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This is problem 2.5.5 part (c) from Richard Haberman applied partial differential equations, 5th edition
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r }\frac {\partial u}{\partial r} + \frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} =0 \]
Inside quarter circle of radius 1 with \(0 \leq \theta \leq \frac {\pi }{2}\) and \(0 \leq r \leq 1\), with following boundary conditions
\begin {align*} u(r,0) &= 0 \\ u(r,\frac {\pi }{2}) &= 0 \\ \frac {\partial u}{\partial r}(1,\theta ) &= f(\theta ) \\ \end {align*}
Mathematica ✗
ClearAll[u, theta, r, f]; pde = D[u[r, theta], {r, 2}] + (1*D[u[r, theta], r]*1*D[u[r, theta], {theta, 2}])/(r*r^2) == 0; bc = {Derivative[1, 0][u][1, theta] == f[theta], u[r, Pi/2] == 0, u[r, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}, Assumptions -> {0 <= r <= 1 && 0 <= theta <= Pi/2}], 60*10]];
\[ \text {Failed} \]
Maple ✓
L:='L'; u:='u'; t:='t'; x:='x';f:='f'; interface(showassumed=0); pde:=diff(u(r,theta),r$2)+ 1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0; bc:=u(r,0)=0,u(r,Pi/2)=0,D[1](u)(1,theta)=f(theta); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta),HINT=boundedseries(r=0)) assuming 0<=theta,theta<=Pi/2,0<=r,r<=1),output='realtime'));
\[ u \left ( r,\theta \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac {\int _{0}^{\pi /2}\!f \left ( \theta \right ) \sin \left ( 2\,n\theta \right ) \,{\rm d}\theta {r}^{2\,n}\sin \left ( 2\,n\theta \right ) }{\pi \,n}} \right ) \]
Hand solution
The Laplace PDE in polar coordinates is \begin {equation} r^{2}\frac {\partial ^{2}u}{\partial r^{2}}+r\frac {\partial u}{\partial r}+\frac {\partial ^{2}u}{\partial \theta ^{2}}=0\tag {A} \end {equation} With boundary conditions \begin {align} u\left ( r,0\right ) & =0\nonumber \\ u\left ( r,\frac {\pi }{2}\right ) & =0\tag {B}\\ u\left ( 1,\theta \right ) & =f\left ( \theta \right ) \nonumber \end {align}
Assuming the solution can be written as \[ u\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right ) \] And substituting this assumed solution back into the (A) gives\[ r^{2}R^{\prime \prime }\Theta +rR^{\prime }\Theta +R\Theta ^{\prime \prime }=0 \] Dividing the above by \(R\Theta \neq 0\) gives\begin {align*} r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+\frac {\Theta ^{\prime \prime }}{\Theta } & =0\\ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R} & =-\frac {\Theta ^{\prime \prime }}{\Theta } \end {align*}
Since each side depends on different independent variable and they are equal, they must be equal to same constant. say \(\lambda \). \[ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}=-\frac {\Theta ^{\prime \prime }}{\Theta }=\lambda \] This results in the following two ODE’s. The boundaries conditions in (B) are also transferred to each ODE. This gives\begin {align} \Theta ^{\prime \prime }+\lambda \Theta & =0\nonumber \\ \Theta \left ( 0\right ) & =0\tag {1}\\ \Theta \left ( \frac {\pi }{2}\right ) & =0\nonumber \end {align}
And\begin {align} r^{2}R^{\prime \prime }+rR^{\prime }-\lambda R & =0\tag {2}\\ \left \vert R\left ( 0\right ) \right \vert & <\infty \nonumber \end {align}
Starting with (1). Consider the Case \(\lambda <0\). The solution in this case will be \[ \Theta =A\cosh \left ( \sqrt {\lambda }\theta \right ) +B\sinh \left ( \sqrt {\lambda }\theta \right ) \] Applying first B.C. gives \(A=0\). The solution becomes \(\Theta =B\sinh \left ( \sqrt {\lambda }\theta \right ) \). Applying second B.C. gives \[ 0=B\sinh \left ( \sqrt {\lambda }\frac {\pi }{2}\right ) \] But \(\sinh \) is zero only when \(\sqrt {\lambda }\frac {\pi }{2}=0\) which is not the case here. Therefore \(B=0\) and hence trivial solution. Hence \(\lambda <0\) is not an eigenvalue.
Case \(\lambda =0\) The ODE becomes \(\Theta ^{\prime \prime }=0\) with solution \(\Theta =A\theta +B\). First B.C. gives \(0=B\). The solution becomes \(\Theta =A\theta \). Second B.C. gives \(0=A\frac {\pi }{2}\), hence \(A=0\) and trivial solution. Therefore \(\lambda =0\) is not an eigenvalue.
