#### 4.1.2 Polar coordinates

4.1.2.6 [311] Haberman 2.5.5 (c)
4.1.2.7 [312] semi-circle
4.1.2.8 [313] Haberman 2.5.8 (b)
4.1.2.9 [314] Circular annulus
4.1.2.10 [315] Outside a disk
4.1.2.11 [316] Outside a disk

##### 4.1.2.1 [306] Laplace PDE inside quarter disk, Neumann BC at edge

problem number 306

Solve Laplace equation in polar coordinates inside quarter disk with $$0<r<1$$ and $$0<\theta <\frac{\pi }{2}$$

Solve for $$u\left ( r,\theta \right )$$ \begin{align*} u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta }=0 \end{align*}

Boundary conditions

\begin{align*} u(r,0) &= 0 \\ u(r,\frac{\pi }{2}) &=0 \\ u_r(1,\theta ) &= f(\theta ) \end{align*}

Mathematica

Failed

Maple

$u \left ( r,\theta \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac{\int _{0}^{\pi /2}\!f \left ( \theta \right ) \sin \left ( 2\,n\theta \right ) \,{\rm d}\theta{r}^{2\,n}\sin \left ( 2\,n\theta \right ) }{\pi \,n}} \right )$

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##### 4.1.2.2 [307] $$r=4$$ and $$u=x^4$$ at boundary of disk

problem number 307

Problem 4.3.25 part c. Peter J. Olver, Introduction to Partial Diﬀerential Equations, 2014 edition.

Solve Laplace equation in polar coordinates inside a disk

Solve $$\nabla ^{2}u=0$$ with $$x^{2} +y^{2}<4$$ and boundary conditions $$u=x^{4},x^{2}+y^{2}=4$$.

Mathematica

$\left \{\left \{u(r,\theta )\to \frac{1}{8} r^4 \cos (4 \theta )+8 r^2 \cos (2 \theta )+96\right \}\right \}$

Maple

$u \left ( r,\theta \right ) =1/4\,{r}^{4} \left ( \cos \left ( 2\,\theta \right ) \right ) ^{2}-1/8\,{r}^{4}+8\,{r}^{2}\cos \left ( 2\,\theta \right ) +96$

Hand solution

Solve the following boundary value problems $$\nabla ^{2}u=0,x^{2}+y^{2}<4,u=x^{4},x^{2}+y^{2}=4$$

Solution

In polar coordinates, where $$x=r\cos \theta ,y=r\sin \theta$$, we need to solve for $$u\left ( r,\theta \right )$$ inside disk of radius $$r_{0}=4$$. The Laplace PDE in polar coordinates is\begin{align*} u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta } & =0\qquad 0<r<r_{0},-\pi <\theta <\pi \\ u\left ( r_{0},\theta \right ) & =f\left ( \theta \right ) =\left ( r_{0}\cos \theta \right ) ^{4}\\ u\left ( -\pi \right ) & =u\left ( \pi \right ) \\ u_{\theta }\left ( -\pi \right ) & =u_{\theta }\left ( \pi \right ) \end{align*}

Let the solution be$u\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right )$ 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 the 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 original PDE are transferred to each ODE which 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$$r^{2}R^{\prime \prime }+rR^{\prime }-\lambda R=0\tag{2}$$ Starting with ODE (1) with periodic boundary conditions.

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 eigenfunctions are $$\{1,\cos \left ( n\theta \right ) ,\sin \left ( n\theta \right ) \}\qquad n=1,2,3,\cdots \tag{5}$$ Now the $$R$$ equation is solved

The case for $$\lambda =0$$ gives from (2)\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*}

The solution to this is$R_{0}\left ( r\right ) =A\ln r+C$ Since $$u$$ is bounded at $$r=0$$ we want $$A=0$$. Hence $$R_{0}\left ( r\right )$$ is just a constant.

Case $$\lambda >0$$ The ODE (2) becomes$r^{2}R^{\prime \prime }+rR^{\prime }-n^{2}R=0\qquad n=1,2,3,\cdots$ 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$R_{n}\left ( r\right ) =C_{n}r^{n}+D_{n}\frac{1}{r^{n}}\qquad n=1,2,3,\cdots$ Since $$u$$ is bounded at $$r=0$$ we want $$D=0$$. Hence $$R_{n}\left ( r\right ) =C_{n}r^{n}$$.

