Dirichlet problem for bounded region

Consider the BVP

$$u_{xx}- u_{yy} = 0~~0< x<L,$$ (3.73)

$$u(0,y) =u(a,y)=u(x,b) = 0,$$ (3.74)

$$u(x,0) = f(x)$$ (3.75)

Solution Assume that the solution can be expressed in separation form as

Figure:

u(x,t)=X(x)T(t)
$\frac{X''}{X} = \frac{Y''}{Y} = \lambda$

1. If $\lambda >0$

   $$X''=\lambda X$$ (3.76)

   $$X(x)=\alpha exp( \sqrt{\lambda}x )+\beta exp(-\sqrt{\lambda}x)$$ (3.77)

   $$X(x)=\alpha cosh( \sqrt{\lambda}x )+\beta sinh(\sqrt{\lambda}x)$$ (3.78)

   using B.C's
   

   $$X(0) = 0,~X(a)=0,$$ (3.79)

   
   which implies that $\alpha=0$ and $\beta=0$ and we get a trivial solution. Thus we can not have positive eigenvalues which means $\lambda \not>0.$

2. If $\lambda=0,$
   $$X''=0\implies X'= \alpha$$
   $$X=\alpha x+\beta$$
   $$X(0)=0\implies \beta=0$$
   and
   $$X(a)=0\implies \alpha=0$$
   $$\implies X=\beta,\beta \neq 0.$$
   For this case we also get trivial solution.

3. If $\lambda < 0,$

   Replace $\lambda$ by $-\lambda$ in Eq. ()
   

   $$X''+\lambda X =$$ 0 (3.80)

   $$X = \alpha \cos(\sqrt{\lambda}x)+\beta \sin(\sqrt{\lambda}x)$$ (3.81)

   
   
   $$X(0)=0\implies \alpha =0$$
   and
   $$X(a)=0\implies \sin(\sqrt{\lambda}a)=0~(\beta\neq 0)$$
   $$\implies\sqrt{\lambda}a=n \pi,~~n=1,2,3,...$$
   $$\lambda_n=\frac{n^2 \pi^2}{a^2}$$
   are the eigenvalues and corresponding eigen-functions are
   $$X_n(x)=\alpha_n \cos(\frac{n\pi}{a}x)$$
   we can combine both the case as
   $$X_n(x)=\alpha_n \cos(\frac{n\pi}{a}x)~~n=0,1,2,...$$

Solving for $Y$:

$$\frac{Y''}{Y}=\lambda$$

$${y_n}^''-\frac{n^2\pi^2}{a^2}y_n=0$$

$$Y_n(y)=\alpha_n\cosh(\frac{n\pi}{a}y)+\beta_n\sinh(\frac{n\pi}{a}y),$$ (3.82)

$y_n(b)=0$ gives

$$\beta_n=-\alpha_n\frac{\cosh(\frac{n\pi}{a}b)}{\sinh(\frac{n\pi}{a}b)}$$

$$Y_n(y)=\alpha_n\frac{1}{\sinh(\frac{n\pi}{a}b)}\bigg[\sinh(\frac{... ...}{a}b)\cosh(\frac{n\pi}{a}y)-\cosh(\frac{n\pi}{a}b)\sinh(\frac{n\pi}{a}y)\bigg]$$

Let

$$\alpha_n\frac{1}{\sinh(\frac{n\pi}{a}b)}=\xi_n$$

$$y_n(y)=\xi_n\sinh(\frac{n\pi}{a}(b-y)).$$

By the principle of superposition of solutions,we get the general solution

$$u(x,t) = \sum_{n=0}^{\infty}u_n(x,t)$$ (3.83)

$$= \sum_{n=0}^{\infty}\xi_n \sin(\frac{n\pi}{a}x)\sinh(\frac{n\pi}{a}(b-y))$$ (3.84)

The unknown coefficients ${\xi_n}$ are found by using the remaining boundary condition $u(x,0)=f(x)$

$$u(x,0)=f(x)=\sum_{n=1}^{\infty}\xi_n \sin(\frac{n\pi}{a}x)\sinh(\frac{n\pi}{a}b).$$ (3.85)

Multiply both sides by $\sin(\frac{m\pi}{a})x$ and integrate both sides $[0,a]$ and using orthogonality, one gets

$$\xi_n=\frac{1}{\sin(\frac{n\pi}{a}b)}\frac{2}{a}\int_{0}^{a}f(x)\sin\sin(\frac{n\pi}{a}x)$$ (3.86)

which all combine given the required solution.