a − b. This
gives a = b and T − (x) = a(1 + x).
We proceed similarly on [0, +1] and we obtain T + (x) = c + dx. The Neumann boundary condition
at x = +1 gives ∂x T + (+1) = 1 = d and T + (x) = c + x.
The interface conditions T − (0) = T + (0) and k − (0)∂x T − (0) = k + (0)∂x T + (0) give
a = c,
In conclusion
and
(
2(1 + x)
T (x) =
2+x
a = 2.
if x ∈ [−1, 0],
if x ∈ [0, +1].
Question 21: Let u solve ∂t u − ∂x ((sin(x) + 2)∂x u) = g(x)e−t , x ∈ (0, L), with ∂x u(0, t) =
sin(L) + 2, ∂x u(L, t) = 2, u(x, 0) = f (x), where f and g are two smooth functions.
RL
d
(a) Compute dt
u(x, t)dx as a function of t.
0
Solution: Integrate the equation over the domain (0, L) and apply the fundamental Theorem of
calculus:
Z
Z L
Z L
Z L
d L
u(x, t)dx =
∂t u(x, t)dx =
∂x ((sin(x) + 2)∂x u)dx + e−t
g(x)dx
dt 0
0
0
0
Z L
−t
= (sin(L) + 2)∂x u(L) − (sin(0) + 2)∂x u(0) + e
g(x)dx
0
= (sin(L) + 2)2 − 2(sin(L) + 2) + e−t
Z
L
g(x)dx
0
= e−t
Z
L
g(x)dx.
0
That is
d
dt
(b) Use (a) to compute
RL
0
Z
0
L
u(x, t)dx = e−t
Z
L
g(x)dx.
0
u(x, t)dx as a function of t.
Math 602
9
Solution: Applying the fundamental Theorem of calculus again gives
L
Z
L
Z
T
Z
u(x, 0)dx +
u(x, T )dx =
0
0
0
u(x, t)dxdt
0
L
Z
g(x)dx.
0
0
(c) What is the limit of
RL
L
Z
f (x)dx + (1 − e−T )
=
L
Z
d
dt
u(x, t)dx as t → +∞?
0
Solution: The above formula gives
L
Z
T →+∞
L
Z
g(x)dx.
0
0
0
L
Z
f (x)dx +
u(x, T )dx =
lim
Question 22: Let u solve ∂t u + ∂x v(x, t)u − µ(x, t)∂x u = g(x)e−t , x ∈ (0, L), t > 0, with
µ(0, t)∂x u(0, t) = 1, µ(L, t)∂x u(L, t) = 1 + 2e−t , u(x, 0) = f (x), where v, µ > 0, f and g are
four smooth functions and v(0) = v(L) = 0.
RL
d
u(x, t)dx as a function of t.
(a) Compute dt
0
Solution: Integrate the equation over the domain (0, L) and apply the Fundamental Theorem of
calculus:
Z
Z L
Z L
Z L
d L
u(x, t)dx =
∂t u(x, t)dx =
∂x (−v(x, t)u + µ(x, t)∂x u)dx + e−t
g(x)dx
dt 0
0
0
0
Z L
= µ(L, t)∂x u(L) − µ(0, t)∂x u(0) + e−t
g(x)dx
0
= 1 + 2e−t − 1 + e−t
Z
L
g(x)dx.
0
That is
d
dt
(b) Use (a) to compute
RL
0
L
Z
L
Z
u(x, t)dx = e (
−t
0
g(x)dx + 2).
0
u(x, t)dx as a function of t.
Solution: Applying the Fundamental Theorem of calculus again (with respect to time this time)
gives
Z
L
Z
L
u(x, T )dx =
0
Z
u(x, 0)dx +
0
Z
=
T
0
L
f (x)dx + (1 − e−T )(
RL
0
g(x)dx + 2).
0
L
Z
g(x)dx + 2).
0
(c) What is the limit of
L
Z
e dt(
−t
0
u(x, t)dx as t → +∞?
Solution: The above formula gives
Z
lim
T →+∞
L
Z
u(x, T )dx =
0
L
Z
f (x)dx +
0
L
g(x)dx + 2.
0
Question 23: Let u solve ∂t u−∂x (µ(x, t)∂x u) = g(x)e−t , x ∈ (0, L), t > 0, with µ(0, t)∂x u(0, t) =
1, µ(L, t)∂x u(L, t) = 1 + 2e−t , u(x, 0) = f (x), where µ > 0, f and g are three smooth functions.
RL
d
(a) Compute dt
u(x, t)dx as a function of t.
0
Math 602
10
Solution: Integrate the equation over the domain (0, L) and apply the fundamental Theorem of
calculus:
Z L
Z
Z L
Z L
d L
g(x)dx
u(x, t)dx =
∂t u(x, t)dx =
∂x (µ(x, t)∂x u)dx + e−t
dt 0
0
0
0
Z L
−t
= µ(L, t)∂x u(L) − µ(0, t)∂x u(0) + e
g(x)dx
0
= 1 + 2e−t − 1 + e−t
Z
L
g(x)dx.
0
That is
d
dt
(b) Use (a) to compute
RL
L
Z
Z
u(x, t)dx = e−t (
0
L
g(x)dx + 2).
0
u(x, t)dx as a function of t.
0
Solution: Applying the fundamental Theorem of calculus again gives
Z L
Z L
Z T
Z L
u(x, T )dx =
u(x, 0)dx +
e−t dt(
g(x)dx + 2).
0
0
0
L
Z
f (x)dx + (1 − e−T )(
=
g(x)dx + 2).
0
(c) What is the limit of
RL
0
0
L
Z
0
u(x, t)dx as t → +∞?
Solution: The above formula gives
Z L
Z
lim
u(x, T )dx =
T →+∞
0
L
Z
f (x)dx +
0
L
g(x)dx + 2.
0
Question 24: Consider the heat equation ∂t T −k∂xx T = f (x), x ∈ [a, b], t > 0, with f (x) = kx,
where k > 0. Compute the steady state solution (i.e., ∂t T = 0) assuming the boundary
conditions: −k∂n T (a) = 0, T (b) = 0 (∂n is the normal derivative)
Solution: At steady state, T does not depend on t and we have ∂xx T (x) = −x, which implies
∂x T (x) = α − 21 x2 , and T (x) = β + αx − 16 x3 , where α, β ∈ R. The two constants α and β
are determined by the boundary conditions. 0 = −∂n T (a) = ∂x T (a) = α − 21 a2 and 0 = T (b) =
β + αb − 61 b3 . We conclude that α = 12 a2 and β = −αb + 61 b3 = − 12 a2 b + 16 b3 . In conclusion
1
1
1
1
1
1
T (x) = − a2 b + b3 + a2 x − x3 = − a2 (b − x) + (b3 − x3 ).
2
6
2
6
2
6
Question 25: Consider the equation ∂t c(x, t) − ∂xx c(x, t) = x, where x ∈ [0, L], t > 0, with
c(x, 0) = f (x), −∂n c(0, t) = 6, −∂n c(L, t) = 5, (∂n is the normal derivative). Compute E(t) :=
RL
c(ξ, t)dξ.
0
Solution: We integrate the equation with respect to x over [0, L]
Z L
Z L
Z L
∂t c(ξ, t)dξ −
∂ξξ c(ξ, t)dξ =
ξdξ.
0
Using that
infer that
RL
0
0
∂t c(ξ, t)dξ = dt
RT
0
0
c(ξ, t)dξ together with the fundamental theorem of calculus, we
1 2
L .
2
The boundary conditions ∂x c(0, t) = −∂n c(0, t) = 6, −∂x c(L, t) = −∂n c(L, t) = 5 give
dt E(t) − ∂x c(L, t) + ∂x c(0, t) =
dt E(t) + 5 + 6 =
1 2
L .
2
Math 602
11
We now apply the fundamental theorem of calculus with respect to t
Z t
1
E(t) − E(0) =
∂τ E(τ )dτ = ( L2 − 11)t.
2
0
In conclusion
L
Z
E(t) =
0
1
f (ξ)dξ + ( L2 − 11)t.
2
Question 26: Consider the heat equation ∂t u(x, t) − 3∂xx u(x, t) = 0, ∂x u(0, t) = 0, ∂x u(1, t) =
P∞
2 2
0, u(x, 0) = u0 (x), t > 0, x ∈ (0, 1). The general solution is u(x, t) = n=0 An cos(nπx)e−3n π t .
Compute the solution corresponding to the initial data u0 (x) = 5 cos(4πx).
Solution: The solution contains one term only, corresponding to n = 4,
2
u(x, t) = 5 cos(4πx)e−48π t .
Question 27: Consider the heat equation ∂t u(x, t) − 2∂xx u(x, t) = 0, u(0, t) = 0, u(1, t) = 0,
P∞
2 2
u(x, 0) = u0 (x), t > 0, x ∈ (0, 1). The general solution is u(x, t) = n=0 An sin(nπx)e−2n π t .
Compute the solution corresponding to the initial data u0 (x) = 3 sin(4πx) − 5 sin(πx).
Solution: The solution contains two terms only, corresponding to n = 1 and n = 4,
2
2
u(x, t) = −5 sin(πx)e−2π t + 3 sin(4πx)e−32π t .
Question 28: Consider the equation ∂t c(x, t) − ∂xx c(x, t) = 6x/L2 , where x ∈ [0, L], t > 0,
with c(x, 0) = f (x), −∂n c(0, t) = 1, −∂n c(L, t) = 2, (∂n is the normal derivative). Compute
RL
E(t) := 0 c(ξ, t)dξ.
Solution: We integrate the equation with respect to x over [0, L]
Z
L
Z
∂t c(ξ, t)dξ −
0
Using that
infer that
RL
0
∂t c(ξ, t)dξ = dt
0
RT
0
L
6
∂ξξ c(ξ, t)dξ = 2
L
Z
L
ξdξ.
0
c(ξ, t)dξ together with the fundamental theorem of calculus, we
dt E(t) − ∂x c(L, t) + ∂x c(0, t) = 3.
The boundary conditions ∂x c(0, t) = −∂n c(0, t) = 1, −∂x c(L, t) = −∂n c(L, t) = 2 give
dt E(t) + 2 + 1 = 3.
We now apply the fundamental theorem of calculus with respect to t
Z t
E(t) − E(0) =
∂τ E(τ )dτ = 0.
0
In conclusion
Z
E(t) =
L
f (ξ)dξ,
∀t ≥ 0.
0
Question 29: Consider the equation ∂t c(x, t)−∂x (1+x2 )∂x c(x, t) = 6x/L2 , where x ∈ [0, L],
2
t > 0, with c(x, 0) = f (x), −∂n c(0, t) = 1, −∂n c(L, t) = 1+L
2 , (∂n is the normal derivative).
RL
Compute E(t) := 0 c(ξ, t)dξ.
Solution: We integrate the equation with respect to x over [0, L]
Z
L
Z
∂t c(ξ, t)dξ −
0
0
L
6
∂ξ (1 + ξ )∂ξ c(ξ, t) dξ = 2
L
2
Z
L
ξdξ.
0
Math 602
Using that
infer that
RL
0
∂t c(ξ, t)dξ = dt
RT
0
12
c(ξ, t)dξ together with the fundamental theorem of calculus, we
dt E(t) − (1 + L2 )∂x c(L, t) + ∂x c(0, t) = 3.
The boundary conditions ∂x c(0, t) = −∂n c(0, t) = 1, −∂x c(L, t) = −∂n c(L, t) =
2
1+L2
give
dt E(t) + 2 + 1 = 3.
We now apply the fundamental theorem of calculus with respect to t
Z t
∂τ E(τ )dτ = 0.
E(t) − E(0) =
0
In conclusion
Z
L
f (ξ)dξ,
E(t) =
∀t ≥ 0.
0
Question 30: Consider the heat equation ∂t u(x, t) − 2∂xx u(x, t) = 0, u(0, t) = 0, u(1, t) = 0,
P∞
2 2
u(x, 0) = u0 (x), t > 0, x ∈ (0, 1). The general solution is u(x, t) = n=0 An sin(nπx)e−2n π t .
Compute the solution corresponding to the initial data u0 (x) = 3 sin(4πx).
Solution: The solution contains one term only, corresponding to n = 4,
2
u(x, t) = 3 sin(4πx)e−32π t .
Question 31: Consider the heat equation ∂t T − k∂xx T = f (x), x ∈ [a, b], t > 0, with f (x) = 0,
where k > 0. Compute the steady state solution (i.e., ∂t T = 0) assuming the boundary
conditions: −k∂n T (a) = 1, T (b) = 0 (∂n is the normal derivative).
Solution: At steady state, T does not depend on t and we have ∂xx T (x) = 0, which implies
∂x T (x) = α, and T (x) = β + αx, where α, β ∈ R. The two constants α and β are determined by
the boundary conditions. 1 = −k∂n T (a) = k∂x T (a) = kα and 0 = T (b) = β + αb. We conclude
that α = k1 and β = −αb = − kb . In conclusion
T (x) =
x−b
.
k
Math 602
3
13
Laplace equation
Question 32: (a) Find a function U (x, y) = a + bx + cy + dxy, such that U (x, 0) = x,
U (1, y) = 1 + y, U (x, 1) = 3x − 1, and U (0, y) = −y.
Solution:
U (x, y) = x − y + 2xy
solves the problem.
(b) Use (a) to solve the PDE uxx + uyy = 0, ∀(x, y) ∈ (0, 1)×(0, 1), with the boundary
conditions u(x, 0) = 3 sin(πx) + x, u(1, y) = 1 + y, u(x, 1) = 3x − 1,and u(0, y) = sin(2πy) − y.
Solution: By setting φ = u − U , we observe that φxx + φyy = 0 and at the boundary of the
domain we have
φ(x, 0) = 3 sin(πx),
φ(1, y) = 0,
φ(x, 1) = 0,
and φ(0, y) = sin(2πy).
It is clear that
φ(x, y) = 3 sin(πx)
sinh(π(1 − y)) sinh(2π(1 − x))
+
sin(2πy)
sinh(π)
sinh(2π)
Then,
u(x, y) = φ(x, y) + U (x, y)
Question 33: Consider the Laplace equation ∆u = 0 in the rectangle x ∈ [0, L], y ∈ [0, H]
with the boundary conditions u(0, y) = 0, u(L, y) = 0, u(x, 0) = 0, u(x, H) = f (x).
(a) Is there any compatibility condition that f must satisfy for a smooth solution to exist?
Solution: f must be such that f (0) = 0 and f (L) = 0, otherwise u would not be continuous at
the two upper corners of the domain.
(b) Solve the Equation.
Solution: Use the separation of variable technique. Let u(x) = φ(x)ψ(y). Then, provided ψ and
00
00
φ are non zero functions, this implies φφ = − ψψ = λ. Observe that φ(0) = φ(L) = 0. The usual
√
√
energy technique implies that λ is negative.
That is to say φ(x) = a cos( λx) + b sin( λx). The
√
boundary conditions imply a = 0√and λL = nπ,
√ i.e., φ(x) = b sin(nπx/L). The fact that λ is
negative implies ψ(y) = c cosh( λy) + d sinh( λy). The boundary condition at y = 0 implies
c = 0. Then
∞
X
nπx
nπy
u(xy) =
An sin(
) sinh(
)
L
L
n=1
Question 34: Solve the PDE (note that the width and the height of the rectangle are not
equal)
∂xx u + ∂yy u = 0,
0 < x < 1, 0 < y < 2,
u(x, 0) = 8 sin(9πx), u(x, 2) = 0,
0 < x < 1,
u(0, y) = sin(2πy), u(1, y) = 0,
0 < y < 2.
Solution: The method of separation of variables tells us that the solution is a sum of terms
like sin(nπx) sinh(nπ(y − 2)) and sin(mπy/2) sinh(mπ(x − 1)/2). By looking at the boundary
conditions we infer that there are two nonzero terms in the expansion: one corresponding to n = 9
and one corresponding to m = 4. This gives
u(x, y) = 8 sin(9πx)
sinh(2π(1 − x))
sinh(9π(2 − y))
+ sin(2πy)
sinh(18π))
sinh(2π))
Math 602
14
Question 35: Consider the Laplace equation ∆u = 0 in the rectangle x ∈ [0, L], y ∈ [0, H]
with the boundary conditions u(0, y) = 0, u(L, y) = 0, u(x, 0) = 0, u(x, H) = f (x).
(a) Is there any compatibility condition that f must satisfy for a smooth solution to exist?
Solution: f must be such that f (0) = 0 and f (L) = 0, otherwise u would not be continuous at
the two upper corners of the domain.
(b) Solve the Equation.
Solution: Use the separation of variable technique. Let u(x) = φ(x)ψ(y). Then, provided ψ and
00
00
φ are non zero functions, this implies φφ = − ψψ = λ. Observe that φ(0) = φ(L) = 0. The usual
√
√
technique implies that λ is negative.
That is to say φ(x) = a cos( λx)+b sin( λx). The boundary
√
conditions imply a = 0√ and λL = √
nπ, i.e., φ(x) = b sin(nπx/L). The fact that λ is negative
implies ψ(y) = c cosh( λy) + d sinh( λy). The boundary condition at y = 0 implies c = 0. Then
the ansatz is
∞
X
nπy
nπx
u(x, y) =
) sinh(
),
An sin(
L
L
n=1
and the usual computation gives
2
An =
L sinh( nπH
L )
Z
L
f (ξ) sin(
0
nπξ
)dξ.
L
Question 36: Consider the Laplace equation ∆u = 0 in the rectangle x ∈ [0, L], y ∈ [0, H]
with the boundary conditions u(0, y) = 0, ∂x u(L, y) = 0, u(x, 0) = 0, u(x, H) = sin( 32 πx/L).
Solve the Equation using the method of separation of variables. (Give all the details.)
Solution: Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies
ψ 00
φ00
φ0 (L) = 0. The usual energy technique implies that
φ = − ψ = λ. Observe that φ(0) = 0 and
√
√
λ is negative. That is to say φ(x) = a cos( λx) + b sin(
at x = 0
√
√ λx). The Dirichlet condition
implies a = 0. The Neumann condition at L implies cos( λL) = 0, which implies λL = (n+ 12 )π,
where n is an integer. This means that φ(x) = b sin((n + 21 )πx/L). The fact that λ is negative
√
√
implies ψ(y) = c cosh( λy) + d sinh( λy). The boundary condition at y = 0 implies c = 0. Then
x
y
u(x, y) = A sin((n + 21 )π ) sinh((n + 21 )π ).
L
L
The boundary condition at y = H gives
3 x
x
H
sin( π ) = A sin((n + 12 )π ) sinh((n + 12 )π ),
2 L
L
L
3 πH
which implies n = 1 and A = sinh−1 2 L , i.e.,
sinh
3
sinh
3
u(x, y) =
2 πy
L
2 πH
L
3
sin( πx/L)
2
Question 37: Consider the equation −∆u = 0 in the rectangle {(x, y); x ∈ [0, L], y ∈ [0, H]}
y
with the boundary conditions u(0, y) = 0, u(L, y) = −5 cos( 23 π H
), ∂y u(x, 0) = 0, u(x, H) = 0.
Solve the equation using the method of separation of variables. (Give all the details.)
Solution: Let u(x) = φ(x)ψ(y). Then, provided ψ and φ are non zero functions, this implies
φ”(x)
ψ”(y)
0
φ(x) = − ψ(y) = λ. Observe that ψ (0) = 0 and ψ(H) = 0. The energy technique applied to
−ψ”(y) = λψ(y) gives
Z H
Z H
Z H
−ψ”(y)ψ(y)dy =
ψ 0 (y)2 dy − ψ 0 (H)ψ(H) + ψ 0 (0)ψ(0) = λ
ψ(y)2 y,
0
0
RH
RH
0
which implies 0 ψ 0 (y)2 dy = λ 0 ψ(y)2 dy since ψ 0 (0) = 0 and ψ(H) = 0. This in turn implies
that λ is nonnegative. Actually λ cannot be zero since it would mean that ψ = 0, which would
Math 602
15
contradict the fact that the solution u is nonzero (λ = 0 ⇒ ψ 0 (y) = 0 ⇒ ψ(y) = ψ(H) = 0 for all
y ∈ [0, H]). As a result λ is positive and
√
√
ψ(y) = a cos( λy) + b sin( λy).
√
The Neumann√condition at y = 0 gives b = 0. The Dirichlet condition at H implies cos( λH) = 0,
y
).
which implies λH = (n+ 12 )π, where n is any integer. This means that ψ(y) = a cos((n+ 21 )π H
√
√
The fact that λ is positive implies φ(x) = c cosh( λx) + d sinh( λx). The boundary condition at
x = 0 implies c = 0. Then
u(x, y) = A cos((n + 21 )π
y
x
) sinh((n + 12 )π ).
H
H
The boundary condition at x = L gives
3 y
y
L
−5 cos( π ) = A cos((n + 12 )π ) sinh((n + 21 )π ),
2 H
H
H
3 πL
which, by identification, implies 1 = 2 and A = −5 sinh−1 2H , i.e.,
sinh
3
sinh
3
u(x, y) = −5
2 πx
H
2 πL
3 y
cos( π ).
2 H
H
Question 38: Does any of the following expressions solve the Laplace equation inside the
rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H, with the following boundary conditions ∂x u(0, y) =
5πy
15π
5πL
H sin( H ) cosh( H ), ∂x u(L, y) = 0, u(x, 0) = 0, u(x, H) = 0? (justify clearly your answer):
5πy
5π(x − L)
) cosh(
),
H
H
5π(x − L)
5πy
) sinh(
),
u3 (x, y) = 3 cos(
H
H
u1 (x, y) = 3 cos(
5πy
5π(x − L)
) cosh(
),
H
H
5πy
5π(x − L)
u4 (x, y) = 3 sin(
) sinh(
).
H
H
u2 (x, y) = 3 sin(
Solution: From class, we know that all the above expressions solve the Laplace equation, hence we
just need to verify the boundary conditions. We observe that u1 and u3 do not satisfy the Dirichlet
boundary conditions u(x, 0) = 0, u(x, H) = 0; therefore u1 and u3 must be discarded.
Both u2 and u4 satify that Dirichlet conditions: u2 (x, 0) = 0, u2 (x, H) = 0, and u4 (x, 0) = 0,
u4 (x, H) = 0. Now we need to check the Neumann conditions.
5πy
Note that u4 is such that ∂x u4 (L, y) = 3 5π
H sin( H ) cosh(0) 6= 0; a result u4 must be discarded as
well.
5πy
5πy
5π
5πL
Finally u2 is such that ∂x u2 (L, y) = 3 5π
H sin( H ) sinh(0) = 0, but ∂x u2 (0, y) = 3 H sin( H ) sinh(− H ),
which shows that u2 is not the solution to our problem either.
In conclusion, none of the proposed solutions solve the problem. The correct solution is
u(x, y) = 3
cosh( 5πL
5πy
5π(x − L)
H )
sin(
) cosh(
).
−5πL
H
H
sinh( H )
Question 39: Does any of the following expressions solve the Laplace equation inside the
rectangle 0 ≤ x ≤ L, 0 ≤ y ≤ H, with the following boundary conditions u(0, y) = 0, u(L, y) =
5πy
3 sinh( 5πL
H ) sin( H ), u(x, 0) = 0, u(x, H) = 0? (justify clearly your answer):
5πy
5π(x − L)
5πy
5π(x − L)
) cosh(
), u2 (x, y) = 3 sin(
) cosh(
),
H
H
H
H
5πy
5πx
5πy
5πx
u3 (x, y) = 3 cos(
) sinh(
), u4 (x, y) = 3 sin(
) sinh(
).
H
H
H
H
u1 (x, y) = 3 cos(
Math 602
16
Solution: From class, we know that all the above expressions solve the Laplace equation, hence we
just need to verify that the boundary conditions are met. We observe that u1 and u3 do not satisfy
the Dirichlet boundary conditions u(x, 0) = 0, u(x, H) = 0; therefore u1 and u3 must be discarded.
Both u2 and u4 satisfy that Dirichlet conditions: u2 (x, 0) = 0, u2 (x, H) = 0, and u4 (x, 0) = 0,
u4 (x, H) = 0. Now we need to check the Neumann conditions.
5π(−L)
) 6= 0, which shows that u2 is not the
Note that u2 is such that u2 (0, y) = 3 sin( 5πy
H ) cosh(
H
solution to our problem either.
5πy
5πL
Finally u4 (0, y) = 3 sin( 5πy
H ) sinh(0) = 0 and u4 (L, y) = 3 sin( H ) sinh( H ); a result u4 is the
solution.
3.1
Cylindrical coordinates
Question 40: Using cylindrical coordinates and the method of separation of variables, solve
the Laplace equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D = {θ ∈ [0, π2 ], r ∈ [0, 1]},
subject to the boundary conditions ∂θ u(r, 0) = 0, u(r, π2 ) = 0, u(1, θ) = cos(3θ).
Solution: We set u(r, θ) = φ(θ)g(r). This means φ00 = −λφ, with φ0 (0) = 0 and φ( π2 ) = 0, and
rdr (rdr g(r)) = λg(r). Then using integration by parts plus the boundary conditions we prove that
λ is non-negative. Then
√
√
φ(θ) = c1 cos( λθ) + c2 sin( λθ).
The boundary condition φ0 (0) = 0 implies c2 = 0. The boundary condition φ( π2 ) = 0 implies
√
√ π
λ 2 = (2n + 1) π2 with n ∈ N. This means λ = (2n + 1). From class we know that g(r) is of
the form rα , α ≥ 0. The equality rdr (rdr rα ) = λrα gives α2 = λ. The condition α ≥ 0 implies
2n + 1 = α. The boundary condition at r = 1 gives cos(3θ) = 12n+1 cos((2n + 1)θ). This implies
n = 1. The solution to the problem is
u(r, θ) = r3 cos(3θ).
Question 41: Using cylindrical coordinates and the method of separation of variables, solve
the Laplace equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D = {θ ∈ [0, π2 ], r ∈ [0, 1]},
subject to the boundary conditions u(r, 0) = 0, u(r, π2 ) = 0, u(1, θ) = sin(2θ). (Give all the
details.)
Solution: We set u(r, θ) = φ(θ)g(r). This means φ00 = −λφ, with φ(0) = 0 and φ( π2 ) = 0,
and rdr (rdr g(r)) = λg(r). The usual energy argument applied to the two-point boundary value
problem
π
φ00 = −λφ,
φ(0) = 0,
φ( ) = 0,
2
implies that λ is non-negative. If λ = 0, then φ(θ) = c1 + c2 θ and the boundary conditions imply
c1 = c2 = 0, i.e., φ = 0, which in turns gives u = 0 and this solution is incompatible with the
boundary condition u(1, θ) = sin(2θ). Hence λ > 0 and
√
√
φ(θ) = c1 cos( λθ) + c2 sin( λθ).
The boundary condition φ(0) = 0 implies c1 = 0. The boundary condition φ( π2 ) = 0 implies
√ π
√
λ 2 = nπ with n ∈ N. This means λ = 2n. From class we know that g(r) is of the form rα ,
α ≥ 0. The equality rdr (rdr rα ) = λrα gives α2 = λ. The condition α ≥ 0 implies 2n = α. The
boundary condition at r = 1 gives sin(2θ) = 12n sin(2nθ) for all θ ∈ [0, π2 ]. This implies n = 1.
The solution to the problem is
u(r, θ) = r2 sin(2θ).
Question 42: Using cylindrical coordinates and the method of separation of variables, solve
the equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D = {θ ∈ [0, π], r ∈ [0, 1]}, subject
to the boundary conditions u(r, 0) = 0, u(r, π) = 0, u(1, θ) = 2 sin(5θ). (Give all the details.)
Math 602
17
Solution: (1) We set u(r, θ) = φ(θ)g(r). This means φ00 = −λφ, with φ(0) = 0 and φ(π) = 0,
d
d
and r dr
(r dr
g(r)) = λg(r).
(2) The usual energy argument applied to the two-point boundary value problem
φ00 = −λφ,
φ(0) = 0,
φ(π) = 0,
implies that λ is non-negative. If λ = 0, then φ(θ) = c1 + c2 θ and the boundary conditions imply
c1 = c2 = 0, i.e., φ = 0, which in turns gives u = 0 and this solution is incompatible with the
boundary condition u(1, θ) = 2 sin(5θ). Hence λ > 0 and
√
√
φ(θ) = c1 cos( λθ) + c2 sin( λθ).
(3)
√ The boundary condition φ(0) = 0 implies
√ c1 = 0. The boundary condition φ(π) = 0 implies
λπ = nπ with n ∈ N \ {0}. This means λ = n, n = 1, 2, . . ..
d
d α
(4) From class we know that g(r) is of the form rα , α ≥ 0. The equality r dr
(r dr
r ) = λrα
2
gives α = λ. The condition α ≥ 0 implies n = α. The boundary condition at r = 1 gives
2 sin(5θ) = c2 1n sin(nθ) for all θ ∈ [0, π]. This implies n = 5 and c2 = 2.
(5) Finally, the solution to the problem is
u(r, θ) = 2r5 sin(5θ).
Question 43: Using cylindrical coordinates and the method of separation of variables, solve
the equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D = {θ ∈ [0, 32 π], r ∈ [0, 3]}, subject
to the boundary conditions u(r, 0) = 0, u(r, 32 π) = 0, u(3, θ) = 18 sin(2θ). (Give all the details.)
Solution: (1) We set u(r, θ) = φ(θ)g(r). This means φ00 = −λφ, with φ(0) = 0 and φ( 32 π) = 0,
d
d
and r dr
(r dr
g(r)) = λg(r).
(2) The usual energy argument applied to the two-point boundary value problem
φ00 = −λφ,
φ(0) = 0,
3
φ( π) = 0,
2
implies that λ is non-negative. If λ = 0, then φ(θ) = c1 + c2 θ and the boundary conditions imply
c1 = c2 = 0, i.e., φ = 0, which in turns gives u = 0 and this solution is incompatible with the
boundary condition u(3, θ) = 18 sin(2θ). Hence λ > 0 and
√
√
φ(θ) = c1 cos( λθ) + c2 sin( λθ).
(3) The boundary condition φ(0) = 0 implies c1 = 0. The boundary condition φ( 32 π) = 0 implies
√ 3
√
λ 2 π = nπ with n ∈ N \ {0}. This means λ = 23 n, n = 1, 2, . . ..
d
d α
(4) From class we know that g(r) is of the form rα , α ≥ 0. The equality r dr
(r dr
r ) = λrα gives
√
α2 = λ. The condition α ≥ 0 implies 32 n = α = λ. The boundary condition at r = 3 gives
2
18 sin(2θ) = c2 3 3 n sin( 32 nθ) for all θ ∈ [0, 32 π]. This implies n = 3 and c2 = 2.
(5) Finally, the solution to the problem is
u(r, θ) = 2r2 sin(2θ).
Question 44: Using cylindrical coordinates and the method of separation of variables, solve
the equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D = {θ ∈ [0, 32 π], r ∈ [0, 3]}, subject
to the boundary conditions ∂θ u(r, 0) = 0, u(r, 32 π) = 0, u(3, θ) = 9 cos(θ). (Give all the details
of all the steps.)
Solution: (1) We set u(r, θ) = φ(θ)g(r). This means φ00 = −λφ, with φ0 (0) = 0 and φ( 32 π) = 0,
d
d
and r dr
(r dr
g(r)) = λg(r).
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18
(2) The usual energy argument applied to the two-point boundary value problem
φ00 = −λφ,
φ0 (0) = 0,
3
φ( π) = 0,
2
implies that λ is non-negative. If λ = 0, then φ(θ) = c1 + c2 θ and the boundary conditions imply
c1 = c2 = 0, i.e., φ = 0, which in turns gives u = 0 and this solution is incompatible with the
boundary condition u(3, θ) = 9 sin(2θ). Hence λ > 0 and
√
√
φ(θ) = c1 cos( λθ) + c2 sin( λθ).
(3) The boundary condition φ0 (0) = 0 implies c2 = 0. The boundary condition φ( 32 π) = 0 implies
√
√
√
that cos( λ 23 π) = 0, i.e., λ 32 π = (2n + 1) π2 with n ∈ N. This means λ = 13 (2n + 1),
n = 0, 1, 2, . . ..
d
d α
(4) From class we know that g(r) is of the form rα , α ≥ 0. The equality r dr
(r dr
r ) = λrα gives
√
α2 = λ. The condition α ≥ 0 implies 31 (2n + 1) = α = λ. The boundary condition at r = 3 gives
1
9 cos(θ) = c1 3 3 (2n+1) cos( 31 (2n + 1)θ) for all θ ∈ [0, 32 π]. This implies n = 1 and c1 = 3.
(5) Finally, the solution to the problem is
u(r, θ) = 3r cos(θ).
Question 45: The solution of the equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D =
{θ ∈ [0, π], r ∈ [0,
subject to the boundary conditions u(r, 0) = 0, u(r, π) = 0, u(2, θ) =
P2]},
∞
g(θ) is u(r, θ) = n=1 bn rn sin(nθ). What is the solution corresponding to g(θ) = 5 sin(2θ) +
2 sin(5θ)? (Give all the details.)
Solution: The only non-zero terms in the expansion are a2 r2 sin(2θ) + b5 r5 sin(5θ). The boundary
condition u(2, θ) = 5 sin(2θ) + 2 sin(5θ) = a2 22 sin(2θ) + a5 25 sin(5θ) is satisfied if a2 = 5/(22 )
and b5 = 2/(25 ), i.e.,
r5
r2
u(r, θ) = 5 2 sin(2θ) + 2 5 sin(5θ).
2
2
Question 46: The solution of the equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D =
{θ ∈ [0, π2 ], r ∈ [0, 2]}, subject to the boundary conditions ∂θ u(r, 0) = 0, u(r, π2 ) = 0, u(2, θ) =
P∞
g(θ) is u(r, θ) = n=0 a2n+1 r(2n+1) cos((2n+1)θ). What is the solution corresponding to g(θ) =
5 cos(3θ) + 2 cos(5θ)? (Give all the details.)
Solution: The only non-zero terms in the expansion are a3 r3 cos(3θ) + a5 r5 cos(5θ). The boundary
condition u(2, θ) = 5 cos(3θ) + 2 cos(5θ) = a3 23 cos(3θ) + a5 25 cos(5θ) is satisfied if a3 = 5/(23 )
and a5 = 2/(25 ), i.e.,
r3
r5
u(r, θ) = 5 3 cos(3θ) + 2 5 cos(5θ).
2
2
Question 47: The solution of the equation, 1r ∂r (r∂r u) + r12 ∂θθ u = 0, inside the domain D =
{θ ∈ [0, π2 ], r ∈ [0, 3]}, subject to the boundary conditions ∂θ u(r, 0) = 0, u(r, π2 ) = 0, u(3, θ) =
P∞
g(θ) is u(r, θ) = n=0 a2n+1 r(2n+1) cos((2n+1)θ). What is the solution corresponding to g(θ) =
5 cos(θ) + 2 cos(3θ)? (Give all the details.)
Solution: The only non-zero terms in the expansion are a1 r cos(θ) + a3 r3 cos(3θ). The boundary
condition u(3, θ) = 5 cos(θ) + 2 cos(3θ) = a1 31 cos(3θ) + a3 33 cos(5θ) is satisfied if a1 = 5/3 and
a3 = 2/(33 ), i.e.,
r
r3
u(r, θ) = 5 cos(θ) + 2 3 cos(3θ).
3
3
Math 602
3.2
19
Variable coefficients
Question 48: Let k : [−1, +1] −→ R be such that k(x) = 1, if x ∈ [−1, 0] and k(x) = 2 if
x ∈ (0, 1]. Solve the boundary value problem −∂x (k(x)∂x T (x)) = 0 with ∂x T (−1) = T (−1)
and ∂x T (1) = 1.
(i) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 1 and k + (0) = 2.
(ii) Solve the problem, i.e., find T (x), x ∈ [−1, +1].
Solution: On [−1, 0] we have k − (x) = 1, which implies ∂xx T − (x) = 0. This in turn implies
T − (x) = a + bx. The Robin boundary condition at x = −1 implies ∂x T −1 (−1) − T − (−1) = 0 =
2b − a. This gives a = 2b and T − (x) = b(2 + x).
We proceed similarly on [0, +1] and we obtain T + (x) = c + dx. The Neumann boundary condition
at x = +1 gives ∂x T + (+1) = 1 = d and T + (x) = c + x.
The interface conditions T − (0) = T + (0) and k − (0)∂x T − (0) = k + (0)∂x T + (0) give
2b = c,
In conclusion
and
b = 2.
(
2(2 + x) if x ∈ [−1, 0],
T (x) =
4+x
if x ∈ [0, +1].
Question 49: Let k : [−1, +1] −→ R be such that k(x) = 6, if x ∈ [−1, 0] and k(x) = 3 if x ∈
(0, 1]. Solve the boundary value problem −∂x (k(x)∂x T (x)) = 0 with 6∂x T (−1) = T (−1) + 13
and T (1) = 5.
(i) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 6 and k + (0) = 3.
(ii) Solve the problem, i.e., find T (x), x ∈ [−1, +1].
Solution: On [−1, 0] we have k − (x) = 1, which implies ∂xx T − (x) = 0. This in turn implies
T − (x) = a + bx. The Robin boundary condition at x = −1 implies 6∂x T − (−1) − T − (−1) = 13 =
6b − (a − b). This gives a = 7b − 13 and T − (x) = 7b − 13 + bx.
We proceed similarly on [0, +1] and we obtain T + (x) = c + dx. The Dirichlet boundary condition
