# continuous dependency estimate for viscosity solutions

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!

• 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. – Hans Engler Nov 25 '14 at 17:58
• 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. – dinosaur Nov 25 '14 at 18:09
• So there is your program of study. Get some facility with working with the various definitions of viscosity solutions first. – Hans Engler Nov 25 '14 at 18:13

## 1 Answer

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