# Unicity of solution of pde

Let the pde $$\dfrac{\partial^2 u}{\partial t^2} - \dfrac{\partial^2 u}{\partial x^2}=f(x)$$ The question is:

Find the limit condition such that this pde admit a unique solution in $[a,b] \times [0,T].$

For this, I suppose the existence of to sulutions $u_1$ and $u_2$ and we put $v = u_1 - u_2.$ Then, $$\dfrac{\partial^2 v}{\partial t^2} - \dfrac{\partial^2 v}{\partial x^2} = 0$$ myltiply this last equation by $\dfrac{\partial v}{\partial t}$ and we integrate over $[a,b]$ and we obtained $$\displaystyle\int_a^b \dfrac{\partial^2 v}{\partial t^2} . \dfrac{\partial v}{\partial t} dx - \int_a^b\dfrac{\partial^2 v} {\partial t^2}. \dfrac{\partial^2 v}{\partial x^2} dx = 0$$

I tried integration by parts, but didn't find the limit conditions to obtain the unicity of this pde. Thanks for any help.

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For the wave equation there is a conserved energy $$E(t)=\int_{-\infty}^\infty (u_t)^2+(u_x)^2 dy.$$ This is obtained multiplying by $u_t$ and integrating by parts (as you did). For $v$ this quantity is zero initially, thus it vanishes for all times. Right?
You have $$\int u_{tt}u_t-\int u_tu_{xx}=\frac{1}{2}\int (u_{t})_t^2+\int u_{xt}u_{x}$$ $$=\frac{1}{2}\int (u_{t})_t^2+\frac{1}{2}\int (u_{x})^2_t=\frac{d}{dt}E(t)=0.$$ – guacho May 27 '13 at 13:02
As $E(0)=0$ you obtain that $v$ is a constant (and thus it vanishes) – guacho May 27 '13 at 13:08
What do you mean by "limit condition"? I think that you refer to the conditions under the functional makes sense, right? In that case you need $H^1$, isn't it? – guacho May 27 '13 at 16:47