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\begin{center}
{\bf ASSIGNED EXERCISES OF MATH544: \\Stability and Bifurcations: set  10\\}
{\it May, 2001}
\end{center}



\vspace{0.3cm}

\noindent
{\bf STABILITY AND BIFURCATIONS: One dimensional case}

\vspace{0.3cm}

\noindent
Classification of points of $f(\mu,u)=0$

\vspace{0.2cm}

\noindent
1.\, {\it Regular point} $P_{0}$ is the point where the implicit function
theorem works: $f_{\mu}(P_{0})=f_{u}(P_{0})=0$. Hence there exists a unique
curve passing through this point. \\
2.\, {\it A regular turning point} $P_{0}$ where the $\alpha$ indicator or
the slope of the tangent at $P_{0}$ changes its sign. This means that
the stability is different on both sides of this point\\
3.\,{\it Singular point} $P_{0}$ is a point at which
$f_{\mu}(P_{0})=f_{u}(P_{0})=0$.\\
4.\,{\it A double point} $P_{0}$ is a singular point through which pass
two and only two branches of $f=0$ possessing distinct tangents.
We shall assume that  $f_{uu}(P_{0})$, $f_{\mu u}(P_{0})$,
and $f_{\mu \mu}(P_{0})$ are not vanishing simultaneously at the double point.\\
5.\,{\it A singular turning (double) point} $P_{0}$ is a double point
at which ${ d\mu \over du}$ changes sign on one branch.\\
6.\,{\it A cusp point} $P_{0}$ is a point of second order contact between two branches
of the curve. The two branches have the same tangent at a cusp point.\\
7.\,{\it A conjugate point}  $P_{0}$ is an isolated point.\\
8.\.{\it higher-order singular point} $P_{0}$ is a singular point
at which all second partial derivatives vanish simultaneously.

\vspace{0.2cm}

\noindent
Summary of what we did so far.

\vspace{0.3cm}

\noindent
{\bf Theorem 1}.\, {\it Let $u_{0}$ be an equilibrium  solution of

\begin{equation}
{du \over dt}=f(\mu, u), ~~~ t>0,
\end{equation}

\noindent
where $\mu$ is a parameter, and assume that
$$f(\mu, u_{0}+\bar{u})=f_{u}(\mu, u_{0}) \bar{u}+ R(u_{0},\bar{u})$$
where the remainder satisfies $|R(u_{0},\bar{u})| \le K |\bar{u}|^2 $ for
$\bar{u}$ sufficiently small where $K$ is a positive constant. Then $u_{0}$
is asymptotically stable if $\alpha=f_{u}(\mu, u_{0}) <0$ and
unstable if $\alpha > 0$.}

\vspace{0.3cm}

\noindent
Hence the indicator $\alpha$ is important in the study of stability. At the
critical points defined above the determination of the $\alpha$ indicator
depend on the slopes of the tangents about these points on the branches of
equilibrium solutions. An important theorem (the implicit function theorem)
for such a purpose is given by

\vspace{0.3cm}

\noindent
{\bf Theorem 2}.\, {\it  Let $f(\mu, u)$ be a continuously differentiable 
function
in a region $U$ of the $\mu u$ plane containing the point
$P_{0}=(\mu_{0}, u_{0})$. If $f(P_{0})=0$ and $f_{u}(P_{0}) \ne 0$, then
there is a rectangle $R$ , defined by $R=\{(\mu,u)| ~ |u-u_{0}| < a,~
|\mu- \mu_{0}| < b \}$ contained in $U$ such that

\noindent
(i)\,\, The equation $f(\mu, u)=0$ has a unique solution $u=u(\mu)$ on $R$.
(ii)\, The function $u(\mu)$ is continuously differentiable on the
interval $|\mu-\mu_{0}| < b$ and its derivative is given by
$${ du \over d\mu}=-{f_{\mu}(\mu,u(\mu)) \over f_{u}(\mu, u(\mu))}$$
}

\vspace{0.3cm}

\noindent
On the regular points of the bifurcating curves we can use above theorems
and find the $\alpha$ indicator and investigate the stability of the model.
Let us now assume that a point $P_{0}$ on the branches is a singular point.
To study the stability about such points we have the following theorem. First
we need the slopes of the curves about these points. For this purpose we have
the following results.