The complete solution for $$R\left ( r\right )$$ is$$R\left ( r\right ) =C_{0}+\sum _{n=1}^{\infty }C_{n}r^{n}\tag{6}$$ 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}r^{n}\right ) \left ( A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \end{align*}

Combining constants to simplify things gives$$u\left ( r,\theta \right ) =\frac{a_{0}}{2}+\sum _{n=1}^{\infty }r^{n}\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right ) \tag{7}$$ When $$r=r_{0}$$ the above becomes$f\left ( \theta \right ) =\frac{a_{0}}{2}+\sum _{n=1}^{\infty }r_{0}^{n}\left ( A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \right )$ Hence \begin{align*} a_{0} & =\frac{1}{\pi }\int _{-\pi }^{\pi }f\left ( \theta \right ) d\theta \\ r_{0}^{n}A_{n} & =\frac{1}{\pi }\int _{-\pi }^{\pi }f\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \\ A_{n} & =\frac{1}{\pi r_{0}^{n}}\int _{-\pi }^{\pi }f\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \end{align*}

And\begin{align*} r_{0}^{n}B_{n} & =\frac{1}{\pi }\int _{-\pi }^{\pi }f\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \\ B_{n} & =\frac{1}{\pi r_{0}^{n}}\int _{-\pi }^{\pi }f\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \end{align*}

Hence (7) becomes$$u\left ( r,\theta \right ) =\frac{1}{2\pi }\int _{-\pi }^{\pi }f\left ( \theta \right ) d\theta +\frac{1}{\pi }\sum _{n=1}^{\infty }\left ( \frac{r}{r_{0}}\right ) ^{n}\left ( \cos \left ( n\theta \right ) \int _{-\pi }^{\pi }f\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta +\sin \left ( n\theta \right ) \int _{-\pi }^{\pi }f\left ( \theta \right ) \sin \left ( n\theta \right ) d\theta \right ) \tag{8}$$ In this problem $$f\left ( \theta \right ) =\left ( r_{0}\cos \theta \right ) ^{4}$$ where $$r_{0}=4$$, hence\begin{align*} A_{0} & =\frac{1}{\pi }\int _{-\pi }^{\pi }256\cos ^{4}\theta d\theta \\ & =\frac{256}{\pi }\int _{-\pi }^{\pi }\cos ^{4}\theta d\theta \\ & =\frac{256}{\pi }\left ( \frac{3\theta }{8}+\frac{1}{4}\sin \left ( 2\theta \right ) +\frac{1}{32}\sin \left ( 4\theta \right ) \right ) _{-\pi }^{\pi }\\ & =\frac{256}{\pi }\left ( \frac{3\pi }{8}+\frac{3\pi }{8}\right ) \\ & =\frac{256}{\pi }\left ( \frac{3\pi }{4}\right ) \\ & =192 \end{align*}

And\begin{align*} A_{n} & =\frac{1}{\pi }\int _{-\pi }^{\pi }256\cos ^{4}\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \\ & =\frac{256}{\pi }\int _{-\pi }^{\pi }\cos ^{4}\left ( \theta \right ) \cos \left ( n\theta \right ) d\theta \end{align*}

To evaluate the above integral, we will start by using the identity $\cos ^{4}\left ( \theta \right ) =\frac{3}{8}+\frac{1}{8}\cos \left ( 4\theta \right ) +\frac{1}{2}\cos \left ( 2\theta \right )$ Therefore the integral now becomes\begin{align} A_{n} & =\frac{256}{\pi }\int _{-\pi }^{\pi }\left ( \frac{3}{8}+\frac{1}{8}\cos \left ( 4\theta \right ) +\frac{1}{2}\cos \left ( 2\theta \right ) \right ) \cos \left ( n\theta \right ) d\theta \nonumber \\ & =\frac{256}{\pi }\left [ \frac{3}{8}\int _{-\pi }^{\pi }\cos \left ( n\theta \right ) d\theta +\frac{1}{8}\int _{-\pi }^{\pi }\cos \left ( 4\theta \right ) \cos \left ( n\theta \right ) d\theta +\frac{1}{2}\int _{-\pi }^{\pi }\cos \left ( 2\theta \right ) \cos \left ( n\theta \right ) d\theta \right ] \tag{1} \end{align}

