= Solution
Let the mass per unit length be $m$ and the constant tension be $T$. Small transverse displacement obeys the <wave equation>
$$
\boxed{m y_{tt}=T y_{xx},\qquad y_{tt}=c^2y_{xx},\quad c=\sqrt{T/m}}.
$$
With characteristic coordinates $\xi=x+ct$ and $\eta=x-ct$, the equation becomes $y_{\xi\eta}=0$. Integrating once in each variable gives <D'Alembert's formula>
$$
\boxed{y(x,t)=f(x+ct)+g(x-ct)}.
$$
For a smooth solution with vanishing endpoint <energy> flux, multiplication by $y_t$ and integration by parts give
$$
\frac{dE}{dt}=\int_0^\infty y_t(y_{tt}-c^2y_{xx})\,dx+[c^2y_ty_x]_0^\infty=0.
$$
The fixed endpoint has $y_t(0,t)=0$. Decay of displacement alone does not guarantee either a pointwise derivative-flux limit or finiteness of the <energy> integral; the conservation statement is understood in the usual finite-<energy> class. The profile formula below proves it directly there without requiring such a pointwise limit at infinity.
The endpoint condition gives $g(-ct)=-f(ct)$, allowing the reflected wave to be expressed through one profile:
$$
y(x,t)=f(x+ct)-f(ct-x).
$$
For forward time one may define the negative-argument part of this profile using the original $g$; no restriction on the independent initial waves is lost. Differentiation gives $y_t=c[f'(x+ct)-f'(ct-x)]$ and $y_x=f'(x+ct)+f'(ct-x)$. Hence
$$
\boxed{E=c^2\int_0^\infty\bigl[f'(x+ct)^2+f'(ct-x)^2\bigr]dx
=c^2\int_{-\infty}^{\infty}f'(s)^2\,ds}.
$$
The two substitutions partition the real line at $ct$, making <conservation of energy> explicit.
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