\vspace{0.3cm}

\noindent
{\bf Theorem 3}.\, {\it Let $P_{0}$ be a double point of $f(\mu, u)=0$.
Then either

\vspace{0.2cm}

\noindent
(i)~~ $f_{\mu \mu}(P_{0}) \ne 0$ and two tangents are given by

\begin{equation}
{d\mu \over du}=-{f_{\mu u} \over f_{\mu \mu}} \pm {1 \over |f_{\mu \mu}|}
\sqrt{\Delta } \label{slp0}
\end{equation}

\noindent
where
$$\Delta={f_{\mu u}}^2-f_{\mu \mu}\, f_{uu}$$
}

\vspace{0.3cm}

\noindent
{\bf Remark}:\, If $\Delta > 0$ then the curves intersecting at $P_{0}$
have different tangents. If $\Delta=0$ then these curves have the same
tangent line at the point $P_{0}$. If $\Delta < 0$ then there exists no
tangent line passing through $P_{0}$ such points are called
{\it isolated points}.

\vspace{0.2cm}

\noindent
{\it (ii)~ If $f_{\mu \mu}=0$ and the two tangents are given by

\begin{equation}
{du \over d\mu}(P_{0})=0, ~~~and ~~ {d\mu \over du}=-{f_{uu}(P_{0}) \over
2\,f_{\mu u}(P_{0}) }
\end{equation}
}
\vspace{0.3cm}

\noindent
For stability analysis we need the $\alpha$ indicator about the point
$P_{0}$. For this purpose, in the neighborhood of a regular turning point 
we have

\vspace{0.3cm}

\noindent
{\bf Theorem 4}.\, {\it Let $P_{0}$ be a regular turning point of $f=0$. Then
equilibrium solutions on one side are stable and on the other side
are not stable.}

\vspace{0.3cm}

\noindent
{\bf Remark}:\, The $\alpha$ indicator, $\alpha=f_{u}(\mu(u),u)=
-{d\mu \over du}(u)\, f_{\mu}(\mu(u),u)$. Since the slope changes
its sign then $\alpha$ changes its sign on different sides of
of the point $P_{0}$.

\vspace{0.3cm}

\noindent
For double points to find the $\alpha$ indicator we consider
two distinct cases.

\vspace{0.2cm}

\noindent
{\bf case 1}.\, The two curves $\mu=\mu^{+}(u)$ and $\mu =\mu^{-}(u)$
passing through (intersecting at) $P_{0}$ have tangents  given by
(\ref{slp0}). Let the corresponding stability indicators are
$\alpha^{+}$ and $\alpha^{-}$. They are given by

\vspace{0.3cm}

\noindent
{\bf Theorem 5}.\, {\it Let $P_{0}$ be a double point with
$f_{\mu \mu}(P_{0}) \ne 0$. Then

\begin{eqnarray}
\alpha^{+}&=&-{d\mu^{+} \over du}(u)\, [{f_{\mu \mu}(P_{0}) \over
|f_{\mu \mu}(P_{0})|} \sqrt{\Delta}\, (u-u_{0}) +o(|u-u_{0}|)],\\
\alpha^{-}&=&{d\mu^{-} \over du}(u) \, [{f_{\mu \mu}(P_{0}) \over
|f_{\mu \mu}(P_{0})| }\, \sqrt{\Delta}\, (u-u_{0})+o(|u-u_{0}|) ]
\end{eqnarray}
}
\vspace{0.3cm}

\noindent
{\bf case 2}.\, One of the curves is $u=u_{1}(\mu)$ with
${du_{1} \over d\mu}(P_{0})=0$ and other is $\mu=\mu_{2}(u)$ with
${d\mu_{2} \over du}=-{f_{uu}(P_{0}) \over 2\,f_{\mu u}(P_{0})}$.
Their corresponding $\alpha$ indicators are given by

\vspace{0.3cm}

\noindent
{\bf Theorem 6}.\, {\it Let $P_{0}$ be a double point with
$f_{\mu \mu }(P_{0})=0$. Then

\begin{eqnarray}
\alpha^{1}(\mu)&=&{f_{\mu u}(P_{0}) \over |f_{\mu u}(P_{0})|}\, \sqrt{\Delta}
\,(u-u_{0})+o(|u-u_{0}|),\\
\alpha^{2}(\mu)&=&-{f_{\mu u}(P_{0}) \over |f_{\mu u}(P_{0})|}\,
 \sqrt{\Delta}\, (u-u_{0})+ o(|u-u_{0}|).
 \end{eqnarray}

