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Return to Part VI of the course APMA0340
Introduction to Linear Algebra
Glossary
Laplace's Equation in Polar Coordinates
For domains whose boundary comprises part of a circle, it is convenient to transform to polar coordinates. We consider Laplace's operator \( \Delta = \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} \) in polar coordinates \( x = r\,\cos \theta \) and \( y = r\,\sin \theta . \) Here x, y are Cartesian coordinates and r, θ are standard polar coordinates on the plane. To determine Laplace's operator in polar coordinates, we use the chain rule
\[
\begin{split}
\frac{\partial}{\partial x} = \frac{\partial r}{\partial x}\,\frac{\partial }{\partial r} + \frac{\partial \theta}{\partial x}\,\frac{\partial }{\partial \theta} ,
\\
\frac{\partial}{\partial y} = \frac{\partial r}{\partial y}\,\frac{\partial }{\partial r} + \frac{\partial \theta}{\partial y}\,\frac{\partial }{\partial \theta} .
\end{split}
\]
To work out these partial derivatives, we need explicit expressions for polar
variables in terms of x and y. Since polar coordinates include
variables r and argument θ (dimensionless because angles are
measured with radians), we need to express Cartesian coordinates
z = (x,y) via polar coordinates (r,θ):
\[
r = \sqrt{x^2 + y^2} > 0, \qquad \theta = \arctan \frac{y}{x}
= \mbox{arg}(z) = \begin{cases}
\left( 1 - \mbox{sign}(x) \right) \frac{\pi}{2} , & \ x\ne 0, \quad y =0 ,
\\
\arctan \left( \frac{y}{x} \right) + \mbox{sign}(y)
\left( 1 - \mbox{sign}(x) \right) \frac{\pi}{2} , & \ x\ne 0, \quad y \ne 0 ,
\\
\frac{\pi}{2}\,\mbox{sign}(y) , & \ x=0, \quad y \ne 0 ,
\end{cases}
\]
where arg(z) is the principal value of the argument (analytic)
function, defined shortly.
Here sign(⋅) is the signum function:
\[
\mbox{sign}(x) = \begin{cases} \phantom{-}1, & \ x> 0 ,
\\
\phantom{-}0, & \ x = 0, \\
-1, & \ x< 0 .
\end{cases}
\]
The analytical (multivaluable) function Arg(⋅) may be introduced as
\[
\mbox{Arg}(z) = \mbox{arg}(z) + 2n\pi , \qquad n \in \mathbb{Z} \ \mbox{ (set of integers)},
\]
with appropriate choices of corresponding branches because both functions
(root and arctangent) are also multi-valued. Note that in polar coordinates,
the root function is always positive. Differentiating with respect to
x and y, we obtain:
\[
\begin{split}
\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \cos \theta , \qquad \frac{\partial r}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} = \sin \theta ,
\\
\frac{\partial \theta}{\partial x} = - \frac{y}{x^2 + y^2} = - \frac{\sin \theta}{r} , \qquad \frac{\partial \theta}{\partial y} = \frac{x}{x^2 + y^2} = \frac{\cos \theta}{r} .
\end{split}
\]
Using subscripts for derivatives, we rewrite the above relations as
\[
\begin{split}
2r\, r_x = 2x, \qquad r_x = \frac{x}{r}, \qquad r_y = \frac{y}{r} ,
\\
\theta_x = \dfrac{\left( - \frac{y}{x^2} \right)}{1 + \left( \frac{y}{x} \right)^2} = - \frac{y}{x^2 + y^2} = -\frac{y}{r^2} , \qquad \theta_y = \frac{x}{r^2} .
\end{split}
\]
Then first derivatives become
\[
\begin{split}
u_x = u_r r_x + u_{\theta} \theta_x = u_r \frac{x}{r} - u_{\theta} \frac{y}{r^2} ,
\\
u_y = u_r r_y + u_{\theta} \theta_y = u_r \frac{y}{r} + u_{\theta} \frac{x}{r^2} .
