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\begin{document}
\title{Solutions, 316-XXV}
\maketitle
{\bf
25( 4/22) Phase plane and stability\\
          12.2<1*,2,3*,4,5,7*,8,9,13*>}\\
CAUTION: there may be errors!!!
\section{Problem 12.2.1}
Classify the critical point at the origin for the system
\begin{eqnarray*}
\frac{dx}{dt} & = & 5x + 6y \\
\frac{dy}{dt} & = &-5x - 8y
\end{eqnarray*}
{\bf Solution:}\\
\begin{eqnarray*}
\det \left( \begin{array}{ll}
5-r & 6 \\ -5 & -8-r
\end{array} \right) = 0 & \Rightarrow &
(r+8)(r-5) +30 =0 \Rightarrow \\
r^2 + 3r - 40 = 0 & \Rightarrow & (r+5)(r-2) = 0  
\end{eqnarray*}
so that the eigenvalues are
\[ r_1 = 2 \ ,\  r_2 = -5 \]
and the origin is an unstable saddle point.
\newpage
\section{Problem 12.2.3}
Classify the critical point at the origin for the system
\begin{eqnarray*}
\frac{dx}{dt} & = & 3x + 5y \\
\frac{dy}{dt} & = &-5x - 5y
\end{eqnarray*}
{\bf Solution:}\\
\begin{eqnarray*}
\det \left( \begin{array}{ll}
3-r & 5 \\ -5 & -5-r
\end{array} \right) = 0 & \Rightarrow &
(r+5)(r-3) +25 =0 \Rightarrow \\
r^2 + 2r + 10 = 0 & \Rightarrow & (r+1)^2+9 = 0  
\end{eqnarray*}
so that the eigenvalues are
\[ r_1 = -1+3i \ ,\  r_2 = -1-3i \]
and the origin is a stable spiral: all solutions tend to zero
as $t\rightarrow \infty$.
\newpage
\section{Problem 12.2.5}
Classify the critical point at the origin for the system
\begin{eqnarray*}
\frac{dx}{dt} & = &-6x + 3y \\
\frac{dy}{dt} & = &  x - 4y
\end{eqnarray*}
{\bf Solution:}\\
\begin{eqnarray*}
\det \left( \begin{array}{ll}
-6-r & 3 \\ 1 & -4-r
\end{array} \right) = 0 & \Rightarrow &
(r+6)(r+4) -3 =0 \Rightarrow \\
r^2 + 10r + 21 = 0 & \Rightarrow & (r+7)(r+3) = 0  
\end{eqnarray*}
so that the eigenvalues are 
\[ r_1 = -3 \ ,\  r_2 = -7 \]
and the origin is a stable improper node: all solutions tend to zero
as $t\rightarrow \infty$.
\newpage
\section{Problem 12.2.7}
Find and classify the critical point of the linear system
\begin{eqnarray*}
\frac{dx}{dt} & = &-4x + 2y + 8\\
\frac{dy}{dt} & = &  x - 2y + 1
\end{eqnarray*}
{\bf Solution:}\\
First we set the rhs equal to zero to determine the critical point:\\
\begin{eqnarray*}
\left. \begin{array}{ccc}
-4x + 2y & = &  -8 \\
  x - 2y & = &  -1 \end{array} \right\}
& \Rightarrow &  
\left. \begin{array}{ccc}
-3x = -9   & \Rightarrow & x =  1 \\
2 y = x +1 & \Rightarrow & y =  1 \end{array} \right\}
\end{eqnarray*}
So, the critical point is $(x,y) = (1,1)$. We introduce variables
\[ u = x-1 \ ,\  v = y - 1 \ . \]
In terms of $(u,v)$ the system becomes
\begin{eqnarray*}
\frac{du}{dt} & = &-4(u+1) + 2(v+1) + 8 = -4u +2v \\
\frac{dv}{dt} & = &  (u+1) - 2(v+1) + 1 =   u -2v
\end{eqnarray*}
We find now the eigenvalues:
\begin{eqnarray*}
\det \left( \begin{array}{ll}
-4-r &  2 \\  1 & -2-r
\end{array} \right) = 0 & \Rightarrow &
(r+2)(r+4) -2 =0 \Rightarrow \\
r^2 + 6r + 6 = 0 & \Rightarrow & (r+3)^2 - 3 = 0  
\end{eqnarray*}
so that the eigenvalues are
\[ r_1 = -3+\sqrt{3} \ ,\  r_2 = -3-\sqrt{3} \]
and the point (1,1) is a stable improper node: all solutions tend to it
as $t\rightarrow \infty$.
\newpage
\section{Problem 12.2.8}
Find and classify the critical point of the linear system
\begin{eqnarray*}
\frac{dx}{dt} & = &-3x -  y - 4\\
\frac{dy}{dt} & = &29x +  y +30
\end{eqnarray*}
{\bf Solution:}\\
First we set the rhs equal to zero to determine the critical point:\\
\begin{eqnarray*}
\left. \begin{array}{ccc}
-3x -  y & = &   4 \\
29x +  y & = & -30 \end{array} \right\}
& \Rightarrow &  
\left. \begin{array}{ccc}
26x = -26 & \Rightarrow & x = -1 \\
y = -3x -4 = -1 \end{array} \right\}
\end{eqnarray*}
So, the critical point is $(x,y) = (-1,-1)$. We introduce variables
\[ u = x+1 \ ,\  v = y + 1 \ . \]
In terms of $(u,v)$ the system becomes 
\begin{eqnarray*}
\frac{du}{dt} & = &-3(u-1) -  (v-1) - 4 = -3u - v \\
\frac{dv}{dt} & = &29(u-1) +  (v-1) +30 = 29u + v
\end{eqnarray*}
We find now the eigenvalues:
\begin{eqnarray*}
\det \left( \begin{array}{ll}
-3-r & -1 \\ 29 & 1-r
\end{array} \right) = 0 & \Rightarrow &
(r-1)(r+3) +29 =0 \Rightarrow \\
r^2 + 2r + 26 = 0 & \Rightarrow & (r+1)^2 + 25 = 0  
\end{eqnarray*}
so that the eigenvalues are 
\[ r_1 = -1+5i \ ,\  r_2 = -1-5i \]
and the point (-1,-1) is a stable spiral: all solutions tend to it
as $t\rightarrow \infty$.

