# Viscosity Solution of Hamilton Jacobi Equation, result from Evans

I have a question about a proof in "Partial Differential Equation by Lawrence C. Evans". We look at the problem

$$(1)\mbox{ }u_t+H(Du,x) = 0 \mbox{ in }\mathbb{R}^n\times (0,T] \mbox{ }$$ and $u=g$ on $\mathbb{R}^n\times \{t=0\}$. On page 546, there is a Lemma called "Extrema at a terminal time", i.e.

Assume $u$ is a viscosity solution of $(1)$ and $u-v$ has a local max at a point $(x_0,t_0)\in \mathbb{R}^n\times (0,T]$. Then $$v_t(x_0,t_0)+H(Dv(x_0,t_0),x_0)\le 0 (\ge 0)$$

So the point is, allowing $t_0=T$.

In the proof, we assume $u-v$ has a local max at $(x_0,T)$. W.l.o.g this is a strict max. Now he defines a new function $$\tilde{v}(x,t):=v(x,t)+\frac{\epsilon}{T-t}$$ for $x\in\mathbb{R}^n$ and $0<t<T$. Now he says: "Then for $\epsilon>0$ small enough, $u-\tilde{v}$ has a local max at a point $(x_\epsilon,t_\epsilon)$, where $0<t_\epsilon <T$ and $(x_\epsilon,t_\epsilon)\to(x_0,T)$."

Two questions:

1. Why does the point $(x_\epsilon,t_\epsilon)$ exists and is convergent to $(x_0,T)$?
2. If we go one step further and define a new equation

$$(2)\mbox{ }u+u_t+H(Du,x) = 0 \mbox{ in }\mathbb{R}^n\times (0,T] \mbox{ }$$

Let the boundary conditions be nice, I think it is not important here. Is there a similar Lemma to the one above and if so, how do we have to choose $\tilde{v}$? The argument, I guess, is then the same.

Thanks

math

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Re: 1. Consider the rectangular box $U=\{(x,t): |x-x_0|\le \delta, T-\delta\le t\le T\}$. The function $u-\tilde v$ is continuous on $U$ in the extended sense (it turns to $-\infty$ when $t=T$), so there is a point $(\tilde x,\tilde t)$ at which it attains $\max_U (u-\tilde v)$. We want to show that $(\tilde x,\tilde t)$ does not belong to $\partial U$, which is the surface of a cylinder. The top of the cylinder $t=T$ does not worry us, but we have to deal with the side and the bottom.
(Side.) choose $\delta$ small so that $(u-v)(x,t)<(u-v)(x_0,t)$ whenever $|x-x_0|= \delta$ and $T-\delta\le t\le T$. This is possible because $(u-v)(x,T)<(u-v)(x_0,T)$ and everything is continuous. Note that we now have $(u-\tilde v)(x,t)<(u-\tilde v)(x_0,t)$ when $|x-x_0|= \delta$ and $T-\delta\le t\le T$, no matter what $\epsilon$ is.
(Bottom.) By continuity, there is $T'<T$ such that $(u-v)(x_0,T')>(u-v)(x,T-\delta)$ whenever $|x-x_0|\le \delta$. Choose $\epsilon$ small so that $(u-v)(x_0,T')-\epsilon/(T-T')>(u-v)(x,T-\delta)-\epsilon/\delta$ whenever $|x-x_0|\le \delta$. The latter means precisely $(u-\tilde v)(x_0,T')>(u-\tilde v)(x,T-\delta)$.
Re: 2. If the equation involves the value of $u$, we must adjust the definition of viscosity solution by requiring the local max/min of $u-v$ to be zero, that is, $u(x_0,t_0)=v(x_0,t_0)$ at the point of max/min. In the above proof, we will now have $u(x_0,T)=v(x_0,T)$. After the perturbation by $\epsilon/(T-t)$ the maximum of $u-\tilde v$ is attained at a nearby point $(x_\epsilon,t_\epsilon)$ but is not necessarily zero. So you define $\widehat{v}=\tilde v+(u-\tilde v)(x_\epsilon,t_\epsilon)$. The only additional point to argue is that the added term tends to $0$ as $\epsilon\to 0$. This should not be hard: for one thing, $(u-\tilde v)(x_\epsilon,t_\epsilon)\le (u-v)(x_\epsilon,t_\epsilon)=0$, for another the maximum of $u-\tilde v$ on $U$ cannot be very negative when $\epsilon>0$ is small.
Thanks for your help. There a lot of things which I do not understand. 1. Why is $u-\tilde{v}$ continuous. For sure, $\tilde{v}$ is continuous in your extended sense, but I think $u$ has not to be. 2. On the side, why is $(u-v)(x,T)<(u-v)(x_0,T)$? –  math Jul 15 '12 at 7:56