But $$\int _{-\pi }^{\pi }\cos \left ( n\theta \right ) d\theta =0$$ and $$\int _{-\pi }^{\pi }\cos \left ( 4\theta \right ) \cos \left ( n\theta \right ) d\theta$$ is not zero, only for $$n=4$$ by orthogonality of cosine functions. Hence \begin{align*} \int _{-\pi }^{\pi }\cos \left ( 4\theta \right ) \cos \left ( n\theta \right ) d\theta & =\int _{-\pi }^{\pi }\cos ^{2}\left ( 4\theta \right ) d\theta \\ & =\pi \end{align*}

And similarly, $$\int _{-\pi }^{\pi }\cos \left ( 2\theta \right ) \cos \left ( n\theta \right ) d\theta$$ is not zero, only for $$n=2$$ by orthogonality of cosine functions. Hence\begin{align*} \int _{-\pi }^{\pi }\cos \left ( 2\theta \right ) \cos \left ( n\theta \right ) d\theta & =\int _{-\pi }^{\pi }\cos ^{2}\left ( 2\theta \right ) d\theta \\ & =\pi \end{align*}

Using these results in (1) gives, for $$n=2$$\begin{align*} A_{2} & =\frac{256}{\pi }\left [ \frac{1}{2}\int _{-\pi }^{\pi }\cos ^{2}\left ( 2\theta \right ) d\theta \right ] \\ & =\frac{256}{\pi }\left ( \frac{\pi }{2}\right ) \\ & =128 \end{align*}

And for $$n=4$$\begin{align*} A_{4} & =\frac{256}{\pi }\left [ \frac{1}{8}\int _{-\pi }^{\pi }\cos ^{2}\left ( 4\theta \right ) d\theta \right ] \\ & =\frac{256}{\pi }\left ( \frac{\pi }{8}\right ) \\ & =32 \end{align*}

And all other $$A_{n}$$ are zero. Now that we found all $$A_{n}$$, and since $$B_{n}=0$$ for all $$n$$ (because $$f\left ( \theta \right )$$ is even function) then the solution (8) becomes\begin{align*} u\left ( r,\theta \right ) & =\frac{192}{2}+a_{2}\left ( \frac{r}{4}\right ) ^{2}\cos \left ( 2\theta \right ) +a_{4}\left ( \frac{r}{4}\right ) ^{4}\cos \left ( 4\theta \right ) \\ & =96+128\left ( \frac{r^{2}}{16}\right ) \cos \left ( 2\theta \right ) +32\frac{r^{4}}{256}\cos \left ( 4\theta \right ) \end{align*}

Therefore$u\left ( r,\theta \right ) =96+8r^{2}\cos \left ( 2\theta \right ) +\frac{1}{8}r^{4}\cos \left ( 4\theta \right )$ Here is plot of the above solution.

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##### 4.1.2.3 [308] $$r=1$$ and $$u_r=x$$ at boundary of disk

problem number 308

Problem 4.3.25 part d, Peter J. Olver, Introduction to Partial Diﬀerential Equations, 2014 edition.

Solve Laplace equation in polar coordinates inside a disk of radius 1.

Solve $$\nabla ^{2}u=0$$ and boundary conditions $$\frac{\partial u}{\partial n}=x$$.

Mathematica

Failed

Maple

sol=()

Hand solution

In polar coordinates, where $$x=r\cos \theta ,y=r\sin \theta$$, we need to solve for $$u\left ( r,\theta \right )$$ inside disk of radius $$r_{0}=1$$. The Laplace PDE in polar coordinates is\begin{align*} u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta } & =0\qquad 0<r<1,-\pi <\theta <\pi \\ u_{r}\left ( 1,\theta \right ) & =f\left ( \theta \right ) =\cos \theta \\ u\left ( -\pi \right ) & =u\left ( \pi \right ) \\ u_{\theta }\left ( -\pi \right ) & =u_{\theta }\left ( \pi \right ) \end{align*}