at x = +1 gives T + (1) = 5 = d + c. This implies c = 5 − d and T + (x) = 5 − d + dx.
The interface conditions T − (0) = T + (0) and k − (0)∂x T − (0) = k + (0)∂x T + (0) give
7b − 13 = 5 − d,
This implies d = 4 and b = 2. In conclusion
(
2x + 1
T (x) =
4x + 1
and
6b = 3d.
if x ∈ [−1, 0],
if x ∈ [0, +1].
Question 50: Let k : [−1, +1] −→ R be such that k(x) = 2, if x ∈ [−1, 0] and k(x) = 3 if
x ∈ (0, 1]. Solve the boundary value problem −∂x (k(x)∂x T (x)) = 0 with ∂x T (−1) = T (−1) + 3
and −∂x T (1) = T (1) − 7.
(i) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 2 and k + (0) = 3.
Math 602
20
(ii) Solve the problem, i.e., find T (x), x ∈ [−1, +1].
Solution: On [−1, 0] we have k − (x) = 1, which implies ∂xx T − (x) = 0. This in turn implies
T − (x) = a + bx. The Robin boundary condition at x = −1 implies ∂x T − (−1) − T − (−1) = 3 =
2b − a. This gives a = 2b − 3 and T − (x) = 2b − 3 + bx.
We proceed similarly on [0, +1] and we obtain T + (x) = c + dx. The Robin boundary condition at
x = +1 gives −∂x T + (+1) − T + (1) = −7 = −2d − c. This implies c = −2d + 7 and T + (x) =
−2d + 7 + dx.
The interface conditions T − (0) = T + (0) and k − (0)∂x T − (0) = k + (0)∂x T + (0) give
2b − 3 = −2d + 7,
This implies d = 2 and b = 3. In conclusion
(
3x + 3
T (x) =
2x + 3
and
2b = 3d.
if x ∈ [−1, 0],
if x ∈ [0, +1].
Question 51: Let k, f : [−1, +1] −→ R be such that k(x) = 2, f (x) = 0 if x ∈ [−1, 0] and
k(x) = 1, f (x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x (k(x)∂x T (x)) = f (x)
with T (−1) = 0 and T (1) = 3.
(a) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 2 and k + (0) = 1.
Question 52: Let k, f : [−1, +1] −→ R be such that k(x) = 2 + x, f (x) = 0 if x ∈ [−1, 0] and
k(x) = 1 + 2x, f (x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x (k(x)∂x T (x)) =
f (x) with T (−1) = 5 and T (1) = −1.
(a) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 2 and k + (0) = 1, i.e., 2∂x T − (0) = ∂x T + (0).
Question 53: Let k, f : [−1, +1] −→ R be such that k(x) = 2, f (x) = 0 if x ∈ [−1, 0] and
k(x) = 1, f (x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x (k(x)∂x T (x)) = f (x)
with T (−1) = −2 and T (1) = 2.
(a) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 2 and k + (0) = 1.
(b) Solve the problem, i.e., find T (x), x ∈ [−1, +1].
Solution: On [−1, 0] we have k − (x) = 2 and f − (x) = 0 which implies −∂xx T − (x) = 0. This
in turn implies T − (x) = ax + b. The Dirichlet condition at x = −1 implies that T − (−1) = −2 =
−a + b. This gives a = b + 2 and T − (x) = (b + 2)x + b.
We proceed similarly on [0, +1] and we obtain −∂xx T − (x) = 2, which implies that T + (x) =
−x2 + cx + d. The Dirichlet condition at x = 1 implies T + (1) = 2 = −1 + c + d. This gives
c = 3 − d and T − (x) = −x2 + (3 − d)x + d.
The interface conditions T − (0) = T + (0) and k − (0)∂x T − (0) = k + (0)∂x T + (0) give b = d and
2(b + 2) = 3 − d, respectively. In conclusion b = − 31 , d = − 13 and
(
5
x− 1
if x ∈ [−1, 0],
T (x) = 3 2 3 10
1
−x + 3 x − 3 if x ∈ [0, 1].
Question 54: Let k, f : [−1, +1] −→ R be such that k(x) = 2, f (x) = 0 if x ∈ [−1, 0] and
k(x) = 1, f (x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x (k(x)∂x T (x)) = f (x)
Math 602
21
with T (−1) = 0 and T (1) = 2.
(a) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 2 and k + (0) = 1.
(b) Solve the problem, i.e., find T (x), x ∈ [−1, +1]. Give all the details.
Solution: On [−1, 0] we have k − (x) = 2 and f − (x) = 0 which implies −∂xx T − (x) = 0. This in
turn implies T − (x) = ax+b. The Dirichlet condition at x = −1 implies that T − (−1) = 0 = −a+b.
This gives a = b and T − (x) = bx + b.
We proceed similarly on [0, +1] and we obtain −∂xx T − (x) = 2, which implies that T + (x) =
−x2 + cx + d. The Dirichlet condition at x = 1 implies T + (1) = 2 = −1 + c + d. This gives
c = 3 − d and T − (x) = −x2 + (3 − d)x + d.
The interface conditions T − (0) = T + (0) and k − (0)∂x T − (0) = k + (0)∂x T + (0) give b = d and
2b = 3 − d, respectively. In conclusion b = 1, d = 1 and
(
x+1
if x ∈ [−1, 0],
T (x) =
−x2 + 2x + 1 if x ∈ [0, 1].
Question 55: Let k, f : [−1, +1] −→ R be such that k(x) = 3, f (x) = −6 if x ∈ [−1, 0] and
k(x) = 1, f (x) = 2 if x ∈ (0, 1]. Consider the boundary value problem −∂x (k(x)∂x T (x)) = f (x)
with T (−1) = 1 and ∂x T (1) = 1.
(a) What should be the interface conditions at x = 0 for this problem to make sense?
Solution: The function T and the flux k(x)∂x T (x) must be continuous at x = 0. Let T − denote
the solution on [−1, 0] and T + the solution on [0, +1]. One should have T − (0) = T + (0) and
k − (0)∂x T − (0) = k + (0)∂x T + (0), where k − (0) = 3 and k + (0) = 1.
(b) Solve the problem, i.e., find T (x), x ∈ [−1, +1]. Give all the details.
Solution: On [−1, 0] we have k − (x) = 3 and f − (x) = −6 which implies −3∂xx T − (x) = −6.
This in turn implies T − (x) = x2 + ax + b. The Dirichlet condition at x = −1 implies that
T − (−1) = 1 = 1 − a + b. This gives a = b and T − (x) = x2 + bx + b.
We proceed similarly on [0, +1] and we obtain −∂xx T − (x) = 2, which implies that T + (x) =
−x2 + cx + d. The Neumann condition at x = 1 implies T + (1) = 1 = −2 + c. This gives c = 3
and T − (x) = −x2 + 3x + d.
The interface conditions T − (0) = T + (0) and k − (0)∂x T − (0) = k + (0)∂x T + (0) give b = d and
3b = 3, respectively. In conclusion b = 1, d = 1 and
(
x2 + x + 1
if x ∈ [−1, 0],
T (x) =
2
−x + 3x + 1 if x ∈ [0, 1].
3.3
Maximum principle
Question 56: Consider the square D = (−1, +1)×(−1, +1). Let f (x, y) = x2 − y 2 − 3. Let
u ∈ C 2 (D) ∩ C 0 (D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) and
max(x,y)∈D u(x, y).
Solution: We use the maximum principle (u is harmonic and has the required regularity). Then
min u(x, y) =
(x,y)∈D
min
f (x, y),
and
(x,y)∈∂D
max u(x, y) =
(x,y)∈D
max f (x, y).
(x,y)∈∂D
A point (x, y) is at the boundary of D if and only if x2 = 1 and y ∈ (−1, 1) or y 2 = 1 and
x ∈ (−1, 1). In the first case, x2 = 1 and y ∈ (−1, 1), we have
f (x, y) = 1 − y 2 − 3,
y ∈ (−1, 1).
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22
The maximum is −2 and the minimum is −3. In the second case, y 2 = 1 and x ∈ (−1, 1), we have
f (x, y) = x2 − 1 − 3,
x ∈ (−1, 1).
The maximum is −3 and the minimum is −4. We finally can conclude
min
(x,y)∈∂D
f (x, y) =
min x2 − 4, = −4,
max f (x, y) = max −2 − y 2 = −2.
−1≤x≤1
(x,y)∈∂D
−1≤y≤1
In conclusion
min u(x, y) = −4,
(x,y)∈D
max u(x, y) = −2
(x,y)∈D
Question 57: (i) Let Ω be an open connected set in R2 . Let u be a real-valued nonconstant
function continuous on Ω and harmonic on Ω. Assume that there exists x0 in Ω such ∇u(x0 ) = 0.
Do we have a minimum, a maximum, or a saddle point at x0 ? (explain)
Solution: The Maximum principle implies that u cannot be either minimum or maximum at x0 .
This point is a saddle point.
(ii) Let Ω = (0, 1) (note that Ω = [0, 1]), and let u : Ω −→ R be such that u(x) = 1 for all
x ∈ Ω, u(0) = 0, and u(1) = −1. Is u harmonic on Ω? Find a point in Ω where u reaches its
maximum? Does this example contradict the Maximum Principle? (explain)
Solution: Yes u is harmonic on Ω since u00 (x) = 0 for all x in Ω. Note however that u is not
continuous on Ω; as a consequence, the hypotheses for the Maximum principle are not satisfied. In
other words, the above example does not contradict the Maximum principle .
Question 58: Let Ω be an open connected set in R2 . Let u be a real-valued nonconstant
function continuous on Ω and harmonic on Ω and of class C 2 in Ω. Assume that there exists x0
in Ω such ∇u(x0 ) = 0. Is the point x0 a minimum, a maximum, or a saddle point? (explain)
Solution: The keyword here is that Ω is open, meaning that the points at the boundary of Ω are
not in Ω (it is not possible to punch a hole around the boundary points). The point x0 is in Ω,
that is x0 is not at the boundary. Since u is continuous on Ω, harmonic on Ω and of class C 2 in Ω,
the maximum principle can be applied. The Maximum principle implies that u cannot have either
a minimum or a maximum at x0 . This means that x0 is a saddle point.
Question 59: Let u be a continuous function on D where D is some open, connected set in
R2 . Explain why, if u is harmonic, it is generally a waste of time to locate a point where u
achieves its maximum by solving ∂x u = 0 and ∂y u = 0 simultaneously.
Solution: From the Maximum Principle, we know that if u is not constant, the maximum of u is
achieved at the boundary of D. The zero gradient condition does not apply for maximums at the
boundary.
√
Question 60: Let D be the open disk of radius 2 centered at (1, 2). Let u ∈ C 2 (D) ∩ C 0 (D)
be a harmonic function in the disk D. Assume that u(x, y) = (x + y)2 on the boundary of disk.
Compute min(x,y)∈D u(x, y) and max(x,y)∈D u(x, y). You can give a geometric answer.
Solution: Since u is in C 2 (D) ∩ C 0 (D) and is harmonic, we can apply the maximum principle
(Theorem). This theorem says that the maximum and minimum of u are attained at the boundary
of the disk. The problem
then amounts to finding the maximum and minimum of (x + y)2 over
√
the circle of radius 2 centered at (1, 2). The iso-values of the function (x + y)2 are parallel
lines of slope −1. These iso-line are perpendicular to the gradient of (x + y)2 which is the vector
(2(x + y), 2(x + y)) = 2(x + y)(1, 1). We must find the two tangent lines to the circle that are
perpendicular to the vector (1, 1). One easily verify that (0, 1) and (2, 3) are the tangent points
(verify that the segment connecting these two points is parallel to the vector (1, 1) and passes
through the center of the circle). As a result
min u(x, y) = 1
(x,y)∈D
and
max u(x, y) = 25.
(x,y)∈D
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23
6
5
4
3
2
1
0
1
2
3
4
5
The dashed lines are iso-lines of (x + y)2 .
Question 61: Consider the disk D centered at (0, 0) of radius 1. Let f (x, y) = x2 − y 2 + 4y − 3.
Let u ∈ C 2 (D) ∩ C 0 (D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) and
max(x,y)∈D u(x, y).
Solution: We use the maximum principle (u is harmonic and has the required regularity). Then
min u(x, y) =
min
f (x, y),
and
(x,y)∈∂D
(x,y)∈D
max u(x, y) =
max f (x, y).
(x,y)∈∂D
(x,y)∈D
A point (x, y) is at the boundary of D if and only if x2 + y 2 = 1; as a result, the following holds
for all (x, y) ∈ ∂D:
f (x, y) = x2 − y 2 + 4y − 3 = 1 − y 2 − y 2 + 4y − 3 = −2(1 − y)2 .
We obtain
f (x, y) = min −2(1 − y)2 = −8,
min
−1≤y≤1
(x,y)∈∂D
max f (x, y) = max −2(1 − y)2 = 0.
(x,y)∈∂D
−1≤y≤1
In conclusion
min u(x, y) = −8,
(x,y)∈D
max u(x, y) = 0
(x,y)∈D
Question 62: Consider the square D = (−1, +1)×(−1, +1). Let f (x, y) = x2 − y 2 − 3. Let
u ∈ C 2 (D) ∩ C 0 (D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) and
max(x,y)∈D u(x, y).
Solution: We use the maximum principle (u is harmonic and has the required regularity). Then
min u(x, y) =
(x,y)∈D
min
f (x, y),
and
(x,y)∈∂D
max u(x, y) =
(x,y)∈D
max f (x, y).
(x,y)∈∂D
A point (x, y) is at the boundary of D if and only if x2 = 1 and y ∈ (−1, 1) or y 2 = 1 and
x ∈ (−1, 1). In the first case, x2 = 1 and y ∈ (−1, 1), we have
f (x, y) = 1 − y 2 − 3,
y ∈ (−1, 1).
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24
The maximum is −2 and the minimum is −3. In the second case, y 2 = 1 and x ∈ (−1, 1), we have
f (x, y) = x2 − 1 − 3,
x ∈ (−1, 1).
The maximum is −3 and the minimum is −4. We finally can conclude
min
f (x, y) =
(x,y)∈∂D
min x2 − 4, = −4,
max f (x, y) = max −2 − y 2 = −2.
−1≤x≤1
(x,y)∈∂D
−1≤y≤1
In conclusion
min u(x, y) = −4,
max u(x, y) = −2
(x,y)∈D
(x,y)∈D
Question 63: Consider the triangular domain D = {(x, y); x ≥ 0, y ≥ 0, 1 − x − y ≤ 0}. Let
f (x, y) = x2 − y 2 − 3. Let u ∈ C 2 (D) ∩ C 0 (D) solve −∆u = 0 in D and u|∂D = f . Compute
min(x,y)∈D u(x, y) and max(x,y)∈D u(x, y).
Solution: We use the maximum principle (u is harmonic and has the required regularity). Then
min u(x, y) =
min
f (x, y),
and
max u(x, y) =
(x,y)∈∂D
(x,y)∈D
max f (x, y).
(x,y)∈∂D
(x,y)∈D
A point (x, y) is at the boundary of D if and only if {x = 0 and y ∈ [0, 1]} or {y = 0 and x ∈ [0, 1]},
or {1 − y − x = 0 and x ∈ [0, 1]}.
(i) In the first case, x = 0 and y ∈ [0, 1], we have
f (x, y) = −y 2 − 3,
y ∈ [0, 1].
The maximum is −3 and the minimum is −4.
(ii) In the second case, y = 0 and x ∈ [0, 1], we have
f (x, y) = x2 − 3,
x ∈ [0, 1].
The maximum is −2 and the minimum is −3.
(iii) In the third case, 1 − x = y and x ∈ [0, 1], we have
f (x, y) = x2 − (1 − x)2 − 3 = 2x − 4,
x ∈ [0, 1].
The maximum is −2 and the minimum is −4.
We finally can conclude
min
f (x, y) = −4,
(x,y)∈∂D
max f (x, y) = −2.
(x,y)∈∂D
In conclusion
min u(x, y) = −4,
(x,y)∈D
max u(x, y) = −2
(x,y)∈D
Question 64: Consider the elliptic domain D = {(x, y); x2 +2y 2 ≤ 2}. Let f (x, y) = x2 −y 2 −3.
Let u ∈ C 2 (D) ∩ C 0 (D) solve −∆u = 0 in D and u|∂D = f . Compute min(x,y)∈D u(x, y) and
max(x,y)∈D u(x, y).
Solution: We use the maximum principle (u is harmonic and has the required regularity). Then
min u(x, y) =
(x,y)∈D
min
(x,y)∈∂D
f (x, y),
and
max u(x, y) =
(x,y)∈D
max f (x, y).
(x,y)∈∂D
A point (x, y) is at the boundary of D if and only if x2 + 2y 2 = 2, meaning that x2 = 2 − 2y 2 . In
conclusion, for any point (x, y) ∈ ∂D we have f (x, y) = x2 − y 2 − 3 = 2 − 2y 2 − y 2 − 3 = −1 − 3y 2
and y ∈ [−1, +1].
(1) The maximum of −1−3y 2 over the interval [−1, +1] is reached for y = 0. Hence max(x,y)∈∂D f (x, y) =
−1.
(2) The minimum of −1−3y 2 over the interval [−1, +1] is reached for y = ±1. Hence min(x,y)∈∂D f (x, y) =
Math 602
−4.
We finally can conclude
min
f (x, y) = −4,
(x,y)∈∂D
max f (x, y) = −1.
(x,y)∈∂D
In conclusion
min u(x, y) = −4,
(x,y)∈D
max u(x, y) = −1.
(x,y)∈D
25
Math 602
4
26
Eigenvalue problems
Question 65: Consider the differential equation
the boundary conditions φ(0) = 0, φ(π) = 0.
(a) What is the sign of λ? Prove your answer.
d2 φ
dt2
+ λφ = 0, t ∈ (0, π), supplemented with
Solution:
(b) Compute all the possible eigenvalues λ for this problem.
Solution:
Question 66: Consider the differential equation
the boundary conditions φ(0) = 0, φ0 (π) = 0.
(a) What is the sign of λ? Prove your answer.
d2 φ
dt2
+ λφ = 0, t ∈ (0, π), supplemented with
Solution: Multiply the equation by φ and integrate over the domain.
Z π
Z π
0
2
0 π
φ2 (t)dt = 0.
(φ (t)) dt + φ φ|0 + λ
−
0
0
Using the BCs, we infer
Z
−
π
0
2
Z
(φ (t)) dt = λ
0
π
φ2 (t)dt,
0
which means that λ is non-negative.
(b) Compute all the possible eigenvalues λ for this problem and compute φ.
Solution: There are two cases: Either λ > 0 or λ = 0. Assume first λ > 0, then
√
√
φ(t) = c1 cos( λt) + c2 sin( λt).
√
The boundary condition φ(0) = 0 implies imply c1 = 0. The other BC implies √
c2 cos( λπ) = 0.
c2 = 0 gives φ = 0, which is not a proper eigenfunction. The other possibility is λπ = (n + 12 )π,
n ∈ N. In conclusion
φ(t) = c2 sin((n + 21 )t),
n ∈ N.
The case λ = 0 gives φ(t) = c1 + c2 t. The BCs imply c1 = c2 = 0, i.e. φ = 0, which is not a
proper eigenfunction.
Question 67: Let Ω = (0, L) and let (λ, u) be an eigenpair of the Laplace equation over Ω with
zero Dirichlet condition. Assume that λ ∈ C and that the function u(x) is complex-valued.
(i) Write the PDE solved by u.
Solution: u and λ are such
−∂xx u(x) = λu(x),
u(0) = 0,
u(L) = 0.
(ii) Let ū be the complex conjugate of u. Write the PDE solved by ū (Hint: take the conjugate
of (i)).
Solution: Take the complex conjugate of (i) we obtain
−∂xx ū(x) = −∂xx u(x) = λu(x) = λ̄ū(x),
ū(0) = 0,
ū(L) = 0
which gives
−∂xx ū(x) = λ̄ū(x),
ū(0) = 0,
ū(L) = 0.
Note that we used the fact that ∂x u = ∂x ū.
(iii) Prove that λ ∈ R (Hint: Use an energy argument with ū in (i) and an energy argument
with u (ii) and conclude that λ = λ̄. Recall u 6= 0 and |z|2 = z z̄ for all z ∈ C).
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27
Solution: Multiply (i) by ū and integrate over Ω
Z
Z
−∂xx u(x)ū(x)dx =
λu(x)ū(x)dx
Ω
Ω
Z
Z
∂x u(x)∂x ū(x)dx − [∂x u(x)ū(x)]L
|u(x)|2 dx
0 =λ
Ω
ZΩ
Z
|u(x)|2 dx
∂x u(x)∂x u(x)dx = λ
Ω
Ω
Z
Z
2
|∂x u(x)| dx = λ
|u(x)|2 dx.
Ω
Ω
Multiply (ii) by u and integrate over Ω
Z
Z
−∂xx ū(x)u(x)dx =
λ̄ū(x)u(x)dx
Ω
Z
Ω
Ω
Z
∂x ū(x)∂x u(x)dx − [∂x ū(x)u(x)]L
|u(x)|2 dx
0 = λ̄
Ω
Z
Z
∂x u(x)∂x u(x)dx = λ̄
|u(x)|2 dx
Ω
Z
ZΩ
2
|∂x u(x)| dx = λ̄
|u(x)|2 dx.
Ω
In conclusion
Z
λ
Ω
|u(x)|2 dx = λ̄
Z
Ω
|u(x)|2 dx,
Ω
which means
Z
(λ − λ̄)
|u(x)|2 dx.
Ω
R
This in turn implies that λ = λ̄ since Ω |u(x)|2 dx is not zero (recall u 6= 0). In conclusion λ is
real. Note in passing that this also prove that λ ≥ 0.
1
1
d
d
(t 2 dt
φ(t)) = λt− 2 φ(t), t ∈ (0, 1), suppleQuestion 68: Consider the eigenvalue problem − dt
mented with the boundary condition φ(0) = 0, ∂t φ(1) = 0.
(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.
Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ and
integrate over the domain. Use the Fundamental Theorem of calculus (i.e., integration by parts) to
obtain
Z 1
Z 1
1
1
1
t 2 (φ0 (t))2 dt − [t 2 φ0 (t)φ(t)]10 = λ
t− 2 φ2 (t)dt.
0
0
Using the boundary conditions, we infer
Z 1
Z
1
t 2 (φ0 (t))2 dt = λ
0
1
1
t− 2 φ2 (t)dt,
0
which means that λ is non-negative since φ is non-zero.
R1 1
(ii) If λ = 0, then 0 t 2 (φ0 (t))2 dt = 0, which implies that φ0 (t) = 0 for all t ∈ (0, 1]. This implies
that φ(t) is constant, and this constant is zero since φ(0) = 0. Hence, φ is zero if λ = 0. Since we
want a nonzero solution, this implies that λ cannot be zero.
(iii) In conclusion, it is necessary that λ be positive for a nonzero smooth solution to exist.
√√
√√
1 d
1
d
(b) The general solution to − dt
(t 2 dt
φ(t)) = λt− 2 φ(t) is φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ)
for λ ≥ 0. Find all the eigenvalues λ > 0 and the associated nonzero eigenfunctions.
Solution: Since λ ≥ 0 by hypothesis, φ is of the following form
√√
√√
φ(t) = φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ).
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28
The boundary
condition
φ(0) = 0 implies c1 = 0. The other boundary condition implies ∂x φ(1) =
√
√
0√
= c2 λ cos(2 λ). The constant c2 cannot be zero since we want φ to be nonzero; as a result,
2 λ = (n + 12 )π, n = 0, 2, . . .. In conclusion
λ = ((2n + 1)π)2 /16,
1 √
φ(t) = c sin((n + )π t).
2
n = 1, 2, . . . ,
1
1
d
d
Question 69: Consider the eigenvalue problem − dt
(t 2 dt
φ(t)) = λt− 2 φ(t), t ∈ (0, 1), supplemented with the boundary condition φ(0) = 0, φ(1) = 0.
(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.
Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ and
integrate over the domain. Use the fundamental Theorem of calculus (i.e., integration by parts) to
obtain
Z 1
Z 1
1
1
1
0
2
0
1
2
2
t− 2 φ2 (t)dt.
t (φ (t)) dt − [t φ (t)φ(t)]0 = λ
0
0
Using the boundary conditions, we infer
Z 1
Z
1
t 2 (φ0 (t))2 dt = λ
0
1
1
t− 2 φ2 (t)dt,
0
which means that λ is non-negative since φ is non-zero.
R1 1
(ii) If λ = 0, then 0 t 2 (φ0 (t))2 dt = 0, which implies that φ0 (t) = 0 for all t ∈ (0, 1]. The
Rt
fundamental theorem of calculus applied between t and 1 implies φ(t) = φ(1) + 1 φ0 (τ )dτ = 0
since φ(1) = 0 and φ0 (τ ) = 0 for all τ ∈ (t, 1]. Hence, φ is zero if λ = 0. Since we want a nonzero
solution, this implies that λ cannot be zero.
(iii) In conclusion, it is necessary that λ be positive for a nonzero smooth solution to exist.
√√
√√
1 d
1
d
(b) The general solution to − dt
(t 2 dt
φ(t)) = λt− 2 φ(t) is φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ)
for λ ≥ 0. Find all the eigenvalues λ ≥ 0 and the associated nonzero eigenfunctions.
Solution: Since λ ≥ 0 by hypothesis, φ is of the following form
√√
√√
φ(t) = φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ).
The boundary
√ condition φ(0) = 0 implies c1 = 0. The other boundary condition implies φ(1) =
0√= c2 sin(2 λ). The constant c2 cannot be zero since we want φ to be nonzero; as a result,
2 λ = nπ, n = 1, 2, . . .. In conclusion
√
λ = (nπ)2 /4, n = 1, 2, . . . ,
φ(t) = c sin(nπ t).
2
d
d
Question 70: Consider the eigenvalue problem − dt
2 φ(t) + 2 dt φ(t) = λφ(t), t ∈ (0, π), suppled
d 2
mented with the boundary condition φ(0) = 0, φ(π) = 0. (Hint: 2φ(t) dt
φ(t) = dt
φ (t).)
(a) Prove that it is necessary that λ be positive for a non-zero solution to exist.
Solution: (i) Let φ be a non-zero solution to the problem. Multiply the equation by φ and integrate
over the domain. Use the fundamental Theorem of calculus and use the hint to obtain
Z π
Z π
Z π
d 2
φ2 (t)dt
(φ0 (t))2 dt − φ0 (π)φ(π) + φ0 (0)φ(0) +
(φ (t))dt = λ
dt
0
0
Z π
Z0 π
0
2
0
0
2
2
(φ (t)) dt − φ (π)φ(π) + φ (0)φ(0) + φ (π) − φ (0) = λ
φ2 (t)dt
0
0
Using the boundary conditions, we infer
Z π
Z
0
2
(φ (t)) dt = λ
0
0
π
φ2 (t)dt,
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29
which means that λ is non-negative since φ is non-zero.
Rπ
(ii) If λ = 0, then 0 (φ0 (t))2 dt = 0 and φ(π)2 = 0, which implies that φ0 (t) = 0. The fundamental
Rt
theorem of calculus implies φ(t) = φ(0) + 0 φ0 (τ )dτ = 0. Hence, φ is zero if λ = 0. Since we
want a nonzero solution, this implies that λ cannot be zero.
(iii) In conclusion, it is necessary that λ be positive for a nonzero solution to exist.
√
√
(b) The general solution to −φ00 + 2φ0 = λφ is φ(t) = et (c1 cos(t λ − 1) + c2 sin(t λ − 1)) for
λ ≥ 1. Find all the eigenvalues λ ≥ 1 and the associated nonzero eigenfunctions.
Solution: Since λ ≥ 1 by hypothesis, φ is of the following form
√
√
φ(t) = et (c1 cos( λ − 1t) + c2 sin( λ − 1t)).
The boundary√condition φ(0) = 0 implies c1 = 0. The other boundary condition implies φ(π) =
0√= eπ c2 sin( λ − 1π). The constant c2 cannot be zero since we want φ to be nonzero; as a result,
λ − 1 = n, n = 1, 2, . . .. In conclusion
λ = n2 + 1,
φ(t) = cet sin(nt)
n = 1, 2, . . . ,
1
1
d
d
Question 71: Consider the eigenvalue problem − dt
(t 2 dt
φ(t)) = λt− 2 φ(t), t ∈ (0, 1), supplemented with the boundary condition φ(0) = 0, φ(1) = 0.
(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.
Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ and
integrate over the domain. Use the fundamental Theorem of calculus (i.e., integration by parts) to
obtain
Z 1
Z 1
1
1
1
t 2 (φ0 (t))2 dt − [t 2 φ0 (t)φ(t)]10 = λ
t− 2 φ2 (t)dt.
0
0
Using the boundary conditions, we infer
Z 1
Z
1
t 2 (φ0 (t))2 dt = λ
0
1
1
t− 2 φ2 (t)dt,
0
which means that λ is non-negative since φ is non-zero.
R1 1
(ii) If λ = 0, then 0 t 2 (φ0 (t))2 dt = 0, which implies that φ0 (t) = 0 for all t ∈ (0, 1]. The
Rt
fundamental theorem of calculus applied between t and 1 implies φ(t) = φ(1) + 1 φ0 (τ )dτ = 0
since φ(1) = 0 and φ0 (τ ) = 0 for all τ ∈ (t, 1]. Hence, φ is zero if λ = 0. Since we want a nonzero
solution, this implies that λ cannot be zero.
(iii) In conclusion, it is necessary that λ be positive for a nonzero smooth solution to exist.
√√
√√
1 d
1
d
(b) The general solution to − dt
(t 2 dt
φ(t)) = λt− 2 φ(t) is φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ)
for λ ≥ 0. Find all the eigenvalues λ ≥ 0 and the associated nonzero eigenfunctions.
Solution: Since λ ≥ 0 by hypothesis, φ is of the following form
√√
√√
φ(t) = φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ).
The boundary
√ condition φ(0) = 0 implies c1 = 0. The other boundary condition implies φ(1) =
0√= c2 sin(2 λ). The constant c2 cannot be zero since we want φ to be nonzero; as a result,
2 λ = nπ, n = 1, 2, . . .. In conclusion
√
λ = (nπ)2 /4, n = 1, 2, . . . ,
φ(t) = c sin(nπ t).
2
Question 72: Consider the differential equation − ddtφ2 = λφ, t ∈ (0, π), supplemented with the
boundary conditions 2φ(0) = φ0 (0), φ(π) = 0.
(a) What should be the sign of λ for a non-zero solution to exist? Prove your answer.
Math 602
30
Solution: Let φ be a non-zero solution to the problem. Multiply the equation by φ and integrate
over the domain.
Z π
Z π
0
2
0
0
(φ (t)) dt − φ (π)φ(π) + φ (0)φ(0) = λ
φ2 (t)dt.
0
0
Using the BCs, we infer
Z
π
(φ0 (t))2 dt + 2φ(0)2 = λ
Z
π
φ2 (t)dt,
0
0
which means that λ is non-negative since φ is non-zero.
√
√
√
(b) Assume that 2 sin( λπ) + λ cos( λπ) 6= 0. Prove that φ = 0 is the only possible solution.
√
√
√
Solution: Observe first that λ cannot be zero, otherwise we would have 2 sin( λπ)+ λ cos( λπ) =
0. As a result, (a) implies that λ is positive. Then φ is of the following form
√
√
φ(t) = c1 cos( λt) + c2 sin( λt).
√
√
The boundary
condition φ0 (0) = 2φ(0) implies 2c1 − λc2 = 0. The other BC implies c1 cos( λπ)+
√
c2 sin( λπ) = 0. The constants c1 and c2 solve the following linear system
(
√
2c1 − λc2 = 0
√
√
c1 cos( λπ) + c2 sin( λπ) = 0
√
√
√
The determinant is equal to 2 sin( λπ) + λ cos( λπ) and is non-zero by hypothesis. Then the
only solution is c1 = c2 = 0, which again gives φ = 0.
√
√
√
In conclusion, the only possible solution to the problem is φ = 0 if 2 sin( λπ) + λ cos( λπ) 6= 0.
1
1
d
d
Question 73: Consider the eigenvalue problem − dt
(t 2 dt
φ(t)) = λt− 2 φ(t), t ∈ (0, 1), supplemented with the boundary condition φ(0) = 0, ∂t φ(1) = 0.
(a) Prove that it is necessary that λ be positive for a non-zero smooth solution to exist.
Solution: (i) Let φ be a non-zero smooth solution to the problem. Multiply the equation by φ and
integrate over the domain. Use the Fundamental Theorem of calculus (i.e., integration by parts) to
obtain
Z 1
Z 1
1
1
1
t 2 (φ0 (t))2 dt − [t 2 φ0 (t)φ(t)]10 = λ
t− 2 φ2 (t)dt.
0
0
Using the boundary conditions, we infer
Z 1
Z
1
t 2 (φ0 (t))2 dt = λ
0
1
1
t− 2 φ2 (t)dt,
0
which means that λ is non-negative since φ is non-zero.
R1 1
(ii) If λ = 0, then 0 t 2 (φ0 (t))2 dt = 0, which implies that φ0 (t) = 0 for all t ∈ (0, 1]. This implies
that φ(t) is constant, and this constant is zero since φ(0) = 0. Hence, φ is zero if λ = 0. Since we
want a nonzero solution, this implies that λ cannot be zero.
(iii) In conclusion, it is necessary that λ be positive for a nonzero smooth solution to exist.
√√
√√
1 d
1
d
(b) The general solution to − dt
(t 2 dt
φ(t)) = λt− 2 φ(t) is φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ)
for λ ≥ 0. Find all the eigenvalues λ > 0 and the associated nonzero eigenfunctions.
Solution: Since λ ≥ 0 by hypothesis, φ is of the following form
√√
√√
φ(t) = φ(t) = c1 cos(2 t λ) + c2 sin(2 t λ).
The boundary
condition
φ(0) = 0 implies c1 = 0. The other boundary condition implies ∂x φ(1) =
√
√
0√
= c2 λ cos(2 λ). The constant c2 cannot be zero since we want φ to be nonzero; as a result,
2 λ = (n + 12 )π, n = 0, 2, . . .. In conclusion
λ = ((2n + 1)π)2 /16,
n = 1, 2, . . . ,
1 √
φ(t) = c sin((n + )π t).
2
Math 602
31
Question 74: Let p, q : [−1, +1] −→ R be smooth functions. Assume that p(x) ≥ 0 and q(x) ≥
q0 for all x ∈ [−1, +1], where q0 ∈ R. Consider the eigenvalue problem −∂x (p(x)∂x φ(x)) +
q(x)φ(x) = λφ(x), supplemented with the boundary conditions φ(−1) = 0 and φ(1) = 0.
(a) Prove that it is necessary that λ ≥ q0 for a non-zero (smooth) solution, φ, to exist. (Hint:
R +1
R +1
q0 −1 φ2 (x)dx ≤ −1 q(x)φ2 (x)dx.)
Solution: As usual we use the energy method. Let (φ, λ) be an eigenpair, then
