I have learned that the IBVP: $$u_{tt}-c^{2}u_{xx}=0,$$ $$u(x,0)=f(x),\,u_{t}(x,0)=g(x),$$ $$u_{x}(0,t)=u_{x}(L,t)=0$$
has the unique solution $\displaystyle u(x,t)=\frac{1}{2}(A_{0}+B_{0}t)+\sum_{n=1}^{\infty}[A_{n}\cos(\omega_{n}t)+B_{n}\sin(\omega_{n})t]\cos\left(\frac{n\pi x}{L}\right)$ with $\displaystyle A_{n}=\frac{2}{L}\int_{0}^{L}f(x)\cos\left(\frac{n\pi x}{L}\right)dx,\,n=0,1,2,\cdots$, $\displaystyle B_{n}=\frac{2}{\omega_{n}L}\int_{0}^{L}g(x)\cos\left(\frac{n\pi x}{L}\right)dx,\,n=1,2,3\cdots$ and $\displaystyle B_{0}=\frac{2}{L}\int_{0}^{L}g(x)dx$ if some assumptions on $f$ and $g$ are met.
Also, by differentiating $\displaystyle e(t)=\frac{1}{2}\int_{0}^{L}[u_{t}^{2}(x,t)+c^{2}u_{x}^{2}(x,t)]dx$, the quantity known as energy, I understood that $e'(t)=0$ hence the energy is conserved for this problem. However, the solution implies that the zeroth-order term is a linear function of $t$ so the solution grows as $t\rightarrow\infty$. This makes me confused about how the conservation of energy should be intuitively (or physically) interpreted (in fact, higher-order terms essentially represent oscillation so I think they are not of problem).
How can the conservation of energy be interpreted when the solution seems to keep increase as time increases?
The same thing applies to an object in inertial movement. Its position in space is $u = u_0 + v t$ which is unbounded, but its kinetic energy $m v^2/2$ is constant. The reason is that only the derivative with respect to time appears in the energy.