Uniqueness of weak solutions of the Hamilton Jacobi Equation

Question

I'm doing a seminar this semester concerning PDE's and their properties. I'm supposed to hold a presentation about the uniqueness of weak solutions of the Hamilton Jacobi Equation. It's a Theorem from Evans' PDE. The Book defines a weak solution to be a function with the following properties:
Weak solutions in Evans

But there are a few things I don't understand in the proof showing uniqueness. The first being, that Evans defines a mollification over both the space and time variables $(x,t)$. How is this even defined? My guess would have been to take the standard mollifier $\eta_{\varepsilon}:\mathbb{R}^{n+1}\longrightarrow \mathbb{R}$ for an $\varepsilon>0$, and interpret the variable $y \in \mathbb{R}^{n+1}$ as $y=(x,t) \in \mathbb{R}^n\times \mathbb{R}$. Then the mollification could be written as (with $\mathcal{L}^n$ being the Lebesgue measure in n dimensions):
\begin{equation}
u_{\varepsilon}(x,t)=(u*\eta_{\varepsilon})(x,t)=\int_{0}^{t}\int_{\mathbb{R}^n}u(y,\tau)\eta_{\varepsilon}(x-y,t-\tau)\ d\mathcal{L}^{n}(y)\ d\mathcal{L}^1(\tau)
\end{equation}
Is this the right definition? Because I am not sure when it comes to integrating over the time domain.

My second struggle is the following inequality, where Evans says that:
\begin{equation}
|\nabla u_{\varepsilon}|\leq \text{Lip}(u)
\end{equation}
Where $\nabla u_{\varepsilon}$ is the gradient and $\text{Lip}(u)$ is the Lipschitz constant of $u$. Now I kind of get why this has to hold, as since $u$ is Lipschitz the partial derivatives $\partial_i u$ have to be bounded by the Lipschitz constant, since
\begin{equation}
|\partial_i u| = \lim_{h \to 0}\frac{|u(x+he_i,t)-u(x,t)|}{|h|}\leq \lim_{h\to 0}\text{Lip}(u)\frac{|he_i|}{|h|}=\text{Lip}(u)
\end{equation}
Thus one would get for the partial derivatives of $u_{\varepsilon}$ the following bound:
\begin{equation}
|\partial_i u_{\varepsilon}|=|(\partial_i u) *\eta_{\varepsilon}|\leq \text{Lip}(u)\cdot\|\eta_{\varepsilon}\|_{L^1}=\text{Lip}(u)
\end{equation}
But then taking the euclidean norm of the gradient would yield
\begin{equation}
|\nabla u_{\varepsilon}|\leq \sqrt{n}\cdot\text{Lip}(u)
\end{equation}
Why is there no factor $\sqrt{n}$ in front of $\text{Lip}(u)$?

The last thing I'm having difficulties with, is that Evans states, that the inequality c) in the definition of weak solution implies that
\begin{equation}
D^2u_{\varepsilon}\leq C\Big(1+\frac{1}{s}\Big)I
\end{equation}
for an appropriate constant $C$, and all $\varepsilon>0, y \in \mathbb{R}^n, s>2\varepsilon$. Now I kind of get, that this just means, that for all $y\in \mathbb{R}^n$ the following holds:
\begin{equation}
y^T(D^2u_{\varepsilon})y \leq C\Big(1+\frac{1}{s}\Big)|y|^2
\end{equation}
and if you divide by $|y|^2$ you have the second directional derivative in the direction of $y$ on the left side of the equation. Then one can just use the definition of the second symmetric derivative which is exactly condition c). Now what I don't understand is where the condition $s>2\varepsilon$ comes from or where exactly it is used. Does it have something to do with the mollification?

Answer 1

1. The right mollification (at least the one consistent with Evans' notation and which works for the proof) should be $$ u^\varepsilon(x,t)=\int_0^\infty\int_{\mathbb{R}^n}\eta_\varepsilon (x-\xi,t-\tau)u(\xi,\tau)d\xi d\tau=\int_0^\infty\int_{\mathbb{R}^n}\eta_\varepsilon (\xi,\tau)u(x-\xi,t-\tau)d\xi d\tau $$
2. Fix a vector $e\in\mathbb{R}^n$ with $|e|=1$. We estimate \begin{align*} |u^\varepsilon(x+he,t)-u^\varepsilon(x,t)|&\leq\int_0^\infty\int_{\mathbb{R}^n}\eta_\varepsilon (\xi,\tau)|u(x+he-\xi,t-\tau)-u(x-\xi,t-\tau)|d\xi d\tau\\ &\leq \int_0^\infty\int_{\mathbb{R}^n}\eta_\varepsilon (\xi,\tau)\operatorname{Lip}(u)hd\xi d\tau=\operatorname{Lip}(u)h \end{align*} so that $$ \frac{u^\varepsilon(x+he,t)-u^\varepsilon(x,t)}{h}\leq \operatorname{Lip}(u). $$ Define the one-variable function $f(h)=u^\varepsilon(x+he,t)$. By the previous line $f'(0)\leq \operatorname{Lip}(u)$, but now also take $e=Du^\varepsilon(x,t)/|Du^\varepsilon(x,t)|$ to get \begin{align} f'(0)=Du^\varepsilon(x,t)\cdot e=|Du^\varepsilon(x,t)|. \end{align}
3. The mollification $u^\varepsilon$ is only defined for $(y,s)$ at least $\varepsilon$ away from the boundary. Although I don't see why you would need to require $s>2\varepsilon$ instead of just $s>\varepsilon$.