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.