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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!

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    $\begingroup$ $u_x$ 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

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