Using separation of variables, let $$u\left ( r,\theta \right ) =R\left ( r\right ) \Theta \left ( \theta \right )$$ the solution is given by$$u\left ( r,\theta \right ) =\frac{a_{0}}{2}+\sum _{n=1}^{\infty }a_{n}r^{n}\cos \left ( n\theta \right ) +b_{n}r^{n}\sin \left ( n\theta \right ) \tag{1}$$ At $$r=r_{0}=1$$ we have that $$\frac{\partial u\left ( r,\theta \right ) }{\partial r}=\cos \theta$$ (since $$x=r\cos \theta$$ but $$r=1$$ at boundary). The above becomes$\cos \theta =\sum _{n=1}^{\infty }na_{n}r^{n-1}\cos \left ( n\theta \right ) +nb_{n}r^{n-1}\sin \left ( n\theta \right )$ Therefore $$n=1$$ is only term that survives in the sum. Hence $$a_{1}=1$$ and all others are zero. The solution (1) becomes$u\left ( r,\theta \right ) =\frac{a_{0}}{2}+r\cos \left ( \theta \right )$ The solution is not unique as there is $$a_{0}$$ arbitrary constant.

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##### 4.1.2.4 [309] Laplace inside disk. General solution

problem number 309

Solve Laplace equation in polar coordinates inside a disk

Solve for $$u\left ( r,\theta \right )$$ \begin{align*} u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta }=0 \end{align*}

With $$0 \leq r\leq a, 0 < \theta \leq 2\pi$$ Boundary conditions \begin{align*} u(a,\theta ) & = f(\theta ) \\ \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

$\left \{\left \{u(r,\theta )\to \underset{n=1}{\overset{\infty }{\sum }}\frac{a^{-n} r^n \left (\cos (n \theta ) \int _{-\pi }^{\pi } \cos (n \theta ) f(\theta ) \, d\theta +\left (\int _{-\pi }^{\pi } f(\theta ) \sin (n \theta ) \, d\theta \right ) \sin (n \theta )\right )}{\pi }+\frac{\int _{-\pi }^{\pi } f(\theta ) \, d\theta }{2 \pi }\right \}\right \}$

Maple

$u \left ( r,\theta \right ) =1/2\,{\frac{1}{\pi } \left ( 2\,\sum _{n=1}^{\infty } \left ({\frac{\cos \left ( n\theta \right ) \int _{-\pi }^{\pi }\!f \left ( \theta \right ) \cos \left ( n\theta \right ) \,{\rm d}\theta +\sin \left ( n\theta \right ) \int _{-\pi }^{\pi }\!f \left ( \theta \right ) \sin \left ( n\theta \right ) \,{\rm d}\theta }{\pi } \left ({\frac{a}{r}} \right ) ^{-n}} \right ) \pi +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \,{\rm d}\theta \right ) }$

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##### 4.1.2.5 [310] Laplace inside disk. Speciﬁc boundary conditions

problem number 310

Solve $$u_{xx}+u_{yy}=0$$ on disk $$x^{2}+y^{2}<1$$ with boundary condition $$xy^{2}$$ when $$x^{2}+y^{2}=a$$. Where $$a=1$$ in this problem. Express solution in $$x,y$$

The ﬁrst step is to convert the boundary condition to polar coordinates. Since $$x=r\cos \theta ,y=r\sin \theta$$, then at the boundary $$u\left ( r,\theta \right ) =r\cos \theta \left ( r\sin \theta \right ) ^{2}$$. But $$r=1$$ (the radius). Hence at the boundary, $$u\left ( 1,\theta \right ) =f\left ( \theta \right )$$ where \begin{align*} f\left ( \theta \right ) & =\cos \theta \sin ^{2}\theta \\ & =\cos \theta \left ( 1-\cos ^{2}\theta \right ) \\ & =\cos \theta -\cos ^{3}\theta \end{align*}

But $$\cos ^{3}\theta =\frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta$$. Therefore the above becomes\begin{align} f\left ( \theta \right ) & =\cos \theta -\left ( \frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta \right ) \nonumber \\ & =\frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta \tag{1} \end{align}

