# MATHEMATICA tutorial, Part 2.6: Laplace in polar coordinates (2023)

\begin{eqnarray*} u(r, \theta ) &=& \frac{1}{\pi} \int_{-\pi}^{\pi} f(\phi )\,{\text d}\phi \left\{ \frac{1}{2} + \sum_{n\ge 1} \frac{r^n}{a^n} \,\cos n(\theta - \phi ) \right\} . \end{eqnarray*}

\begin{eqnarray*} \frac{1}{2} + \sum_{n\ge 1} \frac{r^n}{a^n} \,\cos n(\theta - \phi ) &=& \frac{1}{2} + \Re \,\sum_{n\ge 1} \frac{r^n}{a^n} \,e^{{\bf j} n(\theta - \phi )} \\ &=& \frac{1}{2} + \Re \,\sum_{n\ge 1} \left( \frac{r}{a}\,e^{{\bf j} (\theta - \phi )} \right)^n &=& \frac{1}{2} + \Re \,\dfrac{\frac{r}{a}\, e^{{\bf j} (\theta - \phi )}}{1 -\frac{r}{a}\, e^{{\bf j} (\theta - \phi )} }\\ &=& \frac{1}{2} + \Re \,\dfrac{r\, e^{{\bf j} (\theta - \phi )}}{a - r\,e^{{\bf j} (\theta - \phi )}} , \end{eqnarray*}

$$\Re$$ denotes the real part of a complex number. Multiplying the numerator and denominator of the latter fraction by complex conjugate of the denominator, we obtain

\begin{eqnarray*} \dfrac{r\, e^{{\bf j} (\theta - \phi )}}{a - r\,e^{{\bf j} (\theta - \phi )}} &=& \dfrac{r\, e^{{\bf j} (\theta - \phi )}}{a - r\,e^{{\bf j} (\theta - \phi )}} \, \frac{a - r\,e^{-{\bf j} (\theta - \phi )}}{a - r\,e^{-{\bf j} (\theta - \phi )}} \\ &=& \frac{ra\,e^{-{\bf j} (\theta - \phi )} - r^2 }{a^2 + r^2 -2ar\,\cos (\theta - \phi )}\end{eqnarray*}

Extracting the real part, we have

\begin{eqnarray*} \frac{1}{2} + \sum_{n\ge 1} \frac{r^n}{a^n} \,\cos n(\theta - \phi ) &=& \frac{1}{2} + \frac{ar\,\cos (\theta - \phi ) - r^2}{a^2 + r^2 -2ar\,\cos (\theta - \phi )} \\ &=& \frac{1}{2}\,\frac{a^2 - r^2}{a^2 + r^2 -2ar\,\cos (\theta - \phi )} . \end{eqnarray*}

This results in Poisson’s formula:

$$\label{EqPolar.4} u(r, \theta ) = \frac{1}{2\pi} \,\int_0^{2\pi} f(\phi ) \,P_r (\theta - \phi )\,{\text d}\phi , \qquad P_r (x) = \frac{a^2 - r^2}{a^2 + r^2 -2ar\,\cos x} , \quad r \le a.$$

At r = 0, the integral is easy to compute:

$u(0, \theta ) = \frac{1}{2\pi} \,\int_0^{2\pi} f(\phi ) \,{\text d}\phi = \frac{1}{2\pi} \,\int_0^{2\pi} u(a, \phi ) \,{\text d}\phi .$

Therefore, we can make the following conclusions.
• If u is a solution of the Laplace equation Δu = 0, then the value of u at any point is just the average values ofu on a circle centered on that point. ("Mean value theorem")
• The maximum and minimum values of u are thereforealways on the domain boundary (this is true for any smooth shapedomain).

