Pierson Guthrey
pguthrey@iastate.edu
Part I
Partial Differential Equations
1
Introductory Theory and Notation
1.1
Multiindex notation
A multiindex α = (α1 , α2 , ...., αn ) ∈ Zn+ denotes
Dα =
∂ |α|
(∂x1 )α1 (∂x2 )α2 ...(∂xn )αn
α
x =
n
Y
i
xα
i
i=1
|x| =
q
v
u n
uX
2
2
x1 + ... + xn = t
x2i
i=1
α
α
x f = {y : y = x f (x)}
For multiindices α and β,
• |α| = α1 + α2 + ... + αn
• α!
n
Q
αi !
i=1
• α + β = (α1 + β1 , ..., αn + βn )
• α ≥ β if and only if αi ≥ βi for i = 1, ..., n
1.2
Senses of Solution
Let Lu =
P
aα (x)Dα u were aα (x) ∈ C ∞ (Ω) so LT ∈ D0 (Ω) for T ∈ D0 (Ω)
|α|≤M
Example: If Lu = utt − uxx then the general solution is u(x, t) = F (x + t) + G(x − t)
1.2.0.1 Classical Solution
Lu = f in a classical sense if u ∈ C M (Ω) and u0 (x) = f (x) ∀ x ∈ Ω Example: We require F, G ∈ C 2 (Ω)
1.2.0.2 Weak Solution
Lu = f in a weak sense if u ∈ L1loc and Lu = f in D0 sense.
Classical solutons are always also weak solutions
Example: We require F, G ∈ L1 (Ω)
1.2.0.3 Distributional Solution
Lu = f in a distributional sense if u ∈ D0 and Lu = f in D0 sense.
Classical solutions and weak solutions are always also distributional solutions
Example: We require F, G = δ ∈ D0 (Ω)
1
1.3
Pierson Guthrey
pguthrey@iastate.edu
Distributional Solutions
1.3
1.3.1
Distributional Solutions
Fundamental Solutions
We say E ∈ D0 (RN ) is a fundamental solution of
X
L=
aα (x)Dα
aα (x) ∈ C ∞
|α|≤M
if LE = δ.
• Usually fundamental solutions can be found by taking the FT of LE = δ which is
∧

X

∧
Dα E  = (δ) →
X
(ik)α Ê(k) =
|α|≤M
|α|≤M
1
N
(2π) 2
• Fundamental solutons are never unique since you can add solutions to the homogenous equation H
and the equation L(E + H) = f will still be valid
1.3.1.1 Translational Invariance If E is a fundamental solution to and aα (x) = aα (constant coefficients)
and the domain of E is all of RN , then u = E ∗ f is a solution formula for Lu = f , and E is translationally
invariant which means it commutes with translation. That is, given L : C0∞ (RN ) → C ∞ (RN ),
τh Lφ = Lτh φ
Reason:
∀ φ ∈ C0∞ , ∀ h ∈ RN
Lτh φ = T (τx (τhˇφ)) = T (τx−h φ̌) = (T ∗ φ)(x − h) = Lφ(x − h) = τh Lφ
1.3.1.1.1 Convergence If L : C0∞ (RN ) → C ∞ (RN ) such that L is translation invariant and continuous then there exists a unique T ∈ D0 (RN ) such that Lφ = T ∗ φ
1.3.2
Green’s Function
Given a PDE, a fundamental solution that satisfies the boundary conditions is known as a Green’s Function for that BVP. Given
X
L=
aα (x)Dα
aα (x) ∈ C ∞
|α|≤M
Then E = E(x, y) is a fundamental solution of Lx E(x, y) = δ(x − y) if
ˆ
ˆ
Lu(x) =
Lx E(x, y)f (y)dy =
δ(x − y)f (y)dy = f (x)
RN
RN
E(x, y) can be of the form E(x − y) but need not be in general.
2
1.3
Pierson Guthrey
pguthrey@iastate.edu
Distributional Solutions
1.3.2.1
Example
(
K(x, y) =
y(x − 1) 0 ≤ y ≤ x ≤ 1
x(y − 1) 0 ≤ x ≤ y ≤ 1
|x|
2 ,
IF Lu = u00 with u(0) = u(1) = 0 then Lx K = δ(x − y). If E(x) =
then
and LE(x − y) = δ(x − y)
LE = δ(x)
so they should differ by a solution to the homogenous equation. If H(x, y) = K(x, y) − E(x − y) then
(
y(x − 1) − 21 (x − y) 0 ≤ y ≤ x ≤ 1 n
H(x, y) =
= y − 21 x − 12 y 0 ≤ x, y ≤ 1
1
x(y − 1) − 2 (y − x) 0 ≤ x ≤ y ≤ 1
so Lx H = 0 ∀ y. If u(x) =
´1
K(x, y)f (y)dy then u00 = f but also
0
ˆ
ˆ
1
K(0, y)f (y)dy = 0 and u(1) =
u(0) =
0
1
K(1, y)f (y)dy = 0
0
so we have solved the BVP using the Green’s function K(x, y)
1.3.2.2
Example: For N = 3, ∆
Further, if f ∈
C0∞ (RN )
−1
4π|x|
= δ.
then let u = E ∗ f and so
Lu = L(E ∗ f ) = LE ∗ f = δ ∗ f = f
Example: Let E(x) =
−1
4π|x|
so ∆E(x) = δ and so if ∆u = f then
u = (E ∗ f )(x) =
−1
4π
ˆ
R3
f (y)
dy
|x − y|
which can be check to also be a classical solution since E ∗ f ∈ C ∞ .
1.3.2.3
Example: In Ω = R2 ,
Lu = utt − uxx
u(x, 0) = h(x)
ut (x, 0) = g(x)
Let E(x, t) = 21 H (() t − |x|), where H is the Heaviside function. So
(
0 t < |x|
E(x, t) = 1
(Indicator function for the forward light cone)
t > |x|
2
If f ∈ C0∞ , u = E ∗ f is a soluion of Lu = f .
