MATH 412: Review notes

1. Chapter 1: Introduction

1.1. Some definitions. A PDE is an equation that involves an unknown function (or vector-valued func-tion) u of more than one independent variables, and its partial derivatives.

A few typical examples which we have derived in class from basic principles of physics:

• Simple transport equation:

ut + cux = 0

• Wave equation (describing a vibrating string):

utt = c2uxx

• Diffusion equation (or heat equation):

ut = κuxx

Multi-dimensional analogue of these basic PDEs are

ut + a · ∇u = 0, utt = c2∆u, ut = κ∆u

in which ∇ = (∂x, ∂y, · · · ) and ∆ = ∂2x + ∂2y + · · ·

Some basic definitions: linearity, homogeneity, and order of a PDE.

Moral: A PDE has arbitrary functions in its solution. To determine a unique solution, one needs auxiliaryconditions or constraints: initial or boundary conditions.

Initial conditions are those which are imposed for u or derivatives of u at time t = 0:

u(0, x) = f(x) and/or ut(0, x) = g(x)

Boundary conditions: Three most common kinds are

• Dirichlet boundary condition: u is specified on the boundary.• Neumann boundary condition: the normal derivative of u is specified (that is, we specify the

directional derivative of u in the normal direction of the boundary: ~n · ∇u)• Robin boundary condition: a combination of the above two kinds.

Well-posedness of a PDE problem:

• Existence: There exists at least one solution to the PDE problem.• Uniqueness: There is at most one solution.• Stability: The unique solution depends in a stable manner on the data of the problem.

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1.2. Linear first-order PDE. The most simplest example is the first order PDE:

a(x, y)ux + b(x, y)uy = f(x, y)

for given functions a, b, f . Notice that the PDE reads

V (x, y) · ∇u = f(x, y) with V (x, y) =

(a(x, y)b(x, y)

), ∇u =

(uxuy

)That is, we view the PDE as the directional derivative of u in the direction V (x, y) is equal to some givenfunction f(x, y).

To solve this kind of PDEs, the key idea is to work with the PDE along its characteristic curves definedby

dy

dx=b(x, y)

a(x, y)

Assume that this ODE can be solved, giving solutions y = y(x). We then view the PDE as an ODE alongthese curves {y = y(x)} as follows:

d

dxu(x, y(x)) = ux +

b(x, y)

a(x, y)uy =

f(x, y(x))

a(x, y(x))

By integrating this with respect to x, the general solution is then

u(x, y) = u(0, y(0)) +

∫f(x, y(x))

a(x, y(x))dx,

in which y(0) can be found by following the characteristic curve {y = y(x)} in term of (x, y).

2. Chapter 2: Wave equations

We study the wave problem on the whole lineutt = c2uxx x ∈ R, t ≥ 0

u(0, x) = φ(x), x ∈ Rut(0, x) = ψ(x), x ∈ R

for given functions φ, ψ.

Theorem 2.1. The unique solution to the wave problem is

u(t, x) =1

2

(φ(x+ ct) + φ(x− ct)

)+

1

2c

∫ x+ct

x−ctψ(s) ds

We proceed by two steps. First, find the general solution, and secondly, use the initial data to find theunique solution.

Lemma 2.2. The general solution of the wave equation is of the form:

u(t, x) = u(0, x− ct) +1

2c

∫ x+ct

x−ctf(τ) dτ

for arbitrary functions f(x) and u(0, x)

Proof. Nice trick: factorization. We write utt − c2uxx = (∂t − c∂x)(∂t + c∂x)u, and let v = ut + cux.We get

vt − cvx = 0, ut + cux = v.

The first equation yields v(t, x) = f(x + ct), for arbitrary function f , and the second yields, along thecharacteristic line: x(t) = ct+ x0,

d

dtu(t, x(t)) = v(t, x(t)) = f(x(t) + ct) = f(2ct+ x0).

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Integrating this, we get

u(t, x(t)) = u(0, x0) +

∫ t

0f(2cs+ x0) ds

By making a change of variable in the integral term: τ = 2cs+ x0, this gives

u(t, x(t)) = u(0, x0) +1

2c

∫ x0+2ct

x0

f(τ) dτ.

