Problem 1[edit]
Find the explicit solution, $u=u(t,x_{1},x_{2})$, of
$\partial _{t}u+x_{2}\partial _{x_{1}}ux_{1}\partial _{x_{2}}u=0,\quad t>0,$ subject to $u(t=0,x_{1},x_{2})=e^{x_{1}^{2}}$.

Solution[edit]
Note: For notational purposes, let's put the time variable last. i.e. $u(x_{1},x_{2},t=0)=e^{x_{1}^{2}}$ so that $x_{1}$ is the first variable, $x_{2}$ is the second variable.
We then write our PDE as $0=F(p,z,x)=p_{3}+x_{2}p_{1}x_{1}p_{2}$.
We write the characteristic ODEs $\left\{{\begin{aligned}{\dot {z}}(s)&=D_{p}F\cdot p\\{\dot {x}}(s)&=D_{p}F\end{aligned}}\right.$
This gives $\left\{{\begin{aligned}{\dot {z}}(s)&=x_{2}p_{1}x_{1}p_{2}+p_{3}=0\\{\dot {x}}_{1}(s)&=x_{2}(s);\quad {\dot {x}}_{2}(s)=x_{1}(s);\quad {\dot {t}}(s)=1\end{aligned}}\right.$
Notice that this gives ${\ddot {x}}_{2}=x_{2}$ and ${\ddot {x}}_{1}=x_{1}$ which means that $x_{1}$ and $x_{2}$ must have the following form:
$\left\{{\begin{aligned}x_{1}(s)&=x_{2}^{0}\sin(s)+x_{1}^{0}\cos(s)\\x_{2}(2)&=x_{1}^{0}\sin(s)+x_{2}^{0}\cos(s)\\t(s)=s\end{aligned}}\right.$
where the coefficients are chosen so that $(x_{1}(0),x_{2}(0),t(0))=(x_{1}^{0},x_{2}^{0},0)$.
Also since, ${\dot {z}}(s)=0$, then $z(s)=z^{0}=e^{(x_{1}^{0})^{2}}$.
Now, given any $(x_{1},x_{2},t)\in \mathbb {R} ^{2}\times (0,\infty )$, we need to find $x_{1}^{0},x_{2}^{0},s$ such that $(x_{1},x_{2},t)=(x_{1}(s),x_{2}(s),t(s))$. Clearly, we need $t=s$. This means that we just need to solve the following system for $x_{1}^{0}$
$\left\{{\begin{aligned}x_{1}&=x_{2}^{0}\sin(t)+x_{1}^{0}\cos(t)\\x_{2}&=x_{1}^{0}\sin(t)+x_{2}^{0}\cos(t)\\\end{aligned}}\right.$
Solving the second equation for $x_{2}^{0}$ gives $x_{2}^{0}={\frac {x_{2}+x_{1}^{0}\sin(t)}{\cos(t)}}$. Substitute this into the first equation and we can solve for $x_{1}^{0}$. We should get (after simplifying) $x_{1}^{0}=x_{1}\cos(t)x_{2}\sin(t)$.
Therefore, $u(x_{1},x_{2},t)=z^{0}=e^{(x_{1}^{0})^{2}}=e^{(x_{1}\cos(t)x_{2}\sin(t))^{2}}$.
Problem 2[edit]
Solution[edit]
We perform a change of variables $z={\frac {yx}{r}}$ which gives:
$m_{x}(r)=r^{1N}\int _{\partial B(x,r)}u(y)\,dS(y)==r^{1N}r^{N1}\int _{\partial B(0,1)}u(x+rz)\,dS(z)$.
So then differentiating and the use of Green's Formula gives:
${\begin{aligned}{\frac {d}{dr}}m_{x}(r)&=\int _{\partial B(0,1)}Du(x+rz)\cdot z\,dS(z)\\&=\int _{\partial B(0,1)}{\frac {\partial }{\partial \nu }}u(x+rz)\,dS(z)\\&=\int _{B(0,1)}\Delta u(x+rz)\,dS(z)&=r^{1N}\int _{B(x,r)}\Delta u(y)\,dy.\end{aligned}}$
Notation: I use $\not \int$ to denote the average integral value symbol (dashed integral). The usual symbol used in Evans would not typeset on this wikibook.
Since $u\geq 1$, $\phi (u)\geq 0$. Therefore, $\Delta u=\phi (u)\geq 0$, that is, $u$ is a supersolution to Laplace's equation.
Suppose $u(x_{0})=1$. Then by Part a, ${\frac {dm_{x_{0}}}{dr}}=r^{1N}\int _{B(x_{0},r)}\Delta u(y)\,dy=r^{1N}\int _{B(x_{0},r)}\phi (u)\,dy\leq 0$. So $m_{x_{0}}(r)$ is a decreasing function in $r$.
Now,
${\begin{aligned}m_{x_{0}}(r)&=r^{1N}\int _{\partial B(x_{0},r)}u(y)\,dS(y)&\\&=r^{1N}N\alpha (N)\not \int _{\partial B(x_{0},r)}u(y)\,dS(y)&\\&=r^{1N}C_{N}r^{N1}\not \int _{\partial B(x_{0},r)}u(y)\,dS(y)&{\text{ since }}N\alpha (N)=C_{N}r^{N1}\\&\leq C_{N}u(x_{0})&{\text{ since }}u{\text{ is a supersolution}}.\end{aligned}}$
This estimate must hold for all $r>0$. This necessarily implies $u\equiv u(x_{0})=1$ since nonconstant supersolutions tend to $\infty$ as $r\to \infty$.
Problem 3[edit]
Solution[edit]
Multiply both sides of the PDE by $u$ and integrate.
$\int _{\Pi ^{N}}u(\Delta u)=\int _{\Pi ^{N}}u^{4}+\lambda u^{2}$.
Integrate by parts to obtain:
$\int _{\partial \Pi ^{N}}u{\frac {\partial u}{\partial \nu }}+\int _{\Pi ^{N}}Du^{2}=\int _{\Pi ^{N}}u^{4}+\lambda u^{2}$.
The boundary term vanishes by the periodicity of $u$ in all variables.
Thus $0\leq \int _{\Pi ^{N}}Du^{2}=\int _{\Pi ^{N}}u^{4}+\lambda u^{2}$ implies that $\lambda >0$.
Assuming $\int _{\Pi ^{N}}u_{n}^{2}(x)\,dx=1$ and our result from part a, we get
$\int _{\Pi ^{N}}Du_{n}^{2}=\int _{\Pi ^{N}}u_{n}^{4}+\lambda \,dx$
This gives
${\begin{aligned}{\text{Vol}}(\Pi ^{N})\lambda _{n}&=\int _{\Pi ^{N}}Du_{n}^{2}+u_{n}^{4}\,dx\\&\geq \int _{\Pi ^{N}}u_{n}^{4}\,dx\geq (\int _{\Pi ^{N}}u_{n}^{2}\,dx)^{2}\equiv 1\end{aligned}}$ where the last inequality is due to Jensen's Inequality.
So if $\lambda _{n}\to 0$, this contradicts the above inequality, i.e. we would have $0=\lim \lambda _{n}\geq 1$.