Example 1:: Consider the interior Dirichlet problem

$\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 0 \le r < 2, \qquad u(2, \theta ) = f(\theta ) \equiv \begin{cases} 1, & \ \mbox{if } 0 \le \theta \le \pi , \\ \cos^2 \theta , & \ \mbox{if } \pi \le \theta \le 2\pi . \end{cases} .$

Its solution is known to be

$u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left( \frac{r}{2} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le r \le 2, \quad 0\le \theta \le 2\pi .$

To satisfy the boundary condition

$$u(2, \theta ) = f(\theta ) ,$$ we calculate the coefficients as usual

\begin{eqnarray*} a_0 &=& \frac{1}{\pi} \int_0^{\pi} {\text d}\phi + \frac{1}{\pi} \int_0^{\pi} \cos^2 \phi \,{\text d}\phi = \frac{3}{2} , \\ a_n &=& \frac{1}{\pi} \int_0^{\pi} \cos (n\phi ) \, {\text d}\phi + \frac{1}{\pi} \int_0^{\pi} \cos (n\phi ) \, \cos^2 \phi \,{\text d}\phi = \begin{cases} \frac{1}{4} , & \ \mbox{ if } n=2, \\ 0, & \ \mbox{ if } n \ne 2; \end{cases} \\ b_n &=& \frac{1}{\pi} \int_0^{\pi} \sin (n\phi ) \, {\text d}\phi + \frac{1}{\pi} \int_0^{\pi} \sin (n\phi ) \, \cos^2 \phi \,{\text d}\phi = \begin{cases} \frac{-4}{n\left( n^2 -4 \right)} , & \ \mbox{ if $n$ is odd}, \\ 0, & \ \mbox{ if } n \mbox{ is even}; \end{cases} \end{eqnarray*}

because Mathematica helps

a0 := (1/Pi)*(Integrate[1, {x, 0, Pi}] + Integrate[(Cos[x])^2 , {x, Pi, 2*Pi}])
a[n_] := (1/Pi)*(Integrate[Cos[n*x], {x, 0, Pi}] + Integrate[Cos[n*x]*(Cos[x])^2 , {x, Pi, 2*Pi}])
b[n_] := (1/Pi)*(Integrate[Sin[n*x], {x, 0, Pi}] + Integrate[Sin[n*x]*(Cos[x])^2 , {x, Pi, 2*Pi}])

Therefore, the required solution is

$u(r, \theta ) = \frac{3}{4} + \frac{r^2}{16}\,\cos 2\theta - \frac{4}{\pi}\,\sum_{k\ge 0} \frac{1}{(2k+1) \left[ (2k+1)^2 -4 \right]} \left( \frac{r}{2} \right)^{2k+1} \sin (2k+1)\theta .$

Now we plot the solution using Mathematica . The first step in constructing our table is to define the function and build a table of height values according to the found solution.

u[r_, t_] := )
3/4 + r^2/16*Cos[2*t] - )
4/Pi*Sum[(r/2)^(2*k + 1) /(2*k + 1)/((2*k + 1)^2 - 4) * Sin[(2*k + 1)*t], {k, 0, 20}] )
(* we choose to keep 20 terms in truncated series *)
data = Table[u[r, t], {r, 0, 2, 0.1}, {t, 0, 2*Pi, 0.1}];

Now we can plot the table of data in Cartesian coordinates by using ListPlot3D:

graph = ListPlot3D[data, DataRange -> {{0, 2}, {2, 2*Pi}}]

Next we convert the graph into polar coordinates. First, we define the subroutine:

