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{ \large \textbf{\textsc{Math 5588 -- Homework 7 (Due Thursday March 9)}}}
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Recall a matrix $A=(a_{ij}) \in \R^{n\times n}$ is symmetric if $a_{ij}=a_{ji}$, so that $A^T=A$. We say a symmetric matrix $A$ is \emph{positive definite}, written $A\geq 0$, if
\[v^TAv = \sum_{i=1}^n\sum_{j=1}^n a_{ij}v_iv_j \geq 0 \ \ \text{ for all } v \in \R^n.\]
We write $A\leq B$ whenever $B-A\geq 0$.
\begin{enumerate}
\item A real symmetric matrix $A\in \R^{n\times n}$ has $n$ real eigenvalues $\lambda_1 \leq \lambda_2\leq \cdots \leq \lambda_n$, counted with multiplicity. Show that $A$ is positive definite ($A\geq 0$) if and only if $\lambda_1\geq 0$. [Hint: Recall that
\[\lambda_1 = \min_{v\neq 0} \frac{v^TAv}{v^Tv}.\]
The right hand side above is called a Rayleigh quotient.]
\item Let $A\in \R^{2\times 2}$ be a symmetric matrix. Show that $A$ is positive definite ($A\geq 0$) if and only if
\[\det(A)\geq 0 \ \ \text{ and } \ \ \text{Trace}(A)\geq 0.\]
\item 
\begin{enumerate}
\item Let $A,B\in R^{n\times n}$ be \emph{diagonal} matrices. Show that $A\leq B$ if and only if $a_{ii} \leq b_{ii}$ for all $i\in \{1,\dots,n\}$.
\item Give an example of diagonal matrices $A,B\in \R^{n\times n}$ for which neither $A\leq B$ nor $B\leq A$ hold.
\end{enumerate}
\item Consider a nonlinear PDE in the form
\begin{equation}
H(x,\nabla u,\nabla^2u) + F(x,\nabla u)=0.
\label{eq:pde}
\end{equation}
Assume that $H$ is linear in $\nabla^2u$, that is,
\[\lambda H(x,p,A) + H(x,p,B)=H(x,p,\lambda A+B)\]
for any $\lambda,A,B$. Such a PDE is called \emph{quasilinear}. Show that \eqref{eq:pde} is elliptic if and only if
\[A\leq 0 \implies H(x,p,A) \geq 0.\]
\item Consider the quasilinear PDE
\[-\sum_{i=1}^n\sum_{j=1}^n b_{ij}(x,\nabla u)u_{x_ix_j} + F(x,\nabla u) = 0.\]
Show that this PDE is elliptic if the matrix $B(x,p)=(b_{ij}(x,p))$ is symmetric and positive definite for all $x$ and $p$. [Hint: Note that
\[\sum_{i=1}^n\sum_{j=1}^n b_{ij}(x,\nabla u)u_{x_ix_j} = \text{Trace}(B(x,\nabla u)\nabla^2 u). \]
Use problem 4. You may want to diagonalize $B$ or $A=\nabla^2u$ by an orthogonal transformation and use the property $\text{Trace}(AB)=\text{Trace}(BA)$.
]
\item Let $A=(a_{ij})\in \R^{n\times n}$ be a positive definite symmetric matrix and consider the linear elliptic PDE
\begin{equation}
-\sum_{i=1}^n\sum_{j=1}^n a_{ij}u_{x_ix_j}=f.
\label{eq:lin}
\end{equation}
\begin{enumerate}
\item The fundamental solution of \eqref{eq:lin} satisfies
\[-\sum_{i=1}^n\sum_{j=1}^n a_{ij}u_{x_ix_j} = \delta.\]
Let $\hat{u}(k)$ be the Fourier transform of $u$, and find $S(k)$ so that
\[S(k)\hat{u}(k) = 1.\]
The polynomial $S(k)$ is called the \emph{symbol} of the second order differential operator in \eqref{eq:lin}. 
\item Show that when $n=2$, the equation
\[S(k)=1\]
describes an ellipse in the $k=(k_1,k_2)$-plane. This is the reason for the terminology \emph{elliptic}. [Aside: For $n\geq 3$, $S(k)=1$ describes an $n$-dimensional ellipse.]

\end{enumerate}
\item Consider a functional independent of $u$, that is
\[I(u) = \int_U L(x,\nabla u)\, dx.\]
Show that the corresponding Euler-Lagrange equation
\[-\div\left(\nabla_p L(x,\nabla u) \right) = -\sum_{i=1}^n \frac{\partial }{\partial x_i}\Big( L_{p_i}(x,\nabla u)\Big) = 0\]
is elliptic if $L$ is convex. [Hint: Use the chain rule to expand the divergence and then use problem 5. Recall $L$ is convex if
\[\sum_{i=1}^n\sum_{j=1}^n L_{p_ip_j}(x,p)v_iv_j \geq 0 \ \ \text{ for all } v \in \R^n.\]
]

\end{enumerate}
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