ˆ
ˆ
ˆ
1 ∞ ∞
u(x, t) = E ∗ f =
E(x − y, t − s)dyds =
H(t − s − |x − y|)f (y, s)dyds
2 −∞ −∞
R2
So
u(x, t) =
1
2
ˆ
t
ˆ
x+s−t
f (y, s)dyds
−∞
(Indicator function for backward light cone)
x+t−s
3
1.3
Pierson Guthrey
pguthrey@iastate.edu
Distributional Solutions
If we assume f (x, t) = 0 for t < 0, this becomes
u(x, t) =
1
2
ˆ tˆ
x+s−t
(Integral of cone with vertex at x, y )
f (y, s)dyds
x+t−s
0
Note u(x, 0) = ut (x, 0) = 0
Also (E ∗ g) =
´∞
(x)
and
∂
∂t (E ∗ h)(x, t)
(x)
−∞
E(x − y, t)g(y)dy =
1
2
´ x+t
x−t
g(y)dy (Part of D’Alembert’s Solution Formula)
= 12 (h(x + t) + h(x − t))
so solution of PDE is
u(x, t) = (E ∗ f )(x, y) + (E ∗ g) +
(x)
1.3.3
∂
(E ∗ h)(x, t)
∂t (x)
Symbol of Operator L
Define the Symbol of L to be
N
P (k) = (2π) 2
X
(ik)α
|α|≤M
with P (k)Ê(k) = 1.
• The coefficients of the operator L yield a unique symbol and vice versa
P
N
(ik)α is the principal symbol of L and excludes lower order terms
• Pm (k) = (2π) 2
|α|=M
L is elliptic if Pm (k) = 0 if and only if k = 0 for k ∈ RN : that is Pm (k) has no real roots except 0
• A fundamental solution should satisfy (assuming P (k) 6= 0)
1
1
=⇒ E(x) =
Ê(k) =
N
P (k)
(2π) 2
But this is not clear if
1
p
ˆ
RN
eikx
dk
P (k)
∈
/S
• Malgrange Ehrenpreis Theorem If L 6= 0 everywere then a fundamental solution to Lu = f exists.
• Theorem If L is elliptic, f ∈ C ∞ (RN ), u ∈ D0 (RN ) and Lu = f , then u ∈ C ∞ (RN )
1.3.3.1
N
Example: L = ∆ then P (k) = −(2π) 2
N
P
j=1
kj2 so P (k) = 0 when k1 = k2 = 0 so is therefore
elliptic
1.3.3.2 Example: Lu = utt − uxx then P (k) = 2π(kx2 − kt2 ) so P (k) = 0 when kt = ±kx and is therefore
not elliptic.
1.3.4
Duhamel’s Principle
Duhamel’s principle is a general method for obtaining solutions to inhomogeneous linear evolution equations
(Ex: see heat equation) by convolution of a fundamental solution with the inhomogenous term.
4
1.4
Pierson Guthrey
pguthrey@iastate.edu
Characteristic Curves
1.3.4.1 Regularity of Solutions For ∆u = 0 all solutions are classical, weak, and distributional solutions.
Thus, the regularity of the solution depends on the PDE.
1.4
Characteristic Curves
of a PDE are the lines on which the solution is constant.
1.5
Types of PDEs
1.5.1
Linear PDEs
A PDE of order m is linear if it can be expressed as
X
Lu(x) =
aα (x)Dα u(x) = f (x)
|α|≤m
n
An order k linear PDE over R will have
1.5.2
n+k−1
k−1
distinct terms
Semilinear PDEs
A PDE of order m is semilinear if it can be expressed as
X
aα (x)Dα u(x) + a0 Dk−1 u, ..., Du, u, x = f (x)
|α|=m
1.5.3
Quasilinear PDEs
A PDE of order m is quasilinear if it can be expressed as
X
aα (Dα u(x), ..., Du, u, x) + a0 Dk−1 u, ..., Du, u, x = f (x)
|α|=m
1.5.4
Nonlinear PDEs
A PDE of order m is nonlinear if it depends nonlinearly upon the highest order derivatives.
PDEs are classified by order and type (elliptic, hyperbolic, parabolic)
1.6
First Order Equations (m = 1)
1.6.1
1.6.1.1
Linear Equations
Constant Coefficients
~ · ~θ = ∂u = aux + buy = 0
∇u
∂~θ
A directional derivative vanishes for all points (x, y).
A characteristic curve though point (x0 , 0) obeys ay =
b(x −x0 ), so x0 =
bx−ay
b
If u(x, 0) = f (x), then u(x, y) = u(x0 , 0) = f (x0 ) = f
5
bx−ay
b
1.6
Pierson Guthrey
pguthrey@iastate.edu
First Order Equations (m = 1)
1.6.1.2
Nonconstant Coefficients
a(x, y)ux + b(x, y)uy = 0
A directional derivative at each point vanishes.
