$$u_t = c^2u_{xx} \mid (x,t) \in (0,L) \times (0, \infty)$$ $$u_x(0,t) = u_x(L,t) = 0 \mid t > 0$$ $$u(x,0) = 0 \mid x \in [0,L]$$
I tried energy method based on Pinchover and Rubinstein (*):
$$\text{Let } E(t) = \frac{1}{2} \int_{0}^{L} u^2 dx$$
$$\to E'(t) = \frac{1}{2} \int_{0}^{L} 2uu_t dx$$
$$= c^2 \frac{1}{2} \int_{0}^{L} 2uu_{xx} dx$$
$$= c^2 \int_{0}^{L} uu_{xx} dx$$
$$= c^2 [\underbrace{uu_{x}\mid_{0}^{L}}_{\text{zero}} - \int_{0}^{L} u_x dx] \text{(**)}$$
$$= - c^2 [\int_{0}^{L} u_x dx]$$
I'm suppose to say that $E(t)$ is nonnegative, decreasing and zero when $t=0$, but I think I'm missing another $u_x$ in the integrand in the last line:
(**)
$$\int_{0}^{L} uu_{xx} dx$$
$$ w = u \mid dv = u_{xx} dx$$
$$dw = du \mid v = u_{x}$$
$$= uu_{x} - \int_{0}^{L} u_{x} dx$$
Is it supposed to be that $dw = u_x dx$?
Where did I go wrong? Any other mistakes?
(*)
It's supposed to be $dw = u_xdx$ because the differentiation is w/rt x and not u, like how the integration for the dv part is w/rt x