# UMD PDE Qualifying Exams/Jan2013PDE

## Problem 1

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

### Solution

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

 Let $u\not \equiv 0$ be a $C^{2}(\mathbb {R} ^{N})$ function. Define $m_{x}(r)=r^{1-N}\int _{\partial B(x,r)}u(y)\,dS(y)$ . a). Show that ${\frac {dm_{x}}{dr}}=r^{1-N}\int _{B(x,r)}\Delta u(y)\,dy$ . b). Let $u$ solve $-\Delta u=\phi (u)$ for some continuous $\phi$ . Assume that $u(x)\geq 1$ for every $x\in \mathbb {R} ^{N}$ , and that $\phi (\xi )\geq 0$ for $\xi \geq 1$ . Prove that if $u(x_{0})=1$ at some $x_{0}\in \mathbb {R} ^{N}$ , then $u(x)\equiv 1$ for every $x\in \mathbb {R} ^{N}$ .

### Solution

#### a

We perform a change of variables $z={\frac {y-x}{r}}$  which gives:

$m_{x}(r)=r^{1-N}\int _{\partial B(x,r)}u(y)\,dS(y)==r^{1-N}r^{N-1}\int _{\partial B(0,1)}u(x+rz)\,dS(z)$ .

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^{1-N}\int _{B(x,r)}\Delta u(y)\,dy.\end{aligned}}

#### b

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, ${\frac {dm_{x_{0}}}{dr}}=r^{1-N}\int _{B(x_{0},r)}\Delta u(y)\,dy=r^{1-N}\int _{B(x_{0},r)}-\phi (u)\,dy\leq 0$ . So $m_{x_{0}}(r)$  is a decreasing function in $r$ .

Now,

{\begin{aligned}m_{x_{0}}(r)&=r^{1-N}\int _{\partial B(x_{0},r)}u(y)\,dS(y)&\\&=r^{1-N}N\alpha (N)\not \int _{\partial B(x_{0},r)}u(y)\,dS(y)&\\&=r^{1-N}C_{N}r^{N-1}\not \int _{\partial B(x_{0},r)}u(y)\,dS(y)&{\text{ since }}N\alpha (N)=C_{N}r^{N-1}\\&\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

 Let $u$ solve the nonlinear eigenvalue problem $-\Delta u=-u^{3}+\lambda u.$ Here $u\in C^{2}(\Pi ^{N})$ is a 1-periodic function in all variables (that is, $\Pi ^{N}$ is the $N$ -dimensional torus) with $u\not \equiv 0$ and $\lambda \in \mathbb {R}$ . a. Prove that $\lambda >0$ . b. Prove that there exists no sequence of eigen-solutions $(u_{n},\lambda _{n})$ such that $\lambda _{n}\to 0$ and $\int _{\Pi ^{N}}u_{n}^{2}(x)\,dx=1$ . Hint: Prove b by contradiction.

### Solution

#### a

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 $\lambda >0$ .

#### b

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 $0=\lim \lambda _{n}\geq 1$ .