A characteristic curve parameterized by (x(t), y(t)) satisfying the ODE system
dx
= a(x, y)
dt
dy
= b(x, y)
dt
d
dx
dy
u(x(t), y(t)) = ux
+ uy
= aux + buy = 0
dt
dt
dt
u is a solution of the PDE =⇒ the curves are characteristics =⇒ u is constant along these curves
Let f (x) = u(x, 0), if (x, y) and (x0 , 0) lie on the same characteristic then u(x, y) = u(x0 , 0) = f (x0 )
1.6.1.3
Cauchy Problem A solution u of the PDE is specified by f (s) on a curve γ
• γ can be defined parametrically by x = φ(s) and y = ψ(s) then u(φ(s), ψ(s)) = f (s)
• γ can be nowhere characteristic nor can it be tangent to the characteristic direction (x0 , y 0 ) = (a(x, y), b(x, y))
because if f = 0 anywhere, then f touches a characteritic, which means it is a characteristic
• Solution: Existence isn’t guaranteed. If γ is nowhere tangent to (a, b) and (a, b) 6= (0, 0) then a unique
solution exists locally (near γ)
Solve the ODE for (x(t, s), y(t, s)), a fixed s which is a characteristic through (φ(s), ψ(s))
dx
= a(x, y)
dt
dy
= b(x, y)
dt
x(0, s) = φ(s)
y(0, s) = ψ(s)
Perform a transformation (x(t, s), y(t, s)) → (t(x, y), s(x, y))
Which means u(0, s) = f (s) becomes u(x, y) = f (s(x, y))
Solution is valid at a point (x, y) if it can be conncted to γ by a characteristic curve
1.6.2
Semilinear Equations
a(x, y)ux + b(x, y)uy = c(x, y, u)
So we have an ODE system
dx
= a(x, y)
dt
dy
= b(x, y)
dt
d
(u(x(t), y(t))) = a(x, y)ux + b(x, y)uy = c(x, y, u) = c(x(t), y(t), u(x(t), y(t)))
dt
Parameterize γ (see Cauchy Problem), solve system to get t(x, y) and s(x, y) and solve
u0 = c(x, y, u)
u(s, 0) = f (s)
using previously obtained x(t, s) and s(t, s) to set u(t, s) and substitute to get u = u(x, y)
6
1.7
Pierson Guthrey
pguthrey@iastate.edu
Second Order Equations (m = 2)
1.7
Second Order Equations (m = 2)
Typically made easier using a linear transformation of coordinates
Let η = φ(x, y), ξ = ψ(x, y) be an invertible transformation (x, y) → (η, ξ). We need to find ux and uxx in
terms of the new system. Using chain rule,
ux = uη φx + uξ ψx
uxx = uηη φ2x + uη + uξη ψx φx + uξ + uηξ φx ψx + uη + uξξ ψx2 + uξ
uxx = uηη φ2x + 2uξη ψx φx + uξξ ψx2 + 2uη + 2uξ
Now we must find uy and uyy in the new system.
uy = uη φy + uξ ψy
uyy = uηη φ2y + uη + uξη ψy φy + uξ + uηξ φy ψy + uη + uξξ ψy2 + uξ
uyy = uηη φ2y + 2uξη ψy φy + uξξ ψy2 + 2uη + 2uξ
Now we must compute the mixed derivative uxy based on ux
uxy = uηη φy φx + uη + uξη φy ψx + uη + uηξ ψy φx + uξ + uξξ ψy ψx + uξ
uxy = uηη φy φx + uξη (φy ψx + ψy φx ) + uξξ ψy ψx + 2uη + 2uξ
Adding everything together, a second order equation in the original coordinates
auxx + 2buxy + cuyy + dux + f uy + hu = g(x, y)
becomes in the new coordinate system:
Auηη + Buηξ + Cuξξ + Duη + F uξ + Hu = g(η, ξ)
where
A(η, ξ) =aφ2x + bφx φy + cφ2y
B(η, ξ) =2aφx ψx + b(φy ψx + φx ψy ) + 2cψy φy
C(η, ξ) =aψx2 + bψx ψy + cψy2
D(η, ξ) =dφx + f φy + 2a + 2b + 2c
F (η, ξ) =dψx + f ψy + 2a + 2b + 2c
H(η, ξ) =h
If the transformation η = φ(x, y), ξ = ψ(x, y) is linear, it can be expressed in the form
η
α β
x
=
ξ
γ δ
y
with αδ − βγ 6= 0 to ensure invertibility. This means φx = α, φy = β, ψx = γ, ψy = δ, so
A(η, ξ) =aα2 + bαβ + cβ 2
B(η, ξ) =2aαγ + b(βγ + αδ) + 2cβδ
C(η, ξ) =aγ 2 + bγδ + cδ 2
D(η, ξ) =dα + f β + 2a + 2b + 2c
F (η, ξ) =dγ + f δ + 2a + 2b + 2c
H(η, ξ) =h
7
1.7
Pierson Guthrey
pguthrey@iastate.edu
Second Order Equations (m = 2)
1.7.1
Types of Equations
1.7.1.1 Parabolic Equations b2 −ac = 0 With equations of this type, one can choose b(η, ξ) = c(η, ξ) = 0
to obtain
uηη = 0
or equivalently b(η, ξ) = a(η, ξ) = 0 to get
uξξ = 0
1.7.1.2 Hyperbolic Equations b2 −ac > 0
0 to obtain
With equations of this type, one can choose a(η, ξ) = c(η, ξ) =
uξη + D|1| u = 0
or equivalently b(η, ξ) = 0 to get
uξξ − uηη + D|1| u = 0
Where D|1| u is the lower order terms
If D|1| u = 0, the solutions to these equations are of the form
u(η, ξ) = Φ(η) + Ψ(ξ)
for some functions Φ(η), Ψ(ξ)
1.7.1.3 Elliptic Equations b2 − ac < 0
to obtain
With equations of this type, one can choose a(η, ξ) = c(η, ξ) = 0
uξη + D|1| u = 0
or equivalently b(η, ξ) = 0 to get
uξξ − uηη + D|1| u = 0
Where D|1| u is the lower order terms
1.7.2
Separation of Variables
Seeking separable solutions u(x1 , ..., xn ) = X1 (x1 ) × ... × Xn (xn ) works for many kinds of PDEs in multiple
dimensions. (0 ≤ θ < 2π, r ≥ 0)
1.7.3
Boundary Conditions
For a well-posed mth order PDE, you will have up to m side conditions, usually in the form of
F (u) = 0 on ∂Ω
Boundary conditions are said to be homogenous if they are closed under linear combinations, such as
u = 0 on ∂Ω
The most common boundary conditions are:
1.7.3.1
Dirichlet Conditions (First Type)
u = g on ∂Ω
8
1.7
Pierson Guthrey
pguthrey@iastate.edu
Second Order Equations (m = 2)
1.7.3.2
Neumann (Second Type)
∂u
= ∇u · ~n = g on ∂Ω
∂n
where ~n is the unit outward normal
1.7.3.3
Robin (Third Type), aka Mixed
∂u
+ σu = ∇u · ~n + σu = g on ∂Ω
∂n
1.7.4
Common Well-Posed Problems
The following equations are well-posed and have classic solutions
1.7.4.1
Poisson’s Equation
∆u = f
(Poisson’s Equation)
∆u = 0
(Laplace’s Equation)
Solutions are harmonic functions, typically with Dirichlet, Neumann, or Robin conditions
(rectangular coordinates)
uxx + uyy + uzz = f
1
1
urr + ur + 2 uθθ = f
(polar coordinates)
r
r
Requires product solution families to be 2π periodic in θ:
(
c1 rn cos (nθ) + c2 rn sin (nθ) + c3 r−n cos (nθ) + c4 r−n sin (nθ) n = 1, 2, 3...
