UMD PDE Qualifying Exams/Jan2013PDE

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Problem 1[edit]

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}.


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

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

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)\\

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)\\

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[edit]

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.



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:

\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.


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.


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}.

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[edit]

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.



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.


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

\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.