Toronto Math Forum

You cannot assume that the eigenvalues are integers. By writing the general solution as such a sum, you are assuming that $u=0$ at $r=\pi$, but this is not given. Indeed, there are no boundary conditions for $r>0$.

Just a remark: if a function is different or undefined at a single point (or any finite number of points) it does not change the value of the integral. This is because a finite number of points has Jordan measure zero (https://en.wikipedia.org/wiki/Jordan_measure). So there is no problem in not defining $\theta(0)$.

Introduce $u=X(x)Y(y)$ then applying separation of variables yields
$$\underbrace{\frac{X''}{X}}_{-\lambda_1}+\underbrace{\frac{Y''}{Y}}_{-\lambda_2}=-\lambda$$
Now we have two ODEs, namely $X''+\lambda_1X=0$ and $Y''+\lambda_2Y=0$ both with Neumann B.C. at $0,a$ and $0,b$ respectively. We know the solution to these problems:
$$\begin{cases}
\lambda_{1,0}=0& X_0=\frac{1}{2}\\
\lambda_{1,n}=\frac{n^2\pi^2}{a^2}& X_{n}=\cos(n\pi x/a)\\
\lambda_{2,0}=0& Y_0=\frac{1}{2}\\
\lambda_{2,n}=\frac{m^2\pi^2}{b^2}& Y_{m}=\cos(m\pi x/b)\\
\end{cases}$$
Where $n,m\geq 1$. So the eigenfunctions are
$$u_{n,m}(x,y)=\cos(n\pi x/a)\cos(m\pi x/b)$$
With eigenvalues $\lambda_{m,n}=\lambda_{1,n}+\lambda_{2,n}=\pi^2(n^2/a^2+m^2/b^2)$. We can absorb the case where $m=0$ or $n=0$ by allowing us to plug $m,n=0$ in this equation.

Introduce $u=X(x)T(t)$ and then separation of variables yield $X''+\lambda X=0$ and $T''+\lambda T=0$. Assume that $\lambda=\omega^2$ where $\omega\geq 0$. Then for $\omega\neq 0$, we have solutions $X_1=\sin \omega x$ and $X_2=\cos\omega x$. Applying the boundary conditions to $X_1$ we find
$$0=\sin \omega \pi$$
$$-1=\cos\omega \pi$$
Applying the boundary conditions to $X_2$ yields the same thing. The first equation suggests that $\omega=n$ where $n=1,2,...$. The second equation suggests that $\omega=2n+1$, $n=0,1,2,...$. To satisfy both, we must take the second one. So we have $\omega_n=2n+1$, $n=0,1,2,...$. Now for $\omega=0$ the solution is $Ax+B$. But the second B.C. requires $B=-B$ so $B=0$. Then applying the first B.C. we find $A=-A$ so $A=0$. So we do not have this eigenvalue. Thus our solutions for the spacial part is
$$X_{1,n}=\sin((2n+1)x), X_{2,n}=\cos((2n+1)x)$$
With $\lambda_n=(2n+1)^2$, $n=0,1,2...$. Then the equation for time can be solved immediately: $T=A\sin((2n+1)t)+B\cos((2n+1)t)$. Applying $u_t|_{t=0}=0$ we find $A=0$, and now we may write down the general solution:
$$u(x,t)=\sum_{n\geq 0}[A_n\sin((2n+1)x)+B_n\cos((2n+1)x)]\cos((2n+1)t)$$
Applying the condition $u_{t=0}=1$ we have
$$1=\sum_{n\geq 0}A_n\sin((2n+1)x)+B_n\cos((2n+1)x)$$
At this point the coefficients can be calculated via the standard method using orthogonality:
$$A_n=\frac{2}{\pi}\int_0^\pi \sin((2n+1)x)\,dx=\frac{2}{\pi(2n+1)}\cos((2n+1)x)|_\pi^0=\frac{4}{\pi (2n+1)}$$
$$B_n=\frac{2}{\pi}\int_0^\pi \cos((2n+1)x)\,dx=0$$
Where the second integral is zero because we will be evaluating sine functions at integer values of $\pi$. So we have our solution:
$$u(x,t)=\frac{4}{\pi}\sum_{n\geq 0}\frac{1}{2n+1}\sin((2n+1)x)\cos((2n+1)t)$$

Some more typos in Section 6.5:

1. Equation (6) should contain $h(\theta ')$, not $g(\theta ')$.
2. In the derivation immediately following equation (6), we have a line that says $-\frac{a}{\pi}\mathrm{Re}\log (1-ra^{-1}e^{i(\theta-\theta')})=-\frac{a}{2\pi}\log(a^{-2}(1-ra^{-1}e^{i(\theta-\theta')})(1-ra^{-1}e^{-i(\theta-\theta')}))$. The $a^{-2}$ in the logarithm should not be there at this stage. It should be there in the line right after, so the final expression is correct.
3. At the very bottom of the page where it says "...where for sector $\{r<a, 0<\theta<\alpha\}$ we should set...", the $B_n$ is missing.