The above is also seen as the Fourier series of $$f\left ( \theta \right )$$. The PDE in polar coordinates is$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }=0$

Mathematica

Failed

Maple

$u \left ( r,\theta \right ) =-1/4\,r \left ({r}^{2}\cos \left ( 3\,\theta \right ) -\cos \left ( \theta \right ) \right )$

Hand solution

Solve $$u_{xx}+u_{yy}=0$$ on disk $$x^{2}+y^{2}<1$$ with boundary condition $$xy^{2}$$ when $$x^{2}+y^{2}=a$$. Where $$a=1$$ in this problem. Express solution in $$x,y$$

The ﬁrst step is to convert the boundary condition to polar coordinates. Since $$x=r\cos \theta ,y=r\sin \theta$$, then at the boundary $$u\left ( r,\theta \right ) =r\cos \theta \left ( r\sin \theta \right ) ^{2}$$. But $$r=1$$ (the radius). Hence at the boundary, $$u\left ( 1,\theta \right ) =f\left ( \theta \right )$$ where \begin{align*} f\left ( \theta \right ) & =\cos \theta \sin ^{2}\theta \\ & =\cos \theta \left ( 1-\cos ^{2}\theta \right ) \\ & =\cos \theta -\cos ^{3}\theta \end{align*}

But $$\cos ^{3}\theta =\frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta$$. Therefore the above becomes\begin{align} f\left ( \theta \right ) & =\cos \theta -\left ( \frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta \right ) \nonumber \\ & =\frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta \tag{1} \end{align}

The above is also seen as the Fourier series of $$f\left ( \theta \right )$$. The PDE in polar coordinates is$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }=0$ The solution is known to be$$u\left ( r,\theta \right ) =\frac{c_{0}}{2}+\sum _{n=1}^{\infty }r^{n}\left ( c_{n}\cos \left ( n\theta \right ) +k_{n}\sin \left ( n\theta \right ) \right ) \tag{2}$$ Since the above solution is the same as $$f\left ( \theta \right )$$ when $$r=1$$, then equating (2) when $$r=1$$ to (1) gives$\frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta =\frac{c_{0}}{2}+\sum _{n=1}^{\infty }\left ( c_{n}\cos \left ( n\theta \right ) +k_{n}\sin \left ( n\theta \right ) \right )$ By comparing terms on both sides, this shows by inspection that\begin{align*} c_{0} & =0\\ c_{1} & =\frac{1}{4}\\ c_{3} & =\frac{-1}{4} \end{align*}

And all other $$c_{n},k_{n}$$ are zero. Using the above result back in (2) gives the solution as$$\fbox{u\left ( r,\theta \right ) =\frac{r}{4}\cos \theta -\frac{r^3}{4}\cos 3\theta }\tag{3}$$ This solution is now converted to $$xy$$ using the formula\begin{align*} r^{n}\cos n\theta & =\sum _{\substack{k=0\\even}}^{n}\begin{pmatrix} n\\ k \end{pmatrix} x^{n-k}\left ( -1\right ) ^{\frac{k}{2}}y^{k}\\ & =\sum _{\substack{k=0\\even}}^{n}\frac{n!}{k!\left ( n-k\right ) !}x^{n-k}\left ( -1\right ) ^{\frac{k}{2}}y^{k} \end{align*}

For $$n=1$$ the above gives\begin{align} r\cos \theta & =\frac{1!}{0!\left ( 1-0\right ) !}x^{1-0}\left ( -1\right ) ^{0}y^{0}\nonumber \\ & =x\tag{4} \end{align}

And for $$n=3$$\begin{align} r^{3}\cos 3\theta & =\frac{3!}{0!\left ( 3-0\right ) !}x^{3-0}\left ( -1\right ) ^{0}y^{0}+\frac{3!}{2!\left ( 3-2\right ) !}x^{3-2}\left ( -1\right ) ^{1}y^{2}\nonumber \\ & =x^{3}-3xy^{2}\tag{5} \end{align}