Z +1
Z +1
2
(−∂x (p(x)∂x φ(x))φ(x) + q(x)φ (x))dx = λ
φ2 (x)dx.
−1
−1
After integration by parts and using the boundary conditions, we obtain
Z +1
Z +1
(p(x)∂x φ(x)∂x φ(x) + q(x)φ2 (x))dx = λ
φ2 (x)dx.
−1
−1
which, using the hint, can also be re-written
Z +1
Z
(p(x)∂x φ(x)∂x φ(x) + q0 φ2 (x))dx ≤ λ
−1
+1
φ2 (x)dx.
−1
Then
Z
+1
p(x)(∂x φ(x))2 dx ≤ (λ − q0 )
−1
+1
Z
φ2 (x)dx.
−1
Assume that φ is non-zero, then
R +1
λ − q0 ≥
−1
p(x)(∂x φ(x))2 dx
≥ 0,
R +1
φ2 (x)dx
−1
which proves that it is necessary that λ ≥ q0 for a non-zero (smooth) solution to exist.
(b) Assume that p(x) ≥ p0 > 0 for all x ∈ [−1, +1] where p0 ∈ R+ . Show that λ = q0 cannot
R +1
R +1
be an eigenvalue, i.e., prove that φ = 0 if λ = q0 . (Hint: p0 −1 ψ 2 (x)dx ≤ −1 p(x)ψ 2 (x)dx.)
Solution: Assume that λ = q0 is an eigenvalue. Then the above computation shows that
Z +1
Z +1
p0
(∂x φ(x))2 dx ≤
p(x)(∂x φ(x))2 dx = 0,
−1
−1
R +1
which means that −1 (∂x φ(x))2 dx = 0 since p0 > 0. As a result ∂x φ = 0, i.e., φ(x) = c where c
is a constant. The boundary conditions φ(−1) = 0 = φ(1) imply that c = 0. In conclusion φ = 0
if λ = q0 , thereby proving that (φ, q0 ) is not an eigenpair.
Question 75: Let φ be a non-zero solution to the eigenvalue problem −∂xx φ(x) = λφ(x),
x ∈ (0, π), φ(0) = 0, φ(π) = 0. Determine the sign of λ using the energy method.
Solution: Multiply the equation by φ, integrate over (0, π), and apply the fundamental theorem of
calculus (i.e. integrate by parts):
Z π
Z π
φ(x)∂xx φ(x)dx = −
(∂x (φ(x)∂x φ(x)) − (∂x φ(x))2 )dx
−
0
0
Z π
Z π
2
= −φ(π)∂x φ(π) + φ(0)∂x φ(0) +
(∂x φ(x)) dx = λ
(φ(x))2 dx.
0
In conclusion
Z
0
π
(∂x φ(x))2 dx = λ
Z
0
π
(φ(x))2 dx.
0
Math 602
32
Rπ
Rπ
Assuming that φ is nonzero, we obtain that λ = 0 (∂x φ(x))2 dx/ 0 (φ(x))2 dx ≥ 0, i.e. λ is
non-negative. If λ = 0 then ∂x φ = 0, which implies that φ is constant. The boundary conditions
imply that φ = 0 which contradicts our assumption that φ is non-zero. In conclusion λ is negative.
Question 76: Let φ be a non-zero solution to the eigenvalue problem −∂xx φ(x) = λφ(x),
x ∈ (0, π), φ(0) = 0, ∂x φ(π) + φ(π) = 0. Determine the sign of λ using the energy method.
Solution: Multiply the equation by φ, integrate over (0, π), and apply the fundamental theorem of
calculus (i.e. integrate by parts):
Z π
Z π
Z π
λ
(φ(x))2 dx = −
φ(x)∂xx φ(x)dx = −
(∂x (φ(x)∂x φ(x)) − (∂x φ(x))2 )dx
0
0
0
Z π
= −φ(π)∂x φ(π) + φ(0)∂x φ(0) +
(∂x φ(x))2 dx
0
Z π
(∂x φ(x))2 dx.
= (φ(π))2 +
0
In conclusion
(φ(π))2 +
Z
π
(∂x φ(x))2 dx = λ
Z
0
π
(φ(x))2 dx.
0
Rπ
Rπ
Assuming that φ is nonzero, we obtain that λ = ((φ(π))2 + 0 (∂x φ(x))2 dx)/ 0 (φ(x))2 dx ≥ 0,
i.e. λ is non-negative. If λ = 0 then φ(π) = 0 and ∂x φ = 0, which implies that φ is constant. The
other condition φ(π) = 0 implies that φ = 0 which contradicts our assumption that φ is non-zero.
In conclusion λ is positive.
Question 77: Compute all the positive eigenvalues λ to the above eigenvalue problem.
Solution:
Owing to√λ being non-negative, the general solution to −∂xx φ(x) = λφ(x) is φ(x) =
√
a cos(√ λx) + b sin( λx). The
√ boundary conditions imply that φ(0) = 0 = a and φ(π) =
b sin( λπ) = 0. As a result, λπ = nπ, where n ∈ N, i.e.
λ = n2 ,
n ∈ N.
Question 78: Let φ be a non-zero solution to the eigenvalue problem −∂x (1 + x2 )∂x φ(x) =
λφ(x), x ∈ (0, π), φ(π) = 0, −∂x φ(0) + φ(0) = 0. Determine the sign of λ using the energy
method.
Solution: Multiply the equation by φ, integrate over (0, π), and apply the fundamental theorem of
calculus (i.e. integrate by parts):
Z π
Z π
Z π
λ
(φ(x))2 dx = −
φ(x)∂x (1 + x2 )∂x φ(x) dx = −
(∂x (φ(x)(1 + x2 )∂x φ(x)) − (1 + x2 )(∂x φ(x))2 )dx
0
0
Z0 π
2
= −(1 + π )φ(π)∂x φ(π) + φ(0)∂x φ(0) +
(1 + x2 )(∂x φ(x))2 dx
0
Z π
= (φ(0))2 +
(1 + x2 )(∂x φ(x))2 dx.
0
In conclusion
(φ(0))2 +
Z
0
π
(1 + x2 )(∂x φ(x))2 dx = λ
Z
π
(φ(x))2 dx.
0
Rπ
Rπ
Assuming that φ is nonzero, we obtain that λ = (φ2 (0) + 0 (1 + x2 )(∂x φ(x))2 dx)/ 0 (φ(x))2 dx ≥
0, i.e. λ is non-negative. If λ = 0 then φ(0) = 0 and ∂x φ = 0, which implies that φ is constant. The
other condition φ(0) = 0 implies that φ = 0 which contradicts our assumption that φ is non-zero.
In conclusion λ is positive.
Math 602
5
33
Fourier Series
Here are some formulae that you may want to use:
F S(f )(x) =
+∞
X
an cos(nπx/L) +
n=0
1
a0 =
2L
Z
L
f (x)dx,
−L
1
an =
L
Z
+∞
X
bn sin(nπx/L),
n=1
L
x
f (x) cos(nπ )dx,
L
−L
(1)
1
bn =
L
Z
L
x
f (x) sin(nπ )dx
L
−L
(2)
Question 79: Compute the complex Fourier series of the function f (x) = x defined on [−π, π].
P+∞
Solution: By definition F S(f )(x) = −∞ cn e−inπx/L where L is half the size of the interval on
which f is defined. Here L = π. Hence, by integrating by parts once we obtain
Z π
Z π
1 1
1 1
1
−inx
xe
=−
e−inx +
cn =
(πe−inπ + πeinπ )
2π −π
2π −in −π
2π −in
1 1
2π(−1)n .
=
2π −in
That is cn =
(−1)n−1
in
and
F S(f )(x) =
+∞
X
(−1)n−1
in
−∞
e−inx
Question 80: Let f (x) = x, x ∈ [−L, L].
(a) Sketch the graph of the Fourier series of f for x ∈ (−∞, ∞).
Solution: The Fourier series is equal to the periodic extension of f , except at the points nL, n ∈ Z
where it is equal to 0 = 12 (1 − 1).
(b) Compute the Fourier series of f .
Solution: f is odd, hence the cosine coefficients are zero. The sign coefficients bn are given by
bn =
1
L
Z
L
x sin(
−L
nπx
L 1
)dx =
L
nπ L
Z
L
cos(
−L
nπx
L
)dx − 2
cos(nπ).
L
nπ
L
As a result bn = 2(−1)n+1 nπ
and
F S(f )(x) =
∞
X
2(−1)n
1
L
nπx
sin(
).
nπ
L
Question 81: Let N be a positive integer and let PN be the set of trigonometric polynomials
of degree at most N ; that is, PN = span{1, cos(x), sin(x), . . . , cos(N x), sin(N x)}.
(i) Consider the function f : R −→ R defined by
∞
X
1
f (x) =
sin(7n) cos(2nx).
5
n
n=1
Compute the best L2 -approximation of f in P7 over (0, 2π).
Solution: The best L2 -approximation of f in P7 over (0, 2π) is the truncated Fourier series F S7 (f ).
Clearly
3
X
1
F S7 (f )(x) =
sin(7n) cos(2nx)
n5
n=1
Math 602
34
(ii) Consider the function f : R −→ R defined by
f (x) =
11
X
n=1
n3
1
cos(35n2 ) sin(3nx).
+1
Compute the best L2 -approximation of f in P35 over (0, 2π).
Solution: The best L2 -approximation of f in P35 over (0, 2π) is the truncated Fourier series
F S35 (f ). Observing that f ∈ P33 ⊂ P35 it is clear that F S35 (f ) = f .
Question 82: Let f (x) = x, x ∈ [−L, L].
(a) Sketch the graph of the Fourier series of f for x ∈ (−∞, ∞).
Solution: The Fourier series is equal to the periodic extension of f , except at the points (2n + 1)L,
n ∈ Z where it is equal to 0 = 21 (1 − 1).
(b) Compute the Fourier series of f .
Solution: f is odd, hence the cosine coefficients are zero. The sign coefficients bn are given by
1
bn =
L
Z
L
nπx
L 1
x sin(
)dx =
L
nπ
L
−L
Z
L
cos(
−L
nπx
L
)dx − 2
cos(nπ).
L
nπ
L
L
As a result bn = −2 cos(nπ) nπ
= 2(−1)n+1 nπ
and
∞
X
F S(f )(x) =
2(−1)n
1
nπx
L
sin(
).
nπ
L
Question 83: Let L be a positive real number.
(a) Compute the Fourier series of the function (−L, L) 3 x 7−→ x2 .
Solution: The Function is even; as a result its sine coefficients are zero. We compute the cosine
coefficients as follows:
Z +L
Z +L
2
x cos(mπx/L)dx = am
cos(mπx/L)2 dx,
−L
If m = 0, c0 =
1
2L
R +L
−L
−L
x2 dx = 13 L. Otherwise,
am =
+L
1
L
Z
Z
+L
x2 cos(mπx/L)dx.
−L
Integration by parts two times gives
am = −
=2
1 L
L mπ
2x sin(mπx/L)dx
−L
L
L2
m
x cos(mπx/L)|L
−L = 4 2 2 (−1) .
2
2
m π
m π
(b) For which values of x is the Fourier series equal to x2 ?
Solution: The periodic extension of x2 over R is piecewise smooth and globally continuous. This
means that the Fourier series is equal to x2 over the entire interval [−L, +L].
(c) Using (a) and (b) give the Fourier series of x over [−L, +L] and say where it is equal to x.
Solution: Since we can differentiate cosine series, the Fourier series of x over [−L, +L] is obtained
by differentiating that of x2 ,
∞
F S(x)(x) =
X
d
1
L
nπx
F S( x2 )(x) =
2(−1)n+1
sin(
).
dx
2
nπ
L
1
Math 602
35
We have equality x = F S(x)(x) only on (−L, +L). At −L and +L the Fourier series is zero.
Question 84: Let f : [−1, +1] −→ R be such that f (x) = x, if x ∈ [0, 1] and f (x) = 0 if
x ∈ [−1, 0].
(a) Sketch the graph of the Fourier series of f on (−∞, ∞).
Solution: The Fourier series is equal to the periodic extension of f , except at the points (2n + 1)L,
n ∈ Z, where it is equal to 21 = 12 (1 − 0).
R1
(b) Compute the Fourier coefficients of f (recall a0 = 12 −1 f (x)dx, and for n ≥ 1, an =
R1
R1
f (x) cos(nπx)dx, bn = −1 f (x) sin(nπx)dx. Hint: integrate by parts).
−1
Solution: Clearly a0 = 14 . For n ≥ 1 we have
Z
1
x cos(nπx)dx = −
an =
0
1
nπ
Z
1
sin(nπx) +
0
1
(−1)n − 1
1
cos(nπx)|10 =
x sin(nπx)|10 =
2
nπ
(nπ)
(nπ)2
and
1
Z
bn =
0
1
x sin(nπx)dx =
nπ
Z
1
cos(nπx) −
0
1
(−1)n+1
x cos(nπx)|10 =
.
nπ
nπ
Question 85: Consider f : [−L, L] −→ R, f (x) = x2 . (a) Sketch the graph of the Fourier
series of f .
Solution: F S(f ) is equal to the periodic extension of f (x) over R.
(b) For which values of x is F S(f ) equal to x2 ?
Solution: The periodic extension of f (x) = x2 over R is piecewise smooth and globally continuous.
This means that the Fourier series is equal to x2 over the entire interval [−L, +L].
(c) Is it possible to obtain F S(x) by differentiating 12 F S(f ) term by term?
Solution: Yes it is possible since the periodic extension of f (x) = x2 over R is continuous and
piecewise smooth.
Question 86: Let f (x) = x, x ∈ [−L, L]. (a) Sketch the graph of the Fourier series of f .
Solution: The Fourier series is equal to the periodic extension of f , except at the points (2n + 1)L,
n ∈ Z where it is equal to 0 = 21 (1 − 1).
Rb
R
Rb R
(b) Compute the coefficients of the Fourier series of f . (Hint: a tg(t)dt = [t g]ba − a ( g)(t)dt).
Solution: f is odd, hence the cosine coefficients are zero. The sine coefficients bn are obtained by
integration by parts
bn =
1
L
Z
L
x sin(
−L
nπx
1 L
x
L 1
)dx = −
[x cos(nπ )]+L
+
L
L nπ
L −L nπ L
Z
L
cos(
−L
nπx
)dx.
L
L
L
= 2(−1)n+1 nπ
and
As a result bn = −2 cos(nπ) nπ
∞
2L X (−1)n+1
nπx
F S(f )(x) =
sin(
).
π 1
n
L
Question 87: Let L be a positive real number. Let V = span{1, cos(πt/L), sin(πt/L)} and
R
21
L
consider the norm kf kL2 = −L f (t)2 dt . (a) Compute the best approximation of 1 + t in V
with respect to the above norm.
Solution: We know from class that the truncated Fourier series
F S1 (t) = a0 + a1 cos(πt/L) + b1 sin(πt/L)
Math 602
36
is the best approximation. Now we compute a0 , a1 , a2
Z L
1
(1 + t)dt = 1,
2L −L
Z
1 L
(1 + t) cos(πt/L)dt = 0
a1 =
L −L
Z
Z
1 L
1 L
L
2L
b1 =
(1 + t) sin(πt/L)dt =
t sin(πt/L)dt = −2 cos(π) =
.
L −L
L −L
π
π
a0 =
As a result
F S1 (t) = 1 +
2L
sin(πt/L)
π
(b) Compute the best approximation of 3 + 2 cos(πt/L) − 5 sin(πt/L) in V .
Solution: The function 3 + 2 cos(πt/L) − 5 sin(πt/L) is a member of V ; as a result, The best
approximation is the function itself.
Question 88: Consider f : [−L, L] −→ R, f (x) = x4 . (a) Sketch the graph of the Fourier
series of f .
Solution: F S(f ) is equal to the periodic extension of f (x) over R.
(b) For which values of x ∈ R is F S(f ) equal to x4 ? (Explain)
Solution: The periodic extension of f (x) = x4 over R is piecewise smooth and globally continuous
since f (L) = f (−L). This means that the Fourier series is equal to x4 over the entire interval
[−L, +L].
(c) Is it possible to obtain F S(x3 ) by differentiating 14 F S(x4 ) term by term? Which values this
legitimate? (Explain)
Solution: Yes it is possible since the periodic extension of f (x) = x4 over R is continuous and piecewise smooth. This operation is legitimate everywhere the function F S(x3 ) is smooth, i.e., for all the
points in R\{(2k−1)L, k ∈ Z} (i.e., one needs to exclude the points . . . , −7L, −5L, −3L, −L, +L, +3L, +5L, +7L, . . .
Question 89: Let L be a positive real number. Let P1 = span{1, cos(πt/L), sin(πt/L)} and
R
21
L
consider the norm kf kL2 = −L f (t)2 dt . (a) Compute the best approximation of 2 − 3t in
RL
P1 with respect to the above norm. (Hint: −L t sin(πt/L)dt = 2L2 /π.)
Solution: We know from class that the truncated Fourier series
F S1 (t) = a0 + a1 cos(πt/L) + b1 sin(πt/L)
is the best approximation. Now we compute a0 , a1 , a2
Z L
1
(2 − 3t)dt = 2,
2L −L
Z
1 L
a1 =
(2 − 3t) cos(πt/L)dt = 0
L −L
Z
Z
1 L
1 L
L
6L
(2 − 3t) sin(πt/L)dt =
−3t sin(πt/L)dt = −6 cos(π) = − .
b1 =
L −L
L −L
π
π
a0 =
As a result
F S1 (t) = 2 −
6L
sin(πt/L)
π
(b) Compute the best approximation of h(t) = 2 cos(πt/L) + 7 sin(3πt/L) in P1 .
Solution: The function h(t) − 2 cos(πt/L) = 7 sin(3πt/L) is orthogonal to all the members of P1
since the functions cos(mπt/L) and sin(mπt/L) are orthogonal to both cos(nπt/L) and sin(nπt/L)
Math 602
37
for all m 6= m; as a result, the best approximation of h in P1 is 2 cos(πt/L). (Recall that the best
RL
approximation of h in P1 is such that −L (h(t) − F S1 (h))p(t)dt = 0 for all p ∈ P1 .) In conclusion
F S1 (h) = 2 cos(πt/L).
Question 90: Consider f : [−L, L] −→ R, f (x) = x4 . (a) Sketch the graph of the Fourier
series of f .
Solution: F S(f ) is equal to the periodic extension of f (x) over R.
(b) For which values of x ∈ R is F S(f ) equal to x4 ? (Explain)
Solution: The periodic extension of f (x) = x4 over R is piecewise smooth and globally continuous
since f (L) = f (−L). This means that the Fourier series is equal to x4 over the entire interval
[−L, +L].
(c) Is it possible to obtain F S(x3 ) by differentiating 14 F S(x4 ) term by term? (Explain)
Solution: Yes it is possible since the periodic extension of f (x) = x4 over R is continuous and
piecewise smooth.
Question 91: Let L be a positive real number. Let P1 = span{1, cos(πt/L), sin(πt/L)} and
R
21
L
consider the norm kf kL2 = −L f (t)2 dt . (a) Compute the best approximation of 1 + t in V
RL
with respect to the above norm. (Hint: −L t sin(πt/L)dt = 2L2 /π.)
Solution: We know from class that the truncated Fourier series
F S1 (t) = a0 + a1 cos(πt/L) + b1 sin(πt/L)
is the best approximation. Now we compute a0 , a1 , a2
Z L
1
(1 + t)dt = 1,
a0 =
2L −L
Z
1 L
a1 =
(1 + t) cos(πt/L)dt = 0
L −L
Z
Z
2L
1 L
1 L
L
.
b1 =
(1 + t) sin(πt/L)dt =
t sin(πt/L)dt = −2 cos(π) =
L −L
L −L
π
π
As a result
F S1 (t) = 1 +
2L
sin(πt/L)
π
(b) Compute the best approximation of h(t) = 2 cos(2πt/L) − 5 sin(3πt/L) in P1 .
Solution: The function h(t)2 cos(2πt/L) − 5 sin(3πt/L) is orthogonal to all the members of P1
since the functions cos(mπt/L) and sin(mπt/L) are orthogonal to both cos(nπt/L) and sin(nπt/L)
for all m 6= m; as a result, the best approximation of h in P1 is zero
F S1 (h) = 0.
Question 92: Let L be a positive real number. Let P1 = span{1, cos(πt/L), sin(πt/L)} and
R
21
L
consider the norm kf kL2 = −L f (t)2 dt . (a) Compute the best approximation of 2 − 3t in
RL
P1 with respect to the above norm. (Hint: −L t sin(πt/L)dt = 2L2 /π.)
Solution: We know from class that the truncated Fourier series
F S1 (t) = a0 + a1 cos(πt/L) + b1 sin(πt/L)
Math 602
38
is the best approximation. Now we compute a0 , a1 , a2
Z L
1
a0 =
(2 − 3t)dt = 2,
2L −L
Z
1 L
(2 − 3t) cos(πt/L)dt = 0
a1 =
L −L
Z
Z
1 L
1 L
L
6L
b1 =
(2 − 3t) sin(πt/L)dt =
−3t sin(πt/L)dt = −6 cos(π) = − .
L −L
L −L
π
π
As a result
F S1 (t) = 2 −
6L
sin(πt/L)
π
(b) Compute the best approximation of h(t) = 2 cos(πt/L) + 7 sin(3πt/L) in P1 .
Solution: The function h(t) − 2 cos(πt/L) = 7 sin(3πt/L) is orthogonal to all the members of P1
since the functions cos(mπt/L) and sin(mπt/L) are orthogonal to both cos(nπt/L) and sin(nπt/L)
for all m 6= m; as a result, the best approximation of h in P1 is 2 cos(πt/L). (Recall that the best
RL
approximation of h in P1 is such that −L (h(t) − F S1 (h))p(t)dt = 0 for all p ∈ P1 .) In conclusion
F S1 (h) = 2 cos(πt/L).
Question 93: Consider f : [−L, L] −→ R, f (x) = x4 . (a) Sketch the graph of the Fourier
series of f and the grah of f .
Solution: F S(f ) is equal to the periodic extension of f (x) over R.
(b) For which values of x ∈ R is F S(f ) equal to x4 ? (Explain)
Solution: The periodic extension of f (x) = x4 over R is piecewise smooth and globally continuous
since f (L) = f (−L). This means that the Fourier series is equal to x4 over the entire interval
[−L, +L].
(c) Is it possible to obtain F S(x3 ) by differentiating 14 F S(x4 ) term by term? For which values
is this legitimate? (Explain)
Solution: Yes it is possible since the periodic extension of f (x) = x4 over R is continuous and piecewise smooth. This operation is legitimate everywhere the function F S(x3 ) is smooth, i.e., for all the
points in R\{2k−1, k ∈ Z} (i.e., one needs to exclude the points . . . , −7, −5, −3, −1, +1, +3, +5+
7, . . .
2
Question 94: Let CS(f ) = π6 − 2(cos(x) − cos(2x)
+ cos(3x)
− cos(4x)
. . .) be the Fourier cosine
22
32
42
1 2
series of the function f (x) := 2 x defined over [0, +π].
(a) For which values of x in [0, +π] does this series coincide with f (x)? (Explain).
Solution: The Fourier cosine series coincides with the function f (x) over the entire interval [0, +π]
since f is smooth over [0, +π] (recall that the Fourier cosine series is the Fourier series of the even
extension of f over [−π, +π]).
(b) Compute the Fourier sine series, SS(x), of the function g(x) := x defined over [0, +π].
Solution: We know from class that it is always possible to obtain a Fourier sine series by differentiating term by term a Fourier cosine series, in other words
1
sin(2x) sin(3x) sin(4x)
sin(nx)
SS(x) = ∂x CS( x2 ) = 2 sin(x) −
+
−
...
... .
2
2
3
4
n
Math 602
39
(c) For which values of x ∈ [0, +π] does the Fourier sine series of g coincide with g(x)?.
Solution: The Fourier sine series coincides with the function g(x) := x over the interval [0, +π)
since g is smooth over [0, +π]and g(0) = 0. The Fourier sine series of g is zero at +π, and thus
differs from g(+π).
Question 95: (a) Compute the
of the sine series ofRf (x) = x for x ∈ [0, +π]. (Recall
Pcoefficients
π
+∞
that by definition SS(f )(x) = m=1 bm sin(mx) with bm = π2 0 f (x) sin(mx)dx.)
Solution: The definition of SS(f )(x) implies that
Z
Z
2 π 1
2 π
2
1
x sin(mx)dx = −
− cos(mx)dx + [−x cos(mx)]π0
bm =
π 0
π 0
m
π
m
2
m+1
= (−1)
.
m
P+∞ 2
(−1)m+1 sin(mx).
As a result SS(f )(x) = m=1 m
(b) For which values of x in [0, +π] does the sine series coincide with f (x)? (Explain).
Solution: The sine series coincides with the function f (x) over the entire interval [0, +π) since
f (0) = 0 and f is smooth over [0, +π). The series does not coincide with f (+π) since f (+π) 6= 0.
P+∞
m+1
) sin(mx).
(c) The sine series of x2 over [0, +π] is SS(x2 )(x) = m=1 ( m43 π ((−1)m −1)+ 2π
m (−1)
Compute the sine series of h(x) = x(π − x). (Hint: use (a))
Solution: Let h(x) = x(π − x). Note that by linearity of the sine series we have
SS(h)(x) = SS(πx)(x) − SS(x2 )(x),
as a result bm (h) = πbm (x) − bm (x2 ), i.e.,
bm (h) = π
2
4
2π
4
(−1)m+1 − ( 3 ((−1)m − 1) +
(−1)m+1 ) = 3 (1 + (−1)m+1 ).
m
m π
m
m π
In conclusion
SS(h)(x) =
∞
X
4
(1 + (−1)m+1 ) sin(mx).
3π
m
m=1
(d) Compute the cosine series of the function g(x) := π − 2x defined over [0, +π]. (Hint:
∂x (x(π − x)) = π − 2x.)
Solution: Observe that h(0) = h(π) = 0; as a result the sine series of h is continuous at 0 and
+π. This in turn implies that it is the legitimate to differentiate the sine series of h term by term
to obtain the cosine series of h0 (x) = g(x). In other words,
CS(g)(x) = ∂x SS(h)(x) =
∞
X
4
(1 + (−1)m+1 ) cos(mx).
2π
m
m=1
(e) Compute the sine series of h(x) = sin(x) for x ∈ [0, +π].
Solution: Obviously
SS(h)(x) = sin(x),
∀x ∈ R.
Question 96:
Question 97: Consider f : [−L, L] −→ R, f (x) = |x|x. (a) Sketch the graph of the Fourier
series of f and the graph of f .
Solution: F S(f ) is equal to the periodic extension of f (x) over R except at the points kL, k ∈ Z.
Math 602
40
(b) For which values of x ∈ R is F S(f ) equal to x|x|? (Explain)
Solution: The periodic extension of f (x) = x|x| over R is smooth over each interval [(2k −
1)L, (2k + 1)L], k ∈ Z, but discontinuous at all the points (2k + 1)L, k ∈ Z. This means
that the Fourier series is equal to x|x| over each interval [(2k − 1)L, (2k + 1)L], k ∈ Z. Since
f (−L) + f (+L) = 0, the Fourier series is equal to 0 at all the points (2k + 1)L], k ∈ Z.
(c) Is it possible to obtain F S(2|x|) by differentiating F S(|x|x) term by term? (Explain)
Solution: No, it is not legitimate since the Fourier series is discontinuous. The result would be
wrong.
Question 98: Let L be a positive real number. Let P1 = span{1, cos(πt/L), sin(πt/L)} and
R
21
L
consider the norm kf kL2 := −L f (t)2 dt . (a) Compute the best approximation of h(t) =
3 + π cos(πt/L) + 3 sin(7πt/L) in P1 .
Solution: The function h(t) − 3 − π cos(πt/L) = 3 sin(7πt/L) is orthogonal to all the members
of P1 since the functions cos(mπt/L) and sin(mπt/L) are orthogonal to both cos(nπt/L) and
sin(nπt/L) for all m 6= m; as a result, the best approximation of h in P1 is 3 + π cos(πt/L). (Recall
RL
that the best approximation of h in P1 is such that −L (h(t) − F S1 (h))p(t)dt = 0 for all p ∈ P1 .)
In conclusion
F S1 (h) = 3 + π cos(πt/L).
(b)
of 2 − 3t2 in P1 with respect to the above norm. (Hint:
R 2 Compute the best approximation
2
t cos(t)dt = 2t cos(t) + (t − 2) sin(t).)
Solution: We know from class that the truncated Fourier series
F S1 (t) = a0 + a1 cos(πt/L) + b1 sin(πt/L)
is the best approximation. Now we compute a0 , a1 , a2
Z L
1
2L3
a0 =
(2 − 3t2 )dt = 2 −
= 2 − L2 ,
2L −L
2L
Z
Z
1 L
3 L3 π 2
3 L3
L2
a1 =
(2 − 3t2 ) cos(πt/L)dt = −
t cos(t)dt = −
(−4π) = 12 2
3
3
L −L
L π −π
Lπ
π
Z L
1
b1 =
(2 − 3t2 ) sin(πt/L)dt = 0.
L −L
As a result
F S1 (t) = 2 − L2 +
12L2
cos(πt/L)
π2
Math 602
41
Question 99: Let N be a positive integer and let PN be the set of trigonometric polynomials
of degree at most N ; that is, PN = span{1, cos(x), sin(x), . . . , cos(N x), sin(N x)}. Consider the
P9
function f : [−π, π] −→ R defined by f (x) = n=0 n39+9 sin(9n) cos(4nx). (a) Compute the
Fourier series of f .
Solution: The Fourier series of f is of the following form
F S(f )(x) =
∞
X
∞
∞
∞
X
X
X
x
x
an cos(mπ ) +
bm sin(mπ ) =
am cos(mx) +
bm sin(mx).
π
π
m=0
m=1
n=0
n=1
with
1
a0 =
2L
Z
L
f (x)dx,
am
−L
1
=
L
Z
L
x
f (x) cos(mπ )dx,
L
−L
bm
1
=
L
Z
L
x
f (x) sin(mπ )dx.
L
−L
The orthogonality properties of the cosine and sine families implies that
9
sin(9n),
+9
an = 0, ∀n ≥ 37,
a4n =
n3
bn = 0,
and a4n+1 = a4n+2 = a4n+3 = 0,
0 ≤ n ≤ 9.
∀n ≥ 1.
In conclusion
F S(f )(x) =
9
X
n=0
n3
9
sin(9n) cos(4nx).
+9
Compute the best L2 -approximation of f in P33 over (−π, π).
Solution: We know from class that the best L2 -approximation of f in P33 over (−π, π) is the
truncated Fourier series F S33 (f ):
F S33 (f ) =
8
X
9
sin(9n) cos(4nx).
3+9
n
n=0
x
Question 100: Consider f : [−L, L] −→ R, f (x) = |x| sin(π L
). (a) Sketch the graph of f and
the graph of the Fourier series of f .
Solution: F S(f ) is equal to the periodic extension of f (x) over R including the points kL, k ∈ Z
since f (−) = f (+L).
x
(b) For which values of x ∈ R is F S(f ) equal to |x| sin(π L
)? (Explain)
Math 602
42
x
Solution: The periodic extension of f (x) = |x| sin(π L
) over R is smooth over each interval
[(2k − 1)L, (2k + 1)L], k ∈ Z and is continuous at all the points (2k + 1)L, k ∈ Z. This means
x
) over R.
that the Fourier series is equal to |x| sin(π L
x
(c) What is the derivative of f (x) = |x| sin(π L
)?
Solution: If x ≥ 0 then
x
π
x
|x|
π
x
f 0 (x) = sin(π ) + x cos(π ) = sin(π ) + |x| cos(π ).
L
L
L
L
L
L
If x ≤ 0 then
x
π
x
|x|
π
x
f 0 (x) = − sin(π ) − x cos(π ) = sin(π ) + |x| cos(π ).
L
L
L
L
L
L
(c) Is it possible to obtain F S(f 0 )(x) by differentiating F S(f )(x) term by term? (Explain)
Solution: Yes, it is possible for every x ∈ R since the Fourier series of f and f 0 are continuous.
Question 101: Let f : [0, 2π] −→ R be defined by f (x) = 1 + cos(x). (a)Draw the graph of f ,
the graph of the cosine series of f and the graph of the sine series of f .
Solution: Here are the graph of f , the graph of the cosine series of f and the graph of the sine
series of f (with L = 2π).
(b) Let g : [−2π, 2π] −→ R be defined by g(x) = 1 + cos(x). Is the Fourier series of g equal
to the sum of the cosine and sine series of f . Give all the details (a correct picture would be
enough).
Solution: No. The sine series of f over the interval [−2π, 0] is equal to the odd extension of
f (x) = 1 + cos(x). The odd extension in question is fodd (x) = −1 − cos(x) when x ∈ [−2π, 0].
The cosine series of f over the interval [−2π, 0] is equal to the even extension of f (x) = 1 + cos(x).
The even extension in question is feven (x) = 1 + cos(x) when x ∈ [−2π, 0]. In conclusion the sum
of the cosine and sine series of f is equal to 0 over the interval [−2π, 0], which is obviously different
from g(x) = F S(g)(x) = 1 + cos(x).
Here are the graphs of g, the graphs of F S(g) and the graph of the sum of the cosine and sine
series of f (with L = 2π).
Math 602
6
43
Fourier transform
Here are some formulae that you may want to use:
Z +∞
Z +∞
def 1
iωx
−1
F(f )(ω) =
f (x)e dx,
F (f )(x) =
f (ω)e−iωx dω,
2π −∞
−∞
(3)
F(f ∗ g) = 2πF(f )F(g),
1
2α
α
F(e−α|x| ) =
,
F( 2
)(ω) = e−α|ω| ,
π ω 2 + α2
x + α2
F(f (x − β))(ω) = eiβω F(f )(ω),
2
ω2
1
F(e−αx ) = √
e− 4α
4πα
1
1
−ax
F(H(x)e
)(ω) =
,
H is the Heaviside function
2π a − iω
i
i
1
F(pv( ))(ω) = sign(ω) = (H(ω) − H(−ω))
x
2
2
1
π
1
2
def
F(sech(ax)) =
sech( ω),
sech(ax) =
= x
2a
2a
ch(x)
e + e−x
cos(a) − cos(b) = −2 sin( 12 (a + b)) sin( 21 (a − b))
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
Question 102: (i) Let f be an integrable function on (−∞, +∞). Prove that for all a, b ∈ R,
and for all ξ ∈ R, F([eibx f (ax)])(ξ) = a1 F(f )( ξ+b
a ).
Solution: The definition of the Fourier transform together with the change of variable ax 7−→ x0
implies
Z +∞
1
f (ax)eibx eiξx dx
F[eibx f (ax)])(ξ) =
2π −∞
Z +∞
1
=
f (ax)ei(b+ξ)x dx
2π −∞
Z +∞
(ξ+b) 0
1
1
=
f (x0 )ei a x dx0
2π −∞ a
= a1 F(f )( ξ+b
a ).
2
(ii) Let c be a positive real number. Compute the Fourier transform of f (x) = e−cx sin(bx).
√
Solution: Using the fact that sin(bx) = −i 21 (eibx − e−ibx ), and setting a = c we infer that
f (x) =
=
1 −(ax)2 ibx
(e − e−ibx )
2i e
1 −(ax)2 ibx
1 −(ax)2 i(−b)x
e − 2i
e
e
2i e
and using (i) and (7) we deduce
fˆ(ξ) =
2
− 14 ( ξ+b
1 1 √1
a )
2i a 4π (e
− e− 4 (
1
ξ−b 2
a )
fˆ(ξ) =
1
− 4c
(ξ+b)2
1 √1
2i 4πc (e
− e− 4c (ξ−b) ).
1
2
).
In conclusion
Question 103: (a) Compute the Fourier transform of the function f (x) defined by
(
1 if |x| ≤ 1
f (x) =
0 otherwise
Math 602
44
Solution: By definition
F(f )(ω) =
1
2π
Z
1
1 1 iω
(e − e−iω )
2π iω
1 2 sin(ω)
.
=
2π
ω
eiξω = =
−1
Hence
F(f )(ω) =
1 sin(ω)
.
π ω
(b) Find the inverse Fourier transform of g(ω) =
sin(ω)
ω .
Solution: Using (a) we deduce that g(ω) = πF(f )(ω), that is to say, F −1 (g)(x) = πF −1 (F(f ))(x).
Now, using the inverse Fourier transform, we deduce that F −1 (g)(x) = πf (x) at every point x where
f (x) is of class C 1 and F −1 (g)(x) = π2 (f (x− ) + f (x+ )) at discontinuity points of f . As a result:


