4
$\begingroup$

I have to solve the following eigenvalue problem, i.e. find eigenvalues and eigenfunctions (some of you will notice that this is the Schrödinger equation): $$-\frac{\hbar^2}{2m}\left( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)\Psi \left(x,y \right)=E\cdot \Psi \left( x,y \right)$$

with $$\Psi(x,y)=0, \qquad \left(x,y\right) \notin \mathbf{\Omega}$$

and boundary condition:

$$\Psi(x,y)=0 \quad (x,y)\in \mathbf{\partial \Omega}$$

where $\mathbf{\Omega}$ is shaded region shown in Figure. (I'm new on Stack Exchange so I can only provide a link to figure).

I started with separation of variables: $$ \Psi(x,y)=X(x)\cdot Y(y)$$ and found these general solutions for $X(x)$ and $Y(y)$: $$ X(x)=A\sin \left(k_x x \right)+B \cos \left(k_x x \right)$$ $$Y(y)=C\sin \left(k_y y \right)+D \cos \left(k_y y \right) $$ where $$ k^2_x+k^2_y=\frac{2mE}{\hbar^2}$$

Plugging in boundary conditions: $$ \Psi(x=a,y)=X(x=a)\cdot Y(y)=0 \quad \implies \quad X(x=a)=0$$

$$ X(x=0)=A\sin \left( k_x \cdot 0\right)+B \cos \left( k_x \cdot 0 \right)=B=0 $$

$$ \implies \quad \sin \left( k_x a\right)=0 \quad \implies k_x a=m \pi, \quad \implies k_x=\frac{m\pi}{a},\quad m=1, 2, 3, \ldots $$

$$ \Psi(x,y=0)=X(x)\cdot Y(y=0)=0 \quad \implies \quad Y(y=0)=0 $$

$$ C \sin\left(k_y \cdot 0\right)+D\cos \left( k_y \cdot 0\right)=D=0 $$

$$ Y(y=a^2)= C\sin \left(k_y a^2 \right)+D\cos \left(k_y a^2 \right)=C\sin \left(k_y a^2 \right)=0 $$

$$ \implies \quad \sin \left( k_y a^2\right)=0 \quad \implies k_y a^2=n \pi,\quad k_y=\frac{n\pi}{a^2},\quad n=1, 2, 3, \ldots $$

I'm not sure what to do next. My guess is that solution $\Psi(x,y)$ can be found as a superposition:

$$ \Psi(x,y)=\displaystyle\sum_{m,n}K_{m,n} \sin \left(\frac{m\pi x}{a} \right) \sin \left( \frac{n\pi y}{a^2}\right) $$

where coefficients $K_{m,n}$ are appropriately chosen such that $\Psi(x,y=x^2)=0, \quad \forall x\in[0,a]$, but I have no idea how to find them.

I've searched many textbooks and online resources but I couldn't find anything similar. It might be that my approach is wrong and that this can be solved only using completely different techniques.

Thanks everyone for help!

$\endgroup$
  • $\begingroup$ The boundary is not separable in rectangular coordinates. $\endgroup$ – Mark Viola May 7 '16 at 4:13
  • $\begingroup$ Okay, so separation of variables is not the way to go. Do you have an idea what else can be done? Maybe some transformation of variables so that boundary is separable in new variables? $\endgroup$ – Jack Bauer May 7 '16 at 4:46
  • $\begingroup$ Unfortunately, I don't have a way forward at this time aside from numerical computation. $\endgroup$ – Mark Viola May 7 '16 at 5:09
0
$\begingroup$

You can only find an explicit solution to this eigenvalue problem for particular geometries such as a rectangle or a disk, and maybe a few others. For other domains $\Omega$ such as the one you are investigating, you need to use a numerical approximation, or you can also show some properties of the solutions analytically (corner singularities, nodal sets, symmetries).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.