Are hyperbolic sines and cosines orthogonal in solving higher order PDE's?

[Bleakfacade]

Hello there!
So here's my problem, while you solve the Euler Bernoulli beam Equation by separation of variables, how do I have to prove the separated function of space are orthogonal? If so, are hyperbolic sines and cosines orthogonal when you have a product or a linear combination of them?

The pde is- [; u_{tt}+\alpha^{2} u_{xxxx} = o ;] where [; \alpha ;] is a constant that is material dependent. The separated function of space is [; F(x) = \sum_{n=1}^{\infty} [cosh(\beta_{n}x)-cos(\beta_{n}x)]-[sinh(\beta_{n}x)-sin(\beta_{n}x)] ;] where [; \beta_{n} ;] is some constant.

 

[pasmith]

You are looking for solutions of $X_\lambda^{(4)} = \lambda X_\lambda$ for $\lambda \in \mathbb{R}$. Now for fixed $\lambda$ there is a four-dimensional subspace of solutions, and given an arbitrary inner product on that space there will exist a basis which is orthogonal with respect to that inner product. (The constraint you haven't mentioned is that $X_\lambda$ needs to satisfy the boundary conditions at each end of the beam. That constraint may cause you to reject some of these solutions.) So the difficulty is to find an inner product with respect to which $X_\lambda$ and $X_\mu$ are necessarily orthogonal when $\lambda \neq \mu$.

A sufficient condition for this is that the operator $f \mapsto f^{(4)}$ should be self-adjoint with respect to the inner product, ie. $$(f^{(4)},g) = (f,g^{(4)})$$ for every $f$ and $g$. Now if you take (somewhat arbitrarily) the inner product $$(f,g) = \int_{-L}^L f(x)(g(x))^{*}\,dx$$ where ${}^{*}$ denotes the complex conjugate, then repeatedly integrating $(f^{(4)},g)$ by parts yields $$(f^{(4)},g) = (f,g^{(4)}) + \left[ f'''g^{*} - f''(g^*)' + f'(g^*)'' - f(g^*)''' \right]_{-L}^{L}.$$

 

[Bleakfacade]

You are looking for solutions of $X_\lambda^{(4)} = \lambda X_\lambda$ for $\lambda \in \mathbb{R}$. Now for fixed $\lambda$ there is a four-dimensional subspace of solutions, and given an arbitrary inner product on that space there will exist a basis which is orthogonal with respect to that inner product. (The constraint you haven't mentioned is that $X_\lambda$ needs to satisfy the boundary conditions at each end of the beam. That constraint may cause you to reject some of these solutions.) So the difficulty is to find an inner product with respect to which $X_\lambda$ and $X_\mu$ are necessarily orthogonal when $\lambda \neq \mu$.

A sufficient condition for this is that the operator $f \mapsto f^{(4)}$ should be self-adjoint with respect to the inner product, ie. $$(f^{(4)},g) = (f,g^{(4)})$$ for every $f$ and $g$. Now if you take (somewhat arbitrarily) the inner product $$(f,g) = \int_{-L}^L f(x)(g(x))^{*}\,dx$$ where ${}^{*}$ denotes the complex conjugate, then repeatedly integrating $(f^{(4)},g)$ by parts yields $$(f^{(4)},g) = (f,g^{(4)}) + \left[ f'''g^{*} - f''(g^*)' + f'(g^*)'' - f(g^*)''' \right]_{-L}^{L}.$$

Well, that was my approach, which was not obvious to me to begin with. But the boundary conditions were to replicate a cantilever beam that is suspended at one end. I have a rather monstrous equation which has many hyperbolic sines and cosines in it. I am not quite sure if the hyperbolic terms are orthogonal. I looked at their graphs for some insight but to no avail.
Also, the inner product I have is
[; (f,g) = \int_{0}^{L} f(x)(g(x))*dx ;]