Case \(\lambda >0\) The ODE becomes \(\Theta ^{\prime \prime }+\lambda \Theta =0\) with solution \[ \Theta =A\cos \left ( \sqrt {\lambda }\theta \right ) +B\sin \left ( \sqrt {\lambda }\theta \right ) \] The first B.C. gives \(0=A\). The solution becomes \[ \Theta =B\sin \left ( \sqrt {\lambda }\theta \right ) \] And the second B.C. gives \[ 0=B\sin \left ( \sqrt {\lambda }\frac {\pi }{2}\right ) \] For non-trivial solution \(\sin \left ( \sqrt {\lambda }\frac {\pi }{2}\right ) =0\) or \(\sqrt {\lambda }\frac {\pi }{2}=n\pi \) for \(n=1,2,3,\cdots \). Hence the eigenvalues are\begin {align*} \sqrt {\lambda _{n}} & =2n\\ \lambda _{n} & =4n^{2}\qquad n=1,2,3,\cdots \end {align*}
And the eigenfunctions are \begin {equation} \Theta _{n}\left ( \theta \right ) =B_{n}\sin \left ( 2n\theta \right ) \qquad n=1,2,3,\cdots \tag {3} \end {equation} Now the \(R\) ODE is solved. There is one case to consider, which is \(\lambda >0\) based on the above. The ODE is\begin {align*} r^{2}R^{\prime \prime }+rR^{\prime }-\lambda _{n}R & =0\\ r^{2}R^{\prime \prime }+rR^{\prime }-4n^{2}R & =0\qquad n=1,2,3,\cdots \end {align*}
This is Euler ODE. Let \(R\left ( r\right ) =r^{p}\). Then \(R^{\prime }=pr^{p-1}\) and \(R^{\prime \prime }=p\left ( p-1\right ) r^{p-2}\). This gives\begin {align*} r^{2}\left ( p\left ( p-1\right ) r^{p-2}\right ) +r\left ( pr^{p-1}\right ) -4n^{2}r^{p} & =0\\ \left ( \left ( p^{2}-p\right ) r^{p}\right ) +pr^{p}-4n^{2}r^{p} & =0\\ r^{p}p^{2}-pr^{p}+pr^{p}-4n^{2}r^{p} & =0\\ p^{2}-4n^{2} & =0\\ p & =\pm 2n \end {align*}
Hence the solution is\[ R\left ( r\right ) =Cr^{2n}+D\frac {1}{r^{2n}}\] Applying the condition that \(\left \vert R\left ( 0\right ) \right \vert <\infty \) implies \(D=0\), and the solution becomes\begin {equation} R_{n}\left ( r\right ) =C_{n}r^{2n}\qquad n=1,2,3,\cdots \tag {4} \end {equation} Using (3,4) the solution \(u_{n}\left ( r,\theta \right ) \) is\begin {align*} u_{n}\left ( r,\theta \right ) & =R_{n}\Theta _{n}\\ & =C_{n}r^{2n}B_{n}\sin \left ( 2n\theta \right ) \\ & =B_{n}r^{2n}\sin \left ( 2n\theta \right ) \end {align*}
Where \(C_{n}B_{n}\) was combined into one constant \(B_{n}\). (No need to introduce new symbol). The final solution is\begin {align*} u\left ( r,\theta \right ) & =\sum _{n=1}^{\infty }u_{n}\left ( r,\theta \right ) \\ & =\sum _{n=1}^{\infty }B_{n}r^{2n}\sin \left ( 2n\theta \right ) \end {align*}
Now the nonhomogeneous condition is applied to find \(B_{n}\).\[ \frac {\partial }{\partial r}u\left ( r,\theta \right ) =\sum _{n=1}^{\infty }B_{n}\left ( 2n\right ) r^{2n-1}\sin \left ( 2n\theta \right ) \] Hence \(\frac {\partial }{\partial r}u\left ( 1,\theta \right ) =f\left ( \theta \right ) \) becomes\[ f\left ( \theta \right ) =\sum _{n=1}^{\infty }2B_{n}n\sin \left ( 2n\theta \right ) \] Multiplying by \(\sin \left ( 2m\theta \right ) \) and integrating gives\begin {align} \int _{0}^{\frac {\pi }{2}}f\left ( \theta \right ) \sin \left ( 2m\theta \right ) d\theta & =\int _{0}^{\frac {\pi }{2}}\sin \left ( 2m\theta \right ) \sum _{n=1}^{\infty }2B_{n}n\sin \left ( 2n\theta \right ) d\theta \nonumber \\ & =\sum _{n=1}^{\infty }2nB_{n}\int _{0}^{\frac {\pi }{2}}\sin \left ( 2m\theta \right ) \sin \left ( 2n\theta \right ) d\theta \tag {5} \end {align}
When \(n=m\) then\begin {align*} \int _{0}^{\frac {\pi }{2}}\sin \left ( 2m\theta \right ) \sin \left ( 2n\theta \right ) d\theta & =\int _{0}^{\frac {\pi }{2}}\sin ^{2}\left ( 2n\theta \right ) d\theta \\ & =\int _{0}^{\frac {\pi }{2}}\left ( \frac {1}{2}-\frac {1}{2}\cos 4n\theta \right ) d\theta \\ & =\frac {1}{2}\left [ \theta \right ] _{0}^{\frac {\pi }{2}}-\frac {1}{2}\left [ \frac {\sin 4n\theta }{4n}\right ] _{0}^{\frac {\pi }{2}}\\ & =\frac {\pi }{4}-\left ( \frac {1}{8n}\left ( \sin \frac {4n}{2}\pi \right ) -\sin \left ( 0\right ) \right ) \end {align*}
And since \(n\) is integer, then \(\sin \frac {4n}{2}\pi =\sin 2n\pi =0\) and the above becomes \(\frac {\pi }{4}\).
