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I am looking at the IBVP: $$ u_t+x^2u_{xx}+u_y+f(x,y)=0, x,y \in D\in [0,\infty)\times[0,\infty),t\in(0,T]\ $$$$ u(x,y,0)=F(x,y)\in C^{1,1}\ $$$$ u(x,y,t)=0, x,y\in \partial D $$ and I would like to know the regularity of the solution of this PDE. According to Friedman's book the best I can hope is $C^{1,1,1}$. Any suggestions or references of higher derivatives result? Thanks!

flag	asked Jun 18 at 14:53 Redson 1

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In the $x$-direction you can hope for smoothing away from the origin (if $f$ is smooth). In the $y$-direction it is just a transport equation so you cannot do better than the initial data. – timur Jun 18 at 15:06

thanks. But what about time? I need to do Taylor for $u(t+\tau,x,y)$ and I need to have at least $C^3$ to prove the numerical method for that equation is third order locally... – Redson Jun 18 at 17:44

That means you have to assume your initial data $F$ is more than $C^{1,1}$ at least in the $y$-direction. However these issues are best treated separately: First assume whatever regularity you need and analyze your numerical method, then study when your equation gives solutions meeting your criteria. If the solution is less smooth than required it does not mean your numerical method will fail, in most cases it would mean a reduced order of convergence. – timur Jun 19 at 0:06

thanks, got it. That's what I did, I analysed assuming all the derivatives finite, and there is a third order error in time, so I need to have $u_{ttt}$ finite. I understand how initial value can affect the regularity in space but how initial data affects regularity in time? Even for the plain $u_t=u_xx$ the local error contains $u_{tt}$, and we know that strong solution is only $C^{1,2}$. How to prove I can possibly have more regularity in time? – Redson Jun 20 at 19:11

May be it is $u_t-x^2u_{xx}+\ldots$? As it is written corresponding terms constitutes a backward parabolic equation. – Andrew Jun 29 at 14:47

Smoothness of a solution for degenerate pde

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