 Solution

Take characteristic coordinates $\xi =ct-x$ and $\eta =ct+x$. Then $c^{-2}{y}_{tt}-y_{xx}=4y_{\xi \eta }$. The wave equation is therefore $y_{\xi \eta }=0$, whose general twice-differentiable solution isThe first fixed-end boundary gives $f(s)=-g(s)$. The second gives $g(s+L)=g(s-L)$, hence $g(s+2L)=g(s)$; $f$ has the same period. Thus $y=g(ct+x)-g(ct-x)$ with a $2L$-periodic function $g$.

Let $A=g^{\prime }(ct+x)$ and $B=g^{\prime }(ct-x)$. Then $y_t=c(A-B)$ and $y_x=A+B$, so the quadratic energy density reduces to $\frac{1}{2}[(A-B{)}^2+(A+B{)}^2]=A^2+B^2$. Changing $x$ to $-x$ in the second integral provesThe right side integrates a periodic function over one full period and is independent of its starting point. More explicitly its time derivative is $c[g^{\prime }(ct+L{)}^2-g^{\prime }(ct-L{)}^2]=0$. This is conservation of the fixed-end wave energy.