Proving uniqueness of solution to 1-D Wave equation without energy conservation

We have a homogeneous string of length L fastened at its ends, performing small transverse motion in a vertical plane. The tension in the string is assumed sufficiently large for gravitational forces to be neglected . The motion is then governed by:

• $$c^{2}\frac {\partial^{2} u}{\partial x^{2}} = \frac {\partial^{2} u} {\partial t^{2}}$$

• $$u(0,t)= u(L,t)=0$$, for all $$t\geq 0$$ and $$u(x,0)= f(x) ; \frac {\partial u}{\partial x} (x,0)=g(x)$$

Now, the problem of concern:

Suppose we have found solutions $$u(x,t)$$ and $$v(x,t)$$ to the IBVP. Show uniqueness for the solution, i.e. $$u(x,t)=v(x,t)$$ for all $$x$$ and $$t$$, using the integral:

$$I(t) :=\int_0^L (w(x,t))^2 dx$$,

where $$w(x,t)= u(x,t)-v(x,t)$$.

Further, any insight into where this integral comes from would also be of interest to me.

My argument:

Note that by linearity, $$w(x,t)$$ satisfies the wave equation and all boundary-/initial conditions are homogeneous.

Differentiating twice under the integral (by Leibniz' rule), integrating by parts and applying the wave equation, gives $$I''(t)=0$$. The Identity theorem applies and we have $$I'(t)=const.$$ . But $$I'(0)=0$$, by the initial conditions for $$w(x,t)$$, so $$I'(t)=0$$. The identity theorem applies and we have $$I(t)=const.$$. Using the boundary conditions for $$w(x,t)$$ we obtain $$I(0)=0$$ and hence $$I(t)=0$$. Since the integrand is strictly non-negative ( and assumed to be continuous) we must have $$w(x,t)=0$$. Uniqueness follows immediately and my argument is complete.

I would greatly appreciate, if someone would check that my argument works. Thanking in advance!

• Could you give a bit more detail regarding your conclusion that $I''(t)=0$? I don't agree that it follows immediately from substitution via the wave equation. – jawheele Apr 9 at 19:58

Your proof is flawed. Let $$I(t) = \int_0^L (w(x,t))^2 dx$$, thus $$\frac{dI}{dt} = \int_0^L2w(x, t) w_t(x, t) dx$$. Differentiating again yields $$\frac{d^2I}{dt^2} = \int_0^L 2w(x, t) w_{tt}(x, t) + 2w_{t}^2(x, t) dx$$
Using the wave equation, $$c^2 w_{xx} = w_{tt}$$ will not result in $$I''(t) = 0$$.
Just for completeness and further reference, you can use the energy function $$E(t) = \int_0^L c^2w_x^2 + w_t^2 dx$$ to prove the uniqueness theorem. By differentiating, you can easily see that $$\frac{dE}{dt} = \int_0^L w_t ( w_{tt} - c^2w_{xx}) dx = 0$$ thus $$E = \mathrm{const} = 0$$. It follows that $$w(x, t) = 0$$.