ELLIPTIC PDE LECTURE 1; THE LAPLACE EQUATION.

1. Types of PDE

There are 3 key partial differential equations, that typify to a large extent the main areas of PDE:

Elliptic: Laplace equation: 4u =∑nj=1

∂2

∂x2iu = 0.

Parabolic: Heat equation: ∂∂tu = 4u.

Hyperbolic: Wave equation: ∂2

∂t2u = 4u.

This is a course on elliptic PDE, and hence we will start out by studying the Laplace equation, as well asthe closely related Poisson equation:

4u = f.

Typically one studies these equations on bounded domains U ⊂ Rn, n ≥ 2, with boundary ∂U . The Dirichletproblem asks for finding (suitably regular) u which solves either the Laplace or the Poisson equation andsuch that

u|∂U = g

is a prescribed function, usually assumed continuous, i. e. g ∈ C0(∂U).

2. Fundamental solution of the Laplace equation on domains in Rn; the case n = 2

To begin with, there is a close relation between the Laplace equation on R2 and complex analysis. Recallthat a differentiable function1

f(z) := u(z) + iv(z) : C→ C

is called complex analytic or holomorphic, provided it satisfies the Cauchy-Riemann equations:

∂u

∂x=∂v

∂y,∂u

∂y= −∂v

∂x

It is well-known that such functions are automatically C∞ (in fact, analytic, i. e. can be expanded locally inconvergent power series), and moreover, one easily infers

4u = 4v = 0.

Thus both the real and imaginary parts are solutions of the Laplace equation. Conversely, given a real-valuedsolution u of the Laplace equation on R2, one can find v (unique up to a constant), such that

u+ iv

is a holomorphic function on C. (exercise: show this)

We can use this correspondence to find some basic non-trivial solutions of the Laplace equation on R2, orsubdomains thereof. For example, it is well-known that if one cuts C along the half-axis {(0, x)|x ≤ 0}, thenone can define a branch of the function

log z

on the remaining set. Using polar coordinates (r, θ) on C, one can in fact write

log z = log r + iθ, z = reiθ

1We identify C with R2

1

2 ELLIPTIC PDE LECTURE 1; THE LAPLACE EQUATION.

Then according to the preceding, the function

u(x, y) = log r = log(√x2 + y2)

is a solution of Laplace’s equation on C\{(0, x)|x ≤ 0}, and in fact it is a solution on R2\{(0, 0)}. For reasonsthat will become clear later, one calls the normalised function

φ(x, y) := − 1

2πlog(

√x2 + y2)

the fundamental solution of Laplace’s equation on R2 (strictly speaking, one has to omit the origin, as thefunction blows up there). Observe the remarkable fact that φ(x, y) is a radial function, i. e. it only depends

on r =√x2 + y2, and not the angular variable θ.

3. Fundamental solution of the Laplace equation on domains in Rn; the case n > 2

By analogy to the preceding section we can try to discover radial solutions on Rn, n > 2, which may alsobe singular at the origin. Thus we attempt to find

u(x) = φ(r), r =

√√√√ n∑j=1

x2j ,

solving the Laplace equation 4u = 0 on Rn. This is done by direct computation:

∂

∂xj[φ(r)] =

xjrφ′(r),

∂2

∂x2j[φ(r)] = φ′′(r)

x2jr2

+1

rφ′(r)− φ′(r)

x2jr3

Summing over j, we find the condition

φ′′(r) +n− 1

rφ′(r) = 0

which is an scalar ODE, and is equivalent to

d

dr[rn−1φ′(r)] = 0

For n > 2, this is solved by

φ(r) =C1

rn−2+ C2.

The normalised function

φ(x) =1

ω(n)(n− 2)r2−n, n > 2,

with ω(n) the surface area of the (n − 1)-dimensional sphere in Rn, is called fundamental solution of theLaplace equation there. Again its significance will be explained later.

4. Mean Value property

Solutions of the Laplace equation feature two extremely important properties, which are largely responsiblefor the solution theory of the Laplace equation, and the second of which also for elliptic PDE in general. Theseare the mean value property and the maximum principle. Here we explain the former, which is largelya consequence of the

Proposition 4.1. (Divergence Theorem) Let V(x) : Rn → Rn be a vector field of class C1(Rn), U ⊂ Rn abounded open set with C1-boundary2 ∂U . Then denoting the outward unit normal vector at a point p ∈ ∂Uby n(p), we have the relation ∫

U

divV dx =

∫U

n∑j=1

∂

∂xjVj dx =

∫∂U

V · n dσ

Here dσ denotes the standard surface measure on ∂U induced by Rn.

