Activities for Systems of ODEs

Section 6.4 Activities for Systems of ODEs

Subsection 6.4.1 Phase Plane Diagrams

Activity 6.4.1.

Draw the phase diagram for this system of differential equation, limited to the first quadrant. Label the trajectories of the regions and make predictions about the behaviour of the solutions. (Only include steady states with \(x \gt 0\) and \(y \gt 0\text{.}\))

\begin{align*} \frac{dp}{dt} \amp = p(q-4)\\ \frac{dq}{dt} \amp = q(p-3) \end{align*}

Solution.

First I calcluate the nullclines. The nullclines here are \(p=0\) (the q-axis), \(q=0\) (the p-axis), the horizontal line \(q=4\) and the vertical line \(p=3\text{.}\) Ignoring the axes, the only steady state is at \((4,3)\text{.}\)

There are four regions and four sections of the nullclines to label. I’ll evaluate at points in these regions and on these nullclines. This evaluation will indicate the direction of the trajectory.

\begin{align*} \amp (1,1) \amp \amp \frac{dp}{dt} = -3 \amp \amp \frac{dq}{dt} = -2 \amp \amp \text{down and left}\\ \amp (5,1) \amp \amp \frac{dp}{dt} = -15 \amp \amp \frac{dq}{dt} = 2 \amp \amp \text{up and left}\\ \amp (1,5) \amp \amp \frac{dp}{dt} = 1 \amp \amp \frac{dq}{dt} = -10 \amp \amp \text{down and right}\\ \amp (a,b) \amp \amp \frac{dp}{dt} = 5 \amp \amp \frac{dq}{dt} = 10 \amp \amp \text{up and right}\\ \amp (1,4) \amp \amp \frac{dp}{dt} = 0 \amp \amp \frac{dq}{dt} = -8 \amp \amp \text{down}\\ \amp (5,4) \amp \amp \frac{dp}{dt} = 0 \amp \amp \frac{dq}{dt} = 8 \amp \amp \text{up}\\ \amp (3,1) \amp \amp \frac{dp}{dt} = -9 \amp \amp \frac{dq}{dt} = 0 \amp \amp \text{left}\\ \amp (3,5) \amp \amp \frac{dp}{dt} = 3 \amp \amp \frac{dq}{dt} = 0 \amp \amp \text{right} \end{align*}

Then I draw these trajectories, which are shown in Figure 6.4.1. The steady state looks like a saddle point. Trajectories approach the steady state from up and left of it, and from down and right of it. The trajectories will be deflected away from the steady state and disperse either down to the left or up to the right.

Figure 6.4.1. First Phase Plane Activity

Activity 6.4.2.

\begin{align*} \frac{dp}{dt} \amp = p(9 - p^2 - q)\\ \frac{dq}{dt} \amp = q(q - 4p + 4) \end{align*}

Solution.

First I calcluate the nullclines. In addition to teh two axes, there is a quadratic nullcline \(q = 9 - p^2\) and a linear nullcline \(q = 4p - 4\text{.}\) The intersection point in the first quadrant is at \((-2 + \sqrt{17}, -12 + 4 \sqrt{17})\text{.}\)

There are four regions and four sections of the nullclines to label. I’ll evaluate at points in these regions and on these nullclines. This evaluation will indicate the direction of the trajectory. I haven’t shown the specific points and arithmetic, but the trajectories are sown in Figure 6.4.2. The steady state looks like a saddle point, with trajectories approaches from down and to the left, and from up and to the right. The trajectories then diverge up and to the left, and down and to the right.

Figure 6.4.2. First Phase Plane Activity

Activity 6.4.3.

\begin{align*} \frac{dp}{dt} \amp = q(q - 3)(q - 6)\\ \frac{dq}{dt} \amp = (q - p^2 - 4)(p + 2 - q) \end{align*}

Solution.

First I calcluate the nullclines. In addition to the p-axis, there are four nullclines. There are two horizontal lines: \(q=3\) and \(q=6\text{.}\) There is a quadratic \(q = p^2 + 4\) and a sloped line \(q = p + 2\text{.}\) There are three intersections of these nullclines in the first quadrant (off the axes).

