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I was given the following question, and I think I'm nearly there, I just wanted to ask for some clarification in the last step.

Derive the eigenvalues and functions of the SL problem $\phi_{xx} +\phi_{yy} + \lambda \phi = 0$, with $\phi(x,0) = \phi(x,H) = \phi(0,y) = \phi(L,y) = 0$. If $H = 2L$, show that $\lambda(4,1)$ and $\lambda(2,2)$ are identical, but that $\phi_{\lambda(4,1)}$ and $\phi_{\lambda(2,2)}$ are orthogonal.

So, I've done everything up to showing the orthogonality. Usually, I've just shown orthogonality in one dimension, over a single domain, but I'm a little confused as to how to deal with it in two dimensions.

Ignoring the coefficient, I've basically got this;

$$\phi_{\lambda\left(4,1\right)} = \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi y}{L}\right)$$ $$\phi_{\lambda\left(2,2\right)} = \sin\left(\frac{2 \pi x}{L}\right) \sin\left(\frac{\pi y}{L}\right)$$

Then, I assumed that I needed to multiply these two functions together, and integrate over the domains $x \in [0,L]$ and $y \in [0,2L]$;

$$\int_{0}^{2L} \int_{0}^{L} \sin\left(\frac{\pi y}{L}\right) \sin\left(\frac{2\pi y}{L}\right) \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2 \pi x}{L}\right)\, dx\, dy$$

Now, I know that if I was just examining the integral involving $x$ terms, I'd have zero, and the same would be true for the $y$ terms, but I'm just not too sure how to approach it. Any input would be great, thank you~

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When the integrand in a double integral over a rectangle is of the form $f(x)g(y)$, you are in luck: the integral splits as $$ \int_c^d\int_a^b f(x)g(y)\,dx\,dy = \int_a^b f(x) \,dx \int_c^d g(y) \,dy $$ since when performing integration with respect to $y$, we can move $f(x)$ outside the integral.

So, the double integral you have is the product of $$ \int_{0}^{2L} \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2 \pi x}{L}\right)\, dx $$ and $$\int_{0}^{L} \sin\left(\frac{\pi y}{L}\right) \sin\left(\frac{2\pi y}{L}\right) \, dy$$ Having one of those equal to zero makes the product zero.

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