Now for the case when \(n\neq m\) using \(\sin A\sin B=\frac {1}{2}\left ( \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ) \) then\begin {align*} \int _{0}^{\frac {\pi }{2}}\sin \left ( 2m\theta \right ) \sin \left ( 2n\theta \right ) d\theta & =\int _{0}^{\frac {\pi }{2}}\frac {1}{2}\left ( \cos \left ( 2m\theta -2n\theta \right ) -\cos \left ( 2m\theta +2n\theta \right ) \right ) d\theta \\ & =\frac {1}{2}\int _{0}^{\frac {\pi }{2}}\cos \left ( 2m\theta -2n\theta \right ) d\theta -\frac {1}{2}\int _{0}^{\frac {\pi }{2}}\cos \left ( 2m\theta +2n\theta \right ) d\theta \\ & =\frac {1}{2}\int _{0}^{\frac {\pi }{2}}\cos \left ( \left ( 2m-2n\right ) \theta \right ) d\theta -\frac {1}{2}\int _{0}^{\frac {\pi }{2}}\cos \left ( \left ( 2m+2n\right ) \theta \right ) d\theta \\ & =\frac {1}{2}\left [ \frac {\sin \left ( \left ( 2m-2n\right ) \theta \right ) }{\left ( 2m-2n\right ) }\right ] _{0}^{\frac {\pi }{2}}-\frac {1}{2}\left [ \frac {\sin \left ( \left ( 2m+2n\right ) \theta \right ) }{\left ( 2m+2n\right ) }\right ] _{0}^{\frac {\pi }{2}}\\ & =\frac {1}{4\left ( m-n\right ) }\left [ \sin \left ( \left ( 2m-2n\right ) \theta \right ) \right ] _{0}^{\frac {\pi }{2}}-\frac {1}{4\left ( m+n\right ) }\left [ \sin \left ( \left ( 2m+2n\right ) \theta \right ) \right ] _{0}^{\frac {\pi }{2}}\\ & =\frac {1}{4\left ( m-n\right ) }\left [ \sin \left ( \left ( 2m-2n\right ) \frac {\pi }{2}\right ) -0\right ] -\frac {1}{4\left ( m+n\right ) }\left [ \sin \left ( \left ( 2m+2n\right ) \frac {\pi }{2}\right ) -0\right ] \end {align*}
Since \(2m-2n\frac {\pi }{2}=\pi \left ( m-n\right ) \) which is integer multiple of \(\pi \) and also \(\left ( 2m+2n\right ) \frac {\pi }{2}\) is integer multiple of \(\pi \) then the whole term above becomes zero. Therefore (5) becomes\[ \int _{0}^{\frac {\pi }{2}}f\left ( \theta \right ) \sin \left ( 2m\theta \right ) d\theta =2mB_{m}\frac {\pi }{4}\] Hence\[ B_{n}=\frac {2}{\pi n}\int _{0}^{\frac {\pi }{2}}f\left ( \theta \right ) \sin \left ( 2n\theta \right ) d\theta \] Summary: the final solution is\[ u\left ( r,\theta \right ) =\frac {2}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}\left [ \int _{0}^{\frac {\pi }{2}}f\left ( \theta \right ) \sin \left ( 2n\theta \right ) d\theta \right ] \left ( r^{2n}\sin \left ( 2n\theta \right ) \right ) \]
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Solve Laplace equation \[ \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r }\frac {\partial u}{\partial r} + \frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} =0 \]
Inside semi-circle of radius 1 with \(0 \leq \theta \leq \pi \) and \(0 \leq r \leq 1\), with following boundary conditions
\begin {align*} u(r,0) &= 0 \\ u(r,\pi ) &= 0 \\ u(0,\theta ) &= 0 \\ u(1,\theta ) &= f(\theta ) \end {align*}
Mathematica ✗
ClearAll[u, theta, r, f]; pde = D[u[r, theta], {r, 2}] + (1*D[u[r, theta], r]*1*D[u[r, theta], {theta, 2}])/(r*r^2) == 0; bc = {u[r, 0] == 0, u[r, Pi] == 0, u[0, theta] == 0, u[1, theta] == f[theta]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}, Assumptions -> {0 <= r <= 1 && 0 <= theta <= Pi}], 60*10]];
\[ \text {Failed} \]
Maple ✓
L:='L'; u:='u';f:='f';theta:='theta'; pde:=diff(u(r,theta),r$2) +1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0; bc:=u(r,0)=0,u(r,Pi)=0,u(0,theta)=0,u(1,theta)=f(theta); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta))),output='realtime'));
\[ u \left ( r,\theta \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac {\sin \left ( n\theta \right ) \left ( {r}^{n}\int _{0}^{\pi }\!\sin \left ( n\theta \right ) f \left ( \theta \right ) \,{\rm d}\theta -1/2\,{\it \_C5} \left ( n \right ) \pi \, \left ( {r}^{n}-{r}^{-n} \right ) \right ) }{\pi }} \right ) \]