MyListPolarPlot3D[data_, rRange_, thetaRange_, zRange_] :=
Module[{},
gr1 = ListPlot3D[data,
DataRange -> {{rRange[[1]], rRange[[2]]}, {thetaRange[[1]],
thetaRange[[2]]}}, DisplayFunction -> Identity,
ColorFunction -> "SolarColors", ColorFunction -> Automatic,
MeshFunctions -> {Function[{x, y, z}, x*Cos[y]],
Function[{x, y, z}, x*Sin[y]]}, BoundaryStyle -> None,
ColorFunctionScaling -> True, Mesh -> 30];
substitution = {r_, theta_, z_} -> {r Cos[theta], r Sin[theta], z};
gr2 = gr1 /.
GraphicsComplex[p_List, rest__] :> GraphicsComplex[ReplaceAll[p, substitution], rest];
(* *Retitle the axes and show final graph*)
Return[Show[gr2, AxesLabel -> {"X", "Y", "Z"},
DisplayFunction -> $DisplayFunction, BoxRatios -> {1, 1, 0.8}, PlotRange -> {{-0.65*rRange[[2]], 0.65*rRange[[2]]}, {-0.65 rRange[[2]], 0.65*rRange[[2]]}, {zRange[[1]], zRange[[2]]}}]];] Finally, we plot our data in polar coordinates: MyListPolarPlot3D[data, {0.0, 2.0}, {0.0, 2*Pi}, {0, 1.15}] Example 2:: Consider the interior Dirichlet problem $\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 0 \le r < 3, \qquad u(3, \theta ) = 2\sin 4\theta - 3\,\cos 7\theta .$ Its solution is known to be $u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left( \frac{r}{3} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le r \le 3, \quad 0\le \theta \le 2\pi .$ To satisfy the boundary condition $$u(2, \theta ) = f(\theta ) ,$$ we don't need to use Euler's formalae because we have expansion $u(3, \theta ) = f(\theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le r \le 3, \quad 0\le \theta \le 2\pi ,$ where f is a combination of eigenfunctions. So we know that all coefficients in the above expansion are zeroes except n = 4 and n = 7. Hence, $b_4 =2 \qquad\mbox{and} \qquad a_7 = -3 .$ This yields $u(r, \theta ) = 2 \left( \frac{r}{3} \right)^4 \sin 4\theta -3 \left( \frac{r}{3} \right)^7 \sin 7\theta .$ Dirichlet Problem outside the circle Consider the exterior Dirichlet problem for a circle of radius a: $u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\,u_{\theta\theta} =0, \qquad r> a, \qquad u(a,\theta ) = f(\theta ) ,$ where f is a given function. Its general solution can be expressed as $$\label{EqPolar.5} u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left( \frac{a}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad r>a, \quad 0\le \theta \le 2\pi .$$ The coefficients of the above sum are obtained from the boundary conditions that lead to Poisson’s formula: $u(r, \theta ) = \frac{1}{2\pi} \,\int_0^{2\pi} f(\phi ) \,P_r (\theta - \phi )\,{\text d}\phi , \qquad P_r (x) = \frac{r^2 - a^2}{a^2 + r^2 -2ar\,\cos x} , \quad a \le r.$ Example 3:: Consider the exterior Dirichlet problem $\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 2 \le r , \qquad u(2, \theta ) = f(\theta ) \equiv \begin{cases} 1, & \ \mbox{if } 0 \le \theta \le \pi , \\ \cos^2 \theta , & \ \mbox{if } \pi \le \theta \le 2\pi . \end{cases} .$ Its solution is known to be $u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left( \frac{2}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 2\le r , \quad 0\le \theta \le 2\pi .$ The above series satisfies the boundary condition only when $u(2, \theta ) = f(\theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le \theta \le 2\pi .$ The explicit values of these coefficients were found previously, and we get the required solution: $u(r, \theta ) = \frac{3}{4} + \frac{1}{r^2}\,\cos 2\theta - \frac{4}{\pi}\,\sum_{k\ge 0} \frac{1}{(2k+1) \left[ (2k+1)^2 -4 \right]} \left( \frac{2}{r} \right)^{2k+1} \sin (2k+1)\theta .$ Example 4:: Consider the exterior Dirichlet problem $\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 3 < r, \qquad u(3, \theta ) = 2\sin 4\theta - 3\,\cos 7\theta .$ A solution of Laplace's equation in outside the circle is known to be $u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left( \frac{3}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad r>3, \quad 0\le \theta \le 2\pi .$ Since the input function is a linear combination of eigenfunctions, we conclude that all coefficients in the above series are zeroes except n = 4 and n = 7: $u(r, \theta ) = 2 \left( \frac{3}{r} \right)^4 \sin 4\theta - 3 \left( \frac{3}{r} \right)^7 \sin 7\theta .