I have met a problem while doing the exercises in one PDE book, and the chapter is about prior estimates.
The problem is as follows: Suppose $ u \in C^{2,1}(Q_T)\cap C^{1,0}(\bar{Q_T})$ is the solution to the equations: $$\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2} +c(x,t)u=f(x,t),\quad -1<x<1,0<t<T$$ $$u(-1,t)=u(1,t)=0,\quad 0<t<T$$ $$u(x,0)=0,\quad -1<x<1$$ Where $T>0$, $Q_T=(-1,1)\times (0,T)$, and $c$ is nonnegative and bounded in $Q_T$.
(i) prove that $$ \sup_{Q_T} |u| \leq e^2 \sup_{Q_T}|f|;$$ (ii) prove that $$ \left| \frac{\partial u}{\partial x}(\pm 1,t)\right| \leq e^2 \sup_{Q_T}|f|,\quad 0<t<T; $$
(iii)Judge if (i) and (ii) still hold for the problem: $$\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}+\left(\frac{\partial u}{\partial x}\right)^2u+u^3=f(x,t),\quad -1<x<1,\ 0<t<T,$$ $$ u(-1,t)=u(1,t)=0,\quad 0<t<T, $$ $$ u(x,0)=0,\quad -1<x<1$$
What I have finished: I have solved problem (i) and (ii) by modulus estimate, specifically, with the notation $$\mathscr{L}u=\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}+c(x,t)u, $$ which is a linear operator, we could exert weak maximum or minimum principle on it and get an estimate of those modulus. And the estimate of gradients can be dealt with like that too.
But now I can not figure out where to begin with when doing (iii), since
(1) It is not a linear operator
(2) I don’t know any generalized theorems to deal with it
(3) I have tried to copy the paths of proofs of some theorems for linear operators but none of them worked
(4) I have tried to construct some counter examples but none of them worked
My guesses:
(1) Maybe there are some transformations which can turn it back to the previous problems?
(2) Since it is just an exercise appeared in my textbook, I think it is unlikely to be solved by nonlinear tools. So maybe there is one counter example but I didn’t find it.
Can you help me with this problem? I’ll appreciate it.
I think I have solved the problem. I have thought about this way but somehow gave up and changed my thinking... It is quite obvious but I fooled myself.
The problem i met was the broken properties of nonlinear operators, but if we examine closer at the problem, we can see that it is still in special forms.
We can fix the nonlinear operator by the following procedures:
First we suppose the solution to equations $(iii)$ is $w(x,t)$, and then we have the equation:
$$\frac{\partial w}{\partial t}-\frac{\partial^2 w}{\partial x^2}+\left(\left(\frac{\partial w}{\partial x}\right)^2+w^2\right)w=0$$
If we take the coefficient function $c(x,t)$ to be
$$c(x,t)=\left(\frac{\partial w}{\partial x}\right)^2+w^2$$
The original equation becomes
$$\frac{\partial w}{\partial t}-\frac{\partial^2 w}{\partial x^2}+c(x,t)w=0$$
And then we can just consider about the operator
$$\mathscr{L}u=\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}+c(x,t)u$$
where we can directly use the conclusions we have, and we have an extra condition:
$$\mathscr{L}w=f(x,t)$$
where $w$ is the solution to $(iii)$ as above.