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Rewrite the system in matrix form as
$$\diff{\vc{x}}{t} = A\vc{x}$$
where $\vc{x}=(x,y)$ and the matrix $A$ is
(We call such a matrix a diagonal matrix as all entries that aren't on the diagonal are zero.)
Calculate the eigenvalues of $A$. Enter them in increasing order.
$\lambda_1 =$
, $\lambda_2=$
(It seems that the eigenvalues of a diagonal matrix are easy to compute.)
Calculate the eigenvectors of $A$. Enter them so that they correspond with the above eigenvalues.
$\vc{u}_1 = $
, $\vc{u}_2=$
Since these vectors are so simple, for full credit, enter them as unit vectors (i.e., with length 1) and non-negative entries.
The general solution, which we calculated above, is in terms of these eigenvalues and eigenvectors:
$\vc{x}(t) = c_1 e^{ \lambda_1 t } \vc{u}_1 + c_2 e^{\lambda_2 t} \vc{u}_2$
$= \Biggl($
$,$
$\Biggr)$
Hint
To simplify the general solution, plug in numbers for the eigenvalues and the eigenvectors. Then, you can simplify the result by distributing factors like $c_1 e^{\lambda_1 t}$ onto the vectors so that you can add the vectors.
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In terms of the components of $\vc{x}(t) = (x(t),y(t))$ the solution is
$x(t) = $
$y(t) = $
That answer looks like the one we knew we'd get.
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What is the one equilibrium of the dynamical system?
$\vc{x}(t) = $
Let's imagine that $c_1$ and $c_2$ are small, so that the initial condition $\vc{x}(0) = (c_1,c_2)$ is close to the equilibrium. As $t$ increases, will the solution $\vc{x}(t) = (c_1e^{ -1.4 t}, c_2 e^{ 0.3 t})$ move closer or further away from the equilibrium?
Since the solution $\vc{x}(t)$ will move away from the equilibrium for some initial conditions $\vc{x}(0) = (c_1,c_2)$ that are close to the equilibrium, we say the equilibrium is unstable.
Since, for some special initial conditions, the solution does approach the equilibrium, we classify the equilibrium as a saddle. (The term is an analogy to a saddle you might put on a horse. There is a point in the center of a saddle where, if you set a ball carefully right at that point, the ball wouldn't move. This central point is the equilibrium. If you bumped the ball in most directions, it would fall off the side of saddle. But, you could imagine bumping the ball precisely in the direction where the saddle goes up from the center. If you made such a precise change in the ball's position, you could imagine the possibility of it rolling back to the equilibrium.)
Notice how we got a saddle when one eigenvalue $\lambda_1=-1.4$ was negative and the other eigenvalue $\lambda_2=0.3$ is positive. This means that one term from the general solution is shrinking: the $c_1 e^{ -1.4t } \vc{u}_1$ term. The other term from the general solution is growing: the $c_2 e^{ 0.3t } \vc{u}_2$ term.
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Use the following applet to explore the behavior of this linear system with its saddle. The solution is shown by the thick cyan curves. (To help illustrate the behavior, the solution for negative time is also shown by the thin cyan curves.)
The right panel shows curves that should be familiar. There, $x$ is plotted versus $t$ on the top axes (solid curve) and $y$ is plotted versus $t$ on the bottom axes (dashed curve). With the parameters from this problem (which is the default for this applet), $x(t)$ shows exponential decay (as $x(t)=c_1 e^{ -1.4 t}$) and $y(t)$ shows exponential growth (as $y(t)=c_2e^{ 0.3 t}$). You can move the slider for $t$ (or press the play bottom in the lower left of one of the panels) so that red points illustrating $x(t)$ and $y(t)$ move along the curves.
The left panel shows the phase plane with the state variable $x$ plotted versus the state variable $y$. There is no representation of time $t$ in the phase plane. The cyan solution trajectory
in the phase plane is the curve that the point $(x(t),y(t))$ traces out as $t$ increases. To help visualize time, you can change the value of $t$ and observe how the red point $(x(t),y(t))$ changes.