un (θ, r) =
c1 + c2 log(r)
n=0
Coefficients are found using boundary conditions, initial conditions, and Fourier theory
∂u
1 ∂2u ∂2u
1 ∂
r
+ 2 2 + 2 =0
(cylindrical coordinates)
r ∂r
∂r
r ∂φ
∂z
1 ∂
r2 ∂r
r2
∂u
∂r
+
1
∂2
2
r sin(θ) ∂θ2
sin(θ)
∂u
∂θ
+
r2
1
∂2u
=0
2
sin (θ) ∂φ2
(spherical coordinates)
1.7.4.1.1 Using Distributions
−1
u(x) = 4π|x|
, note u ∈ L1loc (R3 ) since
ˆ
ˆ
R
ˆ
2π
ˆ
π
|u(x)| dx =
R
0
0
−π
1 2
C
r sin(θ)dφdrdθ = R2
4πr
2
ˆ
Claim: ∆u = δ. Reason:
u(∆φ) =
ˆ
u∆φ = lim
→0
u∆φdx
<|x|<R
Since φ has compact support and so for some R φ(|x| > R) = 0. By Green’s Identity this becomes
!
ˆ
ˆ
ˆ
∂u
∂φ
lim
u∆φdx = lim
φ(x)∆udx +
φ
− u ds
→0 <|x|<R
→0
∂n
<|x|<R
|x|= ∂n
9
1.7
Pierson Guthrey
pguthrey@iastate.edu
Second Order Equations (m = 2)
Note that ∆u ∝ ∆ 1r = 0, and
ˆ
ˆ
∂φ ∂φ ∂φ 1
1
u ds ≤
ds =
max 4π2 ≤ C → 0
|x|= ∂n 4π |x|= ∂n 4π
∂n
∂u
since the 4π2 is the surface area of a sphere. We also see that since ∂n
= − ∂u
∂r ,
ˆ
ˆ
ˆ
ˆ
∂u
1
∂u
1
φ ds = −
ds
=
φ ds = −
φ
φds
2
4π2 |x|=
|x|= ∂n
|x|= ∂r
|x|= 4π
which is the´ average of φ over the sphere |x| = which must tend to the value at the center. That is, as
1
φds → φ(0), so
→ 0, 4π
2
|x|=
∆u(φ) = u(∆φ) = φ(0) = δ(φ)
In higher dimensions,
(
u(x) =
−1
2π
AN
log(x) N = 2
1
N ≥3
|x|N −2
Where AN is the surface area of the N -Sphere.
1.7.4.2
Fundamental Solution
E(x) =
 |x|


2


N =1
1
2π log |x|



 CN
|x|N −2
N =2
N ≥3
N
If ∆ is replaced by the positive operator −∆,
Where CN = N (N???
−1)an and an is the volume of B1 (0) in R
the the FS is E− (x) = −E(x)
1.7.4.3
Wave Equation
utt − c2 uxx = 0
Typically with Dirichlet, Neumann, or Robin conditions in space and some
u(~x, 0) =f (~x)
ut (~x, 0) =g(~x)
x∈Ω
η = x + ct and ξ = x − ct, then
u(x, t) = Φ(x + ct) + Ψ(x − ct)
If initial conditions u(x, 0) = f (x) and ut (x, 0) = g(x) are given,
´x
then f (x) = Φ(x) + Ψ(x) and g(x) = Φ0 (x) + Ψ0 (x) and so 0 g(y)dy = Φ(x) + Ψ(x) + C
´
x
Combining, F (x) = 12 f (x) + 0 g(y)dy + C giving us
D’Alembert’s Solution Formula
ˆ x+ct
1
1
u(x, t) = (f (x + ct) + f (x − ct)) +
g(y)dy
2
2c x−ct
10
1.7
Pierson Guthrey
pguthrey@iastate.edu
Second Order Equations (m = 2)
1.7.4.4
Fundamental Solution

1

 2 H(t − |x|) N = 1


 H(t−|x|)
1
N =2
E(x, t) = 2 √t2 −|x|2




 δ(t−|x|)
N =3
4π|x|
Example: For N = 3
ˆ
(E ∗ f ) =
R4
δ(s − |y|)
f (x − y, t − s)dsdy =
4π |y|
Assume f (x, t) < 0 for t < 0, so
ˆ
(E ∗ f ) =
R3 ,|y|<t
1
f (x − y, t − |y|)dy =
4π |y|
ˆ
R3 ,s=|y|
ˆ
B(x,t)
1
f (x − y, t − |y|)dy
4π |y|
1
f (y, t − |x − y|)dy
4π |x − y|
So the solution of utt − ∆u = f , u(x, 0) = h(x), ut (x, 0) = g(x) can be shown to be
∂
(E ∗ h) + (E ∗ f )
∂t (x)
u(x, t) = (E ∗ g) +
(x)
1.7.4.4.1 Uniqueness Showing the uniqueness of solutions to this problem is equivalent to showing
that the homogenous problem
utt − c2 uxx = 0
ut (~x, 0) = u(~x, 0) = 0
with homogenous boundary conditions has only a trivial solution.
