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\begin{document}
\title{Solutions, 316-XXIV}
\maketitle
{\bf
24( 4/17) Real equal eigenvalues; mass-spring systems and forcing\\
          9.5[35*,36], 9.6[19*,20], 9.7[7*,9,11,13*,16]}\\
CAUTION: there may be errors!!!
\section{Problem 9.5.35}
\begin{enumerate}
\item Show that the matrix
\[ {\bf A} = \left( \begin{array}{rr}
1 & -1 \\ 4 & -3 \end{array} \right) \]
has the repeated eigenvalue $r=-1$ and that all the eigenvectors
are of the form ${\bf u} = a \mbox{col}(1,2)$.
\item Use the result of part (1) to obtain a nontrivial solution
${\bf x}_1(t)$ to the system ${\bf x}^\prime = {\bf A} {\bf x}$.
\item To obtain a second linearly independent solution to
${\bf x}^\prime = {\bf A} {\bf x}$, try ${\bf x}_2(t) = te^{-t} {\bf u}_1
+ e^{-t} {\bf u}_2$. [{\it Hint:} Substitute ${\bf x}_2$ into the system
${\bf x}^\prime = {\bf A} {\bf x}$ and derive the relations\\
\[ \left( {\bf A} + {\bf I} \right) {\bf u}_1 =0 \ ,\ 
\left( {\bf A} + {\bf I} \right) {\bf u}_2 = {\bf u}_1 \ .\]
Since ${\bf u}_1$ must be an eigenvector, set ${\bf u}_1 = \mbox{col}(1,2)$
and solve for ${\bf u}_2$. ]
\item What is $\left( {\bf A} + {\bf I} \right)^2 {\bf u}_2$?\\
\end{enumerate}

{\bf Solution:}\\
\begin{enumerate}
\item Eigenvalues:
\[ \left| \begin{array}{rr}
1-r & -1 \\ 4 & -3-r \end{array} \right| =
(r+3)(r-1)+4 = r^2 +2r +1 = (r+1)^2 = 0 \ ,
\]
so that $r = -1$ is a double eigenvalue. The only eigenvector 
associated with this eigenvalue is 
\[ {\bf u}_1 = \left( \begin{array}{r} 
1 \\ 1-r \end{array} \right) = 
\left( \begin{array}{r} 
1 \\ 2 \end{array} \right) \ . \]
\item A solution to the system ${\bf x}^\prime = {\bf A} {\bf x}$ is found as
\[ {\bf x}_1(t) = e^{-t}  {\bf u}_1 
= e^{-t} \left( \begin{array}{r} 
1 \\ 2 \end{array} \right) \ . \]
\item Try now a second solution in the form
\[ {\bf x}_2(t) = te^{-t} {\bf u}_1
+ e^{-t} {\bf u}_2 =
te^{-t} \left( \begin{array}{r} 
1 \\ 2 \end{array} \right) +
e^{-t} \left( \begin{array}{r} 
a \\ b \end{array} \right) \ . \]
Substituting into the system we find
\begin{eqnarray*}
\frac{d{\bf x}_2}{dt} & = & (1-t)e^{-t} {\bf u}_1 - e^{-t} {\bf u}_2  \\
 {\bf A} {\bf x} & = & -te^{-t} {\bf u}_1 + e^{-t} A {\bf u}_2 
\end{eqnarray*}
and setting these equal we find that ${\bf u}_2$ satisfies:
\[ A {\bf u}_2 = (-1) {\bf u}_2 + {\bf u}_1 \]
or
\[ (A - (-1) I) {\bf u}_2 = {\bf u}_1 \]
i.e.
\[ \left( \begin{array}{rr}
2 & -1 \\ 4 & -2 \end{array} \right)
\left( \begin{array}{r} 
a \\ b \end{array} \right) = \left( \begin{array}{r} 
1 \\ 2 \end{array} \right) \ . \]
which is an Underdetermined system (two equations are the same), with a
solution
\[ {\bf u}_2 = \left( \begin{array}{r} 
0 \\ -1 \end{array} \right) \ , \]
while any multiple of ${\bf u}_1$ can be added to it to give another solution.
The general solution is then
\[ {\bf x}_{gen}(t) = C_1 e^{-t} {\bf u}_1 
+ C_2 e^{-t} \left( {\bf u}_2 + t  {\bf u}_1 \right) \ . \]
\item Here we have
\[ (A-(-1)I)^2 {\bf u}_2 = (A-(-1)I) {\bf u}_1 = 0 \ . \]
\end{enumerate}

\section{Problem 9.6.19}
For the coupled mass-spring system governed by the system
\begin{eqnarray*}
m_1 x_1^{\prime \prime} & = & - k_1 x_1 + k_2 \left( x_2 - x_1 \right) \\ 
m_2 x_2^{\prime \prime} & = & - k_2 \left( x_2 - x_1 \right) - k_3 x_2 
\end{eqnarray*}
assume $m_1=m_2 = 1 kg$, $k_1 = k_2 = 2 N/m$ and $k_3 = 3 N/m$. Determine
the normal frequencies for this coupled mass-spring system. \\
({\em Hint:} substitute $x_1 = a e^{i\omega t}$, $x_2 = b e^{i\omega t}$
to arrive at a system of two equations in the unknown coefficients
$a$, $b$; the system involves $\omega^2$ in its coefficients. 
