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# 12.4: Laplace's Equation in Polar Coordinates

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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In Section 12.3 we solved boundary value problems for Laplace’s equation over a rectangle with sides parallel to the $$x,y$$-axes. Now we’ll consider boundary value problems for Laplace’s equation over regions with boundaries best described in terms of polar coordinates. In this case it is appropriate to regard $$u$$ as function of $$(r,\theta)$$ and write Laplace’s equation in polar form as

$\label{eq:12.4.1} u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta\theta}=0,$

where

$r=\sqrt{x^2+y^2}\quad \text{and} \quad \theta=\cos^{-1}\frac{x}{r}=\sin^{-1}\frac{x}{r}. \nonumber$

We begin with the case where the region is a circular disk with radius $$\rho$$, centered at the origin; that is, we want to define a formal solution of the boundary value problem

$\label{eq:12.4.2} \begin{array}{c}{u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta}=0,\quad 0<r<\rho ,\quad -\pi \leq \theta <\pi ,}\\{u(\rho ,\theta )=f(\theta ),\quad -\pi\leq\theta <\pi }\end{array}$

(Figure $$\PageIndex{1}$$). Note that Equation \ref{eq:12.4.2} imposes no restriction on $$u(r,\theta)$$ when $$r=0$$. We’ll address this question at the appropriate time. Figure $$\PageIndex{1}$$: The boundary value problem \ref{eq:12.4.2}

We first look for products $$v(r,\theta)=R(r)\Theta(\theta)$$ that satisfy Equation \ref{eq:12.4.1}. For this function,

$v_{rr}+\frac{1}{r}v_r+\frac{1}{r^2}v_{\theta\theta}= R''\Theta+\frac{1}{r}R'\Theta +\frac{1}{r^2}R\Theta''= 0 \nonumber$

for all $$(r,\theta)$$ with $$r\ne0$$ if

$\frac{r^2R''+rR'}{R}=-\frac{\Theta''}{\Theta}=\lambda, \nonumber$

where $$\lambda$$ is a separation constant. (Verify.) This equation is equivalent to

$\Theta''+\lambda\Theta=0 \nonumber$

and

$\label{eq:12.4.3} r^2R''+rR'-\lambda R=0.$

Since $$(r,\pi)$$ and $$(r,-\pi)$$ are the polar coordinates of the same point, we impose periodic boundary conditions on $$\Theta$$; that is,

$\label{eq:12.4.4} \Theta''+\lambda\Theta=0,\quad \Theta(-\pi)=\Theta(\pi), \quad \Theta'(-\pi)=\Theta'(\pi).$

Since we don’t want $$R\Theta$$ to be identically zero, $$\lambda$$ must be an eigenvalue of Equation \ref{eq:12.4.4} and $$\Theta$$ must be an associated eigenfunction. From Theorem 11.1.6, the eigenvalues of Equation \ref{eq:12.4.4} are $$\lambda_0=0$$ with associated eigenfunctions $$\Theta_0=1$$ and, for $$n=1,2,3,\dots,$$ $$\lambda_n=n^2$$, with associated eigenfunction $$\cos n\theta$$ and $$\sin n\theta$$ therefore,

$\Theta_n=\alpha_n\cos n\theta+\beta_n\sin n\theta \nonumber$

where $$\alpha_n$$ and $$\beta_n$$ are constants.

Substituting $$\lambda=0$$ into Equation \ref{eq:12.4.3} yields the

$r^2R''+rR'=0,\nonumber$

so

$\frac{R_0''}{R_0'}=-\frac{1}{r},\nonumber$

$R_0'=\frac{c_1}{r},\nonumber$

and

$\label{eq:12.4.5} R_0=c_2+c_1\ln r.$

If $$c_1\ne0$$ then

$\lim_{r\to0+}|R_0(r)|=\infty,\nonumber$

which doesn’t make sense if we interpret $$u_0(r,\theta)=R_0(r)\Theta_0(\theta)=R_0(r)$$ as the steady state temperature distribution in a disk whose boundary is maintained at the constant temperature $$R_0(\rho)$$. Therefore we now require $$R_0$$ to be bounded as $$r\to0+$$. This implies that $$c_1=0$$, and we take $$c_2=1$$. Thus, $$R_0=1$$ and $$v_0(r,\theta)=R_0(r)\Theta_0(\theta)=1$$. Note that $$v_0$$ satisfies Equation \ref{eq:12.4.2} with $$f(\theta)=1$$.