Using (4,5) in (3) gives the solution in $$x,y$$$$\fbox{u\left ( x,y\right ) =\frac{1}{4}x-\frac{1}{4}\left ( x^3-3xy^2\right ) }\tag{6}$$ This is now veriﬁed that is satisﬁes the PDE $$u_{xx}+u_{yy}=0$$.\begin{align*} \frac{\partial u}{\partial x} & =\frac{1}{4}-\frac{1}{4}\left ( 3x^{2}-3y^{2}\right ) \\ \frac{\partial ^{2}u}{\partial x^{2}} & =-\frac{6}{4}x \end{align*}

And\begin{align*} \frac{\partial u}{\partial y} & =\frac{6}{4}xy\\ \frac{\partial ^{2}u}{\partial y^{2}} & =\frac{6}{4}x \end{align*}

Therefore $$\frac{\partial ^{2}u}{\partial x^{2}}+\frac{\partial ^{2}u}{\partial y^{2}}=0$$.

Now the boundary conditions $$u\left ( x,y\right ) =xy^{2}$$ are also veriﬁed. This condition applies when $$x^{2}+y^{2}=1$$ or $$y^{2}=1-x^{2}$$. Substituting this into (6) gives$u\left ( x,y\right ) _{@D}=\frac{1}{4}x-\frac{1}{4}\left ( x^{3}-3x\overset{y2}{\overbrace{\left ( 1-x^{2}\right ) }}\right )$ Simplifying gives\begin{align*} u\left ( x,y\right ) _{@D} & =\frac{1}{4}x-\frac{1}{4}\left ( x^{3}-\left ( 3x-3x^{3}\right ) \right ) \\ & =\frac{1}{4}x-\frac{1}{4}x^{3}+\frac{1}{4}\left ( 3x-3x^{3}\right ) \\ & =\frac{1}{4}x-\frac{1}{4}x^{3}+\frac{3}{4}x-\frac{3}{4}x^{3}\\ & =x-x^{3}\\ & =x\left ( 1-x^{2}\right ) \\ & =xy^{2} \end{align*}

Veriﬁed. This is 3D plot of the solution

This is a contour plot

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##### 4.1.2.6 [311] Haberman 2.5.5 (c)

problem number 311

This is problem 2.5.5 part (c) from Richard Haberman applied partial diﬀerential equations, 5th edition

Solve Laplace equation $u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2} u_{\theta \theta } = 0$ 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

Failed

Maple

$u \left ( r,\theta \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac{\int _{0}^{\pi /2}\!f \left ( \theta \right ) \sin \left ( 2\,n\theta \right ) \,{\rm d}\theta{r}^{2\,n}\sin \left ( 2\,n\theta \right ) }{\pi \,n}} \right )$

Hand solution

The Laplace PDE in polar coordinates is  $$r^{2}\frac{\partial ^{2}u}{\partial r^{2}}+r\frac{\partial u}{\partial r}+\frac{\partial ^{2}u}{\partial \theta ^{2}}=0\tag{A}$$ 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 $$\Theta _{n}\left ( \theta \right ) =B_{n}\sin \left ( 2n\theta \right ) \qquad n=1,2,3,\cdots \tag{3}$$ 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$$R_{n}\left ( r\right ) =C_{n}r^{2n}\qquad n=1,2,3,\cdots \tag{4}$$ 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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##### 4.1.2.7 [312] semi-circle

problem number 312

Solve Laplace equation $u_{rr} + \frac{1}{r }u_r + \frac{1}{r^2} u_{\theta \theta } = 0$ 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(1,\theta ) &= f(\theta ) \end{align*}

Mathematica

Failed

Maple

$u \left ( r,\theta \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac{\int _{0}^{\pi }\!f \left ( \theta \right ) \sin \left ( n\theta \right ) \,{\rm d}\theta{r}^{n}\sin \left ( n\theta \right ) }{\pi }} \right )$

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##### 4.1.2.8 [313] Haberman 2.5.8 (b)

problem number 313

This is problem 2.5.8 part (b) from Richard Haberman applied partial diﬀerential equations, 5th edition

Solve Laplace equation $$\nabla ^2 u(r,\theta )=0$$ or $u_{rr} + \frac{1}{r } u_r +\frac{1}{r^2} u_{\theta \theta } =0$ Inside circular annulus $$a<r<b$$ subject to the following boundary conditions \begin{align*} \frac{\partial u}{\partial r}(a,\theta ) &= 0 \\ u(b,\theta ) &= g(\theta ) \end{align*}

