Existence of a maximum $\max_{\mathbb{R}^{2n}\times(0,\infty)^2} \Phi(x,y,t,s)$ (related to viscosity solutions in Evans's PDE)

Question

In Evans's PDE (on page 588), in the proof of a uniqueness result for viscosity solutions of the Hamilton-Jacobi equation, it is claimed that there exists the maximum

$$\max_{\mathbb{R}^{2n}\times (0,\infty)^2} \Phi(x,y,t,s) = \max_{\mathbb{R}^{2n}\times (0,\infty)^2} u(x,t) - v(y,s) - \lambda (t+s) - \frac{1}{\epsilon^2}(|x-y|^2+(t-s)^2) -\epsilon(|x|^2+|y|^2)$$ where $\epsilon>0$, $\lambda < 1$ and $u$ and $v$ are bounded and uniformly continuous (and viscosity solutions of an Hamilton-Jacobi equation).

Can you explain to me in detail the following points?

1. Why does that maximum exist?
2. What's the motivation of the penalization terms?

Answer 1

Let $M$ be a bound for $|u|$ and $|v|$. For $R>0$, if $t, s \geq R$ and $|x|^2+|y|^2 \geq R$, you have that $\Phi(t,s,x,y) \leq 2M - (\lambda + \epsilon)R$, and the r.h.s. goes to $-\infty$ as $R\to +\infty$. Hence there exists $R>0$ such that $$ \Phi(t,s,x,y) \leq \Phi(0,0,0,0)\qquad \forall t,s\geq R,\ \forall (x,y):\ |x|^2 + |y|^2\geq R. $$ In particular, the continuous function $\Phi$ reaches its maximum on the compact set $\{(t,s,x,y):\ t,s\in [0, R],\ |x|^2+|y|^2 \leq R\}$.

This explains the presence of the penalization terms $-\lambda(t+s)$ and $-\epsilon(|x|^2 + |y|^2)$.

The other penalization term $$ -\frac{1}{\epsilon^2} (|x-y|^2 + |t-s|^2) $$ is a standard term in all the proofs based on the Kruzkov's doubling of variables technique. Heuristically, you see that as $\epsilon\to 0$ you need to have $|x_\epsilon - y_\epsilon|$ and $|t_\epsilon - s_\epsilon|$ very small at maximum points $(t_\epsilon, s_\epsilon, x_\epsilon, y_\epsilon)$ of $\Phi$. If you go through Evans' proof, this fact is seen for instance in formula (11).

Answer 2

I'll assume from the notation that $s$ is a fixed value, if it's not, this is not bounded (as a function of $s$).

1) $\Phi(x,y,t,s)$ is bounded from above.

We have that both $u(x,t)$ and $v(y,s)$ are bounded from above.

The term $-\lambda(t+s) - \frac{1}{\epsilon} (t-s)^2$, as a function of $t$, is also bounded from above, as it is a quadratic function where the leading term is associated to a negative value.

The other terms are negative, therefore bounded by $0$ from above.

2) Attains it maximum.

This follows from a general statement. If you have any function $f(x)$ bounded from above that is continuous, then $H(x)=f(x) - c |x|^2$ attains its maximum for any $c>0$.

To see this, you consider the set $A_d=\{ x \in \mathbb{R}^2: H(x)\geq d \}$ for $d < \sup H(x)$. Since $f(x)$ is bounded from above, this set is bounded. Since $H(x)$ is continuous, this set is closed. Since we are in $\mathbb{R}^d$, this set is compact. Therefore, given that $H(x)$ is continuous, it attains its maximum on any compact set.

Now consider $f(x,y,t) = u(x,t)-v(y,s)-\lambda t - \frac{1}{\epsilon} (|x-y|^2 - (t-s)^2) + \epsilon t^2 $, this will be bounded from above as long as $\epsilon <1$ (if this is not the case, you can find a similar way to formulate this, the idea is the same).

by our proposition, we know that $$ f(x,y,t) - \epsilon( |(x,y,t)|^2) = f(x,y,t) - \epsilon( |x|^2 + |y|^2 + t^2) = \Phi (x,y,t,s)$$ attains its maximum, that is what we wanted.

About the reasoning behind the penalization, I don't know, I'm not familiar with what is the author trying to achieve.