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Mathematics LibreTexts

12.4: Laplace's Equation in Polar Coordinates

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    9470
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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.

    clipboard_eac4f46bd50a71270194b918ed07f3b95.png
    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 \(\PageIndex{2}\)).

    clipboard_e9879a59f13103ecb12887895ea24021c.png
    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 \(\PageIndex{3}\)).

    clipboard_e4abc91274160d05b512441339f692c66.png
    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,

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