75042 Partial differential equations

Problem set #3 — suggested solutions

#3.1: The given PDE implies v̇ + cv = 0 where v(t) = u(x + tb, t). Thus v(t) = Ce^{−ct} where the constant C may depend on x. Since the initial data yield v(0) = g(x), we conclude v(t) = g(x)e^{−ct} and therefore u(x + tb, t) = g(x)e^{−ct}. Finally replace x by x − tb to obtain the result u(x, t) = g(x − tb)e^{−ct}.

#3.2: We find

| − ∫_{∂B(x,r)} f(ξ) dξ − f(x)| = | − ∫_{∂B(x,r)} (f(ξ) − f(x)) dξ| ≤ − ∫_{∂B(x,r)} |f(ξ) − f(x)| dξ < ε

provided r ≤ δ and δ > 0 is so that |f(ξ) − f(x)| < ε whenever |ξ − x| < δ.

#3.3: As in the proof on pp. 29–30, define

φ(ρ) = − ∫_{∂B(0,ρ)} u(y) dS(y)

and find

φ′(ρ) = \frac{ρ}{n} − ∫_{B(0,ρ)} Δu(y) dy = −\frac{ρ}{n} − ∫_{B(0,ρ)} f(y) dy

Now that we know φ′ we can integrate to find φ. On one hand (previous problem) φ(0) = u(0) (this is what we want to compute), and on the other hand we can find φ(r) from the given data. So it’s just a question of integrating the derivative:

u(0) = φ(0) = φ(r) − ∫_{0}^{r} φ′(ρ) dρ

= − ∫_{∂B(0,r)} g(y) dS(y) + ∫_{0}^{r} \frac{ρ}{n} \frac{1}{α(n)ρ^{n}} ∫_{B(0,ρ)} f(y) dy

Here we must interchange the order of integration: Note that we integrate over those (y, ρ) ∈ ℝ^{n}×ℝ satisfying |y| < ρ < r.

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

= − ∫_{∂B(0,r)} g(y) dS(y) − \frac{1}{n(n − 2)α(n)} ∫_{B(0,r)} f(y) (\frac{1}{r^{n−2}} − \frac{1}{|y|^{n−2}} ) dy

as desired.

#3.4: Assume u has a local maximum at a point x. Then, applying the usual single variable theory to u as a function of each variable separately, we must conclude u_{x_{i}}(x) = 0 and u_{x_{i}x_{i}}(x) ≤ 0. Adding all the latter inequalities together we get Δu(x) ≤ 0.

Just in case you thought this solution was too simple, here is a more difficult one, relying on the proof of the mean value property: Define φ(ρ) as in the solution to #3.3 above. If Δu(0) > 0 then Δu > 0 in a neigbourhood of 0, and so the integration formula for φ′ shows that φ′(ρ) > 0 for sufficiently small ρ. Therefore, for every sufficiently small ρ > 0, φ(ρ) > φ(0) = u(0). But since φ(ρ) is the average of u over ∂B(0, ρ), there must be some points y ∈ ∂B(0, ρ) with u(y) > u(0), and so 0 cannot be a local maximum (this argument works just as well with any point x replacing 0, of course).

#3.5: If u > 1 somewhere in Ω, u will achieve its maximum at some point x ∈ Ω — compactness guarantees it achieves its maximum in \overline{Ω}, but it cannot happen on the boundary because u = 0 there. At this point we get Δu(x) = u(x)^{3} − u(x) > 0 which is impossible (see #3.4 above). We can argue similarly about the case when u < −1 somewhere, but it is easier to just replace u by −u and use what we already proved, since −u satisfies the same conditions as u does.

75042 PDE/1996: Problem set #3 Page 2

#3.6: In the “more difficult” proof of #3.4 above, we have in fact proved the assertion

v(x) ≤ − ∫_{∂B(x,r)} v dS(y)

for subharmonic v (at least if you replace 0 by x and the strict inequalities by nonstrict ones.) To get the desired inequality is essentially a question of integrating in polar coordinates:

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

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

≥ \frac{v(x)}{r^{n}} ∫_{0}^{r} nρ^{n−1} dρ = v(x)

If v has a maximum point in x ∈ U, the above inequality will imply that v(y) = v(x) for all y in a neighbourhood of x: For when v(y) ≤ v(x) for y ∈ B(x, r) then instead of the above inequality we find the opposite inequality — and it becomes strict if v(y) < v(x) for some y. Then the same connectedness argument as in the proof of the strong maximum princliple shows that u is constant in the component of U containing x, and so u achieves its maximum on the boundary of that component.

To show φ(u) is subharmonic when φ is convex and u is harmonic, is simple: Just compute φ(u)_{x_{i}x_{i}} = φ′′(u)u_{x_{i}}^{2} + φ′(u)u_{x_{i}x_{i}} and add them all together to get Δφ(u) = φ′′(u)|Du|^{2} + φ′(u)Δu ≥ 0 since the first term is nonnegative and the second is zero.

Finally, if u is harmonic then so is u_{x_{i}}, and therefore u_{x_{i}}^{2} is subharmonic by the previous result. Add them all together to get |Du|^{2} subharmonic.

Problem set #4 — for next week

#4.1: Let Ω be an open connected set, and let (u_{n}) be a sequence of positive harmonic functions on Ω. Assume that u_{n}(x) → 0 for some x ∈ Ω. Show that then u → 0 uniformly on compacts [i.e., u_{n}|_{K} → 0 uniformly for each compact K ⊂ Ω].

#4.2: Use the idea from the proof of Harnack’s inequality to prove Liouville’s theorem: A bounded, harmonic function on ℝ^{n} is constant. (Hint: You may assume that u ≥ 0 (why?). Fix two points x and y in ℝ^{n} and set a = |x − y|. Use the inclusion B(y, r) ⊂ B(x, r + a) to compare u(x) with u(y) and let r → ∞.) This actually proves a stronger result than Liouville’s theorem. How much extra can you get out of this proof?

#4.3: Find a nonzero solution to the Dirichlet problem Δu = 0 in Ω, u = 0 on ∂Ω where Ω is the strip {(x, y) ∈ ℝ^{2} : 0 < y < π}. (Hint: Use separation of variables.) What does this example tell you about the maximum principle?

#4.4: Evans problem 6 (p. 81).

#4.5: Assume Ω ⊂ ℝ^{n} is open and bounded, and assume u, v ∈ C(\overline{Ω}) with u subharmonic and v harmonic in Ω. If u ≤ v on ∂Ω, show that u ≤ v in Ω (this is the reason for the term subharmonic).

#4.6: Prove the Hadamard “Three Circles” theorem: Let u(x, y) be subharmonic in a domain containing the annulus r_{1}^{2} ≤ x^{2} + y^{2} ≤ r_{2}^{2}. Then the maximum

M(r) = max_{x^{2}+y^{2}=r^{2}} u(x, y)

of u on any concentric circle satisifies

M(r) ≤ \frac{M(r_{1}) ln \frac{r_{2}}{r} + M(r_{2}) ln \frac{r}{r_{1}}}{ln \frac{r_{2}}{r_{1}}}

where r_{1} < r < r_{2}. (Hint: Any function of the form a ln r + b is harmonic in ℝ^{2}. Compare using suitable choices of a and b.)

(Remark. In the corresponding “Three Spheres” theorem in ℝ^{n} the quantity ln(b/a) is replaced by a^{2−n} − b^{2−n}.)