Toronto Math Forum

Solve
\begin{align*}
& \Delta u:=u_{xx}+u_{yy}=0&& \text{in } r<a\\[3pt]
& u_r|_{r=a}=f(\theta).
\end{align*}
where we use polar coordinates $(r,\theta)$ and $f(\theta)=\left\{\begin{aligned}
&1 &&0<\theta<\pi,\\
-&1 &&\pi<\theta<2\pi.
\end{aligned}\right.$

The expected answer: solution as a series.


Given the condition $r<a$. We can discard the logr and $r^{-n}$ part ,so the general solution is
\begin{gather*}
u(r,\theta) =\frac{1}{2}A_0+\sum_{n=1}^\infty r^{n} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(r,\theta)=\sum_{n=1}^\infty nr^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta))\\
\implies u_r(a,\theta)=\sum_{n=1}^\infty na^{n-1} (A_n {\cos}(n\theta)+B_n {\sin}(n\theta)) = f(\theta).
\end{gather*}
For $n  = 0,1,2,3\ldots$
$$ A_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\cos}(n\theta)d\theta $$
For $n  = 1,2,3\ldots$
$$ B_n = \frac{1}{{\pi}na^{n-1}}\int_{-\pi}^{\pi} f(\theta){\sin}(n\theta)d\theta $$
Since $f(\theta)$ is odd function $\implies$ $A_n$ = 0 and
\begin{gather*}B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi} f(\theta){\sin}(n\theta)d\theta  \\
\implies B_n = \frac{2}{{\pi}na^{n-1}}\int_{0}^{\pi}{\sin}(n\theta)d\theta
\end{gather*}
We integral this get $-\frac{1}{n}\cos(n\theta)$ from 0 to $\pi$
$$\implies B_n = \frac{2}{{\pi}n^{2}a^{n-1}}[1-(-1)^{n}] $$
If we write n = 2m, it becomes 0, so we consider n = 2m+1
\begin{gather*}\implies B_n = \frac{4}{{\pi}(2m+1)^{2}a^{2m}}\\
u(r,\theta) =  \frac{4}{{\pi}}\sum_{m=0}^\infty \frac{r^{2m+1}}{(2m+1)^{2}a^{2m}} \sin(2m+1)\theta 
\end{gather*}

should $\phi(t) = 3cos(-\frac{1}{3}t) + constant $ ? And the answer for -t<x<2t is 3$cos(\frac{x-2t}{3}) -3$

Characteristic Equation :$$\frac{dt}{1} = \frac{dx}{3(t^2-1)} = \frac{du}{6t^2} $$
a). From $$\frac{dt}{1} = \frac{dx}{3(t^2-1)} \\ (3t^2-3)dt = dx \\ \Rightarrow t^3 -3t +D =x  \\ \Rightarrow D = x -t^3 -3t $$ b).From :$$\frac{dt}{1} =  \frac{du}{6t^2}  \\ 6t^2dt = du \\ \Rightarrow 2t^3+A = u\\ \Rightarrow u(x,t) = 2t^3+\phi(x-t^3-3t)$$ c). $$ u(x,0) = \phi(x) = x \\ \Rightarrow u(x,t) = 2t^3+x-t^3-3t  \\ \Rightarrow u(x,t) = t^3+x-3t$$

For the general solution :   \begin{equation} u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]+\frac{1}{2c}
  \int_{x-ct}^{x+ct}h(y)dy
\end{equation}
$$\\
$$for part(a):
\begin{equation}
 u(x,t) =\frac{1}{2}[g(x+ct)+g(x-ct)]
\end{equation}
$$
u(x,t) =
\begin{cases}
0, & \text{$x-ct >1 , x+ct \geq1$}  \\
\frac{1}{2}, & \text{$\vert x-ct \vert <1 , x+ct \geq1$}\\
1, & \text{$\vert x-ct \vert <1 , \vert x+ct \vert<1$}\\
1, & \text{$x-ct <-1 , x+ct \geq1$}\\
\end{cases}
$$for part(b):
\begin{equation}
 u(x,t) =\frac{1}{2c}
  \int_{x-ct}^{x+ct}h(y)dy
\end{equation}
$$
u(x,t) =
\begin{cases}
0, & \text{$x-ct >\frac{\pi}{2} , x+ct >\frac{\pi}{2}$}  \\
\frac{1}{2c}(1-\sin(x-ct), & \text{$\vert x-ct \vert <\frac{\pi}{2} ,x+ct \geq\frac{\pi}{2}$}\\
\frac{1}{2c}(\sin(x+ct)-\sin(x-ct), & \text{$\vert x-ct \vert <\frac{\pi}{2} , \vert x+ct \vert<\frac{\pi}{2}$}\\
\frac{1}{c}, & \text{$x-ct <-\frac{\pi}{2}  , x+ct \geq\frac{\pi}{2} $}\\
\end{cases}
$$