2This means that the boundary can be locally written as the graph of a C1-function

ELLIPTIC PDE LECTURE 1; THE LAPLACE EQUATION. 3

Now we can formulate the fundamental

Theorem 4.2. (Mean Value property) Let U ⊂ Rn be an open set, u ∈ C2(U) satisfy Laplace’s equation

4u = 0, and Br(x0) ⊂ U , r > 0, a closed ball in U . Then we have

u(x0) =1∣∣∂Br(x0)

∣∣ ∫∂Br(x0)

u(y)dσy =1∣∣Br(x0)

∣∣ ∫Br(x0)

u(x) dx

Proof. For r > 0 introduce the function

ψx0(r) :=

1∣∣∂Br(x0)∣∣ ∫

∂Br(x0)

u(y)dσy =1

rn−1ω(n)

∫Sn−1

u(x0 + rω)rn−1dω

=1

ω(n)

∫Sn−1

u(x0 + rω)dω

where now dω denotes the standard surface measure on the sphere Sn−1. Then we compute

ψ′x0(r) =

1

ω(n)

∫Sn−1

∂

∂r[u(x0 + rω)]dω

=1

ω(n)

∫Sn−1

(ω · ∇xu)(x0 + rω)]dω

=1

rn−1ω(n)

∫∂Br(x0)

n · ∇u dσ

=1

rn−1ω(n)

∫Br(x0)

4u(x) dx

= 0,

where in the last step but one we have used the divergence theorem.

It follows that ψx0(r) is in effect independent of r, and can be determined by letting r −→ 0. But the

continuity of u implies thatlimr→0

ψx0(r) = u(x0).

The proof of the Mean-Value property is concluded by observing that upon integration over r, we have∫ r

0

( ∫∂Br̃(x0)

u(y)dσy)dr̃ =

∫Br(x0)

u(x) dx,

from which1∣∣∂Br(x0)

∣∣ ∫∂Br(x0)

u(y)dσy =1∣∣Br(x0)

∣∣ ∫Br(x0)

u(x) dx.

�

5. Maximum principle

We now infer from the mean value property the all important maximum principle. We note that there arevariants of this which are much more broadly applicable to equations of elliptic and parabolic type, but notto hyperbolic equations, which are in some sense the most difficult. We have the following version

Theorem 5.1. (Maximum principle) Let U ⊂ Rn an open bounded set, and u ⊂ C2(U) ∩ C0(U). Also,assume that U is connected, and that

4u = 0.

Then we havemaxx∈U

u(x) = maxx∈∂U

u(x)

In fact, we haveu(x) < max

y∈∂Uu(y)∀x ∈ U

4 ELLIPTIC PDE LECTURE 1; THE LAPLACE EQUATION.

unless u = const is a constant function.

Corollary 5.2. For U, u as above, we have

maxx∈U|u(x)| = max

x∈∂U|u(x)|

Thus the size of u on the boundary precisely controls the size of u inside the region.

The corollary follows by considering u,−u, and applying the theorem.

Proof. (Theorem) It suffices to show that if u attains its maximum at some x0 ∈ U , then u is a constantfunction. Introduce the set

A := {x ∈ U |u(x) = maxy∈U

u(y}

By assumption we have x0 ∈ A, so this set is non-empty. We show that it is closed and open, whence byconnectedness of U we have

A = U

and so u has to be constant. First, by continuity of u, we know that A is closed. So it remains to show it isopen, which follows from the mean-value property. Indeed, if y0 ∈ A, pick r > 0 sufficiently small, such that

Br(y0) ⊂ UThen we have

u(y0) =1∣∣Br(y0)

∣∣ ∫Br(y0)

u(y) dy ≤ maxz∈U

u(z},

with the last inequality being strict unless we have equality

u(y) = maxz∈U

u(z}

for each y ∈ Br(y0). But then Br(y0) ⊂ A, whence the latter set is also open. �

We note here that the maximum principle will be a crucial ingredient later on in the general solution ofthe Dirichlet problem for the Laplace equation along the lines of the Perron method.

Here we mention the following consequence of the maximum principle, which in fact justifies the Dirichletproblem, formulated below

Theorem 5.3. Let u, v ∈ C2(U) ∩ C0(U), with U ⊂ Rn open, bounded. Also, assume that

4u = 4v = 0

as well asu|∂U = v|∂U

Then we haveu = v

The proof is immediate by applying the corollary above to the difference u− v.

According to the preceding theorem, given the boundary data u|∂U = f , there can be at most one solutionof the Laplace equation 4u = 0 with these data. It makes thus sense to formulate the

Definition 5.4. (Dirichlet problem) The Dirichlet problem for the Laplace equation consists in finding forgiven f ∈ C0(∂U) a function u ∈ C2(U) ∩ C0(U) such that

4u = 0

on U , and we haveu|∂U = f

Note that the boundary restriction u|∂U makes sense since we assume u ∈ C0(U)