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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!

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    $\begingroup$ 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. $\endgroup$ – jawheele Apr 9 at 19:58
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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 $w(x, t) = 0$.

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