π if |x| < 1
−1
F (g)(x) = π2 at |x| = 1


0 otherwise
Question 104: Find the inverse Fourier transform of
Solution: Since F(Sλ (x)) =
1 sin(λω
π
ω .
1 sin(λω)
,
π
ω
the inverse Forier transform theorem implies that


1 if |x| < λ
1 sin(λω)
−1
F
(x) = 21 if |x| = λ

π
ω

0 otherwise.
Question 105: Use the Fourier transform technique to solve the following PDE:
∂t u(x, t) + c∂x u(x, t) + γu(x, t) = 0,
for all x ∈ (−∞, +∞), t > 0, with u(x, 0) = u0 (x) for all x ∈ (−∞, +∞).
Solution: By taking the Fourier transform of the PDE, one obtains
∂t F(u) − iωcF(y) + γF(y) = 0.
The solution is
F(u)(ω, t) = c(ω)eiωct−γt .
The initial condition implies that c(ω) = F(u0 )(ω):
F(u)(ω, t) = F(u0 )(ω)eiωct e−γt .
The shift lemma in turn implies that
F(u)(ω, t) = F(u0 (x − ct))(ω)e−γt = F(u0 (x − ct)e−γt )(ω).
Applying the inverse Fourier transform gives:
u(x, t) = u0 (x − ct)e−γt .
Question 106: Solve by the Fourier transform technique the following equation: ∂xx φ(x) −
2∂x φ(x) + φ(x) = H(x)e−x , x ∈ (−∞, +∞), where H(x) is the Heaviside function. (Hint:
use the factorization iω 3 + ω 2 + iω + 1 = (1 + ω 2 )(1 + iω) and recall that F(f (x))(−ω) =
F(f (−x))(ω)).
Math 602
45
Solution: Applying the Fourier transform with respect to x gives
(−ω 2 + 2iω + 1)F(φ)(ω) = F(H(x)e−x )(ω) =
1
1
.
2π 1 − iω
where we used (8). Then, using the hint gives
1
1
1
1
=
2π (1 − iω)(−ω 2 + 2iω + 1)
2π iω 3 + ω 2 + iω + 1
1
1
1
=
.
2π 1 + ω 2 1 + iω
F(φ)(ω) =
We now use again (8) and (5) to obtain
F(φ)(ω) = π
1
1
1 1
= πF(e−|x| )(ω)F(H(x)e−x )(−ω).
π 1 + ω 2 2π 1 − i(−ω)
Now we use F(H(x)e−x )(−ω) = F(H(−x)ex )(ω) and we finally have
F(φ)(ω) = πF(e−|x| )(ω)F(H(−x)ex )(ω).
The Convolution Theorem (4) gives
F(φ)(ω) = π
1
F(e−|x| ∗ (H(−x)ex ))(ω).
2π
We obtain φ by using the inverse Fourier transform
φ(x) =
i.e.,
Z
1 −|x|
e
∗ (H(−x)ex ),
2
+∞
φ(x) =
−∞
1 −|x−y|
e
H(−y)ey dy
2
and recalling that H is the Heaviside function we finally have
(
Z
1 −x
e
1 0 y−|x−y|
φ(x) =
e
dy = 41
2 −∞
( 4 − x)ex
if x ≥ 0
if x ≤ 0.
Question 107: Use the Fourier transform technique to solve the following ODE y 00 (x) − y(x) =
f (x) for x ∈ (−∞, +∞), with y(±∞) = 0, where f is a function such that |f | is integrable over
R.
Solution: By taking the Fourier transform of the ODE, one obtains
−ω 2 F(y) − F(y) = F(f ).
That is
F(y) = −F(f )
1
.
1 + ω2
and the convolution Theorem, see (4), together with (5) gives
1
F(y) = −πF(f )F(e−|x| ) = − F(f ∗ e−|x| ).
2
Applying F −1 on both sides we obtain
1
1
y(x) = − f ∗ e−|x| = −
2
2
Z
∞
−∞
e−|x−z| f (z)dz
Math 602
That is
1
y(x) = −
2
Z
46
∞
e−|x−z| f (z)dz.
−∞
Question 108: Use the Fourier transform method to compute the solution of utt − a2 uxx = 0,
where x ∈ R and t ∈ (0, +∞), with u(0, x) = f (x) := sin2 (x) and ut (0, x) = 0 for all x ∈ R.
Solution: Take the Fourier transform in the x direction:
F(u)tt + ω 2 a2 F(u) = 0.
This is an ODE. The solution is
F(u)(t, ω) = c1 (ω) cos(ωat) + c2 (ω) sin(ωat).
The initial boundary conditions give
F(u)(0, ξ) = F(f )(ω) = c1 (ω)
and c2 (ω) = 0. Hence
F(t, ω) = F(f )(ω) cos(ωat) =
1
F(f )(ω)(eiaωt + e−iaωt ).
2
Using the shift lemma, we infer that
F(t, ω) =
1
(F(f (x − at))(ω) + F(f (x + at))(ω)).
2
Usin the inverse Fourier transform, we finaly conclude that
u(t, x) =
1
1
(f (x − at) + f (x + at)) = (sin2 (x + at) + sin2 (x − at)).
2
2
Note that this is the D’Alembert formula.
Question 109: Use the Fourier transform method to compute the solution of utt − a2 uxx = 0,
where x ∈ R and t ∈ (0, +∞), with u(0, x) = f (x) := cos2 (x) and ut (0, x) = 0 for all x ∈ R.
Solution: Take the Fourier transform in the x direction:
F(u)tt + ω 2 a2 F(u) = 0.
This is an ODE. The solution is
F(u)(t, ω) = c1 (ω) cos(ωat) + c2 (ω) sin(ωat).
The initial boundary conditions give
F(u)(0, ξ) = F(f )(ω) = c1 (ω)
and c2 (ω) = 0. Hence
F(t, ω) = F(f )(ω) cos(ωat) =
1
F(f )(ω)(eiaωt + e−iaωt ).
2
Using the shift lemma (i.e., formula (6)) we obtain
u(t, x) =
1
1
(f (x − at) + f (x + at)) = (cos2 (x + at) + cos2 (x − at)).
2
2
Note that this is the D’Alembert formula.
Question 110: Solve the integral equation: f (x) +
x ∈ (−∞, +∞).
1
2π
R +∞
f (y)
dy
−∞ (x−y)2 +1
=
1
x2 +4
+
1
x2 +1 ,
for all
Math 602
47
Solution: The equation can be re-written
f (x) +
1
1
1
1
f∗ 2
= 2
+
.
2π
x +1
x + 4 x2 + 1
We take the Fourier transform of the equation and we apply the Convolution Theorem (see (4))
F(f ) +
1
1
1
1
2πF( 2
)F(f ) = F( 2
) + F( 2
).
2π
x +1
x +4
x +1
Using (5), we obtain
1
1
1
F(f ) + e−|ω| F(f ) = e−2|ω| + e−|ω| ,
2
4
2
which gives
1
1
1
F(f )(1 + e−|ω| ) = e−|ω| ( e−|ω| + 1).
2
2
2
We then deduce
1 −|ω|
e
.
2
Taking the inverse Fourier transform, we finally obtain f (x) =
F(f ) =
1
x2 +1 .
Question 111: Solve the following integral equation (Hint: solution is short):
Z +∞
√ Z +∞ − y2
x2
e 2π f (x − y)dy = −2πe− 4π .
f (y)f (x − y)dy − 2 2
∀x ∈ R.
−∞
−∞
Solution: This equation can be re-written using the convolution operator:
√
x2
x2
f ∗ f − 2 2e− 2π ∗ f = −2πe− 4π .
We take the Fourier transform and use the convolution theorem (4) together with (7) to obtain
√
−ω 2 11
−ω 2 11
1
1
4
4
2π = −2π q
4π
e
e
2πF(f )2 − 2π2 2F(f ) q
1
1
4π 2π
4π 4π
F(f )2 − 2F(f )e−ω
2π
2
+ e−ω
(F(f ) − e
−ω 2 π
2
2
π
=0
)2 = 0.
This implies
F(f ) = e−ω
2π
2
.
Taking the inverse Fourier transform, we obtain
f (x) =
√
x2
2e− 2π .
Question 112: Solve the integral equation: f (x) +
(−∞, +∞).
3
2
R +∞
−∞
e−|x−y| f (y)dy = e−|x| , for all x ∈
Solution: The equation can be re-written
3
f (x) + e−|x| ∗ f = e−|x| .
2
We take the Fourier transform of the equation and apply the Convolution Theorem (see (4))
3
F(f ) + 2πF(e−|x| )F(f ) = F(e−|x| ).
2
Using (5), we obtain
F(f ) + 3π
1 1
1 1
F(f ) =
,
2
π1+ω
π 1 + ω2
Math 602
48
which gives
F(f )
ω2 + 4
1 1
=
.
1 + ω2
π 1 + ω2
We then deduce
1 1
1
= F(e−2|x| ).
2
π4+ω
2
Taking the inverse Fourier transform, we finally obtain f (x) = 21 e−2|x| .
R +∞
2
1 2
Question 113: Solve the following integral equation −∞ e−(x−y) g(y)dy = e− 2 x for all x ∈
(−∞, +∞), i.e., find the function g that solves the above equation.
F(f ) =
Solution: The left-hand side of the equation is a convolution; hence,
2
1
2
e−x ∗ g(x) = e− 2 x .
By taking the Fourier transform, we obtain
1 2
1 2
1
1
2π √ e− 4 ω Fg(ω) = √ e− 2 ω .
4π
2π
That gives
1 2
1
F(g)(ω) = √ e− 4 ω .
2π
By taking the inverse Fourier transform, we deduce
r
√
4π −x2
2 −x2
e
.
g(x) = √ e
=
π
2π
Question 114: Solve the integral equation:
How many solutions did you find?
R +∞
−∞
f (y)f (x−y)dy =
4
x2 +4 ,
for all x ∈ (−∞, +∞).
Solution: The equation can be re-written
f ∗f =
4
.
x2 + 4
We take the Fourier transform of the equation and apply the Convolution Theorem (see (4))
2πF(f )2 = F(
x2
4
)
+4
Using (5), we obtain
2πF(f )2 = e−2|ω| .
which gives
1
F(f ) = ± √ e−|ω| .
2π
Taking the inverse Fourier transform, we finally obtain
1
2
f (x) = ± √
.
2
2π x + 1
We found two solutions: a positive one and a negative one.
Question 115: Use the Fourier transform method to solve the equation ∂t u +
u(x, 0) = u0 (x), in the domain x ∈ (−∞, +∞) and t > 0.
Solution: We take the Fourier transform of the equation with respect to x
2t
∂x u)
1 + t2
2t
= ∂t F(u) +
F(∂x u)
1 + t2
2t
= ∂t F(u) − iω
F(u).
1 + t2
0 = ∂t F(u) + F(
2t
1+t2 ∂x u
= 0,
Math 602
49
This is a first-order linear ODE:
∂t F(u)
d
2t
= iω (log(1 + t2 ))
= iω
2
F(u)
1+t
dt
The solution is
2
F(u)(ω, t) = K(ω)eiω log(1+t ) .
Using the initial condition, we obtain
F(u0 )(ω) = F(u)(ω, 0) = K(ω).
The shift lemma (see formula (6)) implies
F(u)(ω, t) = F(u0 )(ω)eiω log(1+t
2
)
= F(u0 (x − log(1 + t2 ))),
Applying the inverse Fourier transform finally gives
u(x, t) = u0 (x − log(1 + t2 )).
Question 116: Solve the integral equation:
Z +∞ Z +∞ √
√ − y2
1
2 − (x−y)2
2π
f (y) − 2e
−
f (x−y)dy = −
e 2π dy,
2
2
1+y
−∞
−∞ 1 + y
∀x ∈ (−∞, +∞).
(Hint: there is an easy factorization after applying the Fourier transform.)
Solution: The equation can be re-written
f ∗ (f −
√
x2
2e− 2π −
√
x2
1
1
)=−
∗ 2e− 2π .
2
2
1+x
1+x
We take the Fourier transform of the equation and apply the Convolution Theorem (see (4))
√
√
x2
x2
1
1
2πF(f ) F(f ) − 2F(e− 2π ) − F(
)
= −2πF(
) 2F(e− 2π )
2
2
1+x
1+x
Solution 1: Using (5), (7) we obtain
√
x2
2F(e− 2π ) =
F(
√
2
− ω1
πω 2
1
2q
e 4 2π = e− 2
1
4π 2π
1
1
) = e−|ω| ,
2
1+x
2
which gives
πω 2
πω 2
1
1
F(f ) F(f ) − e− 2 − e−|ω| = − e−|ω| e− 2 .
2
2
This equation can also be re-written as follows
F(f )2 − F(f )e−
πω 2
2
πω 2
1
1
− F(f ) e−|ω| + e−|ω| e− 2 = 0,
2
2
and can be factorized as follows:
(F(f ) − e−
This means that either F(f ) = e−
finally obtain two solutions
πω 2
2
f (x) =
πω 2
2
1
)(F(f ) − e−|ω| ) = 0.
2
or F(f ) = 21 e−|ω| . Taking the inverse Fourier transform, we
√
x2
2e− 2π ,
or
f (x) =
1
.
1 + x2
Math 602
50
Solution 2: Another solution consists of remarking that the equation with the Fourier transform can
be rewritten as follows:
√
x2
F(f )2 − F( 2e− 2π )F(f ) − F(
√
x2
1
1
)F(f ) + F( 2e− 2π )F(
) = 0,
2
1+x
1 + x2
which can factorized as follows:
√
x2
(F(f ) − F( 2e− 2π ))(F(f ) − F(
1
)) = 0.
1 + x2
√
x2
1
The either F(f ) = F( 2e− 2π )) or (F(f ) = F( 1+x
2 ). The conclusion follows easily.
Question 117: Use the Fourier transform technique to solve ∂t u(x, t) + sin(t)∂x u(x, t) + (2 +
3t2 )u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x). (Use the shift lemma: F(f (x − β))(ω) =
R +∞
1
f (x)eiωx dx)
F(f )(ω)eiωβ and the definition F(f )(ω) := 2π
−∞
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + sin(t)(−iω)F(u)(ω, t) + (2 + 3t2 )F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= iω sin(t) − (2 + 3t2 ).
F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −iω(cos(t) − 1) − (2t + t3 ).
This implies
3
F(u)(ω, t) = F(u0 )(ω)e−iω(cos(t)−1) e−(2t+t ) .
Then the shift lemma gives
3
F(u)(ω, t) = F(u0 (x + cos(t) − 1)(ω)e−(2t+t ) .
This finally gives
3
u(x, t) = u0 (x + cos(t) − 1)e−(2t+t ) .
Question 118: Solve the following integral equation (Hint: x2 − 3xa + 2a2 = (x − a)(x − 2a)):
Z
+∞
√
y2
x2
(f (y) − 3 2e− 2π )f (x − y)dy = −4πe− 4π .
∀x ∈ R.
−∞
Solution: This equation can be re-written using the convolution operator:
√
x2
x2
f ∗ f − 3 2e− 2π ∗ f = −4πe− 4π .
We take the Fourier transform and use (7) to obtain
√
−ω 2 11
−ω 2 11
1
1
4
4
2π = −4π q
4π
2πF(f )2 − 2π3 2F(f ) q
e
e
1
1
4π 2π
4π 4π
F(f )2 − 3F(f )e−ω
(F(f ) − e
−ω 2 π
2
2π
2
+ 2e−ω
)(F(f ) − 2e
2
−ω 2 π
2
π
=0
) = 0.
This implies
either
F(f ) = e−ω
2π
2
,
or
F(f ) = 2e−ω
2π
2
.
Math 602
Taking the inverse Fourier transform, we obtain
√
x2
either f (x) = 2e− 2π ,
or
51
√
x2
f (x) = 2 2e− 2π .
Question 119: Use the Fourier transform technique to solve ∂t u(x, t)−∂xx u(x, t)+sin(t)∂x u(x, t)+
(2 + 3t2 )u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x). (Hint: use the definition F(f )(ω) :=
R +∞
2
ω2
1
1
iωx
dx), the result F(e−αx )(ω) = √4πα
e− 4α , the convolution theorem and the
2π −∞ f (x)e
shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ . Go slowly and give all the details.)
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + ω 2 F(u)(ω, t) + sin(t)(−iω)F(u)(ω, t) + (2 + 3t2 )F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= −ω 2 + iω sin(t) − (2 + 3t2 ).
F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −ω 2 t − iω(cos(t) − 1) − (2t + t3 ).
This implies
2
3
F(u)(ω, t) = F(u0 )(ω)e−ω t e−iω(cos(t)−1) e−(2t+t ) .
2
Using the result F(e−αx )(ω) =
ω2
√ 1 e− 4α
4πα
where α =
1
4t ,
this implies that
r
3
x2
π
F(u0 )(ω)F(e− 4t )(ω)e−iω(cos(t)−1) e−(2t+t )
t
r
3
x2
π
=
F(u0 ∗ e− 4t )(ω)e−iω(cos(t)−1) e−(2t+t ) .
t
F(u)(ω, t) =
x2
Then setting g = u0 ∗ e− 4t the convolution theorem followed by the shift lemma gives
r
3
1
π
F(u)(ω, t) =
F(g(x + cos(t) − 1))(ω)e−(2t+t ) .
2π t
This finally gives
r
r
Z +∞
(x+cos(t)−1−y)2
1
1
−(2t+t3 )
−(2t+t3 )
4t
u(x, t) =
g(x + cos(t) − 1)e
=e
u0 (y)e−
dy.
4πt
4πt −∞
Question 120: Consider the telegraph equation ∂tt u + 2α∂t u + α2 u − c2 ∂xx u = 0 with α ≥ 0,
u(x, 0) = 0, ∂t u(x, 0) = g(x), x ∈ R, t > 0 and boundary condition at infinity u(±∞, t) =
0. Solve the equation by the Fourier transform technique. (Hint: the solution to the ODE
φ00 (t) + 2αφ0 (t) + (α2 + λ2 )φ(t) = 0 is φ(t) = e−αt (a cos(λt) + b sin(λt))
Solution: Applying the Fourier transform with respect to x to the equation, we infer that
0 = ∂tt F(u)(ω, t) + 2α∂t F(u)(ω, t) + α2 F(u)(ω, t) − c2 (−iω)2 F(u)(ω, t)
= ∂tt F(u)(ω, t) + 2α∂t F(u)(ω, t) + (α2 + c2 ω 2 )F(u)(ω, t)
Using the hint, we deduce that
F(u)(ω, t) = e−αt (a(ω) cos(ωct) + b(ω) sin(ωct)).
The initial condition implies that a(ω) = 0 and F(g)(ω) = ωcb(ω); as a result, b(ω) = F(g)(ω)/(ωc)
and
sin(ωct)
.
F(u)(ω, t) = e−αt F(g)
ωc
Math 602
Then using (??), we have
F(u)(ω, t) =
52
π −αt
e F(g)F(Sct ).
c
The convolution theorem implies that
u(x, t) = e
−αt
1
1
g ∗ Sct = e−αt
2c
2c
Z
∞
g(y)Sct (x − y)dy.
−∞
Finally the definition of Sct implies that Sct (x − y) is equal to 1 if −ct < x − y < ct and is equal
zero otherwise, which finally means that
Z x+ct
−αt 1
g(y)dy.
u(x, t) = e
2c x−ct
Question 121: Use the Fourier transform technique to solve ∂t u(x, t)−∂xx u(x, t)+cos(t)∂x u(x, t)+
(1 + 2t)u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x). (Hint: use the definition F(f )(ω) :=
R +∞
2
ω2
1
1
iωx
dx), the result F(e−αx )(ω) = √4πα
e− 4α , the convolution theorem and the
2π −∞ f (x)e
shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ . Go slowly and give all the details.)
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + ω 2 F(u)(ω, t) + cos(t)(−iω)F(u)(ω, t) + (1 + 2t)F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= −ω 2 + iω cos(t) − (1 + 2t).
F(u)(ω, t)
Then applying the fundamental theorem of calculus between 0 and t, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = −ω 2 t + iω sin(t) − (t + t2 ).
This implies
2
2
F(u)(ω, t) = F(u0 )(ω)e−ω t eiω sin(t) e−(t+t ) .
2
Using the result F(e−αx )(ω) =
ω2
√ 1 e− 4α
4πα
where α =
1
4t ,
this implies that
r
2
x2
π
F(u0 )(ω)F(e− 4t )(ω)eiω sin(t) e−(t+t )
t
r
2
x2
π
=
F(u0 ∗ e− 4t )(ω)eiω sin(t) e−(t+t ) .
t
F(u)(ω, t) =
x2
Then setting g = u0 ∗ e− 4t the convolution theorem followed by the shift lemma gives
r
2
1
π
F(u)(ω, t) =
F(g(x − sin(t)))(ω)e−(t+t ) .
2π t
This finally gives
r
u(x, t) =
2
2
1
g(x − sin(t))e−(t+t ) = e−(t+t )
4πt
r
1
4πt
Z
+∞
u0 (y)e−
(x−sin(t)−y)2
4t
dy.
−∞
Question 122: Solve the following integral equation (Hint: (a + b)2 = a2 + 2ab + b2 ):
Z
+∞
Z
+∞
f (y)f (x − y)dy − 2
−∞
−∞
2
4
f (x − y)dy + 2π 2
=0
y2 + 1
x +4
∀x ∈ R.
Math 602
53
Solution: This equation can be re-written using the convolution operator:
f ∗ f − 2(
2
4
) ∗ f + 2π 2
= 0.
x2 + 1
x +4
We take the Fourier transform and use the convolution theorem (??) together with (??) to obtain
2πF(f )2 − 4πF(f )e−|ω| − 2πe−2|ω| = 0
F(f )2 − 2F(f )e−|ω| + e−2|ω| = 0
(F(f ) − e−|ω| )2 = 0
This implies
F(f ) = e−|ω| .
Taking the inverse Fourier transform, we obtain
f (x) =
x2
2
.
+1
Question 123: State the shift lemma (Do not prove).
Solution: Let f be an integrable function over R (in L1 (R)), and let β ∈ R. Using the definitions
above, the following holds:
F(f (x − β))(ω) = F(f )(ω)eiωβ ,
∀ω ∈ R.
Question 124: Use the Fourier transform technique to solve ∂t u(x, t)+t∂x u(x, t)+2u(x, t) = 0,
x ∈ R, t > 0, with u(x, 0) = u0 (x).
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + t(−iω)F(u)(ω, t) + 2F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= iωt − 2.
F(u)(ω, t)
Then applying the fundamental theorem of calculus we obtain
1
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = iω t2 − 2t.
2
This implies
1 2
F(u)(ω, t) = F(u0 )(ω)eiω 2 t e−2t .
Then the shift lemma gives
1
F(u)(ω, t) = F(u0 (x − t2 )(ω)e−2t .
2
This finally gives
1
u(x, t) = u0 (x − t2 )e−2t .
2
Question 125: Prove the shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ , ∀ω ∈ R.
Solution: Let f be an integrable function over R (in L1 (R)), and let β ∈ R. Using the definitions
above, the following holds:
Z +∞
1
F(f (x − β))(ω) =
f (x − β)eiωx dx
2π −∞
Math 602
54
Upon making the change of variable z = x − β, we obtain
Z +∞
Z +∞
1
1
f (z)eiω(z+β) dz = eiωβ
f (z)eiωz dz
F(f (x − β))(ω) =
2π −∞
2π −∞
= eiωβ F(f )(ω).
which proves the lemma.
Question 126: Use the Fourier transform technique to solve ∂t u(x, t) + cos(t)∂x u(x, t) +
sin(t)u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = u0 (x).
Solution: Applying the Fourier transform to the equation gives
∂t F(u)(ω, t) + cos(t)(−iω)F(u)(ω, t) + sin(t)F(u)(ω, t) = 0
This can also be re-written as follows:
∂t F(u)(ω, t)
= iω cos(t) − sin(t).
F(u)(ω, t)
Then applying the fundamental theorem of calculus, we obtain
log(F(u)(ω, t)) − log(F(u)(ω, 0)) = iω sin(t) + cos(t) − 1.
This implies
F(u)(ω, t) = F(u0 )(ω)eiω sin(t) ecos(t)−1 .
Then the shift lemma gives
F(u)(ω, t) = F(u0 (x − sin(t))(ω)ecos(t)−1 .
This finally gives
u(x, t) = u0 (x − sin(t))ecos(t)−1 .
Math 602
7
55
Wave equation
Question 127: Consider the following wave equation
∂tt w − c2 ∂xx w = 0,
w(x, 0) = f (x),
x > 0, t > 0
x > 0,
∂t w(x, 0) = 0,
x > 0,
and w(0, t) = 0,
t > 0.
(a) Solve the equation (Hint: recall that the solution can always be put in the form F (x − ct) +
G(x + ct))
Solution: If x − ct > 0, we can apply D’Alembert’s formula u(x, t) = F (x − ct) + G(x + ct)
F (z) =
1
1
f (z) +
2
2c
Z
0
g(τ )dτ,
G(z) =
z
1
1
f (z) +
2
2c
Z
z
g(τ )dτ,
0
where g(x) = ∂t w(x, 0) = 0. In other words
w(x, t) =
1
(f (x − ct) + f (x + ct)),
2
if x ≥ ct.
If x − ct < 0 we apply the boundary condition at x = 0
∀t ≥ 0.
w(0, t) = 0 = F (−ct) + G(ct),
This means f (−z) = −G(z) = − 21 f (z) for all z ≥ 0. In other words we have obtained
w(x, t) =
1
(−f (−x + ct) + f (x + ct)),
2
if x ≤ ct.
(b) We now set c = 1, f (x) = x, if x ∈ [0, 1], f (x) = 2 − x, if x ∈ [1, 2], and f (x) = 0 otherwise.
Draw the graph of the solution at t = 0, t = 1, and t = 2 (draw three different graphs).
Solution:
0
1
2
0
1
2
3
1
2
3
4
5
Question 128: Let u solve the wave equation ∂tt u − c2 ∂xx u = 0, x ∈ [0, L], t ≥ 0, with
u(x, 0) = f (x), ut (x, 0) = 0. Let F (x), x ∈ (−∞, +∞) be the odd periodic extension of f .
Prove that u(x, t) = 21 (F (x + ct) + F (x − ct)) is the solution (do not spend to much time proving
u(L, t) = 0 if you do not remember how to do it).
Solution:
Clearly u(x, t) solve the PDE ∂tt u − c2 ∂xx u = 0, x ∈ [0, L], t ≥ 0. Moreover
1
u(x, 0) = 2 (F (x) + F (x)) = f (x), for all x ∈ [0, L]. Moreover ∂t u(x, 0) = 12 (cF (x) − cF (x)) = 0,
for all x ∈ [0, L]. Moreover u(0, t) = 12 (F (ct) + F (−ct)) = 0 since F is odd. Let us compute
u(L, t) = 21 (F (L + ct) + F (L − ct)). Observe −(L + ct) = −L − ct = L − ct − 2L, that is to say
F (−(L + ct)) = F (L − ct − 2L) = F (L − ct), since F is 2L-periodic. But since F is odd, we have
−F (L + ct) = F (−(L + ct)) = F (L − ct), i.e., u(L, t) = 0.
Question 129: Let u be a solution of the wave equation ∂tt u − c2 ∂xx u = 0, x ∈ [0, L], t ≥ 0.
RL
2 RL
Let E = 21 0 (∂t u)2 dx + c2 0 (∂x u)2 dx.
(a) Compute the time derivative of E.
Solution: Multiply the equation by ∂t u and integrate over the domain. It is clear that
dE
= ∂x u(x, t)∂t u(x, t)|L
0.
dt
Math 602
56
(b) In addition to being a solution of the wave equation, assume that u(0, t) = 0, ∂x u(L, t) = 0,
and u(x, 0) = f (x), ∂t u(x, 0) = g(x), where f and g are two given functions. Prove that if a
solution to this problem exists, then it is unique.
Solution: Let u1 , u2 be two solutions and let φ = u1 − u2 . Then φ solves the homogeneous
problem. From (a) we infer E(t) = E(0) = 0 where E is the energy for φ. Hence ∂x φ = 0, which
means φ(x, t) = ψ(t). But φ(0, t) = ψ(t) = 0. In conclusion φ = 0, i.e., u1 = u2 .
Question 130: Solve the PDE
utt − a2 uxx = 0,
u(x, 0) = sin(x),
− ∞ < x < +∞,
0 ≤ t,
− ∞ < x < +∞.
ut (x, 0) = a cos(x),
Solution: Apply D’Alembert’s Formula. u(x, t) = sin(x + at).
Question 131: Solve the PDE
utt − a2 uxx = 0,
− ∞ < x < +∞,
u(x, 0) = x3 ,
0 ≤ t,
0 ≤ t,
u(0, t) = 0,
− ∞ < x < +∞.
ut (x, 0) = 0,
Solution: Use the odd extension of x3 , that is, x3 . Than, apply D’Alembert’s Formula. u(x, t) =
1
3
3
2 [(x + at) + (x − at) ].
Question 132: Solve the PDE
utt − a2 uxx = 0,
u(x, 0) = cos(x),
− ∞ < x < +∞,
ut (x, 0) = −a sin(x),
0 ≤ t,
− ∞ < x < +∞.
Solution: Apply D’Alembert’s Formula.
u(x, t) =
1
2 (cos(x
− at) + cos(x + at)) −
1
2a
Z
x+at
a sin(ξ)dξ
x−at
= 12 (cos(x − at) + cos(x + at)) + 12 (cos(x + at) − cos(x − at))
= cos(x + at).
Hence u(x, t) = cos(x + at).
Question 133: Solve the PDE
utt − uxx = 0,
0 < x < +∞,
u(0, t) = 0, u(1, t) = 0
u(x, 0) = sin(πx), ut (x, 0) = 0,
0 < t,
0 < x < +∞.
0 < t,
Solution: We have to define the odd extension of sin(πx) on (−1, +1). Clearly sin(πx) is the
odd extension. Now we define the periodic extension of sin(πx) over the entire real line. Clearly
sin(πx) is the extension in question. The D’Alembert formula, which is valid on the entire real line,
gives
u(x, t) = 21 (sin(π(x − t)) + sin(π(x + t))
= 12 ((cos(πt) sin(πx) − sin(πt) cos(πx)) + 12 ((cos(πt) sin(πx) + sin(πt) cos(πx))
= cos(πt) sin(πx).
Hence u(x, t) = cos(πt) sin(πx) for all x ∈ (0, 1), t > 0.
Question 134: Consider the PDE
utt − uxx = 0,
0 ≤ x ≤ 2,
u(0, t) = 0,
0 < t,
u(2, t) = 0
(
ut (x, 0) = 0,
u(x, 0) = f (x) :=
x
0 ≤ x ≤ 1,
2 − x 1 ≤ x ≤ 2.
0 < x < +2.
0 < t,
Math 602
57
(a) Give u(x,t) for all x ∈ [0, +2], t > 0. (Hint use an extension technique).
Solution: We notice first that the wave speed is 1. We define fo to be the odd extension of f
over (−2, +2), the we define fop to be the periodic extension of fo over (−∞, +∞) with period 4.
From class we know that the solution to the above problem is given by the D’Alembert formula
u(x, t) =
1
(fop (x − t) + fop (x + t)).
2
(b) Using (a), compute u(x, 12 ), for all x ∈ [0, +2].
Solution: We have to compute fop (x − 21 ) and fop (x + 12 ).
Case 1: 0 ≤ x ≤ 21 . Then − 12 ≤ x − 21 ≤ 0 and by definition of fop , fop (x − 12 ) = −f (−x + 21 ) =
x − 12 . We also have 12 ≤ x + 21 ≤ 1, which means fop (x + 12 ) = f (x + 12 ) = x + 12 . Finally
u(x, 21 ) = 12 (x − 12 + x + 12 ) = x for all x ∈ [0, 21 ].
Case 2: 12 ≤ x ≤ 32 . Then 0 ≤ x − 12 ≤ 1 and fop (x − 12 ) = f (x − 21 ) = x − 12 . We also
have 1 ≤ x + 12 ≤ 2, which means fop (x + 12 ) = f (x + 12 ) = 2 − (x + 12 ) = −x + 32 . Finally
u(x, 12 ) = 12 (x − 12 − x + 32 ) = 21 for all x ∈ [ 21 , 32 ].
Case 3: 32 ≤ x ≤ 2. Then 1 ≤ x − 12 ≤ 23 and fop (x − 21 ) = f (x − 12 ) = 2 − (x − 12 ) = 52 − x. We
also have 2 ≤ x+ 21 ≤ 52 , which means by periodicity that fop (x+ 21 ) = fop (x+ 21 −4) = fop (x− 72 ).
Now we observe that −2 ≤ x − 27 ≤ − 32 , which means fop (x + 12 ) = fop (x − 72 ) = −f ( 72 − x) =
−(2 − ( 72 − x)) = −(− 32 + x) = 32 − x. In conclusion u(x, 12 ) = 12 ( 25 − x + 32 − x) = 2 − x for all
x ∈ [ 23 , 2].
Conclusion: We now put everything together


x ∈ [0, 21 ],
x,
1
1
u(x, ) = 2 ,
x ∈ [ 12 , 32 ], .