$ Neumann Problem inside the circle Consider the interior Neumann problem for a circle of radius a: $u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\,u_{\theta\theta} =0, \qquad 0 \le r< a, \qquad \left. \frac{\partial u}{\partial r} \right\vert_{r=a} = g(\theta ) , \qquad \int_0^{2\pi} g(\theta )\,{\text d}\theta = 0 ,$ where g is a given function. The general solution of Laplace's equation inside a circle of radius a is $$\label{EqPolar.6} u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left( \frac{r}{a} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0 \le r<a, \quad 0\le \theta \le 2\pi .$$ Its derivative with respect to r becomes (assuming uniform convergence of the above series) $\frac{\partial u}{\partial r} = \sum_{n\ge 1} \frac{n}{r} \left( \frac{r}{a} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0 \le r<a, \quad 0\le \theta \le 2\pi .$ Setting r equals to a yields $\left. \frac{\partial u}{\partial r} \right\vert_{r=a} = g(\theta ) = \sum_{n\ge 1} \frac{n}{a} \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le \theta \le 2\pi .$ The coefficients of the Fourier series are calculated according to Euler's formulae \begin{eqnarray*} a_0 &=& \frac{1}{\pi} \int_0^{2\pi} g(\phi )\,{\text d}\phi =0 , \\ \frac{n}{a}\,a_n &=& \frac{1}{\pi} \int_0^{2\pi} g(\phi )\,\cos (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots ; \\ \frac{n}{a}\,b_n &=& \frac{1}{\pi} \int_0^{2\pi} g(\phi )\, \sin (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots . \end{eqnarray*} Note that the coefficient a0 must be zero because the corresponding Fourier series for $\left. \frac{\partial u}{\partial r} \right\vert_{r=a} = \sum_{n\ge 1} \frac{n}{a} \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le \theta \le 2\pi ;$ does not contain a free term. Therefore, a Neumann problem has a solution if and only if the integral over the boundary vanishes: $\int_0^{2\pi} g(\phi )\,{\text d}\phi =0 .$ Then the general solution to a Neumann problem is not unique but up to an arbitrary additive constant: \begin{eqnarray*} u(r, \theta ) &=& C + \sum_{n\ge 1} \left( \frac{r}{a} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] \\ &=& C + \sum_{n\ge 1} \left( \frac{r}{a} \right)^n \left[ \frac{a}{n\pi} \,\int_0^{2\pi} g(\phi )\,\cos n\phi \,{\text d}\phi \, \cos n\theta + \frac{a}{n\pi} \,\int_0^{2\pi} g(\phi )\,\sin n\phi \,{\text d}\phi \, \sin n\theta \right] \\ &=& C + \frac{a}{\pi} \,\sum_{n\ge 1} \left( \frac{r}{a} \right)^n \int_0^{2\pi} {\text d}\phi \, \cos n \left( \theta - \phi \right) \frac{1}{n} \\ &=& C + \frac{a}{\pi} \,\Re\,\sum_{n\ge 1} \frac{1}{n} \left( \frac{r}{a}\, e^{{\bf j} (\theta - \phi )} \right)^n \\ &=& C - \frac{a}{\pi} \,\Re\,\ln \left( 1 - \frac{r}{a}\, e^{{\bf j} (\theta - \phi )} \right) \end{eqnarray*} since according to Mathematica Sum[z^n /n, {n, 1, Infinity} ] -Log[1 - z] We are not going to simplify the above formula because it requires a solid knowledge of functions of a complex variable. Example 5:: Consider the interior Dirichlet problem $\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 0 \le r < 5, \qquad \left. \frac{\partial u}{\partial r} \right\vert_{r=5} = g(\theta ) \equiv \begin{cases} -1/2, & \ \mbox{if } 0 \le \theta \le \pi , \\ \cos^2 \theta , & \ \mbox{if } \pi \le \theta \le 2\pi . \end{cases} .$ The general solution of the Neumann problem for Laplace's equation is known to be $u(r, \theta ) = C + \sum_{n\ge 1} \left( \frac{r}{5} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le r \le 5, \quad 0\le \theta \le 2\pi ;$ where C is an arbitrary constant. The given problem has a solution only if the integral of g over the boundary is zero. Mathematica helps Integrate[(Cos[x])^2, {x, Pi, 2*Pi}]/Pi - Integrate[1/2, {x, 0, Pi}]/Pi Next we calculate coefficients in the Fourier series \begin{eqnarray*} a_n &=& \frac{5}{n\pi} \int_0^{2\pi} g(\phi )\,\cos (n\phi ) \, {\text d}\phi = \begin{cases} 0, & \ \mbox{ for } n \ne 2, \\ \frac{5}{8} , & \ \mbox{ for } n=2; \end{cases} \qquad n=1,2,\ldots ; \\ b_n &=& \frac{5}{n\pi} \int_0^{2\pi} g(\phi )\, \sin (n\phi ) \, {\text d}\phi = \begin{cases} 0, & \ \mbox{ for &n& even}, \\ -\frac{5 \left( 3n^2 -8 \right)}{n^2 \left( n^2 -4 \right) \pi} , & \ \mbox{ for$n\$ odd}; \end{cases} \qquad n=1,2,\ldots . \end{eqnarray*}

