75042 Partielle differensialligninger

Oppgavesett #9 (til 1997–11–04 ) — løsningsforslag

Evans 2.5–13: With v(x, t) = e^{ct}u(x, t) we see that v satisfies the equation v_{t} − Δv = e^{ct}f with the initial value v(x, 0) = g(x). There is an explicit formula for the solution to this problem on the top of p. 53. Just replace u by v and f(y, s) by e^{cs}f(y, s) in that formula, and finally set u(x, t) = e^{−ct}v(x, t). I omit the full detailed formula here.

Evans 2.5–14: Extend v(x, t) = u(x, t) − h(t) to x < 0 by odd reflection: v(−x, t) = −v(x, t). Clearly, v is continuous at x = 0 (because of the boundary condition u(0, t) = h(t)). When x > 0 we find v_{t} − v_{xx} = u_{t} − u_{xx} − h′(t) = −h′(t). The same relation, with a sign reversal, will hold for x < 0 by antisymmetry, so we have

v_{t} − v_{xx} = \begin{cases}−h′(t) & x > 0, \\ h′(t) & x < 0. \end{cases}

Now, the solution to the problem v_{t} − v_{xx} = f(x, t) with v(x, 0) = 0 is given by Duhamel’s principle (where we replace f(y, s) by the righthand side of the above equation of course):

v(x, t) = ∫_{0}^{t} \frac{1}{\sqrt{4π(t − s)}} ∫_{ℝ} exp [−\frac{(x − y)^{2}}{4(t − s)} ] f(y, s) dy ds

= ∫_{0}^{t} \frac{1}{\sqrt{4π(t − s)}} (∫_{−∞}^{0} − ∫_{0}^{∞}) exp [−\frac{(x − y)^{2}}{4(t − s)} ] dy h′(s) ds

= ∫_{0}^{t} \underbrace{\frac{1}{\sqrt{π}} (∫_{x/\sqrt{4(t−s)}}^{∞} − ∫_{−∞}^{x/\sqrt{4(t−s)}}) exp[−z^{2}] dz}_{k(s)} h′(s) ds

= k(t)h(t) − k(0)h(0) − ∫_{0}^{t} k′(s)h(s) ds

where we note that k(t) = 0, h(0) = 0, and k′(s) is easily computed as

k′(s) = −\frac{x}{\sqrt{4π}(t − s)^{3/2}} exp [−\frac{x^{2}}{4(t − s)}] .

From this we obtain the required formula, which is as shown in the problem with the term h(t) added (clearly there is a misprint).

Evans 2.5–16: It is probably a good idea to extend our given function g ∈ C([0, 1]) to have compact support, say g(x) = 0 when x ∉ [−1, 2], and let g vary linearly from 0 to g(0) on the interval [−1, 0] and linearly from g(1) to 0 on [1, 2]. We let

u(x, t) = ∫_{ℝ} Φ(y, t)g(x − y) dy = ∫_{x−2}^{x+1} Φ(y, t)g(x − y) dy  (1)

for t > 0, and u(x, 0) = g(x). Here Φ is the standard heat kernel. That u does in fact solve the heat equation is of no use to us here, but its continuity in t ≥ 0 is useful: Not only does u(x, t) → g(x) as t ↘ 0, but u is uniformly continuous on every compact subset of the closed upper half plane t ≥ 0, and therefore the function x ↦ u(x, t) converges uniformly to g on every bounded interval — and in particular it does so on [0, 1].

In other words, given ε > 0 we can find δ > 0 so that |u(x, t) − g(x)| whenever 0 ≤ x ≤ 1 and 0 < t < δ. From now on, we fix one such t. It only remains to approximate u(x, t) by a polynomial.

Now we use the explicit form of the heat kernel:

Φ(y, t) = \frac{1}{\sqrt{4πt}} exp (−\frac{y^{2}}{4t} )

75042 PDE 1997: Oppgavesett #9 Side 2

The plan is to replace the exponential in Φ by a partial sum of the Taylor series exp(z) = ∑_{k=0}^{∞} z^{k}/k!. We are interested in u(x, t) for x ∈ [0, 1]. We see that the integral in (1) only uses values of Φ(y, t) with |y| ≤ 3, which implies y^{2}/(4t) ≤ 9/(4t).

For our particular fixed t, let n be so large that |e^{z}−∑_{k=0}^{n} z^{k}/k!| < ε\sqrt{4πt} whenever |z| ≤ 9/(4t) (just make sure that ∑_{k=n+1}^{∞} 9^{k}/(k!(4t)^{k}) < ε\sqrt{4πt}). Let

Ψ(y, t) = \frac{1}{\sqrt{4πt}} ∑_{k=0}^{n} \frac{y^{2k}}{k!(4t)^{k}}

so that |Ψ(y, t) − Φ(y, t)| < ε whenever |y| ≤ 3 (and t is still fixed). Let

P (x) = ∫_{x−2}^{x+1} Ψ(y, t)g(x − y) dy = ∫_{−1}^{2} Ψ(x − y, t)g(y) dy

P is a polynomial, for Ψ(x − y, t) is a polynomial of degree 2n in x − y, and so differentiation of the above integral more than 2n times yields zero. Whenever x ∈ [0, 1] we get |g(x) − P (x)| ≤ |g(x) − u(x, t)| + |u(x, t) − P (x)| < ε + 4ε max |g| which shows that P can be made to approximate g arbitrarily well in the uniform norm.

#9–1: 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.

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.

Oppgavesett #10 (til 1997–11–11 )

Fra Evans side 84: 20, 22, 23. Det er en tyrkleif i oppgave 20: Det skal stå Ũ = rU (hvor U(r, t) er definert nederst side 68).