Before we begin, we need one result: $\int x^2e^{-x^2/2}\,dx$. Changing variables, this can be written as
$$-\int x\,d(e^{-x^2/2})$$
Now integrate by parts, set $u=x$, $du=dx$, $dv=d(e^{-x^2/2})$, $v=e^{-x^2/2}$, we obtain
$$-xe^{-x^2/2}+\int e^{-x^2/2}\,dx$$
Now we begin the problem.
By D'Alembert's formula and Duhamel integral, the solution is
$$u(x,t)=\frac{1}{2}\left(-e^{(x-t)^2/2}-e^{(x+t)^2/2}\right)+\frac{1}{2}\int_0^t\int_{x-(t-t')}^{x+(t-t')} x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'dt'$$
Now consider the inside integral
$$\int_{x-(t-t')}^{x+(t-t')} x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(*)$$
Let $a=x-(t-t')$ and $b=x+(t-t')$ for notational simplicity.
Using the result above, the first term can be integrated:
$$\int_{a}^{b} x'^2e^{-x'^2/2}\,dx'=ae^{-a^2/2}-be^{-b^2/2}+\int_a^b e^{-x'^2/2}\,dx'$$
But note that the second integral here cancels with the integral of the second term in $(*)$. So we just have
$$\int_a^b x'^2e^{-x'^2/2}-e^{-x'^2/2}\,dx'=ae^{-a^2/2}-be^{-b^2/2}$$
Now we need to integrate this with respect to $t'$, from $0$ to $t$. Consider the first term first. Note that $da=dt'$. When $t'=0$, $a=x-t$, When $t'=t$, $a=x$. Thus the first term integrates
$$\int_{x-t}^x ae^{-a^2/2}\,da=e^{-(x-t)^2/2}-e^{-x^2/2}$$
Now for the second term, note that $db=-dt'$. When $t'=0$, $b=x+t$. When $t'=t$, $b=x$. Thus we have
$$-\int_{x+t}^x be^{-b^2/2}\,db=e^{-x^2/2}-e^{-(x+t)^2/2}$$
Now putting together all the results:
$$u(x,t)=\frac{1}{2}\left(-e^{(x-t)^2/2}-e^{(x+t)^2/2}\right)+\frac{1}{2}\left(e^{-(x-t)^2/2}-e^{-x^2/2}-e^{-x^2/2}+e^{-(x+t)^2/2}\right) =-e^{-x^2/2}$$
Since this does not depend on $t$, we have $\lim_{t\to\infty} u(x,t)=-e^{-x^2/2}$.

Before we begin, we need one result:
$$\int z^2 e^{-z^2}\, dz$$
Changing variables, this can be written as
$$-\frac{1}{2}\int z\,d(e^{-z^2})$$
Integrate by parts, set $s=z$, $ds=dz$, $dv=d(e^{-z^2})$, $v=e^{-z^2}$ we have
$$-\frac{1}{2}ze^{-z^2}+\frac{1}{2}\int e^{-z^2}\,dz=\frac{\sqrt{\pi}}{4}erf(z)-\frac{1}{2} ze^{-z^2}$$

Now we begin the problem. The solution is
$$u(x,t)=\frac{1}{\sqrt{4\pi t}} \int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}(1-y^2)\,dy=\frac{1}{\sqrt{4\pi t}} \left( \int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}\,dy- \int_{-1}^1 y^2e^{-\frac{(x-y)^2}{4t}}\,dy\right)$$
From now on, define $a=(x+1)/\sqrt{4t}$ and $b=(x-1)/\sqrt{4t}$ for notational simplicity.
In the first integral, substitute $z=(x-y)/\sqrt{4t}$ so that $dz=-dy/\sqrt{4t}$, so that
$$\int_{-1}^1 e^{-\frac{(x-y)^2}{4t}}\,dy=-\sqrt{4t}\int_a^b e^{-z^2}\,dz=\frac{\sqrt{4\pi t}}{2}\left(erf(a)-erf(b)\right)$$
In the second integral, substitute the same thing. Note that $y^2=x^2-2xz\sqrt{4t}+4tz^2$. So we have
$$ \int_{-1}^1 y^2e^{-\frac{(x-y)^2}{4t}}\,dy=-\sqrt{4t}\int_a^b x^2e^{-z^2}-2x\sqrt{4t} ze^{-z^2}+4tz^2 e^{-z^2}\,dz$$
Now integrate the three terms. The first one can be done using error function, the second one can be integrated directly, the third one using the result we obtained above:
$$ \int_{-1}^1 y^2e^{-\frac{(x-y)^2}{4t}}\,dy=-\sqrt{4t}\left[ \frac{x^2\sqrt{\pi}}{2}\left(erf(b)-erf(a)\right)+x\sqrt{4t}\left(e^{-b^2}-e^{-a^2}\right)+4t\left(\frac{\sqrt{\pi}}{4}erf(b)-\frac{\sqrt{\pi}}{4}erf(a)-\frac{1}{2}\frac{x-1}{\sqrt{4t}}e^{-b^2}+\frac{1}{2}\frac{x+1}{\sqrt{4t}}e^{-a^2}\right)\right]$$
Now all that is left is to combine the two and simplify. In the end, one obtains
$$u(x,t)=\left(\frac{1}{2}-\frac{x^2}{2}-t\right)erf\left(\frac{x+1}{\sqrt{4t}}\right)+\left(\frac{x^2}{2}-\frac{1}{2}+t\right)erf\left(\frac{x-1}{\sqrt{4t}}\right)+\sqrt{\frac{t}{\pi}}\left[(x+1)e^{-\frac{(x-1)^2}{4t}}+(1-x)e^{-\frac{(x+1)^2}{4t}}\right]$$