\newpage
\section{Problem 12.2.13}
Classify the critical point at the origin for the system
and sketch the phase plane diagram
\begin{eqnarray*}
\frac{dx}{dt} & = &  x - 6y \\
\frac{dy}{dt} & = & 2x - 7y
\end{eqnarray*}
{\bf Solution:}\\
We find the eigenvalues:
\begin{eqnarray*}
\det \left( \begin{array}{ll}
1-r & -6 \\ 2 & -7-r
\end{array} \right) = 0 & \Rightarrow &
(r+7)(r-1)+ 12 = 0   \Rightarrow \\
r^2 +6r  + 5 = 0 & \Rightarrow & (r+5)(r+1) = 0  
\end{eqnarray*}
so that the eigenvalues are
\[ r_1 = -1 \ ,\  r_2 = -5 \]
and the origin is an unstable saddle point. Now the eigenvectors:
\[ {\bf A} {\bf u}_i = r_i {\bf u}_i \ :\]
or
\[ \left( \begin{array}{rr}
1-r & -6 \\ 2 & -7-r \end{array} \right)
\left( \begin{array}{r} x \\ y \end{array} \right) =
\left( \begin{array}{r} 0 \\ 0 \end{array} \right) \ . \]
\begin{enumerate}
\item Eigenvalue $r_1 = -1$:\\
\[ \begin{array}{rcrcrcr}
 2x   &-&  6y &=& 0 \ ,\\
 2x   &-&  6y &=& 0 \ .
\end{array} \]
The solution is
\[ {\bf u}_1  =  \left( \begin{array}{r} 3 \\ 1 \end{array} \right)\ . \]

\item Eigenvalue $r_2 = -5$:\\
\[ \begin{array}{rcrcrcr}
 6x   &-&  6y &=& 0 \ ,\\
 2x   &-&  2y &=& 0 \ .
\end{array} \]
The solution is
\[ {\bf u}_2  =  \left( \begin{array}{r} 1 \\ 1 \end{array} \right)\ . \]
\end{enumerate}
\input hw25_13.tex
\newpage
\section{Problem 12.2.15}
Classify the critical point at the origin for the system
and sketch the phase plane diagram
\begin{eqnarray*}
\frac{dx}{dt} & = &  x + 2y \\
\frac{dy}{dt} & = & 5x - 2y
\end{eqnarray*}
{\bf Solution:}\\
We find the eigenvalues:
\begin{eqnarray*}
\det \left( \begin{array}{ll}
1-r & 2 \\ 5 & -2-r
\end{array} \right) = 0 & \Rightarrow &
(r+2)(r-1)- 10 = 0   \Rightarrow \\
r^2 + r  - 12 = 0 & \Rightarrow & (r+4)(r-3) = 0  
\end{eqnarray*}
so that the eigenvalues are
\[ r_1 = 3 \ ,\  r_2 = -4 \]
and the origin is an unstable saddle point. Now the eigenvectors:
\[ {\bf A} {\bf u}_i = r_i {\bf u}_i \ :\]
or
\[ \left( \begin{array}{rr}
1-r & 2 \\ 5 & -2-r \end{array} \right)
\left( \begin{array}{r} x \\ y \end{array} \right) =
\left( \begin{array}{r} 0 \\ 0 \end{array} \right) \ . \]
\begin{enumerate}
\item Eigenvalue $r_1 = 3$:\\
\[ \begin{array}{rcrcrcr}
-2x   &+&  2y &=& 0 \ ,\\
 5x   &-&  5y &=& 0 \ .
\end{array} \]
The solution is
\[ {\bf u}_1  =  \left( \begin{array}{r} 1 \\1 \end{array} \right)\ . \]

\item Eigenvalue $r_2 = -4$:\\
\[ \begin{array}{rcrcrcr}
 5x   &+&  2y &=& 0 \ ,\\
 5x   &+&  2y &=& 0 \ .
\end{array} \]
The solution is
\[ {\bf u}_2  =  \left( \begin{array}{r} 2 \\-5 \end{array} \right)\ . \]
\end{enumerate}
\input Phase25a.tex
\end{document}