Substituting $$\lambda=n^2$$ into Equation \ref{eq:12.4.3} yields the Euler equation

$\label{eq:12.4.6} r^2R_n''+rR_n'-n^2 R_n=0$

for $$R_n$$. The indicial polynomial of this equation is

$s(s-1)+s-n^2=(s-n)(s+n),\nonumber$

so the general solution of Equation \ref{eq:12.4.6} is

$\label{eq:12.4.7} R_n=c_1r^n+c_2r^{-n},$

by Theorem 7.4.3. Consistent with our previous assumption on $$R_0$$, we now require $$R_n$$ to be bounded as $$r\to0+$$. This implies that $$c_2=0$$, and we choose $$c_1=\rho^{-n}$$. Then $$R_n(r)=r^n/\rho^n$$, so

$v_n(r,\theta)=R_n(r)\Theta_n(\theta)=\frac{r^n}{\rho^n}(\alpha_n\cos n\theta+\sin n\theta).\nonumber$

Now $$v_n$$ satisfies Equation \ref{eq:12.4.2} with

$f(\theta)=\alpha_n\cos n\theta+\beta_n\sin n\theta.\nonumber$

More generally, if $$\alpha_0$$, $$\alpha_1$$,…, $$\alpha_m$$ and $$\beta_1$$, $$\beta_2$$, …, $$\beta_m$$ are arbitrary constants then

$u_m(r,\theta)=\alpha_0+\sum_{n=1}^m\frac{r^n}{\rho^n}(\alpha_n\cos n\theta+\beta_n\sin n\theta)\nonumber$

satisfies Equation \ref{eq:12.4.2} with

$f(\theta) =\alpha_0+\sum_{n=1}^m(\alpha_n\cos n\theta+\beta_n\sin n\theta).\nonumber$

This motivates the next definition.

Theorem $$\PageIndex{1}$$

The bounded formal solution of the boundary value problem Equation \ref{eq:12.4.2} is

$\label{eq:12.4.8} u(r,\theta)=\alpha_0+\sum_{n=1}^\infty\frac{r^n}{\rho^n}(\alpha_n\cos n\theta+\beta_n\sin n\theta),$

where

$F(\theta) =\alpha_0+\sum_{n=1}^\infty(\alpha_n\cos n\theta+\beta_n\sin n\theta)\nonumber$

is the Fourier series of $$f$$ on $$[-\pi,\pi]$$; that is,

$\alpha_0=\frac{1}{2\pi}\int_{-\pi}^\pi f(\theta)\,d\theta,\nonumber$

and

$\alpha_n=\frac{1}{\pi}\int_{-\pi}^\pi f(\theta)\cos n\theta\,d\theta \quad \text{and} \quad \beta_n=\frac{1}{\pi}\int_{-\pi}^\pi f(\theta)\sin n\theta\,d\theta, \quad n=1,2,3,\dots.\nonumber$

Since $$\sum_{n=0}^\infty n^k(r/\rho)^n$$ converges for every $$k$$ if $$0< r<\rho$$, Theorem 12.1.2 can be used to show that if $$0< r<\rho$$ then Equation \ref{eq:12.4.8} can be differentiated term by term any number of times with respect to both $$r$$ and $$\theta$$. Since the terms in Equation \ref{eq:12.4.8} satisfy Laplace’s equation if $$r>0$$, Equation \ref{eq:12.4.8} satisfies Laplace’s equation if $$0 $F(\theta)=f(\theta),\quad -\pi\le\theta<\pi.\nonumber$ From Theorem 11.2.4, this is true if \(f$$ is continuous and piecewise smooth on $$[-\pi,\pi]$$ and $$f(-\pi)=f(\pi)$$.