You can change the initial condition $\vc{x}(0)=(x(0),y(0))$ by dragging a cyan point in either the phase plane (to change both $x(0)$ and $y(0)$ simultaneously) or in the plots versus time (to change either $x(0)$ or $y(0)$ separately).
Eigenvalues and eigenvectors
The eigenvalues of \[{}\] are `{}` and `{}`.
The corresponding eigenvectors are `{}` and `{}`.
(The applet calculates only real eigenvectors. If the eigenvectors are complex or there is only one eigenvector, it will display a “?” in place of an eigenvector.)
(Hide)
Solution
For the linear dynamical system \[{}\] the solution is \[{}\]
(Hide)
The applet also automatically calculates the eigenvalues and eigenvectors, as well as show the formula for the solution, given the initial condition determined by the cyan points (or via the control panel).
Click the “show eigenvectors” box to reveal how the eigenvectors $\vc{u}_1$ and $\vc{u}_2$ point in the direction of $x$-axis and the $y$-axis, respectively. Arrows on the eigenvectors point toward the origin if the corresponding eigenvalue is positive.
Click the “show vector” box to show a vector from the origin to the point $(x(t),y(t))$. In this way, you can see the direction of the point even if it moves off the window. For most initial conditions, in what direction does the vector $(x(t),y(t))$ point after $t$ gets large? In the direction of the vector
. (You can increase $t$ as large as 10 using the slider. If you want to try even larger values of $t$, you can change this bound in the control panel.)
Can you find an initial condition for which $(x(t),y(t))$ tends toward the equilibrium as $t$ gets large? One such initial condition is $(x(0),y(0)) =$
. (You can't enter the origin $(0,0)$, as that's too easy and misses the point.)
You should have found that the solution usually ends up pointing in the direction of the eigenvector $\vc{u}_2$ with the positive eigenvalue. Only if the initial condition is exactly along the eigenvector $\vc{u}_1$ (which corresponds to $c_2=0$) does the solution head to the origin.
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What changes if you change the sign of the lower right entry of $A$? Then
$$A = \begin{bmatrix} -1.4& 0\\0 & -0.3\end{bmatrix},$$
and the system of equations becomes
\begin{align*}
\diff{x}{t} &= -1.4x\\
\diff{y}{t} &= -0.3y
\end{align*}
with solution
$x(t) = $
$y(t) = $
(Use $c_1$ and $c_2$ for the constants, as before.)
The eigenvectors remain the same and the eigenvalues become $\lambda_1=$
and $\lambda_2=$
. Since both eigenvalues are negative, will the solution $\vc{x}(t)$ move closer or further away from the equilibrium at the origin as $t$ increases?
Since for all nearby initial conditions, the solution moves toward the origin, we say that the origin is stable. Since the solution moves directly toward the origin (as opposed to spiral into the origin), we classify the equilibrium as a stable node. (Some folks call it a sink.)
Be sure to familiarize yourself with the behavior of the stable node by changing the matrix in the above applet using the control panel.
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We will see yet different behavior if you make both diagonal entries of $A$ be positive. For the matrix
$$A = \begin{bmatrix} 1.4& 0\\0 & 0.3\end{bmatrix},$$
the system of equations becomes
\begin{align*}
\diff{x}{t} &= 1.4x\\
\diff{y}{t} &= 0.3y
\end{align*}
with solution
$x(t) = $
$y(t) = $
(Use $c_1$ and $c_2$ for the constants, as before.)
Now the eigenvalues are $\lambda_1=$
and $\lambda_2=$
. Since both eigenvalues are positive, will the solution $\vc{x}(t)$ that starts near the equilibrium move closer or further away from the equilibrium at the origin as $t$ increases?
Since for all nearby initial conditions, the solution moves away from the origin, we say that the origin is unstable. Since the solution moves directly away from the origin (as opposed to spiral away from the origin), we classify the equilibrium as an unstable node. (Some folks call it a source.)
Be sure to familiarize yourself with the behavior of the unstable node by changing the matrix in the above applet using the control panel.