Chapter 11 Lesson 36: Systems of ODEs I

11.1 Objectives

  1. Understand that a system of ODEs is a collection of multiple ODEs with a single input and multiple outputs.

  2. Understand that the solution of a system of ODEs is a vector of functions, rather than a single function.

  3. Given a system of two first‐order ODEs and a proposed solution, determine if the proposed solution satisfies the system.

  4. Given a system of two first‐order ODEs, find all equilibrium solutions of the system algebraically by hand or using R.

11.2 Note

The official syllabus says that we should find equilibrium solutions by hand. That is a typo – using R is fine also.

11.4 In Class

  1. Systems of ODEs. Introduce systems of ODEs. Just like with systems of equations, you want functions that satisfy both equations.

For example, if a system of equations is \[x+y=10\] \[x-y=0\] you want to find numbers, \(x\) and \(y\) with the property that both \(x+y=10\) and \(x-y=0\).

Similarly, if the system of ODEs is \[x'=y\] and \[y'=x+4\] you are looking for two functions \(x\) and \(y\), that satisfy both equations.

  1. General Solutions. Once we move to systems of ODEs, our general solutions will have more constants in them. This is a fact, you can try to motivate that fact or not – I would not. In general, there are as many constants in the general solution as there are equations in the system. We will stick to two equations in the system.

  2. Verifying Proposed Solutions. Spend some time verifying that proposed solutions (which are now pairs of functions) are indeed solutions to the system of ODEs.

  3. Initial conditions. Now that we have multiple constants in our general solution, we’ll need to provide multiple initial conditions in order to pin down a specific solution. It is fine if they use findZeros to solve for the constants needed to solve the initial conditions.

  4. Equilibrium Solutions. Very analogous so equilibrium solutions for scalar ODEs, equilibrium solutions are pairs of functions that solve the system of ODEs and are both constant. Spend some time finding equilibrium solutions. It is completely fine to rely on findZeros here.

11.5 R Commands

findZeros

11.6 Problems and Activities

This is a pretty full lesson, you’ll need to spend your time wisely.

  1. Start with a system of ODEs, and verifying a proposed solution. Examples 1-4 provide this aplenty. Only spend enough time here to make sure that they get the idea that solutions to systems of ODEs are pairs of functions. Perhaps 2 examples.

  2. Initial Conditions. Again, do just barely enough here. This doesn’t even align with an objective, but it does help remind the students that we are now dealing with pairs of functions. Example 5 is a good one, or add some initial conditions to any of Example 1-4. I would do 1-2 examples here.

  3. Now move on to equilibrium solutions. Analogously to finding equilibrium solutions for scalar ODE, you are now looking for solutions to the system (so, pairs of functions) that are both constant. If both functions are constant, then both of their derivatives are zero.

    For instance, consider the system of ODEs \[\begin{aligned} x'&=y-4\\ y'&=x^2-9\end{aligned}\]

    If we have two constant functions (equilibrium solutions) that solve this system, then \[\begin{aligned} 0&=y-4\\ 0&=x^2-9\end{aligned}\]

    This one is pretty easy to solve, there are two solutions: \(x=3,y=4\) and \(x=-3 ,y=4\). However, it is easy to make up much harder problems. It is fine to use findZeros to solve these problems.

    findZeros(c(y-4,x^2-9)~x&y)
    ##    x y
## 1 -3 4
## 2  3 4

  4. Now move on to more examples. Example #6 is a predator-prey example. Focus on finding the equilibrium solutions, and then interpreting the equilibrium solutions as populations. If the population ever manages to hit these states, then the population will never change. Example 7 is a competing species example. Interpret it similar to above.

  5. Example 8 is a classic pendulum problem. I particularly like this one, though I expect it is more difficult for students to interpret. Don’t shy away! For this problem I would start by asking students, based on physical intuition, what they expect the equilibrium solutions to be. That is, what configuration could the pendulum get into such that the pendulum will never move from the configuration? Someone will surely say hanging straight down and at rest. That gets you \(a=0\), \(v=0\). Next, ask about \(a=2\pi\), \(v=0\), is that just as good? As far as the equations are concerned, yes, that’s just as good (but not a physically different configuration). Finally, after some prompting, hopefully someone will say that the pendulum might be balanced upside down. That’s correct, and corresponds to \(a=\pi\), \(v=0\). Finally, use findZeros (or trigonometry) to solve for the equilibrium solutions.

  6. Example 9 or any of the exercises at the end of the chapter, as time allows.