Mathematica

Failed

Maple

$u \left ( r,\theta \right ) ={\it invfourier} \left ({\frac{{\it fourier} \left ( g \left ( \theta \right ) ,\theta ,s \right ){{\rm e}^{s \left ( \ln \left ( b \right ) +\ln \left ( r \right ) -2\,\ln \left ( a \right ) \right ) }}}{{{\rm e}^{2\, \left ( \ln \left ( b \right ) -\ln \left ( a \right ) \right ) s}}+1}},s,\theta \right ) +{\it invfourier} \left ({\frac{{\it fourier} \left ( g \left ( \theta \right ) ,\theta ,s \right ){a}^{s}{r}^{-s}{{\rm e}^{ \left ( \ln \left ( b \right ) -\ln \left ( a \right ) \right ) s}}}{{{\rm e}^{2\, \left ( \ln \left ( b \right ) -\ln \left ( a \right ) \right ) s}}+1}},s,\theta \right ) -{\it invfourier} \left ({\frac{{\it fourier} \left ( g \left ( \theta \right ) ,\theta ,s \right ){{\rm e}^{2\, \left ( \ln \left ( b \right ) -\ln \left ( a \right ) \right ) s}}}{{{\rm e}^{2\, \left ( \ln \left ( b \right ) -\ln \left ( a \right ) \right ) s}}+1}},s,\theta \right ) -{\it invfourier} \left ({\frac{{\it fourier} \left ( g \left ( \theta \right ) ,\theta ,s \right ) }{{{\rm e}^{2\, \left ( \ln \left ( b \right ) -\ln \left ( a \right ) \right ) s}}+1}},s,\theta \right ) +g \left ( \theta \right )$ But has unresolved Invfourier and Fourier calls

Hand solution

The Laplace PDE in polar coordinates is  $$r^{2}\frac{\partial ^{2}u}{\partial r^{2}}+r\frac{\partial u}{\partial r}+\frac{\partial ^{2}u}{\partial \theta ^{2}}=0\tag{A}$$ 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$$\Theta =A_{0}+\sum _{n=1}^{\infty }A_{n}\cos \left ( n\theta \right ) +B_{n}\sin \left ( n\theta \right ) \tag{5}$$ 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$$R\left ( r\right ) =C_{0}+\sum _{n=1}^{\infty }C_{n}\left ( r^{n}+\frac{a^{2n}}{r^{n}}\right ) \tag{6}$$ 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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##### 4.1.2.9 [314] Circular annulus

problem number 314

Solve Laplace equation $u_{rr} + \frac{1}{r } u_r + \frac{1}{r^2} u_{\theta \theta } =0$

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

$\left \{\left \{u(r,\theta )\to \begin{array}{cc} \{ & \begin{array}{cc} \frac{2 \left (r^2-1\right ) \sin (\theta )}{3 r} & 1\leq r\leq 2 \\ \text{Indeterminate} & \text{True} \\\end{array} \\\end{array}\right \}\right \}$

Maple

$u \left ( r,\theta \right ) =2/3\,{\frac{\sin \left ( \theta \right ) \left ({r}^{2}-1 \right ) }{r}}$

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##### 4.1.2.10 [315] Outside a disk

problem number 315

Solve Laplace equation in polar coordinates outside a disk

Solve for $$u\left ( r,\theta \right )$$ \begin{align*} u_{rr}+\frac{1}{r} u_r +\frac{1}{r^2} u_{\theta \theta } & =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

Failed

Maple

$u \left ( r,\theta \right ) =1/2\,{\frac{1}{\pi } \left ( 2\,\sum _{n=1}^{\infty } \left ({\frac{\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \cos \left ( n\theta \right ) \,{\rm d}\theta \cos \left ( n\theta \right ) +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \sin \left ( n\theta \right ) \,{\rm d}\theta \sin \left ( n\theta \right ) }{\pi } \left ({\frac{r}{a}} \right ) ^{-n}} \right ) \pi +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \,{\rm d}\theta \right ) }$

