# Laplace's equation

In electrostatics, a potential function can be always defined for a given electric field function, and it has to satisfy Poisson's equation

$\nabla^2 V = -\frac{\rho}{\epsilon_0}$

and in case there is no charge distributed in the space of interest, following Laplace's equation holds

$\nabla^2 V = 0$.

## 1D Laplace's equation

In one-dimension, Laplace's equation becomes

$\frac{d^2V}{dx^2} = 0$

and its general solution is $V(x) = mx + b$. This solution of course satisfies the property $V(x) = \frac12 [ V(x-a) + V(x+a) ]$.

## 2D Laplace's equation

In two-dimension, Laplace's equation becomes

$\frac{\partial^2V}{\partial x^2}+\frac{\partial^2V}{\partial y^2} = 0$

which is a partial differential equation.

Following properties should hold for harmonic functions in 2-dim.

• If you draw a circle of any radius R about the point (x,y), following equation should hold
$V(x,y) = \frac{1}{2\pi R} \oint_{circle} V dl$
• V has no local maxima or minima.
• Harmonic function in 2-dim minimizes the surface area spanning the given boundary line.

## 3D Laplace's equation

$\frac{\partial^2V}{\partial x^2}+\frac{\partial^2V}{\partial y^2}+\frac{\partial^2V}{\partial z^2} = 0$
• If you imagine a sphere of any radius R about the point $\textbf{r}$, following equation should hold
$V(\textbf{r}) = \frac{1}{4\pi R^2} \oint_{sphere} V da$
• Therefore, V can have no local maxima or minima.

### Separation of variables

#### Cartesian coordinates

Use $V(\mathbf{r})=X(x)Y(y)Z(z)$, then Laplace's equation becomes

$\nabla^2 V(\mathbf{r}) = YZ\frac{dX}{dx} + XZ\frac{dY}{dy} + XY\frac{dZ}{dz}=0$

which, after dividing by V, becomes

$\frac1X\frac{dX}{dx} + \frac1Y\frac{dY}{dy} + \frac1Z\frac{dZ}{dz}=0$

which can be only true when each term becomes constant.

#### Spherical coordinates

Use $V(\mathbf{r})=R(r)\Theta(\theta)\Phi(\phi)$, then Laplace's equation becomes

$\frac1{r^2}\frac{\partial}{\partial r}\left( r^2\frac{\partial V}{\partial r}\right)+\frac1{r^2\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial V}{\partial\theta}\right) + \frac1{r^2\sin^2\theta}\frac{\partial^2V}{\partial\phi^2}=0$
$\Theta\Phi\frac{d}{dr}\left( r^2\frac{dR}{dr}\right) + \frac{R\Phi}{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \frac{R\Theta}{\sin^2\theta}\frac{d^2\Phi}{d\phi^2}=0$

which, after dividing by V, becomes

$\underbrace{\frac1R\frac{d}{dr}\left( r^2\frac{dR}{dr}\right)}_{\alpha} + \underbrace{\frac1{\Theta}\frac1{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)}_{-\alpha+\frac{\beta}{\sin^2\theta}} + \underbrace{\frac1{\Phi}\frac1{\sin^2\theta}\frac{d^2\Phi}{d\phi^2}}_{-\frac{\beta}{\sin^2\theta}}=0$

where $\alpha$ and $\beta$ are arbitrary constants.

The radial equation is usually formulated as below:

$\frac{d}{dr}\left( r^2\frac{dR}{dr}\right) = l(l+1)R$

where we replaced $\alpha$ with l(l+1) and l is integer. Then it has general solution of

$R(r)=Ar^l + \frac{B}{r^{l+1}}$
##### Angular equation

The angular part reduces to following differential equation, which is Associated Legendre Equation:

$\frac1{\sin\theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right)+\left(l(l+1)-\frac{m^2}{\sin^2\theta}\right)\Theta=0$
##### Azimuthal equation

The azimuthal part becomes simple ODE whose solution is an oscillation with m nodes:

$\frac{d^2\Phi}{d\phi^2} + m^2\Phi = 0$