There are eight regions and ten sections of the nullclines to label. I’ll evaluate at points in these regions and on these nullclines. This evaluation will indicate the direction of the trajectory. I haven’t shown the specific points and arithmetic, but the trajectories are sown in Figure 6.4.3. The steady state at \((4,6)\) looks like a centre; the steady state at \((\sqrt{2},6)\) looks like a saddle point; and the steady state at \((1,3)\) also looks like a saddle point. The trajectories diverge away from the saddle points. If the trajectories start large enough to grow away from the saddle points, it looks like they will eventuall get trapped into circles around \((4,6)\text{.}\) Otherwise, it looks like the trajectories decay down to one of the axes.

Figure 6.4.3. First Phase Plane Activity

Activity 6.4.4.

\begin{align*} \frac{dp}{dt} \amp = q(pq - 5)(pq - 10)\\ \frac{dq}{dt} \amp = p(4p - q)(2p - q) \end{align*}

Solution.

First I calcluate the nullclines. In addition to the axes, there are four nullclines. There are two lines through the origin: \(q = 4p\) and \(q = 2p\text{.}\) There are two reciprocal graphs: \(q = \frac{10}{p}\) and \(q = \frac{5}{p}\text{.}\) There are four intersections of these nullclines in the first quadrant: \((\frac{\sqrt{5}}{2}, 2\sqrt{5})\) \((\sqrt{\frac{5}{2}}, \sqrt{10})\) \((\sqrt{\frac{5}{2}}, 2\sqrt{10})\) and \((\sqrt{5}, 2\sqrt{5})\text{.}\)

There are nine regions and twelve sections of the nullclines to label. I’ll evaluate at points in these regions and on these nullclines. This evaluation will indicate the direction of the trajectory. I haven’t shown the specific points and arithmetic, but the trajectories are sown in Figure 6.4.4. It’s getting harder to make judgements about the steady states from all these arrows, but it looks like \(\left( \sqrt{\frac{5}{2}}, \sqrt{10} \right)\) could be a centre; \(\left( \frac{\sqrt{5}}{2}, 2\sqrt{5} \right)\) could be a saddle point; \(\left( \sqrt{\frac{5}{2}}, 2\sqrt{10} \right)\) could be an unstable focus; and \(\left( \sqrt{5}, 2\sqrt{5} \right)\) could be another saddle point. The trajecoties in and around the saddle points get pretty confusing. I can clearly say that there are growth trajectories up and to the right of all the steady state and divergenc away from the saddle points and the focus is mostly growth to large values instead of decay. Even starting with small values, the diagram shows growth, though the exact path around the steady states is hard to determine.

Figure 6.4.4. First Phase Plane Activity

Activity 6.4.5.

\begin{align*} \frac{dp}{dt} \amp = p((p-4)^2 + (q-4)^2 - 9) \\ \frac{dq}{dt} \amp = q(p - 3)(p - 5) \end{align*}

Solution.

First I calcluate the nullclines. In addition to the axes, there are two vertical lines \(p=3\) and \(p=5\text{.}\) There is also a circle of radius 3 centred at \((4,4)\) given by the equation \((p-4)^2 + (q-4)^2 = 9\text{.}\) There are four steady states where the vertical lines meet the cirlce: \((3, 4 + 2\sqrt{2})\text{,}\) \((5, 4 + 2\sqrt{2})\text{,}\) \((3, 4 - 2\sqrt{2})\text{,}\) and \((5, 4 - 2\sqrt{2})\text{.}\)

There are eight regions and ten sections of the nullclines to label. I’ll evaluate at points in these regions and on these nullclines. This evaluation will indicate the direction of the trajectory. I haven’t shown the specific points and arithmetic, but the trajectories are sown in Figure 6.4.5. The steady state at \((3, 4 - 2\sqrt{2})\) looks like a center and the steady state at \((3, 4 + 2\sqrt{2})\) looks like a steady state. The other two steady states don’t resemble any of the six types, at lest not to my eye. It’s possible that the non-linearity here is severe enough that they patterns around these two steady states are something more exotic. For trajectories, it looks like anything that starts at small values will get caught in a circle around \((3, 4 - 2\sqrt{2})\text{.}\) For anything larger, after some slightly strange pathing near the steady states, all the trajectories seems to show unbounded growth.