Hand solution
The Laplace PDE in polar coordinates is \begin {equation} r^{2}\frac {\partial ^{2}u}{\partial r^{2}}+r\frac {\partial u}{\partial r}+\frac {\partial ^{2}u}{\partial \theta ^{2}}=0\tag {A} \end {equation} With\begin {align} \frac {\partial u}{\partial r}\left ( a,\theta \right ) & =0\nonumber \\ u\left ( b,\theta \right ) & =g\left ( \theta \right ) \tag {B} \end {align}
Assuming the solution can be written as \[ u\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right ) \] And substituting this assumed solution back into the (A) gives\[ r^{2}R^{\prime \prime }\Theta +rR^{\prime }\Theta +R\Theta ^{\prime \prime }=0 \] Dividing the above by \(R\Theta \) gives\begin {align*} r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}+\frac {\Theta ^{\prime \prime }}{\Theta } & =0\\ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R} & =-\frac {\Theta ^{\prime \prime }}{\Theta } \end {align*}
Since each side depends on different independent variable and they are equal, they must be equal to same constant. say \(\lambda \). \[ r^{2}\frac {R^{\prime \prime }}{R}+r\frac {R^{\prime }}{R}=-\frac {\Theta ^{\prime \prime }}{\Theta }=\lambda \] This results in the following two ODE’s. The boundaries conditions in (B) are also transferred to each ODE. This results in\begin {align} \Theta ^{\prime \prime }+\lambda \Theta & =0\tag {1}\\ \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \nonumber \\ \Theta ^{\prime }\left ( -\pi \right ) & =\Theta ^{\prime }\left ( \pi \right ) \nonumber \end {align}
And\begin {align} r^{2}R^{\prime \prime }+rR^{\prime }-\lambda R & =0\tag {2}\\ R^{\prime }\left ( a\right ) & =0\nonumber \end {align}
Starting with (1)
Case \(\lambda <0\) The solution is\[ \Theta \left ( \theta \right ) =A\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\theta \right ) +B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }\theta \right ) \] First B.C. gives\begin {align*} \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \\ A\cosh \left ( -\sqrt {\left \vert \lambda \right \vert }\pi \right ) +B\sinh \left ( -\sqrt {\left \vert \lambda \right \vert }\pi \right ) & =A\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) +B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) \\ A\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) -B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) & =A\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) +B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) \\ 2B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) & =0 \end {align*}
But \(\sinh =0\) only at zero and \(\lambda \neq 0\), hence \(B=0\) and the solution becomes\begin {align*} \Theta \left ( \theta \right ) & =A\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\theta \right ) \\ \Theta ^{\prime }\left ( \theta \right ) & =A\sqrt {\lambda }\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\theta \right ) \end {align*}
Applying the second B.C. gives\begin {align*} \Theta ^{\prime }\left ( -\pi \right ) & =\Theta ^{\prime }\left ( \pi \right ) \\ A\sqrt {\left \vert \lambda \right \vert }\cosh \left ( -\sqrt {\left \vert \lambda \right \vert }\pi \right ) & =A\sqrt {\left \vert \lambda \right \vert }\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) \\ A\sqrt {\left \vert \lambda \right \vert }\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) & =A\sqrt {\left \vert \lambda \right \vert }\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) \\ 2A\sqrt {\left \vert \lambda \right \vert }\cosh \left ( \sqrt {\left \vert \lambda \right \vert }\pi \right ) & =0 \end {align*}
But \(\cosh \) is never zero, hence \(A=0\). Therefore trivial solution and \(\lambda <0\) is not an eigenvalue.
