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let the following problem $$ \begin{cases} \dfrac{\partial u}{\partial t}= \dfrac{\partial^2 u}{\partial x^2}, 0<x<1, t>0\\ u(0,t)=0, u(1,t)+ \dfrac{\partial u}{\partial x}(1,t)=0\\ u(x,0)=f(x) \end{cases} $$ How we prove that this problem admits a unique solution? I try to set $v=u_1-u_2$ where $u_1$ and $u_2$ are two solutions of the problem, but i don't have idea for the following. Thank's for the help.

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  • $\begingroup$ You are missing an initial condition $u(x,0)=f(x)$. $\endgroup$ Commented Apr 28, 2016 at 11:15
  • $\begingroup$ Sorry, i edit my message. $\endgroup$
    – rosa
    Commented Apr 28, 2016 at 11:19

2 Answers 2

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Setting $v=u_1-u_2$, define $I(t)=\frac{1}{2}\int_0^1v(x,t)^2dx$. Then

$$\frac{dI}{dt}=\int_0^1v(x,t)v_t(x,t)dx=\int_0^1vv_{xx}dx$$

Integrating by parts, this becomes:

$$\frac{dI}{dt}=\left[vv_x\right]_0^1-\int_0^1v_x^2dx$$

Now, note:

  • $vv_x\vert_{x=0}=0,vv_x\vert_{x=1}=-v(1,t)^2$ by the conditions given, so:

$$\frac{dI}{dt}=-v(1,t)^2-\int_0^1v_x^2dx\le0$$

But clearly $I\ge0$, and we can verify that $I(0)=0$, so if $\frac{dI}{dt}<0$ , we would end up with a negative value for $I$. It must thus be the case that:

  • $\frac{dI}{dt}=0\implies v=0\implies$ solution is unique.
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always with the same problem, i want to calculate the solution with separate variable methode. That what i try: we put $$u(x,t)= X(x) T(t) \neq 0$$ then we have $$\dfrac{T'(t)}{T(t)}= \dfrac{X"(x)}{X(x)}$$ So we obtain twe equations $$ T'(t) + \lambda T(t)= 0 $$ and $$ X''(x)+ \lambda X(x)=0, X(0)=0, X(1)+ X'(1)=0 $$ this second equation admits eigenvalues $\lambda=\alpha^2$ where $\alpha$ is solution of equation $\tg x = - \dfrac{1}{x}$, and the eigenfunctions associated are $y_{\alpha}(x)= C \sin(\alpha x)$. The solutions of the first equations are $$T_{\alpha}(t)= e^{-\alpha^2 t}$$ Then we have $$u(x,t)= \sum_{\alpha} C_{\alpha} e^{- \alpha t} \sin(\alpha x)$$ My question is:my solution is correct? And what about $\alpha?$ he must to be in $\Z$. No? Thank's to the help.

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