Laplace’s Equations and Harmonic Functions

Fundamental Solutions of Harmonic Functions

Definition Let $u(x_1,\cdots,x_n): U \subset \mathbb{R}^n \to \mathbb{R}$ be a unknown function, the Laplacian of the function is defined by \(\Delta u = \sum_{i=1}^n \frac{\partial^2 u}{\partial x_i^2}.\) Laplace equation is of the form \(\Delta u = 0\) for all $x \in U$, any function satisfying the above is also called harmonic.

Lemma Laplace’s equation is invariant under rotation, i.e if $P$ is an orthogonal matrix with entries $p_{ij}$, then $\Delta u(Px) = 0$ for all $x\in U$.

Proof By Chain rule, we have

and similarly

Since $P$ is orthogonal, thus $p_{ik} p_{il} = 1$ iff $k=l$ and hence

QED.

Given this, we seek for radial solutions which takes the form $u(x) = v(r)$ where we denote $r = (x_1^2+\cdots+x_n^2)^{1/2}$, let $x\neq 0$, then we see that

and by Chain rule we have

using the fact that $x_1^2+\cdots+x_n^2 = r^2$, we would have

by letting $\Delta u = 0$, we obtain the following ODE:

Now if $v’(r) \neq 0$, we see that

and we integrate with respect to $r$, we have

which simplifies to $v’(r) = C_0 / r^{n-1}$ for some constant $C_0$, hence the general solution $v(r)$ takes the form

where $A,B,C,D$ are all constants. This motivates the fundamental solution to Laplace’s equation:

Definition The function defined by

is the fundamental solution to Laplace’s equation $\Delta u = 0$ for all $x \in \mathbb{R}^n / { 0 }$ where $\alpha(n)$ denotes the volume of the unit ball in $\mathbb{R}^n$.

Laplace’s equation is a special class of Poisson’s equations, we define Poisson’s equation as

Definition For a given $f: \mathbb{R}^n \to \mathbb{R}$, and the unknown $u(x_1,\cdots,x_n) : U \subset \mathbb{R}^n \to \mathbb{R}$, the Poisson’s equation takes the form

By construction, we know $\Phi(x)$ is harmonic since it solves $\Delta u = 0$. We shift the origin to a new point $y$, and under invariant we still have $\Delta u = 0$ and now $\Phi(x-y)$ is harmonic in $x$. Consider a given function $f: \mathbb{R}^n \to \mathbb{R}$ and that $\Phi(x-y) f(y)$ is also harmonic when $x\neq y$, by taking all possible $y$, it suggests the following convolution integral:

We may want to compute $\Delta u$ in this case but we see that it is hard to deal with at when $x=y$, the next theorem will formally prove $u(x)$ is indeed the fundamental solution to the Poisson’s equation:

Theorem Define $\displaystyle{u(x) = \int_{\mathbb{R}^n} \Phi(x-y) f(y) dy}$ and assume that $f \in C_C^2(\mathbb{R}^n)$, then:

1. $ u \in C^2(\mathbb{R}^n)$;\

2. $-\Delta u = f$ in $\mathbb{R}^n$.

We will prove both results separately:

For 1:

Proof:

Use the property of a convolution, we have \(u(x) = \int_{\mathbb{R}^n} \Phi(y) f(x-y) dy\)

then denote $e_i$ as the unit vector in $i$th position of $x = (x_1,\cdots,x_n)$, and abusing the notion of interchanging limits and integrals (we actually could do that), we have

Similarly one can show

Since $f \in C_C^2(\mathbb{R}^n)$, the second partial derivative is continuous w.r.t $x$, hence $u \in C^2(\mathbb{R}^n)$.

QED.