NOW, recall that x(t) = ct+ x0. We get from the above

u(t, x) = u(0, x− ct) +1

2c

∫ x+ct

x−ctf(τ) dτ,

which proves the lemma. �

Proof of the theorem. Now use the lemma and the initial data to find the unique solution to the waveproblem. We only need to find the function f in the general formula. We have

ψ(x) = ut(0, x) = −cux(0, x) +1

2

[f(x) + f(x)

].

Hence, f(x) = ψ(x) + cφ′(x). Putting this back to the formula found in the above lemma, we get

u(t, x) = φ(x− ct) +1

2c

∫ x+ct

x−ct

[ψ(τ) + cφ′(τ)

]dτ

By using that the integration of φ′ is equal to φ, the theorem is proved. �

Conservation of Energy The total energy

E(t) =1

2

∫R

(u2t + c2u2x

)dx

is conserved, that is, constant in time t, as long as the integral is well-defined. Check this by directlytaking the time derivative of the energy.

Domain of influence: the initial value at the position x0 only affects the solution at the latter time t > 0in the cone:

{x0 − ct < x < x0 + ct.}

Domain of dependence: The solution at (x, t) only depends on the initial data at positions within theinterval

[x− ct, x+ ct].

Examples: Review the examples of plucked string. Assume that the initial displacement is non zero onlyon the interval [−a, a] and that the initial velocity is identically zero. What is the solution to the waveequation after the time t = a/c ?

3. Chapter 2, cont’d: Diffusion equations

Let I be the whole line R, or the half-line: [0,∞), or a bounded interval [a, b]. We study the followingdiffusion problem: {

ut = uxx x ∈ I, t ≥ 0

u(0, x) = φ(x), x ∈ Ifor a given function φ, together with appropriate boundary conditions in the case the interval I has endpoint(s).

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3.1. Energy decay. Unlike the wave equation, the energy of the heat equation is decaying in time. Indeed,let the energy be defined by

E(t) :=1

2

∫I|u(x, t)|2 dx

By taking time-derivative of E(t) and by using the equation ut = uxx, we get

d

dtE(t) =

∫Iuut dx =

∫Iuuxx dx = −

∫Iu2x(x, t) dx+ (uux)

∣∣∣end-points of I

in which the last identity was obtained by taking integration by parts. The last term accounts for theboundary values: assume that I = (a, b), where a, b could be infinite. We have

(uux)∣∣∣end-points of I

= (uux)|x=b− (uux)|x=a

.

Hence, if we assume either Dirichlet or Neumann boundary conditions at x = a, b, or in case a or b isinfinite, assume that u and its derivatives vanish at infinities, the above boundary term is equal to zero.

Thus, the energy is decreasing in time, since

d

dtE(t) = −

∫Iu2x(x, t) dx ≤ 0, ∀ t > 0.

3.2. Uniqueness and Stability. As direct consequences of the above Energy Method, we immediatelyhave the uniqueness and stability of the solution. Indeed, consider two solutions u1, u2 to the heat problem,corresponding to two initial values φ1, φ2. The difference w = u1 − u2 solves the same heat equation, andhence satisfies the energy decay principle:

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2

∫I|u1(x, t)− u2(x, t)|2 dx ≤

1

2

∫I|u1(x, 0)− u2(x, 0)|2 dx =

1

2

∫I|φ1(x)− φ2(x)|2 dx.

This proves that the heat solution is stable with respect to the L2 norm, and moreover, the solution isunique (that is, u1 = u2 if φ1 = φ2).

3.3. Diffusion on the whole line. Consider the heat problem:{ut = uxx t ≥ 0, −∞ < x <∞,

u(x, 0) = φ(x), −∞ < x <∞.

3.3.1. The fundamental solution or the Green function. The Green function of the diffusion on the wholeline is defined by

G(x, t) =1√4πt

e−x2/4t.

Fact 1: G(x, t) ∈ C∞(R× (0,∞)). Check it directly.

Fact 2: ∂tG(x, t) = ∂xxG(x, t). This is by our construction.

Fact 3:∫RG(x− y, t) dy = 1, for any x. Indeed, we get∫

RG(x− y, t) dy =

1√4πt

∫Re−|x−y|

2/4t dy =1√π

∫Re−p

2dp = 1

in which we have made a change of variable p = (x− y)/√

4t in the integration.

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Theorem 3.1. Assume u0(x) ∈ C0⋂L∞(R). We define

u(x, t) =

∫RG(x− y, t)u0(y) dy.