1.7.4.4.2 Four Point Property for solutions to the wave equation uxx − utt = 0 on Ω ⊂ R2 containing
the tilted rectangle with vertices (x, t), (x + h − k, t + h + k), (x + h, t + h), u(x − k, t + k),
u(x, t) + u(x + h − k, t + h + k) = u(x + h, t + h) + u(x − k, t + k)
1.7.4.4.3 Using Distributions
Let F ∈ L1loc (R), then u(x, t) = F (x + t) is one solution to the wave equation in D0 (R2 ) since
ˆ ˆ
T (φ) = (utt − uxx )(φ) = u(φtt − φxx ) =
F (x + t)(φtt (x, t) − φxx (x, t))dxdt
R2
By change of coordinates, ξ = x + t, η = x − t, φtt − φxx = −4φξη , dxdt =
ˆ
ˆ
∞
=2
∞
F (η)
−∞
φξη dξdη = 0
−∞
since φ has compact support
11
∂(x,t)
∂(ξ,η)
= − 21 dξdη
1.7
Pierson Guthrey
pguthrey@iastate.edu
Second Order Equations (m = 2)
1.7.4.5
Heat Equation
ut − ∆u = 0
(~x, t) ∈ Ω × [0, ∞)
Typically with Dirichlet, Neumann, or Robin conditions in space and some
u(~x, 0) = f (~x), x ∈ Ω
Look for separable solutions of the form u(t, x) = Φ(t)Ψ(x) where
Φ(t) = c1 ekt + c2 e−kt
Ψ(x) = c1 sin(kx) + c2 cos(kx)
Solutions will usually involve k = kn so the set of product solutions is
n
o∞
2
2
2
2
∞
{Φn (t)Ψn (x)}0 = c1 ekn t sin(kn x) + c2 ekn t cos(kn x) + c3 e−kn t sin(kn x) + c4 e−kn t sin(kn x)
0
Where the c1 , ..., c4 is chosen to satify the boundary and initial conditions using Fourier theory.
1.7.4.5.1 Uniqueness Showing the uniqueness of solutions to this problem is equivalent to showing
that the homogenous problem
ut − ∆u = 0
ut (~x, 0) = u(~x, 0) = 0
with homogenous boundary conditions has only a trivial solution.
1.7.4.6
Fundamental Solution
E(x, t) =
1.7.4.7
H(t)
(2πt)
N
2
e
−|x|2
4t
Reason:
1.7.4.7.1
∧
Using FT For Lu = ut − uxx with u(x, 0) = f (x) taking the partial FT in x yields
ˆ
1
∧
u(x, t)e−ikx dx
and (ut ) = (û)t
û(k, t) =
N
N
2
(2π)
R
2
2
So (−∆u) = |k| û so ût + |k| û = 0, which is an ODE for û for fixed k. Solving,
∧ !
ˆ
|x−y|2
1
1
−|k|2 t
−1
ˆ
û(k, t) = f e
=⇒ u(x, t) =
F
(f
∗
g)
=
e− 4t f (y)dy
N
N
(2π) 2
(4πt) 2 RN
2
where ĝ(k) = e−|k|
t
so g(x) =
|x|2
4t
N
(2t) 2
e−
. This is valid for f ∈ Lp (RN ) for some 1 ≤ p ≤ ∞, t > 0, and
lim u(x, t) = f (x) a.e.
t→0+
12
1.7
Pierson Guthrey
pguthrey@iastate.edu
Second Order Equations (m = 2)
1.7.4.7.2 Using FT and Duhamel’s Principle By Duhamel’s Principle, we can do the same for Lu =
´t
h. Let v(x, t, s) satisfy vt − ∆v = 0 x ∈ RN , t > 0, v(x, 0, s) = h(x, s) and let u(x, t) = 0 v(x, t − s, s)ds.
ˆ
ˆ
t
t
vt (x, t − s, s)ds = h(x, t) +
ut = v(x, 0, t) +
0
∆v(x, t − s, s)ds = h(x, t) + ∆u
0
So for ut − ∆u = h(x, t) for t > 0 with u(x, 0) = 0, we have
v(x, t, s) =
where E =
ˆ
1
(2πt)
N
2
−|x|2
4t
N
(2πt) 2
H(t)e
e
−|x−y|2
4t
ˆ tˆ
1
h(y, s)dy =⇒ u(x, t) =
RN
0
RN
(2π(t − s))
N
2
e
−|x−y|2
4(t−s)
h(y, s)dyds = E ∗ h
and h(x, t) = 0 for t < 0. So our solution is
u(x, t) = (E ∗ h)(x, t) + (E ∗ f )(x, t)
(x)
N F √xt if
So E is the fundamental solution to the Heat Equation. Note: for t > 0, E(x, t) = √1t
|x|2
´
−
F (x) = e 4N . Here F ≥ 0, F ∈ L1 , and RN F (x)dx = 1. By previous discussion of approximate identities
let k =
(2π)
1
√
t
2
k N F (kx) → δ as k → ∞ =⇒ E(x, t) → δ ∈ D0 as t → 0+
so (E ∗ f )(x, t) = E(·, t) ∗ f → f as t → 0+ (similar to approximate identities)
(x)
1.7.4.8 Maximum Principle The solution u is bounded by the extremes of the initial condition and the
Dirichlet boundary conditions.
1.7.4.9
Shrodinger Equation
ut − i∆u = 0
(~x, t) ∈ Ω × [0, ∞)
Typically with Dirichlet, Neumann, or Robin conditions in space and some u(~x, 0) = f (~x), x ∈ Ω
1.7.4.10
Fundamental Solution
E(x, t) =
1.7.4.11
H(t)
(2πt)
N
2
−|x|2
4it
Helmholtz Equation
∆u − k 2 u = 0
1.7.4.12
π
ei(N −2) 4 e
(~x, t) ∈ Ω × [0, ∞)
Fundamental Solution
(
E(x, t) =
1
2π K0 (k |x|)
−e−k|x|
4π|x|
N =2
N =3
Where Ki is the ith order Modified Bessel Function.
Note as k → 0, E(x) tends toward the E(x) corresponding to L = ∆
13
Pierson Guthrey
pguthrey@iastate.edu
1.7.4.13
Euler-Tricomi Equation
uxx − xuyy = 0
Hyperbolic in the half plane x > 0, parabolic at x = 0, elliptic in x < 0
2
Characteristics are along y ± 23 x 3 = C
Particular solutions are
u =c1 xy + c2 x + c3 y + c4
u =c1 3y 2 + x3 + c2 y 3 + x3 y + c3 6xy 2 + x4
1.7.4.14
Biharmonic Oscillator
∆2 u = 0
(~x, t) ∈ Ω × [0, ∞)
1.7.4.15
Fundamental Solution ??
