# UMD PDE Qualifying Exams/Jan2013PDE

## Problem 1

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

### Solution

Note: For notational purposes, let's put the time variable last. i.e. ${\displaystyle u(x_{1},x_{2},t=0)=e^{-x_{1}^{2}}}$ so that ${\displaystyle x_{1}}$ is the first variable, ${\displaystyle x_{2}}$ is the second variable.

We then write our PDE as ${\displaystyle 0=F(p,z,x)=p_{3}+x_{2}p_{1}-x_{1}p_{2}}$.

We write the characteristic ODEs {\displaystyle \left\{{\begin{aligned}{\dot {z}}(s)&=D_{p}F\cdot p\\{\dot {x}}(s)&=D_{p}F\end{aligned}}\right.}

This gives {\displaystyle \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 ${\displaystyle {\ddot {x}}_{2}=-x_{2}}$ and ${\displaystyle {\ddot {x}}_{1}=-x_{1}}$ which means that ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ must have the following form:

{\displaystyle \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 ${\displaystyle (x_{1}(0),x_{2}(0),t(0))=(x_{1}^{0},x_{2}^{0},0)}$.

Also since, ${\displaystyle {\dot {z}}(s)=0}$, then ${\displaystyle z(s)=z^{0}=e^{-(x_{1}^{0})^{2}}}$.

Now, given any ${\displaystyle (x_{1},x_{2},t)\in \mathbb {R} ^{2}\times (0,\infty )}$, we need to find ${\displaystyle x_{1}^{0},x_{2}^{0},s}$ such that ${\displaystyle (x_{1},x_{2},t)=(x_{1}(s),x_{2}(s),t(s))}$. Clearly, we need ${\displaystyle t=s}$. This means that we just need to solve the following system for ${\displaystyle x_{1}^{0}}$

{\displaystyle \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 ${\displaystyle x_{2}^{0}}$ gives ${\displaystyle x_{2}^{0}={\frac {x_{2}+x_{1}^{0}\sin(t)}{\cos(t)}}}$. Substitute this into the first equation and we can solve for ${\displaystyle x_{1}^{0}}$. We should get (after simplifying) ${\displaystyle x_{1}^{0}=x_{1}\cos(t)-x_{2}\sin(t)}$.

Therefore, ${\displaystyle u(x_{1},x_{2},t)=z^{0}=e^{-(x_{1}^{0})^{2}}=e^{-(x_{1}\cos(t)-x_{2}\sin(t))^{2}}}$.

## Problem 2

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

### Solution

#### a

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

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

So then differentiating and the use of Green's Formula gives:

{\displaystyle {\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 ${\displaystyle \not \int }$ to denote the average integral value symbol (dashed integral). The usual symbol used in Evans would not typeset on this wikibook.

Since ${\displaystyle u\geq 1}$, ${\displaystyle \phi (u)\geq 0}$. Therefore, ${\displaystyle -\Delta u=\phi (u)\geq 0}$, that is, ${\displaystyle u}$ is a supersolution to Laplace's equation.

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

Now,

{\displaystyle {\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 ${\displaystyle r>0}$. This necessarily implies ${\displaystyle u\equiv u(x_{0})=1}$ since nonconstant supersolutions tend to ${\displaystyle -\infty }$ as ${\displaystyle r\to \infty }$.

## Problem 3

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

### Solution

#### a

Multiply both sides of the PDE by ${\displaystyle u}$ and integrate.

${\displaystyle \int _{\Pi ^{N}}u(-\Delta u)=\int _{\Pi ^{N}}-u^{4}+\lambda u^{2}}$.

Integrate by parts to obtain:

${\displaystyle -\int _{\partial \Pi ^{N}}u{\frac {\partial u}{\partial \nu }}+\int _{\Pi ^{N}}|Du|^{2}=\int _{\Pi ^{N}}-u^{4}+\lambda u^{2}}$.

The boundary term vanishes by the periodicity of ${\displaystyle u}$ in all variables.

Thus ${\displaystyle 0\leq \int _{\Pi ^{N}}|Du|^{2}=\int _{\Pi ^{N}}-u^{4}+\lambda u^{2}}$ implies that ${\displaystyle \lambda >0}$.

#### b

Assuming ${\displaystyle \int _{\Pi ^{N}}u_{n}^{2}(x)\,dx=1}$ and our result from part a, we get

${\displaystyle \int _{\Pi ^{N}}|Du_{n}|^{2}=\int _{\Pi ^{N}}-u_{n}^{4}+\lambda \,dx}$

This gives

{\displaystyle {\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 ${\displaystyle \lambda _{n}\to 0}$, this contradicts the above inequality, i.e. we would have ${\displaystyle 0=\lim \lambda _{n}\geq 1}$.