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Chapter 8
Partial Differential Equations
8.1 Introduction
Definition. PDE: the generic notation f (Dαu) = 0
where f : function space → R ( or C). |α| is called the order of differential equation equation.
Example. ∆u = 0 in Rn, u : Rn^ → C. L = ∆ =
∑n i=
∂^2 ∂x^2 i^ is a linear operator.
Most interesting examples of PDE tend to be nonlinear systems of PDE. PDE courses in general start with scalar linear PDE’s:
- Elliptic: Laplace’s equation ∆u = 0 / Poisson’s equation ∆u = f.
- Parabolic: Heat equation ∂tu − ∆u = 0.
- Hyperbolic: Wave equation ∂^2 t u − ∆u = 0.
Differential operators are unbounded operators. That is, taking derivatives does not preserve continuity, differentiability, etc. Properties of the function spaces changes. Methods for dealing with unbounded operators:
- Weak solutions (distributions)
- Green’s functions.
8.2 Green’s Functions
Associated to a differential operator is a natural domain. Let A be a differential operator. D(A) is the subset of the function space that we want to consider.
Example. Aku = u′′^ (ODE), k = 1, 2 , 3 , 4. D(A) may be many different things.
- D(A 1 ) =
u ∈ C^2 ([0, 1]) : u(0) = u(1) = 0
2. D(A 2 ) = C^2 ([0, 1]).
3. D(A 3 ) = H^2 ((0, 1)).
4. D(A 4 ) =
u ∈ H^2 ((0, 1)) : u(0) = u(1) = 0
Green’s function act as an inverse to certain differential operators. Work now where D(A) ⊂ H, H for Hilbert space like L^2 (Ω), Hk(Ω). Want D(A) to be density defined in H. If D(A) is not dense in H, modify A = 0 on D(A)⊥.
Definition. A : D(A) ⊂ H → H is a densely defined unbounded linear operator on H if its adjoint A∗^ : D(A∗) ⊂ H → H is the operator with
D(A∗) = {y ∈ H : there exists z ∈ H with 〈Ax, y〉 = 〈x, z〉 for all x ∈ D(A)}.
If y ∈ D(A∗) define A∗y = z, where z is the unique element such that 〈Ax, y〉 = 〈x, z〉 for all x ∈ D(A).
Just like dual space: Note that boundary conditions for A come from D(A) and D(A∗) gives boundary conditions for A∗. We call A is self-adjoint if A∗^ = A, i.e., A∗x = Ax for all x ∈ D(A) and also D(A) = D(A∗); symmetric if A∗Ax = Ax for all x ∈ D(A), A∗^ is an extension of A if D(A) ⊂ D(A∗).
Example. Computing adjoints formally. Let
Au = au′′^ + bu′^ + cu
where a ∈ C^2 ([0, 1]), a ∈ C^1 ([0, 1]), a ∈ C([0, 1]). Let 〈·, ·〉 be the usual inner product.
A∗v = (av)′′^ − (bv)′^ + cv.
For all u, v ∈ C^2 ([0, 1]),
〈v, Au〉 − 〈Av, u〉 =
a(vu′^ − v′u) + (b − a)∇u
1 0
just proof by integration by parts
〈v, Au〉 =
0
v(au′′^ + bu′^ + cu)dx.
Possible domain: H^2 ((0, 1)). If the above a, b, c are smooth enough, then the above formula (Green’s formula holds for all u, v ∈ H^2 ((0, 1)), not formal anymore. The proof is: If u, v ∈ H^2 ((0, 1)), there exists {un} , {vn} ∈ C∞((0, 1)) such that un → u and vn → v in H^2 ((0, 1)). Formula holds for un, vn:
〈vn, Aun〉 − 〈Avn, un〉 = (a(vnu′ n − v n′un) + (b − a)∇u)
1 0
pointwise converges since, by Sobolev embedding, un, vn are C^1.
Assume Au = f , u(0) = u(1) = 0, and given f : [0, 1] → C is continuous. We want to find
u(x) =
0
g(x, y)f (y)dx
where g is called the Green;s function of the above system. Let
D(A) =
u ∈ C^2 ([0, 1]) | u(0) = u(1) = 0
then G : C([0, 1]) → D(A) defined by
Gf (x) =
0
g(x, y)f (y)dy
is the inverse of A. (idea: Au = f , Gf = u) Write the equation for the green’s function in terms of δ(x), i.e., g(x, y) must solve
Ag(x, y) = δ(x − y), g(0, y) = g(1, y) = 0
Formally
Au(x) =
0
Ag(x, y)f (y)dy =
0
δ(x − y)f (y)dy = f (x),
and boundary condition
u(0) =
0
g(0, y)f (y) = 0, u(1) =
0
g(1, y)f (y) = 0.
