# Use energy method prove wave equation has a unique of solution.

Problem: Use energy method prove wave equation has a unique of solution. \begin{align}U_{tt}+U_t-C^2U_{xx}=0 \\ U(a,t)=U(b,t)=0 \\ U(x,0)=f(x) \\ u_t(x,0)=g(x) \end{align} \quad \text{with} \quad a\le x\le b ,t\ge 0.

I try assume problem has two solution $U_1\neq U_2$ and set $W=U_1-U_2$, define $\displaystyle E(t)=\dfrac{1}{2}\int\limits_{a}^{b} \left((W_t)^2-c^2(W_x)^2\right)dx$.

I can't prove $(E't)=0$ .

## 1 Answer

I think you need a $+$ in your energy rather than a minus, to account for an integration by parts.

As you said, suppose $U_1$ and $U_2$ are two solutions and take the difference $w = U_1 - U_2$. Then $w$ solves a very similar problem $$w_{tt} + w_t - c^2w_{xx} = 0$$ $$w(a,t) = w(b,t)=0$$ $$w(x,0) = 0$$ $$w_t(x,0) = 0$$

Notice that as a consequence of the boundary conditions $w_t(a,t) = 0$ and $w_t(b,t) = 0$, since $w(a, \cdot)$ (resp. $b$) is always $0$. Notice also for my suggested energy $E(t) = \int_a^b (w_t)^2 + c^2(w_x)^2 \, dx$, if $E(t) = 0$ for all $t$ then $w_t =0$ for all $t$, thus $w(x, \cdot)$ is constant and $w(x,0)=0$, therefore $w = 0$.

Differentiate $E$, $$E'(t) = \frac{1}{2}\int_a^b \left( 2 w_t w_{tt} + 2 c^2 w_x w_{x,t} \right) \, dx$$ We would like to use the equation $w$ satisfies to make this $0$ or $\leq 0$. The obstacles are that the $x$-derivative terms are not quite right, and the absence of an "independent" $w_t$ term.

Integrate $\int w_x w_{x,t} \, dx$ by parts, using the fact that $w_t(a,t)=w_t(b,t)=0$ to dispose of the boundary terms leaving only $-\int w_{xx} w_t \, dx$. This places us at

$$E'(t) = \int_a^b w_{tt} w_t - c^2 w_{xx} w_t \, dx$$

Add and subtract $(w_t)^2$ inside the integral,

$$E'(t) = \int_a^b (w_{tt} w_t + w_t w_t - c^2 w_{xx} w_t) - (w_t)^2 \, dx$$ Using $w_tt + w_t - c^2 w_{xx} = 0$, this ends up at $E'(t) = \int_a^b - (w_t)^2 \, dx$, which is a negative quantity. Therefore $E$ is decreasing. Observe that $E(0) = 0$ and $E(t) \geq 0$ to conclude that $E(t)$ is constantly $0$.