Theory of PDE Homework 2

Adrienne Sands

March 22, 2017

1. Let Φ be the fundamental solution of the Laplace equation in Rn, n ≥ 2. Prove that −∆Φ = δ0

in the sense of distributions, that is for each ψ ∈ C∞c (Rn) (smooth functions with compactsupport), one has

−ˆRn

Φ(x)∆ψ(x)dx = ψ(0)

Proof. Since Φ blows up at 0, we isolate the singularity inside a ball of fixed radius ε > 0.ˆRn

Φ(x)∆ψ(x) dx =

ˆRn−B(0,ε)

Φ(x)∆ψ(x) dx+

ˆB(0,ε)

Φ(x)∆ψ(x) dx

=: Iε + Jε

We claim Iε → −ψ(0) and Jε → 0 as ε→ 0+. First, apply Green’s formula to Iε.

Iε =

ˆRn−B(0,ε)

∆Φ(x)ψ(x) dx+

ˆ∂(Rn−B(0,ε))

Φ(x)∂ψ(x)

∂νdS −

ˆ∂(Rn−B(0,ε))

ψ(x)∂Φ(x)

∂νdS

The first term vanishes since ∆Φ(x) = 0 away from 0, and we need only consider the boundary∂B(0, ε) for ψ compactly supported. Thus,

Iε =

ˆ∂B(0,ε)

Φ(x)∂ψ(x)

∂νdS −

ˆ∂B(0,ε)

ψ(x)∂Φ(x)

∂νdS

=: I1 + I2

We claim I1 → 0 and I2 → −ψ(0) as ε→ 0+. By definition of the fundamental solution,

ˆ∂B(0,ε)

|Φ(x)| dS =

1

2π

´∂B(0,ε) |log |x|| dS = | log ε|

2π

´∂B(0,ε) dS = ε| log ε| (n = 2)

1n(n−2)α(n)

´∂B(0,ε) |x|

2−ndS = ε2−n

n(n−2)α(n)

´∂B(0,ε) dS = ε

n−2 (n ≥ 3)

For ψ continuous and compactly supported,

|I1| =

∣∣∣∣∣ˆ∂B(0,ε)

Φ(x)∂ψ(x)

∂νdS

∣∣∣∣∣ ≤ ‖Dψ‖L∞ˆ∂B(0,ε)

|Φ(x)| dS ≤

{Cε| log ε| (n = 2)

Cε (n ≥ 3)→ 0

Next we compute the (inward) normal derivative∂Φ

∂νat x ∈ ∂B(0, ε)

∂Φ

∂ν= DΦ(x) · ν =

−xnα(n)εn

· −xε

=|x|2

nα(n)εn+1=

1

nα(n)εn−1

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Math 8583 - Sands

By continuity of ψ(x),

I2 = −ˆ∂B(0,ε)

ψ(x)∂Φ

∂νdS =

−1

nα(n)εn−1

ˆ∂B(0,ε)

ψ(x)dS = −−ˆ∂B(0,ε)

ψ(x)dS → −ψ(0)

Consider the final term Jε. For ε sufficiently close to 0 and n = 2, we have the following:∣∣∣∣∣ˆB(0,ε)

Φ(x)dx

∣∣∣∣∣ =1

2π

∣∣∣∣∣ˆB(0,ε)

log |x|dx

∣∣∣∣∣ =1

2π

∣∣∣∣ˆ 2π

0

ˆ ε

0r log rdrdθ

∣∣∣∣=

∣∣∣∣ˆ ε

0r log rdr

∣∣∣∣ =

∣∣∣∣2ε2 log ε− ε2

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∣∣∣∣ ≤ ε2| log ε|2

By a similar computation for n ≥ 3,∣∣∣∣∣ˆB(0,ε)

Φ(x)dx

∣∣∣∣∣ =1

n(n− 2)α(n)

∣∣∣∣∣ˆB(0,ε)

|x|2−ndx

∣∣∣∣∣=

1

n(n− 2)α(n)

∣∣∣∣∣ˆ ε

0r2−ndr

ˆ∂B(0,r)

dS

∣∣∣∣∣=

1

n− 2

∣∣∣∣ˆ ε

0rdr

∣∣∣∣ =ε2

2(n− 2)

Thus,

|Jε| =

∣∣∣∣∣ˆB(0,ε)

Φ(x)∆ψ(x) dx

∣∣∣∣∣ ≤ ‖∆Φ(x)‖L∞

∣∣∣∣∣ˆB(0,ε)

Φ(x)dx

∣∣∣∣∣ ≤{Cε2| log ε| (n = 2)