Figure 6.4.5. First Phase Plane Activity

Subsection 6.4.2 Linear Systems

Activity 6.4.6.

Solve this linear system using Laplace transforms and classify the steady state. The initial conditions are \(x(0) = 1\) and \(y(0) = 1\text{.}\)

\begin{align*} \frac{dx}{dt} \amp = 5x\\ \frac{dy}{dt} \amp = x + 4y \end{align*}

Solution.

I apply the Laplace transform to the system.

\begin{align*} sX - 1 \amp = 5X\\ sY - 1 \amp = X + 4Y \end{align*}

The I solve for algebraic system for \(X\) and \(Y\) (I haven’t shown the algebra for this, but it is a straightforward isolate and replace of one of the variables.)

\begin{align*} X \amp = \frac{1}{s-5} \\ Y \amp = \frac{1}{s-5} \end{align*}

Then I reverse the transform.

\begin{align*} x(t) \amp = e^{5t}\\ y(t) \amp = e^{5t} \end{align*}

There is outward exponenial growth in both variables with no rotation terms. The steady state at the origin is an unstable focus or a saddle point. Since I’ve only chosen on initial condition, I can’t distinguish: the exponential growth from this point might fit either steady state. Mostly likely this is an unstable focus, since it would be quite unlikely to have chosen initial conditions right along the growth direction of a saddle point, but it is possible.

Activity 6.4.7.

Solve this linear system using Laplace transforms and classify the steady state. The initial conditions are \(x(0) = 1\) and \(y(0) = 1\text{.}\)

\begin{align*} \frac{dx}{dt} \amp = 3x - 6y\\ \frac{dy}{dt} \amp = x + y \end{align*}

Solution.

I apply the Laplace transform to the system.

\begin{align*} sX - 1 \amp = 3X - 6Y\\ sY - 1 \amp = X + Y \end{align*}

The I solve for algebraic system for \(X\) and \(Y\) (I haven’t shown the algebra for this. In solving, there may be linear terms \((s-1)\) which cancel off, depending on how you do the replacement.)

\begin{align*} X \amp = \frac{s-7}{s^2 - 4 + 9} \\ Y \amp = \frac{s-2}{s^2 - 4s + 9} \end{align*}

I’ll adjust the terms to fit the patterns of inverse Laplace transforms.

\begin{align*} X \amp = \frac{s-2}{(s-2)^2 + 5} - \sqrt{5} \frac{\sqrt{5}}{(s-2)^2 + 5} \\ Y \amp = \frac{s-2}{(s-2)^2 + 5} \end{align*}

Then I reverse the transform.

\begin{align*} x(t) \amp = e^{2t} \cos (\sqrt{5}t) - \sqrt{5} e^{2t} \sin (\sqrt{5}t) \\ y(t) \amp = e^{2t} \cos (\sqrt{5}t) \end{align*}

There is outward exponential growth, but there are also rotation terms. The steady state is an unstable node.

Activity 6.4.8.

Solve this linear system using Laplace transforms and classify the steady state. The initial conditions are \(x(0) = 1\) and \(y(0) = 1\text{.}\)

\begin{align*} \frac{dx}{dt} \amp = -4x - 2y \\ \frac{dy}{dt} \amp = 4x + 3y \end{align*}

Solution.

