We search to prove that the following problem admits a unique solution. $$ \begin{cases} \dfrac{\partial^2 u}{\partial t^2}= a^2 \dfrac{\partial^2 u}{\partial x^2}\\ u(0,t)=u(l,t)=0\\ u(x,0)=f(x)\\ \dfrac{\partial u}{\partial t}(x,0)=g(x) \end{cases} $$ For this, we suppose that there exist two solutions $u_1$ and $u_2$, et put $v(x,t)=u_1(x,t)-u_2(x,t)$, and we prouve that $v(x,t)=0$ for all $(x,t)$. $v$ satisfies the equation: $$\dfrac{\partial^2 v}{\partial t^2} - a^2 \dfrac{\partial^2 v}{\partial x^2}=0$$ By multipliynig by $\dfrac{\partial v}{\partial t}$, and integring on x, we found $$ \displaystyle\int_0^l \dfrac{\partial^2 v}{\partial t^2} \dfrac{\partial v}{\partial t} dx - a^2 \displaystyle\int_0^l \dfrac{\partial^2 v}{\partial x^2} \dfrac{\partial v}{\partial t} dx=0 $$ By integration the term $\displaystyle\int_0^l \dfrac{\partial^2 v}{\partial x^2} \dfrac{\partial v}{\partial f} dx$, on trouve $$ [\dfrac{\partial v}{\partial t}(l,t) \dfrac{\partial v}{\partial x}(l,t)]_0^l - \displaystyle\int_0^l \dfrac{\partial v}{\partial x} \dfrac{\partial}{\partial t}(\dfrac{\partial v}{\partial x}) dx $$
What we can do after this? Thank's for the help.
It's not clear why you decided to perform these manipulations, but it seems you are considering the energy method. The energy for this equation is $$E(t)=\int_0^l (v_t^2+a^2v_x^2)\,dx$$ which is basically the sum of kinetic and potential energy. Differentiate with respect to time, apply the PDE, and integrate one of the terms by parts: you will get zero. This shows the energy is constant. Since it was zero initially, it stays equal to zero. This implies the function $v$ is constant, as desired.