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Consider the heat initial value problem on $\mathbb{R}$: \begin{align*}\tag{1} \begin{cases} \frac{\partial u}{\partial t}(t,x)&=\frac{\partial^2 u}{\partial x^2}(t,x) \qquad t>0, x\in\mathbb{R}, \\ u(0,x)&=f(x),\qquad x\in\mathbb{R} \end{cases} \end{align*} We know that $u$ is continuous on $[0,T]\times\mathbb{R}$. Then we proved that $u$ is a solution to (1) in classical sense by showing all its required derivatives exists and $u$ itself satisfies (1). In particular we use a continuity argument similar to Evans PDE, see for instance p.64-65 in enter link description here

Now, my question is, how can I show continuity for $\frac{\partial u}{\partial x}$, $\frac{\partial u^2}{\partial x^2}$, and $\frac{\partial u}{\partial t}$.

I think I just need to modify Evans' continuity argument for $u$, but I am not sure how.To be more specific, what is the general strategy to prove continuity of derivatives of solution? Given we know the solution itself is continuous.

Any help is appreciated.

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  • $\begingroup$ Use the explicitly known Green's function $G(x-y,t)$, $u(x,t)=\int dyG(x-y,t)u(y,0)$ to investigate these properties. $\endgroup$
    – Urgje
    Commented Nov 9, 2015 at 14:07
  • $\begingroup$ Could you elaborate a bit more? A bit too general I'm sorry but I'm not sure what u mean $\endgroup$
    – math101
    Commented Nov 9, 2015 at 14:21

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The Green's function is usually obtained by Fourier transforming. In the square integrable case the procedure is rather elegant. Suppose $u\in L^{2}(% \mathbb{R},dx)$. Then $\partial _{x}^{2}$ extends to a self-adjoint operator $-p^{2}$, $p^{2}\geqslant 0$. The solution is \begin{equation*} u(x,t)=\exp [-p^{2}t]u(x,0) \end{equation*} \begin{eqnarray*} &<&x|\exp [-p^{2}t]|y>=\int dp<x|p>\exp [-p^{2}t]<p|y> \\ &=&\frac{1}{2\pi }\int dp\exp [i(x-y)p]\exp [-p^{2}t] \end{eqnarray*} \begin{eqnarray*} \int dp\exp [iap]\exp [-p^{2}t] &=&\int dp\exp [-p^{2}t+iap]=\int dp\exp [-t(p^{2}-i\frac{a}{t}p)] \\ &=&\int dp\exp [-t\{(p-i\frac{a}{2t})^{2}+(\frac{a}{2t})^{2}\}] \\ &=&\exp [-\frac{a^{2}}{2t}]\int dp\exp [-tp^{2}]=\sqrt{\frac{\pi }{t}}\exp [- \frac{a^{2}}{2t}] \end{eqnarray*} Obviously the same Green's function is obtained in other situations.

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  • $\begingroup$ Thanks. I will come back to questions... $\endgroup$
    – math101
    Commented Nov 10, 2015 at 7:32
  • $\begingroup$ What you had there is correct. But my question is to show continuity for $\frac{\partial u}{\partial x}$, $\frac{\partial u^2}{\partial x^2}$, and $\frac{\partial u}{\partial t}$. $\endgroup$
    – math101
    Commented Nov 12, 2015 at 23:01
  • $\begingroup$ Since $u(x,t)=∫dyG(x−y,t)u(y,0)$ and we know $G$ and $u(y,0)$ these properties can be checked directly. In the $L{^2}$-case we have $u(t)$ is (strongly) continuous in $t$ as an $L{^2}$-function. The spatial derivatives of $u(t)$ need not exist but do if they exist for $u(0)$. $\endgroup$
    – Urgje
    Commented Nov 13, 2015 at 12:04

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