Krdytkyu

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

$$(1)mbox{ }u_t+H(Du,x) = 0 mbox{ in }mathbb{R}^ntimes (0,T] mbox{ }$$
and $u=g$ on $mathbb{R}^ntimes {t=0}$.
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 $(x_0,t_0)in mathbb{R}^ntimes (0,T]$. 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 $xinmathbb{R}^n$ 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

   
   
   

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

   
   
   

   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

Re: 1. Consider the rectangular box $U={(x,t): |x-x_0|le delta, T-deltale tle T}$. 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 $T-deltale tle T$. 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 $T-deltale tle T$, 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 $epsilonto 0$. 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.