Chapter 3
Laplace Equation

One of the most important PDEs is the Laplace equation

The corresponding inhomogeneous PDE is Poisson’s equation

Both equations are linear PDEs of second order with the unknown function $u:{ℝ}^n\to ℝ$. A function that solves Laplace’s equation is called harmonic. As is typical with linear inhomogeneous equations, the sum of a solution of Poisson’s equation and a harmonic function is again a solution to Poisson’s equation.

These equations show up in many situations. In physics they describe for example the potential $u$ (also called the voltage) of an electric field in the vacuum with some distribution of charges $f$. To give some more detail, perhaps you are familiar with Coulomb’s law: if we have a particle with charge $Q$ at the origin and another particle with charge $q$ at $x$, then the force on the second particle is

where $k_e$ is an empirical constant. (If you haven’t seen this before, it is very much like Newton’s equation for gravity.) If the charges have the same sign the force pushes the second particle in the $\widehat{x}$ direction (repulsion); if the opposite sign the force is in the $-\widehat{x}$ direction (attraction). We interpret the bracket as the electric field of the first particle. Then this same rule could be stated that a positively-charged particle moves in the direction of the electric field and a negatively-charged particle in the opposite direction. And in fact this vector field is a gradient

For historical reasons, the potential is defined $E=-\nabla <!--\; FUNCTION\; APPLICATION\; -->u$. So then we could say that a positively-charged particle tries to decrease the electric potential, like a ball rolling down a hill. The steeper the change in potential, the stronger the force. We will use this example of electric potential to give an interpretation of some of our results. Indeed, much of this theory was developed first by physicists and some techniques seem strange if one does not know the physics motivation!

3.1 Fundamental Solution

The Laplace equation is invariant with respect to all rotations and translations of the Euclidean space ${ℝ}^n$. Therefore we first look for solutions which are invariant with respect to all rotations. These solutions depend only on the length $r=|x|=\sqrt{x\cdot x}$ of the position vector $x$. For such functions $u(x)=v(r)=v(\sqrt{x\cdot x})$ we calculate:

Hence the Laplace equation simplifies to an ODE

Let us solve this ODE:

We see two things here. The space of solutions is two dimensional, with one solution being just the constant solution $u=C^{″}$. The other solution is not a solution on all of ${ℝ}^n$ because is has a singularity at the origin. Never-the-less these are important ‘solutions’ to consider!

This solution lies in the space of radially symmetric solutions. And notice that for $n=3$ it is the electric potential of a single particle. We have chosen $C^{″}=0$, which makes the solution tend to zero for large $x$. The constant $C^{\prime }$ is chosen in such a way that the following theorem holds:

In general, a fundamental solution of a constant coefficient linear PDE $\mathit{𝐿𝑢}=f$ has the property that $L\mathrm{\Phi }=\delta$ in the sense of distribution. We make these assumptions on $L$ so that $L$ is just the real-linear combination of partial derivatives, and so interacts well with convolution. In particular, if we apply $L$ to the convolution of $f$ and the fundamental solution

This shows that the convolution $\mathrm{\Phi }\ast f$ solves the inhomogeneous PDE as long as it is well defined and the derivative rule for convolutions holds.

To give the physics explanation, the fundamental solution is the potential of a single particle with unit charge. The charge of a particle is described by the delta distribution because it is only at a point but the total amount is finite. Consider the situation with two particles $f=Q_1{\delta }_p+Q_2{\delta }_q$. This formula (pretending that $\delta$ is a function) says that their potential is

The interpretation is that if you have charges described by $f$, then treat them as a sum (or integral) of particles. Each particle produces an electric potential $Q_1\mathrm{\Phi }(x-p)$, and the total potential $u$ is the sum (or integral).

Fundamental solutions are not usually unique however. Consider the present case of the Laplace equation. If we have any harmonic function $v$ then $△(\mathrm{\Phi }+v)=△\mathrm{\Phi }+△v=\delta +0$ shows that $\mathrm{\Phi }+v$ is also a fundamental solution. The difference between two fundamental solutions solves the Laplace equation, so this is the only possibility for other fundamental solutions. Different fundamental solutions can produce different solutions to the PDE. We shall see that the fundamental solution we have chosen is the only one that vanishes at infinity, which makes it in some sense the best one.