This system has the trivial solution $a = b = 0$; to find
a nontrivial solution the determinant of the coefficients of this system must
vanish. Find the value(s) of $\omega^2$ that make the determinant vanish.
Their square roots should be real and they give the normal frequencies
of the system.) \\
{\bf Solution:}\\
Substituting $x_1 = a e^{i\omega t}$ and $x_2 = b e^{i\omega t}$
(and dividing both sides by $ e^{i\omega t}$) we find the system
\begin{eqnarray*}
- m_1 a \omega^2  & = & - k_1 a + k_2 \left( b - a \right) \\
- m_2 b \omega^2  & = & - k_2 \left( b - a \right) - k_3 b
\end{eqnarray*}
or
\[
\left( \begin{array}{rr}
m_1\omega^2-(k_1+k_2) & k_2 \\ k_2 & m_2\omega^2-(k_2+k_3) \end{array} \right)
\left( \begin{array}{r} 
a \\ b \end{array} \right) = 
\left( \begin{array}{rr}
\omega^2-4 & 2 \\ 2 & \omega^2-5 \end{array} \right)
\left( \begin{array}{r}
a \\ b \end{array} \right) =
0
\]
which has a nontrivial solution provided the determinant of the 
coefficient matrix vanishes:
\[ (\omega^2-4)(\omega^2-5)-4 = 0 \Rightarrow
\omega^4 - 9 \omega^2 + 16 = 0 \Rightarrow \omega^2 = 
\frac{9 \pm \sqrt{17}}{2} \ ,\]
so that the characteristic frequencies of the system are (multiply
by $2\pi$ to get angular frequencies):
\[ \omega_1 = \sqrt{\frac{9 + \sqrt{17}}{2}} \ ,\ 
   \omega_2 = \sqrt{\frac{9 - \sqrt{17}}{2}}  \ . \]
\newpage

\section{Problem 9.7.7}
Determine the form of the particular solution for the system
${\bf x}^\prime (t) = {\bf A x}(t) + {\bf f}(t)$ by the method of undetermined
coefficients if ${\bf A}$ and ${\bf f}(t)$ are given by
\[ {\bf A} = \left[ \begin{array}{rr} 0 & 1\\2 & 0 \end{array} \right]\ ,
{\bf f}(t) = \left[ \begin{array}{r} \sin{3t} \\ t \end{array} \right] \ . \]
{\bf Solution:}\\
We need the homogeneous solution first: 
\[ r^2-2 =0 \Rightarrow r_1 = \sqrt{2} \ , \  r_2 = -\sqrt{2} \]
with eigenvectors:
\[ {\bf u}_1 = \left( \begin{array}{r} 1 \\ \sqrt{2} \end{array} \right) \ ,\ 
   {\bf u}_2 = \left( \begin{array}{r} 1 \\-\sqrt{2} \end{array} \right) \ .\]
Then
\[ {\bf x}_{hom}(t) = C_1 e^{\sqrt{2}t} {\bf u}_1 
                   +  C_2 e^{-\sqrt{2}t} {\bf u}_2 \ . \]
Since none of the forcing terms are resonant, the particular solution will have
the form
\[ {\bf x}_{part}(t) =
\left( \begin{array}{r}
A_1 \cos{t} + B_1 \sin{t} + C_1 + D_1 t \\
A_2 \cos{t} + B_2 \sin{t} + C_2 + D_2 t \end{array} \right) \, \]
and substituting into the system will give equations for determining the 
unknown coefficients $A_1,\ldots,D_2$.
\newpage

\section{Problem 9.7.13}
Determine the general solution for the system
${\bf x}^\prime (t) = {\bf A x}(t) + {\bf f}(t)$ by the method of 
variation of parameters if ${\bf A}$ and ${\bf f}(t)$ are given by
\[ {\bf A} = \left[ \begin{array}{rr} 2 & 1\\-3 & -2 \end{array} \right]\ ,
{\bf f}(t) = \left[ \begin{array}{r} 2e^t \\ 4e^t \end{array} \right] \ . \]
\begin{small}
({\bf Instructions:}\\
Use the formula:
\[ {\bf x}_gen(t) = {\bf X}(t) {\bf c} +
                    {\bf X}(t) \int_{t_0}^t {\bf X}^{-1}(s) {\bf f}(s) ds 
\ . \]
where ${\bf X}(t)$ is a fundamental matrix for the homogeneous system
and ${\bf c}$ a constant vector. \\
{\bf NOTE:} To find ${\bf X}^{-1}(s)$ you can use the formula given in class,
but there is a {\bf catch}: as mentioned in class,
\[ {\bf X}(t - s) = {\bf X}(t) {\bf X}( - s) \ . \]
However, this means
\[ {\bf X}(t - t) = {\bf X}(0) =  {\bf X}(t) {\bf X}( - t) \]
so that
\[  {\bf X}( - t) =  {\bf X}^{-1}(t) {\bf X}(0)  \]
i.e.
