75042 Partial differential equations

Problem set #6 — suggested solutions

#6.1: From d’Alembert’s formula

u(x, t) = \frac{1}{2}(g(x + t) + g(x − t)) + \frac{1}{2} ∫_{x−t}^{x+t} h(y) dy

we immediately get

u_{t} = \frac{1}{2}(g′(x + t) − g′(x − t) + h(x + t) + h(x − t)) = F (x + t) − G(x − t),
u_{x} = \frac{1}{2}(g′(x + t) + g′(x − t) + h(x + t) − h(x − t)) = F (x + t) + G(x − t)

where

F = \frac{1}{2}(g′ + h) and G = \frac{1}{2}(g′ − h).

Substituting this into the kinetic and potential energies

k(t) = \frac{1}{2} ∫_{−∞}^{∞} u_{t}^{2} dx and p(t) = \frac{1}{2} ∫_{−∞}^{∞} u_{x}^{2} dx

and then adding, we get

k(t) + p(t) = \frac{1}{2} ∫_{−∞}^{∞} [(F (x + t) − G(x − t))^{2} + (F (x + t) + G(x − t))^{2}]dx
= ∫_{−∞}^{∞} F (x + t)^{2} dx + ∫_{−∞}^{∞} G(x − t)^{2} dx = ∫_{−∞}^{∞} F (x)^{2} dx + ∫_{−∞}^{∞} G(x)^{2} dx

which is clearly independent of t. Similarly, subtracting we get

k(t) − p(t) = \frac{1}{2} ∫_{−∞}^{∞} [(F (x + t) − G(x − t))^{2} − (F (x + t) + G(x − t))^{2}]dx
= −2 ∫_{−∞}^{∞} F (x + t)G(x − t) dx

which is zero for large t, since F and G both have compact support and therefore F (x+t)G(x−t) = 0 for all x when t is sufficiently large.

#6.2: It is helpful to note that a subharmonic function u (in the present sense) satisfies u(x) ≤ − ∫_{B(x,r)} u dy when B(x, r) ⊂ Ω, as we can deduce from integration in “polar” coordinates:

− ∫_{B(x,r)} u dy = \frac{1}{α(n)r^{n}} ∫_{0}^{r} ∫_{∂B(x,ρ)} u dS dρ ≥ \frac{1}{α(n)r^{n}} ∫_{0}^{r} nα(n)ρ^{n−1}u(x) dρ = u(x)

If u is subharmonic and v is harmonic then w = u − v is subharmonic, since v satisfies the mean value property. Also w ≤ 0 on ∂Ω, and we have to prove that w ≤ 0 in Ω.

Now a subharmonic function satisfies the strong maximum principle, for if w achieves its maximum M at a point x ∈ Ω then

M = w(x) ≤ − ∫_{B(x,r)} w dy ≤ M

where the second inequality becomes strict if w < M anywhere within B(x, r). That is obviously absurd, so w must equal M in all of B(x, r). It follows, just as in the proof of the maximum principle for harmonic functions, that u must be constant on the component of Ω containing x.

From the strong maximum principle follows the weak maximum principle, that is w achieves its maximum on ∂Ω. Since w ≤ 0 there, w ≤ 0 everywhere — and so we have proved u ≤ v in Ω.

75042 PDE/1996: Problem set #6 Page 2

For the last question, note that if u ∈ C^{2}(Ω) then

φ(r) = − ∫_{∂B(x,r)} u dS yields φ′(r) = \frac{r}{n} − ∫_{B(x,r)} Δu dy

(see the proof of the mean value formula on p. 30 of Evans’ notes.)

Clearly, if −Δu ≤ 0 in Ω we get φ′ ≥ 0, and since φ(r) → u(x) when r ↘ 0, we find φ(r) ≥ u(x)

— and so u is subharmonic.

On the other hand, if −Δu > 0 at some point in Ω the inequality holds in an open set, since Δu is continuous. But if we place x in this open set, φ′(r) < 0 for small r, and hence φ(r) < u(x) for small r. Hence u is not subharmonic.

We conclude that, for C^{2} functions, the new and old definitions of “subharmonic” coincide.

Problem set #7 — for next week

#7.1: Evans problem 20 (p. 84).

#7.2: Evans problem 23 (p. 84).

#7.3: Let A be a bounded, open subset of ℝ^{n}, and let u: A → ℝ^{m} be an integrable function. Let φ: ℝ^{m} → ℝ be convex. Prove Jensen’s inequality:

φ( − ∫_{A} u(x) dx) ≤ − ∫_{A} φ(u(x)) dx.

(Hint: Let ū = − ∫_{A} u(x) dx. By the convexity of φ, there exists some a ∈ ℝ^{m} so that φ(v) ≥ φ(ū) + a · (v − ū) for all v ∈ ℝ^{m}. This is true even if φ is not differentiable at ū.)

Explain why the same result holds if the integral is replaced by a surface integral over some surface of finite area.

#7.4: Prove that, if u: Ω → ℝ^{m} is harmonic (i.e., each component function is harmonic) and φ: ℝ^{m} → ℝ is convex (but not necessarily differentiable) then φ ∘ u is subharmonic (as defined in problem #6.2). The composition φ ∘ u is defined by φ ∘ u(x) = φ(u(x)).