2

2 − x, x ∈ [ 32 , 2].
Question 135: Solve the wave equation on the semi-infinite domain (0, +∞),
∂tt w − 4∂xx w = 0,
2 −1
w(x, 0) = (1 + x )
x ∈ (0, +∞), t > 0
,
x ∈ (0, +∞);
x ∈ (0, +∞);
∂t w(x, 0) = 0,
and ∂x w(0, t) = 0,
(Hint: Consider a particular extension of w over R)
Solution: We define f (x) = (1 + x2 )−1 and its even extension fe (x) on −∞, +∞. Let we be the
solution to the wave equation over the entire real line with fe as initial data:
∂tt we − 4∂xx we = 0,
we (x, 0) = fe (x),
x ∈ R, t > 0
x > 0,
∂t we (x, 0) = 0,
x ∈ R.
The solution to this problem is given by the D’Alembert formula
we (x, t) =
1
(fe (x − 2t) + fe (x + 2t)),
2
for all x ∈ R and t ≥ 0.
Let x be positive. Then w(x, t) = we (x, t) for all x ∈ (0, +∞), since by construction ∂x we (0, t) = 0
for all times.
Case 1: If x − 2t > 0, fe (x − 2t) = f (x − 2t); as a result
w(x, t) =
1
(f (x − 2t) + f (x + 2t),
2
If x − 2t > 0.
Case 2: If x − 2t < 0, fe (x − 2t) = f (−x + 2t); as a result
w(x, t) =
1
(f (−x + 2t) + f (x + 2t)),
2
If x − 2t < 0.
t > 0.
Math 602
58
Note that actually fe (x) = (1 + x2 )−1 ; as a result, the solution can also be re-written as follows:
1
1
1
+
.
w(x, t) =
2 1 + (x − 2t)2
1 + (x + 2t)2
Question 136: Solve the PDE
utt − uxx = 0,
0 ≤ x ≤ 1,
∂x u(0, t) = 0,
∂x u(1, t) = 0
u(x, 0) = cos(πx),
0 < t,
0 < t,
ut (x, 0) = 0,
0 < x < +1.
(Hint: Consider the periodic extension over R of a particular extension of u over [−1 + 1]).
Solution: The even extension of u over [−1+1], say ue , satisfies the PDE and the initial conditions,
and always satisfies ∂x ue (0, t) = 0, ∂x ue (1, t) = 0. Since ∂ue (1, t), we deduce ∂ue (−1, t) = 0.
This means that the periodic extension of ue , says up , is smooth and also satisfies the PDE plus the
initial conditions. By construction ∂x up (0, t) = 0 and ∂x ue (1, t) = 0. As a result, we can obtain
u by computing the solution of the wave equation on R using the periodic extension over R of the
even extension of the initial data over [−1 + 1], i.e., u = up |[0,1]
We have to define the even extension of cos(πx) on (−1, +1). Clearly cos(πx) is the even extension.
Now we define the periodic extension of cos(πx) over the entire real line. Clearly cos(πx) is the
extension in question. The D’Alembert formula, which is valid on the entire real line, gives
u(x, t) = 21 (cos(π(x − t)) + cos(π(x + t))
= 12 ((cos(πt) cos(πx) + sin(πt) sin(πx)) + 12 ((cos(πt) cos(πx) − sin(πt) sin(πx))
= cos(πt) cos(πx).
Hence u(x, t) = cos(πt) cos(πx) for all x ∈ (0, 1), t > 0.
Question 137: Consider the following wave equation
∂tt w − 4∂xx w = 0,
x > 0, t > 0
2 −1
w(x, 0) = x(1 + x )
,
x > 0,
∂t w(x, 0) = 0,
x > 0,
and w(0, t) = 0,
t > 0.
(a) Solve the equation.
Solution: We define f (x) = x(1 + x2 )−1 and its odd extension fo (x). Let w0 be the solution to
the wave equation over the entire real line with fo as initial data:
∂tt wo − 4∂xx wo = 0,
wo (x, 0) = fo (x),
x ∈ R, t > 0
x > 0,
∂t wo (x, 0) = 0,
x ∈ R.
The solution to this problem is given by the D’Alembert formula
wo (x, t) =
1
(fo (x − 2t) + fo (x + 2t)),
2
for all x ∈ R and t ≥ 0.
Let x be positive. Then w(x, t) = wo (x, t) (since by construction wo (0, t) = 0 for all times).
Case 1: If x − 2t > 0, fo (x − 2t) = f (x − 2t); as a result
w(x, t) =
1
(f (x − 2t) + f (x + 2t).
2
Case 2: If x − 2t < 0, fo (x − 2t) = −f (−x + 2t); as a result
w(x, t) =
1
(−f (−x + 2t) + f (x + 2t)).
2
Note that actually f0 (x) = x(1 + x2 )−1 ; as a result, the solution can also be re-written as follows:
w(x, t) =
1
((x − 2t)(1 + (x − 2t)2 )−1 + (x + 2t)(1 + (x + 2t)2 )−1 ).
2
Math 602
59
Question 138: Consider the wave equation
∂tt w − ∂xx w = 0,
w(x, 0) = f (x),
x < 0, t > 0
x < 0,
∂t w(x, 0) = 0,
x < 0,
and w(0, t) = 0,
t > 0.
where f (x) = −x, if x ∈ [−1, 0], f (x) = 2 + x, if x ∈ [−2, −1], and f (x) = 0 otherwise. Give a
graphical solution to the problem at t = 0, t = 1, and t = 2 (draw three different graphs and
explain what you do)
Solution:
5
4
3
2
1
3
2
1
0
2
1
0
Question 139: Solve the PDE
utt − a2 uxx = 0,
− ∞ < x < +∞,
u(x, 0) = sin(x),
0 ≤ t,
− ∞ < x < +∞.
ut (x, 0) = a cos(x),
Solution: Apply D’Alembert’s Formula.
1
(sin(x + at) + sin(x − at)) +
2
1
= (sin(x + at) + sin(x − at)) +
2
= sin(x + at).
u(x, t) =
Z x+at
1
a cos(ξ)ξ
2a x−at
1
(sin(x + at) − sin(x − at))
2
Question 140: Solve the PDE
utt − uxx = 0,
0 ≤ x ≤ 1,
u(0, t) = 0,
0 < t,
u(1, t) = 0
u(x, 0) = sin(πx),
ut (x, 0) = 0,
0 < t,
0 < x < +∞.
(Hint: Consider the periodic extension over R of the odd extension of u over [−1 + 1]).
Solution: The odd extension of u over [−1+1], say uo , satisfies the PDE and the initial conditions,
and always satisfies uo (0, t) = 0, uo (1, t) = 0, uo (−1, t) = 0. Since uo (1, t) = uo (−1, t) = 0, the
periodic extension, says up , is smooth and also satisfies the PDE plus the initial conditions. As a
result, we can obtain u by computing the solution of the wave equation on R using the periodic
extension over R of the odd extension of the initial data over [−1 + 1], i.e., u = up |[0,1]
We have to define the odd extension of sin(πx) on (−1, +1). Clearly sin(πx) is the odd extension.
Now we define the periodic extension of sin(πx) over the entire real line. Clearly sin(πx) is the
extension in question. The D’Alembert formula, which is valid on the entire real line, gives
u(x, t) = 12 (sin(π(x − t)) + sin(π(x + t))
= 12 ((cos(πt) sin(πx) − sin(πt) cos(πx)) + 12 ((cos(πt) sin(πx) + sin(πt) cos(πx))
= cos(πt) sin(πx).
Hence u(x, t) = cos(πt) sin(πx) for all x ∈ (0, 1), t > 0.
Question 141: Solve utt − 4uxx = 0, x ∈ (−∞, +∞) and t ≥ 0, with u(x, 0) = sin2 (x),
2x
∂t u(x, 0) = − (1+x
2 )2 .
Math 602
60
Solution: The speed is 2. We apply the D’Alembert formula.
Z
1 x+2t
1
2ξ
2
2
u(x, t) =
−
dξ,
sin (x − 2t) + sin (x + 2t) +
2
4 x−2t
(1 + ξ 2 )2
x+2t
1
1
1 =
sin2 (x − 2t) + sin2 (x + 2t) +
2
4 1 + ξ 2 x−2t
1
1
1
1
2
2
=
sin (x − 2t) + sin (x + 2t) +
−
.
2
4 1 + (x + 2t)2
1 + (x − 2t)2
Question 142: Consider the wave equation ∂tt w − ∂xx w = 0, x ∈ (0, 4), t > 0, with
x ∈ (0, 4),
w(x, 0) = f (x),
∂t w(x, 0) = 0,
x ∈ (0, 4),
and w(0, t) = 0,
w(4, t) = 0,
t > 0.
where f (x) = x − 1, if x ∈ [1, 2], f (x) = 3 − x, if x ∈ [2, 3], and f (x) = 0 otherwise. Give a
simple expression of the solution in terms of an extension of f . (a) Give a simple expression of
the solution in terms of an extension of f .
Solution: We know from class that with Dirichlet boundary conditions, the solution to this problem
is given by the D’Alembert formula where f must be replaced by the periodic extension (of period
8) of its odd extension, say fo,p , where
fo,p (x + 8) = fo,p (x),
∀x ∈ R
(
f (x)
if x ∈ [0, 4]
fo,p (x) =
−f (−x) if x ∈ [−4, 0)
The solution is
u(x, t) =
1
(fo,p (x − t) + fo,p (x + t)).
2
(b) Give a graphical solution to the problem at t = 0, t = 1, t = 2, and t = 3 (draw four
different graphs and explain).
Solution: I draw on the left of the figure the graph of fo,p . Half the graph moves to the right
with speed 1 (solid line in left panel of the figure), the other half moves to the left with speed 1
(dashed line in left panel of the figure). The graph of the solution at times t = 0, 1, 2, 3 is shown in
the right panel of the figure.
1
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
1/2
1/2
1/2
Initial data + periodic extension of the odd extension at
t = 0, 1, 2, 3.
Solution in domain [0, 4]
at t = 0, 1, 2, 3
Question 143: Solve utt −4uxx = 0, x ∈ (0, 1) and t ≥ 0, with u(0, t) = u(1, t) = 0, u(x, 0) = 0,
∂t u(x, 0) = g(x) := 2π sin(πx). (Hint: use an extension technique).
Math 602
61
Solution: We notice first that the wave speed is 2. We define go to be the odd extension of g
over (−1, +1). Clearly go (x) = 2π sin(πx) since sin(πx) is odd. We define gop to be the periodic
extension of go over (∞, +∞) with period 2. Clearly, gop (x) = 2π sin(πx) since 2 is a period for
sin(πx). From class we know that the solution to the above problem is given by the restriction of
the D’Alembert formula to the interval [0, 1]:
u(x, t) =
1
4
Z
x+2t
gop (ξ)dξ =
x−2t
1
4
Z
x+2t
2π sin(πξ)dξ,
x−2t
1
= − (cos(π(x + 2t)) − cos(π(x − 2t)))
2
π
π
= sin( (2x)) sin( (4t))
2
2
= sin(πx) sin(2πt),
∀x ∈ [0, 1], ∀t ≥ 0.
Question 144: Consider the wave equation ∂tt w − ∂xx w = 0, x ∈ (0, 4), t > 0, with
x ∈ (0, 4),
w(x, 0) = f (x),
∂t w(x, 0) = 0,
x ∈ (0, 4),
and ∂x w(0, t) = 0,
∂x w(4, t) = 0,
where f (x) = x − 1, if x ∈ [1, 2], f (x) = 3 − x, if x ∈ [2, 3], and f (x) = 0 otherwise. Give a
simple expression of the solution in terms of an extension of f . Give a graphical solution to the
problem at t = 0, t = 1, t = 2, and t = 3 (draw four different graphs and explain).
Solution: We know from class that with homogeneous Neumann boundary conditions, the solution
to this problem is given by the D’Alembert formula where f must be replaced by the periodic
extension (of period 8) of its even extension, say fe,p , where
fe,p (x + 8) = fe,p (x),
∀x ∈ R
(
f (x)
if x ∈ [0, 4]
fe,p (x) =
f (−x) if x ∈ [−4, 0)
The solution is
1
(fe,p (x − t) + fe,p (x + t)).
2
I draw on the left of the figure the graph of fo,p . Half the graph moves to the right with speed 1,
the other half moves to the left with speed 1.
u(x, t) =
1
1
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
1/2
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
0
1
2
3
4
1/2
1/2
1/2
1/2
1/2
(a) Initial data + periodic extension of the
even extension at t = 0, 1, 2, 3. Solid line
waves move to the right, dotted line waves
move to the left
(b) Solution in domain (0, 4) at t =
0, 1, 2, 3
t > 0.
Math 602
62
Question 145: Consider the wave equation ∂tt w − ∂xx w = 0, x ∈ (−∞, +∞), t > 0, with
1
2x
initial data w(x, 0) = 1+x
2 , ∂t w(x, 0) = (1+x2 )2 . Compute the solution w(x, t).
Solution: The wave speed is 2. The solution is given by the D’Alembert formula,
Z
1
1
1 x+t
2τ
1
w(x, t) =
+
+
dτ
2 1 + (x − t)2
1 + (x + t)2
2 x−t (1 + τ 2 )2
After integration, we obtain
1
=
2
1
1
+
2
1 + (x − t)
1 + (x + t)2
x+t
1
1
−
,
2 (1 + τ 2 ) x−t
which finally gives
w(x, t) =
1
.
1 + (x − t)2
Question 146: Solve utt − 4uxx = 0, x ∈ (0, 1) and t ≥ 0, with u(0, t) = 0, ∂x u(1, t) = 0,
u(x, 0) = 0, ∂t u(x, 0) = g(x) := 2π sin( π2 x). (Hint: Pay attention to the boundary conditions.
Use three extensions.)
Solution: To be able to apply the d’Alembert formula, we need to extend the above problem
to the (−∞, +∞). the Dirichlet condition a x = 0 requires an odd extension and the Neumann
condition requires an even extension.
We define go to be the odd extension of g over (−1, +1) to account for the Dirichlet boundary
condition at x = 0.
(
g(x)
for all x ∈ (0, 1),
go (x) =
−g(−x) for all x ∈ (−1, 0)
Clearly go (x) = 2π sin( π2 x) since sin( π2 x) is odd. More precisely,
(
g(x) = 2π sin( π2 x)
go (x) =
−g(−x) = −2π sin( π2 (−x)) = 2π sin( π2 x)
for all x ∈ (0, 1),
for all x ∈ (−1, 0).
Now we need to consider the even extension of go about the point x = 1 to account for the Neumann
boundary condition at x = 1. Let us denote goe (x) this extension. The function goe (x) is such that
(
goe (x) = go (x)
for all x ∈ (−1, 1),
goe (x) =
goe (x) = go (2 − x) for all x ∈ (1, 3)
Now we observe that sin( π2 (2 − x)) = sin(π − π2 x) = sin( π2 x), which means that goe (x) =
2π sin( π2 x). More precisely,
(
goe (x) = go (x) = 2π sin( π2 x)
for all x ∈ (−1, 1),
goe (x) =
π
π
goe (x) = go (2 − x) = 2π sin( 2 (2 − x)) = 2π sin( 2 x) for all x ∈ (1, 3).
Now we consider the periodic extension of goe of period 4, say goep . Clearly goep = 2π sin( π2 x),
since 42π sin( π2 x) is periodic of period 4. See Figure
Math 602
63
We notice finally that the wave speed is 2. From class we know that the solution to the above
problem is given by the restriction of the D’Alembert formula to the interval [0, 1]:
Z
x+2t
1
4
goep (ξ)dξ =
1
4
Z
x+2t
π
2π sin( ξ)dξ,
2
x−2t
x−2t
π
π
= − cos( (x + 2t)) − cos( (x − 2t))
2
2
π
π
= 2 sin( (x)) sin( (2t))
2
2
π
= 2 sin( x) sin(πt),
∀x ∈ [0, 1], ∀t ≥ 0.
2
u(x, t) =
Question 147: Consider the wave equation ∂tt w − 4∂xx w = 0, x ∈ (−∞, +∞), t > 0, with
4x
1
initial data u(x, 0) = 1+x
2 , ∂t u(x, 0) = − (1+x2 )2 . Compute the solution w(x, t).
Solution: The wave speed is 2. The solution is given by the D’Alembert formula,
Z
1
1
1
1 x+2t
4τ
w(x, t) =
+
−
dτ
+
2 1 + (x − 2t)2
1 + (x + 2t)2
4 x−2t
(1 + τ 2 )2
After integration, we obtain
=
1
2
1
1
+
1 + (x − 2t)2
1 + (x + 2t)2
+
x+2t
1
2
,
4 (1 + τ 2 ) x−2t
which finally gives
w(x, t) =
1
.
1 + (x + 2t)2
Math 602
8
64
Energy method
Question 148: Let u solve the following equation: ∂tt u+∂t u−a2 ∂xx u = 0 for all x ∈ (−∞, +∞)
and t > 0 with the boundary and initial conditions u(±∞, t) = 0, u(x, 0) = u0 (x), ∂t u(x, 0) =
u1 (x).
(i) Assuming u is smooth, prove that the following holds for all T > 0:
2
1
2 k∂t u(·, T )kL2
+ 12 a2 k∂x u(·, T )k2L2 +
Z
0
T
k∂t u(·, t)k2L2 dt = 12 ku1 k2L2 + 21 a2 k∂x u0 k2L2
Solution: We multiply by ∂t u and integrate over space to obtain
Z +∞
0=
∂tt u∂t u + (∂t u)2 − a2 ∂xx u∂t u dx
−∞
+∞
Z
1
∂t (∂t u)2 + (∂t u)2 + a2 ∂x u∂t (∂x u)dx
−∞ 2
1 d
1 d
k∂t u(·, t)k2L2 + k∂t u(·, t)k2L2 + a2
k∂x u(·, t)k2L2 .
=
2 dt
2 dt
=
Now we integrate over time between 0 and T ,
0=
1
k∂t u(·, T )k2L2 +
2
T
Z
0
1
1
1
k∂t u(·, t)k2L2 dt + a2 k∂x u(·, t)k2L2 − k∂t u(·, 0)k2L2 − a2 k∂x u(·, 0)k2L2 .
2
2
2
Z
T
This gives
1
k∂t u(·, T )k2L2 +
2
1
1
1
k∂t u(·, t)k2L2 dt + a2 k∂x u(·, t)k2L2 = ku1 k2L2 + a2 k∂x u0 k2L2 .
2
2
2
0
Question 149: Consider the quasilinear Klein-Gordon equation: ∂tt φ(x, t) − c2 ∂xx φ(x, t) +
m2 φ(x, t)+β 2 φ3 (x, t) = 0, x ∈ R, t > 0, with φ(x, 0) = f (x), ∂t φ(x, 0) = g(x) and φ(±∞, t) = 0,
∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0. Find an energy E(t) which is invariant with respect to time
1
(Hint: test with ∂t φ(x, t) and use φp φ0 = ( p+1
φp+1 )0 .)
Solution: Testing with ∂t φ(x, t) and integrating over R and using the property ∂t φ(±∞, t) = 0,
∂x φ(±∞, t) = 0, we obtain
Z +∞
Z +∞
Z +∞
1
1 2
1
2
2
2
2
0=
∂t ( (∂t φ) )dx − c
∂xx φ∂t φdx + m
∂t ( φ )dx + β
∂t ( φ4 )dx
2
2
4
−∞
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞ 2
2
m 2 β 4
1
= dt
(∂t φ)2 dx + c2
∂x φ∂t ∂x φdx + dt
(
φ +
φ )dx
2
4
−∞
−∞
−∞ 2
Z +∞ 2
Z +∞ 2
Z +∞
1
c
m 2 β2 4
= dt
(∂t φ)2 dx + dt
(∂x φ)2 dx + dt
(
φ +
φ )dx
2
4
−∞ 2
−∞ 2
−∞
Z +∞ 1
c2
m2 2 β 2 4
2
2
= dt
(∂t φ) + (∂x φ) +
φ +
φ dx.
2
2
2
4
−∞
Z
+∞
Introduce
Z
+∞
E(t) =
−∞
c2
m2 2 β 2 4
1
2
2
(∂t φ) + (∂x φ) +
φ +
φ )dx.
2
2
2
4
Then
dt E(t) = 0.
The fundamental Theorem of calculus gives
E(t) = E(0).
Math 602
65
In conclusion the quantity E(t) is invariant with respect to time, as requested.
Question 150: Consider (1 + x)φ(x) + ∂x φ(x) − ∂x ((2 + x2 )∂x φ) = f (x), x ∈ (0, 1) with
φ(0) = 1, ∂x φ(1) = 2. Assume that φ1 and φ2 are two smooth solutions (φ1 ∈ C 2 ([0, 1]) and
φ2 ∈ C 2 ([0, 1])). Use the energy argument to prove that φ1 = φ2 . (Hint: 2φ(x)∂x φ(x) =
∂x φ2 (x).)
Solution: The difference φ := φ1 − φ2 satisfies
(1 + x)φ(x) + ∂x φ(x) − ∂x ((2 + x2 )∂x φ) = 0,
x ∈ (0, 1),
and φ(0) = 1,
∂x φ(1) = 0.
Multiply this equation by φ and integrate over the domain to obtain
Z 1
Z 1
Z 1
0=
(1 + x)φ(x)2 dx +
φ(x)∂x φ(x)dx −
φ(x)∂x ((2 + x2 )∂x φ(x))dx
0
Z
=
0
0
1
1
(1 + x)φ(x) dx +
2
2
0
1
Z
2
Z
∂x (φ(x) )dx +
0
1
(2 + x2 )∂x φ(x)∂x φ(x)dx − [(2 + x2 )∂x φ(x)]|10 .
0
The boundary term [(2 + x2 )∂x φ(x)]|10 is zero owing to the boundary conditions. We apply the
fundamental Theorem of calculus one more time to obtain
Z 1
Z 1
1
0=
(1 + x)φ(x)2 dx + [φ(x)2 ]|10 +
(2 + x2 )[∂x φ(x)]2 dx
2
0
0
Z 1
Z 1
1
(2 + x2 )[∂x φ(x)]2 dx.
=
(1 + x)φ(x)2 dx + φ(1)2 +
2
0
0
Since all the terms are non negative, this means that all the terms are zero:
Z 1
Z 1
1
0=
(2 + x2 )[∂x φ(x)]2 dx.
(1 + x)φ(x)2 dx, 0 = φ(1), 0 =
2
0
0
This means in particular that
0 = (1 + x)φ(x)2 ,
for all x ∈ (0, 1).
Since 1 + x ≥ 1 in the interval (0, 1), this implies that φ(x) = 0 for all x ∈ (0, 1). In conclusion
φ1 = φ2 .
Question 151: Prove that if there exists a smooth solution to the Klein-Gordon equation then
it is unique: ∂tt u(x, t) − c2 ∂xx u(x, t) + m2 u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = f (x),
∂t u(x, 0) = g(x) and u(±∞, t) = 0, ∂t u(±∞, t) = 0, ∂x u(±∞, t) = 0. (Hint: test with ∂t u(x, t)
and use φφ0 = ( 21 φ2 )0 .)
Solution: Let u1 and u2 be two solutions. Then setting φ = u1 − u2 , we obtain that φ solves the
homogeneous problem ∂tt φ − c2 ∂xx φ + m2 φ = 0, x ∈ R, t > 0, with φ(x, 0) = φ(x), ∂t φ(x, 0) = 0
and φ(±∞, t) = 0, ∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0. Testing with ∂t φ(x, t) and integrating over
R and using the property ∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0, we obtain
Z +∞
Z +∞
1
1
∂t ( (∂t φ)2 )dx − c2
∂xx φ∂t φdx + m2
∂t ( φ2 )dx
2
2
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞
1
1 2
= dt
(∂t φ)2 dx + c2
∂x φ∂t ∂x φdx + m2 dt
φ dx
2
2
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞
1
1
1 2
2
2
2
2
= dt
(∂t φ) dx + c dt
(∂x φ) dx + m dt
φ dx
2
2
−∞
−∞
−∞ 2
Z +∞
Z +∞
Z +∞
1
1 2
1
2
2
2
2
(∂t φ) dx + c
(∂x φ) dx + m
φ dx
= dt
−∞ 2
−∞ 2
−∞ 2
Z +∞
= dt
(∂t φ)2 + c2 (∂x φ)2 + m2 φ2 dx .
Z
0=
+∞
−∞
Math 602
Introduce
Z
66
+∞
(∂t φ)2 + c2 (∂x φ)2 + m2 φ2 dx.
E(t) =
−∞
Then
dt E(t) = 0.
The fundamental Theorem of calculus gives
E(t) = E(0) = 0.
This means
Z
+∞
(∂t φ(x, t))2 + c2 (∂x φ(x, t))2 + m2 φ2 (x, t) dx = 0,
for all t ≥ 0.
−∞
This implies
0 = (∂t φ(x, t))2 + c2 (∂x φ(x, t))2 + m2 φ2 (x, t),
for all t ≥ 0, x ∈ R,
i.e. φ = 0, thereby proving the uniqueness.
Question 152: Consider the quasilinear Klein-Gordon equation: ∂tt φ(x, t) − c2 ∂xx φ(x, t) +
m2 φ(x, t)+β 2 φ3 (x, t) = 0, x ∈ R, t > 0, with φ(x, 0) = f (x), ∂t φ(x, 0) = g(x) and φ(±∞, t) = 0,
∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0. Find an energy E(t) which is invariant with respect to time
1
(Hint: test with ∂t φ(x, t) and use φp φ0 = ( p+1
φp+1 )0 .)
Solution: Testing with ∂t φ(x, t) and integrating over R and using the property ∂t φ(±∞, t) = 0,
∂x φ(±∞, t) = 0, we obtain
Z +∞
Z +∞
Z +∞
1
1
1 2
2
2
2
2
0=
∂t ( (∂t φ) )dx − c
∂xx φ∂t φdx + m
∂t ( φ4 )dx
∂t ( φ )dx + β
2
2
4
−∞
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞ 2
2
1
m 2 β 4
= dt
(∂t φ)2 dx + c2
φ +
φ )dx
∂x φ∂t ∂x φdx + dt
(
2
4
−∞ 2
−∞
−∞
Z +∞ 2
Z +∞ 2
Z +∞
c
m 2 β2 4
1
(∂t φ)2 dx + dt
(∂x φ)2 dx + dt
φ +
φ )dx
= dt
(
2
4
−∞ 2
−∞
−∞ 2
Z +∞ c2
m2 2 β 2 4
1
2
2
= dt
(∂t φ) + (∂x φ) +
φ +
φ dx.
2
2
2
4
−∞
Z
+∞
Introduce
Z
+∞
E(t) =
−∞
1
c2
m2 2 β 2 4
2
2
(∂t φ) + (∂x φ) +
φ +
φ )dx.
2
2
2
4
Then
dt E(t) = 0.
The fundamental Theorem of calculus gives
E(t) = E(0).
In conclusion the quantity E(t) is invariant with respect to time, as requested.
Question 153: Prove that if there exists a smooth solution to the Klein-Gordon equation then
it is unique: ∂tt u(x, t) − c2 ∂xx u(x, t) + m2 u(x, t) = 0, x ∈ R, t > 0, with u(x, 0) = f (x),
∂t u(x, 0) = g(x) and u(±∞, t) = 0, ∂t u(±∞, t) = 0, ∂x u(±∞, t) = 0. (Hint: test with ∂t u(x, t)
and use φφ0 = ( 21 φ2 )0 .)
Solution: Let u1 and u2 be two solutions. Then setting φ = u1 − u2 , we obtain that φ solves the
homogeneous problem ∂tt φ − c2 ∂xx φ + m2 φ = 0, x ∈ R, t > 0, with φ(x, 0) = φ(x), ∂t φ(x, 0) = 0
Math 602
67
and φ(±∞, t) = 0, ∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0. Testing with ∂t φ(x, t) and integrating over
R and using the property ∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0, we obtain
Z +∞
Z +∞
Z +∞
1
1
∂xx φ∂t φdx + m2
∂t ( φ2 )dx
0=
∂t ( (∂t φ)2 )dx − c2
2
2
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞
1
1 2
(∂t φ)2 dx + c2
∂x φ∂t ∂x φdx + m2 dt
φ dx
= dt
2
2
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞
1
1
1 2
2
2
2
2
= dt
(∂t φ) dx + c dt
(∂x φ) dx + m dt
φ dx
2
2
−∞
−∞
−∞ 2
Z +∞
Z +∞
Z +∞
1
1
1 2
2
2
2
2
= dt
(∂t φ) dx + c
(∂x φ) dx + m
φ dx
−∞ 2
−∞ 2
−∞ 2
Z +∞
(∂t φ)2 + c2 (∂x φ)2 + m2 φ2 dx .
= dt
−∞
Introduce
Z
+∞
(∂t φ)2 + c2 (∂x φ)2 + m2 φ2 dx.
E(t) =
−∞
Then
dt E(t) = 0.
The fundamental Theorem of calculus gives
E(t) = E(0) = 0.
This means
Z
+∞
(∂t φ(x, t))2 + c2 (∂x φ(x, t))2 + m2 φ2 (x, t) dx = 0,
for all t ≥ 0.
−∞
This implies
0 = (∂t φ(x, t))2 + c2 (∂x φ(x, t))2 + m2 φ2 (x, t),
for all t ≥ 0, x ∈ R,
i.e. φ = 0, thereby proving the uniqueness.
Question 154: Consider the quasilinear Klein-Gordon equation: ∂tt φ(x, t) − c2 ∂xx φ(x, t) +
m2 φ(x, t)+β 2 φ3 (x, t) = 0, x ∈ R, t > 0, with φ(x, 0) = f (x), ∂t φ(x, 0) = g(x) and φ(±∞, t) = 0,
∂t φ(±∞, t) = 0, ∂x φ(±∞, t) = 0. Find an energy E(t) which is invariant with respect to time
1
(Hint: test with ∂t φ(x, t) and use φp φ0 = ( p+1
φp+1 )0 .)
Solution: Testing with ∂t φ(x, t) and integrating over R and using the property ∂t φ(±∞, t) = 0,
∂x φ(±∞, t) = 0, we obtain
Z +∞
Z +∞
Z +∞
Z +∞
1
1
1
∂t ( φ4 )dx
∂xx φ∂t φdx + m2
∂t ( φ2 )dx + β 2
0=
∂t ( (∂t φ)2 )dx − c2
2
2
4
−∞
−∞
−∞
−∞
Z +∞
Z +∞
Z +∞ 2
2
1
m 2 β 4
= dt
(∂t φ)2 dx + c2
∂x φ∂t ∂x φdx + dt
(
φ +
φ )dx
2
2
4
−∞
−∞
−∞
Z +∞
Z +∞ 2
Z +∞ 2
1
c
m 2 β2 4
2
2
= dt
(∂t φ) dx + dt
(∂x φ) dx + dt
(
φ +
φ )dx
2
4
−∞ 2
−∞ 2
−∞
Z +∞ 1
c2
m2 2 β 2 4
= dt
(∂t φ)2 + (∂x φ)2 +
φ +
φ dx.
2
2
2
4
−∞
Introduce
Z
+∞
E(t) =
−∞
1
c2
m2 2 β 2 4
2
2
(∂t φ) + (∂x φ) +
φ +
φ )dx.
2
2
2
4
Then
dt E(t) = 0.
Math 602
68
The fundamental Theorem of calculus gives
E(t) = E(0).
In conclusion the quantity E(t) is invariant with respect to time, as requested.
Question 155: Consider the wave equation with variable coefficients
m(x)∂tt u(x, t) − ∂x (µ(x)∂x u(x, t)) = 0,
x ∈ (0, L),
t > 0,
with boundary condition u(0, t) = 0, ∂x u(L, t) = 0, where m (density of the material) and µ,
(elasticity of the material) and smooth positive functions. Assuming that the solution u(x, t) is
RL
smooth, prove that the quantity 0 12 m(x)|∂t u(x, t)|2 + 12 µ(x)|∂x u(x, t)|2 dx is independent
of t. (Hint: energy argument with ∂t u(x, t) and use a(t)∂t a(t) = ∂t ( 12 a(t)2 ).)
Solution: Multiply the equation by ∂t (ux, t) and integrate over (0, L).
Z
L
(m(x)∂tt u(x, t)∂t u(x, t) − ∂x µ(x)∂x u(x, t)∂t u(x, t)) dx
0=
0
Integrate by parts, use the boundary conditions, and use twice the formula a(t)∂t a(t) = ∂t ( 12 a(t)2 ),
L
1
2
0=
m(x)∂t |∂t u(x, t)| + µ(x)∂x u(x, t)∂x ∂t u(x, t) dx
2
0
Z L
1
=
m(x)∂t |∂t u(x, t)|2 + µ(x)∂x u(x, t)∂t ∂x u(x, t) dx
2
0
Z L
1
1
=
m(x)∂t |∂t u(x, t)|2 + µ(x)∂t |∂x u(x, t)|2 dx.
2
2
0
Z
Using the fact that m and µ are time-independent, we also have
Z L
1
1
0=
∂t ( m(x)|∂t u(x, t)|2 ) + ∂t ( µ(x)|∂x u(x, t)|2 ) dx,
2
2
0
Z L
d
1
1
=
m(x)|∂t u(x, t)|2 + µ(x)|∂x u(x, t)|2 dx,
dt 0
2
2
which proves the statement.
Question 156: Consider the vibrating beam equation ∂tt u(x, t) + ∂xxxx u(x, t) = 0, x ∈
(−∞, +∞), t > 0 with u(±∞, t) = 0, ∂x u(±∞, t) = 0, ∂xx u(±∞, t) = 0. Use the energy method
R +∞
to compute ∂t −∞ ([∂t u(x, t)]2 + [∂xx u(x, t)]2 )dx. (Hint: test the equation with ∂t u(x, t)).
Solution: Using the hint we have
Z +∞
0=
(∂tt u(x, t)∂t u(x, t) + ∂xxxx u(x, t)∂t u(x, t))dx
−∞
Using the product rule, a∂t a = 12 ∂t a2 where a = ∂t u(x, t), and integrating by parts two times (i.e.,
applying the fundamental theorem of calculus) we obtain
Z +∞
1
0=
( ∂t (∂t u(x, t))2 − ∂xxx u(x, t)∂t ∂x u(x, t))dx
2
−∞
Z +∞
1
=
(∂t (∂t u(x, t))2 + ∂xx u(x, t)∂t ∂xx u(x, t))dx.
2
−∞
We apply again the product rule a∂t a = 21 ∂t a2 where a = ∂xx u(x, t),
Z
+∞
0=
−∞
1
1
(∂t (∂t u(x, t))2 + ∂t (∂xx u(x, t))2 )dx.
2
2
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69
Switching the derivative with respect to t and the integration with respect to x, this finally gives
Z +∞
1
([∂t u(x, t)]2 + [∂xx u(x, t)]2 )dx.
0 = ∂t
2
−∞
Question 157: Consider the telegraph equation ∂tt u+(α+β)∂t u+αβu−c2 ∂xx u = 0 with α, β ≥
0, u(x, 0) = f (x), ∂t u(x, 0) = g(x), x ∈ R, t > 0 and boundary condition at infinity u(±∞, t) =
R +∞
0. Show that E(t) := −∞ 12 (∂t u(x, t))2 + c2 (∂x u(x, t))2 + αβ(u(x, t))2 dx decreases in time.
(Hint: Energy argument with ∂t u.)
Solution: We test the equation with ∂t u and integrate over R,
Z
Z 1
1
∂t ( (∂t u)2 ) + (α + β)(∂t u)2 + αβ∂t ( u2 ) dx − c2 ∂xx u(x, t)∂t u(x, t)dx = 0,
2
2
R
R
where we used the product rule, 2ψ(γ)∂γ ψ(γ) = ∂γ (ψ 2 (γ)). Integration by parts in the last integral
gives
Z
Z 1
1
∂t ( (∂t u)2 ) + (α + β)(∂t u)2 + αβ∂t ( u2 ) dx + c2 ∂x u(x, t)∂t ∂x u(x, t)dx = 0,
2
2
R
R
where we used the boundary condition at infinity u(±∞, t) = 0. This means that
Z 1
1
1
∂t ( (∂t u)2 ) + (α + β)(∂t u)2 + αβ∂t ( u2 ) + c2 ∂t ( (∂x u)2 ) dx = 0.
2
2
2
R
After exchanging the time derivative and the space integral, we have
Z Z
1
1 2
2
21
2
∂t
(∂t u) + αβ u + c (∂x u) dx = −(α + β) (∂t u)2 dx,
2
2
2
R
R
which means
Z
∂t E(t) = −(α + β)
(∂t u)2 dx ≤ 0,
R
i.e., E(t) decreases in time.
2+|x|
Question 158: Consider the vibrating beam equation ∂tt u(x, t) + ∂xx 1+|x|
∂xx u(x, t) = 0,
x ∈ (−∞, +∞), t > 0 with u(±∞, t) = 0, ∂x u(±∞, t) = 0, ∂xx u(±∞, t) = 0. Use the energy
R +∞
2+|x|
[∂xx u(x, t)]2 )dx. Give all the details. (Hint: test
method to compute ∂t −∞ ([∂t u(x, t)]2 + 1+|x|
the equation with ∂t u(x, t)).
Solution: Using the hint we have
Z +∞
2 + |x|
∂xx u(x, t) )dx
0=
(∂tt u(x, t)∂t u(x, t) + ∂xx
1 + |x|
−∞
Using the product rule, a∂t a = 12 ∂t a2 where a = ∂t u(x, t), and integrating by parts two times (i.e.,
applying the fundamental theorem of calculus) we obtain
Z +∞
1
2 + |x|
0=
( ∂t (∂t u(x, t))2 − ∂x
∂xx u(x, t) ∂t ∂x u(x, t))dx
1 + |x|
−∞ 2
Z +∞
1
2 + |x|
2
=
(∂t (∂t u(x, t)) +
∂xx u(x, t)∂t ∂xx u(x, t))dx.
2
1 + |x|
−∞
We apply again the product rule a∂t a = 12 ∂t a2 where a = ∂xx u(x, t),
Z
+∞
0=
−∞
1
1 2 + |x|
(∂t (∂t u(x, t))2 +
∂t (∂xx u(x, t))2 )dx.
2
2 1 + |x|
Math 602
70
Switching the derivative with respect to t and the integration with respect to x, this finally gives
Z +∞
1
2 + |x|
0 = ∂t
([∂t u(x, t)]2 +
[∂xx u(x, t)]2 )dx.
2
1 + |x|
−∞
Question 159: Let µ and c be two positive numbers. Consider the following system of coupled
partial differential equations with dependent variables ρ(x, t) and u(x, t): ∂t ρ + ∂x u = 0, ∂t u −
µ∂xx u + c2 ∂x ρ = 0, x ∈ (0, L), t > 0, with initial data ρ(x, 0) = ρ0 , u(x, 0) = u0 , and boundary
RL
conditions u(0, t) = u(L, t) = 0. (i) Show that the quantity E(t) = 0 21 (ρ2 (x, t)+c−2 u(x, t))dx
decreases in time. (Hint: Use the energy argument: Use ρ for the first equation, u/c2 for the
second, add the results, do not panic and proceed as usual.)
Solution: We proceed as in the hint. We multiply the first equation by ρ and integrate over the
domain,
Z L
Z L
1
0=
(ρ(x, t)∂t ρ(x, t) + ρ(x, t)∂x u(x, t))dx =
(∂t ( ρ2 (x, t)) + ρ(x, t)∂x u(x, t))dx
2
0
0
Z
Z L
d L1 2
ρ (x, t)dx +
ρ(x, t)∂x u(x, t)dx.
=
dt 0 2
0
We proceed similarly for the second equation. We multiply the second equation by c−2 u and
integrate over the domain,
Z
L
(c−2 u(x, t)∂t u(x, t) − µc−2 u(x, t)∂xx u(x, t) + u(x, t)∂x ρ(x, t))dx
0=
0
RL
RL
We integrate by parts the viscous term and we obtain − 0 µc−2 u(x, t)∂xx u(x, t)dx = 0 µc−2 (∂x u(x, t))2 dx,
owing to the boundary conditions u(0, t) = u(L, t) = 0,
Z
L
1
(c−2 ∂t ( u2 (x, t)) + µc−2 (∂x u(x, t))2 + u(x, t)∂x ρ(x, t))dx
2
0
Z
Z L
d L 1 −2 2
=
c u (x, t)dx +
(µc−2 (∂x u(x, t))2 + u(x, t)∂x ρ(x, t))dx.
dt 0 2
0
0=
Adding the two equations together and using the product rule gives
d
0=
dt
Z
d
=
dt
Z
d
=
dt
Z
L
0
L
0
0
L
Z
Z L
d L 1 −2 2
ρ(x, t)∂x u(x, t)dx +
c u (x, t)dx +
(µc−2 (∂x u(x, t))2 + u(x, t)∂x ρ(x, t))dx
dt 0 2
0
0
Z L
Z L
1 2
(ρ (x, t) + c−2 u2 (x, t))dx +
µc−2 (∂x u(x, t))2 dx +
(ρ(x, t)∂x u(x, t) + u(x, t)∂x ρ(x, t))dx
2
0
0
Z L
Z L
1 2
(ρ (x, t) + c−2 u2 (x, t))dx +
µc−2 (∂x u(x, t))2 dx +
∂x (ρ(x, t)u(x, t))dx.
2
0
0
1 2
ρ (x, t)dx +
2
Z
L
RL
The Fundamental Theorem of Calculus gives 0 ∂x (ρ(x, t)u(x, t))dx = 0, owing to the boundary
conditions on u. In conclusion
Z
Z L
d L1 2
(ρ (x, t) + c−2 u2 (x, t))dx = −
µc−2 (∂x u(x, t))2 dx ≤ 0,
dt 0 2
0
which proves that the quantity E(t) =
RL
1 2
(ρ (x, t)
0 2
+ c−2 u(x, t))dx decreases in time.
(ii) What how does E(t) behaves when µ = 0?
Solution: The above computation shows that
in time when µ = 0.
d
dt E(t)
= 0 when µ = 0; as a result, E(t) is constant
2
Question 160: Consider the vibrating beam equation ∂tt u(x, t) + ∂xx x +cos(x)
