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{ \large \textbf{\textsc{Math 5588 -- Homework 4 (Due Thursday February 9)}}}
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\begin{enumerate}
\item Consider the following version of the isoperimetric problem:
\[\max_{\substack{u:[-1,1]\to \R \\ u(-1)=0=u(1)}} \int_{-1}^1 u(x) \, dx \ \ \text{subject to} \int_{-1}^1 \sqrt{1+u'(x)^2}\,dx =L,\]
where $L>2$. Show that the optimal curve $C(x)=(x,u(x))$ must be a segment of a circle. [Hint: Proceed in a similar fashion to the Brachistochrone problem. In particular, after reducing the Euler-Lagrange equation to a first order ODE, you should solve it by reparamaterizing the curve $C$ in terms of the angle $\theta$ between the tangent vector to $C$ and the $x$-axis.]
\item Consider the constrained problem
\[\min_{\substack{u:U \to \R \\ u=0 \text{ on } \partial U} } \int_U |\nabla u|^2 \, dx \ \ \text{subject to} \ \ \int_U u^2 \, dx = 1.\]
Show that any minimizer is a solution of the eigenvalue problem
\[\begin{cases}
-\Delta u = \lambda u,& \text{in } U\\
u =0,&\text{on } \partial U\end{cases}\]
where $\lambda > 0$ is given by
\[\lambda = \int_U |\nabla u|^2\, dx.\]
 [Hint: To show that $\lambda > 0$ and verify the formula for $\lambda$, multiply by $u$ on both sides of the PDE and integrate by parts.]
\item Show that the plane
\[u(x) = a\cdot x + b\]
solves the minimal surface equation on $U=\R^n$, where $a\in \R^n$ and $b \in \R$.
\item Show that for $n=2$ the Scherk surface
\[u(x) = \log\left(\frac{\cos(x_1)}{\cos(x_2)} \right)\]
solves the minimal surface equation on the box $U=(-\tfrac{\pi}{2},\tfrac{\pi}{2})^2$.
\end{enumerate}

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