\[ {\bf X}^{-1}(t) = {\bf X}( - t) {\bf X}^{-1}(0) \ .\]
In class I was assuming (but did not mention it!) that
\[ {\bf X}(0) = I\]
which is not true in general. Also, it turns out that these matrices commute, i.e.
\[ {\bf X}(t) {\bf X}( - t) = {\bf X}( - t) {\bf X}(t) = {\bf X}(0) \]
and \[ {\bf X}^{-1}(t) = {\bf X}( - t) {\bf X}^{-1}(0)
= {\bf X}^{-1}(0) {\bf X}( - t) \]
so you can multiply them in any order you prefer!) \\
\end{small}
{\bf Solution:}\\
\begin{enumerate}
\item Eigenvalues:
\[ \left| \begin{array}{rr}
 2-r & 1 \\ -3 & -2 -r \end{array} \right| 
= 0 \ \Rightarrow \  r^2 -1 = 0 \]
so that
\[ r_1 = 1 \ \ ,\ \ r_2 = -1 \ .\]
\item Eigenvectors:
\[ {\bf v}_1 = 
\left(\begin{array}{r}
1 \\ r_1 - 2
\end{array} \right)
=
\left(\begin{array}{r}
1 \\ -1 \end{array} \right)
\ ,\  
{\bf v}_2  = 
\left(\begin{array}{r}
1 \\ r_2 - 2
\end{array} \right)
=
\left(\begin{array}{r}
1 \\ -3 \end{array} \right) \ . \]
\item Homogeneous solution:\\
\[ {\bf x}_{hom} = C_1 e^{t} \left(\begin{array}{r}
1 \\ -1 \end{array} \right) + C_2 e^{-t} \left(\begin{array}{r}
1 \\ -3 \end{array} \right) \ . \]
\item Fundamental matrix:\\
\[ {bf X}(t) =
\left(\begin{array}{rr}
e^{t} & e^{-t} \\ -e^{t} & -3e^{-t}
\end{array} \right) \ . \]
\item Determinant of fundamental matrix (Wronskian):
\[ \left|\begin{array}{rr}
e^{t} & e^{-t} \\ -e^{t} & -3e^{-t}
\end{array} \right| = -3 + 1 = -2 \ . \]
\item Inverse of fundamental matrix:
\[ {\bf X}^{-1}(t) =
\frac{-1}{2}
\left(\begin{array}{rr}
-3e^{-t} & -e^{-t} \\ e^{t} & e^{t} 
\end{array} \right) \ . \]
\item Integral:
\begin{eqnarray*}
 \int_{t_0}^t {\bf X}^{-1}(s) {\bf f}(s) ds & =  &
\frac{-1}{2}
\int_{0}^t \left(\begin{array}{rr}
-3e^{-s} & -e^{-s} \\ e^{s} & e^{s}
\end{array} \right) 
\left( \begin{array}{r} 2e^s \\ 4e^s \end{array} \right) ds \\
& = &
\int_{0}^t 
\left( \begin{array}{r} 5 \\ -3e^{2s} \end{array} \right) ds \\
& = &
\left( \begin{array}{r} 5t \\ -\frac{3}{2}e^{2t}+\frac{3}{2} \end{array} \right)\end{eqnarray*}
\item Particular solution:
\begin{eqnarray*} {\bf X}(t) \int_{t_0}^t {\bf X}^{-1}(s) {\bf f}(s) ds & = &
\left(\begin{array}{rr}
e^{t} & e^{-t} \\ -e^{t} & -3e^{-t}
\end{array} \right) 
\left( \begin{array}{r} 5t \\ -\frac{3}{2}e^{2t} +\frac{3}{2} \end{array} \right)
\\ & = &
\left( \begin{array}{r}
5t e^{t} - \frac{3}{2}e^{t} +\frac{3}{2}e^{-t} \\
-5t e^{t} +\frac{9}{2}e^{t} - \frac{9}{2}e^{-t}
\end{array} \right)
\end{eqnarray*}
\item General solution:
\[ {\bf x}_{gen} =
C_1 e^{t} \left(\begin{array}{r}
1 \\ -1 \end{array} \right) + C_2 e^{-t} \left(\begin{array}{r}
1 \\ -3 \end{array} \right)
+\left( \begin{array}{r}
5t e^{t} - \frac{3}{2}e^{t} +\frac{3}{2}e^{-t} \\
-5t e^{t} +\frac{9}{2}e^{t} - \frac{9}{2}e^{-t}
\end{array} \right) \]
\end{enumerate}
Note: this answer is equivalent to the answer in the book. They differ by
a homogeneous solution (I did not simplify here to make it obvious how
I got everything!).
\end{document}