Case \(\lambda =0\) The solution is \(\Theta =A\theta +B\). Applying the first B.C. gives\begin {align*} \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \\ -A\pi +B & =\pi A+B\\ 2\pi A & =0\\ A & =0 \end {align*}
And the solution becomes \(\Theta =B_{0}\). A constant. Hence \(\lambda =0\) is an eigenvalue.
Case \(\lambda >0\)
The solution becomes\begin {align*} \Theta & =A\cos \left ( \sqrt {\lambda }\theta \right ) +B\sin \left ( \sqrt {\lambda }\theta \right ) \\ \Theta ^{\prime } & =-A\sqrt {\lambda }\sin \left ( \sqrt {\lambda }\theta \right ) +B\sqrt {\lambda }\cos \left ( \sqrt {\lambda }\theta \right ) \end {align*}
Applying first B.C. gives\begin {align} \Theta \left ( -\pi \right ) & =\Theta \left ( \pi \right ) \nonumber \\ A\cos \left ( -\sqrt {\lambda }\pi \right ) +B\sin \left ( -\sqrt {\lambda }\pi \right ) & =A\cos \left ( \sqrt {\lambda }\pi \right ) +B\sin \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ A\cos \left ( \sqrt {\lambda }\pi \right ) -B\sin \left ( \sqrt {\lambda }\pi \right ) & =A\cos \left ( \sqrt {\lambda }\pi \right ) +B\sin \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ 2B\sin \left ( \sqrt {\lambda }\pi \right ) & =0 \tag {3} \end {align}
Applying second B.C. gives\begin {align} \Theta ^{\prime }\left ( -\pi \right ) & =\Theta ^{\prime }\left ( \pi \right ) \nonumber \\ -A\sqrt {\lambda }\sin \left ( -\sqrt {\lambda }\pi \right ) +B\sqrt {\lambda }\cos \left ( -\sqrt {\lambda }\pi \right ) & =-A\sqrt {\lambda }\sin \left ( \sqrt {\lambda }\pi \right ) +B\sqrt {\lambda }\cos \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ A\sqrt {\lambda }\sin \left ( \sqrt {\lambda }\pi \right ) +B\sqrt {\lambda }\cos \left ( \sqrt {\lambda }\pi \right ) & =-A\sqrt {\lambda }\sin \left ( \sqrt {\lambda }\pi \right ) +B\sqrt {\lambda }\cos \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ A\sqrt {\lambda }\sin \left ( \sqrt {\lambda }\pi \right ) & =-A\sqrt {\lambda }\sin \left ( \sqrt {\lambda }\pi \right ) \nonumber \\ 2A\sin \left ( \sqrt {\lambda }\pi \right ) & =0 \tag {4} \end {align}
Equations (3,4) can be both zero only if \(A=B=0\) which gives trivial solution, or when \(\sin \left ( \sqrt {\lambda }\pi \right ) =0\). Therefore taking \(\sin \left ( \sqrt {\lambda }\pi \right ) =0\) gives a non-trivial solution. Hence\begin {align*} \sqrt {\lambda }\pi & =n\pi \qquad n=1,2,3,\cdots \\ \lambda _{n} & =n^{2}\qquad n=1,2,3,\cdots \end {align*}
Hence the solution for \(\Theta \) is\begin {equation} \Theta =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \tag {5} \end {equation} Now the \(R\) equation is solved
The case for \(\lambda =0\) gives\begin {align*} r^{2}R^{\prime \prime }+rR^{\prime } & =0\\ R^{\prime \prime }+\frac {1}{r}R^{\prime } & =0\qquad r\neq 0 \end {align*}
As was done in last problem, the solution to this is\[ R\left ( r\right ) =A\ln \left \vert r\right \vert +C \] Since \(r>0\) no need to keep worrying about \(\left \vert r\right \vert \) and is removed for simplicity. Applying the B.C. gives\[ R^{\prime }=A\frac {1}{r}\] Evaluating at \(r=a\) gives\[ 0=A\frac {1}{a}\] Hence \(A=0\), and the solution becomes\[ R\left ( r\right ) =C_{0}\] Which is a constant.