The difference between the first and second claim of the theorem is the assumption of regularity of $f$: twice continuously differentiable or smooth respectively. In fact it is possible to generalise this theorem further: the convolution of $f$ with $\mathrm{\Phi }$ is defined for continuous functions $f\in L1({ℝ}^n)$ and belongs to $L1({ℝ}^n)$. In this case the result of the convolution may not be differentiable but it is a solution of Poisson’s equation in the sense of distributions. However, if one assumes that $f$ is Lipschitz continuous and belongs to $L1({ℝ}^n)$ then $u$ is twice differentiable (in the usual sense) and solves the PDE. This situation is typical of the delicate questions of regularity of the solution.

3.2 Mean Value Property

In the previous section we constructed a solution to the inhomogeneous equation. Any other solution must differ from the constructed one by a harmonic function. We should therefore understand harmonic functions in order to understand the space of solutions. In this section we shall prove the following property of a harmonic function $u$ on an open domain $\mathrm{\Omega }\subset {ℝ}^n$: the value $u(x)$ of $u$ at the center of any ball $B(x,r)$ with compact closure in $\mathrm{\Omega }$ is equal to the mean of $u$ on the boundary of the ball. Conversely, if this holds for all balls with compact closure in $\mathrm{\Omega }$, then $u$ is harmonic. This relation is called mean value property and has many important consequences.

Let us introduce some notation. Given a function $u$ let

be its spherical mean. Here ${\omega }_n$ denotes the volume of the unit ball in Euclidean space ${ℝ}^n$ and equality follows from Lemma 2.9(iv) using $\mathit{𝑟𝑃}(z)=x+\mathit{𝑟𝑧}$. We write $𝒮(r)$ when the function and center point are clear.

The ball mean or of $u$ on the ball $B(x,r)$ is

using the co-area formula, Lemma 2.11. Many statements can therefore be made either in terms of ball means or spherical means.

The spherical mean, and means generally, have several nice properties. First note that the normalisation constant in the definition ensures that $𝒮[1]=1$ and likewise for any other constant. The mean is real-linear in the function: $𝒮[\mathit{𝑎𝑢}+\mathit{𝑏𝑣}]=a𝒮[u]+b𝒮[v]$, which just follows from linearity of the integral. Likewise it follows from the monotonicity of the integral that if $u\le v$ then $𝒮[u]\le 𝒮[v]$. From these basic properties follows continuity at the center:

Particularly important is the relationship between the spherical mean and the Laplacian of $u$. Differentiating the spherical mean with respect to the radius and using the divergence theorem gives

Therefore if $u$ is harmonic then $𝒮(r)$ is constant. With these important properties of means prepared, we are ready to fully prove our claim.

Keeping with our theme of distributions, we might wonder how we can reinterpret the mean value property for distributions. As is typical for extending definitions to distributions, we first develop a formula for regular distributions. Suppose that $u:\mathrm{\Omega }\to ℝ$ is continuous and $\overline{B(a,R)}\subset \mathrm{\Omega }$. For each point $a\in \mathrm{\Omega }$, we view the spherical mean as a function $r\mapsto 𝒮[u](a,r)$ on $(0,R)$. Therefore $F_{𝒮[u](a,r)}\in {𝒟}^{\prime }((0,R))$. For any test function $\psi \in 𝒟((0,R))$ we compute

Therefore we make the following definition for any distribution $F\in {𝒟}^{\prime }(\mathrm{\Omega })$. For any $a\in \mathrm{\Omega }$ there is a ball $\overline{B(a,R)}\subset \mathrm{\Omega }$. The spherical mean $S_a[F]$ of $F$ around $a$ is the distribution on $(0,R)\subset ℝ$ with the formula

This is well-defined for two reasons. First, the support of $\psi$ excludes $0$, so ${\tilde{\psi }}_a$ is identically zero on a neighborhood of $a$. In particular, dividing by $|x-a{|}^{n-1}$ does not produce a singularity. And second, the support of ${\tilde{\psi }}_a$ is contained in $B(a,R)$. This shows that it is a test function on $\mathrm{\Omega }$.

