7.5 Applying guess and check more broadly

As noted earlier, the guess and check method may seem to be trivial - if you have a differential equation, and solution, why verify it? Well, this method helps to introduce a useful solution technique to a differential equation, and one that we can build up through direct verification.

We are going to revisit the lynx hare model, but simplified a little bit. Here we are going to assume that lynx and hares both decline at a rate proportional to the population size, but the lynx population increases according to the rate of hare decline:

\[\begin{equation} \begin{split} \frac{dH}{dt} &= -b H \\ \frac{dL}{dt} & = b H - d L \end{split} \end{equation}\]

Based on these simplified assumptions a good approach is to assume a solution that is exponential for both \(H\) and \(L\):

\[\begin{equation} \begin{split} \tilde{H}(t) &= C_{1} e^{\lambda t} \\ \tilde{L}(t) &= C_{2} e^{\lambda t} \end{split} \end{equation}\]

The form of this solution has three unknowns: \(C_{1}\), \(C_{2}\), and \(\lambda\). If you have taken a course in Linear Algebra, you may recognize that we are assuming the solution is a vector of the form \(\vec{v} = \vec{C} e^{\lambda t}\) . Let’s apply Guess and Check to solve these equations. By differentiation, we have the following:

\[\begin{equation} \begin{split} \frac{d\tilde{H}}{dt} &= \lambda C_{1} e^{\lambda t} \\ \frac{d\tilde{L}}{dt} &= \lambda C_{2} e^{\lambda t}. \end{split} \end{equation}\]

Comparing to our differential equation we can show that

\[\begin{equation} \begin{split} \lambda C_{1} e^{\lambda t} &= - b C_{1} e^{\lambda t} \rightarrow (\lambda +\delta) C_{1} e^{\lambda t} = 0\\ \lambda C_{2} e^{\lambda t} &= b C_{1} e^{\lambda t} - d C_{2} e^{\lambda t} \end{split} \end{equation}\]

Let’s rearrange this expression a little bit:

\[\begin{equation} \begin{split} (\lambda + b) C_{1} e^{\lambda t} &= 0\\ (\lambda + d) C_{2} e^{\lambda t} &=b C_{1} e^{\lambda t} \end{split} \end{equation}\]

Notice that for the second equation we can solve for \(C_{1} e^{\lambda t}\), or \(\displaystyle C_{1} e^{\lambda t} = \frac{(\lambda + d)}{b} C_{2} e^{\lambda t}\).

This allows for something neat to happen. We can substitute this expression for \(C_{1} e^{\lambda t}\) into the first equation:

\[\begin{equation} (\lambda + b) \frac{(\lambda + d)}{b} C_{2} e^{\lambda t} = 0 \end{equation}\]

If we assume that \(b \neq 0\), then we have the following simplified expression:

\[\begin{equation} (\lambda +b) (\lambda +d) C_{2} e^{\lambda t} = 0 \end{equation}\]

Because the exponential function never equals zero, with this new equation, the only possibility is that \((\lambda + b)(\lambda + d)=0\), or that \(\lambda = -b\) or \(\lambda = -d\). Remember: if expressions multiply to zero, then the only possibility is that at least one of them is zero. The process outlined here finds the eigenvalues and eigenvectors of a system of equations. We will study these concepts in Section 18.

Next we need to determine values of \(C_{1}\) and \(C_{2}\). We can do this by going back to the equation \((\lambda + d) C_{2} e^{\lambda t} =b C_{1} e^{\lambda t}\), or \((\lambda + d) C_{2} e^{\lambda t} -b C_{1} e^{\lambda t}=0\) rearranged.

Let’s analyze this equation for each of the values of \(\lambda\):

7.5.1 Case \(\lambda = -d\)

For this situation, we have

\[(-d +d) C_{2} e^{-d t} - b C_{1} e^{-d t} =0 \rightarrow -b C_{1} e^{-d t} =0.\]

The only way for this equation to be consistent and remain zero is if \(C_{1}=0\). We don’t have any restrictions on \(C_{2}\), so the general solution will be

\[\begin{align} \tilde{H}(t) &=0 \\ \tilde{L}(t) &= C_{2} e^{-d t}. \end{align}\]

7.5.2 Case \(\lambda = -d\)

For this situation, we have \((-d +b) C_{2} e^{-d t} - d C_{1} e^{-d t} =0\) which leads to the following equation: \[\begin{equation} \left( (-d +b) C_{2} - d C_{1} \right) e^{-d t} =0 \end{equation}\]

The only way for this equation to be consistent and remain zero is if \(\left( (-d +b) C_{2} - d C_{1} \right)=0\), or if \(\displaystyle C_{2} = \left( \frac{d}{-d + b} \right) C_{1}\). In this case, the general solution will be

\[\begin{align} \tilde{H}(t) &= C_{1} e^{-d t} \\ \tilde{L}(t) &= \left( \frac{d}{-d + b} \right) C_{1} e^{-d t}, \end{align}\]

The parameter \(C_{2}\) can be determined by the initial condition. Notice that we need to have \(d \neq b\) or our solution will be undefined.

Now we can write down a general solution to the system by combining our two solutions together. Here we can you the fact that two solutions can be added together (superposition) to generate a solution.

\[\begin{align} H(t) &= C_{1} e^{-d t} \\ L(t) &= \left( \frac{d}{-d + b} \right) C_{1} e^{-d t} + C_{2} e^{-b t} \end{align}\]

This method only works on linear differential equations (i.e. it wouldn’t work if there was a term such as \(kHL\) in our dynamics. Later on in the course we will look a more systematic method (i.e eigenvalues) to determine solutions to linear systems of equations.