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UNM Dept. of Mathematics and Statistics \\
Math. 570, Singular Perturbations \\
Problem Set I \\[.2in]
\end{center}
Spring 2011 \\[.2in]
\begin{enumerate}

\item 
Find an expansion for the roots in powers of $\epsilon$ as $\epsilon\rightarrow 0$.
\begin{enumerate}
\item $\epsilon x^4 - x^2 + 1 = 0$.
\item $\epsilon x^4 - x^2 + 2 x - 1 = 0$.
\end{enumerate}

\item 
Find an approximation to the solution, correct to ${\cal O}(\epsilon)$
in the entire interval $\left[ 0,1\right]$ for the B.V.P.:
\begin{eqnarray}
\epsilon y^{\prime \prime} + y^\prime - 2 y = 0 & , & y(0)= 1\ ,\ y(1)=2 \\
\epsilon y^{\prime \prime} -3y^\prime +   y = 0 & , & y(0)=-1\ ,\ y(1)=1 \ .
\end{eqnarray}

\item 
Consider the equation
\[ y^{\prime \prime} + y - \epsilon y^3 = 0\ ,\  y(0)= A\ ,\ y^\prime(0)=0 \ .\]
For $\epsilon = 0$ this problem has the (periodic) solution
$y(t) = A\cos{t}$, whose period is independent of $A$. By introducing a time
scale
\[ t^* = \omega(\epsilon) t = \left(1+\omega_1\epsilon +\omega_2\epsilon^2+\cdots
\right)t \]
and looking for a periodic solution in the form
\[ y(t\ ;\epsilon) = y_0(t^*) +\epsilon y_1(t^*) +{\cal O}(\epsilon^2) \]
show that for $\epsilon \ne 0$ the period depends on the amplitude.\\
({\em Hint}: use $\cos^3{a} = (3/4)\cos{a} +(1/4) \cos{3a}$.)\\
{\em Answer}: $ y(t\ ;\epsilon) =  A\cos{\left(1 - (3/8)A^2\epsilon+\cdots\right)t}
+{\cal O}(\epsilon)$.


\end{enumerate}
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