1.7.4.16
Klein Gordon Equation
Lu = utt − uxx + u
1.7.4.17
Fundamental Solution
E(x, t) =
2
(~x, t) ∈ Ω × [0, ∞)
p
1
H(t − |x|)J0 ( t2 − x2 )
2
Distribution Theory in DE
If f 0 (x) = 0 for a ≤ x ≤ b then f (x) = c classically If T 0 = 0 ∈ D0 (a, b) then T = c
´b
´b
Reason: Choose φ0 ∈ C0∞ with a φ0 (x)dx = 1. If φ ∈ C0∞ (a, b) let ψ(x) = φ(x) − a φ(y)dyφ0 (x). Note
´b
´x
ψ(x)dx = 0. Let ζ = a ψ(s)ds so ζ 0 = ψ(x) since ζ ∈ C ∞ (a, b) and further ζ ∈ C0∞ since
a
´ ζ(a) = ζ(b) = 0
b
0
0
0
and ζ = 0 for x < a or x > b. Then 0 = T (ζ) = −T (ζ ) = −T (ψ) = −T (φ) + a φydy T (φ0 ) so
´b
´b
T (φ) = a T (φ0 )φ(y)dy = a cφ(y)dy = c in D0 sense.
2.1
ODEs in D0 sense
Let g(x) =
´x
a
T0 = f
f (s)ds (antiderivative of f ). Claim g 0 = f in D0 (a, b). Reason:
ˆ
Tg0 (φ)
f ∈ L1loc (R)
0
ˆ
b
b
ˆ
0
= −Tg (φ ) = −
ˆ
x
0
g(x)φ (x)dx = −
f (s)dsφ (x)dx = −
a
a
a
b
a
So the general solution of T 0 = f in D0 (a, b) is
ˆ
T =
x
f (s)ds + C
a
14
f (s)φ(s)ds = Tf (φ)
b
φ0 (x)dxds
f (s)
a
Using FTC and compact support of φ,
ˆ b
ˆ
0
Tg (φ) = −
f (s)(φ(b) − φ(s))ds =
a
ˆ
b
s
Pierson Guthrey
pguthrey@iastate.edu
3
Advanced Theory
4
New Notes
Notation Let A ∈ Cn×m , B ∈ Cn×m , then the Hadamar product
A:B=
m X
n
X
aij bij
i=1 i=1
2
kAk = |A : A|
Sn are the symmetric n × n matrices. Sn ⊂ Cn×n .
B(x, r) is the unit ball in Rn centered at x B(x, 1) is the unit ball in Rn centered at x. B(0, 1) is the unit
ball in Rn centered at the origin.
S n−1 is the unit sphere in Rn . S n−1 = ∂B(0, 1)
ei is the ith coordinate vector in Rn .
Rn+ = {y ∈ Rn : yn > 0} is the upper half space of Rn .
Let U, V, W be open subsets of Rn . V ⊂⊂ U means V is compactly contained in U . That is V ⊂ U , V is
compact. ∂U is the boundary of U . U = ∂U ∪ U UT = U × (0, T ]
n
ΓT = U T − UT (parabolic boundary) α(n) = Γ πn 2+1 is the volume of the B(0, 1) in RN . nα(n) is the
(2 )
surface area of ∂B(0, 1) in RN \0 = \ ∪ {0}
Du is the Gradient of u D2 uP
is the Hessian of u
n
The Laplacian of u is ∆u = i=1 uxi xi = tr (D2 u)
Airy eqn -3rd order linear
ut + uxxx = g(x)
KdV eqn -3rd order semilinear
ut + uux + uxxx = g(u)
Monge-Ampere eqn - fully nonlinear, second order
det(D2 u) = f (u)
Transport equation solutions
ut + b · Du = f, (~x, t) ∈ Rn × (0, ∞)
Laplace Poisson eqn
∆u = f
Heat Equation
ut − ∆u = f
Wave Eqn
utt − ∆u = f
Wave Eqn
ˆ
ˆ
Φ(y)∆x f (x − y)dy +
∆u =
Φ(y)∆x f (x − y)dy
RN \B(0,)
B(0,)
The first part vanishes by FTC, and the second part becomes
ˆ
ˆ
ˆ
∂
∂
=
∆y Φ(y)∆x f (x − y)dy −
Φ(y)f (x − y)dS(y) +
Φ(y) f (x − y)dS(y)
∂n
∂n
RN \B(0,)
∂B(0,)
∂B(0,)
15
Pierson Guthrey
pguthrey@iastate.edu
∂
∂n f (x
− y) = 0 away from the origin. We also note that
d
−1
1
−Dn Φ(y) = − −
log(r) |r= =
dr
2π
2π
ˆ
So
∆u(x) = −
∂B(0,)
f (x − y)
dS(y) = −f (x)
2π
ˆ
∂B(x,)
1
dS(y) = −f (x)
2π
So ∆u(x) → f (x) uniformly as → 0.
If u ∈ C 2 (U ) is Harmonic, then it satisfies the Mean Value Property
´
udS
∂B(x,r)
u(x) =
udS = ´
dS
∂B(x,r)
∂B(x,r)
For all B(x, r) ⊂ U
Proof. Set
´
ϕ(ν) =
v(ν)dS(y) =
∂B(0,1)
v(x + rz)dS(z)rn−1
=
k∂B(0, 1)k
∂B(x,r)
v(x + rz)dS(z)
∂B(0,1)
Since k∂B(0, 1)k = nα(n)rn−1
The average value of a Harmonic function over a sphere is the value at the center of the sphere.
ϕ0 (r) =
∂B(0,1)
Since
y−x
r
=
z
n
Du(y) ·
Du(x + rz)zdS(z) =
∂B(x,r)
y−x
dS(z)
r
(?) it is the outward unit normal
´
=
∂B(x,r)
∂u
dS =
∂n
∂B(x,r)
∆uds
|∂B(x, r)|
=0
This implies ϕ(r) is a constant.
lim ϕ() = lim ?/ =⇒ φ(r) =
→0
ffl
Corollary: Same holds on B(x, r). u(x) = B(x,r) u(y)dy. Proof. Use the shell method
!