a[n_] := 5*Integrate[Cos[n*x]*(Cos[x])^2, {x, Pi, 2*Pi}]/Pi /n - 5*Integrate[1/2*Cos[n*x], {x, 0, Pi}]/Pi/n
b[n_] := 5*Integrate[Sin[n*x]*(Cos[x])^2, {x, Pi, 2*Pi}]/Pi /n - 5*Integrate[1/2*Sin[n*x], {x, 0, Pi}]/Pi/n

Therefore, the solution becomes

$u(r, \theta ) = C + \frac{r^2}{40}\,\cos 2\theta - \sum_{k\ge 0} \frac{3(2k+1)^2 -8}{(2k+1)^2 \left( (2k+1)^2 -4 \right)} \left( \frac{r}{5} \right)^{2k+1} \sin (2k+1)\theta , \qquad 0\le r \le 5, \quad 0\le \theta \le 2\pi .$

Example 6:: Consider the interior Neumann problem

$\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 0 \le r < 5, \qquad \left. \frac{\partial u}{\partial r} \right\vert_{r=5} = f(\theta ) \equiv 2\sin 4\theta - 3\,\cos 7\theta .$

First, we check with Mathematica that the given Neumann problem has a solution:

Integrate[2*Sin[4*x] - 3*Cos[7*x], {x,0,2*Pi}]

Then we substitute the general solution

$u(r, \theta ) = C + \sum_{n\ge 1} \left( \frac{r}{5} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right]$

into the boundary condition

$\left. \frac{\partial u}{\partial r} \right\vert_{r=5} = \sum_{n\ge 1} \left. \left. \frac{n}{5} \left( \frac{r}{5} \right)^n \right\vert_{r=5} \left[ a_n \cos n\theta + b_n \sin n\theta \right] = 2\sin 4\theta - 3\,\cos 7\theta .$

Comparison of both sides leads to conclusion that all coefficients of the Fourier series are zeroes except two indices:

$\frac{7}{5} \, a_7 = -3, \qquad\mbox{and} \qquad \frac{4}{5} \, b_4 = 2 .$

Hence, the required solution becomes

$u(r, \theta ) = C + \frac{5}{2} \left( \frac{r}{5} \right)^4 \sin 4\theta - \frac{15}{7} \left( \frac{r}{5} \right)^7 \sin 7\theta .$

Neumann Problem outside the circle

Consider the exterior Neumann problem for a circle of radius 𝑎:

$u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\,u_{\theta\theta} =0, \qquad r> a, \qquad \left. \frac{\partial u}{\partial r} \right\vert_{r=a} = g(\theta ) , \qquad \int_0^{2\pi} g(\theta )\,{\text d}\theta = 0 ,$

where g is a given function. Note that the Neumann problem has no solution if the integral of the given function g over the circumference (which is the boundary of the circle) is not zero. The general solution of Laplace's equation outside a circle of radius 𝑎 is

$u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left( \frac{a}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad r>a, \quad 0\le \theta \le 2\pi .$