Then, the followings holda. u is very smooth: u(x, t) ∈ C∞(R× (0,∞))b. u solves the heat equation: ut = uxx.c. u satisfies the initial data: limt→0+ u(x, t) = u0(x), for all x ∈ R.

Example 1: u0(x) = δ(x), that is the delta function centered at x = 0. The solution to the heat problemis

u(x, t) =1√4πt

e−x2/4t.

see Figure 1, page 50, Section 2.4, for an illustration of the solution when time changes.

Example 2: The solution with initial data

u0(x) =

{1 x ∈ [−1, 1],

0 elsewhere

is of the form

u(x, t) =

∫ 1

−1

1√4πt

e−|x−y|2/4t dy

for all x ∈ R, t > 0. Notice: infinite speed of propagation !

3.4. Maximum Principle. We study the diffusion equation on a bounded interval:

ut = uxx, 0 ≤ x ≤ L, t ≥ 0.

Theorem 3.2 (Maximum principle). For each T > 0, let Q = [0, T ]× [0, L] and let ∂Q the boundary of Qwithout the top boundary at t = T . The solution to the Heat equation satisfies

minQ

u(x, t) = min∂Q

u(x, t), maxQ

u(x, t) = max∂Q

u(x, t)

(In the other words, the maximum and minimum are attained on the boundary).

Proof. As will be seen shortly, it’s useful to study the function v = e−λtu(x, t), for arbitrary λ > 0, whichsolves

vt = vxx − λv.We will show that the maximum principle holds for v. Indeed, assume that v(x, t) attains its maximumat (x0, t0) in the interior of Q and assume that the maximum value is positive (or else, simply add a bigconstant to u(x, t), which would change no thing in the equation and the equalities that we’re proving). Itthen follows that

v(x0, t0) > 0, vt(x0, t0) = vx(x0, t0) = 0, vxx(x0, t0) ≤ 0,

which is a contradiction, since 0 = vt(x0, t0) = vxx(x0, t0)− λv(x0, t0) < 0. In the case that the maximumis attained on the upper boundary, we still have limt→T vt(x0, t) ≥ 0 and the contradiction is still reached.Hence, v(x, t) attains its maximum on the boundary ∂Q, that is

v(x, t) ≤ max∂Q

v(x, t),

for all (x, t) ∈ Q. Now, plugging v(x, t) = e−λtu(x, t), we get

u(x, t) ≤ eλt max∂Q

e−λtu(x, t) ≤ eλt max∂Q

u(x, t),

for all λ > 0. By taking the limit of λ→ 0, the above proves the last equality as claimed in the theorem.The equality for minimum follows by applying the maximum principle to the function −u(x, t). �

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Remark 3.3 (Comparison Principle). The above proof in particular shows that if u satisfies ut − uxx ≤ 0on Q and u ≤ 0 on ∂Q, then u ≤ 0 on Q.

3.4.1. Uniqueness via Maximal Principle. Consider the heat problem on a bounded interval:

(3.1)

ut = uxx + f(x, t) 0 ≤ x ≤ L, t ≥ 0

u(x, 0) = φ(x), 0 ≤ x ≤ Lu(0, t) = g(t), t ≥ 0

u(L, t) = h(t), t ≥ 0

for given functions f, φ, g, h.

Theorem 3.4 (Uniqueness). There is at most one solution to the above heat problem.

Proof. Indeed, let u1, u2 be two solutions to the same problem, and let w = u1 − u2 be the difference. Itthen follows that w solve the heat equation with zero boundary conditions and zero initial data. By theMaximum Principle,

min∂Q

w ≤ w ≤ max∂Q

w.

Since w vanishes identically on the boundary ∂Q, w = 0 everywhere in Q, which proves that u1 = u1.Hence, the solution is unique. �

3.4.2. Stability via Maximal Principle. We have the following theorem.

Theorem 3.5 (Stability). Let u1, u2 be the two solutions to the heat problem with two different initialdata: φ1, φ2, respectively. Then, there holds the stability inequality:

maxQ|u1(x, t)− u2(x, t)| ≤ max

0≤x≤L|φ1(x)− φ2(x)|.

Proof. The proof follows from the Maximum Principle as done above:

min∂Q

w ≤ w ≤ max∂Q

w

with w = u1 − u2. �