UMD PDE Qualifying Exams/Jan2013PDE

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, $$u(x_1,x_2,t)=z^0=e^{-(x_1^0)^2}=e^{-(x_1 \cos(t)-x_2\sin(t))^2}$$.

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 $$r\to\infty$$.

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 $$0=\lim\lambda_n\geq 1$$.