Its derivative with respect to r becomes (assuming uniform convergence of the above series)

$\frac{\partial u}{\partial r} = - \sum_{n\ge 1} \frac{n}{r} \left( \frac{a}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad r>a, \quad 0\le \theta \le 2\pi .$

To satisfy the boundary condition, we have to check validity of

$\lim_{r\to a} \,\frac{\partial u}{\partial r} = - \sum_{n\ge 1} \frac{n}{a} \left[ a_n \cos n\theta + b_n \sin n\theta \right] = g(\theta ), \qquad 0\le \theta \le 2\pi .$

Since the last identity is just the Fourier series for the given function g, we find the coefficients according to Euler's formulae:

\begin{eqnarray*} a_0 &=& \frac{1}{\pi} \int_0^{2\pi} g(\phi )\,{\text d}\phi =0 , \\ -\frac{n}{a}\,a_n &=& \frac{1}{\pi} \int_0^{2\pi} g(\phi )\,\cos (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots ; \\ -\frac{n}{a}\,b_n &=& \frac{1}{\pi} \int_0^{2\pi} g(\phi )\, \sin (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots . \end{eqnarray*}

This allows us to determine the solution of exterior Neumann problem outside the circle:

$u(r, \theta ) = C + \sum_{n\ge 1} \left( \frac{a}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad r>a, \quad 0\le \theta \le 2\pi ,$

where C is an arbitrary constant and

\begin{eqnarray*} a_n &=& -\frac{a}{n\pi} \int_0^{2\pi} g(\phi )\,\cos (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots ; \\ b_n &=& -\frac{a}{n\pi} \int_0^{2\pi} g(\phi )\, \sin (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots . \end{eqnarray*}

Example 7:: Consider the exterior Neumann problem

$\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 5 < r, \qquad \left. \frac{\partial u}{\partial r} \right\vert_{r=5} = f(\theta ) \equiv \begin{cases} -1/2, & \ \mbox{if } 0 \le \theta \le \pi , \\ \cos^2 \theta , & \ \mbox{if } \pi \le \theta \le 2\pi . \end{cases} .$

Its solution is known to be

$u(r, \theta ) = C + \sum_{n\ge 1} \left( \frac{5}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad r \ge 5, \quad 0\le \theta \le 2\pi ,$

where C is an arbitrary constant. We determine the values of coefficients from known formulas:

\begin{eqnarray*} a_n &=& -\frac{5}{n\pi} \int_0^{2\pi} g(\phi )\,\cos (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots ; \\ b_n &=& -\frac{5}{n\pi} \int_0^{2\pi} g(\phi )\, \sin (n\phi ) \, {\text d}\phi , \qquad n=1,2,\ldots ; \end{eqnarray*}

that were evaluated in the previous example. Thus, we get

$u(r, \theta ) = C - \frac{r^2}{40}\,\cos 2\theta + \sum_{k\ge 0} \frac{3(2k+1)^2 -8}{(2k+1)^2 \left( (2k+1)^2 -4 \right)} \left( \frac{5}{r} \right)^{2k+1} \sin (2k+1)\theta , \qquad r \ge 5, \quad 0\le \theta \le 2\pi .$

Example 8:: Consider the exterior Neumann problem

$\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad r > 4, \qquad \left. \frac{\partial u}{\partial r} \right\vert_{r=4} = g(\theta ) \equiv 2\sin 4\theta - 3\,\cos 7\theta .$

Substituting the general solution (which is assumed to be represented through a uniformly convergent series)

$u(r, \theta ) = C + \sum_{n\ge 1} \left( \frac{4}{r} \right)^n \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad r \ge 4, \quad 0\le \theta \le 2\pi ,$

where C is an arbitrary constant, into the boundary condition

$\lim_{r\to 4} \, \frac{\partial u}{\partial r} = - \sum_{n\ge 1} \frac{n}{4} \left[ a_n \cos n\theta + b_n \sin n\theta \right] = 2\sin 4\theta - 3\,\cos 7\theta ,$

we see that all coefficients must be zeroes, but two of them not:

$b_4 = -2 \qquad\mbox{and} \qquad a_7 = \frac{12}{7} .$

So

$u(r, \theta ) = C + \frac{12}{7} \left( \frac{4}{r} \right)^7 \cos 7\theta - 2 \left( \frac{4}{r} \right)^4 \sin 4\theta , \qquad r \ge 4, \quad 0\le \theta \le 2\pi .$

Laplace Equation in a Donut

We consider boundary value problems inside the domain bounded by two circles, which is usually called a donut. In Mathematica, we plot a donut as follows.

RegionPlot[1 < Abs[x + I y] < 2, {x, -2, 2}, {y, -2, 2}, ImagePadding -> 1, PlotStyle -> Blue]

Example: Consider the mixed boundary value problem

$\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} = 0 , \qquad 2 \le r < 5, \qquad u(2, \theta ) = f(\theta ) \equiv \begin{cases} 1, & \ \mbox{if } 0 \le \theta \le \pi , \\ \cos^2 \theta , & \ \mbox{if } \pi \le \theta \le 2\pi ; \end{cases} \qquad \left. \frac{\partial u}{\partial r} \right\vert_{r=5} = g(\theta ) \equiv 2\sin 4\theta - 3\,\cos 7\theta .$

The general solution to Laplace's equation is known to be

$u(r, \theta ) = \frac{a_0}{2} + d_0\ln r + \sum_{n\ge 1} \left( \frac{r}{2} \right)^n \left[ a_n \cos \left( n \theta \right) + b_n \sin \left( n \theta \right) \right] + \sum_{n\ge 1} \left( \frac{5}{r} \right)^n \left[ c_n \cos \left( n \theta \right) + d_n \sin \left( n \theta \right) \right] .$

This function will be a required solution if the following boundary conditions hold:

\begin{eqnarray*} u(2, \theta ) &=& \frac{a_0}{2} + d_0\ln 2 + \sum_{n\ge 1} \left[ a_n \cos \left( n \theta \right) + b_n \sin \left( n \theta \right) \right] + \sum_{n\ge 1} \left( \frac{5}{2} \right)^n \left[ c_n \cos \left( n \theta \right) + d_n \sin \left( n \theta \right) \right] \\ &=& \frac{a_0}{2} + d_0\ln 2 + \sum_{n\ge 1} \left[ a_n + \left( \frac{5}{2} \right)^n c_n \right] \cos \left( n \theta \right) + \sum_{n\ge 1} \left[ b_n + \left( \frac{5}{2} \right)^n d_n \right] \sin \left( n \theta \right) = f(\theta ) \equiv \begin{cases} 1, & \ \mbox{if } 0 \le \theta \le \pi , \\ \cos^2 \theta , & \ \mbox{if } \pi \le \theta \le 2\pi ; \end{cases} \\ \left. \frac{\partial u}{\partial r} \right\vert_{r=5} &=& \frac{d_0}{5} + \sum_{n\ge 1} \left[ \frac{n}{5} \left( \frac{5}{2} \right)^n a_n - \frac{n}{5}\, c_n \right] \cos \left( n \theta \right) + \sum_{n\ge 1} \left[ \frac{n}{5} \left( \frac{5}{2} \right)^n b_n - \frac{n}{5}\, d_n \right] \sin \left( n \theta \right) = 2\sin 4\theta - 3\,\cos 7\theta . \end{eqnarray*}