ˆ
ˆ
ˆ
→0
r
u(y)dy =
B(x,r)
udS
0
ds = ... = u(x)α(n)rn = u(x) |∂B(x, r)|
∂B(x,s)
ffl
Theorem. Converse. If u ∈ C 2 (U ) satisfies u(x) = ∂B(x,r) uds ∀ B(x, r) ⊂ U then u is harmonic. Proof.
If not, there exists r > 0 : ∆u > 0 on B(x0 , r). As before,
´
∆udy
B(x0 ,r)
0
ϕ (r) = 0 =
> 0 ⇒⇐
|∂B(x0 , r)|
Strong Maximum Principle Suppose u ∈ C 2 (U ) ∩ C(U ) is harmonic on U , and assume U is bounded.
Let M = maxU u.
1. max∂U = M
2. If U is connected and ∃ x0 ∈ U : u(x0 ) = M , then u = M .
Corollary, Strong Minimum principle also holds. Suppose u ∈ C 2 (U ) ∩ C(U ) is harmonic on U , and assume
U is bounded. Let m = minU u.
16
Pierson Guthrey
pguthrey@iastate.edu
1. min∂U = m
2. If U is connected and ∃ x0 ∈ U : u(x0 ) = m, then u = m.
ffl
Proof of Strong Maximum Principle If ∃ x0 ∈ U : u(x0 ) = M then if B(x0 , r) ⊂ U , M = B(x0 ,r) u(y)dy ≤ M
with equality if and only if u = M on B(x0 , r). This implies u = M on B(x0 , r). This implies if S =
{x ∈ U : u(x) = M } =⇒ S is relatively open in U . u is continuous u−1 (M ) is closed in U =⇒ S is
relatively open, closed. If U is connected, nonempty then U = S. So (2) holds. The continuity implies that
(1) holds.
Corollary. Consider
(
∆u = 0, x ∈ U
v=
g, x ∈ ∂U
Assume U is connected, u ∈ C 2 (U ) ∩ C(U ).
• If g ≥ 0, then u ≥ 0.
• If also ∃ s0 ∈ ∂U : g(s0 ) > 0, then u > 0 in U .
Theorem 5. This problem has at most one solution. If u1 , u2 are solutions, then y = u1 − u2 solves this
problem with g = 0 (the homogenous case). Maximum principle implies y ≥ 0, minimum principle implies
y ≤ 0, so y = 0 =⇒ u1 = u2 .
Mollifiers
The standard mollifier
(
−1
2
ce(|x| −1)
|x| < 1
η( x) =
0
|x| ≥ 1
η( x) ∈ C0∞ (RN )
´
If f : U → R ∈ L1loc , define f = η ∗ f = U η (x − y)f (y)dy in U . η =
η (x) is a delta sequence
´
• RN η dx = 1
1
n η
x
• supp η = B(0, ) ⊂ B(0, 1)
• η ≥ 0.
Theorem p 714
• f ∈ C ∞ (U ), U = {x ∈ U : dist(x, ∂U ) > }
• f → f a.e. as → 0
• If f ∈ C(U ) then f → f uniformly on compact subsets of U
p
• f ∈ LP
loc (U ) =⇒ f → f in Lloc (U )
Theorem If U ∈ C(U ) satisfies the mean value property on each B(x, r) ⊂ U then u ∈ C ∞ by showing
ˆ
ˆ
1
|x − y|
U (x) =? (x), x ∈ U (x) =
η (x − y)u(y)dy = n
η
u(y)dy
B(x,)
U
1
= n
=
1
n
ˆ
η
r ˆ
0
ˆ
η
0
r
!
U ds dr
∂B(x,r)
U (x) |∂B(x, r)| dr
17
Pierson Guthrey
pguthrey@iastate.edu
ˆ
= U (x)
η dy = U (x)
B(0,)
If u is harmonic on U then |Dα u(x0 )| ≤
n+1
ck
kukL1 (B(,x0 ,r))
r n+k
for all B(x0 , r) ⊂ U . |α| = k. where C0 =
1
α(n)
k
nk)
and Ck = (2 α(n)
Proof k = 0
|U (x0 )| = ´
|u| dx
C0
B(x0 ,r)
u(x)dx ≤
= n kU kL1 (B(x0 ,r)
n
α(n)r
r
B(x0 ,r)
Proof k = 1
´
´
u · ni dx
ds
r
B(x
,
)
B(x0 ,( r )
2n
(
)
0
|Uxi (x0 )| = uxi (x)dx ≤
≤
kuk
n
∞
r = kU kL1 (B(x0 ,r)) = kU kL1 (B(x0 ,r)) r
r
B(x0 , r )
B(x0 , )
α(n) ...
Liouville’s Theorem. Suppose u : Rn → R is harmonic and bounded. Then u ≡ const. .
Proof. Let B = B(x0 , r)
√
C1
C
nC1
|D(x0 )| ≤ n+1 kU kL1 (B) ≤ n+1 kU kL∞ (B) |B| ≤ kU kL∞ (R) ∀ r
r
r
r
=⇒ |D(x0 )| → 0r → ∞ =⇒ |Du| = 0 =⇒ u ≡ const.
Theorem 9. Let f ∈ C02 (Rn ), n ≥ 3. Then any bdd solution of −∆u = f in Rn satisfies
ˆ
u=
Φ(x − y)f (y)dy + C
Rn
´
Proof. For n ≥ 3 Φ is bdd, Φ → 0 as |x| → ∞ =⇒ Rn Φ(x − y)f (y)dy is bounded. Let v be another bdd
solution to ∆v = f , then y = u − v satisfies ∆y = 0, y bdd. so y ≡ const. .
Theorem 10. u harmonic in U implies u analytic P
on U . u is analytic on U if for all x0 ∈ U , u(x) is given by
Dα
α
its Taylor series in a neighborhood of x0 . Ie u(x) =
u (x0 )α!(x − x0 ) . Idea of pf, use Cauchy estimates
α
to show that (the Taylor remainder term) RN (x0 , x) → 0 as N → ∞.
