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Poisson's Equation

Related Works

Stanford course Math 220B

1. Problem Description

We want to solve the Laplace's equation

\[ \Delta u = 0 \]

and it's inhomogeneous versions, Poisson's equation

\[ - \Delta u = f \]

We say a function u satisfying Laplace's equation is a harmonic function.

2. The Fundamental Solution

Consider Laplace's equation in \(\mathbb{R}^n\),

\[ \Delta u = 0 \quad x \in \mathbb{R}^n \]

Clearly, there are a lot of functions u which satisfy this equation. In particular, any constant function is harmonic. In addition, any function of the form \(u(x)=a_1x_1 + ... + a_nx_n\) for constants \(a_i\) is also a solution. Of course, we can list a number of others. Here, however, we are intrested in finding a particular solution of Laplace's equation which will allow us to solve Poisson's equation.

Give the symmertic nature of Laplace's equation, we look for a radial solution. That is, we look for a harmonic funtion \(u\) on \(\mathbb{R}^n\) such that \(u(x)=v(|x|)\). In addition, to being a natural choice due to the symmetry of Laplace's equation, radial solutions are natural to look for because they reduce a PDE to an ODE, which is generally easier to solve. Therefore, we look for a radial solution.

If \(u(x)=v(|x|)\), then

\[ u_{x_i} = \frac{x_i}{|x|}v'(|x|) \quad |x| \neq 0 \]

which implies

\[ u_{x_ix_i} = \frac{1}{|x|}v'(|x|) - \frac{x_i^2}{|x|^3}v'(|x|) + \frac{x_i^2}{|x|^2}v''(|x|) \quad |x| \neq 0 \]

Therefore,

\[ \Delta u = \frac{n-1}{|x|}v'(|x|) + v''(|x|) \]

Letting \(r = |x|\), we see that \(u(x)=v(|x|)\) is a radial solution of Laplace's equation implies \(v\) satisfies

\[ \frac{n-1}{r}v'(r)+v''(r) = 0 \]

And then, from some calculations, we see that for any constants \(c1\), \(c2\), the function

\[ u(x) \equiv \begin{cases} c_1 \ln |x| + c_2, & \text{if } n = 2 \\ \frac{c_1}{(2 - n)|x|^{n - 2}} + c_2, & \text{if } n \geq 3 \end{cases} \tag{3.1} \]

for \(x \in \mathbb{R}^n\), \(|x| \neq 0\) is a solution of Laplace's equation in \(\mathbb{R}^n - {0}\). We notice that the function \(u\) defined in (3.1) satisfies \(\Delta u(x)=0\) for \(x \neq 0\), but at \(x=0\), \(\Delta u(0)\) is undefined. We claim that we can choose constants \(c_1\) and \(c_2\) appropriately so that

\[ - \Delta_x u = \delta_0 \]

in the sense of distributions. Recall that \(\delta_0\) is the distribution which is defined as follows. For all \(\phi \in \mathbb{D}\),

\[ (\delta_0, \phi) = \phi(0) \]

Claim 1

For \(\Phi\) defined in (3.3), \(\Phi\) satisfies

\[ -\Delta_x\Phi = \delta_0 \]

in the sense of distributions. That is, for all \(g\in \mathcal{D}\),

\[ -\int_{\mathbb{R}^n}\Phi(x)\Delta_xg(x)\,dx=g(0) \]
Proof

The proof of this claim is a little complex. You can reference to this handouts: Stanford course Math 220B laplace

For now, though, assume we can prove this. That is, assume we can find constants \(c_1\),\(c_2\) such that \(u\) defined in (3.1) satisfied

\[ - \Delta_x u = \delta_0 \tag{3.2} \]

Let \(\Phi\) denote the solution of (3.2). Then, define

\[ v(x) = \int_{\mathbb{R}^n} \Phi(x - y) f(y) \, dy \]

\(Formally\), we compute the Laplacian of \(v\) as follows,

\[ \begin{aligned} - \Delta_x v &= - \int_{\mathbb{R}^n} \Delta_x \Phi(x - y) f(y) \, dy \\ &= - \int_{\mathbb{R}^n} \Delta_y \Phi(x - y) f(y) \, dy \\ &= \int_{\mathbb{R}^n} \delta_x f(y) \, dy = f(x) \end{aligned} \]

That is, \(v\) is a solution of Poisson's equation! Of course, this set of equalities above is entirely formal. We have not prove anything yet. However, we have motivated a solution formula for Poisson's equation from a solution to (3.2). We now return to using the radial solution (3.1) to find a solution of (3.2).

