75042 Partial differential equations

Problem set #4 — suggested solutions

#4.1: By Harnack’s inequality, if K̃ ⊂ Ω is a connected compact set then there is a constant C so that maxK̃ u ≤ C minK̃ u whenever u > 0 is harmonic in Ω.

It seems quite plausible that, given x ∈ Ω and a compact K ⊂ Ω, there exists a connected compact set K̃ ⊂ Ω with K ∪ {x} ⊂ K̃. Let C correspond to this set K̃ like in the previous paragraph. Then, if (u_{n}) is a sequence of positive harmonic functions on Ω and x ∈ Ω with u_{n}(x) → 0, we find

max_{K} u_{n} ≤ max_{K̃} u_{n} ≤ C min_{K̃} u_{n} ≤ u_{n}(x) → 0 (n → ∞)

and so u_{n}|_{K} → 0 uniformly.

It remains to prove the “plausible” statement of the previous paragraph. First, replace K by K ∪ {x}. If K has only a finite number of components, it is easy: Just make a path between each pair of components and let K̃ be the union of K and all these paths. However K may have infinitely many components. Our first task, then, is to extend K to a compact subset of Ω with finitely many components. But the open balls U(x, δ(x)) where x ∈ K and δ(x) = dist(x, ∂Ω)/2 form an open cover of K, and so by compactness a finite number of these also cover K. Taking closures, we get

K ⊂ ⋃_{i=1}^{m} B(x_{i}, δ(x_{i}))

Now, for i = 1, 2, …, m − 1 let γ_{i} be a path in Ω from x_{i} to x_{i+1} (here we use the connectedness of Ω); i.e., γ_{i}: [0, 1] → Ω is continuous with γ_{i}(0) = x_{i} and γ_{i}(1) = x_{i+1}. Now let

K̃ = ⋃_{i=1}^{m} B(x_{i}, δ(x_{i})) ∪ ⋃_{i=1}^{m−1} γ_{i}[0, 1]

which is connected, compact, and K ∪ {x} ⊂ K ⊂ Ω.

#4.2: If u is bounded and harmonic in ℝ^{n} then u + b ≥ 0 in ℝ^{n} if b is a lower bound for u. To prove that u is constant is clearly equivalent with proving u + b constant. Thus we will replace u by u + b, and therefore assume u ≥ 0 throughout.

Fix two points x and y in ℝ^{n} and set a = |x − y|. The inclusion B(y, r) ⊂ B(x, r + a) together with the mean value theorem proves

u(x) = − ∫_{B(x,r+a)} u(z) dz = \frac{1}{α(n)(r + a)^{n}} ∫_{B(x,r+a)} u(z) dz

≥ \frac{1}{α(n)(r + a)^{n}} ∫_{B(y,r)} u(z) dz = \frac{α(n)r^{n}}{α(n)(r + a)^{n}} − ∫_{B(y,r)} u(z) dz

= (\frac{r}{r + a})^{n} u(y) → u(y) as r → ∞.

Thus u(x) ≥ u(y). Exchanging x and y we get the opposite inequality, and so u(x) = u(y). Therefore u is constant.

We have only used the existence of a lower bound on u: So a harmonic function which is bounded below (or above) on ℝ^{n} is constant.

We may get the same conclusion just based on a suitable growth condition on u. Subtracting the two sides of the inequality in the above computation we have

\frac{1}{α(n)(r + a)^{n}} ∫_{B(x,r+a)\B(y,r)} u(z) dz ≥ 0

but the same conclusion will hold if the lefthand side is greater than some quantity which tends to 0 as r → ∞. This will hold if |u(z)| = o(|z|) as |z| → ∞. (Note that the domain of integration in the above integral is contained in the region {z: r − |y| ≤ |z| ≤ r + |x| − a} to see this. This result is best possible, since u(x) = x_{1} is an example of a nonconstant harmonic function.)

75042 PDE/1996: Problem set #4 Page 2

#4.3: The standard separation of variables procedure tells us to look for solutions to Δu = 0 on the form u(x, y) = X(x)Y (y) with Y (0) = Y (π) = 0. The PDE becomes X′′/X + Y′′/Y = 0 (after dividing by XY), so each term must be constant, and we get Y (y) = sin ky (k = 1, 2, …) because of the conditions on Y. Then X(x) = Ae^{kx} + Be^{−kx}, and we get unbounded solutions. The example tells us that the maximum principle may fail for unbounded domains.

#4.4: When |x| < r and |y| = r we get r − |x| ≤ |y − x| ≤ r + |x| (by the triangle inequality).

Poisson’s formula (Evans formula (45) on p. 43) gives us (and here we must rely on g ≥ 0 together with the above inequality, plus the identity r^{2} − |x|^{2} = (r − |x|)(r + |x|)):

u(x) = \frac{r^{2} − |x|^{2}}{nα(n)r} ∫_{∂B(0,r)} \frac{g(y)}{|x − y|^{n}} dS(y) ≤ \frac{r^{2} − |x|^{2}}{nα(n)r} ∫_{∂B(0,r)} \frac{g(y)}{(r − |x|)^{n}} dS(y)

= \frac{(r + |x|)nα(n)r^{n−1}}{nα(n)r(r − |x|)^{n−1}} − ∫_{∂B(0,r)} g(y) dS(y) = r^{n−2} \frac{r + |x|}{(r − |x|)^{n−1}} u(0)

which is what we wanted. The other inequality is shown similarly, using the other half of the inequality of the first paragraph.

#4.5: If u is subharmonic and v is harmonic, then u − v is subharmonic. By Evans problem 4b (p. 81, our problem #3.6) then max\overline{Ω}(u − v) ≤ max_{∂Ω}(u − v) ≤ 0, so u ≤ v in Ω.

#4.6: We take the hint, and determine constants a and b so that M(r_{i}) = a ln r_{i} + b for i = 1, 2. Then u(x, y) ≤ a ln r_{i} + b when x^{2} + y^{2} = r_{i}^{2}; that is, u ≤ v on the boundary of the annulus, where v(x, y) = a ln \sqrt{x^{2} + y^{2}} + b. By the previous exercise then u ≤ v in the entire annulus, and so M(r) ≤ a ln r + b.

It only remains to determine a and b, which are given by a pair of linear equations:

[\begin{matrix}a \\ b \end{matrix}] = [\begin{matrix}ln r_{2} & 1 \\ ln r_{1} & 1 \end{matrix}]^{−1} [\begin{matrix}M(r_{2}) \\ M(r_{1}) \end{matrix}] = \frac{1}{ln r_{2} − ln r_{1}} [\begin{matrix}1 & −1 \\ − ln r_{1} & ln r_{2} \end{matrix}] [\begin{matrix}M(r_{2}) \\ M(r_{1}) \end{matrix}]

= \frac{1}{ln(r_{2}/r_{1})} [\begin{matrix}M(r_{2}) − M(r_{1}) \\ −M(r_{2}) ln r_{1} + M(r_{1}) ln r_{2} \end{matrix}]

and inserting these values of a and b into M(r) ≤ a ln r + b we get the desired inequality.

Problem set #5 — for next week

#5.1: Evans problem 9 (pp. 81–82).

#5.2: Evans problem 11 (p. 82).

#5.3: Evans problem 13 (p. 82).

#5.4: Evans problem 14 (p. 82).

#5.5: Evans problem 16 (p. 83).