Example $$\PageIndex{1}$$

Find the bounded formal solution of Equation \ref{eq:12.4.2} with $$f(\theta)=\theta(\pi^2-\theta^2)$$.

Solution

From Example 11.2.6,

$\theta(\pi^2-\theta^2)=12\sum_{n=1}^\infty\frac{(-1)^n}{n^3}\sin n\theta,\quad -\pi\le\theta\le\pi,\nonumber$

so

$u(r,\theta)=12\sum_{n=1}^\infty\frac{r^n}{\rho^n}\frac{(-1)^n}{n^3}\sin n\theta,\quad 0\le r\le \rho,\quad -\pi\le\theta\le\pi.\nonumber$

Example $$\PageIndex{2}$$

Define the formal solution of

$\label{eq:12.4.9} \begin{array}{c}{u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta\theta}=0,\quad \rho _{0}<r<\rho ,\quad -\pi \leq\theta <\pi , }\\{u(\rho _{0},\theta )=0,\quad u(\rho, \theta )=f(\theta ),\quad -\pi\leq\theta <\pi ,}\end{array}$

where $$0<\rho_0<\rho$$ (Figure $$\PageIndex{2}$$). Figure $$\PageIndex{2}$$: The boundary value problem \ref{eq:12.4.9}

Solution

We use separation of variables exactly as before, except that now we choose the constants in Equation \ref{eq:12.4.5} and Equation \ref{eq:12.4.7} so that $$R_n(\rho_0)=0$$ for $$n=0$$, $$1$$, $$2$$,…. In view of the nonhomogeneous Dirichlet condition on the boundary $$r=\rho$$, it is also convenient to require that $$R_n(\rho)=1$$ for $$n=0$$, $$1$$, $$2$$,…. We leave it to you to verify that

$R_0(r)=\frac{\ln r/\rho_0}{\ln\rho/\rho_0} \quad \text{and} \quad R_n=\frac{\rho_0^{-n}r^n-\rho_0^nr^{-n}} {\rho_0^{-n}\rho^n-\rho_0^n\rho^{-n}},\quad n=1,2,3,\dots\nonumber$

satisfy these requirements. Therefore

$v_0(\rho,\theta)=\frac{\ln r/\rho_0}{\ln\rho/\rho_0}\nonumber$

and

$v_n(r,\theta)=\frac{\rho_0^{-n}r^n-\rho_0^nr^{-n}} {\rho_0^{-n}\rho^n-\rho_0^n\rho^{-n}}(\alpha_n\cos n\theta+\beta_n\sin n\theta), \quad n=1,2,3,\dots,\nonumber$

where $$\alpha_n$$ and $$\beta_n$$ are arbitrary constants.

If $$\alpha_0$$, $$\alpha_1$$,…, $$\alpha_m$$ and $$\beta_1$$, $$\beta_2$$, …, $$\beta_m$$ are arbitrary constants then

$u_m(r,\theta)=\alpha_0\frac{\ln r/\rho_0}{\ln\rho/\rho_0}+ \sum_{n=1}^m \frac{\rho_0^{-n}r^n-\rho_0^nr^{-n}} {\rho_0^{-n}\rho^n-\rho_0^n\rho^{-n}} (\alpha_n\cos n\theta+\beta_n\sin n\theta)\nonumber$

satisfies Equation \ref{eq:12.4.9}, with

$f(\theta) =\alpha_0+\sum_{n=1}^m(\alpha_n\cos n\theta+\beta_n\sin n\theta).\nonumber$

This motivates us to define the formal solution of Equation \ref{eq:12.4.9} for general $$f$$ to be

$u(r,\theta)=\alpha_0\frac{\ln r/\rho_0}{\ln\rho/\rho_0}+ \sum_{n=1}^\infty \frac{\rho_0^{-n}r^n-\rho_0^nr^{-n}} {\rho_0^{-n}\rho^n-\rho_0^n\rho^{-n}} (\alpha_n\cos n\theta+\beta_n\sin n\theta),\nonumber$

where

$F(\theta) =\alpha_0+\sum_{n=1}^\infty(\alpha_n\cos n\theta+\beta_n\sin n\theta)\nonumber$

is the Fourier series of $$f$$ on $$[-\pi,\pi]$$.

