| Home |
| Click on the diagram for further details. |
If $\tau=\delta=0$, there are two possibilities. The most
obvious one is that $A$ is just the zero matrix, in which case the differential
equation is just $\dot{x}=0$ so the solutions are constant and the solution
curves are just points.
The other possibility is that $A$ is a matrix like $A=\left[\begin{array}{cc} 0&1 \\ 0&0\end{array}\right]$. The only eigenvalue is $0$, and the only eigenvectors are multiples of $v_1=\left[\begin{array}{cc} 1 \\ 0\end{array}\right]$. We cannot find two linearly independent eigenvectors, so the matrix cannot be diagonalized. The differential equation $\left[\begin{array}{cc} \dot{x} \\\dot{y}\end{array}\right] = \left[\begin{array}{cc} 0&1 \\ 0&0\end{array}\right] \left[\begin{array}{cc} x \\ y\end{array}\right] $ gives $\dot{x}=y$ and $\dot{y}=0$, so the solutions are like $x=x_0+y_0t$ and $y=y_0$ for some constants $x_0$ and $y_0$. We call this a shear flow.
The matrix $A$ gives a centre if $\tau=0$ and
$\delta<0$, so the eigenvalues are of the form $\lambda_1=i\omega$
and $\lambda_2=-i\omega$ for some real number $\omega>0$. If so, the
solution curves are closed ellipses centred at the origin. They run
anticlockwise if $b>0$, and clockwise if $b<0$.
If $\tau<0$ and $\delta=\tau^2/4$, then there are two
possibilities. The most obvious one is that $A=\lambda I$ for some $\lambda<0$, so
$\tau=2\lambda$ and
$$ \text{det}(A)=\lambda^2=(2\lambda)^2/4=\tau^2/4. $$
The differential equation
$\left[\begin{array}{cc} \dot{x} \\\dot{y}\end{array}\right] =
A \left[\begin{array}{cc} x \\ y\end{array}\right]
$ is just $\dot{x}=\lambda x$ and $\dot{y}=\lambda y$, so
$\left[\begin{array}{cc} x \\ y\end{array}\right] =
e^{\lambda t}\left[\begin{array}{cc} x \\ y\end{array}\right]
$. The solution curves are straight lines running from infinity in to the
origin. We call this an improper stable node.
The other possibility is that $A$ is a matrix like $A=\left[\begin{array}{cc} \lambda&1 \\ 0&\lambda\end{array}\right]$, where again $\lambda<0$. The only eigenvalue is $\lambda$, and the only eigenvectors are multiples of $v_1=\left[\begin{array}{cc} 1 \\ 0\end{array}\right]$. We cannot find two linearly independent eigenvectors, so the matrix cannot be diagonalized. The differential equation $\left[\begin{array}{cc} \dot{x} \\\dot{y}\end{array}\right] = \left[\begin{array}{cc} \lambda&1 \\ 0&\lambda\end{array}\right] \left[\begin{array}{cc} x \\ y\end{array}\right] $ gives $\dot{x}=\lambda x+y$ and $\dot{y}=\lambda y$. One can check that the solutions are like $x=(x_0+y_0t)e^{\lambda t}$ and $y=y_0e^{\lambda t}$. We call this a degenerate stable node.
If $\tau>0$ and $\delta=\tau^2/4$, then there are two
possibilities. The most obvious one is that $A=\lambda I$ for some $\lambda>0$, so
$\tau=2\lambda$ and
$$ \text{det}(A)=\lambda^2=(2\lambda)^2/4=\tau^2/4. $$
The differential equation
$\left[\begin{array}{cc} \dot{x} \\\dot{y}\end{array}\right] =
A \left[\begin{array}{cc} x \\ y\end{array}\right]
$ is just $\dot{x}=\lambda x$ and $\dot{y}=\lambda y$, so
$\left[\begin{array}{cc} x \\ y\end{array}\right] =
e^{\lambda t}\left[\begin{array}{cc} x \\ y\end{array}\right]
$. The solution curves are straight lines running from the origin out to infinity.
We call this an improper unstable node.
The other possibility is that $A$ is a matrix like $A=\left[\begin{array}{cc} \lambda&1 \\ 0&\lambda\end{array}\right]$, where again $\lambda>0$. The only eigenvalue is $\lambda$, and the only eigenvectors are multiples of $v_1=\left[\begin{array}{cc} 1 \\ 0\end{array}\right]$. We cannot find two linearly independent eigenvectors, so the matrix cannot be diagonalized. The differential equation $\left[\begin{array}{cc} \dot{x} \\\dot{y}\end{array}\right] = \left[\begin{array}{cc} \lambda&1 \\ 0&\lambda\end{array}\right] \left[\begin{array}{cc} x \\ y\end{array}\right] $ gives $\dot{x}=\lambda x+y$ and $\dot{y}=\lambda y$. One can check that the solutions are like $x=(x_0+y_0t)e^{\lambda t}$ and $y=y_0e^{\lambda t}$. We call this a degenerate unstable node.
If $\tau<0$ and $\delta<\tau^2/4$ then we have a
stable node. The eigenvalues
$\lambda_1=(\tau+\sqrt{\tau^2-4\delta})/2$ and
$\lambda_2=(\tau-\sqrt{\tau^2-4\delta})/2$ are both
negative real numbers, and we can find corresponding eigenvectors $v_1$
and $v_2$ that have real entries. Let $V_1$ be the set of multiples of $v_1$,
and let $V_2$ be the set of multiples of $V_2$, so $V_1$ and $V_2$ are lines
through the origin. The solution curves move quickly towards the line $V_1$,
and then move more slowly along $V_1$ towards the origin. The formulae
involve $e^{\lambda_1 t}$ and $e^{\lambda_2 t}$; these tend to zero as $t$
increases, because $\lambda_1$ and $\lambda_2$ are negative.