Cε2 (n ≥ 3)→ 0

Therefore as ε→ 0+

ˆRn

Φ(x)∆ψ(x) dx = Iε + Jε → 0− ψ(0) + 0 = −ψ(0)

2. Let U = {(x1, x2) ∈ R2 : −1 < x2 < 1}.

(a) Prove that there is no nontrivial bounded solution of the boundary value problem

∆u = 0 in U

u = 0 on ∂U

Remark: Here we are talking about classical solutions, that is, solutions in C2(U)∩C(U)Hint: One way to prove this is by using a reflection principle as in [Evans, 2.5/10]

Proof. Suppose u is a bounded solution to this boundary value problem, and let

v(x1, x2) =

{u(x1, x2) |x2| ≤ 1

−u(x1, 2− x2) 1 < x2 < 2

By construction, u(x1, 1) = −u(x1, 1) = 0, so v ∈ C(V ). We claim v is a harmonicextension of u to the larger strip V = {(x1, x2) : −1 ≤ x2 < 2}, and it suffices to show

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v satisfies the mean value property in V . Since u is harmonic in U , v satisfies the meanvalue property for |x2| < 1 and 1 < x2 < 2. Let B+ and B− denote the upper and lowerhalves of the ball. For x = (x1, 1) arbitrary, we have

−ˆB(x,r)

v(y)dy = −ˆB−(x,r)

u(y1, y2)dy −−ˆB+(x,r)

u(y1, 2− y2)dy

= −ˆB−(x,r)

u(y1, y2)dy −−ˆB−(x,r)

u(y1, y2)dy = 0 = v(x)

Thus v is harmonic in V . Reflecting infinitely to larger strips, we can extend u to a(bounded) harmonic function v on R2 which is 0 on the boundary of each strip. ByLiouville v is trivial, and since u = v

∣∣U

by construction, u is trivial.

(b) Is there a nontrivial solution of this problem if the boundedness requirement is removed?

Proof. Consider u(x, y) = eπx sinπy on U . By an easy computation,

∆u =∂2u

∂x2+∂2u

∂y2= π2eπx sinπy − π2eπx sinπy = 0

Since u is harmonic and u(x,±1) = 0, u solves the boundary value problem.

3. The Kelvin transform of a function u on Rn is defined by

Ku(x) = |x|2−nu(

x

|x|2

)(a) Prove that if u is harmonic in B(0, 1), then its Kelvin transform is harmonic in Rn −

B(0, 1) (cp. [Evans, 2.5/11]).

Proof. To simplify the proof, we use polar coordinates such that

u = u(r, θ) and ∆ =

(∂2

∂r2+n− 1

r

∂

∂r+

1

r2∆Sn−1

)Since x 7→ x

|x|2 is conformal and scales r to 1r , we write the Kelvin transform as follows:

Ku(r, θ) = r2−nu

(1

r, θ

)= rn−2u(y), y =

1

r

(reasonably) suppressing θ for convenience. By the chain rule,

n− 1

r

∂(Ku)

∂r= (2− n)(n− 1)r−nu(y)− (n− 1)y−n−1u′(y)

∂2(Ku)

∂r2= (2− n)(1− n)r−nu(y) + 2(n− 1)r−n−1u′(y) + r−n−2u′′(y)

Thus,

∆(Ku) =∂2(Ku)

∂r2+n− 1

r

∂(Ku)

∂r+ ∆Sn−1(Ku)

= r−n−2(u′′(y) + r(n− 1)u′(y) + r2∆Sn−1u(y)

)= yn+2

(u′′(y) +

(n− 1)

yu′(y) +

1

y2∆Sn−1u(y)

)= 0

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Math 8583 - Sands

(b) Using this, find a solution of the exterior problem

∆u = 0, in Rn −B(0, 1)

u = g, on ∂B(0, 1)

where g is a continuous function on ∂B(0, 1).

Proof. By Poisson’s formula for the ball,

v(x) =1− |x|2

nα(n)

ˆ∂B(0,1)

g(y)

|x− y|ndS(y)

is harmonic in B(0, 1) and satisfies v = g on ∂B(0, 1). We claim u(x) = Kv(x) solvesthe exterior problem. By part (a) u is harmonic on Rn −B(0, 1), and for x ∈ ∂B(0, 1),

Kv(x) = |x|2−nu(x

|x|2) = u(x) = g(x)

4. [Evans 2.5/5] We say v ∈ C2(U) is subharmonic if

−∆v ≤ 0 in U

(a) Prove for subharmonic v that

v(x) ≤ −ˆB(x,r)

v(y) dy for all B(x, r) ⊂ U

Proof. We adapt the proof of the mean value formula for harmonic functions. Define

φ(r) = −ˆ∂B(x,r)

v(y)dS(y) = −ˆ∂B(0,1)

v(x+ rz)dS(z)