\noindent
where $\alpha^{1}(\mu)=f_{u}(\mu, u_{1}(\mu)$ and
$\alpha^{2}(u)=f_{u}(\mu_{2}(u),u)$ are the stability indicators of the curves
$u_{1}(\mu)$ and $\mu_{2}(u)$}

\vspace{0.3cm}

\noindent
In addition to the {\bf solved problems and exercises given in Logan
page 434-436} solve also the following problems

\vspace{0.3cm}

\noindent
{\bf QUESTIONS}

\vspace{0.2cm}

\noindent
{\bf 1}.\,Check the stability of all equilibrium solutions of the
equation (\ref{slp0}) with

  $$f=u(9-\mu u)(\mu +2u-u^2)[(\mu-10)^2 +(u-3)^2-1]$$

and draw a branching (bifurcation) diagram using the dashed lines
for unstable equilibria.


\vspace{0.2cm}


\noindent
{\bf 2}.\, Consider the integral equation
$$\lambda y(t)={2 \over \pi}\, \int_{0}^{\pi}\, (3 \sin \tau\, \sin t+2 \sin 2\tau \sin 2t)\,
(y(\tau)+y^3 (\tau)) d\tau$$

Solutions $y(t,\lambda)$ of this integral equation depend on the independent
 variable $t$ and on the parameter $\lambda$.

 \vspace{0.2cm}

\noindent
(a)~~ Show that every solution is of the form
$$y=A(\lambda) \sin t+ B(\lambda) \sin 2 t$$
(b)~~ Are there nontrivial solutions for $\lambda=0$?\\
(c)~~Calculate all solutions for $\lambda > 0$ (for $\lambda=10$ there are
nine different solutions).\\
(d)~~ For each solution, calculate
$$[y]=\int_{0}^{\pi}\, y^2(t, \lambda) dt$$
and sketch a branching diagram

\vspace{0.2cm}

\noindent
{\bf 3}.\,{\it Cusp point Bifurcation} At the cusp points it is not possible
in general to obtain $u$ as a function of $\mu$ or $\mu$ as a function of $u$.
We then introduce a new parameter $\eta$ in order to obtain a parametric 
representations $u(\eta)$ and $\mu(\eta)$. Let $P_{0}=(0,0)$ be a singular
point
$$f(P_{0})=0,~~f_{\mu}(P_{0})=f_{u}(P_{0})=0 ~~ and~~ \Delta=0 $$
We assume that the second partial derivatives of $f$ are not vanishing
simultaneously at the point $P_{0}$. Here we assume that $f_{uu}(P_{0}) \ne 0$.
Show that in the neighborhood of the point $P_{0}$ we have

\begin{eqnarray}
\mu= \eta^2 ~~~ or ~~~ \mu=-\eta^2,\\
u={1 \over 2} u_{\eta \eta}\, \eta^2+{1 \over 6}\, u_{\eta \eta \eta}\, \eta^3
O(\eta^4),
\end{eqnarray}

\noindent
{\bf Hint}: The above representations can be obtained by differentiating
$f(\mu(\eta), u(\eta))=0$ with respect $\eta$ in the neighborhood of the 
point $P_{0}$ and letting $u(0)=\eta(0)=0$. For instance differentiating
twice we obtain
$$f_{uu}\, (u_{\eta})^2 + 2 f_{\mu u}\, u_{\eta}\, \mu_{\eta}+
f_{\mu \mu}\, (\mu_{\eta})^2=0$$
and so on.

\vspace{0.3cm}

\noindent
{\bf STABILITY AND BIFURCATIONS: Two dimensional case }

\vspace{0.2cm}

\noindent
Let $x$ and $y$ be functions of $t \in I$ satisfying the system of ODEs

\begin{equation}
{dx \over dt}=P(x,y),~~~ {dy \over dt}=Q(x,y),~~ t\in I \label{two0}
\end{equation}

\noindent
where $P$ and $Q$ are functions of $x,y$ having continuous partial
derivatives of all orders in some domain of the $xy$ plane which
is called the {\it phase plane}.   Let us consider the following
initial value problem

\begin{eqnarray}
\dot{x}=P(x,y),~~~ \dot{y}=Q(x,y), ~~~t \in I\\
x(t_{0})=x_{0},~~ y(t_{0})=y_{0},~~t_{0} \in I
\end{eqnarray}