\end{split}
\]
Finally, we obtain the Laplacian in polar coordinates:
\begin{eqnarray*}
\Delta &=& \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} = \left( \cos\theta \,\frac{\partial}{\partial r} - \frac{\sin \theta}{r} \, \frac{\partial}{\partial \theta} \right)^2 + \left( \sin\theta \,\frac{\partial}{\partial r} + \frac{\cos \theta}{r} \, \frac{\partial}{\partial \theta} \right)^2
\\
&=& \cos^2 \theta \,\frac{\partial^2}{\partial r^2} - \cos\theta\,\frac{\partial}{\partial r} \, \frac{\sin \theta}{r} \, \frac{\partial}{\partial \theta} - \frac{\sin \theta}{r} \, \frac{\partial}{\partial \theta} \, \cos\theta \,\frac{\partial}{\partial r} + \frac{\sin \theta}{r} \, \frac{\partial}{\partial \theta} \,\frac{\sin \theta}{r} \, \frac{\partial}{\partial \theta}
\\
&\quad& + \sin^2 \theta \,\frac{\partial^2}{\partial r^2} + \sin\theta \,\frac{\partial}{\partial r} \, \frac{\cos \theta}{r} \, \frac{\partial}{\partial \theta} + \frac{\cos \theta}{r} \, \frac{\partial}{\partial \theta} \, \sin\theta \,\frac{\partial}{\partial r} + \frac{\cos \theta}{r} \, \frac{\partial}{\partial \theta} \, \frac{\cos \theta}{r} \, \frac{\partial}{\partial \theta}
\\
&=& \frac{\partial^2}{\partial r^2} + 2\cos\theta \, \frac{\sin \theta}{r^2} \, \frac{\partial}{\partial \theta} - 2\cos\theta \, \frac{\sin \theta}{r} \, \frac{\partial^2}{\partial r\,\partial \theta} + \frac{\sin^2 \theta}{r} \, \frac{\partial}{\partial r} + \frac{\sin^2 \theta}{r^2}\, \frac{\partial^2}{\partial \theta^2}
\\
&\quad& - 2\cos\theta \,\frac{\sin \theta}{r^2} \, \frac{\partial}{\partial \theta} + 2\cos\theta \,\frac{\sin \theta}{r}\, \frac{\partial^2}{\partial r\partial \theta} + \frac{\cos^2 \theta}{r}\, \frac{\partial}{\partial r} + \frac{\cos^2 \theta}{r^2} \,\frac{\partial^2}{\partial \theta^2}
\\
&=& \frac{\partial^2}{\partial r^2} + \frac{1}{r}\, \frac{\partial}{\partial r} + \frac{1}{r^2} \, \frac{\partial^2}{\partial \theta^2} .
\end{eqnarray*}
Now we apply separation of variables to find the general solution of the Laplace equation in polar coordinates:
\begin{equation} \label{EqPolar.1}
\Delta u =0 \qquad \iff \qquad \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\, \frac{\partial u}{\partial r} + \frac{1}{r^2} \, \frac{\partial^2 u}{\partial \theta^2} =0 .
\end{equation}
We seek partial nontrivial solutions of the Laplace equation represented as a product \( u(r, \theta ) = R(r)\,\Theta (\theta ) . \) Substituting this product into the Laplace equation, we get upon multiplication by
\[
R'' (r)\, \Theta (\theta ) + \frac{1}{r}\, R' (r)\, \Theta (\theta ) + \frac{1}{r^2} \,R(r)\, \Theta'' (\theta ) =0 \qquad \iff \qquad \frac{r^2 R'' (r)}{R(r)} + \frac{r\,R' (r)}{R(r)} + \frac{\Theta'' (\theta )}{\Theta (\theta )} =0 .
\]
Separation of variables yields
\[
\frac{r^2 R'' (r)}{R(r)} + \frac{r\,R' (r)}{R(r)} =- \frac{\Theta'' (\theta )}{\Theta (\theta )} = \lambda ,
\]
where, as usual, we denote a constant by λ. Then we get two ordinary differential equations for each unknown function:
\[
\begin{split}
r^2 R'' (r) + r\,R' (r) - \lambda\, R(r) &=0 ,
\\
\Theta'' (\theta ) + \lambda \Theta (\theta ) &=0 .
\end{split}
\]
We want to find solutions that are periodic:
\[
\Theta'' (\theta ) + \lambda \Theta (\theta ) =0 , \qquad \Theta (\theta ) = \Theta (\theta + 2\pi ) .