Case \(\lambda >0\) The ODE in this case is \[ r^{2}R^{\prime \prime }+rR^{\prime }-n^{2}R=0\qquad n=1,2,3,\cdots \] Let \(R=r^{p}\), the above becomes\begin {align*} r^{2}p\left ( p-1\right ) r^{p-2}+rpr^{p-1}-n^{2}r^{p} & =0\\ p\left ( p-1\right ) r^{p}+pr^{p}-n^{2}r^{p} & =0\\ p\left ( p-1\right ) +p-n^{2} & =0\\ p^{2} & =n^{2}\\ p & =\pm n \end {align*}
Hence the solution is\[ R_{n}\left ( r\right ) =Cr^{n}+D\frac {1}{r^{n}}\qquad n=1,2,3,\cdots \] Applying the boundary condition \(R^{\prime }\left ( a\right ) =0\) gives\begin {align*} R_{n}^{\prime }\left ( r\right ) & =nC_{n}r^{n-1}-nD_{n}\frac {1}{r^{n+1}}\\ 0 & =R_{n}^{\prime }\left ( a\right ) \\ & =nC_{n}a^{n-1}-nD_{n}\frac {1}{a^{n+1}}\\ & =nC_{n}a^{2n}-nD_{n}\\ & =C_{n}a^{2n}-D_{n}\\ D_{n} & =C_{n}a^{2n} \end {align*}
The solution becomes\begin {align*} R_{n}\left ( r\right ) & =C_{n}r^{n}+C_{n}a^{2n}\frac {1}{r^{n}}\qquad n=1,2,3,\cdots \\ & =C_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \end {align*}
Hence the complete solution for \(R\left ( r\right ) \) is\begin {equation} R\left ( r\right ) =C_{0}+\sum _{n=1}^{\infty }C_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \tag {6} \end {equation} Using (5),(6) gives\begin {align*} u_{n}\left ( r,\theta \right ) & =R_{n}\Theta _{n}\\ u\left ( r,\theta \right ) & =\left [ C_{0}+\sum _{n=1}^{\infty }C_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \right ] \left [ A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ] \\ & =D_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) C_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( n\theta \right ) C_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \end {align*}
Where \(D_{0}=C_{0}A_{0}\). To simplify more, \(A_{n}C_{n}\) is combined to \(A_{n}\) and \(B_{n}C_{n}\) is combined to \(B_{n}\). The full solution is\[ u\left ( r,\theta \right ) =D_{0}+\sum _{n=1}^{\infty }A_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }B_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \sin \left ( n\theta \right ) \] The final nonhomogeneous B.C. is applied.\begin {align*} u\left ( b,\theta \right ) & =g\left ( \theta \right ) \\ g\left ( \theta \right ) & =D_{0}+\sum _{n=1}^{\infty }A_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }B_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \sin \left ( n\theta \right ) \end {align*}
For \(n=0\), integrating both sides give\begin {align*} \int _{-\pi }^{\pi }g\left ( \theta \right ) d\theta & =\int _{-\pi }^{\pi }D_{0}d\theta \\ D_{0} & =\frac {1}{2\pi }\int _{-\pi }^{\pi }g\left ( \theta \right ) d\theta \end {align*}
For \(n>0\), multiplying both sides by \(\cos \left ( m\theta \right ) \) and integrating gives\begin {align*} \int _{-\pi }^{\pi }g\left ( \theta \right ) \cos \left ( m\theta \right ) d\theta & =\int _{-\pi }^{\pi }D_{0}\cos \left ( m\theta \right ) d\theta \\ & +\int _{-\pi }^{\pi }\sum _{n=1}^{\infty }A_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \cos \left ( m\theta \right ) \cos \left ( n\theta \right ) d\theta \\ & +\int _{-\pi }^{\pi }\sum _{n=1}^{\infty }B_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \cos \left ( m\theta \right ) \sin \left ( n\theta \right ) d\theta \end {align*}
Hence\begin {align} \int _{-\pi }^{\pi }g\left ( \theta \right ) \cos \left ( m\theta \right ) d\theta & =\int _{-\pi }^{\pi }D_{0}\cos \left ( m\theta \right ) d\theta \nonumber \\ & +\sum _{n=1}^{\infty }A_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \int _{-\pi }^{\pi }\cos \left ( m\theta \right ) \cos \left ( n\theta \right ) d\theta \nonumber \\ & +\sum _{n=1}^{\infty }B_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \int _{-\pi }^{\pi }\cos \left ( m\theta \right ) \sin \left ( n\theta \right ) d\theta \tag {7} \end {align}
But \begin {align*} \int _{-\pi }^{\pi }\cos \left ( m\theta \right ) \cos \left ( n\theta \right ) d\theta & =\pi \qquad n=m\neq 0\\ \int _{-\pi }^{\pi }\cos \left ( m\theta \right ) \cos \left ( n\theta \right ) d\theta & =0\qquad n\neq m \end {align*}