\noindent
Form the theory of ODEs we have

\vspace{0.3cm}

\noindent
{\bf Lemma 6}.\, {\it If $P$ and $Q$ have continuous partial derivatives
in some domain $D$ the the initial value problem stated above has unique
solutions}

\vspace{0.3cm}
Some definitions

\vspace{0.3cm}

\noindent
1. {\it Phase Plane}. The xy plane.\\
2. {\it Phase variables or state variables }: x and y\\
3. {\it Trajectory or path}. A Curve $C$ defined by
the solution $x=x(t), y=y(t) $ where $t \in I$ of (\ref{two0})\\
4. {\it Phase portrait}. The totality of all paths and critical points
of (\ref{two0}) graphed in the phase plane.\\
5. {\it Critical Points}. All solution of $P(x_{0},y_{0})=Q(x_{0},y_{0})=0$.
Critical points are the equilibrium or steady state solutions of (\ref{two0}).

\vspace{0.3cm}

\noindent
{\bf A corollary of Lemma 6}. {\it There exists only one path passing through
each point of the phase plane.}

\vspace{0.3cm}

\noindent
This means that the paths corresponding to different initial values do not
intersect each other. Critical points are also solutions of (\ref{two0})
which are called the equilibrium solutions. According to the above
Corollary of Lemma 6 none of the trajectories pass through such critical
points. Critical points are called isolated it there exists a neighborhood
which contains only this critical point. Without proof we state
the following results

\vspace{0.3cm}

\noindent
{\bf (a)}. If a path approaches a critical point then $t \rightarrow
\pm \infty$\\
{\bf (b)}.  As $t \rightarrow \pm \infty$ a path approaches
{\it a critical point}, {\it moves on a closed path}, {\it approaches a
closed path}, or {\it leaves every bounded set}

\vspace{0.3cm}

\noindent
{\bf Stability Analysis}. In two dimensional case there stability definitions
are given as follows

\vspace{0.3cm}

\noindent
{\bf Definition}( of stability). Let $u=(x(t),y(t))$ define the state vector
for $t \in I$. Let $P_{0}=(x_{0},y_{0})$ be a critical point and
$u(t_{0})=(x(t_{0}),y(t_{0}),~~ t_{0} \in I$ be the initial value
of the state vector. Then for given positive number $\varepsilon > 0$
there exists a positive number $\delta_{\varepsilon}$ such that
$||u(t)-u_{0}|| \le 0$ for all $t \in I$ whenever
$||u(0)-u_{0}|| \le \delta_{\varepsilon}$

\vspace{0.3cm}

\noindent
This implies that any path starting sufficiently closer to the critical point
remains closer to the point.

\vspace{0.3cm}

\noindent
{\bf Definition}(of asymptotical stability). A critical point is asymptotically
stable if it is stable and if
all paths starting sufficiently closer to this point approach this
point asymptotically (as $t \rightarrow \infty$). It means
$\lim_{t \rightarrow \infty}\,||u(t)-u_{0}||=0$.

\vspace{0.2cm}

\noindent
This means that a critical point is {\it asymptotically stable} if all paths
starting sufficiently closer to it asymptotically reaches that point.
There are four types of Critical point, {\it center}, {\it node},
{\it saddle}, and {\it spiral}.

\vspace{0.3cm}

\noindent
{\bf Remark}. The critical points of center type are stable
but not asymptotically stable


\vspace{0.3cm}

\noindent
{\bf Linearized Stability}. Let $P_{0}=(x_{0},y_{0})$ be a critical point.
Stability of this critical point is investigated by using linear perturbations
about this equilibrium solution, $x=x_{0}+\bar{x}, y=y_{0}+\bar{y}$.
Let us choose $P_{0}=(0,0)$ for simplicity then by using the
differentiability properties of $P$ and $Q$ at all orders we get

\begin{equation}
\dot{x}=a x + b y, ~~~ \dot{y}=c x + d y ,  \label{two2}
\end{equation}

\noindent
where $a,b,c,d$ are arbitrary constants which are given by
$$a=P_{x}(0,0),~~b=P_{y}(0,0),~~c=Q_{x}(0,0),~~d=Q_{y}(0,0)$$
This system of equations (\ref{two2}) may also be written by

\begin{equation}
\dot{u}=A u, \label{two3}
\end{equation}

\noindent
where $A$ is the coefficient matrix in (\ref{two2}) and $u$ is the
column state vector. Let $\lambda_{1}$ and $\lambda_{2}$ be eigenvectors
of the matrix $A$ and $v_{1}$ and $v_{2}$ be the corresponding eigenvectors.
For the solution of (\ref{two3}) we two distinct cases

\vspace{0.3cm}

\noindent
{\bf A}.\, Two eigenvalues are different ($\lambda_{1} \ne \lambda_{2}$).
then the solution is given by

\begin{equation}
u(t)=c_{1} v_{1} e^{\lambda_{1}\,t} + c_{2} v_{2} e^{\lambda_{2}\,t}
\label{two4}
\end{equation}

\noindent
where $c_{1}$ and $c_{2}$ are arbitrary constants to be determined by the
initial conditions.

