UMD PDE Qualifying Exams/Jan2013PDE

Problem 1

Find the explicit solution, $u=u(t,x_1,x_2)$, of ${\partial }_{t}u+x_2{\partial }_{x_1}u-x_1{\partial }_{x_2}u=0,t>0,$ 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 $\{\begin{array}{cc}\dot{z}(s)& =D_{p}F\cdot p\\ \dot{x}(s)& =D_{p}F\end{array}$

This gives $\{\begin{array}{cc}\dot{z}(s)& =x_2{p}_1-x_1{p}_2+p_3=0\\ {\dot{x}}_1(s)& =x_2(s);{\dot{x}}_2(s)=-x_1(s);\dot{t}(s)=1\end{array}$

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:

$\{\begin{array}{cc}{x}_1(s)& =x_2^0\sin (s)+x_1^0\cos (s)\\ x_2(2)& =-x_1^0\sin (s)+x_2^0\cos (s)\\ t(s)=s\end{array}$

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 $(x_1,x_2,t)\in {ℝ}^2\times (0,\mathrm{\infty })$, 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 $x_1^0$

$\{\begin{array}{cc}{x}_1& =x_2^0\sin (t)+x_1^0\cos (t)\\ x_2& =-x_1^0\sin (t)+x_2^0\cos (t)\\ \end{array}$

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 $u\equiv ̸0$ be a $C^2({ℝ}^N)$ function. Define $m_x(r)=r^{1-N}\underset{\partial B(x,r)}{\int }u(y)dS(y)$. a). Show that $\frac{d{m}_x}{dr}=r^{1-N}\underset{B(x,r)}{\int }\mathrm{\Delta }u(y)dy$. b). Let $u$ solve $-\mathrm{\Delta }u=\varphi (u)$ for some continuous $\varphi$. Assume that $u(x)\ge 1$ for every $x\in {ℝ}^N$, and that $\varphi (\xi )\ge 0$ for $\xi \ge 1$. Prove that if $u(x_0)=1$ at some $x_0\in {ℝ}^N$, then $u(x)\equiv 1$ for every $x\in {ℝ}^N$.

Solution

a

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

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

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

$\begin{array}{cc}\frac{d}{dr}{m}_x(r)& =\underset{\partial B(0,1)}{\int }Du(x+rz)\cdot zdS(z)\\ & =\underset{\partial B(0,1)}{\int }\frac{\partial }{\partial \nu }u(x+rz)dS(z)\\ & =\underset{B(0,1)}{\int }\mathrm{\Delta }u(x+rz)dS(z)& =r^{1-N}\underset{B(x,r)}{\int }\mathrm{\Delta }u(y)dy.\end{array}$

b

Notation: I use $\int ̸$ to denote the average integral value symbol (dashed integral). The usual symbol used in Evans would not typeset on this wikibook.

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

Suppose $u(x_0)=1$. Then by Part a, $\frac{d{m}_{x_0}}{dr}=r^{1-N}\underset{B(x_0,r)}{\int }\mathrm{\Delta }u(y)dy=r^{1-N}\underset{B(x_0,r)}{\int }-\varphi (u)dy\le 0$. So $m_{x_0}(r)$ is a decreasing function in $r$.

Now,

$\begin{array}{ccc}{m}_{x_0}(r)& =r^{1-N}\underset{\partial B(x_0,r)}{\int }u(y)dS(y)& \\ & =r^{1-N}N\alpha (N)\underset{\partial B(x_0,r)}{\int }u(y)dS(y)& \\ & =r^{1-N}{C}_N{r}^{N-1}\underset{\partial B(x_0,r)}{\int }u(y)dS(y)& \text{ since }N\alpha (N)=C_N{r}^{N-1}\\ & \le C_{N}u(x_0)& \text{ since }u\text{ is a supersolution}.\end{array}$

This estimate must hold for all $r>0$. This necessarily implies $u\equiv u(x_0)=1$ since nonconstant supersolutions tend to $-\mathrm{\infty }$ as $r\to \mathrm{\infty }$.

Problem 3

Let $u$ solve the nonlinear eigenvalue problem $-\mathrm{\Delta }u=-u^3+\lambda u.$ Here $u\in C^2({\mathrm{\Pi }}^N)$ is a 1-periodic function in all variables (that is, ${\mathrm{\Pi }}^N$ is the $N$-dimensional torus) with $u\equiv ̸0$ and $\lambda \in ℝ$. 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 $\underset{{\mathrm{\Pi }}^N}{\int }{u}_n^2(x)dx=1$. Hint: Prove b by contradiction.

Solution

a

Multiply both sides of the PDE by $u$ and integrate.

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

Integrate by parts to obtain:

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

The boundary term vanishes by the periodicity of $u$ in all variables.

Thus $0\le \underset{{\mathrm{\Pi }}^N}{\int }|Du{|}^2=\underset{{\mathrm{\Pi }}^N}{\int }-u^4+\lambda u^2$ implies that $\lambda >0$.

b

Assuming $\underset{{\mathrm{\Pi }}^N}{\int }{u}_n^2(x)dx=1$ and our result from part a, we get

$\underset{{\mathrm{\Pi }}^N}{\int }|D{u}_n{|}^2=\underset{{\mathrm{\Pi }}^N}{\int }-u_n^4+\lambda dx$

This gives

$\begin{array}{cc}\text{Vol}({\mathrm{\Pi }}^N){\lambda }_n& =\underset{{\mathrm{\Pi }}^N}{\int }|D{u}_n{|}^2+u_n^{4}dx\\ & \ge \underset{{\mathrm{\Pi }}^N}{\int }{u}_n^{4}dx\ge (\underset{{\mathrm{\Pi }}^N}{\int }{u}_n^{2}dx{)}^2\equiv 1\end{array}$ 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 $0=\lim {\lambda }_n\ge 1$.