📂Partial Differential Equations

Fundamental Solution of the Laplace Equation

Buildup¹

Laplace’s equation is invariant under rotation transformations, so consider changing $u(x)$’s variables to radii. This allows simplifying the differential equation as follows.

Let’s assume that $u=u(x)$ is a solution to Laplace’s equation.

$$ \Delta u = 0 $$

And let’s set $r=|x|=(x_{1}^{2} + \cdots + x_{n}^{2})^{1/2}$ and assume $v\in C^2$ and $u(x) = v(|x|) = v(r) (x\in \mathbb{R}^{n} \setminus \left\{ 0 \right\})$.

$$ \begin{align*} v(r) &= u(x) \\ \Delta v &= 0 \end{align*} $$

Now calculate the following derivatives to represent Laplace’s equation for $u$.

$$ \begin{equation*} \begin{aligned} \dfrac{\partial r}{\partial x_{i}} &= \dfrac{\partial}{\partial x_{i}} (x_{1}^{2} + \cdots x_{n}^{2} )^{1/2} \\ &= \dfrac{1}{2(x_{1}^{2} + \cdots x_{n}^{2} )^{1/2}}2x_{i} \\ &= \dfrac{x_{i}}{(x_{1}^{2} + \cdots x_{n}^{2} )^{1/2}} \\ &= \dfrac{x_{i}}{r} \end{aligned} \quad \text{and} \quad \begin{aligned} u_{x_{i}}(x) &= \dfrac{\partial }{\partial x_{i}} v(r) \\ &= \dfrac{d v(r)}{d r}\dfrac{\partial r}{\partial x_{i}} \\ &= v^{\prime}(r) \dfrac{x_{i}}{r} \end{aligned} \end{equation*} $$

$$ \begin{align*} u_{x_{i}x_{i}} &= \dfrac{\partial }{\partial x_{i} } \left( v^{\prime}(r)\dfrac{x_{i}}{r} \right) \\ &= \dfrac{\partial v^{\prime}(r)}{\partial x_{i}} \dfrac{x_{i}}{r} + v^{\prime}(r)\dfrac{\partial}{\partial x_{i}}\left(\dfrac{x_{i}}{r}\right) \\ &= \dfrac{d v^{\prime}(r)}{dr} \dfrac{\partial r}{\partial x_{i}} \dfrac{x_{i}}{r} + v^{\prime}(r) \left( \dfrac{1}{r} + x_{i}\dfrac{\partial}{ \partial x_{i}} \left(\dfrac{x_{i}}{r}\right) \right) \\ &= v^{\prime \prime}(r) \dfrac{x_{i}^{2}}{r^{2}} + v^{\prime}(r) \left( \dfrac{1}{r} + x_{i}\dfrac{d}{dr} \left( \dfrac{1}{r} \right) \dfrac{\partial r}{\partial x_{i}} \right) \\ &= v^{\prime \prime}(r) \dfrac{x_{i}^{2}}{r^{2}} + v^{\prime}(r)\left(\dfrac{1}{r} + x_{i}\left(-\dfrac{1}{r^{2}}\right) \dfrac{x_{i}}{r} \right) \\ &= v^{\prime \prime}(r) \dfrac{x_{i}^{2}}{r^{2}} + v^{\prime}(r)\left(\dfrac{1}{r}-\dfrac{x_{i}^{2}}{r^{3}} \right) \end{align*} $$

Then, the Laplace equation is as follows.

$$ \begin{align*} \Delta u &= \sum \limits_{i}^{n} u_{x_{i} x_{i}} \\ &= \dfrac{v^{\prime \prime}(r)}{r^{2}}\left( x_{1}^{2} + \cdots x_{n}^{2} \right) + v^{\prime}(r) \left( \dfrac{n}{r} - \dfrac{x_{1}^{2} + \cdots x_{n}^{2}}{r^{3}} \right) \\ &= \dfrac{v^{\prime \prime}(r)}{r^{2}}r^{2} + v^{\prime}(r) \left( \dfrac{n}{r} - \dfrac{r^{2}}{r^{3}} \right) \\ &= v^{\prime \prime}(r) + \dfrac{n-1}{r} v^{\prime}(r) & x\ne 0 \end{align*} $$

Therefore, the following two equations are identical.

$$ \Delta u =0 \text{ in }\mathbb{R}^n\setminus\left\{ 0\right\} \iff v^{\prime \prime}(r ) + \dfrac{n-1}{r}v^{\prime}(r)=0\quad (r>0) $$

Solving Laplace’s equation for $u(x)$ has reduced to solving a second-order ordinary differential equation for $v(r)$.

Now, assume $v^{\prime \prime}+\dfrac{n-1}{r}v^{\prime}=0 \text{ in } (0, \infty)$ and $v^{\prime}\ne 0 \text{ in } (0, \infty)$.

Then, summarizing, we obtain the following equation.

$$ \dfrac{v^{\prime \prime}}{v^{\prime}} = \dfrac{1-n}{r} $$

Substituting $w=v^{\prime} \in C^{1}(0, \infty)$, we obtain the following.

$$ \dfrac{w^{\prime}}{w} = \dfrac{1-n}{r} $$

Integrating the left side gives the following.

$$ \int_{1}^{s} \dfrac{w^{\prime}}{w} dr = \log |w(s)| - \log |w(1)| = \log \left| \dfrac{w(s)}{w(1)} \right| $$

Integrating the right side gives the following.

$$ \int_{1}^{s} \dfrac{1-n}{r} dr = (1-n) \left[ \log s - \log 1 \right] = (1-n)\log s =\log s^{1-n} $$

Therefore, we get the following.

$$ \dfrac{|w(s)|}{|w(1)|}=s^{1-n} \implies |w(s)|=|w(1)|s^{1-n} (s>0) $$

Expressing it again in terms of $v$ gives the following.

$$ v^{\prime}(s)=w(s)=w(1)s^{1-n}=v^{\prime}(1)s^{1-n} $$

At this point, applying the Fundamental Theorem of Calculus to $v(r)-v(1)$, and then substituting the above equation, we get the following.

$$ \begin{align*} v(r)-v(1) &= \int_{1}^{r} v^{\prime}(s) ds = v^{\prime}(1)\int_{1}^{r} s^{1-n} ds \\ &= \begin{cases} v^{\prime}(1)\log r & n=2 \\ v^{\prime}(1) \dfrac{1}{2-n}(r^{2-n}-1) & n \ge 3 \end{cases} \end{align*} $$

Rearranging for $v(r)$ gives the following.

$$ v(r) = \begin{cases} v^{\prime}(1)\log r + v^{\prime}(1) & n=2 \\ v^{\prime}(1)\dfrac{1}{2-n} \dfrac{1}{r^{n-2}} + \left(v(1)+\dfrac{v^{\prime}(1)}{n-2} \right) & n \ge 3 \end{cases} $$

Expressing the constant part as $b, c$ simplifies as follows.

$$ v(r) = \begin{cases} b\log r + c & n=2 \\ \dfrac{b}{r^{n-2}} + c & n \ge 3 \end{cases} $$

From these results, we define the fundamental solution to the Laplace equation.

Definition

$x \in \mathbb{R}^{n}$, and for $x \ne 0$, the function $\Phi$ below is defined as the fundamental solution to Laplace’s equation.

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

Here, $\alpha (n)$ is the volume of the unit ball $B(0,\ 1)$ in $\mathbb{R}^{n}$. $n\alpha (n)$ is the surface area of the unit ball in $\mathbb{R}^{n}$.