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\centerline{{\sc Problem set 5}}
\centerline{Math 207A, Fall 2011}
\centerline{Due: Wed., Nov. 2}


\bigskip\noindent
{\bf 1.} Solve the ODE
\[
\left(\begin{array}{c} x \\ y\end{array}\right)_t = 
\left(\begin{array}{cc} \mu & -\omega \\ \omega & \mu\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right),
\qquad
\left(\begin{array}{c} x(0) \\ y(0)\end{array}\right) = \left(\begin{array}{c} x_0 \\ y_0\end{array}\right).
\]
If $\omega > 0$, sketch the phase plane for $\mu > 0$, $\mu =0$, and $\mu < 0$. Classify the equilibrium $\vec{x}=0$ in each case.

\bigskip\noindent
{\bf 2.} Solve the ODE
\[
\left(\begin{array}{c} x \\ y\end{array}\right)_t =
\left(\begin{array}{cc} \lambda & 1 \\ 0 & \lambda\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right),
\qquad
\left(\begin{array}{c} x(0) \\ y(0)\end{array}\right) = \left(\begin{array}{c} x_0 \\ y_0\end{array}\right).
\]
Sketch the phase plane for $\lambda > 0$, $\lambda =0$, and $\lambda < 0$. Classify the equilibrium $\vec{x}=0$ in each case.

\bigskip\noindent
{\bf 3.} (a) Use the power series definition of the exponential of a linear map
\[
e^{tA} = \sum_{k=0}^\infty \frac{t^k}{k!} A^k
\]
to show that if $AB = BA$ then $e^{tA} e^{tB} = e^{t(A+B)}$.
(You can assume that the series can be multiplied term by term and rearranged.)


\smallskip\noindent
(b) If
\[
A = \left(\begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array}\right)
=
\left(\begin{array}{cc} \lambda & 0 \\ 0 & \lambda \end{array}\right)
+
\left(\begin{array}{cc} 0 & 1 \\ 0 & 0\end{array}\right)
\]
use the series definition and the result of (a) to compute $e^{tA}$. Compare your answer with the solution of Problem 2.

\smallskip\noindent
(c) If $A$, $B$ need not commute, show that as $t\to 0$
\[
e^{tA} e^{tB} = e^{t(A+B)} + \frac{1}{2} t^2 [A,B] + O(t^3)
\]
where $[A,B] = AB - BA$ is the commutator of $A$ and $B$.
 


\bigskip\noindent
{\bf 4.} Let $\vec{f} : \Rl \to \Rl^d$ be a continuous vector-valued function and $A : \Rl^d \to \Rl^d$ a linear map. Show that the
solution $\vec{x} : \Rl \to \Rl^d$ of the autonomous, nonhomogeneous system
\[
\vec{x}_t = A\vec{x} + \vec{f}(t), \qquad \vec{x}(0) = \vec{x}_0
\] 
is given by
\[
\vec{x}(t) = e^{tA} \vec{x}_0 + \int_0^t e^{(t-s)A}\vec{f}(s)\, ds.
\]
(This expression for the solution of the nonhomogeneous equation in terms of the solution of the homogeneous equation
is called Duhamel's formula.)
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