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.
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.