The mean value property is that the spherical means of the function are constant in the radius. Hence the corresponding property of distributions should require $S_a[F]$ to be ‘constant’ in a suitable sense. We will prove in an exercise that a distribution $G$ corresponds to a constant function if and only if

Together we have

What is the relationship of the weak mean value property to the (strong) mean value property? Suppose $U=F_u$ for a continuous function $u\in C(\mathrm{\Omega })$. If $u$ has the mean value property, then we observe that

In other words, for each $a\in \mathrm{\Omega }$ the distribution ${𝒮}_a[F_u]$ corresponds to the constant function $u(a)$. Thus $F_u$ has the weak mean value property. Conversely, suppose that $F_u$ has the weak mean value property: For each point $a\in \mathrm{\Omega }$ there is a constant $c$ such that $F_c={𝒮}_a[F_u]=F_{𝒮[u](a,r)}$. But we may use the fundamental lemma of the calculus of variations, Lemma 2.15, to conclude that $c=𝒮[u](a,r)$. Hence the spherical mean of $u$ is constant in the radius. In summary:

The functions ${\tilde{\psi }}_a$ may look a little scary, but in fact they are actually friendly once you get to know them. They are smooth functions characterised by two properties:

1.

they are radially symmetric around $a$, and

2.

they have compact support in ${ℝ}^n\setminus \{a\}$.

It is clear that any ${\tilde{\psi }}_a$ has these two properties. If a smooth function $\phi$ has Property 1, then it is a function of the distance $|x-a|$. Another way to state Property 2 is to say that the support is contained in an annulus centered at $a$. Because it vanishes in a neighborhood of $a$, there are no issues with the non-smoothness of $|x-a|$ at $x=a$. So define $\psi (|x-a|)=n{\omega }_n|x-a{|}^{n-1}\phi (x)$ to get the function $\psi \in C_0^{\infty }((0,R))$ with the relation $\phi ={\tilde{\psi }}_a$.

These functions also behave well under convolution, so long as its the convolution of a ‘big annulus’ with a ‘little annulus’. By this we mean the following. Consider ${\tilde{\chi }}_b,{\tilde{\psi }}_a$. Further suppose that ${\tilde{\chi }}_b$ is identically zero on $B(b,R)$ and the support of ${\tilde{\psi }}_a$ lies in $B(a,r)$ for $r<R$. Then ${\tilde{\chi }}_b\ast {\tilde{\psi }}_a$ also obeys Property 1 and 2. Let us demonstrate this now. First, due to Lemma 2.13 we know that ${\tilde{\chi }}_b\ast {\tilde{\psi }}_a$ is rotationally symmetric around $b+a$. Second, the convolution has compact support in ${ℝ}^n$ by the addition formula for supports. It remains to show that it vanishes in a neighbourhood of $b+a$. But this too follows from the addition formula for the support of a convolution, since $a+b\notin ({ℝ}^n\setminus B(b,R))+B(a,r)$.

There is a final point to be made about the total integral of these functions. Recall the formula $F_u({\tilde{\psi }}_a)=F_{𝒮[u](a,r)}(\psi )$. We apply this to the function $u\equiv 1$, which has the mean value property, to get $F_1({\tilde{\psi }}_a)=F_1(\psi )$. Writing this out as integrals shows

In particular, the integral of ${\tilde{\psi }}_a$ is zero if and only if the integral of $\psi$ is zero. And as a reminder, when we introduced convolutions we noted that the integral of ${\tilde{\chi }}_b\ast {\tilde{\psi }}_a$ is the product of the integral of each function. Important to the proof below is that if ${\tilde{\chi }}_b$ has total integral zero, so too does ${\tilde{\chi }}_b\ast {\tilde{\psi }}_a$. In particular, the weak mean value property applies to it.

Now we ready to prove that a distribution has the weak mean value property if and only if it is a harmonic distribution. This should be seen as a generalisation of Theorem 3.5. Something stronger comes out of this proof, a famous result known as Weyl’s lemma. It tells us that weak solutions of the Laplace equations coincide with the strong solutions, and all solutions are smooth.

To conclude this section we show that the mean value property leads to a growth estimate.