For 2:

Proof

We split the integral into two parts to avoid the blow-up of $\Phi$ near $0$, $\forall \epsilon>0$ fixed, denote $B(0,\epsilon)$ as the ball centerted at $0$ with radius $\epsilon$, write

hence

Denote the first term as $I_{\epsilon}$, second term as $J_{\epsilon}$. Our first goal is to bound $I_{\epsilon}$, proceed as follows:

When $n=2$, we have

for some constant $C$, and we further have

so we have

for some constant $C$ . When $n\geq 3$, similarly we can show that

hence the term $I_{\epsilon}$ is bounded in terms of arbitrary $\epsilon$. Now for $J_{\epsilon}$, first note that $\Delta_x f(x-y) = \Delta_y f(x-y)$, we have

To compute this integral, we need some technical lemmas:

Lemma (Gauss-Green Theorem) Let $u(x_1,\cdots,x_n) \in C^1(\overline{U})$, let $\nu = (\nu_1,\cdots,\nu_n)$ be the outward pointing unit normal vector field of $U$ defined on $\partial U$ then

We can use this lemma to further derive an integration by parts formula:

Lemma (Integration by Parts) Let $u,v \in C^1(\overline{U})$, then

Then we may rewrite $J_\epsilon$ as

where $\nu$ is the inward pointing unit normal vector field. We now denote the first term by $K_\epsilon$ and the second term by $L_\epsilon$, then using the same technique as we did for $I_\epsilon$, we can check that

hence $L_{\epsilon}$ is also bounded in terms of $\epsilon$. We again perform integration by parts in $K_{\epsilon}$, and we have

Since

when

on $\partial B(0,\epsilon)$, so we will have

also $n\alpha(n)\epsilon^{n-1}$ is the surface area of the ball $\partial B(0,\epsilon)$, so

as $\epsilon \to 0$. Since we have shown that $I_\epsilon, L_\epsilon$ are all bounded by terms of $\epsilon$ so by setting $\epsilon \to 0$ they will vanish as well. Hence we have

QED.

Properties of Harmonic Functions

Mean Value Property

For a harmonic function $u$, its value $u(x)$ is equal to the average $u$ over both the boundary $\partial B(x,r)$ and the ball $B(x,r)$ as long as $B(x,r) \in U$.

Theorem Let $U \in \mathbb{R}^n$ open, if $u(x_1,\cdots,x_n) \in C^2(U)$ is harmonic then for each ball $B(x,r) \in U$,

where

$\alpha(n)$ is the volume of the unit ball in $\mathbb{R}^n$ and $n\alpha(n)$ is the surface area of the unit ball in $\mathbb{R}^n$.

Proof Let $u(x_1,\cdots,x_n) \in C^2(U)$ be harmonic, and define

If $u$ is a smooth function (we later will show all harmonic functions are smooth), then $\lim_{r \to 0} \phi(r) = u(x)$ and hence $\phi$ is continuous. Note that by change of variables, we have

and we take the derivative of $\phi(r)$:

which means $\phi$ is a constant function, and hence we have

Furthermore we have

QED.

Another result is the converse to mean-value property:

Theorem If $u \in C^2(U)$ satisfies

for each ball $B(x,r) \in U$, then $u$ is harmonic.

Proof Suppose $\Delta u \neq 0$, then $\exists B(x,r) \in U$ such that $\Delta u > 0$ say, but then

\(\phi'(r) = \frac{1}{n\alpha(n)r^{n-1}} \int_{\partial B(x,r)}\Delta u dS = \frac{r}{n}⨍_{B(x,r)} \Delta u(y) dy >0\) which contradicts the fact that $\phi$ is a constant function.

QED.

Maximum Principle

A nice property of harmonic functions is that if $u$ is harmonic on a bounded domain $\Omega$, then $u$ attains its maximum value on the boundary of $\Omega$.

Theorem (Strong Maximum Principle) Suppose $u \in C^2(U) \cap C(\overline{U})$ is harmonic within $U$, then

1. $\max_{\overline{U}} u = \max_{\partial U} u$;\

2. Furthermore, if $U$ is connected and there exists a point $x_0 \in U$ such that $u(x_0) = \max_{\overline{U}} u$, then $u$ is constant within $U$.

