75042 Partial differential equations

Problem set #5 — suggested solutions

#5.1: In case you got nowhere on this, here is a hint: Consider v on the boundary of the ball and apply the Poisson integral formula. Maybe you should try again? The full solution is at the end.

#5.2: That u_{λ}(x, t) = u(λx, λ^{2}t) solves u_{t} − Δu = 0 if u does so, is trivial, since the chain rule yields u_{λ,t} − Δu_{λ} = λ^{2}(u_{t} − Δu).

(In cases like this where not all subscripts mean differentiation, is is handy to introduce a comma to separate those subscripts that indicate a derivative from those that do not.)

Now differentiate u_{λ} with respect to λ. Equality of mixed partials, i.e., u_{λ,λt} = u_{λ,tλ} and Δu_{λ,λ} = (Δu_{λ})_{λ} guarantee that u_{λ,λ} satisfies the heat equation. Put λ = 1 to obtain the solution v = x · Du + 2tu_{t}.

#5.3: 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.

#5.4: 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).

#5.5: 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)

75042 PDE/1996: Problem set #5 Page 2

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} )

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.

#5.1: Extend u to a function v on the full ball by antisymmetry in the x_{n} variable, as explained in the problem. The function

w(x) = ∫_{U} K(x, y)v(y) dS(y)

is then continuous in \overline{U} (v is certainly continuous on ∂U, since u = 0 on x_{n} = 0) and harmonic in U. The Poisson kernel K(x, y) is clearly symmetric in y_{n} when x_{n} = 0 (see its formula!) and therefore, when x_{n} = 0 the integrand above is antisymmetric in y_{n}. Thus w(x) = 0 when x_{n} = 0. Since w is harmonic in U^{+} with the same boundary values as u, w = u in all of U^{+}. Further, w(x) is antisymmetric in x_{n} and therefore w = v. This antisymmetry can be seen directly (note that K(x, y) in unchanged if x_{n} and y_{n} both are replaced by −x_{n} and −y_{n} respectively).

Alternatively, one can note that the function −w(x_{1}, …, x_{n−1}, −x_{n}) is harmonic with the same boundary values as w to get the antisymmetry (and in particular the vanishing of w(x) when x_{n} = 0).

Problem set #6 — for next week

#6.1: Evans problem 22 (p. 84).

#6.2: We say a continuous function u is subharmonic in an open set Ω if u(x) ≤ − ∫_{∂B(x,r)} u dS whenever B(x, r) ⊂ Ω. Note that we do not require u to be differentiable. Prove that, if u is subharmonic and v is harmonic in Ω and both functions are continuous on \overline{Ω} with u ≤ v on ∂Ω, then u ≤ v in Ω. Also prove that, if u ∈ C^{2}(Ω), then u is subharmonic in the present sense if and only if −Δu ≤ 0 in Ω.