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##### 4.1.2.11 [316] Outside a disk

problem number 316

Laplace PDE in polar coordinates outside a disk

Solve $$u_{xx}+u_{yy}=0$$ outside disk $$x^{2}+y^{2}>1$$ with boundary condition $$xy^{2}$$ when $$x^{2}+y^{2}=a$$. Where $$a=1$$ in this problem. Express solution in $$x,y$$

The ﬁrst step is to convert the boundary condition to polar coordinates. Since $$x=r\cos \theta ,y=r\sin \theta$$, then at the boundary $$u\left ( r,\theta \right ) =r\cos \theta \left ( r\sin \theta \right ) ^{2}$$. But $$r=1$$ (the radius). Hence at the boundary, $$u\left ( 1,\theta \right ) =f\left ( \theta \right )$$ where \begin{align*} f\left ( \theta \right ) & =\cos \theta \sin ^{2}\theta \\ & =\cos \theta \left ( 1-\cos ^{2}\theta \right ) \\ & =\cos \theta -\cos ^{3}\theta \end{align*}

But $$\cos ^{3}\theta =\frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta$$. Therefore the above becomes\begin{align} f\left ( \theta \right ) & =\cos \theta -\left ( \frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta \right ) \nonumber \\ & =\frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta \tag{1} \end{align}

The above is also seen as the Fourier series of $$f\left ( \theta \right )$$. The PDE in polar coordinates is$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }=0$

Mathematica

Failed

Maple

$u \left ( r,\theta \right ) =1/4\,{\frac{{r}^{2}\cos \left ( \theta \right ) -\cos \left ( 3\,\theta \right ) }{{r}^{3}}}$

Hand solution

The ﬁrst step is to convert the boundary condition to polar coordinates. Since $$x=r\cos \theta ,y=r\sin \theta$$, then at the boundary $$u\left ( r,\theta \right ) =r\cos \theta \left ( r\sin \theta \right ) ^{2}$$. But $$r=1$$ (the radius). Hence at the boundary, $$u\left ( 1,\theta \right ) =f\left ( \theta \right )$$ where \begin{align*} f\left ( \theta \right ) & =\cos \theta \sin ^{2}\theta \\ & =\cos \theta \left ( 1-\cos ^{2}\theta \right ) \\ & =\cos \theta -\cos ^{3}\theta \end{align*}

But $$\cos ^{3}\theta =\frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta$$. Therefore the above becomes\begin{align} f\left ( \theta \right ) & =\cos \theta -\left ( \frac{3}{4}\cos \theta +\frac{1}{4}\cos 3\theta \right ) \nonumber \\ & =\frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta \tag{1} \end{align}

The above is also seen as the Fourier series of $$f\left ( \theta \right )$$. The PDE in polar coordinates is$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^{2}}u_{\theta \theta }=0$ The solution is known to be$$u\left ( r,\theta \right ) =\frac{c_{0}}{2}+\sum _{n=1}^{\infty }r^{-n}\left ( c_{n}\cos \left ( n\theta \right ) +k_{n}\sin \left ( n\theta \right ) \right ) \tag{2}$$ Since the above solution is the same as $$f\left ( \theta \right )$$ when $$r=1$$, then equating (2) when $$r=1$$ to (1) gives$\frac{1}{4}\cos \theta -\frac{1}{4}\cos 3\theta =\frac{c_{0}}{2}+\sum _{n=1}^{\infty }\left ( c_{n}\cos \left ( n\theta \right ) +k_{n}\sin \left ( n\theta \right ) \right )$ By comparing terms on both sides, this shows by inspection that\begin{align*} c_{0} & =0\\ c_{1} & =\frac{1}{4}\\ c_{3} & =\frac{-1}{4} \end{align*}

And all other $$c_{n},k_{n}$$ are zero. Using the above result back in (2) gives the solution as \begin{align} u\left ( r,\theta \right ) & =\frac{r^{-1}}{4}\cos \theta -\frac{r^{-3}}{4}\cos 3\theta \nonumber \\ & =\frac{r^{2}\cos \left ( \theta \right ) -\cos \left ( 3\theta \right ) }{4r^{3}}\tag{3} \end{align}

This is 3D plot of the solution

This is a contour plot