Proof We will prove the second statement since the first followed from the second. Suppose $\exists x_0 \in U$ such that $u(x_0) = \max_{\bar{U}} u(x) \equiv M$. Then $\forall r \in (0,\text{dist}(x_0,\partial U))$, the mean-value property states that

by taking the ‘‘average’’ around the maximum $x_0$, we would except

and such will hold only if $u \equiv M$ within $B(x_0,r)$, and $u(y) = M$.

QED.

An important application of maximum principle is establishing the uniqueness of solutions to certain boundary value problems for Poisson’s equation.

Theorem Let $g \in C(\partial U), f\in C(U)$, then there exists at most one solution $u \in C^2(U) \cap C(\bar{U})$ of the boundary value problem

Proof Assume $u_1, u_2$ both solves the boundary value problem, then let $w = u_1 - u_2$, then we have \(\begin{cases} -\Delta w = 0 &\text{ in $U$} \\ w = 0 &\text{ in $\partial U$} \end{cases}\)

then by maximum principle, $\max_{\partial U} w = \max_{\partial \bar{U}} w = 0$, i.e $u_1=u_2$.

QED.

Smoothness

A nice thing about harmonic functions is that they are smooth!

Theorem If $u \in C(U)$ satisfies the mean-value property for each ball $B(x,r) \in U$, then $u \in C^\infty(U)$.

The proof of this theorem requires some knowledge on mollifiers, so before the proof we shall spend some time on mollifiers.

Definition Define a smooth function $\eta \in C^\infty(\mathbb{R}^n)$ by

for some constant $C>0$ so that $\int_{\mathbb{R}^n} \eta dx=1$. Then for each $\epsilon>0$, set

We call $\eta$ the standard mollifier, the functions $\eta_\epsilon$ are $C^\infty$ as well.

Definition Denote $U_\epsilon = { x \in U | \text{ dist}(x,\partial U) > \epsilon }$, if a function $f: U \to \mathbb{R}$ is locally integrable, its mollification is defined by

for $x\in U_\epsilon$.

Now we prove the theorem:

Proof Let $\eta$ be the standard mollifier, set $u^\epsilon = \eta_\epsilon * u $ in $U_\epsilon$, we first show $u^\epsilon \in C^\infty$, which is left as an exercise to the reader. Then to show $u$ is smooth, we will show in fact $u \equiv u^\epsilon$ on $U_\epsilon$ for each $\epsilon>0$. Let $x \in U_\epsilon$, then we know the support of $\eta_\epsilon(x-y)$ as a function of $y$ is given by ${ y: |x-y| < \epsilon }$, and trhus we have

so $u \in C^\infty(U_\epsilon)$ for all $\epsilon>0$.

QED.

Liouville’s Theorem

Theorem If $u : \mathbb{R}^n \to \mathbb{R}$ is harmonic and bounded (either from above or below), then $u$ is constant.

To prove Liouville’s theorem, we first introduce a technical lemma for estimating the derivatives of harmonic functions:

Lemma Assume $u$ is harmonic in $U$, then for multi-index $\alpha = 1$,

\(| D^\alpha u(x_0) | \leq \frac{n}{\alpha(n)} \cdot\left( \frac{2}{r} \right)^{n+1} \cdot||u||_{L^1(B(x_0,r))}.\) for each ball $B(x_0,r) \in U$.

Proof Let $x_0 \in U$ and $B(x,r) \subset U$. Since $u_{x_i}$ is also harmonic, then use mean-value property we have

Now for $u(x_0)$, note that

Hence we have

QED.

We now prove Liouville’s theorem:

Proof Fix $x_0 \in \mathbb{R}^n$ and $r>0$, we have \(|Du(x_0)| \leq \frac{n}{\alpha(n)} \left( \frac{2}{r} \right)^{n+1} \cdot || u || \leq \frac{n}{\alpha(n)} \left( \frac{2}{r} \right)^{n+1} \cdot ||u||_{L^\infty(\mathbb{R}^n)} \overset{r \to \infty}{\to} 0.\) which means $u$ is a constant.

QED.

Theorem Let $f \in C_C^2(\mathbb{R}^n)$ and $n \geq 3$, then any bounded solution of $-\Delta u =f$ in $\mathbb{R}^n$ has the form \(u(x) = \int_{\mathbb{R}^n} \Phi(x-y) f(y) dy + C\) for some constant $C$.

