# Solve the Laplace equation on a circular disk of radius 3 with boundary condition f ( theta ) =...

## Question:

Solve the Laplace equation on a circular disk of radius 3 with boundary condition {eq}f(\theta) = \sin(\theta) + \cos^3(\theta) {/eq}.

Laplace's equation in two dimensions with the function {eq}u(x, y) {/eq} is

{eq}u_{xx}(x, y) + u_{yy}(x, y) = 0. {/eq}

A solution to Laplace's equation is called harmonic. If we wish to rewrite the equation in polar coordinates then it becomes

{eq}\displaystyle\frac1{r} \: \frac{\partial}{\partial r} \left( r \displaystyle\frac{\partial u}{\partial r} \right) + \frac{1}{r^2} \: \frac{\partial^2 u}{\partial \theta^2} = 0. {/eq}

We will use this second form and separation of variables to solve this problem.

We are asked to solve the Laplace equation on a circular disk of radius 3 with boundary condition {eq}f(\theta) = \sin(\theta) + \cos^3(\theta) {/eq}. Assume that {eq}u(r, \theta) = g(r) h(\theta). {/eq} Then {eq}u_r = g'(r) h(\theta) {/eq} and {eq}u_\theta = g(r) h''(\theta). {/eq} Substituting these into the second form of Laplace's equation, we get

{eq}\begin{eqnarray*}\displaystyle\frac1{r} \: \frac{\partial}{\partial r} \left( r g'(r) h(\theta) \right) + \frac{1}{r^2} \: g(r) h''(\theta) & = & 0 \\ \\ \displaystyle\frac{1}{r} \left( g'(r) h(\theta) + r g''(r) h(\theta) \right) + \frac1{r^2} g(r) h''(\theta) & = & 0 \\ \\ r g'(r) h(\theta) + r^2 g''(r) h(\theta) + g(r) h''(\theta) & = & 0 \\ \\ (r g'(r) + r^2 g''(r) ) h(\theta) & = & -g(r) h''(\theta) \\ \\ \displaystyle\frac{h''(\theta)}{h(\theta)} & = & -\displaystyle\frac{r g'(r) + r^2 g''(r)}{g(r)} \end{eqnarray*} {/eq}

Now, the left hand side of the equation is a function of {eq}\theta {/eq} alone, and the right hand side is a function of {eq}r {/eq} alone. Since they are equal to each other, they must both be equal to the same constant, call it {eq}C. {/eq} Therefore

{eq}\begin{eqnarray*}\displaystyle\frac{h''(\theta)}{h(\theta)} & = & C \\ \\ h''(\theta) & = & C h(\theta) \\ \\ h''(\theta) - C h(\theta) & = & 0 \end{eqnarray*} {/eq}

If {eq}C > 0 {/eq} then we have the general solution {eq}h(\theta) = c_1 e^{-\theta \sqrt{C}} + c_2 e^{\theta \sqrt{C}}. {/eq} However, the boundary condition is periodic, and this function is not. Therefore we will not use this solution.

If {eq}C = 0 {/eq} then {eq}h(\theta) = A \theta + B, {/eq} where {eq}A {/eq} and {eq}B {/eq} are constant. Again, this is not periodic.

Assume {eq}C < 0, {/eq} and define {eq}C = -\lambda^2, {/eq} where {eq}\lambda > 0. {/eq} Then {eq}h(\theta) = c_1 \cos \lambda \theta + c_2 \sin \lambda \theta. {/eq}

Now we move on to the function of {eq}r. {/eq} We have

{eq}\begin{eqnarray*}-\displaystyle\frac{r g'(r) + r^2 g''(r)}{g(r)} & = & C \\ \\ r g'(r) + r^2 g''(r) & = & \lambda^2 g(r) \\ \\ r^2 g''(r) + r g'(r) - \lambda^2 g(r) & = & 0 \end{eqnarray*} {/eq}

This is a Cauchy-Euler equation. We assume that {eq}g(r) = r^n {/eq} for some undetermined value(s) of {eq}n. {/eq} Then {eq}g'(r) = n r^{n - 1} {/eq} and {eq}g''(r) = n(n - 1) r^{n - 2}. {/eq} This gives

{eq}\begin{eqnarray*}r^2 \left( n(n - 1)r^{n - 2} \right) + r \left( n r^{n - 1} \right) - \lambda^2 r^n & = & 0 \\ \\ r^n (n(n - 1) + n - \lambda^2) & = & 0 \\ \\ r^n (n^2 - \lambda^2) & = & 0 \end{eqnarray*} {/eq}

This equation must hold true for all values of {eq}r, {/eq} so therefore {eq}n = \pm \lambda. {/eq} This gives the general solution {eq}g(r) = d_1 r^{-\lambda} + d_2 r^{\lambda}. {/eq} The first term is not defined at {eq}r = 0. {/eq} Therefore the general solution to Laplace's equation becomes

{eq}u(r, \theta) = r^{\lambda} \left( c_1 \cos \lambda \theta + c_2 \sin \lambda \theta \right). {/eq}

We will assume that {eq}\lambda {/eq} is a positive integer, so a family of solutions to the equation is {eq}\{ r^n \cos n \theta, \: r^n \sin n \theta \}, \: n > 0. {/eq}

The boundary condition is {eq}u(3, \theta) = \sin \theta + \cos^3 \theta = \sin \theta + \displaystyle\frac14 \cos 3 \theta + \frac34 \cos \theta {/eq} Therefore

{eq}u(r, \theta) = c_1 r \sin \theta + c_2 r \cos \theta + c_3 r^3 \cos 3\theta. {/eq}

Substituting {eq}r = 3, {/eq} we get

{eq}3 c_1 \sin \theta + 3 c_2 \cos \theta + 27 c_3 \cos 3\theta = \sin \theta + \displaystyle\frac14 \cos 3 \theta + \frac34 \cos \theta, {/eq}

so {eq}c_1 = \displaystyle\frac13, \: c_2 = \frac14, \: c_3 = \frac1{108}, {/eq} and

{eq}u(r, \theta) = \displaystyle\frac13 r \sin \theta + \frac14 r \cos \theta + \frac1{108} r^3 \cos 3\theta. {/eq}