And\[ \int _{-\pi }^{\pi }\cos \left ( m\theta \right ) \sin \left ( n\theta \right ) d\theta =0\qquad \] And\[ \int _{-\pi }^{\pi }D_{0}\cos \left ( m\theta \right ) d\theta =0 \] Then (7) becomes\begin {align} \int _{-\pi }^{\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta & =\pi A_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \nonumber \\ A_{n} & =\frac {1}{\pi }\frac {\int _{-\pi }^{\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta }{b^{n}+\frac {a^{2n}}{b^{n}}} \tag {8} \end {align}
Again, multiplying both sides by \(\sin \left ( m\theta \right ) \) and integrating gives\begin {align*} \int _{-\pi }^{\pi }g\left ( \theta \right ) \sin \left ( m\theta \right ) d\theta & =\int _{-\pi }^{\pi }D_{0}\sin \left ( m\theta \right ) d\theta \\ & +\int _{-\pi }^{\pi }\sum _{n=1}^{\infty }A_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \sin \left ( m\theta \right ) \cos \left ( n\theta \right ) d\theta \\ & +\int _{-\pi }^{\pi }\sum _{n=1}^{\infty }B_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \sin \left ( m\theta \right ) \sin \left ( n\theta \right ) d\theta \end {align*}
Hence\begin {align} \int _{-\pi }^{\pi }g\left ( \theta \right ) \sin \left ( m\theta \right ) d\theta & =\int _{-\pi }^{\pi }D_{0}\sin \left ( m\theta \right ) d\theta \nonumber \\ & +\sum _{n=1}^{\infty }A_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \int _{-\pi }^{\pi }\sin \left ( m\theta \right ) \cos \left ( n\theta \right ) d\theta \nonumber \\ & +\sum _{n=1}^{\infty }B_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \int _{-\pi }^{\pi }\sin \left ( m\theta \right ) \sin \left ( n\theta \right ) d\theta \tag {9} \end {align}
But \begin {align*} \int _{-\pi }^{\pi }\sin \left ( m\theta \right ) \sin \left ( n\theta \right ) d\theta & =\pi \qquad n=m\neq 0\\ \int _{-\pi }^{\pi }\sin \left ( m\theta \right ) \sin \left ( n\theta \right ) d\theta & =0\qquad n\neq m \end {align*}
And\[ \int _{-\pi }^{\pi }\sin \left ( m\theta \right ) \cos \left ( n\theta \right ) d\theta =0 \] And\[ \int _{-\pi }^{\pi }D_{0}\sin \left ( m\theta \right ) d\theta =0 \] Then (9) becomes\begin {align*} \int _{-\pi }^{\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta & =\pi B_{n}\left ( b^{n}+\frac {a^{2n}}{b^{n}}\right ) \\ B_{n} & =\frac {1}{\pi }\frac {\int _{-\pi }^{\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta }{b^{n}+\frac {a^{2n}}{b^{n}}} \end {align*}
This complete the solution. Summary\begin {align*} u\left ( r,\theta \right ) & =D_{0}+\sum _{n=1}^{\infty }A_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \cos \left ( n\theta \right ) +\sum _{n=1}^{\infty }B_{n}\left ( r^{n}+\frac {a^{2n}}{r^{n}}\right ) \sin \left ( n\theta \right ) \\ D_{0} & =\frac {1}{2\pi }\int _{-\pi }^{\pi }g\left ( \theta \right ) d\theta \\ A_{n} & =\frac {1}{\pi }\frac {\int _{-\pi }^{\pi }g\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta }{b^{n}+\frac {a^{2n}}{b^{n}}}\\ B_{n} & =\frac {1}{\pi }\frac {\int _{-\pi }^{\pi }g\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta }{b^{n}+\frac {a^{2n}}{b^{n}}} \end {align*}
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This is problem 2.5.8 part (b) from Richard Haberman applied partial differential equations, 5th edition
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r }\frac {\partial u}{\partial r} + \frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} =0 \]
Inside circular annulus \(a<r<b\) subject to the following boundary conditions
\begin {align*} \frac {\partial u}{\partial r}(a,\theta ) &= 0 \\ u(b,0) &= g(\theta ) \end {align*}
Mathematica ✗
ClearAll[u, a, theta, r, g]; pde = D[u[r, theta], {r, 2}] + (1*D[u[r, theta], r])/r + (1*D[u[r, theta], {theta, 2}])/r^2 == 0; bc = {Derivative[1, 0][u][a, theta] == 0, u[b, theta] == g[theta]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}, Assumptions -> Inequality[a, Less, r, LessEqual, b]], 60*10]];