\noindent
{\bf B}.\, Eigenvalues are the same ($\lambda_{1}=\lambda_{2}$). We have two
sub cases. $(B_{1})$:\, $A=\lambda_{1} I$, then

\begin{equation}
u(t)=[c_{1} v_{1}+c_{2} v_{2}] e^{\lambda_{1}\,t} . \label{two5}
\end{equation}

\noindent
Here $v_{1}$ and $v_{2}$ correspond to two different eigenvectors of
corresponding to the same eigenvalue $\lambda_{1}$. $(B_{2})$:\,
$(A-\lambda_{1} I)^2=0$, then

\begin{equation}
u(t)=[v t+ w] e^{\lambda_{1}\,t}, \label{two6}
\end{equation}

\noindent
where $v=Aw-\lambda_{1}\,w$.

\vspace{0.3cm}

\noindent
The classification of the critical point $(0,0)$ is achieved by  the
eigenvalues of the matrix $A$

\vspace{0.3cm}

\noindent
(1).\, Eigenvalues have the same sign. Critical point is a {\bf node}.
Node is stable if the eigenvalues are negative and unstable if the 
eigenvalues are
positive. Solution is given (\ref{two4}).

\noindent
(2).\, Eigenvalues have opposite sign. Critical point is a
{\bf saddle point}. Solution is given (\ref{two4}).

\noindent
(3).\, Eigenvalues are equal. It is a {\bf stable node} if the
eigenvalue is negative. It is  an {\bf unstable node} otherwise.
The solutions is given in (\ref{two5}) and (\ref{two6}).

\noindent
(4).\, Eigenvalues are complex. Let $\lambda_{1}=\alpha+i\beta$
then $\lambda_{2}=\alpha-i \beta$. Let the eigenvector $v_{1}$
corresponding to eigenvalue be $v_{1}=w+i v$. We have two types.
Path about the critical point is a {\bf spiral} if $\alpha \ne 0$
and {\bf center} type of critical points if $\alpha=0$ where the path
is a closed curve (periodic solutions). The solutions are found from
(\ref{two4}) by replacing the eigenvalues and taking the real part.
We obtain $u(t)=c_{1} u_{1}+c_{2} u_{2}$ where
$$u_{1}=e^{\alpha\,t}(w \cos \beta t - v \sin \beta t)$$
$$u_{2}=e^{\alpha\,t}(w \cos \beta t +v \sin \beta t)$$

\vspace{0.3cm}

\noindent
As a summary  we have the following theorem

\vspace{0.3cm}

\noindent
{\bf Theorem 7}.\, {\it The critical point of the system (\ref{two2})
is stable if and only if , the eigenvalues of $A$ have non positive 
real parts, it is asymptotically stable if and only if the eigenvalues 
have negative real parts.} 

\vspace{0.3cm}

\noindent
So far we investigated the stability of the linear systems (\ref{two2}).
For the nonlinear systems we let

\begin{eqnarray}
\dot{x}&=&a x + b y + f(x,y),\\
\dot{y}&=&c x + d y + g(x,y)   \label{two7}
\end{eqnarray}

\noindent
where $a,b,c,d$ are constant with (i). \,$ad-bc \ne 0$ and (ii).\,
$$\lim_{(x,y) \rightarrow (0,0)}\, {f(x,y) \over \sqrt{x^2+y^2}}=0,~~
\lim_{(x,y) \rightarrow (0,0)} \, {g(x,y) \over \sqrt{x^2+y^2}}=0$$
It is obvious that $(0,0)$ is the critical point. Then we have the following
theorem for the general case with the above conditions

\vspace{0.3cm}

\noindent
{\bf Theorem 8}.\, (Poincare' and Liapunov) {\it Let $(0,0)$ be a critical 
point of the linear
system (\ref{two2}) and the nonlinear system (\ref{two7}) with the
conditions (i) and (ii) stated above. The critical point $(0,0)$ of the
nonlinear system is the same type as of the linear case if

\vspace{0.1cm}

\noindent
a)\, The eigenvalues of $A$ are real, distinct , and have the same
sign (node).\\
b)\, The eigenvalues of $A$ are real and have opposite signs (saddle).\\
c)\, The eigenvalues of $A$ are complex, but not purely imaginary (spiral).

