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{ \large \textbf{\textsc{Math 5588 -- Homework 8 (Due Thursday March 23)}}}
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\begin{enumerate}
\item Use an argument based on the maximum principle to show that if $\max f \geq 0$,
\[u + |\nabla u|^2 - \Delta u = f(x) \ \ \ \text{for } |x|\leq 1\]
and $u(x)=0$ for $|x|=1$ then
\[u(x) \leq \max f \ \ \ \text{ for all } |x| \leq 1.\]
What happens if $\max f < 0$?
\item Use the maximum principle to show that if
\[-\Delta u(x) =f(x) \ \ \ \text{for } |x| < 1\]
and $u(x)=0$ for $|x|=1$  then 
\[u(x) \leq \frac{\max f}{2n}(1-|x|^2) \ \ \ \text{ for all } |x| \leq 1.\]
[Hint: Note that $-\Delta v = 2Cn$ for $v(x) =C(1-|x|^2)$. Choose $C$ so that $-\Delta v \geq -\Delta u$ and use the maximum (or comparison) principle.]
\item Assume $u:\R\to \R$ is a continuous \emph{increasing} function, but not necessarily differentiable. Show that $u$ is a viscosity supersolution of $u'(x)=0$. [Hint: You need to show that if $u-\phi$ has a local minimum at $x_0$ then $\phi'(x_0)\geq 0$, where $\phi$ is a smooth test function. Since $u-\phi$ has a local minimum at $x_0$ there exists $r>0$ such that
\[u(x) - \phi(x) \geq u(x_0)-\phi(x_0)  \ \ \text{ for } |x-x_0| \leq r.\]
Choose $x=x_0-h$ for $h>0$ and send $h\to 0$ to show that $\phi'(x_0)\geq 0$. Also note that a similar argument shows $u$ is a viscosity subsolution of $-u'(x)=0$.]
\item Show that $u(x)=-|x|$ is a viscosity solution of $u'(x)^2 - 1 = 0$. [Hint: The only point you need to check is $x=0$. You need to show that if $u-\phi$ has a local maximum at $x=0$ then $-1 \leq \phi'(x)\leq 1$. Use an argument similar to the previous question. For the supersolution property at $x=0$, show that there are no smooth test functions $\phi$ for which $u-\phi$ has a local minimum at $x=0$ by showing that any such function would satisfy $\phi'(0)\geq 1$ and $\phi'(0)\leq -1$ (draw a picture of a smooth function touching from below at $x=0$).]
\end{enumerate}
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