1+x2 ∂xx u(x, t) = 0,
u(x, 0) = f (x), ∂t u(x, 0) = g(x), x ∈ (−∞, +∞), t > 0 with u(±∞, t) = 0, ∂x u(±∞, t) = 0,
Math 602
71
R +∞
2
2
∂xx u(±∞, t) = 0. Use the energy method to compute −∞ ([∂t u(x, t)]2 + x +cos(x)
1+x2 [∂xx u(x, t)] )dx
in terms of f and g. Give all the details. (Hint: test the equation with ∂t u(x, t)).
Solution: Using the hint we have
2
Z +∞
x + cos(x)
(∂tt u(x, t)∂t u(x, t) + ∂xx
0=
∂
u(x,
t)
)dx
xx
1 + x2
−∞
Using the product rule, a∂t a = 21 ∂t a2 where a = ∂t u(x, t), and integrating by parts two times (i.e.,
applying the fundamental theorem of calculus) we obtain
2
Z +∞
1
x + cos(x)
( ∂t (∂t u(x, t))2 − ∂x
0=
∂
u(x,
t)
∂t ∂x u(x, t))dx
xx
1 + x2
−∞ 2
2
Z +∞
1
x + cos(x)
2
(∂t (∂t u(x, t)) +
∂xx u(x, t)∂t ∂xx u(x, t))dx.
=
2
1 + x2
−∞
We apply again the product rule a∂t a = 12 ∂t a2 where a = ∂xx u(x, t),
Z
+∞
0=
−∞
1 x2 + cos(x)
1
(∂t (∂t u(x, t))2 +
∂t (∂xx u(x, t))2 )dx.
2
2 1 + x2
Switching the derivative with respect to t and the integration with respect to x, this finally gives
Z +∞
1
x2 + cos(x)
[∂xx u(x, t)]2 )dx.
0 = ∂t
([∂t u(x, t)]2 +
2
2
1
+
x
−∞
In other words,
Z +∞
Z +∞
x2 + cos(x)
x2 + cos(x)
2
([∂t u(x, t)]2 +
[∂
u(x,
t)]
)dx
=
(g(x)2 +
[∂x f (x)]2 )dx.
xx
2
2
1
+
x
1
+
x
−∞
−∞
Question 161: Assume that the following equation has a smooth solution: −∂x ((1+x2 )∂x T (x))+
∂x T (x) + T (x) = 2x − 1, T (a) = 1, T (b) = π, x ∈ [a, b], t > 0, where k > 0. Prove that this
solution is unique by using the energy method. (Hint: Do not try to simplify −∂x ((1 + x2 )∂x T ).
Solution: Assume that there are two solutions T1 and T2 . Let φ = T2 − T1 . Then
−∂x ((1 + x2 )∂x φ(x)) + ∂x φ(x) + φ(x) = 0,
φ(a) = 0,
φ(b) = 0
Multiply the PDE by φ, integrate over (a, b), and integrate by parts (i.e. apply the fundamental
theorem of calculus):
Z
b
− ∂x ((1 + x2 )∂x φ(x))φ(x) + (∂x φ(x))φ(x) + (φ(x))2 dx
0=
a
Z
b
1
− ∂x (φ(x)(1 + x2 )∂x φ(x)) + (1 + x2 )(∂x φ(x))2 + ∂x ( φ(x)2 ) + (φ(x))2 dx
2
(1 + x2 )(∂x φ(x))2 + (φ(x))2 dx
=
a
Z
b
=
a
This implies
Rb
a
(φ(x))2 dx = 0, i.e. φ = 0, meaning that T2 = T1 .
Question 162: Assume that the following equation has a smooth solution: −∂x ((1+x2 )∂x T (x))+
∂x T (x) + T (x) = cos(x), T (a) = 1, T (b) = π, x ∈ [a, b], t > 0, where k > 0. Prove that this
solution is unique by using the energy method. (Hint: Do not try to simplify −∂x ((1 + x2 )∂x T ).
Solution: Assume that there are two solutions T1 and T2 . Let φ = T2 − T1 . Then
−∂x ((1 + x2 )∂x φ(x)) + ∂x φ(x) + φ(x) = 0,
φ(a) = 0,
φ(b) = 0
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72
Multiply the PDE by φ, integrate over (a, b), and integrate by parts (i.e. apply the fundamental
theorem of calculus):
Z
b
− ∂x ((1 + x2 )∂x φ(x))φ(x) + (∂x φ(x))φ(x) + (φ(x))2 dx
0=
a
Z
b
1
− ∂x (φ(x)(1 + x2 )∂x φ(x)) + (1 + x2 )(∂x φ(x))2 + ∂x ( φ(x)2 ) + (φ(x))2 dx
2
(1 + x2 )(∂x φ(x))2 + (φ(x))2 dx
=
a
Z
b
=
a
This implies
Rb
a
(φ(x))2 dx = 0, i.e. φ = 0, meaning that T2 = T1 .
Question 163: Assume that the following equation has a smooth solution: −∂x ((1+x2 )∂x T (x))−
5∂x T (x) + (1 + b − x)T (x) = cos(x), T (a) = 1, T (b) = π, x ∈ [a, b], t > 0, where k > 0.
Prove that this solution is unique by using the energy method. (Hint: Do not try to simplify
−∂x ((1 + x2 )∂x T ).
Solution: Assume that there are two solutions T1 and T2 . Let φ = T2 − T1 . Then
−∂x ((1 + x2 )∂x φ(x)) − 5∂x φ(x) + (1 + b − x)φ(x) = 0,
φ(a) = 0,
φ(b) = 0
Multiply the PDE by φ, integrate over (a, b), and integrate by parts (i.e. apply the fundamental
theorem of calculus):
Z
b
− ∂x ((1 + x2 )∂x φ(x))φ(x) − 5(∂x φ(x))φ(x) + (1 + b − x)(φ(x))2 dx
0=
a
Z
b
− ∂x (φ(x)(1 + x2 )∂x φ(x)) + (1 + x2 )(∂x φ(x))2 − 5∂x
=
a
Z
=
b
(1 + x2 )(∂x φ(x))2 + (1 + b − x)(φ(x))2 dx ≥
a
since 1 + b − x ≥ 1 for all x ∈ [a, b]. This implies
T2 = T1 .
Z
1
2
φ(x)2 + (1 + b − x)(φ(x))2 dx
b
φ2 (x)dx,
a
Rb
a
(φ(x))2 dx = 0, i.e., φ = 0, meaning that
Math 602
9
73
Method of characteristics
Question 164: (a) Show that the PDE uy = 0 in the half plane {x > 0} has no solution which
is C 1 and satisfies the boundary condition u(y 2 , y) = y.
Solution: The PDE implies that u(x, y) = φ(x) where φ is any C 1 function. The boundary condition
implies φ(1) = u(1, −1) = −1 and φ(1) = u(1, 1) = 1, which is impossible. The reason for this
happening is that the characteristics lines (x = c) cross the boundary curve (the parabola of equation
x = y 2 ) twice.
(b) Find the C 1 function that solves the above PDE in the quadrant {x > 0, 0 > y} (beware
the sign of y).
Solution: The PDE implies u(x, y) = φ(x) and the boundary
condition implies φ(y 2 ) = u(y 2 , y) =
√
y = −|y| since y is negative. Then u(x, y) = φ(x) = − x.
Question 165: Let Ω = {x > 0, y > 0} be the first quadrant of the plane. Let Γ be the line
defined by the following parameterization Γ = {x = s, y = 1/s, s > 0}. Solve the following
PDE:
xux + 2yuy = 0,
u(x, y) = x
in Ω,
on Γ.
Solution: The characteristics are X(τ, s) = seτ , Y (τ, s) = s−1 e2τ . Upon setting u(X(τ, s), Y (τ, s)) =
w(τ, s), we obtain w(τ, s) = w(0, s). Then the boundary condition implies w(0, s) = u(s, 1s ) = s.
In other words u(x, y) = (x2 y −1 )1/3 .
Question 166: (a) Solve the quasi-linear PDE 3u2 ux + 3u2 uy = 1 in the plane by using the
method of Lagrange (that is, show that u solves the nonlinear equation c(a(x, y, u), b(x, y, u)) =
0 where c is an arbitrary function and a, b are polynomials of degree 3 that you must find.)
Solution: The auxiliary equation is 3z 2 φx + 3z 2 φy + φz = 0. Define the plane Γ = {x = s, y =
s0 , z = 0} and enforce φ(x, y, z) = φ0 (s, s0 ) on Γ, where φ0 is an arbitrary C 1 function. The
characteristics are X(τ, s, s0 ) = τ 3 + s Y (τ, s, s0 ) = τ 3 + s0 , Z(τ, s, s0 ) = τ . Then φ(x, y, z) =
φ0 (s, s0 ) where s = x − z 3 and s0 = y − z 3 . Then φ(x, y, z) = φ0 (x − z 3 , y − z 3 ). Hence, u solves
φ0 (x − u3 , y − u3 ) = 0.
(b) Find a solution to the above PDE that satisfies the boundary condition u(x, 2x) = 1.
Solution: We want φ0 (x−1, 2x−1) = 0. Take φ0 (α, β) = 2α−β+1. Then 2(x−u3 )−(y−u3 )+1 =
0, that is u(x, y) = (1 + 2x − y)1/3 .
Question 167: We want to solve the following PDE:
∂t w + 3∂x w = 0,
x > −t, t > 0
w(x, t) = wΓ (x, t), for all (x, t) ∈ Γ where
Γ = {(x, t) ∈ R2 s.t. x = −t, x < 0} ∪ {(x, t) ∈ R2 s.t. t = 0, x ≥ 0}
and wΓ is a given function.
(a) Draw a picture of the domain Ω where the PDE must be solved, of the boundary Γ, and of
the characteristics.
Solution:
(b) Define a one-to-one parametric representation of the boundary Γ.
Solution: For negative s we set xΓ (s) = s and tΓ (s) = −s; clearly we have xΓ (s) = −tΓ (s) for all
s < 0. For positive s we set xΓ (s) = s and tΓ (s) = 0. The map R ∈ s 7→ (xΓ (s), tΓ (s)) ∈ Γ is
one-t-one.
(c) Give a parametric representation of the characteristics associated with the PDE.
Solution: (i) We use t and s to parameterize the characteristics. The characteristics are defined
by
∂t X(t, s) = 3,
with x(tΓ (s), s) = xΓ (s).
Math 602
74
This yields the following parametric representation of the characteristics
X(t, s) = 3(t − tΓ (s)) + xΓ (s),
where t ≥ 0 and s ∈ (−∞, +∞).
(d) Give an implicit parametric representation of the solution to the PDE.
Solution: (i) Now we set φ(t, s) = w(X(t, s), t) and we insert this ansatz in the equation. This
gives dφ
dt (t, s) = 0, i.e., φ(t, s) does not depend on t. In other words
w(X(t, s), t) = φ(t, s) = φ(0, s) = w(x(0, s), t(0, s)) = wΓ (xΓ (s), tΓ (s))
A parametric representation of the solution is given by
X(t, s) = 3(t − tΓ (s)) + xΓ (s),
w(X(t, s), t) = wΓ (xΓ (s), tΓ (s)).
(e) Give an explicit representation of the solution.
Solution: (i) We have to find the inverse map (x, t) 7→ (s, t), where x − 3t = xΓ (s) − 3tΓ (s).
Then, there are two cases depending on the sign of s.
case 1: If s < 0, then xΓ (s) = s and tΓ (s) = −s. That means x − 3t = 4s, which in turns implies
s = 14 (x − 3t). Then
w(x, t) = wΓ ( 41 (x − 3t), − 41 (x − 3t)),
if x − 3t < 0.
case 2: If s ≥ 0, then xΓ (s) = s and tΓ (s) = 0. That means x − 3t = s. Then
w(x, t) = wΓ (x − 3t, 0),
if x − 3t ≥ 0.
Note that the explicit representation of the solution does not depend on the choice of the parameterization.
Question 168: Solve the following PDE by the method of characteristics:
∂t w + 3∂x w = 0,
w(x, 0) = f (x),
x > 0, t > 0
x > 0,
and w(0, t) = h(t),
t > 0.
Solution: First we parameterize the boundary of Ω by setting Γ = {x = xΓ (s), t = tΓ (s); s ∈ R}
with
(
(
0 if s < 0,
−s if s < 0,
xΓ (s) =
and
tΓ (s) =
s, if s ≥ 0.
0, if s ≥ 0.
The we define the characteristics by
∂t X(s, t) = 3,
with
X(s, tΓ (s)) = xΓ (s).
The general solution is X(s, t) = 3(t − tΓ (s)) + xΓ (s). Now we make the change of variable
φ(s, t) = w(X(s, t), t) and we compute ∂t φ(s, t),
∂t φ(s, t) = ∂t w(X(s, t), t) + ∂x w(X(s, t), t)∂t X(s, t) = ∂t w(X(s, t), t) + 3∂x w(X(s, t), t) = 0.
This means that φ(s, t) = φ(s, tΓ (s)). In other words
w(X(s, t), t) = w(X(s, tΓ (s)), tΓ (s)) = w(xΓ (s), tΓ (s)).
Case 1: If s < 0, then X(s, t) = 3(t − tΓ (s)). This implies tΓ (s) = t − X/3. The condition s < 0
and the definition tΓ (s) = −s imply t − X/3 ≥ 0. Moreover we have
w(X, t) = w(0, tΓ (s)) = h(tΓ (s)).
Math 602
75
In conlusion
w(X, t) = h(t − X/3),
if
3t > X.
Case 2: If s ≥ 0, then X(s, t) = 3t + xΓ (s). This implies xΓ (s) = X − 3t. The condition s ≥ 0
and the definition xΓ (s) = s imply X − 3t ≥ 0. Moreover we have
w(X, t) = w(xΓ (s), 0) = f (xΓ (s)).
In conlusion
w(X, t) = f (X − 3t),
X ≥ 3t.
if
Question 169: Let Ω = {(x, t) ∈ R2 ; x + 2t ≥ 0}. Solve the following PDE in explicit form
with the method of characteristics:
∂t u(x, t) + 3∂x u(x, t) = u(x, t),
in Ω,
and u(x, t) = 1 + sin(x), if x + 2t = 0.
Solution: (i) First we parameterize the boundary of Ω by setting Γ = {x = xΓ (s), t = tΓ (s); s ∈
R} with xΓ (s) = −2s and tΓ (s) = s . This choice implies
u(xΓ (s), tΓ (s)) := uΓ (s) := 1 + sin(−2s).
(ii) We compute the characteristics
∂t X(t, s) = 3,
X(tΓ (s), s) = xΓ (s).
The solution is X(t, s) = 3(t − tΓ (s)) + xΓ (s).
(iii) Set Φ(t, s) := u(X(t, s), t) and compute ∂t Φ(t, s). This gives
∂t Φ(t, s) = ∂t u(X(t, s), t) + ∂x u(X(t, s), t)∂t X(t, s)
= ∂t u(X(t, s), t) + 3∂x u(X(t, s), t) = u(X(t, s), t) = Φ(t, s).
The solution is Φ(t, s) = Φ(tΓ (s), s)et−tΓ (s) .
(iv) The implicit representation of the solution is
X(t, s) = 3(t − tΓ (s)) + xΓ (s) u(X(t, s)) = uΓ (s)et−tΓ (s) .
(v) The explicit representation is obtained by using the definitions of −tΓ (s), xΓ (s) and uΓ (s).
X(s, t) = 3(t − s) − 2s = 3t − 5s,
which gives
s=
1
(3t − X).
5
The solution is
1
u(x, t) = (1 + sin( 52 (x − 3t)))et− 5 (3t−x)
= (1 + sin( 2(x−3t)
))e
5
x+2t
5
.
Question 170: Let Ω = {(x, t) ∈ R2 ; x ≥ 0, t ≥ 0}. Solve the following PDE in explicit form
∂t u(x, t) + t∂x u(x, t) = 2u(x, t),
in Ω,
and u(0, t) = t, u(x, 0) = x.
Solution: (i) First we parameterize the boundary of Ω by setting Γ = {x = xΓ (s), t = tΓ (s); s ∈
R} with xΓ (s) = s and tΓ (s) = 0 if s > 0 and xΓ (s) = 0 and tΓ (s) = −s if s ≤ 0. This choice
implies
(
s
if s > 0
u(xΓ (s), tΓ (s)) := uΓ (s) :=
.
−s if s ≤ 0
Math 602
76
(ii) We compute the characteristics
∂t X(t, s) = t,
X(tΓ (s), s) = xΓ (s).
The solution is X(t, s) = 21 t2 − 12 t2Γ (s) + xΓ (s).
(iii) Set Φ(t, s) := u(X(t, s), t) and compute ∂t Φ(t, s). This gives
∂t Φ(t, s) = ∂t u(X(t, s), t) + ∂x u(X(t, s), t)∂t X(t, s)
= ∂t u(X(t, s), t) + t∂x u(X(t, s), t) = 2u(X(t, s), t) = 2Φ(t, s).
The solution is Φ(t, s) = Φ(tΓ (s), s)e2(t−tΓ (s)) .
(iv) The implicit representation of the solution is
1
1
X(t, s) = t2 − t2Γ (s) + xΓ (s),
2
2
2(t−tΓ (s))
u(X(t, s)) = uΓ (s)e
,
(
s
if s > 0
.
uΓ (s) =
−s if s ≤ 0
(v) We distinguish two cases to get the explicit form of the solution:
Case 1: Assume s > 0, then tΓ (s) = 0 and xΓ (s) = s. This implies X(t, s) = 21 t2 + s, meaning
s = X − 21 t2 . The solution is
1
u(x, t) = (x − t2 )e2t ,
2
if x >
1 2
t .
2
Case 2: Assume s ≤ 0, then tΓ (s) = −s and xΓ (s) = 0. This implies X(t, s) = 12 t2 − 12 s2 , meaning
√
s = − t2 − 2X. The solution is
√
p
1
2
u(x, t) = t2 − 2x e2(t− t −2x) , if x ≤ t2 .
2
Question 171: Let Ω = {(t, x) ∈ R2 : t > 0, x ≥ t}. Let Γ be defined by the following
parameterization Γ = {x = xΓ (s), t = tΓ (s), s ∈ R}, with xΓ (s) = −s and tΓ (s) = −s if s ≤ 0,
xΓ (s) = s and tΓ (s) = 0 if s ≥ 0. Solve the following PDE (give the implicit and explicit
representations):
(
1 if t = 0
ut + 3ux + 2u = 0, in Ω,
u(x, t) = uΓ (x, t) :=
for all (x, t) in Γ.
2 if x = t
Solution: We define the characteristics by
dx(t, s)
= 3,
dt
x(tΓ (s), s) = xΓ (s).
This gives x(t, s) = xΓ (s) + 3(t − tΓ (s)). Upon setting φ(t, s) = u(x(t, s), t), we observe that
∂t φ(t, s) + 2φ(t, s) = 0, which means
φ(t, s) = ce−2t .
The initial condition implies φ(tΓ (s), s) = uΓ (xΓ (s), tΓ (s)); as a result c = uΓ (xΓ (s), tΓ (s))e2tΓ (s) .
φ(t, s) = uΓ (xΓ (s), tΓ (s))e2(tΓ (s)−t) .
The implicit representation of the solution is
u(x(t, s), t) = uΓ (xΓ (s), tΓ (s))e2(tΓ (s)−t) ,
x(t, s) = xΓ (s) + 3(t − tΓ (s)).
Now we give the explicit representation.
Case 1: If s ≤ 0, xΓ (s) = −s, tΓ (s) = −s, and uΓ (xΓ (s), tΓ (s)) = 2. This means x(t, s) =
−s + 3(t + s) and we obtain s = 21 (x − 3t), which means
1
u(x, t) = 2e−2( 2 (x−3t)−t) = 2et−x ,
if x − 3t < 0.
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77
Case 2: If s ≥ 0, xΓ (s) = s, tΓ (s) = 0, and uΓ (xΓ (s), tΓ (s)) = 1. This means x(t, s) = s + 3t and
we obtain s = x − 3t, which means
u(x, t) = e−2t ,
if x − 3t > 0.
√
Question 172: Let Ω = {(t, x) ∈ R2 : t > 0, x ≥ − t}. Let Γ be defined by the following
parameterization Γ = {x = xΓ (s), t = tΓ (s), s ∈ R}, with xΓ (s) = s and tΓ (s) = s2 if s ≤ 0,
xΓ (s) = s and tΓ (s) = 0 if s ≥ 0. Solve the following PDE (give the implicit and explicit
representations):
ut + 2ux + 3u = 0,
in Ω,
and u(xΓ (s), tΓ (s)) := e−tΓ (s)−xΓ (s) ,
∀s ∈ (−∞, +∞).
Solution: We define the characteristics by
dX(t, s)
= 2,
dt
X(tΓ (s), s) = xΓ (s).
This gives X(t, s) = xΓ (s) + 2(t − tΓ (s)). Upon setting φ(t, s) = u(X(t, s), t), we observe that
∂t φ(t, s) + 3φ(t, s) = 0, which means
φ(t, s) = ce−3t .
The initial condition implies φ(tΓ (s), s) = u(xΓ (s), tΓ (s)) = e−tΓ (s)−xΓ (s) = ce−3tΓ (s) ; as a result
c = e2tΓ (s)−xΓ (s) and
φ(t, s) = e2tΓ (s)−xΓ (s)−3t .
The implicit representation of the solution is
u(X(t, s), t) = e2tΓ (s)−xΓ (s)−3t ,
X(t, s) = xΓ (s) + 2(t − tΓ (s)).
Now we give the explicit representation.
We observe the following:
2tΓ (s) − xΓ (s) = 2t − X(t, s),
which gives
u(X(t, s), t) = e2t−X(t,s)−3t = e−X(t,s)−t .
In conclusion, the explicit representation of the solution to the problem is the following:
u(x, t) = e−x−t .
Question 173: Let Ω = {(t, x) ∈ R2 : t > 0, x ≥ −t}. Let Γ be defined by the following
parameterization Γ = {x = xΓ (s), t = tΓ (s), s ∈ R}, with xΓ (s) = s and tΓ (s) = −s if s ≤ 0,
xΓ (s) = s and tΓ (s) = 0 if s ≥ 0. Solve the following PDE (give the implicit and the explicit
representations):
(
1 if x > 0
ut + 2ux + u = 0, in Ω,
u(x, t) = uΓ (x, t) :=
for all (x, t) in Γ.
2 if x < 0
Solution: We define the characteristics by
dx(t, s)
= 2,
dt
x(tΓ (s), s) = xΓ (s).
This gives x(t, s) = xΓ (s) + 2(t − tΓ (s)). Upon setting φ(t, s) = u(x(t, s), t), we observe that
∂t φ(t, s) + φ(t, s) = 0, which means
φ(t, s) = ce−t .
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78
The initial condition implies φ(tΓ (s), s) = uΓ (xΓ (s), tΓ (s)); as a result c = uΓ (xΓ (s), tΓ (s))etΓ (s) .
φ(t, s) = uΓ (xΓ (s), tΓ (s))etΓ (s)−t .
The implicit representation of the solution is
u(x(t, s), t) = uΓ (xΓ (s), tΓ (s))etΓ (s)−t ,
x(t, s) = xΓ (s) + 2(t − tΓ (s)).
Now we give the explicit representation.
Case 1: If s ≤ 0, xΓ (s) = s, tΓ (s) = −s, and uΓ (xΓ (s), tΓ (s)) = 2. This means x(t, s) = s+2(t+s)
and we obtain s = 31 (x − 2t), which means
1
u(x, t) = 2e− 3 (x−2t)−t ,
if x − 2t < 0.
Case 2: If s ≥ 0, xΓ (s) = s, tΓ (s) = 0, and uΓ (xΓ (s), tΓ (s)) = 1. This means x(t, s) = s + 2t and
we obtain s = x − 2t, which means
u(x, t) = e−t ,
if x − 2t > 0.
Question 174: Let Ω = {(x, t) ∈ R2 | t > 0, x ≥ 1t }. Solve the following PDE in explicit form
1
with the method of characteristics: (Solution: u(x, t) = (2 + cos(s))e s −t with s = 21 [(x − 2t) +
p
(x − 2t)2 + 8])
∂t u(x, t) + 2∂x u(x, t) = −u(x, t),
in Ω,
and u(x, t) = 2 + cos(x), if x = 1/t.
Solution: (i) First we parameterize the boundary of Ω by setting Γ = {x = xΓ (s), t = tΓ (s); s ∈
R} with xΓ (s) = s and tΓ (s) = 1s . This choice implies
u(xΓ (s), tΓ (s)) := uΓ (s) := 2 + cos(s).
(ii) We compute the characteristics
∂t X(t, s) = 2,
X(tΓ (s), s) = xΓ (s).
The solution is X(t, s) = 2(t − tΓ (s)) + xΓ (s).
(iii) Set Φ(t, s) := u(X(t, s), t) and compute ∂t Φ(t, s). This gives
∂t Φ(t, s) = ∂t u(X(t, s), t) + ∂x u(X(t, s), t)∂t X(t, s)
= ∂t u(X(t, s), t) + 2∂x u(X(t, s), t) = u(X(t, s), t) = −Φ(t, s).
The solution is Φ(t, s) = Φ(tΓ (s), s)e−t+tΓ (s) .
(iv) The implicit representation of the solution is
X(t, s) = 2(t − tΓ (s)) + xΓ (s),
u(X(t, s)) = uΓ (s)e−t+tΓ (s) .
(v) The explicit representation is obtained by using the definitions of −tΓ (s), xΓ (s) and uΓ (s).
2
1
X(s, t) = 2(t − ) + s = 2t − + s
s
s
which gives the equation
The solutions are s±
positive one:
s2 − s(X − 2t) − 2 = 0
p
= 21 (X − 2t) ± (X − 2t)2 + 8 . The only legitimate solution is the
s=
p
1
(X − 2t) + (X − 2t)2 + 8
2
Math 602
79
The solution is
1
u(x, t) = (2 + cos(s))e s −t
p
1
with s = ((x − 2t) + (x − 2t)2 + 8)
2
Question 175: Solve the following PDE by the method of the characteristics:
1
∂x w = 0, x > 1, t > 0
2x
w(x, 0) = f (x), x > 1, and w(0, t) = h(t),
∂t w +
t > 0.
Solution: First we parameterize the boundary of Ω by setting Γ = {x = xΓ (s), t = tΓ (s); s ∈ R}
with
(
(
−s if s < 0,
1
if s < 0,
and
tΓ (s) =
xΓ (s) =
0, if s ≥ 0.
1 + s, if s ≥ 0.
We define the family of characteristics X(s, t) by
∂t X(s, t) =
1
,
2X(s, t)
with
X(s, tΓ (s)) = xΓ (s).
Then ∂t X(s, t)2 = 1. The general solution is X(s, t)2 = xΓ (s)2 + t − tΓ (s). Now we make the
change of variable φ(s, t) = w(X(s, t), t) and we compute ∂t φ(s, t),
∂t φ(s, t) = ∂t w(X(s, t), t)+∂x w(X(s, t), t)∂t X(s, t) = ∂t w(X(s, t), t)+
1
∂x w(X(s, t), t) = 0.
2X(s, t)
This means that φ(s, t) = φ(s, tΓ (s)). In other words
w(X(s, t), t) = w(X(s, tΓ (s)), tΓ (s)) = w(xΓ (s), tΓ (s)).
Case 1: If s < 0, then X(s, t)2 = 1 + t + s. This implies s = X 2 − 1 − t. The condition s < 0
implies X 2 < 1 + t. Moreover we have
w(X, t) = w(0, tΓ (s)) = h(tΓ (s)) = h(−s).
In conclusion
w(X, t) = h(1 + t − X 2 ),
X 2 < 1 + t.
√
Case 2: If s ≥ 0, then X(s, t)2 = (1 + s)2 + t. This implies s = X 2 − t − 1. The condition s ≥ 0
implies that X 2 ≥ 1 + t. Moreover we have
if
w(X, t) = w(xΓ (s), 0) = f (xΓ (s)) = f (1 + s).
In conclusion
p
w(X, t) = f ( X 2 − t),
if
X 2 ≥ 1 + t.
Solve the PDE ∂t u + 4∂x u + 12u = 0 in Ω = {(x, t) ∈ R2 | x ≥ 0, x + 11t ≥ 0} with boundary
conditions u(x, 0) = x + 4, u(−11t, t) = t + 4 for t > 0.
Solution: (1) The boundary of Ω, say Γ, is parametrized as follows: Γ = {(xΓ (s), tΓ (s)) | s ∈ R}
where
(
(
11s if s < 0,
−s if s < 0,
xΓ (s) =
tΓ (s) =
s
if s > 0.
0
if s > 0.
(2) The characteridtics are defined by the ODE
∂t X(s, t) = 4,
X(s, tΓ (s)) = xΓ (s).
Math 602
The solution is X(s, t) = 4(t − tΓ (s)) + xΓ (s).
(3) We make a change of variable: φ(s, t) = u(X(s, t), t) and compute ∂t φ(s, t),
∂t φ(s, t) = ∂t u(X(s, t), t) + ∂x u(X(s, t), t)∂t X(s, t) = ∂t u(X(s, t), t) + 4∂x u(X(s, t), t)
= −12u(X(s, t), t) = −12φ(s, t),
with initial data φ(s, tΓ(s) ) = u(xΓ(s) , tΓ(s) ). The solution is
φ(s, t) = φ(s, tΓ(s) )e−12(t−tΓ(s) ) = u(xΓ(s) , tΓ(s) )e−12(t−tΓ(s) ) .
i.e., the implicit representation of the solution is
u(X(s, t), t) = u(xΓ(s) , t)e−12(t−tΓ(s) ) ,
X(s, t) = 4(t − tΓ (s)) + xΓ (s).
(4) The explicit representation is obtained as follows:
Case 1 (s < 0): X(s, t) = 4(t + s) + 11s, then s = (X − 4t)/15 and
u(x, t) = u(11s, −s)e−12(t+s) = (−s + 4)e−12(t+s) = (4 −
= (4 +
(4t − x) −4 11t+x
5
)e
,
15
(x − 4t) −12(t+ x−4t )
15
)e
15
if x < 4t.
Case 2 (s > 0): X(s, t) = 4t + s, then s = X − 4t, and
u(x, t) = u(s, 0)e−12t = (s + 4)e−12t = (4 + x − 4t)e−12t
if 4t < x.
80
Math 602
10
81
Conservation equations
The implicit representation of the solution to the equation ∂t v + ∂x q(v) = 0, v(x, 0) = v0 (x), is
X(s, t) = q 0 (v0 (s))t + s;
v(X(s, t), t) = v0 (s).
(12)
Question 176: Consider the following conservation equation
(
x ∈ (−∞, +∞), t > 0,
∂t ρ + ∂x (q(ρ)) = 0,
ρ(x, 0) = ρ0 (x) :=
1
6
1
3
if x < 0,
if x > 0,
where q(ρ) = ρ(2 − 3ρ) (and ρ(x, t) is the conserved quantity). Solve this problem using the
method of characteristics. Do we have a shock or an expansion wave here?
Solution: The characteristics are defined by
dX(s, t)
= q 0 (ρ) = 2(1 − 3ρ(X(s, t), t)),
dt
X(s, t) = s.
Set φ(s, t) = ρ(X(s, t), t), then we obtain that φ is constant, i.e., ρ is constant along the characteristics: ρ(X(s, t), t) = ρ(s, 0) = ρ0 (s). As a result we can integrate the equation defining the
characteristics and we obtain X(t) = 2(1−3ρ0 (s))t+s. We then have two cases depending whether
s is positive or negative.
1. s < 0, then ρ0 (s) =
1
6
and X(s, t) = t + s. This means
ρ(x, t) =
2. s > 0, then ρ0 (s) =
1
3
1
6
if
x < t.
and X(s, t) = s. This means
ρ(x, t) =
1
3
if
x > 0.
We see that the characteristics cross in the region {t > x > 0}. This implies that there is a shock.
The Rankin-Hugoniot relation gives the speed of this shock:
s=
q+ − q−
=
ρ+ − ρ−
1
13
62 − 3
1
1
6 − 3
=
1
1
6= .
12
2
In conclusion
1
,
6
1
ρ= ,
3
ρ=
t
,
2
t
x> .
2
x<
Question 177: Consider the following conservation equation
(
∂t ρ + ∂x (q(ρ)) = 0,
x ∈ (−∞, +∞), t > 0,
ρ(x, 0) = ρ0 (x) :=
1
6
1
3
if x < 0,
if x > 0,
where q(ρ) = ρ(2 − 3ρ) (and ρ(x, t) is the conserved quantity). (i) Given that this initial data
produces a shock, give the speed of the shock.
Solution: The Rankin-Hugoniot relation gives the speed of this shock:
dxs
q+ − q−
= +
=
dt
ρ − ρ−
13
1
62 − 3
1
1
6 − 3
=
1
1
6= .
12
2
Math 602
82
(ii) Give the solution to the problem.
Solution: In conclusion the location of the shock is xs (t) = 21 t and the explicit representation of
the solution is
1
,
6
1
ρ= ,
3
ρ=
t
= xs (t),
2
t
if x > = xs (t).
2
if x <
Question 178: Consider the following conservation equation
∂t ρ + ∂x (q(ρ)) = 0,
x ∈ (−∞, +∞), t > 0,
(
3 if x < 0,
ρ(x, 0) = ρ0 (x) :=
1 if x > 0,
where q(ρ) = ρ(2 + ρ) (and ρ(x, t) is the conserved quantity). Solve this problem using the
method of characteristics. Do we have a shock or an expansion wave here?
Solution: The characteristics are defined by
dX(t)
= q 0 (ρ) = 2(1 + ρ(x(t), t)),
dt
X(0) = X0 .
Set φ(t) = ρ(X(t), t), then we obtain that φ is constant, i.e., ρ is constant along the characteristics:
ρ(X(t), t) = ρ(X0 , 0) = ρ0 (X0 ). As a result we can integrate the equation defining the characteristics and we obtain X(t) = 2(1 + ρ0 (X0 ))t + X0 . We then have two cases depending whether X0
is positive or negative.
1. X0 < 0, then ρ0 (X0 ) = 3 and X(t) = 2(1 + 3)t + X0 = 8t + X0 . This means
ρ(x, t) = 3 if
x < 8t.
2. X0 > 0, then ρ0 (X0 ) = 1 and X(t) = 2(1 + 1)t + X0 = 4t + X0 . This means
ρ(x, t) = 1 if
x > 4t.
We see that the characteristics cross in the region {8t > x > 4t}. This implies that there is a shock.
The Rankin-Hugoniot relation gives the speed of this shock:
dxs (t)
q+ − q−
15 − 3
= +
= 6,
=
−
dt
ρ −ρ
3−1
xs (0) = 0.
In conclusion, xs (t) = 6t and
ρ = 3,
x < xs (t) = 6t,
ρ = 1,
x > xs (t) = 6t.
Question 179: Solve the conservation equation ∂t ρ + ∂x q(ρ) = 0, x ∈ (∞, +∞), t > 0 with
flux q(ρ) = ρ2 + ρ, and with the initial condition ρ(x, 0) = −1, if x < 0, ρ(x, 0) = 1, if x > 0.
Do we have a shock or an expansion wave here?
Solution: The solution is given by the implicit representation
ρ(X(s, t), t) = ρ0 (s),
X(s, t) = s + (2ρ0 (s) + 1)t.
Case 1: s < 0. Then ρ0 (s) = −1 and X(s, t) = s + (−2 + 1)t. This means s = X + t. The solution
is
ρ(x, t) = −1, if x < t.
Math 602
83
Case 2: s < 0. Then ρ0 (s) = 1 and X(s, t) = s + (2 + 1)t. This means s = X − 3t. The solution is
ρ(x, t) = 1,
if 3t < x.
We have a expansion wave. We need to consider the case ρ0 ∈ [−1, 1] at s = 0.
Case 3: s = 0 and ρ0 ∈ [−1, 1]. Then X(s, t) = s + (2ρ0 + 1)t = (2ρ0 + 1)t. This means
ρ0 = (X/t − 1)2. In conclusion
1 x
ρ(x, t) =
− 1 , if − t < x < 3t.
2 t
Question 180: Solve the conservation equation ∂t ρ + ∂x q(ρ) = 0, x ∈ (∞, +∞), t > 0 with
flux q(ρ) = ρ4 + 2ρ, and with the initial condition ρ(x, 0) = 1, if x < 0, ρ(x, 0) = −1, if x > 0.
Do we have a shock or an expansion wave here?
Solution: The solution is given by the implicit representation
ρ(X(s, t), t) = ρ0 (s),
X(s, t) = s + (4ρ0 (s)3 + 2)t.
We then have two cases depending whether s is positive or negative.
Case 1: s < 0, then ρ0 (s) = 1 and X(s, t) = (4 + 2)t + s = 6t + s. This means
ρ(x, t) = 1
if
x < 6t.
Case 2: s > 0, then ρ0 (s) = −1 and X(s, t) = (−4 + 2)t + s = −2t + s. This means
ρ(x, t) = −1
if
x > −2t.
We see that the characteristics cross in the region {6t > x > −2t}. This implies that there is a
shock. The Rankin-Hugoniot relation gives the speed of this shock with ρ− = 1 and ρ+ = −1:
q+ − q−
−1 − 3
dxs (t)
= +
= 2,
=
dt
ρ − ρ−
−1 − 1
xs (0) = 0.
In conclusion the location of the shock is xs (t) = 2t and the solution is as follows:
ρ = 1,
x < xs (t) = 2t,
ρ = −1,
x > xs (t) = 2t.
Question 181: Consider the following conservation equation
(
∂t ρ + ∂x (q(ρ)) = 0,
x ∈ (−∞, +∞), t > 0,
ρ(x, 0) = ρ0 (x) :=
1
2
1
if x < 0,
if x > 0,
where q(ρ) = ρ(2 − ρ) (and ρ(x, t) is the conserved quantity). Solve this problem using the
method of characteristics. Do we have a shock or an expansion wave here?
Solution: The characteristics are defined by
dX(t)
= q 0 (ρ) = 2(1 − ρ(x(t), t)),
dt
X(0) = X0 .
Set φ(t) = ρ(X(t), t), then we obtain that φ is constant, i.e., ρ is constant along the characteristics:
ρ(X(t), t) = ρ(X0 , 0) = ρ0 (X0 ). As a result we can integrate the equation defining the characteristics and we obtain X(t) = 2(1 − ρ0 (X0 ))t + X0 . We then have two cases depending whether X0
is positive or negative.
1. X0 < 0, then ρ0 (X0 ) =
1
2
and X(t) = t + X0 . This means
ρ(x, t) =
1
2
if
x < t.
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84
2. X0 > 0, then ρ0 (X0 ) = 1 and X(t) = X0 . This means
ρ(x, t) = 1
if
x > 0.
We see that the characteristics cross in the region {t > x > 0}. This implies that there is a shock.
The Rankin-Hugoniot relation gives the speed of this shock:
s=
q+ − q−
=
ρ+ − ρ−
3
4
1
2
−1
1
= .
2
−1
In conclusion
ρ=
1
,
2
ρ = 1,
t
,
2
t
x> .
2
x<
Question 182: Consider the following conservation equation
∂t ρ + ∂x (q(ρ)) = 0,
x ∈ (−∞, +∞), t > 0,
(
2 if x < 0,
ρ(x, 0) = ρ0 (x) :=
1 if x > 0,
where q(ρ) = ρ(2 − ρ) (and ρ(x, t) is the conserved quantity). Solve this problem using the
method of characteristics. Do we have a shock or an expansion wave here?
Solution: The characteristics are defined by
dX(t, x0 )
= q 0 (ρ) = 2(1 − ρ(X(t, x0 ), t)),
dt
X(0, x0 ) = x0 .
Set φ(t) = ρ(X(t, x0 ), t) and insert in the equation. We obtain that ∂t φ(t, x0 ) = 0; meaning that
φ(t, x0 ) = φ(0, x0 ), i.e., ρ is constant along the characteristics: ρ(X(t, x0 ), t) = ρ(x0 , 0) = ρ0 (x0 ).
As a result we can integrate the equation defining the characteristics and we obtain X(t, x0 ) =
2(1 − ρ0 (x0 ))t + x0 . The implicit representation of the solution is
X(t, x0 ) = 2(1 − ρ0 (x0 ))t + x0 ;
ρ(X(t, x0 ), t) = ρ0 (x0 )
We then have two cases depending whether x0 is positive or negative.
Case 1: x0 < 0, then ρ0 (x0 ) = 2 and X(t, x0 ) = 2(1 − 2)t + x0 = −2t + x0 . This means
x0 = X(t, x0 ) + 2t and
ρ(x, t) = 2 if x < −2t.
Case 2: x0 > 0, then ρ0 (x0 ) = 1 and X(t, x0 ) = 2(1 − 1)t + x0 = x0 . This means x0 = X(t, x0 )
and
ρ(x, t) = 1 if 0 < x.
We see that there is a gap in the region {−2t < x < 0}. This implies that there is an expansion
wave. We have to consider a third case x0 = 0 and ρ0 ∈ (1, 2).
0)
Case 3: x0 = 0, then X(t, x0 ) = 2(1 − ρ0 )t, i.e., ρ0 = 1 − X(t,x
. This means that
2t
ρ(x, t) = 1 −
x
,
2t
if
− 2t < x < 0.
Question 183: Assume u1 > u2 ≥ u3 ≥ 0 and consider the following conservation equation


