# Trick to prove orthogonality for eigenfunctions?

Is there a magic trick to see that $$\int_0^LX_m(x)X_n(x)dx=0$$ for $$X_n(x) = c_n\Big[\big[\sin(\omega_n L) - \sinh(\omega_n L)\big]\big[\cos(\omega_n x) - \cosh(\omega_n x)\big] \\ - \big[\cos(\omega_n L) - \cosh(\omega_n L)\big]\big[\sin(\omega_n x) - \sinh(\omega_n x)\big]\Big]$$ ?

Where is assumed $X_n\neq X_m,c_n$ and $L$ are constants ,$w_n$ has the eigenvalue implicit $w_n=(\lambda_n/k)^{1/4}$

and $w_n$ solves $\det M=0$ from $\begin{bmatrix}\cos\omega L-\cosh\omega L&\sin\omega L-\sinh\omega L\\-\sin\omega L-\sinh\omega L&\cos\omega L-\cosh\omega L\end{bmatrix}\begin{bmatrix}A\\B\end{bmatrix} = M\begin{bmatrix}A\\B\end{bmatrix} = 0$

(This question has roots from Understanding solution of PDE using method: separation of variables.).

I tried to do it by brute force, i.e. substitution and then to expand the product between $X_m$ and $X_n$ but each calculation it's getting worst since a lot of weird integrals appear.

• You need to include the additional condition that $\omega_n$ solves $\det M = 0$ where $M$ is the coefficient matrix, from your previous questions. That context is important. – Dylan May 12 '18 at 19:21
• Imagine asking about the orthogonality of $\sin (\omega_n x)$ but leaving out $\omega = n\pi$. It's the same here. – Dylan May 13 '18 at 7:36
Write $X_n$ as a linear combination of exponentials, including complex exponentials. The integral will be easy then.