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Assume $u^1, u^2$ are two weak solutions of the initial value problems. \begin{align*} u_t^i+H(Du^i)&=0 \ \text{in} \ \ \mathbb{R}^n\times (0,\infty)\\ u^i&=g^i \ \text{on} \ \ \mathbb{R}^n\times\{t=0\} \ (i=1,2)\end{align*} for $H$ as in section 3.3 (that is $H$ is convex and $\lim_{|p|\rightarrow\infty}H(p)/|p|=+\infty$. Prove the $L^\infty$ contraction inequality: $$\sup_{\mathbb{R}}|u^1(\cdot,t)-u^2(\cdot ,t)|\leq \sup_{\mathbb{R}}|g^1-g^2| \ \ (t>0)$$

This is an exercise in Evans. And this is what we have so far

By Hopf lax, we there exists $y,y'\in \mathbb{R}^n$, we have $$u_1(x,t)=tL(\frac{x-y}{t})+g^1(y)$$ and $$u_2(x,t)=tL(\frac{x-y'}{t})+g^2(y')$$ So $$|u^1(x-y',t)-u^2(x-y,t)|=|g^1(y)-g^2(y')|\leq \sup_{\mathbb{R}}|g^1-g^2|$$ From here, is it safe to conclude that $|u^1(z,t)-u^2(w,t)|\leq \sup_{\mathbb{R}}|g^1-g^2|$ for all $z,w\in\mathbb{R}^n$? The left hand side still depends on $y,y'$ so I think the answer is negative. Is there any other ways to solve this problem? Hints? Thank you very much

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  • $\begingroup$ Using your notation, you can actually show that $u^1(x,t)-u^2(x,t)\ge g^1(y)-g^2(y)$, and $u^1(x,t)-u^2(x,t)\le g^1(y')-g^2(y')$. This is because Hopf Lax gives us $u^2(x,t)\le tL(x-y/t) + g^2(y)$, and similarly with $u^1$. $\endgroup$ – Joey Zou Nov 29 '15 at 4:01
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    $\begingroup$ Right, good point, I left out the easiest inequality I could use out of Hopf Lax. Thanks for the hint, I truly appreciated $\endgroup$ – nerd Nov 29 '15 at 5:32

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