0
if x ≤ 0,





u1 x if 0 < x ≤ 1,
∂t u + u∂x u = 0, x ∈ (−∞, +∞), t > 0,
u(x, 0) = u0 (x) := u1
if 1 < x ≤ 2,


u2
if 2 < x ≤ 3,



u
if 3 ≤ x.
3
Math 602
85
(i) Assume u2 = u3 . Solve until the expansion catches up the shock. When does it happen?
Solution: The characteristics are defined by
dX(t, s)
= u(X(t, s), t),
dt
X(0, s) = s.
From class we know that u(X(t, s), t) does not depend on time, that is to say
X(t, s) = u(X(0, s), 0)t + s = u(s, 0)t + x0 = u0 (s)t + s.
Case 1: If s ≤ 0, we have u0 (s) = 0 and X(t, s) = s; as a result, s = X(t, s), and
u(x, t) = 0,
if x ≤ 0.
Case 2: If 0 < s ≤ 1, we have u0 (s) = u1 s and X(t, x0 ) = u1 st + s; as a result s = X/(1 + u1 t),
and
u(x, t) = u1 x/(1 + u1 t),
if 0 < x ≤ 1 + u1 t.
case 3: If 1 < s ≤ 2, we have u0 (s) = u1 and X(t, x0 ) = u1 t + s; as a result s = X(t, s) − u1 t,
which implies
u(x, t) = u1 ,
if 1 + u1 t < x ≤ 2 + u1 t.
Case 4: If 2 < s, we have u0 (s) = 0 and X(t, s) = u2 t + s; as a result s = X(t, s), which implies
u(x, t) = 0
if 2 < x.
We have a shock at x = 2 and t = 0. The speed of the shock is given by the Rankin-Hugoniot
formula
1 2
u − 1 u2
1
dx1
= 2 1 2 2 = (u1 + u2 ).
dt
u1 − u2
2
As a result x1 (t) = 2 + 21 (u1 + u2 )t. This implies that the solution is

0,



u x/(1 + u t),
1
1
u(x, t) =
u1 ,



u2 ,
if
if
if
if
x ≤ 0,
0 < x ≤ 1 + u1 t,
1 + u1 t < x ≤ 2 + 21 (u1 + u2 )t,
2 + 12 (u1 + u2 )t < x.
The time T when the expansion wave catches up the shock is defined by
1
2 + (u1 + u2 )T = 1 + u1 T,
2
that is to say
T =
2
.
u1 − u2
(ii) Draw the characteristics corresponding to the situation (i) with u1 = 2 and u2 = 1.
Solution:
(iii) Assume now that u1 > u2 > u3 = 0. When does the first shock catches the second one?
Solution: The speed of the first shock (starting at x = 2 when t = 0) is given by the RankinHugoniot formula
1 2
u − 1 u2
dx1
1
= 2 1 2 2 = (u1 + u2 ).
dt
u1 − u2
2
As a result x1 (t) = 2 + 21 (u1 + u2 )t. The speed of the second shock (starting at x = 3 when t = 0)
is given by the Rankin-Hugoniot formula
1 2
u
dx2
1
= 2 2 = u2 .
dt
u2
2
Math 602
86
t
t=1
x
x=0
x=1
x=2
x=3
x=4
As a result x2 (t) = 3 + 12 u2 t.
The time T 0 when the two shocks are at the same location is such that x1 (T 0 ) = x2 (T 0 ); that is to
say,
1
1
2 + (u1 + u2 )T 0 = 3 + u2 T 0 ,
2
2
which gives
2
T0 =
.
u1
Note that T > T 0 for all u2 > 0. This means that the first shock catches up the
the fans catches the first shock.
Question 184: Consider the following conservation equation

0



x
∂t u + u∂x u = 0, x ∈ (−∞, +∞), t > 0,
u(x, 0) = u0 (x) :=
2 − x



0
second one before
if
if
if
if
x ≤ 0,
0 ≤ x ≤ 1,
1≤x≤2
2≤x
(i) Solve this problem using the method of characteristics for 0 ≤ t < 1.
Solution: The characteristics are defined by
dX(t, x0 )
= u(X(t, x0 ), t),
dt
X(0, x0 ) = x0 .
From class we know that u(X(t, x0 ), t) does not depend on time, that is to say
X(t, x0 ) = u(X(0, x0 ), 0)t + x0 = u(x0 , 0)t + x0 = u0 (x0 )t + x0 .
Case 1: If x0 ≤ 0, we have u0 (x0 ) = 0 and X(t, x0 ) = x0 ; as a result, x0 = X(t, x0 ), and
u(x, t) = 0,
if x ≤ 0.
Case 2: If 0 ≤ x0 ≤ 1, we have u0 (x0 ) = x0 and X(t, x0 ) = tx0 + x0 ; as a result x0 = X/(1 + t),
and
u(x, t) = x/(1 + t),
if 0 ≤ x ≤ 1 + t.
case 3: If 1 ≤ x0 ≤ 2, we have u0 (x0 ) = 2 − x0 and X(t, x0 ) = t(2 − x0 ) + x0 ; as a result
x0 = (X(t, x0 ) − 2t)/(1 − t), which implies
u(x, t) = 2 − (x − 2t)/(1 − t) = (2 − x)/(1 − t),
if 1 + t ≤ x ≤ 2.
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87
Case 4: If 2 ≤ x0 , we have u0 (x0 ) = 0 and X(t, x0 ) = x0 ; as a result x0 = X(t, x0 ), which implies
if 2 ≤ x.
u(x, t) = 0
(ii) Draw the characteristics for all t > 0 and all x ∈ R.
Solution:
t
t=1
x
x=0
x=1
x=2
(iii) There is a shock forming at t = 1 and x = 2. Let xs (t) be the location of the shock as a
function of t. Compute xs (t) for t > 1.
Solution: Let u− (t) be the value of u at the left of the shock. Conservation of mass implies
1 −
u (t)xs (t) =
2
Z
+∞
u0 (x)dx = 1.
−∞
The Rankin-Hugoniot formula gives
ẋs (t) =
1
−
2
2 (u (t))
−
u (t)
=
1 −
1
u (t) =
.
2
xs (t)
This implies
1 d
(xs (t)2 ) = 1,
2 dt
The Fundamental Theorem of Calculus implies
xs (t)ẋs (t) =
with
xs (1) = 2.
xs (t)2 − 22 = 2(t − 1),
which in turn implies xs (t) =
√
2t + 2, for all t ≥ 1.
(iv) Write the solution for t > 1.
Solution: In conclusion
u(x, t) =


0
x
 1+t

0
if x < 0,
√
if 0 ≤ x < xs (t) = 2t + 2,
√
if 2t + 2 = xs (t) ≤ x.
Math 602
88
Question 185: Consider the following conservation equation


if x ≤ 0,
1
u(x, 0) = u0 (x) := 1 − x if 0 ≤ x ≤ 1,


0
if 1 ≤ x.
x ∈ (−∞, +∞), t > 0,
∂t u + u∂x u = 0,
(i) Solve this problem using the method of characteristics for 0 ≤ t ≤ 1.
Solution: The characteristics are defined by
dX(t, x0 )
= u(X(t, x0 ), t),
dt
X(0, x0 ) = x0 .
From class we know that u(X(t, x0 ), t) does not depend on time, that is to say
X(t, x0 ) = u(X(0, x0 ), 0)t + x0 = u(x0 , 0)t + x0 = u0 (x0 )t + x0 .
Case 1: If x0 ≤ 0, we have u0 (x0 ) = 1 and X(t, x0 ) = t + x0 ; as a result, x0 = X − t, and
if x ≤ t.
u(x, t) = 1,
Case 2: If 0 ≤ x0 ≤ 1, we have u0 (x0 ) = 1 − x0 and X(t, x0 ) = t(1 − x0 ) + x0 ; as a result
x0 = (X − t)/(1 − t), and
u(x, t) = 1 − (t − x)/(t − 1),
if 0 ≤ x − t ≤ 1 − t,
which can also be re-written
u(x, t) =
x−1
,
t−1
if t ≤ x ≤ 1.
case 3: If 1 ≤ x0 , we have u0 (x0 ) = 0 and X(t, x0 ) = x0 ; as a result
if 1 ≤ x.
u(x, t) = 0,
(ii) Draw the characteristics for all t > 0 and all x ∈ R.
Solution:
t=1
x=0
x=1
(iii) At t = 1 we have u(x, 1) = 1 if x < 1 and u(x, 1) = 0 if x > 1. Solve the problem for t > 1.
Solution: Denote by u1 (x) the solution at t = 1. The characteristics are X(t, x0 ) = u1 (x0 )(t −
1) + x0 .
Case 1: If x0 < 1, u1 (x0 ) = 1 and X(t, x0 ) = t − 1 + x0 ; as a result,
u(x, t) = 1,
If x < t.
Math 602
89
Case 2: If 1 < x0 , u1 (x0 ) = 0 and X(t, x0 ) = x0 ; as a result,
u(x, t) = 0,
If 1 < x.
The characteristics cross in the domain {1 < x < t}; as a result we have a shock. The speed of the
shock is given by the Rankin-Hugoniot relation (recall that q(u) = u2 /2):
1/2 − 0
q+ − q−
1
dxs (t)
=
= +
= ,
dt
u − u−
1−0
2
xs (1) = 1,
Which gives xs (t) = 21 (t + 1). In conclusion,
(
1 If t > 1 and x < 21 (t + 1),
u(x, t) =
0 If t > 1 and 21 (t + 1) < x.
Question 186: Give an explicit solution to the equation ∂t u + ∂x (u4 ) = 0, where x ∈
1
(−∞, +∞), t > 0, with initial data u0 (x) = 0 if x < 0, u0 (x) = x 3 if 0 < x < 1, and
u0 (x) = 0 if 1 < x.
Solution: The implicit representation of the solution is
X(s, t) = s + 4u0 (s)3 t.
u(X(s, t), t) = u0 (s),
Case 1: s < 0, then u0 (s) = 0 and X(s, t) = s. This means
u(x, t) = 0 if x < 0.
1
Case 2: 0 < s < 1, then u0 (s) = s 3 and X(s, t) = s + 4st. This means s = X/(1 + 4t)
u(x, t) =
x
1 + 4t
13
if 0 < x < 1 + 4t.
Case 3: 1 < s, then u0 (s) = 0 and X(s, t) = s. This means
u(x, t) = 0 if 1 < x.
There is a shock starting at x = 1 (this is visible when one draws the characteristics).
Solution 1: The speed of the shock is given by the Rankin-Hugoniot formula
u4 − u4−
dxs (t)
= +
,
dt
u+ − u−
where u+ (t) = 0 and u− (t) =
xs (t)
1+4t
13
and xs (0) = 1,
. This gives
dxs (t)
xs (t)
= u− (t)3 =
,
dt
1 + 4t
which we re-write as follows:
d log(xs (t))
1
1 d log(1 + 4t)
=
=
.
dt
1 + 4t
4
dt
Applying the fundamental of calculus between 0 and t gives
log(xs (t)) − log(1) =
This give
1
(log(1 + 4t) − log(1)).
4
1
xs (t) = (1 + 4t) 4 .
Math 602
90
Solution 2: Another (equivalent) way of solving this problem, that does not require to solve the
Rankin-Hugoniot relation, consists of writing that the value of u− is such that the total mass is
conserved:
Z xs (t)
Z xs (0)
Z 1
1
3
u(x, t)dx =
u0 (x)dx =
x 3 dx =
4
0
0
0
1
i.e., using the fact that u(x, t) = (x/(1 + 4t)) 3 for all 0 ≤ x ≤ xs (t), we have
1
3
= (1 + 4t)− 3
4
Z
xs (t)
0
1
1 3
4
x 3 dx = (1 + 4t)− 3 xs (t) 3 .
4
This again gives
1
xs (t) = (1 + 4t) 4 .
Conclusion: The solution is finally expressed as follows:

0
if x < 0



13
1
x
u(x, t) =
if 0 < x < (1 + 4t) 4
1+4t


1

0
if (1 + 4t) 4 < x
Question 187: Consider the equation ∂t u + ∂x (u4 ) = 0, where x ∈ (−∞, +∞), t > 0, with
1
initial data u0 (x) = 0 if x < 0, u0 (x) = x 3 if 0 < x < 1, and u0 (x) = 0 if 1 < x. There is
1
x
a shock moving to the right. The solution is u(x, t) = ( 1+4t
) 3 on the left of the shock and
u(x, t) = 0 on the right. Give the position of the shock as a function of t.
Solution: There is a shock starting at x = 1 (this is visible when one draws the characteristics).
Solution 1: The speed of the shock is given by the Rankin-Hugoniot formula
u4 − u4−
dxs (t)
= +
,
dt
u+ − u−
where u+ (t) = 0 and u− (t) =
xs (t)
1+4t
13
and xs (0) = 1,
. This gives
dxs (t)
xs (t)
= u− (t)3 =
,
dt
1 + 4t
which we re-write as follows:
1
1 d log(1 + 4t)
d log(xs (t))
=
=
.
dt
1 + 4t
4
dt
Applying the fundamental of calculus between 0 and t gives
log(xs (t)) − log(1) =
This give
1
(log(1 + 4t) − log(1)).
4
1
xs (t) = (1 + 4t) 4 .
Solution 2: Another (equivalent) way of solving this problem, that does not require to solve the
Rankin-Hugoniot relation, consists of writing that the value of u− is such that the total mass is
conserved:
Z xs (t)
Z xs (0)
Z 1
1
3
u0 (x)dx =
x 3 dx =
u(x, t)dx =
4
0
0
0
1
i.e., using the fact that u(x, t) = (x/(1 + 4t)) 3 for all 0 ≤ x ≤ xs (t), we have
1
3
= (1 + 4t)− 3
4
Z
0
xs (t)
1
1 3
4
x 3 dx = (1 + 4t)− 3 xs (t) 3 .
4
Math 602
This again gives
91
1
xs (t) = (1 + 4t) 4 .
Conclusion: The solution is finally expressed as follows:

0
if x < 0



13
1
x
u(x, t) =
if 0 < x < (1 + 4t) 4
1+4t


1

0
if (1 + 4t) 4 < x
Question 188: Consider the conservation equation ∂t ρ + ∂x (sin( π2 ρ)) = 0, x ∈ R, t > 0, with
initial data ρ0 (x) = 0 if x < 0 and ρ0 (x) = 1 if x > 0. Draw the characteristics and give the
explicit representation of the solution.
Solution: The implicit representation of the solution to the equation ∂t ρ + ∂x q(ρ) = 0, ρ(x, 0) =
ρ0 (x), is
X(s, t) = q 0 (ρ0 (s))t + s; ρ(X(s, t), t) = ρ0 (s).
(13)
The explicit representation is obtained by expressing s in terms of X and t.
Case 1: s < 0, we have ρ0 (s) = 0, q 0 (ρ0 (s)) = π2 cos(0) = π2 , X = π2 t+s, which means s = X − π2 t.
Then
π
ρ(x, t) = 0 if x < t.
2
Case 2: 0 < s, we have ρ0 (s) = 1, q 0 (ρ0 (s)) = π2 cos( π2 ) = 0, X = s. Then
ρ(x, t) = 1 if 0 < x.
The characteristics cross in the region 0 < x <
π
2 t;
this means that there is a shock in this region.
The Rankin-Hugoniot formula gives the speed of the shock
sin( π2 ) − sin(0)
dxs (t)
=
= 1.
dt
1−0
The equation of the trajectory of the shock is xs (t) = t. Finally
(
0 if x < t,
ρ(x, t) =
1 if t < x.
(14)
Math 602
92
Question 189: Consider the conservation equation with flux q(ρ) = ρ3 . Assume that the initial
data is ρ0 (x) = 2, if x < 0, ρ0 (x) = 1, if 0 < x < 1, and ρ0 (x) = 0, if 1 < x. (i) Draw the
characteristics
Solution: There are three families of characteristics.
Case 1: s < 0, X(s, t) = 12t + s. In the x-t plane, these are lines with slope
1
12 .
Case 2: 0 < s < 1, X(s, t) = 3t + s. In the x-t plane, these are lines with slope 13 .
Case 3: 1 < s, X(s, t) = s. In the x-t plane, these are vertical lines.
One shock forms between the two black characteristics and another forms between the two red
characteristic (see figure).
(ii) Describe qualitatively the nature of the solution.
Solution: We have two shocks moving to the right. One shock forms between the two black
characteristics and another forms between the two red characteristic (see figure).
(iii) When and where does the left shock catch up with the right one?
Solution: The speeds of the shocks are
23 − 1
dx1 (t)
=
= 7,
dt
2−1
and
dx2 (t)
1−0
=
= 1.
dt
1−0
The location of the left shock at time t is x1 (t) = 7t and that of the right shock is x2 (t) = t + 1.
The two shocks are at the same location when 7t = t + 1, i.e., t = 61 ; the two shocks merge at
x = 76 .
(iv) What is the speed of the shock once the two shocks have merged and what is the position
of the shock as a function of time?
Solution: When the shocks have merged the left state is ρ = 2 and the right state is ρ = 0; as a
result the speed of the shock is
dx3 (t)
23 − 0
=
= 4,
dt
2−0
and the shock trajectory is x3 (t) = 4(t − 16 ) + 76 .
(v) Draw precisely all the characteristics of the solution.
Solution: The three shocks are shown in color.
Question 190: Consider the conservation equation with flux q(ρ) = ρ4 . Assume that the initial
data is ρ0 (x) = 2, if x < 0, ρ0 (x) = 1, if 0 < x < 1, and ρ0 (x) = 0, if 1 < x. (i) Draw the
characteristics
Math 602
93
Solution: There are three families of characteristics.
Case 1: s < 0, X(s, t) = 32t + s. In the x-t plane, these are lines with slope
1
32 .
Case 2: 0 < s < 1, X(s, t) = 4t + s. In the x-t plane, these are lines with slope 14 .
Case 3: 1 < s, X(s, t) = s. In the x-t plane, these are verical lines.
One shock forms between the two black characteristics and another forms between the two red
characteristic (see figure).
(ii) Describe qualitatively the nature of the solution.
Solution: We have two shocks moving to the right. One shock forms between the two black
characteristics and another forms between the two red characteristic (see figure).
(iii) When does the left shock catch up with the right one?
Solution: The speeds of the shocks are
dx1 (t)
24 − 1
=
= 15,
dt
2−1
and
dx2 (t)
1−0
=
= 1.
dt
1−0
The location of the left shock at time t is x1 (t) = 15t and that of the right shock is x2 (t) = t + 1.
1
The two shocks are at the same location when 15t = t + 1, i.e., t = 14
.
(iv) What is the speed of the shock when the two shocks have merged and what is the position
of the shock as function of time?
Solution: When the shocks have merged the left state is ρ = 2 and the right state is ρ = 0; as a
result the speed of the shock is
dx3 (t)
24 − 0
=
= 8,
dt
2−0
and the shock trajectory is x3 (t) = 8(t −
1
14 )
+
15
14 .
Question 191: Consider the following conservation equation
(
∂t ρ + ∂x (q(ρ)) = 0,
x ∈ (−∞, +∞), t > 0,
ρ(x, 0) = ρ0 (x) :=
1
2
1
if x < 0,
if x > 0,
where q(ρ) = ρ(2 − sin(ρ)) (and ρ(x, t) is the conserved quantity). What is the wave speed for
this problem?
Solution: The wave speed if the quantity q 0 (ρ) = 2 − sin(ρ) − ρ cos(ρ).
Question 192: Consider the following conservation equation
(
∂t ρ + ∂x (q(ρ)) = 0,
x ∈ (−∞, +∞), t > 0,
ρ(x, 0) = ρ0 (x) :=
1
2
1
if x < 0,
if x > 0,
Math 602
94
where q(ρ) = 2ρ + cos(ρ2 ) (and ρ(x, t) is the conserved quantity). What is the wave speed for
this problem?
Solution: The wave speed is the quantity q 0 (ρ) = 2 − 2ρ sin(ρ2 ).
Question 193: Consider the following conservation equation
(
∂t ρ + ∂x (q(ρ)) = 0,
where q(ρ) = 41 ρ2 −
q 00 (ρ)?
1
8
x ∈ (−∞, +∞), t > 0,
ρ(x, 0) = ρ0 (x) :=
3π
4
π
4
if x < 1,
if x > 1,
cos(2ρ) (and ρ(x, t) is the conserved quantity). (i) What is the sign of
Solution: Note that q 0 (ρ) = 12 ρ + 14 sin(2ρ) and q 00 (ρ) = 12 + 12 cos(2ρ). This implies that
q 00 (ρ) ≥ 21 − 12 ≥ 0. Actually we also have q 00 (ρ) = cos2 (ρ), which clearly proves that q 00 (ρ) ≥ 0,
i.e., q is convex.
(ii) Given that the given initial data produces a shock, give the speed of the shock.
Solution: Using ρ+ =
π
4
and π − =
3π
4 ,
the Rankin-Hugoniot relation gives the speed of this shock:
dxs
q(ρ+ ) − q(ρ− )
1 π 2 − 9π 2
8π 2
π
=
=
=
= .
+
−
dt
ρ −ρ
16 π − 3π
32π
4
(iii) Give the solution to the problem.
Solution: In conclusion we have dxs (t)/dt = π4 and xs (0) = 1. This means that the location of
the shock is xs (t) = π4 t + 1 and the explicit representation of the solution is
3π
,
4
π
ρ= ,
4
ρ=
πt
+ 1 = xs (t),
4
πt
if x >
+ 1 = xs (t).
4
if x <
Question 194: Solve the conservation equation ∂t ρ + ∂x q(ρ) = 0, x ∈ (∞, +∞), t > 0 with
flux q(ρ) = ρ4 + 2ρ, and with the initial condition ρ(x, 0) = 1, if x < 0, ρ(x, 0) = 2, if x > 0. Is
it a shock or an expansion wave?
Solution: The solution is given by the implicit representation
ρ(X(s, t), t) = ρ0 (s),
X(s, t) = s + (4ρ0 (s)3 + 2)t.
We then have two cases depending whether s is positive or negative.
Case 1: s < 0, then ρ0 (s) = 1 and X(s, t) = (4 + 2)t + s = 6t + s. This means
ρ(x, t) = 1
if
x < 6t.
Case 2: s > 0, then ρ0 (s) = 2 and X(s, t) = (4 × 8 + 2)t + s = 34t + s. This means
ρ(x, t) = 2 if
x > 34t.
Case 3: We see that there are no characteristics in the region {34t > x > 6t}. This implies that
there is an expansion wave in this region. The characteristics are given by setting s = 0,
X = (4ρ30 + 2)t,
Then
1 ≤ ρ0 ≤ 2.
with
31
1 X
ρ0 =
−2
.
4 t
Note that the condition 1 ≤ ρ0 ≤ 2 implies that 6t ≤ X ≤ 34t. In conclusion
ρ(x, t) =
31
1 x
−2
,
4 t
if
6t ≤ X ≤ 34t.
Math 602
95
Question 195: Consider the conservation equation with flux q(ρ) = ρ4 . Assume that the initial
data is ρ0 (x) = 2, if x < 0, ρ0 (x) = 1, if 0 < x < 1, and ρ0 (x) = 0, if 1 < x. (i) Draw the
characteristics
Solution: There are three families of characteristics.
Case 1: s < 0, X(s, t) = 32t + s. In the x-t plane, these are lines with slope
1
32 .
Case 2: 0 < s < 1, X(s, t) = 4t + s. In the x-t plane, these are lines with slope 14 .
Case 3: 1 < s, X(s, t) = s. In the x-t plane, these are verical lines.
One shock forms between the two black characteristics and another forms between the two red
characteristic (see figure).
(ii) Describe qualitatively the nature of the solution.
Solution: We have two shocks moving to the right. One shock forms between the two black
characteristics and another forms between the two red characteristic (see figure).
(iii) When does the left shock catch up with the right one?
Solution: The speeds of the shocks are
dx1 (t)
24 − 1
=
= 15,
dt
2−1
and
dx2 (t)
1−0
=
= 1.
dt
1−0
The location of the left shock at time t is x1 (t) = 15t and that of the right shock is x2 (t) = t + 1.
1
The two shocks are at the same location when 15t = t + 1, i.e., t = 14
.
(iv) What is the speed of the shock when the two shocks have merged and what is the position
of the shock as function of time?
Solution: When the shocks have merged the left state is ρ = 2 and the right state is ρ = 0; as a
result the speed of the shock is
dx3 (t)
24 − 0
=
= 8,
dt
2−0
and the shock trajectory is x3 (t) = 8(t −
1
14 )
+
15
14 .
Math 602
11
96
Green’s function
Question 196: Let Ω be a three-dimensional domain and consider the PDE
∇2 u = f (x),
x ∈ Ω,
with u(x) = h(x)
on the boundary of Ω, say Γ.
Let G(x, x0 ) be the Green’s function of this problem (the exact expression of G does not matter;
just assume that G is known). Give a representation1 of u(x) in terms of G, f and h.
Solution: By definition
∇2x G(x, x0 ) = δ(x − x0 ),
x ∈ Ω,
with G(x, x0 ) = 0
x ∈ Γ.
Then using the integration by parts formula, we obtain
Z
Z
Z
Z
u(x)∇2x (G(x, x0 ))dx =
∇2x (u(x))G(x, x0 )dx+ u(x)∂n (G(x, x0 ))dx− ∂n (u(x))G(x, x0 )dx.
Ω
Ω
Γ
Γ
which can also be rewritten
Z
u(x0 ) =
Z
f (x)G(x, x0 )dx +
Ω
h(x)∂n (G(x, x0 ))dx.
Γ
Question 197: Let f be a smooth function in [0, 1]. Consider the PDE
u − ∂xx u = f (x),
x ∈ (0, 1),
∂x u(1) + u(1) = 2, −∂x u(0) + u(0) = 1.
What PDE and which boundary conditions must satisfy the Green function, G(x, x0 ), (DO
NOT compute the Green function)? Give the integral representation of u assuming G(x, x0 ) is
known. Fully justify your answer.
Solution: Multiply the equation by G(x, x0 ) and integrate over (0, 1):
Z
1
1
Z
(u(x) − ∂xx u(x))G(x, x0 )dx
f (x)G(x, x0 )dx =
0
0
1
Z
u(x)G(x, x0 ) + ∂x u(x)∂x G(x, x0 )dx − ∂x u(1)G(1, x0 ) + ∂x u(0)G(0, x0 )
=
0
1
Z
u(x)(G(x, x0 ) − ∂xx G(x, x0 ))dx + u(1)∂x G(1, x0 ) − u(0)∂x G(0, x0 )
=
0
− ∂x u(1)G(1, x0 ) + ∂x u(0)G(0, x0 )
Z 1
=
u(x)(G(x, x0 ) − ∂xx G(x, x0 ))dx + u(1)∂x G(1, x0 ) − u(0)∂x G(0, x0 )
0
(u(1) − 2)G(1, x0 ) + (u(0) − 1)G(0, x0 )
Z 1
=
u(x)(G(x, x0 ) − ∂xx G(x, x0 ))dx
0
+ u(1)(G(1, x0 ) + ∂x G(1, x0 )) + u(0)(G(0, x0 ) − ∂x G(0, x0 )) − 2G(1, x0 ) − G(0, x0 )
If we define G(x, x0 ) so that
G(x, x0 ) − ∂xx G(x, x0 ) = δ(x − x0 ),
G(1, x0 ) + ∂x G(1, x0 ) = 0, G(0, x0 ) − ∂x G(0, x0 ) = 0,
then u(x0 ), x0 ∈ (0, 1), has the following representation
1
Z
u(x0 ) =
f (x)G(x, x0 )dx + 2G(1, x0 ) + G(0, x0 ),
0
1 Hint:
use
R
Ω
ψ∇2 (φ) =
R
Ω
∇2 (ψ)φ +
R
Γ
ψ∂n (φ) −
R
Γ
∂n (ψ)φ
∀x0 ∈ (0, 1).
Math 602
97
Question 198: Consider the equation u0 (x) + u = f (x) for x ∈ (0, 1) with u(0) = a. Let
G(x, x0 ) be the associated Green’s function. (Pay attention to the number of derivatives).
(a) Give the equation and boundary condition defining G and give an integral representation
of u(x0 ) in terms of G, f and the boundary data a. (Do not compute G.)
Solution: The Green’s function is defined by
−G0 (x, x0 ) + G(x, x0 ) = δ(x − x0 ),
G(1, x0 ) = 0.
We multiply the equation by u and we integrate over (0, 1) (in the distribution sense),
Z 1
Z 1
−G0 (x, x0 )u(x)dx +
G(x, x0 )u(x)dx = u(x0 ).
0
0
We integrates by parts and we obtain,
Z 1
G(x, x0 )(u0 (x) + u(x))dx − G(1, x0 )u(1) + G(0, x0 )u(0)
u(x0 ) =
0
Then, using the fact that u0 + u = f and using the boundary conditions for G and u, we obtain
Z 1
u(x0 ) =
G(x, x0 )f (x)dx + 2G(0, x0 ). ∀x0 ∈ (0, 1).
0
(b) Compute G(x, x0 ).
Solution: For x < x0 and x0 > x we have
−G0 (x, x0 ) + G(x, x0 ) = 0.
The solution is
(
αex
G(x, x0 ) =
βex
for x < x0
for x > x0 .
The boundary condition G(1, x0 ) = 0 implies β = 0.
For every > 0 we have
Z
x0 +
1=
(−G0 (x, x0 ) + G(x, x0 ))dx
x0 −
Z
x0 +
= G(x0 − , x0 ) − G(x0 + , x0 ) +
G(x, x0 )dx
x0 −
The term R =
R x0 +
x0 −
G(x, x0 )dx can be bounded as follows:
|R | ≤ 2 max |G(x, x0 )| = 2αex0 .
x∈[0,1]
Clearly R goes to 0 with . As a result we obtain the jump condition
+
x0
1 = G(x−
0 , x0 ) − G(x0 , x0 ) = αe .
This implies
α = e−x0 .
Finally
(
ex−x0
G(x, x0 ) =
0
for x < x0
for x > x0 .
Question 199: Consider the equation −∂x (x∂x u(x)) = f (x) for all x ∈ (1, 2) with u(1) = a
and u(2) = b. Let G(x, x0 ) be the associated Green’s function.
Math 602
98
(i) Give the equation and boundary conditions satisfied by G and give the integral representation
of u(x0 ) for all x0 ∈ (1, 2) in terms of G, f , and the boundary data. (Do not compute G in this
question).
Solution: We have a second-order PDE and the operator is clearly self-adjoint. The Green’s function
solves the equation
−∂x (x∂x G(x, x0 )) = δ(x − x0 ),
G(1, x0 ) = 0,
G(2, x0 ) = 0.
We multiply the equation by u and integrate over the domain (1, 2) (in the distribution sense).
Z 2
hδ(x − x0 ), ui = u(x0 ) = −
∂x (x∂x G(x, x0 ))u(x)dx.
1
We integrate by parts and we obtain,
Z 2
x∂x G(x, x0 )∂x u(x)dx − [x∂x G(x, x0 )u(x)]21
u(x0 ) =
1
Z
2
G(x, x0 )∂x (x∂x u(x))dx − 2∂x G(2, x0 )u(2) + ∂x G(1, x0 )u(1).
=−
1
Now, using the boundary conditions and the fact that −∂x (x∂x u(x)) = f (x), we finally have
Z 2
u(x0 ) =
G(x, x0 )f (x)dx − 2∂x G(2, x0 )b + ∂x G(1, x0 )a.
1
(ii) Compute G(x, x0 ) for all x, x0 ∈ (1, 2).
Solution: For all x 6= x0 we have
−∂x (x∂x G(x, x0 )) = 0.
The solution is
(
G(x, x0 ) =
a log(x) + b if 1 < x < x0
c log(x) + d if x0 < x < 2
The boundary conditions give b = 0 and d = −c log(2); as a result,
(
a log(x)
if 1 < x < x0
G(x, x0 ) =
c log(x/2) if x0 < x < 2
G must be continuous at x0 ,
a log(x0 ) = c log(x0 ) − c log(2)
and must satisfy the gap condition
Z x0 +
−
∂x (x∂x G(x, x0 ))dx = 1,
∀ > 0.
x0 −
This gives
−
−x0 ∂x G(x+
0 , x0 ) − G(x0 , x0 ) = 1
a
c
−x0 ( − ) = 1
x0
x0
This gives
a − c = 1.
In conclusion log(x0 ) = −c log 2 and
c = − log(x0 )/ log(2),
a = 1 − log(x0 )/ log(2) = log(2/x0 )/ log(2).
Math 602
This means
( log(2/x
G(x, x0 ) =
0)
log(2) log(x)
log(x0 )
log(2) log(2/x)
99
if 1 < x < x0
if x0 < x < 2
Question 200: Consider the equation ∂xx u(x) = f (x), x ∈ (0, L), with u(0) = a and ∂x u(L) =
b.
(a) Compute the Green’s function of the problem.
Solution: Let x0 be a point in (0, L). The Green’s function of the problem is such that
∂xx G(x, x0 ) = δx0 ,
G(0, x0 ) = 0,
∂x G(L, x0 ) = 0.
The following holds for all x ∈ (0, x0 ):
∂xx G(x, x0 ) = 0.
This implies that G(x, x0 ) = ax + b in (0, x0 ). The boundary condition G(0, x0 ) = 0 gives b = 0.
Likewise, the following holds for all x ∈ (x0 , L):
∂xx G(x, x0 ) = 0.
This implies that G(x, x0 ) = cx + d in (x0 , L). The boundary condition ∂x G(L, x0 ) = 0 gives
c = 0. The continuity of G(x, x0 ) at x0 implies that ax0 = d. The condition
Z ∂xx G(x, x0 )dx = 1,
∀ > 0,
−
−
gives the so-called jump condition: ∂x G(x+
0 , x0 ) − ∂x G(x0 , x0 ) = 1. This means that 0 − a = 1,
i.e., a = −1 and d = −x0 . In conclusion
(
−x if ≤ x ≤ x0 ,
G(x, x0 ) =
−x0 otherwise.
(b) Give the integral representation of u using the Green’s function.
Solution: Let x0 be a point in (0, L). The definition of the Dirac measure at x0 is such that
u(x0 ) = hδx0 , ui = h∂xx G(·, x0 ), ui
Z L
L
=−
∂x G(x, x0 )∂x u(x)dx + [∂x G(x, x0 )u(x)]0
Z
0
L
0
Z
L
L
G(x, x0 )∂xx u(x)dx − [G(x, x0 )∂x u(x)]0 + [∂x G(x, x0 )u(x)]0
=
L
G(x, x0 )f (x)dx − G(L, x0 )∂x u(L) + G(0, x0 )∂x u(0) + ∂x G(L, x0 )u(L) − ∂x G(0, x0 )u(0).
=
0
This finally gives the following representation of the solution:
Z
L
G(x, x0 )f (x)dx − G(L, x0 )b − ∂x G(0, x0 )a
u(x0 ) =
0
Question 201: Consider the equation u00 (x) = f (x) for x ∈ (0, 1) with u(0) = 1 and u0 (1) = 1.
Let G(x, x0 ) be the associated Green’s function.
(a) Give an expression of u(x) in terms of G, f and the boundary data.
Solution: The Green’s function is defined by
G00 (x, x0 ) = δ(x − x0 ),
G(0, x0 ) = 0,
G0 (1, x0 ) = 0.
Math 602
100
We multiply the equation by u and we integrate (in the distribution sense),
Z 1
G00 (x, x0 )u(x)dx = u(x0 ).
0
We integrates by parts twice and we obtain,
Z 1
G0 (x, x0 )u0 (x)dx + G0 (1, x0 )u(1) − G0 (0, x0 )u(0)
u(x0 ) = −
0
Z
=
1
G(x, x0 )u00 (x)dx − G(1, x0 )u0 (1) + G(0, x0 )u0 (0) + G0 (1, x0 )u(1) − G0 (0, x0 )u(0).
0
Then, using the boundary conditions for G and u, we obtain
Z 1
u(x0 ) =
G(x, x0 )f (x)dx − G0 (0, x0 ) − G(1, x0 ),
∀x0 ∈ (0, 1).
0
(b) Compute G(x, x0 ).
Solution: For x < x0 we have
G(x, x0 ) = ax + b.
The boundary condition G(0, x0 ) = 0 implies b = 0. For x0 < x we have
G(x, x0 ) = cx + d.
0
The boundary condition G (1, x0 ) = 0 implies c = 0. Moreover we have
Z 0 −
1 = lim
G00 (x, x0 )dx = G0 (x+
0 , x0 ) − G (x0 , x0 ) = −a,
→0
−
meaning a = −1. The continuity of G at x0 implies
ax0 = d,
implying d = −x0 . As a result,
(
−x, if 0 ≤ x ≤ x0 ,
G(x, x0 ) =
−x0 , if x0 ≤ x ≤ 1.
Question 202: Consider the equation ∂x
1
1+3x2 ∂x u(x)
= f (x), x ∈ (0, 1), ∂x u(0) = a,
u(1) = b. Let G(x, x0 ) be the associated Green’s function.
(i) Give the equation and boundary conditions satisfied by G.
Solution: The operator is clearly self-adjoint. Then for all x 6= x0 we have
1
∂x
∂
G(x,
x
)
= δx−x0 , ∂x G(0, x0 ) = 0, G(1, x0 ) = 0.
x
0
1 + 3x2
(ii) Give the integral representation of u(x0 ) for all x0 ∈ (0, 1) in terms of G, f , and the
boundary data. (Do not compute G in this question).
Solution: Multiply the equation defining G by u and integrate over (0, 1),
Z 1 1
hδx−x0 , ui = u(x0 ) =
∂x
∂
G(x,
x
)
u(x)dx.
x
0
1 + 3x2
0
We integrate by parts and we obtain
1
Z 1
1
1
u(x0 ) = −
∂
G(x,
x
)∂
u(x)dx
+
∂
G(x,
x
)u(x)
0 x
x
0
2 x
1 + 3x2
0 1 + 3x
0
1
Z 1
1
1
1
=
G(x, x0 )∂x
∂x u(x) dx + ∂x G(1, x0 )u(1) − G(x, x0 )
∂x u(x)
1 + 3x2
4
1 + 3x2
0
0
Z 1
1
1
=
G(x, x0 )∂x
∂ u(x) dx + ∂x G(1, x0 )u(1) + G(0, x0 )∂x u(0).
2 x
1
+
3x
4
0
Math 602
101
Now, using the boundary conditions and the fact that ∂x ((1 + 3x2 )−1 ∂x u(x)) = f (x), we finally
have
Z 1
1
u(x0 ) =
G(x, x0 )f (x)dx + ∂x G(1, x0 )b + G(0, x0 )a.
4
0
(iii) Compute G(x, x0 ) for all x, x0 ∈ (0, 1).
Solution: For all x 6= x0 we have
1
∂x
∂x G(x, x0 ) = δx−x0 ,
1 + 3x2
∂x G(0, x0 ) = 0,
G(1, x0 ) = 0.
The generic solution is
(
a(x + x3 ) + b
G(x, x0 ) =
c(x + x3 ) + d
if 0 ≤ x < x0
if x0 < x ≤ 1.
The boundary conditions give
∂x G(0, x0 ) = 0 = a,
As a result
G(1, x0 ) = 0 = 2c + d.
(
b
if 0 ≤ x < x0
G(x, x0 ) =
3
c(x + x ) − 2c if x0 < x ≤ 1.
G must be continuous at x0 ,
b = c(x0 + x30 ) − 2c
and must satisfy the gap condition
Z x0 + 1
∂x
∂
G(x,
x
)
dx = 1,
x
0
1 + 3x2
x0 −
∀ > 0.
This gives
1
−
(∂x G(x+
0 , x0 ) − ∂x G(x0 , x0 )) = 1,
1 + 3x20
2
3
3
i.e. ∂x G(x+
0 , x0 ) = 1 + 3x0 = c(1 + 3x0 ). In conclusion c = 1 and b = x0 + x0 − 2. In other words,
(
x0 + x30 − 2 if 0 ≤ x < x0
G(x, x0 ) =
x + x3 − 2 if x0 < x ≤ 1.
Question 203: Consider the equation −∂x ((1 + x)∂x u(x)) + ∂x u(x) = f (x), x ∈ (0, 1) with
u(0) = α and −2∂x u(1) + u(1) = β. Let G(x, x0 ) be the Green’s function. (i) Give the integral
representation of u(x0 ) for all x0 ∈ (0, 1) in terms of G, f , α and β and give the equation
and boundary conditions that G must satisfy. Do not compute G at this question. (Hint: The
differential operator is not self-adjoint. You should find G(0, x0 ) = 0, ∂x G(1, x0 ) = 0).
Solution: We multiply the PDE by G(x, x0 ) and integrate by parts,
Z 1
Z 1
f (x)G(x, x0 )dx =
((1 + x)∂x u(x)∂x G(x, x0 ) − u(x)∂x G(x, x0 )) dx + [(−(1 + x)∂x u(x) + u(x))G(x, x0 )]10
0
0
Z
1
((1 + x)∂x u(x)∂x G(x, x0 ) − u(x)∂x G(x, x0 )) dx + βG(1, x0 ) − (−∂x u(0) + α)G(0, x0 ).
=
0
Since ∂x u(0) is not known, we must have G(0, x0 ) = 0. Then
Z 1
Z 1
f (x)G(x, x0 )dx =
(−u(x)∂((1 + x)∂x G(x, x0 )) − u(x)∂x G(x, x0 )) dx + βG(1, x0 ) + [(1 + x)u(x)∂x G(x, x0 )]10
0
0
Z
1
−u(x) (∂((1 + x)∂x G(x, x0 )) + ∂x G(x, x0 )) dx + βG(1, x0 ) + 2u(1)∂x G(1, x0 ) − α∂x G(0, x0 )
=
0
Math 602
102
Since u(1) is not known, we must have ∂x G(1, x0 ) = 0. Finally G must satisfy
−∂x ((1 + x)∂x G(x, x0 )) − ∂x G(x, x0 ) = δ(x − x0 ),
G(0, x0 ) = 0,
∂x G(1, x0 ) = 0,
and we have the following representation for u(x0 ),
Z 1
f (x)G(x, x0 )dx − βG(1, x0 ) + α∂x G(0, x0 ).
u(x0 ) =
0
(ii) Compute G(x, x0 ) such that −∂x ((1 + x)∂x G(x, x0 )) − ∂x G(x, x0 ) = δ(x − x0 ), G(0, x0 ) = 0,
∂x G(1, x0 ) = 0, for all x, x0 ∈ (0, 1). (Hint: observe that (1 + x)φ0 (x) + φ(x) = ((1 + x)φ(x))0 .)
Solution: For all x 6= x0 we have
−∂((1 + x)∂x G(x, x0 )) − ∂x G(x, x0 ) = −∂((1 + x)∂x G(x, x0 ) + G(x, x0 )) = 0
Using the hint, this implies that
(1 + x)∂x G(x, x0 ) + G(x, x0 ) = ∂x ((1 + x)G(x, x0 )) = a,
In conclusion
(
G(x, x0 ) =
ax+b
1+x
cx+d
1+x
if x ≤ x0
if x0 ≤ x.
The boundary condition at 0 gives
G(0, x0 ) = 0 = b,
i.e., b = 0, G(x, x0 ) =
ax
1+x
if x ≤ x0 . The boundary condition at 1 gives
∂x G(0, x0 ) =
c(1 + 0) − (c × 0 + d) × 1
= 0,
(1 + 0)2
i.e., c = d. G(x, x0 ) = c, if x0 ≤ x. We need to impose the continuity of G(x, x0 ) at x0 ,
ax0
= c,
1 + x0
which gives ax0 = c(1 + x0 ). The jump condition gives
Z x0 +
+
1 = lim
(−∂x ((1 + x)∂x G(x, x0 )) + ∂x G(x, x0 )) dx = (1 + x0 )(∂x G(x−
0 , x0 ) − ∂x G(x0 , x0 ))
→0
x0 −
= (1 + x0 )(
a
a(1 + x0 ) − ax0
)=
.
(1 + x0 )2
1 + x0
In conclusion a = (1 + x0 ) and c = x0 , Then
(
G(x, x0 ) =
x(1+x0 )
1+x
x0
if x ≤ x0
if x0 ≤ x.
Question 204: Let Ω be a five-dimensional domain with boundary Γ, and consider the PDE
u − ∆u = f (x),
x ∈ Ω,
with ∂n u(x) = h(x)
on the boundary Γ.
Let G(x, x0 ) be the Green’s function of this problem (the exact expression of G(x, x0 ) does not
matter; just assume that G(x, x0 ) is known). (a) what is the PDE solved by G(x, x0 )?
Solution: Either you remember from class that the operator u 7−→ u − ∆u with zero Neumann
boundary condition is self-adjoint, or you redo the computation. Multiply the PDE by G(x, x0 ),
integrate over Ω, and integrate by parts (using the hint):
Z
Z
G(x, x0 )f (x)dx =
G(x, x0 )(u(x) − ∆u(x))dx
Ω
ZΩ
Z
Z
Z
=
G(x, x0 )u(x)dx −
u(x)∆G(x, x0 )dx − G(x, x0 )∂n u(x)dσ + u(x)∂n G(x, x0 )dσ
Ω
Γ
Γ
ZΩ
Z
Z
= (G(x, x0 ) − ∆G(x, x0 ))u(x)dx − G(x, x0 )h(x)dσ + u(x)∂n G(x, x0 )dσ
Ω
Γ
Γ
Math 602
103
We then define G(x, x0 ) so that
where δx−x0
G(x, x0 ) − ∆G(x, x0 ) = δx−x0 , ∂n G(x, x0 )|Γ = 0,
R
is the Dirac measure: δx−x0 ϕ = ϕ(x0 ) for all ϕ ∈ C 0 (R5 ). This means that
Z
Z
G(x, x0 )f (x)dx = u(x0 ) − G(x, x0 )h(x)dσ.
Ω
Γ
(b) Give a representation of u(x) in terms of G, f and h.
Solution: The above computation shows that
Z
Z
u(x0 ) =
f (x)G(x, x0 )dx + h(x)G(x, x0 )dx.
Ω
Γ
Question 205: Let Ω be a seven-dimensional domain with boundary Γ, and consider the PDE
u − ∆u = f (x),
x ∈ Ω,
with ∂n u(x) + 2u(x) = h(x)
on the boundary Γ.
Let G(x, x0 ) be the Green’s function of this problem (the exact expression of G(x, x0 ) does not
matter; just assume that G(x, x0 ) is known). (a) what is the PDE solved by G(x, x0 )?
Solution: Either you remember from class that the operator u 7−→ u − ∆u with zero Neumann
boundary condition is self-adjoint, or you redo the computation. Multiply the PDE by G(x, x0 ),
integrate over Ω, and integrate by parts (using the hint):
Z
Z
G(x, x0 )f (x)dx =
G(x, x0 )(u(x) − ∆u(x))dx
Ω
Ω
Z
Z
Z
Z
=
G(x, x0 )u(x)dx −
u(x)∆G(x, x0 )dx − G(x, x0 )∂n u(x)dσ + u(x)∂n G(x, x0 )dσ
Ω
Γ
ZΩ
Z
ZΓ
= (G(x, x0 ) − ∆G(x, x0 ))u(x)dx + G(x, x0 )(2u(x) − h(x))dσ + u(x)∂n G(x, x0 )dσ
Γ
ZΩ
ZΓ
Z
= (G(x, x0 ) − ∆G(x, x0 ))u(x)dx + u(x)(2G(x, x0 ) + ∂n G(x, x0 ))dσ − G(x, x0 )h(x)dσ
Ω
Γ
We then define G(x, x0 ) so that
where δx−x0
G(x, x0 ) − ∆G(x, x0 ) = δx−x0 , 2G(x, x0 ) + ∂n G(x, x0 )|Γ = 0,
R
is the Dirac measure: δx−x0 ϕ = ϕ(x0 ) for all ϕ ∈ C 0 (R7 ). This means that
Z
Z
G(x, x0 )f (x)dx = u(x0 ) − G(x, x0 )h(x)dσ.
Ω
Γ
(b) Give a representation of u(x) in terms of G, f and h.
Solution: The above computation shows that
Z
Z
u(x0 ) =
f (x)G(x, x0 )dx + h(x)G(x, x0 )dx.
Ω
Γ
Γ
Math 602
12
104
Fredholm alternative
Question 206: (a) Compute the solution set of the equation
u00 + u = 0, x ∈ (0, 2π)
u(0) = u(2π), u0 (0) = u0 (2π).
with
Solution: clearly u(x) = a cos(x) + b sin(x) where a and b are arbitrary numbers. The solution set
is a two-dimensional vector space spanned by cos(x) and sin(x).
(b) Do the following equation have solution(s)? (Hint: think of the Fredholm alternative)
u00 (x) + u(x) = sin(2x), ∀x ∈ (0, 2π)
with u(0) = u(2π), u0 (0) = u0 (2π).
Solution: We are in the second case of the Fredholm alternative, i.e., the null space of the operator
is not {0}. We have to verify that sin(2x) is orthogonal to cos(x) and sin(x), which is clearly true.
In conclusion the equation has solutions.
(c) Do the following equation have solution(s)? (Hint: think of the Fredholm alternative)
u00 (x) + u(x) = sin(x), ∀x ∈ (0, 2π)
with
u(0) = u(2π), u0 (0) = u0 (2π).
Solution: We are in the second case of the Fredholm alternative, i.e., the null space of the operator
is not {0}. We have to verify that sin(x) is orthogonal to cos(x) and sin(x), which is clearly wrong.
In conclusion the equation has no solution.
Question 207: (i) Use the energy method to compute the null space of the self-adjoint operator
L : {v ∈ C 2 [0, 1]; v 0 (0) = v 0 (1) = 0} −→ C 0 [0, 1] defined by L(v) := −v 00 .
Solution: Let v be a member of the null space. Then L(v) = 0 if and only if
−v 00 (x) = 0,
∀x ∈ [0, 1],
v 0 (0) = 0,
v 0 (1) = 0.
Multiply the equation by v and integrate over [0, 1]:
Z
0=−
1
Z
00
0
0
0
0
Z
v (x)v (x)dx − v (1)v(1) + v (0)v(0) =
v (x)v(x)dx =
0
1
0
1
(v 0 (x))2 dx.
0
This implies that v 0 (x) = 0 for all x ∈ [0, 1], which in turns implies that v(x) = a where a is an
arbitrary constant, i.e., Null(L) ⊂ span(1) . Conversely it is clear that v(x) = a is in the null space
of L, i.e., span(1) ⊂ Null. In conclusion
Null(L) = span(1).
(ii) Apply the Fredholm alternative to deduce whether the following equation has a solution,
and if it does whether it is unique: −u00 (x) = 21 − x, where x ∈ [0, 1] and u0 (0) = 0, u0 (1) = 0.
Solution: The problem consists of finding u in H := {v ∈ C 2 [0, 1]; v 0 (0) = v 0 (1) = 0} so that
Lu = 21 − x. From (i) we infer that the null space of L is not reduced to {0}, this means
that
There exists a solution if and only if
R 1 1 we are in the second case of the Fredholm alternative.
t
( − x)v(x)dx = 0 for all v in the null space of L = L. Let v be in the null space of L. We
0 2
have seen in (i) that v(x) = a, where a ∈ R; this implies
Z
0
1
1
( − x)v(x)dx = a
2
R1
Z
0
1
1
1 1
( − x)dx = a( − ) = 0.
2
2 2
In conclusion the condition 0 ( 12 − x)v(x)dx = 0 for all v ∈ Null(L) is satisfied. This means that
the problem has a solution but the solution is not unique.
Math 602
105
Question 208: Consider the equation −u00 (x) = f (x) for x ∈ (0, 1) with u0 (0) = 2 and u(1) = 1.
Let G(x, x0 ) be the associated Green’s function. (Pay attention to the minus sign).
(a) Give an expression of u(x) in terms of G, f and the boundary data.
Solution: The Green’s function is defined by
−G00 (x, x0 ) = δ(x − x0 ),
G0 (0, x0 ) = 0,
G(1, x0 ) = 0.
We multiply the equation by u and we integrate over (0, 1) (in the distribution sense),
Z
1
−G00 (x, x0 )u(x)dx = u(x0 ).
0
We integrates by parts twice and we obtain,
Z 1
G0 (x, x0 )u0 (x)dx − G0 (1, x0 )u(1) + G0 (0, x0 )u(0)
u(x0 ) =
0
Z
=−
1
G(x, x0 )u00 (x)dx + G(1, x0 )u0 (1) − G(0, x0 )u0 (0) − G0 (1, x0 )u(1) + G0 (0, x0 )u(0).
0
Then, using the boundary conditions for G and u, we obtain
Z 1
u(x0 ) =
G(x, x0 )f (x)dx − 2G(0, x0 ) − G0 (1, x0 ),
∀x0 ∈ (0, 1).
0
(b) Compute G(x, x0 ).
Solution: For x < x0 we have
G(x, x0 ) = ax + b.
The boundary condition G0 (0, x0 ) = 0 implies a = 0; hence, G(x, x0 ) = b. For x0 < x we have
G(x, x0 ) = c(x − 1) + d.
The boundary condition G(1, x0 ) = 0 implies d = 0; hence, G(x, x0 ) = c(x − 1). Moreover we have
Z
1=−
1
G00 (x, x0 )dx = −G0 (1, x0 ) + G0 (0, x0 ) = −c,
0
meaning c = −1. The continuity of G at x0 implies
b = 1 − x0 .
As a result,
(
1 − x0 , if 0 ≤ x ≤ x0 ,
G(x, x0 ) =
1 − x, if x0 ≤ x ≤ 1.
1
2
1
Question 209: Consider the operator L : φ 7−→ −∂x (x 2 ∂x φ(x)) − π4 x− 2 φ(x), with domain
D = {v ∈ C 2 (1, 4); v(1) = 0, v(4) = 0}.
1
− 21
2
(i) What is the Null
of L? (Hint:
√
√ √The general solution to −∂x (x ∂x φ(x))−λx φ(x) = 0
√ space
is φ(x) = c1 cos(2 x λ) + c2 sin(2 x λ) for all λ ≥ 0.)
Solution: Let φ be a member of the null space of L, say N(L). Then
π2 − 1
x 2 φ(x) = 0.
4
√ √
√ √
In other words, using the hint, φ(x) = c1 cos(2 x λ)+c2 sin(2 x λ) with λ =
conditions imply that
1
−∂x (x 2 ∂x φ(x)) −
φ(1) = 0 = −c1 ,
and φ(4) = 0 = c2 sin(2π).
π2
4 .
The boundary
Math 602
106
√
In conclusion N(L) = span{sin(π x)}, i.e., N(L) is a one-dimensional vector space.
2
1
1
1
(ii) Consider the problem −∂x (x 2 ∂x φ(x)) − π4 x− 2 φ(x) = 12 x− 2 , x ∈ (1, 4), with φ(1) = 0,
1
1
φ(4) = 0. Does this problem have a solution? (Hint: d(x 2 ) = 21 x− 2 dx.)
Solution: We are in the second case of the Fredholm alternative, since the null
√ space of the operator
1
L is not reduced to {0}. We must verify that 12 x− 2 is orthogonal to sin(π x). Using the hint and
1
the change of variable x 2 = z, we have
Z 4
Z 4
Z 2
√ 1 1
1
1
1
2
2
sin(π x) x− 2 dx =
sin(πx 2 )d(x 2 ) =
sin(πz)dz = − [cos(πz)]1 = − .
2
π
π
1
1
1
Hence
R4
1
√
1
sin(π x) 12 x− 2 dx 6= 0, which means that the above problem does not have a solution.
Question 210: Consider the following problem: Let f be a smooth function in [0, π], find u
such that u + ∂xx u = f (x), x ∈ (0, π), u(π) = 0, u(0) = 0. (i) Under which condition does this
problem have a solution.
Solution: Let us compute the null space of the operator L : {v ∈ C 2 (0, π) : v(π) = 0, v(0) =
0} 3 u 7−→ u(x) + ∂xx u(x) ∈ C 0 (0, π). Let N (L) be the null space. Let v ∈ N (L), then
v + ∂xx v = 0
which means that v = a cos(x) + b sin(x). The boundary conditions imply that a = 0; as a result
N (L) = span(sin(x)), i.e., N (L) is the one-dimensional vector space spannedRby the function sin(x).
π
Fredholm’s alternative implies that the above problem has a solution only if 0 sin(x)f (x)dx = 0.
(ii) Does the above problem have a solution for f (x) = cos(x)?
Rπ
Solution: We need to compute 0 sin(x) cos(x)dx,
Z
0
π
1
sin(x) cos(x)dx =
2
Z
π
sin(2x)dx = 0.
0
The Fredholm alternative implies that the problem has a solution.
Question 211: Consider the following Cauchy-Euler problem: Let f be a smooth function in
π
π
[0, π2 ] and find u such that 13u − 5x∂x u + x2 ∂xx u = f (x), x ∈ (1, e 2 ), u(1) = 0, u(e 2 ) = 0. (i)
Does this problem have a unique solution. (Hint: The solution to the homogeneous equation
(i.e., f = 0) is v(x) = ax3 cos(2 log(|x|)) + bx3 sin(2 log(|x|)).)
Solution: Let us compute the null space of the operator
π
L : {v ∈ C 2 (0, π)| v( ) = 0, v(0) = 0} 3 u 7−→ 13u − 5x∂x u + x2 ∂xx u ∈ C 0 (0, π).
2
Let N (L) be the null space. Let v ∈ N (L), then
13v − 5x∂x v + x2 ∂xx v = 0
which means that v = ax3 cos(2 log(|x|)) + bx3 sin(2 log(|x|)). The boundary conditions imply that
0 = a cos(2 log(1)) + bx3 sin(2 log(1)) = a,
π
0 = b(π/2)3 sin(2 log(e 2 )) = b(π/2)3 sin(π),
a = 0 and b is an arbitrary real number; as a result N (L) = span(x3 sin(2 log(|x|))), i.e., N (L)
is the one-dimensional vector space spanned by the function x3 sin(2 log(|x|)). Assuming that the
problem has a solution, it cannot be unique. We are in the second case of Fredholm’s alternative.
Rπ
Once N (LT ) is charaterized, the above problem has a solution only if 12 vf (x)dx = 0 for all
v ∈ N (LT ). Question: What is N (LT )?
Question 212: Use the Fredholm alternative to determine whether the following problem has
a unique solution: Let f be a smooth function in [0, 1], find u such that
u − ∂xx u = f (x),
x ∈ (0, 1),
∂x u(1) + u(1) = 0, −∂x u(0) + u(0) = 0
Math 602
107
Fully justify your answer.
Solution: Let us compute the null space of the operator L : {v ∈ C 2 (0, 1) : ∂x v(1) + v(1) =
0, −∂x v(0) + v(0) = 0} 3 u 7−→ u(x) − ∂xx u(x) ∈ C 0 (0, 1). Let N (L) be the null space. Let
v ∈ N (L). The energy argument implies that
Z
0=
1
(v 2 + (∂x v)2 )dx − [v∂x v]10 =
1
Z
0
(v 2 + (∂x v)2 )dx + v(1)2 + v(0)2 .
0
This proves that v = 0, i.e., N (L) = {0}. Fredholm’s alternative implies that the above problem
has a unique solution.
1
2
1
Question 213: Consider the equation −∂x (x 2 ∂x u(x)) − x− 2 u(x) = 1 + x for all x ∈ ( π4 , π 2 )
2
with u( π4 ) = 0 and u(π 2 ) = 0. Use the Fredholm alternative to determine whether the above
1
− 21
2
problem has a √solution. (Hint:
√ The general solution to −∂x (x ∂x φ(x)) − x φ(x) = 0 is
φ(x) = c1 cos(2 x) + c2 sin(2 x).)
2
2
Solution: Let us compute the null space of the operator L : {v ∈ C 2 ( π4 , π 2 ) : v( π4 ) = 0, v(π 2 ) =
2
1
1
0} 3 u 7−→ −∂x (x 2 ∂x u(x)) − x− 2 u(x) ∈ C 0 ( π4 , π 2 ). Let N (L) be the null space and consider
v ∈ N (L). By using the hint, we infer that
√
√
v(x) = c1 cos(2 x) + c2 sin(2 x)
2
2
The boundary condition v( π4 ) = 0 implies c1 = 0. The second
√ boundary condition v(π ) = 0
is automatically satisfied. In conclusion N (L) = span{sin(2 x}. We are in the second case
of Fredholm’s alternative. The problem has solutions (infinitely
many) if the right-hand
side is
√
√
orthogonal to N (L∗ ). But L = L∗ , i.e., N (L∗ ) = span{sin(2 x)}. Let v(x) = λ sin(2 x), λ ∈ R,
then
Z π2
Z π2
√
(1 + x)v(x)dx = λ 2 (1 + x) sin(2 x)dx.
2
π
4
π
4
The above integral is nonzero unless λ = 0 since
problem does not have any solution.
R π2
π2
4
√
(1 + x) sin(2 x)dx > 0. As a result, the
Math 602
13
108
Classification of PDEs
Question 214: Consider the following PDE’s:
∂y u(x, y) + 3∂xx u(x, y) = f (x, y),
y > 0,
u(x, 0) = 1,
∂y u(x, y) − 3∂xx u(x, y) = f (x, y),
u(0, y) = 3,
y > 0,
u(x, 0) = 1,
∂y u(x, y) − 3∂xx u(x, y) = f (x, y),
u(x, 0) = 1,
u(x, 0) = 1,
∂yy u(x, y) − 3∂xx u(x, y) = f (x, y),
u(x, 0) = 1,
∂yy u(x, y) + 3∂xx u(x, y) = f (x, y),
u(x, 0) = 1,
−∂yy u(x, y) − ∂xx u(x, y) = f (x, y),
y ∈ (0, H),
∂y u(x, y) + 3∂x u(x, y) = f (x, y),
u(x, 0) = 1,
∂y u(x, y) + 3∂x u(x, y) = f (x, y),
y>0
• Heat equation?
• Laplace equation?
9
13
• Transport equation?
• Wave equation?
18
11
(20)
∂x u(L, y) = 3
(21)
∂x u(L, y) = 3 (22)
∂x u(0, y) = 3
(23)
x ∈ (0, L)
u(L, y) = 2,
∂x u(0, y) = 3
(24)
x ∈ (0, L)
∂x u(L, y) = 3
(25)
x ∈ (0, L)
u(0, y) = 2
Which one is the
∂x u(L, y) = 3
x ∈ (0, L)
u(0, y) = 2,
u(x, 0) = 1,
u(0, y) = 2,
∂x u(x, H) = 2,
u(x, H) = 1,
y>0
(19)
x ∈ (0, L)
u(L, y) = 2,
u(x, 0) = 1,
∂x u(L, y) = 3
u(0, y) = 2,
u(x, H) = 1,
y ∈ (0, H),
∂y u(x, y) + 3∂x u(x, y) = f (x, y),
u(0, y) = 2,
x ∈ (0, L)
u(x, H) = 1,
u(x, 0) = 1,
(18)
x ∈ (0, L)
y ∈ (0, H),
−∂yy u(x, y) − ∂xx u(x, y) = f (x, y),
(17)
∂x u(L, y) = 2
∂y u(x, 0) = 1,
u(x, 0) = 1,
(16)
x ∈ (0, L)
y ∈ (0, H),
u(x, 0) = 1,
∂x u(L, y) = 2
∂x u(L, y) = 2
∂y u(x, 0) = 1,
y > 0,
−∂yy u(x, y) − ∂xx u(x, y) = f (x, y),
∂x u(0, y) = 2,
x ∈ (0, L)
u(0, y) = 3,
y > 0,
(15)
x ∈ (0, L)
u(0, y) = 3,
y > 0,
∂x u(L, y) = 2
x ∈ (0, L)
u(0, y) = 3,
y > 0,
∂yy u(x, y) − 3∂xx u(x, y) = f (x, y),
x ∈ (0, L)
(26)