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Notes, books, and calculators are not authorized. Show all your work in the blank space you are given on the exam
sheet. Answers with no justification will not be graded.
Question 1: 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.)
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 withRzero divergence and zero
R normal component at the boundary of
Ω. Let p be a smooth scalar field in Ω. Compute Ω p(x)∇·u(x)dx and Ω u(x)·∇p(x)dx.
(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
where ∂Ω denotes the boundary of Ω. Since u·n = 0 at the boundary, we finally infer that
R
(ii) Note finally that Ω p∇·udx = 0 since ∇·u = 0.
R
Ω
u·∇pdx = 0.
2
Mid-Term TEST, February, 25, 2014
Question 2: 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.
Using the product rule, we have
∇φ·∂i (∇φ) =
d
X
d
∂j φ∂i (∂j φ) = ∂i (
j=1
p
(b) Let > 0. Show that ∇·( k∇φk2 + ei ) = √
1X
1
(∂j φ)2 ) = ∂i ( k∇φk2 ) = k∇φk∂i k∇φk.
2 j=1
2
∇φ
·∂i (∇φ)
k∇φk2 +
where ei is the unit vector in the direction i.
Using the chain rule, we have
p
p
1
1
∂i (k∇φk2 + )
∇·( k∇φk2 + ei ) = ∂i ( k∇φk2 + ) = p
2 k∇φk2 + =p
k∇φk
k∇φk2 + ∂i k∇φk.
Then using (a) we infer that
p
∇φ
∇·( k∇φk2 + ei ) = p
·∂i (∇φ).
k∇φk2 + (c) Assume that φ(x) = 0 for all kxk ≥ R where R is a positive real number. Compute
R
Rd
√
∇φ
·∂i (∇φ)dx,
k∇φk2 +
for all i = 1, . . . , d. Give all the details.
Using (b) we have
Z
Rd
∇φ
p
·∂i (∇φ)dx =
k∇φk2 + Z
p
∇·( k∇φk2 + ei )dx,
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
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2+|x|
1+|x| ∂xx u(x, t)
= 0, x ∈ (−∞, +∞), t > 0
R +∞
with u(±∞, t) = 0, ∂x u(±∞, t) = 0, ∂xx u(±∞, t) = 0. Use the energy method to compute ∂t −∞ ([∂t u(x, t)]2 +
Question 3: Consider the vibrating beam equation ∂tt u(x, t) + ∂xx
2+|x|
2
1+|x| [∂xx u(x, t)] )dx.
Give all the details. (Hint: test the equation with ∂t u(x, t)).
Using the hint we have
Z
+∞
(∂tt u(x, t)∂t u(x, t) + ∂xx
0=
−∞
2 + |x|
∂xx u(x, t) )dx
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 +∞
2 + |x|
1
2
∂xx u(x, t) ∂t ∂x u(x, t))dx
0=
( ∂t (∂t u(x, t)) − ∂x
1 + |x|
−∞ 2
Z +∞
1
2 + |x|
=
(∂t (∂t u(x, t))2 +
∂xx u(x, t)∂t ∂xx u(x, t))dx.
2
1 + |x|
−∞
We apply again the product rule a∂t a = 21 ∂t a2 where a = ∂xx u(x, t),
Z
+∞
0=
−∞
1 2 + |x|
1
(∂t (∂t u(x, t))2 +
∂t (∂xx u(x, t))2 )dx.
2
2 1 + |x|
Switching the derivative with respect to t and the integration with respect to x, this finally gives
Z +∞
1
2 + |x|
0 = ∂t
[∂xx u(x, t)]2 )dx.
([∂t u(x, t)]2 +
2
1 + |x|
−∞
4
Mid-Term TEST, February, 25, 2014
Question 4: 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) = 2.
(a) What should be the interface conditions at x = 0 for this problem to make sense?
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.
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].
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5
Question 5:
Compute the coefficients of
R πthe sine series of f (x) = x for x ∈ [0, +π]. (Recall that by definition
P(a)
+∞
SS(f )(x) = m=1 bm sin(mx) with bm = π2 0 f (x) sin(mx)dx.)
