0
$\begingroup$

Let $\Omega\in\mathbb{R}^{n}$ be a bounded connected open set. I have the following partial differential Equation; \begin{align} \nabla\cdot\left(-D(x)\nabla \psi\right)&=F\quad \text{in}\quad \Omega\times(0,\,T)\\ \psi(x,0)&=0\quad \text{in}\quad \Omega\\ \psi(0,t)&=0\quad \text{in}\quad \partial\Omega\\ \nabla\psi(\ell,t)&=0\quad \text{on}\quad \partial \Omega\times(0,\,T) \end{align} I would like to show the uniqueness of the solutions. I assume existence of two different solutions $\psi_{1}$ and $\psi_{2}$ and defined the difference of the solutions $\omega=\psi_{2}-\psi_{1}$ clearly $\omega$ satisfies the equation \begin{align} \nabla\cdot\left(-D(x)\nabla \omega\right)&=0\quad \text{in}\quad \Omega\times(0,\,T)\\ \omega(x,0)&=0\quad \text{in}\quad \Omega\\ \omega(0,t)&=0\quad \text{in}\quad \partial\Omega\\ \nabla\omega(\ell,t)&=0\quad \text{on}\quad \partial \Omega\times(0,\,T) \end{align} I considered the following integral \begin{align} J=\int\limits_{\Omega}\omega\nabla\cdot\left(-D(x)\nabla \omega\right) dV=0 \end{align} Assuming that $D(x)>0$, I want to show that $J>0$ so that I conclude that if $J>0$ then all the terms inside the inside the integration are zero. Thus $\psi_{1}=\psi_{2}$. Am looking for nay identities to help on this. Rhanks a lot

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.