Since the function f has the Fourier expansion

$f(\theta ) = \frac{3}{4} + \frac{1}{4}\,\cos 2\theta - \sum_{k\ge 0} \frac{4}{(2k+1) \left( (2k+1)^2 -4 \right)} \,\sin (2k+1)\theta ,$

and the function g is itself the Fourier series, we get the following system of algebraic equations:

\begin{eqnarray*} \frac{3}{4} &=& \frac{a_0}{2} + d_0\ln 2 , \\ \frac{1}{4}\,\delta_{2,n} &=& a_n + \left( \frac{5}{2} \right)^n c_n , \\ - \frac{4}{(2k+1) \left( (2k+1)^2 -4 \right)} &=& b_{2k+1} + \left( \frac{5}{2} \right)^{2k+1} c_{2k+1} , \\ 0 &=& b_{2k} + \left( \frac{5}{2} \right)^{2k} c_{2k} , \\ \frac{d_0}{5} &=& 0 , \\ \frac{n}{5} \left( \frac{5}{2} \right)^n a_n - \frac{n}{5}\, c_n &=& -3\,\delta_{n,4}\\ \frac{n}{5} \left( \frac{5}{2} \right)^n b_n - \frac{n}{5}\, d_n &=& 2\,\delta_{7,n} , \end{eqnarray*}

where δ is the delta-symbol of Kronecker:

$\delta_{i,j} = \begin{cases} 0, & \ \mbox{if } i \ne j, \\ 1, & \ \mbox{if } i=j . \end{cases}$

The initial coefficients are $$c_0 = \frac{3}{2} , \quad d_0 =0 .$$ Some other coefficients could be related by

$a_n = - c_n \left( \frac{5}{2} \right)^{2k} c_{2k} , \quad n \ne 2, \qquad b_{2k} = - \left( \frac{5}{2} \right)^{2k} c_{2k} , \qquad c_n = a_n \left( \frac{5}{2} \right)^{n} \quad n \ne 4, \qquad d_n = b_n \left( \frac{5}{2} \right)^{n} \quad n \ne 7.$

For other coefficients, we ask Mathematica for help:

Example: Consider the Dirichlet problem for the Laplace's equation inside the annulus

$u_{xx} + u_{yy} = 0 \quad (0 < x^2 + y^2 < 1), \qquad \left. u \right\vert_{\partial\Omega} = f = \begin{cases} 0, & \ \mbox{ on }\quad x^2 + y^2 = 1, \\ 1, & \ \mbox{ at }\quad (0,0) . \end{cases}$

In this case, the condition that f on the boundary ((x² + y² = 1) ∪ (0,0)) is continuous is relaxed somewhat (17. page 138--139)

By symmetry and uniqueness, it is obvious that is the given problem has a solution, it must have a rotational symmetry (that is, the solution must be a function of $$\displaystyle r = \sqrt{x^2 + y^2} ,$$ independent of the polar angle θ). )

It is clear that the solution of the given problem must be of the form

$u = u(r) = a\,\ln r + b,$

where costants 𝑎 and b are to be determiined from the bpundary conditions:

$u (r=1) = b = 0 .$

If 𝑎 ≠ 0, then u is unbounded near the origin. In any way, the second boundary condition at the origin cannot be satisfied. ■

Dirichlet Problem inside a cylinder

In cylindrical coordinates, Laplace's equation is written

$\frac{1}{r} \, \frac{\partial}{\partial r} \left( r\,\frac{\partial u}{\partial r} \right) + \frac{1}{r^2 } \, \frac{\partial^2 u}{\partial \theta^2} + \frac{\partial^2 u}{\partial z^2} = 0 , \qquad 0 \le r < a, \quad 0 < z < \ell .$

We will specify Dirichlet boundary conditions later. We we concentrate on finding the general solution using separation of variables:

$u(r, \theta , z ) = R(r)\, \Theta (\theta )\,Z(z) .$

Substitution of the above form into the Laplace equation yields

$\begin{split} \frac{{\text d}^2 Z}{{\text d} z^2} - k^2 Z &= 0 , \\ \frac{{\text d}^2 \Theta}{{\text d}\theta^2} + \lambda^2 \Theta &= 0 , \\ \frac{{\text d}^2 R}{{\text d} r^2} + \frac{1}{r}\,\frac{{\text d}R}{{\text d}r} + \left( k^2 - \frac{\lambda^2}{r^2} \right) R(r) &=0 . \end{split}$

For Θ we get the Sturm--Liouville problem

$\Theta'' + \lambda^2 \Theta (\theta ) =0, \qquad \Theta (\theta ) = \Theta (\theta + 2\pi ) .$

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