📂Partial Differential Equations

Poisson's Equation Fundamental Solution

Buildup¹

Fundamental Solution to the Laplace Equation

Let’s define the function $\Phi$ as the fundamental solution to the Laplace equation, where $x \in \mathbb{R}^{n}$ and $x \ne 0$.

$$\Phi (x) := \begin{cases} -\frac{1}{2\pi}\log |x| & n=2 \\ \frac{1}{n(n-2)\alpha (n)} \frac{1}{|x|^{n-2}} & n \ge 3 \end{cases}$$

Consider a function that maps as $x \mapsto \Phi (x)$. It is harmonic at places where $x \ne 0$. Suppose we symmetrize the origin from $0$ to $y\in \mathbb{R}^{n}$. Then, the function $x \mapsto \Phi (x-y)$ is harmonic at places where $x\ne y$. Now, let’s say an arbitrary function $f : \mathbb{R}^{n} \to \mathbb{R}$ is given. Then, the following function, since $f$ is a function of $y$, is still harmonic with respect to the variable $x$.

$$x \mapsto \Phi (x-y)f(y)$$

Then, since the above function is harmonic for each $y_{k}\in \mathbb{R}^{n}$, summing them all up remains harmonic as well.

$$x \mapsto \sum _{k=1}^{N}\Phi (x-y_{k})f(y_{k})\text{ is harmonic in } \mathbb{R}^{n}\setminus \left\{ y_{1},\dots,y_{N} \right\}$$

From here, let’s define the function $u$ with the sense of increasing $N$ as follows.

Definition

Let’s regard $\Phi$ as the fundamental solution to the Laplace equation. Then, $u$ defined as follows is known as the fundamental solution to the Poisson equation.

$$\begin{equation} \begin{aligned} u(x) &= \int_{\mathbb{R}^n} \Phi (x-y) f(y)dy = \Phi \ast f (x) \\ &= \begin{cases} \displaystyle -\dfrac{1}{2\pi} \int_{\mathbb{R}^2} \log (|x-y|) f(y) dy & (n=2) \\ \displaystyle \dfrac{1}{n(n-2)\alpha (n) }\int_{\mathbb{R}^n} \dfrac{f(y)}{|x-y|^{n-2}}dy & (n \ge 3) \end{cases} \end{aligned} \end{equation}$$

Here, $\ast$ represents convolution.

Explanation

$$\begin{equation} \Delta u = f \end{equation}$$

Now, we expect that $u$ satisfies the Poisson Equation $(2)$. If appropriate conditions are given to $f$, we can see that $u$ is well-defined and also satisfies the Poisson equation. Those conditions are that $f$ has a compact support and is twice continuously differentiable.

$$f \in C^{2}_{c}$$

• Well-defined

  Let $f \in C_{c}(\mathbb{R}^n)$. Then, there exists an open ball $B(x,r_{x})$ satisfying the following.

  $$\text{supp}f \subset B(x, r_{x}),\quad r_{x}>0$$

  From the following calculation, we can see that $u$ is well-defined.

  $$\begin{align*} \left| u(x) \right| &\le \int_{ \mathbb{R}^{n} } \left| \Phi (x-y) \right| \left| f(y) \right| dy \\ &= \int_{ B(x, r_{x}) } \left| \Phi (x-y) \right| \left| f(y) \right| dy \\ &\le \max \left| f \right| \int_{ B(x, r_{x}) } \left| \Phi (x-y) \right| dy \\ &= \max \left| f \right| \int_{ B(0, r_{x}) } \left| \Phi (y) \right| dy < \infty \end{align*}$$

Theorem

Let $f \in C^{2}_{c}(\mathbb{R}^{n})$. Suppose $u$ is as follows in $(1)$. Then, the following holds.

$\text{(i)}$ $u\in C^2 (\mathbb{R}^n)$

$\text{(ii)}$ $-\Delta u=f\quad \text{ in } \mathbb{R}^n$

Proof

$\text{(i)}$

Given a fixed $x \in \mathbb{R}^n$ and $0 \ne h \in \mathbb{R}$, $i\in \left\{ 1,\cdots,n\right\}$, the following holds.

$$\dfrac{u(x+he_{i})-u(x) }{h} =\int_{\mathbb{R}^n} \Phi (y)\dfrac{f(x+he_{i}-y) -f(x-y)}{h}dy$$

Here, $e_{i}=(0,\cdots ,1,\cdots, 0)$ is a vector with the $i$th component as $1$ and the others as $0$. Then, since $f$ is differentiable, by the Mean Value Theorem (MVT), for any $y \in \mathbb{R}^n$ and $h^{\prime} \in (0,h)$, the following holds.

$$\dfrac{f(x+he_{i}-y) - f(x-y)}{h}=f_{x_{i}}(x+h^{\prime}e_{i}-y)$$

Given the assumption that $f_{x_{i}} \in C_{c}^1 (\mathbb{R}^n)$ and since a continuous function in a compact space is uniformly continuous, $f_{x_{i}}$ is uniformly continuous in $\mathbb{R}^n$. Therefore, for a given $\epsilon >0$, there exists a $\delta >0$ satisfying $|z-w|<\delta \implies |f_{x_{i}}(z)-f_{x_{i}}(w)|<\epsilon$. If $0<|h|<\delta$, then for all $y\in \mathbb{R}^n$, the following holds.

$$|(x+h^{\prime}e_{i}-y)-(x-y)|=|h^{\prime}|<|h|<\delta$$

Thus, the following holds.

$$\begin{align*} &&|(x+h^{\prime}e_{i}-y)-(x-y)|=|h^{\prime}|<|h| &< \delta \\ \implies && |f_{x_{i}}(x+h^{\prime}e_{i}-y)-f_{x_{i}}(x-y)| &< \epsilon \\ \implies && \sup \limits_{y\in \mathbb{R}^n} \left| \dfrac{ f(x+he_{i}-y)-f(x-y)}{h}-f_{x_{i}}(x-y) \right| &< \epsilon \end{align*}$$

Therefore, the following holds.

$$\dfrac{f(x+he_{i}-y) - f(x-y)}{h} \rightrightarrows f_{x_{i}}(x-y)\quad \mathrm{as}\quad h\rightarrow 0$$

Hence, we obtain the following.

$$\begin{align*} u_{x_{i}}(x) &= \lim \limits_{h \rightarrow 0}\dfrac{ u(x+he_{i})-u(x)}{h} \\ &= \int_{\mathbb{R}^n} \Phi (x) f_{x_{i}}(x-y)dy \quad (i=1,\cdots, ) \end{align*}$$

Similarly, the following holds.

$$\begin{align*} u_{x_{i}x_{j}}(x) &= \lim \limits_{h \rightarrow 0}\dfrac{ u_{x_{i}}(x+he_{j})-u_{x_{i}}(x)}{h} \\ &= \int_{\mathbb{R}^n} \Phi (x) f_{x_{i}x_{j}}(x-y)dy \quad (i,j=1,\cdots, ) \end{align*}$$

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