Example $$\PageIndex{3}$$

Define the bounded formal solution of

$\label{eq:12.4.10} \begin{array}{c}{u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta\theta}=0,\quad 0<r<\rho,\quad 0<\theta <\gamma , }\\{u(\rho ,\theta )= f(\theta ),\quad 0\leq \theta\leq\gamma ,}\\ {u(r,0)=0,\quad u(r,\gamma )=0,\quad 0<r<\rho ,} \end{array}$

where $$0<\gamma<2\pi$$ (Figure $$\PageIndex{3}$$). Figure $$\PageIndex{3}$$: The boundary value problem \ref{eq:12.4.10}

Solution

Now $$v(r,\theta)=R(r)\Theta(\theta)$$, where

$\label{eq:12.4.11} r^2R''+rR'-\lambda R=0$

and

$\label{eq:12.4.12} \Theta''+\lambda\Theta=0,\quad \Theta(0)=0,\quad \Theta(\gamma)=0.$

From Theorem 11.1.2, the eigenvalues of Equation \ref{eq:12.4.12} are $$\lambda_n=n^2\pi^2/\gamma^2$$, with associated eigenfunction $$\Theta_n=\sin n\pi\theta/\gamma$$, $$n=1$$, $$2$$, $$3$$,…. Substituting $$\lambda=n^2\pi^2/\gamma^2$$ into Equation \ref{eq:12.4.11} yields the Euler equation

$r^2R''+rR_n'-\frac{n^2\pi^2}{\gamma^2} R=0.\nonumber$

The indicial polynomial of this equation is

$s(s-1)+s-\frac{n^2\pi^2}{\gamma^2}=\left(s-\frac{n\pi}{\gamma}\right) \left(s+\frac{n\pi}{\gamma}\right),\nonumber$

so

$R_n=c_1r^{n\pi/\gamma}+c_2r^{-n\pi/\gamma},\nonumber$

by Theorem 7.4.3. To obtain a solution that remains bounded as $$r\to0+$$ we let $$c_2=0$$. Because of the Dirichlet condition at $$r=\rho$$, it is convenient to have $$r(\rho)=1$$; therefore we take $$c_1=\rho^{-n\pi/\gamma}$$, so

$R_n(r)=\frac{r^{n\pi/\gamma}}{\rho^{n\pi/\gamma}}.\nonumber$

Now

$v_n(r,\theta)=R_n(r)\Theta_n(\theta)=\frac{r^{n\pi/\gamma}} {\rho^{n\pi/\gamma}}\sin\frac{n\pi\theta}{\gamma}\nonumber$

satisfies Equation \ref{eq:12.4.10} with

$f(\theta)=\sin\frac{n\pi\theta}{\gamma}.\nonumber$

More generally, if $$\alpha_1$$, $$\alpha_2$$, …, $$\alpha_m$$ and are arbitrary constants then

$u_m(r,\theta)=\sum_{n=1}^m\alpha_n\frac{r^{n\pi/\gamma}}{\rho^{n\pi/\gamma}} \sin\frac{n\pi\theta}{\gamma}\nonumber$

satisfies Equation \ref{eq:12.4.10} with

$f(\theta) =\sum_{n=1}^m\alpha_n\sin\frac{n\pi\theta}{\gamma}.\nonumber$

This motivates us to define the bounded formal solution of Equation \ref{eq:12.4.10} to be

$u_m(r,\theta)=\sum_{n=1}^\infty\alpha_n\frac{r^{n\pi/\gamma}}{\rho^{n\pi/\gamma}} \sin\frac{n\pi\theta}{\gamma},\nonumber$

where

$S(\theta)=\sum_{n=1}^\infty\alpha_n \sin\frac{n\pi\theta}{\gamma}\nonumber$

is the Fourier sine expansion of $$f$$ on $$[0,\gamma]$$; that is,

$\alpha_n=\frac{2}{\gamma}\int_0^\gamma f(\theta)\sin\frac{n\pi\theta}{\gamma}\,d\theta. \nonumber$