# 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 $0=F(p,z,x)=p_3+x_2p_1-x_1p_2$.

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

This gives \left\{ \begin{align} \dot z(s)&= x_2 p_1-x_1p_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{align}\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{align} 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{align}\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{align} 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{align}\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{align} \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{align}

#### 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{align} 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{align}

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 $\lambda_n\to0$ 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{align} \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{align} 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 $0=\lim\lambda_n\geq 1$.