Find the solution of the Dirichlet problem:
$$\Delta u(r,\phi)=0, r<1, u(1,\phi)=f(\phi)$$
where $x=r\cos\phi$ and $y=r\sin\phi$ and
$$f(\phi)=\sin^3(\phi).$$
I start by doing the following:
Enter the polar coordinates $x=r\cos\phi$ and $y=r\sin\phi$. Deriving:
$$\begin{gather} u_r=u_x\cos\phi+u_y\sin\phi \tag{1}\\ u_\phi=-ru_x\sin\phi+ru_y\cos\phi \tag{2}\\ u_{rr}=u_{xx}\cos^2(\phi)+2u_{xy}\cos\phi \sin\phi+u_{yy}\sin^2\phi \tag{3}\\ u_{\phi\phi}=r^2u_{xx}\sin^2\phi-ru_{x}\cos\phi \tag{4}\\ -2r^2u_{xy}\sin\phi \cos\phi+r^2u_{yy}\cos^2\phi-ru_y\sin\phi \tag{5} \end{gather}$$
Adding equation $(3)$ and equation $(5)$ divided by $r^2$, we obtain
$$u_{rr}+\frac{1}{r^2}u_{\phi\phi}=u_{xx}+u_{yy}-\frac{1}{r}u_x\cos\phi-\frac{1}{r}u_y\sin\phi$$
Using $(1)$, we obtain
$$\Delta u=\frac{1}{r} (ru_r)_r+\frac{1}{r^2}u_{\phi\phi}$$
Therefore in polar coordinates, the problem takes the form:
$$\frac{1}{r}(ru_r(r,\phi))_r+\frac{1}{r^2}u_{\phi\phi}(r,\phi)=0 \text{ with } u(a,\phi)=f(\phi).$$
My biggest problem is to reach the conclusion that the solution is
$$r(3\sin\phi-r^2\sin 3\phi)/4.$$
$\endgroup$2 Answers
$\begingroup$Here you should use the Fourier expansion of $f\left(\phi\right)=\frac{a_{0}}{2}+\sum\limits^{+\infty}_{n=1}{a_{n}\cos(n\phi)+b_{n}\sin(n\phi)}$. General solution of Laplas problem in circle is
$$ u(r,\phi)=\sum\limits^{+\infty}_{n=1}{r^{n}(A_{n}\cos(n\phi)+B_{n}\sin(n\phi))} $$
So, when you expand $f(\phi)$ to Fourier series, you get
$$ \frac{a_{0}}{2}+\sum\limits^{+\infty}_{n=1}{a_{n}\cos(n\phi)+b_{n}\sin(n\phi)}=\sum\limits^{+\infty}_{n=1}{1^{n}(A_{n}\cos(n\phi)+B_{n}\sin(n\phi))} $$
Because of the fact that expressions on the left and right must coincide you'll get system of equations of coefficients $\sin(n\phi),\cos(n\phi)$. So you'll get your solution.
$\endgroup$ 1 $\begingroup$Use the fact that $\sin^3{\phi} = \frac{3}{4} \sin{\phi} - \frac14 \sin{3 \phi}$, and that the general solution to the interior Dirichlet problem in polar coordinates is
$$u(r,\phi) = \frac{a_0}{2} + \sum_{k=1}^{\infty} r^k (a_k \cos{k \phi} + b_k \sin{k \phi})$$
Now, $u(1,\phi) = \sin^3{\phi} = \frac{3}{4} \sin{\phi} - \frac14 \sin{3 \phi}$ means that we may simply conclude $a_k=0$ for all $k$ and $b_k=0$ for all $k$ expect $k=1$ and $k=3$, where $b_1=\frac{3}{4}$ and $b_3=-\frac14$. This is true because the sines and cosines form an orthogonal basis set over the unit circle. Thus, the solution to this interior Dirichlet problem is
$$u(r,\phi) = \frac{3}{4} r \sin{\phi} -\frac14 r^3 \sin{3 \phi}$$
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