We claim φ(r) is monotonically increasing. Differentiating,

φ′(r) = −ˆ∂B(0,1)

z ·Dv(x+ rz)dS(z) = −ˆ∂B(x,r)

y − xr·Dv(y)dS(y)

For y ∈ ∂B(x, r), the (outward) unit normal vector is ν = y−xr , so we are integrating

the directional derivative ν ·Dv(y) = ∂v(y)∂ν . By Green’s formula,

φ′(r) = −ˆ∂B(x,r)

∂v(y)

∂νdS(y) =

n

r

1

α(n)rn

ˆ∂B(x,r)

∂v(y)

∂νdS(y)

=n

r

1

α(n)rn

ˆB(x,r)

∆v dy =n

r−ˆB(x,r)

∆v dy ≥ 0

for v subharmonic. Since φ(r) is monotonically increasing,

v(x) = limr→0+

φ(r) ≤ φ(r) = −ˆ∂B(x,r)

v(y) dS(y) =1

nα(n)rn−1

ˆ∂B(x,r)

v(y) dS(y)

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Polar coordinates yield our result:

−ˆB(x,r)

v(y)dy =1

α(n)rn

ˆB(x,r)

v(y)dy =1

α(n)rn

ˆ r

0

ˆ∂B(x,ρ)

v(y)dS(y)dρ

≥ 1

α(n)rn

ˆ r

0nα(n)ρn−1v(x)dρ =

v(x)

rn

ˆ r

0nρn−1dρ = v(x)

(b) Prove that therefore maxU v = max∂U v.

Proof. Since v ∈ C2(U), v achieves a maximum on U . Suppose there is some x0 ∈ Usuch that v(x0) = max

Uv = M . For all 0 < r < dist(x0, ∂U), part (a) asserts

M = v(x0) ≤ −ˆB(x0,r)

v(y)dy ≤ −ˆB(x0,r)

Mdy = M

By equality of the terms above, v ≡ M in every B(x0, r) ⊂ U and {x ∈ U : v(x) = M}is relatively open in U . For continuous v, this set is also relatively closed in U andtherefore equal to U by connectedness (i.e. v ≡M in U). Finally, take any x ∈ ∂U andsequence {xk} ⊂ U converging to x. By continuity of v,

v(x) = limxk→x

v(xk) = limxk→x

M = M

Thus, v ≡M in U and maxU

v = max∂U

.

(c) Let φ : R→ R be smooth and convex. Assume u is harmonic and v := φ(u). Prove v issubharmonic.

Proof. Since φ′′(u) ≥ 0 for convex φ, our result follows by direct computation:

D2v = φ′′(u)|Du|2 + φ′(u)D2u

∆v = tr(D2v) = φ′′(u)|Du|2 + φ′(u)∆u

= φ′′(u)|Du|2 ≥ 0

(d) Prove v := |Du|2 is subharmonic, whenever u is harmonic.

Proof. Since u is harmonic, uxi is harmonic. Thus φ(uxi) = (uxi)2 is subharmonic by

(c) and |Du|2 =∑i≤n

(uxi)2 is subharmonic as a sum of subharmonic functions.

5. [Evans 2.5/7] Use Poisson’s formula for the ball to prove

rn−2 r − |x|(r + |x|)n−1

u(0) ≤ u(x) ≤ rn−2 r + |x|(r − |x|)n−1

u(0)

whenever u is positive and harmonic in B◦(0, r). This is an explicit form of Harnack’s in-equality.

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Math 8583 - Sands

Proof. Let x ∈ B◦(0, r). We freely use Poisson’s formula for the ball:

u(x) =r2 − |x|2

nα(n)r

ˆ∂B(0,r)

u(y)

|x− y|ndS(y) = rn−2(r2 − |x|2)−

ˆ∂B(0,r)

u(y)

|x− y|ndS(y)

By the triangle inequality (with |x| < |y| = r and n ≥ 1),

(r − |x|)n ≤ |x− y|n ≤ (r + |x|)n ⇐⇒ 1

(r + |x|)n≤ 1

|x− y|n≤ 1

(r − |x|)n

Notice1

(r ± |x|)nare independent of y. By Poisson’s formula and the mean value formula for

u harmonic, we have our result:

rn−2 r2 − |x|2

(r + |x|)n−ˆ∂B(0,r)

u(y)dS(y) ≤ u(x) ≤ rn−2 r2 − |x|2

(r − |x|)n−ˆ∂B(0,r)

u(y)dS(y)

m

rn−2 r − |x|(r + |x|)n−1

u(0) ≤ u(x) ≤ rn−2 r + |x|(r − |x|)n−1

u(0)

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