Proof If $u’$ is another solution, then $u - u’$ is a constant by Liouville’s theorem.

QED.

Harnack’s Inequality

We write $V\subset\subset U$ to denote $V \subset \bar{V} \subset U$ and $\bar{V}$ is compact, we say $V$ is compactly contained in $U$.

Theorem For each connected open set $V \subset\subset U$, there exists a positive constant $C$ depending only on $V$ such that

for all non-negative harmonic functions $u$ in $U$. In particular

for all $x,y \in V$.

Proof Consider a ball $B(a,r) \subset U$, and let $x,y \in B(a,\frac{1}{4}r)$, we first claim that $B(x,\frac{1}{4}r) \subset B(y,\frac{3}{4}r) \subset B(a,r)$. Then we use mean value property and we have

Now since $V$ is compact so there exists a finite cover of balls ${ B_i }, i = 1,\cdots, M$ where $B_i \equiv B(a_i,r_i) \subset U$. Then for any $x,y \in V$, we may find an ordered subcover $B_{i1},\cdots B_{i\ell}$ such that $x \in B_{i1}$, $y \in B_{i\ell}$ and $B_{ij} \cap B_{ij+1} \neq \varnothing$, and the previous conclusion holds. Then we have a chain of inequalities:

\(u(x) \leq 3^n u(x_1) \leq \cdots \leq 3^{n\ell} u(y) \leq 3^{nM} u(y)\) for all $x,y \in V$, and hence we have \(\sup_V u \leq C \inf_V u\) where $C$ is a constant only depend on $V$.

QED.

Convergence Theorems

In this section we shall discuss some convergence theorems for harmonic functions:

Theorem (Weierstrass Convergence Theorem) If a sequence of harmonic functions ${ h_n }$ converges locally uniformly on $U$, then the limit $h = \lim_{n\to\infty} h_n$ is also harmonic on $U$, furthermore the derivatives converge locally uniformly as well, i.e $\lim_{n\to\infty} D_{x_i} h_n = D_{x_i} h$.

Proof Let $K \subset U$ be compact, we choose a larger compact set $\tilde{K}$ and $r:= \text{dist}(K,\mathbb{R}^n/\tilde{K}) >0$, we use the derivative estimate to get

and from the uniform convergence of ${ h_n }$ on $\tilde{K}$, and we see that $ {D^\alpha h_n }$ is a Cauchy sequence on $K$, thus uniformly convergent on $K$, and locally uniformly convergent on $U$. We can also show that $\lim_{n\to\infty} D^\alpha h_n = D^\alpha h$ locally uniformly on $U$ for all index $\alpha$. In particular we have $\Delta h = \lim_{n \to \infty} \Delta h_n$ and hence $h$ is harmonic.

QED.

Theorem (Harnack Convergence Theorem) Consider an increasing sequence of harmonic functions ${ h_n }$ on $U \subset \mathbb{R}^n$, then either $\lim_{n\to\infty} h_n = \infty$ or it converge locally uniformaly to a harmonic function. In particular, if $\exists x_0 \in U$ such that $\lim_{n\to\infty} h_n(x_0) \neq \infty$, then we may conclude locally uniform convergence.

Proof Assume $x_0 \in U$ such that $\lim_{n\to\infty} h(x_0) \neq \infty$. For all $k \leq l$, $K$ compact, and $x_0 \in K \in U$, we have $h_l - h_k$ as a non-negative harmonic function, and hence by Harnack’s inequality, \(\max_K (h_l - h_k) \leq C \min_{K} (h_l - h_k) \leq C(h_l(x_0) - h_k(x_0))\) which implies the uniform Cauchy property of ${ h_n }$ on $K$, and hence locally uniform convergence is achieved. We now let $\lim_{n \to \infty} h_n = h$ and it remains to show $h$ is harmonic, which follows from Weierstrass convergence theorem.

QED.

Recommended citation: Jiajun Zhang, (2025). Harmonic Functions
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