So u(x, t) = ( û(ξ, t))∨^ is what we mean by the notation. FACT2: Consider ∂^2 t u − c^2 ∆u = 0 with speed c = 1 for us. If f, g are supported in {x ∈ Rn^ : |x| ≤ M }, then u(·, t) is supported in {x ∈ Rn^ : |x| ≤ M + t}. look first at f part of the solution: f has support in {|x| ≤ M }
u(x, t) = cos(Dt)f = (cos(̂ Dt)f )∨^ = (cos(2π |ξ| t)f̂ (ξ))∨
=
e^2 πi|ξ|t^ + e−^2 πi|x|t 2
f
τ̂−tf (ξ) + τ̂tf (ξ)
(f (x + t) + f (x − t))
So if supp f ⊂ {|x| ≤ M }, then supp f (x ± t) ⊂ {|x| ≤ M + t}. Now see g part for n = 1, and obtain in the general solution in n = 1. Not in Fourier form. By Fourier inversion formula, ∫ (^) ∞
−∞
f (ξ)
e^2 πi(x+t)|ξ|^ + e^2 πi(x−t)|ξ| 2
dξ =
(f (x + t) + f (x − t)).
g part
∫ (^) ∞
−∞
g(ξ) e^2 πi(x+t)|ξ|^ − e^2 πi(x−t)|ξ| 4 πiξ
dξ =
−∞
g(k)
∫ (^) x+t
x−t
e^2 πisk 2
dk
∫ (^) x+t
x−t
−∞
g(k)e^2 πiskdkds =
∫ (^) x+t
x−t
g(s)ds.
D’Alembert’s solution to wave equation: (n = 1)
u(x, t) =
(f (x + t) + f (x − t)) +
∫ (^) x+t
x−t
g(s)ds
FACT3: Define the energy for wave equation:
E(t) =
Rn
|∂tu|^2 + |∇u|^2 dt.
Claim E(t) = E(0) where
E(0) =
Rn
|∂tu(x, 0)|^2 + |∇u(x, 0)|^2 dx =
Rn
|g|^2 + |∇f |^2 dx.
We will prove that for all t, E(t) = E(0), furthermore
lim t→∞
Rn
|∂tu|^2 (x, t)dx =
E(0)
Equipartition of energy: E(0) 2
|∂tu|^2 ,
E(0)
|∇u|^2
The equipartition of energy will follow from our Fourier representation of the solution and a modification of the Riemann-Lebesgue lemma (Brodsky) Energy method very standard technique for understanding linear/nonlinear PDE. Idea in general, take equation, multiply by some number of derivatives of u (or just u) and integrate: (Assume u ∈ C 0 ∞ (Rn) )
Rn
∂ t^2 u − ∆u
∂tu dx =
Rn
(∂^2 t u)∂tu dx −
Rn
(∆u)∂tu dx,
where ∫
Rn
(∂^2 t u)∂tu dx =
Rn
∂t(∂tu)^2 dx =
d dt
Rn
(∂tu)^2 dx,
Rn
∆u∂tu dx = +
Rn
∂xi u∂t∂xi u dx =
d dt
Rn
|∂xi u|^2 dx =
d dt
Rn
|∇u|^2 dx.
So
0 =
Rn
∂^2 t u − ∆u
u dx =
d dt
Rn
(∂tu)^2 + |∇u|^2 dx =
d dt E(t).
Thus E(t) = E(0), i.e., energy for wave equation conserved. Also,
E(t) = E(0) =
|g|^2 + |∇f |^2 dx.
Now we want to show
lim t→∞
Rn
|∂tu|^2 (x, t)dx =
E(0)
(Recall Riemann-Lebesgue Lemma: f ∈ L^1 , then f̂ is continuous and f̂ (ξ) → 0 as |ξ| → ∞. Recall
Plancherel: ‖f ‖L 2 =
∥̂f
L^2
Note that ̂u is in FACT1,
∂t ̂u = − 2 π |k| sin(2π |k| t)f̂ (k) + cos(2π |k| t)̂g(k),
and
(∂t ̂u)^2 = 4π^2 |k|^2 sin^2 (2π |k| t)f̂ (k)^2 + cos^2 (2π |k| t)̂g(k)^2 − 4 π |k| sin(2π |k| t) cos(2π |k| t)f̂ (k)̂g(k).
Apply triangular double-angle identity, and apply Plancherel
∫ (∂tu)^2 dx =
(∂t û)^2 dk
= 12 ∫^ |g|^2 + |∇f |^2 dx ︷ ︸︸ ︷ 1 2
4 π^2 |k|^2 f̂ (k)^2 +
g(k)^2 dk
cos(4π |k| t)
g(k)^2 − 4 π^2 |k|^2 f̂ 2
dk −
4 π |k| sin(2π |k| t) cos(2π |k| t)f̂ (k)̂g(k)dk
Since f, g ∈ L^2 , we have f ,̂ ĝ ∈ L^2 and f̂ 2 , ̂g^2 , f̂ ̂g ∈ L^1. Hence
∫ cos(4π |k| t)
∈ L^1 ︷ ︸︸ ︷̂ g(k)^2 − 4 π^2 |k|^2 f̂ 2
dk → 0
as t → ∞.
8.4 Weak Solution
Weak form of the wave equation. Assume u ∈ C^2 (Rn) so all the derivatives in wave equation make sense
∂^2 t u − ∆u = F (x, t), x ∈ Rn, t > 0 u(x, 0) = f (x) ∂u ∂t
(x, 0) = g(x)
Now take φ ∈ C 0 ∞ (Rn+1) and φ = φ(x, t) has compact support in x (i.e., φ(0, x) is not necessarily zero). Multiply φ by F and integrate over space and time: ∫
t> 0
Rn
φF dxdt =
t> 0
Rn
φ
∂^2 t u − ∆u
dxdt.