\]
Since the differential equation has a periodic solution only when constant λ is not negative,
\[
\begin{split}
\Theta (\theta ) &= A\,\cos \left( \sqrt{\lambda} \,\theta \right) + B\,\sin \left( \sqrt{\lambda} \, \theta \right) \qquad\mbox{if } \lambda \mbox{ is positive},
\\
\Theta (\theta ) &= A + B\,\theta \qquad\mbox{if } \lambda \mbox{ is zero} ,
\end{split}
\]
we conclude λ must be an integer; its square root branch is convenient (but not necessarily) to choose nonnegative. Therefore, we get a sequence of eigenvalues
\[
\begin{split}
\Theta_n (\theta ) &= A_n \cos \left( n \theta \right) + B_n \sin \left( n \theta \right) \qquad\mbox{for } n=1,2,\ldots ,
\\
\Theta_n (\theta ) &= A_0 \qquad\mbox{for } n=0 ,
\end{split}
\]
with some arbitrary constants A's and B's.
The above two formulae could be united into one
\[
\Theta_n (\theta ) = A_n \cos \left( n \theta \right) + B_n \sin \left( n \theta \right) \qquad n=0,1,2,\ldots ,
\]
because
\[
r^2 R'' (r) + r\,R' (r) - n^2 R(r) =0 .
\]
The above equation is an example of Euler's equation; so we seek its solution in the form
\[
\alpha (\alpha -1) + \alpha - n^2 =0 \qquad \iff \qquad \alpha^2 - n^2 =0 .
\]
Thus, we have two distinct solutions
\[
\begin{split}
R_n (r) &= C_n r^n + D_n r^{-n} \qquad\mbox{for } n=1,2,3,\dots ,
\\
R_0 (r) &= C_0 + D_0 \ln r \qquad\mbox{for } n=0.
\end{split}
\]
Now we conclude that the most general solution of the Laplace partial differential equation would be a sum of the solutions found for all separation constants:
\begin{eqnarray*}
u(r, \theta ) &=& C_0 + D_0 \ln r + \sum_{n\ge 1} \left[ A_n \cos \left( n \theta \right) + B_n \sin \left( n \theta \right) \right] \left( C_n r^n + D_n r^{-n} \right)
\\
&=& \frac{a_0}{2} + d_0\ln r + \sum_{n\ge 1} \left( \frac{r}{a} \right)^n \left[ a_n \cos \left( n \theta \right) + b_n \sin \left( n \theta \right) \right] + \sum_{n\ge 1} \left( \frac{a}{r} \right)^n \left[ c_n \cos \left( n \theta \right) + d_n \sin \left( n \theta \right) \right] .
\end{eqnarray*}
Next, we give some example of boundary value problems for domains that are naturally described in polar coordinates. In each example, we first present the general solutions and then show some particular cases.
Dirichlet Problem inside the circle
Consider the interior Dirichlet problem for a circle of radius a:
\[
u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\,u_{\theta\theta} =0, \qquad 0\le r < a, \qquad u(a,\theta ) = f(\theta ) ,
\]
where f is a given function.
circle = Graphics[{Orange, Disk[{0, 0}, 1]}]
arrow1 = Graphics[{Red, Arrowheads[0.05], Arrow[{{0, 0}, {1.3, 0}}]}]
arrow2 = Graphics[{Red, Arrowheads[0.05], Arrow[{{0, 0}, {0, 1.25}}]}]
arrow = Graphics[{Blue, Arrowheads[0.05], Arrow[{{0, 0}, {1/Sqrt[2], 1/Sqrt[2]}}]}]
text = Graphics[ Text[Style[" \[CapitalDelta]u = 0", FontSize -> 14, Black], {-0.0, -0.35}]]
text1 = Graphics[Text[Style["x", FontSize -> 14, Red], {1.25, 0.1}]]
text2 = Graphics[Text[Style["y", FontSize -> 14, Red], {0.1, 1.25}]]
textt = Graphics[ Text[Style["\[Theta]", FontSize -> 14, Black], {0.25, 0.1}]]
texta = Graphics[ Text[Style["a", FontSize -> 14, Black], {0.55, 0.66}]]
textf = Graphics[ Text[Style["f(\[Theta])", FontSize -> 14, Black], {-0.75, 0.86}]]
Show[circle, arrow, arrow1, arrow2, text, text1, text2, textt, texta, textf]
arrow1 = Graphics[{Red, Arrowheads[0.05], Arrow[{{0, 0}, {1.3, 0}}]}]
arrow2 = Graphics[{Red, Arrowheads[0.05], Arrow[{{0, 0}, {0, 1.25}}]}]
arrow = Graphics[{Blue, Arrowheads[0.05], Arrow[{{0, 0}, {1/Sqrt[2], 1/Sqrt[2]}}]}]
text = Graphics[ Text[Style[" \[CapitalDelta]u = 0", FontSize -> 14, Black], {-0.0, -0.35}]]
text1 = Graphics[Text[Style["x", FontSize -> 14, Red], {1.25, 0.1}]]
text2 = Graphics[Text[Style["y", FontSize -> 14, Red], {0.1, 1.25}]]
textt = Graphics[ Text[Style["\[Theta]", FontSize -> 14, Black], {0.25, 0.1}]]
texta = Graphics[ Text[Style["a", FontSize -> 14, Black], {0.55, 0.66}]]
textf = Graphics[ Text[Style["f(\[Theta])", FontSize -> 14, Black], {-0.75, 0.86}]]
Show[circle, arrow, arrow1, arrow2, text, text1, text2, textt, texta, textf]
\begin{equation} \label{EqPolar.2}
u(r, \theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left[ a_n \cos \left( n \theta \right) + b_n \sin \left( n \theta \right) \right] \frac{r^n}{a^n} ,
\end{equation}
where arbitrary constants should be determined from the Dirichlet boundary condition:
\[
u(a, \theta ) = a_0 /2 + \sum_{n\ge 1} \left[ a_n \cos \left( n \theta \right) + b_n \sin \left( n \theta \right) \right] = f(\theta ) .
