A model for the spruce budworm population

Question

A model for the spruce budworm population $u(t)$ is governed by $$\frac{du}{dt}=ru\left(1-\frac{u}{q}\right)-\frac{u^2}{1+u^2}$$ where $r,q$ are positive dimensionless parameters. The nonzero stedy states are thus given by the intersection of the two curve: $$U(u)=r\left(1-\frac{u}{q}\right), \ \ \ V(u)=\frac{u}{1+u^2}$$

Show, using the conditions for a double root, that the curve in $r,q$ space which divides it into regions where there are $1$ or $3$ positive steady states is given parametrically by: $$r=\frac{2a^3}{(1+a^2)^2}, \ \ \ q=\frac{2a^3}{a^2-1}$$

What mean "conditions of double roots"? How can I find the parametrization?

Thank you for your help.

Answer 1

We have the system:

$$\tag 1 \frac{du}{dt}=r u\left(1-\frac{u}{q}\right)-\frac{u^2}{1+u^2}$$

where $r, q$ are positive dimensionless parameters.

To find the equilibrium points, we set $(1)$ equal to zero and we see $u=0$ is one and the other two occur where the death rate equals the birth rate (intersection) and is given by:

$$\tag 2 r \left[1 - \dfrac{u}{q} \right] = \left[ \dfrac{u}{1+u^2} \right] $$

As $q$ remains fixed and $r$ increases, the model undergoes stages where the steady states vary from one, two or three equilibrium solutions. We can see this in the following plot (the lines are various values of $r$ and the plot is the RHS of $(2)$). Notice the number of intersections between the curve and each line ($r$ value).

However, the semi-stable solutions occur when the reproduction rate reaches a value where those two curves are tangent. In order to determine these values of tangency, we take the derivative of each curve and find where they are equal. So we have:

$$ \dfrac{d}{du} \left[ r \left(1 - \dfrac{u}{q} \right) \right] = \dfrac{d}{du} \left[ \dfrac{u}{1+u^2} \right]$$

Differentiating and solving for $r$ yields:

$$r = \dfrac{q(u^2-1)}{(1+u^2)^2}$$

Substituting this back into $(2)$, yields

$$q = \dfrac{2u^3}{u^2-1}$$

Substituting this back into $(2)$ again, yields:

$$r = \dfrac{2u^3}{(1+u^2)^2}$$

We can parameterize this (let $u$ be the parameter $a$) as:

$$r=\dfrac{2a^3}{(1+a^2)^2}, \ \ \ q=\dfrac{2a^3}{a^2-1}$$

I will do one example, but you should try others for various values of $r$. We will let $r = \dfrac{1}{2}, q = 10$, and our analysis above tells us we should have three equilibrium.

The equilibrium points are given by:

$$ \dfrac{1}{2}\left(1 - \dfrac{u}{10} \right) = \dfrac{u}{1 + u^2} $$

This yields:

$$u = 4 - \sqrt{11}, ~~2, ~~4 + \sqrt{11}$$

If we draw a direction field plot, we should see these three points (two stable and one unstable).

As another example, for $r = 0.38$, we get a stable equilibrium solution and a semi-stable equilibrium solution. A direction field plot shows these.

If you draw a figure of the qr-plane, you will see that it is divided into regions where we have exactly one or three equilibrium points based on $r$ and $q$ for the paramaterization.

Answer 2

The equation $U(u)=V(u)$ leads to an equation $P(u)=0$ where $P$ is a polynomial of degree three. There are here cases:

1. There are three different real roots, and therefore three steady states
2. There are two different real roots, one simple and one doble
3. There is one real root and two complex conjugate roots, and only one steady state.

The values of $q$ and $r$ such that the second case happens delimit the different regions of parameters with one or three steady states. To fin them you use the conditions of double roots:

$x$ is a double root of the equation $P(u)=0$ if $P(x)=P'(x)=0$ and $P''(x)\ne0$.

$P'$ is of degree two, so you will be able to solve $P'(u)=0$. Plug one of the solutions (you will have to decide which one) into $P(u)=0$ and you will get an equation in $r$ and $q$, whose solution you will have to parametrize.