Define the function \(\Phi\) as follows. For \(|x| \neq 0\), let

\[ \Phi(x) = \begin{cases} -\dfrac{1}{2\pi} \ln |x|, \quad \text{if } n = 2\\ \dfrac{1}{n(n-2)\alpha(n)} \cdot \dfrac{1}{|x|^{n-2}}, \quad \text{if } n \geq 3 \end{cases} \tag{3.3} \]

where \(\alpha (n)\) is the volume of the unit ball in \(\mathbb{R}^n\). We see that \(\Phi\) satisfies Laplace's equation on \(\mathbb{R}^n-{0}\). As we will show in the following claim, \(\Phi\) satisfies \(-\Delta_x \Phi = \delta_0\). For this reason, we call \(\Phi\) the fundamental solution of Laplace's equation.

3. Solving Poisson's Equation

We now return to solving Poisson's Equation.

\[ - \Delta u = f, \quad x \in \mathbb{R}^n \]

From our discussion before, we expect the function

\[ v(x) \equiv \int_{\mathbb{R}^n} \Phi(x-y) f(y)\, dy \]

to give us a solution of Poisson's Equation. We now prove that this is in fact true. First, we make a remark.

Remark. If we hope that the function \(v\) defined above solves Poisson's equation, we must first verify that integral actually converges. If we assume \(f\) has compact support on some bounded set \(K\) in \(\mathbb{R}^n\), then we see that

\[ \int_{\mathbb{R}^n} \Phi(x-y)f(y) \, dy \leq \|f\|_{L^\infty} \int_{K} \big| \Phi(x-y) \big| \, dy \]

If we additionally assume that \(f\) is bounded, then \(\|f\|_{L^\infty} \leq C\). It is left as an exercise to verify that

\[ \int_{K} \big| \Phi(x-y) \big| \, dy \le + \infty \]

on any compact set \(K\).

Theorem 2.

Assume \(f \in C^2(\mathbb{R}^n)\) and has compact support. Let

\[ u(x) \equiv \int_{\mathbb{R}^n} \Phi(x-y)f(y) \, dy \]

where \(\Phi\) is the fundamental solution of Laplace's equation. Then

  1. \(u \in C^2(\mathbb{R}^n)\)
  2. \(- \Delta u = f \quad in \quad \mathbb{R}^n\)
Proof
  1. By a change of variables, we write
\[ u(x)=\int_{\mathbb{R}^n}\Phi(x-y)f(y)\,dy = \int_{\mathbb{R}^n}\Phi(y)f(x-y)\,dy \]

Let \(e_i=(...,0,1,0,...)\) be the unit vector in \(\mathbb{R}^n\) with a \(1\) in the \(i^{th}\) slot. Then

\[ \frac{u(x+he_i)-u(x)}{h} = \int_{\mathbb{R}^n}\Phi(y)\left[\frac{f(x+he_i-y)-f(x-y)}{h}\right]\,dy \]

Now \(f\in C^2\) implies

\[ \frac{f(x+he_i-y)-f(x-y)}{h} \rightarrow \frac{\partial f}{\partial x_i}(x-y) \quad \text{as} \quad h \rightarrow 0 \]

uniformly on \(\mathbb{R}^n\). Therefore,

\[ \frac{\partial u}{\partial x_i}(x) = \int_{\mathbb{R}^n}\Phi(y)\frac{\partial f}{\partial x_i}(x-y)\,dy \]

Similarly,

\[ \frac{\partial^2 u}{\partial x_i\partial x_j}(x) = \int_{\mathbb{R}^n}\Phi(y)\frac{\partial^2 f}{\partial x_i\partial x_j}(x-y)\,dy \]

This function is continuous because the right-hand side is continuous.

  1. By the above calculations and Claim \(1\), we see that
\[ \begin{aligned} \Delta_xu(x)&=\int_{\mathbb{R}^n}\Phi(y)\Delta_xf(x-y)\,dy \\ &=\int_{\mathbb{R}^n}\Phi(y)\Delta_yf(x-y)\,dy \\ &=-f(x) \end{aligned} \]

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