1
$\begingroup$

Given $\omega$ is a bounded domain in $R^n$. Assume that $N(x,y,t)$ is a function such that $u(x,t) = \int_{\omega} N(x,y,t) u_{0}(y)\ dy$ is a solution to the heat equation: $u_t - \triangle u = 0$ in $\omega$, $\frac{\partial{u}}{\partial v} = 0$ on $\partial{\omega}$, and $u = u_0$ at $t=0$.

Show that $u(x,t) = \int_{0}^{t} \int_{\partial {\omega}} N(x,y,t-s)g(y,s)\ dy\ ds$ is a solution to the PDE: $u_t - \triangle u = 0$ for $x\in \omega$ and $t > 0$, $\frac{\partial{u}}{\partial v} = g$ for $x\in \partial{\omega}$ and $u=0$ at $t=0$.

My attempt: The $3$rd condition $u = 0$ at $t=0$ is trivial to verify. For the $1$st condition, I only get: $\frac{\partial u}{\partial t} = u_{t} = \int_{0}^{t} \int_{\partial {\omega}} \frac{\partial N(x,y,t-s)}{\partial t}g(y,s)\ dy\ ds$. Since the inner integral occurs on $\partial \omega$, to replace $\frac{\partial N(x,y,t-s)}{\partial t}\ $ by $\ \frac{\partial^2 N(x,y,t-s)}{\partial t^2}$, I applied the Green's 2nd identity and obtained:

$\int_{0}^{t} \int_{\partial {\omega}} \frac{\partial N(x,y,t-s)}{\partial t}g(y,s)\ dy\ ds = \int_{0}^{t}N_{xx}(x,y,t-s)\triangle u\ dV$ (since $g(y,s) = \frac{\partial u}{\partial v}$). But I can't see how $\int_{0}^{t} \triangle N(x,y,t-s)\triangle u\ dV = u_{xx}$.

Can anyone please help with this part, and the part of verifying $u(x,t)$ satisfies the Neumann condition? Any help would really be appreciated.

$\endgroup$
  • $\begingroup$ Can anyone please give this problem a try:( $\endgroup$ – user177196 Nov 2 '15 at 2:03

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.