\]
This is a Fourier series for f, and its coefficients are known to be
\begin{equation} \label{EqPolar.3}
\begin{split}
a_n &= \frac{1}{\pi} \int_0^{2\pi} f(\phi )\,\cos (n\phi )\,{\text d}\phi , \qquad n=0,1,2,\ldots ;
\\
b_n &= \frac{1}{\pi} \int_0^{2\pi} f(\phi )\,\sin (n\phi )\,{\text d}\phi , \qquad n=1,2,\ldots .
\end{split}
\end{equation}
Inserting the Fourier coefficient formulas into the general
solution, we obtain
\begin{eqnarray*}
u(r, \theta ) &=& a_0 /2 + \sum_{n\ge 1} \left[ a_n \cos \left( n \theta \right) + b_n \sin \left( n \theta \right) \right] \frac{r^n}{a^n}
\\
&=& \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\phi )\,{\text d}\phi + \frac{1}{\pi} \sum_{n\ge 1} \frac{r^n}{a^n} \left[ \cos \left( n \theta \right) \int_{-\pi}^{\pi} f(\phi )\,\cos (n\phi )\,{\text d}\phi + \sin \left( n \theta \right) \int_{-\pi}^{\pi} f(\phi )\,\sin (n\phi )\,{\text d}\phi \right]
\\
&=& \frac{1}{\pi} \int_{-\pi}^{\pi} f(\phi )\,{\text d}\phi \left\{ \frac{1}{2} + \sum_{n\ge 1} \frac{r^n}{a^n} \left[ \cos (n\theta )\,\cos n\phi ) + \sin (n\theta )\,\sin n\phi ) \right] \right\} .
\end{eqnarray*}
Using the identity
\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*}
We consider the sum, which is the geometric series:
\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*}
where
\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:
\begin{equation} \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.
\end{equation}
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 of u on a circle centered on that point. ("Mean value theorem")
- The maximum and minimum values of u are therefore always on the domain boundary (this is true for any smooth shape domain).
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
\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
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}])
\[
u(r, \theta ) = \frac{3}{4} + \frac{r^2}{16}\,\cos 2\theta - \frac{4}{\pi}\,\sum_{k\ge 0} \frac{1}{(2k+1) \left[ (2k+1)^2 -4 \right]} \left( \frac{r}{2} \right)^{2k+1} \sin (2k+1)\theta .
\]
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:
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}];
graph = ListPlot3D[data, DataRange -> {{0, 2}, {2, 2*Pi}}]
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:
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]]}}]];]
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
\[
u(3, \theta ) = f(\theta ) = \frac{a_0}{2} + \sum_{n\ge 1} \left[ a_n \cos n\theta + b_n \sin n\theta \right] , \qquad 0\le r \le 3, \quad 0\le \theta \le 2\pi ,
\]
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
\begin{equation} \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 .
\end{equation}
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
\begin{equation} \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 .
\end{equation}
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
0
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
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
\[
u(r, \theta ) = C + \frac{r^2}{40}\,\cos 2\theta - \sum_{k\ge 0} \frac{3(2k+1)^2 -8}{(2k+1)^2 \left( (2k+1)^2 -4 \right)} \left( \frac{r}{5} \right)^{2k+1} \sin (2k+1)\theta , \qquad 0\le r \le 5, \quad 0\le \theta \le 2\pi .
\]
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}]
0
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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