3.3 Maximum Principle

We have already mentioned the intuition that if a harmonic function is increasing in some direction then it must decreasing in another. This would imply that a harmonic function cannot have a local extremum, and this is indeed the case. Suppose a harmonic function $u$ has a maximum at a point $x$ of an open connected domain $\mathrm{\Omega }\subset {ℝ}^n$. The mean value property implies on all balls $B(x,r)\subset \mathrm{\Omega }$

By the fundamental lemma of the calculus of variations (or a standard argument from continuity), we conclude that $u(y)=u(x)$ for all $y\in B(x,r)$. Hence $u$ takes the maximum on all these balls $B(x,r)\subset \mathrm{\Omega }$. This shows that the set $\{y\in \mathrm{\Omega }\mid u(y)=u(x)\}$ is open. But it is also the preimage of a single value, and therefore closed. It is non-empty since by assumption $u$ does have a maximum. By the definition of connected, this set must be all of $\mathrm{\Omega }$.

There is a more geometric proof in the case that $\mathrm{\Omega }$ is path connected. We again begin with showing that $u$ takes its maximum on every ball centered at $x$ in the domain. Since $\mathrm{\Omega }$ is path-connected every other point $y\in \mathrm{\Omega }$ is connected with $x$ by a continuous path $\gamma :[0,1]\to \mathrm{\Omega }$ with $\gamma (0)=x$ and $\gamma (1)=y$. The compact image $\gamma [0,1]$ is covered by finitely many balls $B(\gamma (t_1),r_1),\dots ,B(\gamma (t_N),r_N)\subset \mathrm{\Omega }$ with $0\le t_1<\dots t_N\le 1$ and $r_1,\dots ,r_N>0$. Supplementing the balls if necessary, we can assume that the center of each ball belongs to the previous ball. Then repeating the argument Inductively, $u$ is constantly $u(x)$ on all these balls too and hence $u\equiv u(x)$ along $\gamma$, and on $\mathrm{\Omega }$ since this is true for all $y\in \mathrm{\Omega }$.

A practical consequence is the following

Since the negative of a harmonic function is harmonic the same conclusion holds for minima.

The triumph of the Maximum Principle is that it generalises to many elliptic operators (Definition 2.1), unlike the mean value property. It really goes to the heart of ellipticity.

The negative of the functions $u$ in the theorem obey $\mathit{𝐿𝑢}\le 0$ and take a minimum on the boundary. In particular, the solutions $u$ of $\mathit{𝐿𝑢}=0$ take the maximum and the minimum on the boundary.

Now let us see why maximum principles are so important. We consider the following very natural boundary value problem:

The condition that $u$ extends continuously to the boundary is necessary for the boundary value problem to be meaningful. Otherwise the values on the boundary could be complete unrelated to the rest of the function. We say that a function $u$ is $m$ times continuously differentiable on the closure $\overline{\mathrm{\Omega }}$ of an domain, if it is $m$ times continuously differentiable on $\mathrm{\Omega }$ and all partial derivatives of order at most $m$ extend continuously to $\partial \mathrm{\Omega }$.

Let $\mathrm{\Omega }\subset {ℝ}^n$ be an open and bounded domain and suppose that there are two solutions $u_1$ and $u_2$ to the Dirichlet problem for the Poisson equation with inhomogeneous term $f$ and boundary value $g$. Then the difference $v:=u_2-u_1$ solves the homogeneous problem, i.e. it is harmonic, and $v\equiv 0$ on $\partial \mathrm{\Omega }$. Therefore by the weak maximum principle we know that both the maximum and minimum of $v$ on every connected component of $\mathrm{\Omega }$ is $0$. The only possibility is that $v\equiv 0$ on all of $\mathrm{\Omega }$. This shows that solutions to the Dirichlet problem are unique.

Putting this another way, we can uniquely determine a harmonic function if we know its values on the boundary of its domain. This gives us a way to understand the space of harmonic functions.

3.4 Green’s Function

We just saw that the solution to the Dirichlet problem is unique, if a solution exists. In this section we try to find some conditions which ensure the existence.

First we prepare some well known formulas, which hopefully you have already proved as an exercise. In first formula we apply the Divergence Theorem to $x\mapsto v(x)\nabla <!--\; FUNCTION\; APPLICATION\; -->u(x)$: