Consider the heat equation on the interval $a \leq x \leq b$ with Newmman boundary conditions: $$u_{t}(x,t)-u_{xx}(x,t)=0 \quad \textit{ for } a \leq x \leq b, t>0$$ $$u(x,0)=g(x) \quad \textit{ for } a \leq x \leq b$$ $$u_{x}(a,t)=0, u_{x}(b,t)=0 \quad \textit{ for } t>0$$

Show that if $g(x)$ is an increasing, differentiable function, then $u(x,t)$ is lso an increasing function of $x$ for each fixed $t>0$

I have no idea how to do this one since there is no solution for heat equation with interval domain [a,b].

Any help would be apprecitated!

New contributor
Ben is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
  • 2
    $\begingroup$ $u_x$ satifies the heat equation, with Dirichlet boundary conditions and is nonnegative for $t=0$, hence It Is nonnegative, by the maximum principle. $\endgroup$ – Giorgio Metafune 2 days ago
  • $\begingroup$ I don't understand how $u_{x}$ satisfies the heat equation. How does $(u_{x})_{t} = (u_{x})_{xx}$? $\endgroup$ – Ben yesterday
  • $\begingroup$ For $t>0$ the solution is $C^\infty$, so just differentiate the equality $u_t-u_{xx}=0$ with respect to $x$. The only problem is the continuity of $u_x$ up to $t=0$. This is true if $g$ is regular enough (expand the solution in Fourier series). If $g$ is only continuously differentiable, you can solve the analogous of your problem with $g_x$ instead of $g$ and Dirichlet b.c. If $v$ is the solution, you get $u$ integrating $v$ with respect to $x$. $\endgroup$ – Giorgio Metafune 12 hours ago

Your Answer

Ben is a new contributor. Be nice, and check out our Code of Conduct.

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.