Given that kinetic energy is given as $T=\frac{1}{2}\int\limits_0^l\rho u_t^2dx$, where $\rho$ is a uniform linear density constant, potential energy is given as $U=\frac{1}{2}\int\limits_0^l \mu u_{xx}^2dx$, where $\mu>0$ is flexural rigidity. Also given: $u=u(x,t)$, $u=u_x=0$ at $x=0$ and $x=l$.
How does one derive the equation of motion $\rho u_{tt}+\mu u_{xxxx}=0$?
I took the generalized coordinates to be $u$ and $x$ and obtained $$\rho\int\limits_0^l u_{tt}=0.$$
With the second part, I tried integration by parts, but did not obtain the desired result. Would appreciate some input.
Hamilton's principle states that the first variation of the Lagrangian should be zero. That is,
$$\partial (T-U) = 0.$$
To calculate the first variation, consider a test function $v\in C^\infty_c$ and compute
$$\lim_{h\to 0}\frac{1}{2h}\left(\int \rho (u_t+hv_t)^2 - \mu (u_{xx}+hv_{xx})^2 -\rho u_t^2 + \mu u_{xx}^2 \right) = \frac{1}{2} \int 2\rho u_tv_t-2\mu u_{xx}v_{xx}.$$
Integration by parts (note that we integrate by parts 2 times for the space derivatives) gives
$$0= \partial (T-U) = \int - \rho u_{tt} v - \mu u_{xxxx} v = - \int (\rho u_{tt}+\mu u_{xxxx})v.$$
Since $v$ is arbitrary we deduce that $\rho u_{tt} +\mu u_{xxxx}=0$, as desired.