If $\tau>0$ and $\delta<\tau^2/4$ then we have an
unstable node. The eigenvalues
$\lambda_1=(\tau+\sqrt{\tau^2-4\delta})/2$ and
$\lambda_2=(\tau-\sqrt{\tau^2-4\delta})/2$ are both
positive real numbers, and we can find corresponding eigenvectors $v_1$
and $v_2$ that have real entries. The solution curves run from the origin out
to infinity. The formulae involve $e^{\lambda_1 t}$ and $e^{\lambda_2 t}$;
these tend to infinity as $t$ increases, because $\lambda_1$ and $\lambda_2$
are positive.
If $\tau<0$ and $\delta=0$ then we say that the system is
semistable. One eigenvalue is $\lambda_1=0$, the other is
$\lambda_2=\tau<0$, and we can find corresponding eigenvectors
$v_1$ and $v_2$. Let $V_1$ be the set of multiples of $v_1$,
and let $V_2$ be the set of multiples of $V_2$, so $V_1$ and $V_2$ are lines
through the origin. The solution curves are straight lines parallel to $V_2$.
They approach $V_1$ as $t$ tends to infinity, but they never actually reach $V_1$.
The formulae have some constant terms and some terms involving $e^{\lambda_2 t}$.
The $e^{\lambda_2t}$ terms tend to zero as $t$ increases, because $\lambda_2<0$.
If $\tau>0$ and $\delta=0$ then we say that the system is
semiunstable. One eigenvalue is $\lambda_1=0$, the other is
$\lambda_2=\tau>0$, and we can find corresponding eigenvectors
$v_1$ and $v_2$. Let $V_1$ be the set of multiples of $v_1$,
and let $V_2$ be the set of multiples of $V_2$, so $V_1$ and $V_2$ are lines
through the origin. The solution curves are straight lines parallel to $V_2$
that run out to infinity away from $V_1$. The formulae have some constant terms
and some terms involving $e^{\lambda_2 t}$. The $e^{\lambda_2t}$ terms tend to
infinity as $t$ increases, because $\lambda_2>0$.
If $\delta<0$ then we have a saddle. The
eigenvalue $\lambda_1=(\tau+\sqrt{\tau^2-4\delta})/2$
is a positive real number, and the eigenvalue
$\lambda_2=(\tau-\sqrt{\tau^2-4\delta})/2$ is a
negative real numbers. We can find corresponding eigenvectors $v_1$
and $v_2$ that have real entries. Let $V_1$ be the set of multiples of $v_1$,
and let $V_2$ be the set of multiples of $V_2$, so $V_1$ and $V_2$ are lines
through the origin. There are two solution curves that run out from the origin
to infinity along $V_1$, one in each direction. There are two more solution
curves that run in from infinity to the origin along $V_2$. Between $V_1$ and
$V_2$ there are solution curves that come in from infinity and then turn around
and head back out to infinity in a different direction. The formulae involve
$e^{\lambda_1 t}$ and $e^{\lambda_2 t}$. As $t$ tends to $-\infty$, the
$e^{\lambda_1 t}$ terms tend to zero and the $e^{\lambda_2 t}$ terms tend to
infinity. As $t$ tends to $+\infty$, the
$e^{\lambda_2 t}$ terms tend to zero and the $e^{\lambda_1 t}$ terms tend to
infinity.
If $\tau<0$ and $\delta>\tau^2/4$ then we have a
stable focus. The eigenvalues are $\lambda+i\omega$ and $\lambda-i\omega$,
where $\lambda=\tau/2<0$ and
$\omega=\sqrt{\delta-\tau^2/4}>0$. The corresponding eigenvectors
have complex entries, so we cannot draw them on our usual diagram. The formulae
for the solution involve the expression
$$ e^{\lambda t\pm i\omega t} =
e^{\lambda t}e^{i\omega t} =
e^{\lambda t}(\cos(\omega t) + i\sin(\omega t)).
$$
When we work out all the details we find that the imaginary parts cancel out.
We can therefore write the solution in terms of
$e^{\lambda t}\cos(\omega t)$ and $e^{\lambda t}\sin(\omega t)$, with real
coefficients. As $\lambda<0$, these expressions tend to zero as $t\to\infty$.
The solution curves spiral in from infinity to the origin. They turn
clockwise if $b>0$, and anticlockwise if $b<0$.
If $\tau>0$ and $\delta>\tau^2/4$ then we have an
unstable focus. The eigenvalues are $\lambda+i\omega$ and $\lambda-i\omega$,
where $\lambda=\tau/2>0$ and
$\omega=\sqrt{\delta-\tau^2/4}>0$. The corresponding eigenvectors
have complex entries, so we cannot draw them on our usual diagram. The formulae
for the solution involve the expression
$$ e^{\lambda t\pm i\omega t} =
e^{\lambda t}e^{i\omega t} =
e^{\lambda t}(\cos(\omega t) + i\sin(\omega t)).
$$
When we work out all the details we find that the imaginary parts cancel out.
We can therefore write the solution in terms of
$e^{\lambda t}\cos(\omega t)$ and $e^{\lambda t}\sin(\omega t)$, with real
coefficients. As $\lambda>0$, these expressions tend to zero as $t\to\infty$.
The solution curves spiral out to infinity from the origin. They turn
clockwise if $b>0$, and anticlockwise if $b<0$.
|