\vspace{0.1cm}

\noindent
In addition if $(0,0)$ is asymptotically stable for the linear system
(\ref{two2}), then it is asymptotically stable for the nonlinear system
(\ref{two7})
}

\vspace{0.3cm}

\noindent
To understand whether the path is a closed curve we have the theorem 
of Bendixon and Dulac

\vspace{0.3cm}

\noindent
{\bf Theorem 9}.\, (Bendixon \& Dulac) {\it If $P_{x}+Q_{y}$ has a fixed sign
in a region of phase plane then (\ref{two7}) can not have a closed path in that region}

\vspace{0.3cm}

\noindent
Assuming there exists a closed a path $C$ and consider the form 
$P dy-Q dy$ and integrating over $C$ one obtains a contardiction.
We have another useful theorems due to Poincare' and Bendixon.

\vspace{0.3cm}

\noindent
{\bf Theorem 10}.\, (Poincare') {\it A closed path of the system (\ref{two7}) 
surrounds at least one critical point of the system.}

\vspace{0.3cm}

\noindent
{\bf Theorem 11}.\, (Poncare' and Bendixon) {\it Let $R$ be a closed and 
bounded region in the phase plane containing no critical points of
 (\ref{two7}). If $C$ is a path of (\ref{two7}) that lies in $R$ for some time 
$t_{0}$ and remains in $R$ for all $t > t_{0}$, then $C$ is either a closed
path or spirals toward a closed path as $t \rightarrow \infty$}

\noindent
{\bf Exercises. In addition to the assigned and solved exercises in
Logan solve also the following problems.}

\vspace{0.2cm}

\noindent
{\bf 1}.\, (Duffing equation). Let $u$ be a function of $t \in I$ satisfying the
second order equation
$$\ddot{u}+\dot{u}-u+u^3=0$$
which is called the Duffing equation. The stability analysis of this equation
is easily studied by letting $x=u,~~y=\dot{u}$. We then obtain the following
nonlinear system

\begin{eqnarray}
\dot{x}&=&P(x,y)=y,\\
\dot{y}&=&Q(x,y)=x-x^3-y
\end{eqnarray}

\noindent
We have the following critical points
$(0,0)$ , $(1,0)$ and $(-1,0)$. For the first critical point we
have $a=0, b=1, c=1, d=-1$ then the eigenvalues are $\lambda_{1,2}=
{1 \over 2}\,(-1 \pm \sqrt{5})$ (calculate the eigenvectors).
This critical point is a {\bf saddle } point. For the other
critical points $(\pm 1, 0)$ we have $a=0, b=1, c=-2, d=-1$. Hence
$\lambda_{1,2}={1 \over 2}\,(-1 \pm i \sqrt{7})$. Hence $ \alpha < 0$,
then these critical points are {\it stable foci} ({\bf spiraling paths})

\vspace{0.3cm}

\noindent
{\bf 2}.\,  Discuss the stability of the equilibrium solutions of the
Van der Pol equation
$$\ddot{u}-\lambda (1-u^2) + \dot{u}+u=0$$
for all possible values of $\lambda$.

\vspace{0.3cm}

\noindent
{\bf 3}.\,Consider the system of ODEs

\begin{eqnarray}
\dot{x}&=&-y+x {1-x^2-y^2 \over \sqrt{x^2+y^2}},\\
\dot{y}&=&x+y {1-x^2-y^2 \over \sqrt{x^2+y^2}}
\end{eqnarray}

\noindent
Show that the unit circle in the pats plane is a stable cycle.
{\bf Hint}. Use polar coordinates.

\vspace{0.3cm}

\noindent
{\bf 4}.\, Consider the ordinary differential system

\begin{eqnarray}
\dot{x}&=&-y+x (x^2+y^2-1),\\
\dot{y}&=&x+y (x^2+y^2-1).
\end{eqnarray}

\noindent
Show that the periodic path is unstable.

\end{document}