\[ \text {Failed} \]
Maple ✓
a:='a'; u:='u'; r:='r'; theta:='theta';g:='g'; interface(showassumed=0); pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0; bc:=D[1](u)(a,theta)=0,u(b,theta)=g(theta); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta)) assuming a<r,r<b),output='realtime'));
\[ u \left ( r,\theta \right ) ={\it invfourier} \left ( {\frac {{\it fourier} \left ( g \left ( \theta \right ) ,\theta ,s \right ) {{\rm e}^{s \left ( 2\,\ln \left ( a \right ) -\ln \left ( r \right ) -\ln \left ( b \right ) \right ) }}}{{{\rm e}^{2\, \left ( \ln \left ( a \right ) -\ln \left ( b \right ) \right ) s}}+1}},s,\theta \right ) +{\it invfourier} \left ( {\frac {{\it fourier} \left ( g \left ( \theta \right ) ,\theta ,s \right ) {{\rm e}^{s \left ( \ln \left ( r \right ) -\ln \left ( b \right ) \right ) }}}{{{\rm e}^{2\, \left ( \ln \left ( a \right ) -\ln \left ( b \right ) \right ) s}}+1}},s,\theta \right ) \] But has unresolved Invfourier and Fourier calls
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Solve Laplace equation \[ \frac {\partial ^2 u}{\partial r^2} + \frac {1}{r }\frac {\partial u}{\partial r} + \frac {1}{r^2} \frac {\partial ^2 u}{\partial \theta ^2} =0 \]
Inside circular annulus \(1<r<2\) subject to the following boundary conditions
\begin {align*} u(1,\theta ) &= 0 \\ u(2,\theta ) &= \sin \theta \end {align*}
Mathematica ✓
ClearAll[u, r, theta]; pde = D[u[r, theta], {r, 2}] + (1*D[u[r, theta], r])/r + (1*D[u[r, theta], {theta, 2}])/r^2 == 0; bc = {u[1, theta] == 0, u[2, theta] == Sin[theta]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}], 60*10]];
\[ \left \{\left \{u(r,\theta )\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {2 \left (r^2-1\right ) \sin (\theta )}{3 r} & 1\leq r\leq 2 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \} \]
Maple ✓
u:='u'; r:='r'; theta:='theta'; pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0; bc:=u(1,theta)=0,u(2,theta)=sin(theta); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(r,theta))),output='realtime'));
\[ u \left ( r,\theta \right ) =2/3\,{\frac {\sin \left ( \theta \right ) \left ( {r}^{2}-1 \right ) }{r}} \]
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Solve Laplace equation in polar coordinates outside a disk
Solve for \(u\left ( r,\theta \right ) \)
\begin {align*} \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\\ a & \leq r \\ 0 & <\theta \leq 2\pi \end {align*}
Boundary conditions
\begin {align*} u\left ( a,\theta \right ) & =f\left ( \theta \right ) \\ \left \vert u\left ( 0,\theta \right ) \right \vert & <\infty \\ u\left ( r,0\right ) & =u\left ( r,2\pi \right ) \\ \frac {\partial u}{\partial \theta }\left ( r,0\right ) & =\frac {\partial u}{\partial \theta }\left ( r,2\pi \right ) \end {align*}
Mathematica ✗
ClearAll[u, theta, r, a, f]; pde = D[u[r, theta], {r, 2}] + (1*D[u[r, theta], r])/r + (1*D[u[r, theta], {theta, 2}])/r^2 == 0; bc = {u[a, theta] == f[theta], u[r, -Pi] == u[r, Pi], Derivative[0, 1][u][r, -Pi] == Derivative[0, 1][u][r, Pi]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}, Assumptions -> {a > 0, r > a}], 60*10]];
\[ \text {Failed} \]
Maple ✓
r:='r'; theta:='theta'; a:='a'; r:='r';f:='f'; interface(showassumed=0); pde := (diff(r*(diff(u(r, theta), r)), r))/r+(diff(u(r, theta), theta, theta))/r^2 = 0; bc := u(a, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], u(r, theta), HINT = boundedseries(r=infinity))),output='realtime'));
\[ u \left ( r,\theta \right ) =1/2\,{\frac {1}{\pi } \left ( 2\,\sum _{n=1}^{\infty } \left ( {\frac {\int _{-\pi }^{\pi }\!\sin \left ( n\theta \right ) f \left ( \theta \right ) \,{\rm d}\theta \sin \left ( n\theta \right ) +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \cos \left ( n\theta \right ) \,{\rm d}\theta \cos \left ( n\theta \right ) }{\pi } \left ( {\frac {r}{a}} \right ) ^{-n}} \right ) \pi +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \,{\rm d}\theta \right ) } \]