I'm trying to solve the Euler-Bernoulli equation for a beam for a construction with a solution in the form of $w(\zeta,t) = f(\zeta)g(t)$ (separation of variables). However, it is mentioned that this can still be tricky, but since the beam is undamped, we can assume $ g(t) = e^{i \lambda t}$ with $\lambda$ real.
My first question is; why is the above choice logical to assume?
(does this have anything to do with $e^{i \lambda t} = cos(\lambda t) + i sin(\lambda t)$ for $\lambda \in \mathbb{R}$)
I made a beginning with the separation of variable technique:
The Euler-Bernoulli equation:
$\rho\frac{\partial^{2}w}{\partial t^{2}}(\zeta,t) = -EI\frac{\partial^{4}w}{\partial \zeta^{4}}(\zeta,t),\text{ }\text{ }\text{ }\zeta \in [0,1], t \geq 0$
With boundary conditions:
$w(0,t) = w(1,t), \frac{\partial w}{\partial \zeta}(0,t) = \frac{\partial w}{\partial \zeta}(1,t)$
Substituting $w(\zeta,t) = f(\zeta)g(t)$ in the Euler-Bernoulli equation gives:
$\rho\frac{\partial^{2}f(\zeta)g(t)}{\partial t^{2}}(\zeta,t) = -EI\frac{\partial^{4}f(\zeta)g(t)}{\partial \zeta^{4}}(\zeta,t)$
$f(\zeta)\rho\frac{\partial^{2}g(t)}{\partial t^{2}}(\zeta,t) = -EIg(t)\frac{\partial^{4}f(\zeta)}{\partial \zeta^{4}}(\zeta,t)$
Since space and time are in this case independent of each other, both sides have to be equal to the same constant -$\lambda$. This gives us a set of three qualities:
$ 1. \frac{d^2g(t)}{dt^2} = -g(t)\lambda \\ 2. \frac{d^4f(\zeta)}{d\zeta^4} = \rho f(t)\frac{\lambda}{EI} \\ 3. w(0,t) = w(1,t) \rightarrow f(0) = f(1) $
The general solution of g(t) (if positive) is:
$ g(t) = C_{1} * e^{-i \sqrt{\lambda}t} + C_{2} * e^{i \sqrt{\lambda}t} $
Using Euler's formula, g(t) becomes:
$ g(t) = C_1 cos(\sqrt{\lambda}t) + C_2 sin(\sqrt{\lambda}t) $
From here, I am pretty much stuck to find a proposed solution for $f(\zeta)$ and to determine the solution to this problem.
The second question now is how to determine a proposed solution for $f(\zeta)$ and how to determine the 3 lowest (strictly positive) $\lambda$'s such that $w(\zeta,t) = f(\zeta)e^{i\lambda t}$ is a non-zero solution of the Euler-Bernoulli equation?
Thank you for all of your help.
Hint.
After solving for $f(\xi)$ we have
$$ f(\xi) = \sum_{k=1}^4 C_kf_k(\xi,\lambda) $$
A typical $f_k(\xi,\lambda)$ is
$$ f_1(\xi,\lambda) = e^{\sqrt[4]{-\frac{\lambda}{EI}}\xi} $$
Here $C_k$ are integration constants, so to accomplish with the boundary conditions we should solve the linear system
$$ f(0,\lambda)-f(1,\lambda) = 0\\ f'(0,\lambda)-f'(1,\lambda) = 0 $$
or
$$ M(\lambda)C = 0 $$
with
$$ C=\left(C1,C_2,C_3,C_4\right)^{\dagger} $$
thus to have a non trivial solution for $C$ we need that $\det\left(M(\lambda)\right) = 0$ and the roots for this equation furnish the eigenvalues $\lambda_j$