The definition of SS(f )(x) implies that
Z
Z
2 π
2 π 1
2
1
bm =
x sin(mx)dx = −
− cos(mx)dx + [−x cos(mx)]π0
π 0
π 0
m
π
m
2
= (−1)m+1 .
m
P+∞ 2
As a result SS(f )(x) = m=1 m (−1)m+1 sin(mx).
(b) For which values of x in [0, +π] does the sine series coincide with f (x)? (Explain).
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
(c) The sine series of x2 over [0, +π] is SS(x2 )(x) = m=1 ( m43 π ((−1)m − 1) + 2π
) sin(mx). Compute the
m (−1)
sine series of h(x) = x(π − x). (Hint: use (a))
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.)
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,
∞
X
4
CS(g)(x) = ∂x SS(h)(x) =
(1 + (−1)m+1 ) cos(mx).
2π
m
m=1
6
Mid-Term TEST, February, 25, 2014
Question 6: Using cylindrical coordinates and the method of separation of variables, solve the equation, 1r ∂r (r∂r u)+
3
1
r 2 ∂θθ u = 0, inside the domain D = {θ ∈ [0, 2 π], r ∈ [0, 3]}, subject to the boundary conditions u(r, 0) = 0,
3
u(r, 2 π) = 0, u(3, θ) = 18 sin(2θ). (Give all the details.)
d
d
(r dr
g(r)) = λg(r).
(1) We set u(r, θ) = φ(θ)g(r). This means φ00 = −λφ, with φ(0) = 0 and φ( 32 π) = 0, and r dr
(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 λ 32 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
√
2
α ≥ 0 implies 32 n = α = λ. The boundary condition at r = 3 gives 18 sin(2θ) = c2 3 3 n sin( 23 nθ) for all θ ∈ [0, 23 π].
This implies n = 3 and c2 = 2.
(5) Finally, the solution to the problem is
u(r, θ) = 2r2 sin(2θ).
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7
Question 7: 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.
R +1
(a) Prove that it is necessary that λ ≥ q0 for a non-zero (smooth) solution, φ, to exist. (Hint: q0 −1 φ2 (x)dx ≤
R +1
q(x)φ2 (x)dx.)
−1
As usual we use the energy method. Let (φ, λ) be an eigenpair, then
Z
+1
(−∂x (p(x)∂x φ(x))φ(x) + q(x)φ2 (x))dx = λ
Z
−1
+1
φ2 (x)dx.
−1
After integration by parts and using the boundary conditions, we obtain
Z +1
Z
(p(x)∂x φ(x)∂x φ(x) + q(x)φ2 (x))dx = λ
−1
+1
φ2 (x)dx.
−1
which, using the hint, can also be re-written
Z
Z +1
(p(x)∂x φ(x)∂x φ(x) + q0 φ2 (x))dx ≤ λ
+1
φ2 (x)dx.
−1
−1
Then
Z
+1
Z
2
+1
p(x)(∂x φ(x)) dx ≤ (λ − q0 )
−1
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 be an eigenvalue, i.e.,
R +1
R +1
prove that φ = 0 if λ = q0 . (Hint: p0 −1 ψ 2 (x)dx ≤ −1 p(x)ψ 2 (x)dx.)
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.
8
Mid-Term TEST, February, 25, 2014
Question 8: 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,
R +∞
1
f (x)eiωx dx), the result
x ∈ R, t > 0, with u(x, 0) = u0 (x). (Hint: use the definition F(f )(ω) := 2π
−∞
ω2
2
1
F(e−αx )(ω) = √4πα
e− 4α , the convolution theorem and the shift lemma: F(f (x − β))(ω) = F(f )(ω)eiωβ . Go
slowly and give all the details.)
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
u(x, t) =
3
3
1
g(x + cos(t) − 1)e−(2t+t ) = e−(2t+t )
4πt
r
1
4πt
Z
+∞
−∞
u0 (y)e−
(x+cos(t)−1−y)2
4t
dy.