0
$\begingroup$

Let $u^i$, $i=1,2$, be viscosity solutions of \begin{align*} u_t^i + H(Du^i,x) & = 0\quad\mathrm{in}\ \mathbb R^n\times (0,\infty)\\ u^i & = g^i\quad\mathrm{on}\ \mathbb R^n\times \{t=0\} \end{align*} where $H:\mathbb R^{2n}\to\mathbb R$ is a function such that \begin{align*} |H(p,x) - H(q,x)| & \leq c|p-q| \\ |H(p,x) - H(p,y)| & \leq c|x-y|(1+|p|) \end{align*}

for $x,y,p,q\in\mathbb R^n$ and $c\geq 0$ constant.

Then for $t\geq 0$ the estimate $$\|u^1(\cdot,t)-u^2(\cdot,t)\|_{L^\infty}\leq \|g^1-g^2\|_{L^\infty}$$ holds.

This is an exercise from Evans, Partial Differential Eq. Chapter 10. I really don't know where to start here. I have done a similar proof for the case $H=H(Du)$ convex using the Hopf-Lax formula and tried to take a similar approach, but I got nowhere with this.

Can anyone help me here? Thanks!

$\endgroup$
  • $\begingroup$ Try to redo the proof for the $x$-independent case using another approach (e.g. the definition of viscosity solutions, not Hopf-Lax). Then generalize to the case at hand. You can also look at the simpler $r$-independent case $u + H(Du,x) = 0$ for insight. $\endgroup$ – Hans Engler Nov 25 '14 at 17:58
  • $\begingroup$ Well my problem here is that I don't really know how to get to use the differential equation for this. I know the definition of a viscosity solution, but I don't know how to use it. I thought about picking a test function $v$ and estimating $\|u^1-u^2\|\leq \|u^1-v\| + \|u^2-v\|$, which yields the two functions $u^1-v$, $u^2-v$ we find in the definition of viscosity solutions, but I have no idea how to proceed from this. $\endgroup$ – dinosaur Nov 25 '14 at 18:09
  • $\begingroup$ So there is your program of study. Get some facility with working with the various definitions of viscosity solutions first. $\endgroup$ – Hans Engler Nov 25 '14 at 18:13
1
$\begingroup$

Actually this problem is proved on Evans and Lions' paper, so it is not trivial. enter link description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.