I apply the Laplace transform to the system.

\begin{align*} sX - 1 \amp = -4X - 2Y\\ sY - 1 \amp = 4X + 3Y \end{align*}

The I solve for algebraic system for \(X\) and \(Y\) (I haven’t shown the algebra for this. Depending on how you do the isolation and replacement, there may be a \((s-3)\) linear term which will cancel off.)

\begin{align*} X \amp = \frac{s-5}{s^2 + s - 4}\\ Y \amp = \frac{s+8}{s^2 + s - 4} \end{align*}

I use partial fractions to separate the rational function denominator. I’ll adjust the terms to fit the patterns of inverse Laplace transforms.

\begin{align*} X \amp = \frac{s + \frac{1}{2}}{ \left( s + \frac{1}{2} \right)^2 + \frac{17}{4}} - \frac{11}{\sqrt{17}} \frac{\frac{\sqrt{17}}{2}}{ \left( s + \frac{1}{2} \right)^2 + \frac{17}{4}} \\ Y \amp = \frac{s + \frac{1}{2}}{ \left( s + \frac{1}{2} \right)^2 + \frac{17}{4}} - \frac{15}{\sqrt{17}} \frac{\frac{\sqrt{17}}{2}}{ \left( s + \frac{1}{2} \right)^2 + \frac{17}{4}} \end{align*}

Then I reverse the transform.

\begin{align*} x(t) \amp = e^{\frac{-1}{2}t} \cos \left( \frac{\sqrt{17}}{2} t \right) - \frac{11}{\sqrt{17}} e^{\frac{-1}{2}t} \sin \left( \frac{\sqrt{17}}{2} t \right) \\ y(t) \amp = e^{\frac{-1}{2}t} \cos \left( \frac{\sqrt{17}}{2} t \right) - \frac{15}{\sqrt{17}} e^{\frac{-1}{2}t} \sin \left( \frac{\sqrt{17}}{2} t \right) \end{align*}

There is exponential decay and sinusoidal terms giving rotation. The steady state must be a stable node.

Activity 6.4.9.

Solve this linear system using Laplace transforms and classify the steady state. The initial conditions are \(x(0) = 1\) and \(y(0) = 1\text{.}\)

\begin{align*} \frac{dx}{dt} \amp = -2x + 3y \\ \frac{dy}{dt} \amp = x + 2y \end{align*}

Solution.

I apply the Laplace transform to the system.

\begin{align*} sX - 1 \amp = -2X + 3Y \\ sY - 1 \amp = X + 2Y \end{align*}

The I solve for algebraic system for \(X\) and \(Y\) (I haven’t shown the algebra for this, but it is a straightforward isolate and replace of one of the variables.)

\begin{align*} X \amp = \frac{s + 1}{s^2 - 7} \\ Y \amp = \frac{s + 3}{s^2 - 7} \end{align*}

I use partial fractions to separate the rational function denominator. I’ll adjust the terms to fit the patterns of inverse Laplace transforms.

\begin{align*} X \amp = \left( \frac{\sqrt{7} - 1}{2 \sqrt{7}} \right) \frac{1}{s + \sqrt{7}} + \left( \frac{ 1 + \sqrt{7}}{2\sqrt{7}} \right) \frac{1}{s - \sqrt{7}} \\ Y \amp = \left( \frac{\sqrt{7} - 3}{2 \sqrt{7}} \right) \frac{1}{s + \sqrt{7}} + \left( \frac{ 3 + \sqrt{7}}{2\sqrt{7}} \right) \frac{1}{s - \sqrt{7}} \end{align*}

Then I reverse the transform.

\begin{align*} x(t) \amp = \left( \frac{\sqrt{7} - 1}{2\sqrt{7}} \right) e^{-\sqrt{7}t} + \left( \frac{1 + \sqrt{7}}{2\sqrt{7}} \right) e^{\sqrt{7}t} \\ y(t) \amp = \left( \frac{\sqrt{7} - 3}{2\sqrt{7}} \right) e^{-\sqrt{7}t} + \left( \frac{3 + \sqrt{7}}{2\sqrt{7}} \right) e^{\sqrt{7}t} \end{align*}

There is a mixture of exponential growth and exponential decay in this solution. I can conclude that the steady state must be a saddle point.

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