Harmek’s theorem. For all connected open sets V ⊂⊂ U , for all nonnegative Harmonic functions u
defined on U , there exists C > 0 for which (C = CV )
sup u ≤ C inf u
V
V
In particular,
1
u(y) ≤ u(x) ≤ Cu(y) ∀ x, y ∈ V
C
Proof. Let r =
dist(V,∂U )
.
4
Let x, y ∈ V , |x − y| ≤ r.
ˆ
1
1
u(x) =
≥
udz = n
n
α(n)(2r)
2
B(x,2r)
B(y,r)
udz =
B(y,r)
1
u(y)
2n
so 21n u(y) ≤ u(x) ≤ 2n u(y), where the upper inequality is obtained by symmetry of the first inequality. Then
cover V with a finite set of balls of radius 2r (since V is compact and connected), then apply this theorem to
N
each ball {Bi }i=1 with Bi ∩ Bi−1 6= ∅, i = 2, 3, ...N .
18
Pierson Guthrey
pguthrey@iastate.edu
Green’s function for −∆u = f
5
−∆u = f ∀ x ∈ U, u = g ∀ x ∈ ∂U
for some open bounded U where ∂U ∈ C 1 .
Let V = U − B(x, ), x ∈ U . Apply Green’s formula to Φ(y − x), u(y).
ˆ
ˆ
∂
∂
u(y)∆Φ(y − x) − Φ(y − x)∆(y) =
u(y) Φ(y − x) − Φ(y − x) u(y)dS(y)
∂n
∂n
V
∂V
Which as → 0, we obtain
ˆ
ˆ
∂
∂
Φ(y − x)f (x)dy
u(x) =
Φ(y − x) u(y) − g(y) Φ(y − x)dS(y) +
∂n
∂n
U
∂U
Corrector function. Suppose we can find φ∗ (y) that solves
∆y u = 0Φ(y − x) ∈ U, φ∗ (y) = Φ(y − x) ∈ ∂U
Apply Green’s function to u1 φ∗ :
ˆ
ˆ
−
φ∗ (y)∆u(y)dy =
U
∂U
u(y)
∂ ∗
∂
φ (y) − Φ(y − x) u(y)dS(y)
∂n
∂n
Combining this with the above equation,
ˆ
ˆ
∂ ∗
∂
u(x) =
u(y)
φ (y) −
Φ(y − x) dS(y) +
(ψ ∗ (y) − Φ(y − x)) ∆u(y)dS(y)
∂n
∂n
∂U
U
So we have the Green’s function G(x, y) = Φ(y − x) − φ∗ (y) x, y ∈ U x 6= y. So
ˆ
ˆ
∂
u(x) = −
u(y) G(x, y)dS(y) −
G(x, y)∆u(y)dy, x ∈ U
∂n
∂U
U
Theorem 12. If u ∈ C 2 (U ) solves the Poisson problem, then
ˆ
ˆ
∂
u(x) = −
g(y) G(x, y)dS(y) −
G(x, y)f (y)dy, x ∈ U
∂n
∂U
U
That is, IF a solution exists and if φ∗ can be found, the above is the solution.
5.1
Properties of Green’s functions
• G(x, y) = G(y, x)
Green’s functions on a half-space. Let Rn+ = {(x1 , ..., xn ) : xn > 0}. If x ∈ Rn+ , define The reflection of x
as x̃ = (x1 , ..., −xn ). Define φ∗ (y) = Φ(y − x̃), x, y ∈ Rn+ . Then
∆φ∗ = 0, x ∈ R∗n , φ∗ (y)|∂Rn+ = Φ(y − x) = Φ(y − x̃)
Consequently, G(x, y) = Φ(y − x) − Φ(y − x̃) is the Green’s function.
So a solution to the Poisson Problem is
1
yn − xn
yn + xn
∂
G = −Gyn (x, y) = − (yn Φ(y − x) − yn Φ(y − x̃)) =
−
n
∂n
nα(n) |y − x|n
|y − x̃|
19
5.1
Pierson Guthrey
pguthrey@iastate.edu
Properties of Green’s functions
so
−2xn
∂
G|∂Rn+ =
∂n
nα(n) |x − y n |
Define the Poisson Kernel
K(x, y) =
2xn
nα(n) |x − y n |
ˆ
then
ˆ
u(x) =
K(x, y)g(y)dy +
yn =0
Assume g ∈ C(Rn−1 ) ∩ L∞ (Rn−1 ) then u =
´
??
G(x, y)f (y)dy
U
K(x, y)g(y)dy satisfies
n−1
∞
• u = C ∞ (Rn−1
+ ) ∩ L (R+ )
• ∆u = 0 in Rn+
• limx→x0 u(x) = g(x0 ) ∀ x0 ∈ ∂Rn+
5.1.1
Green’s function on the unit ball
If x ∈ Rn \ {0} define x̃ =
x
.
|x|2
x̃ is the dual point to x with respect to ∂B(0, 1). Note x, x̃ are parallel vectors,
1
= |x|
.
∗
and |x̃|
We say R : x → x̃ is the inversion map.
Find φ (y) such that
∆φ∗ (y) = 0 ∈ B 0 (0, 1), φ∗ (y) = Φ(y − x) ∈ ∂B(0, 1)
Note that Φ(y − x̃) is harmonic in y in B 0 (0, 1) for all x ∈ B(0, 1) so Φ(|x| (y − x̃)) is harmonic in y in B 0 (0, 1).
Define the Green’s function for the unit Ball
G(x, y) = Φ(y − x) − Φ(|x| (y − x̃)) = Φ(y − x) − φ∗ (y)
where
2
−1 1 − |x|
∂
G(x, y) =
∂n
nα(n) |x − y|n
Define Poisson Kernel for B(0, 1) by
2
K(x, y) =
Then u =
´
∂B(0,1)
1 1 − |x|
nα(n) |x − y|n
K(x, y)u(y)dy and so ∆u = 0 ∈ B(0, 1), and u = g ∈ ∂B(0, 1).
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