Chapter 3.7, Problem 1E

Interpretation Introduction

Interpretation:

To show there is always unstable fixed point at ${\text{x}}^{\text{*}}\text{ = 0}$ for the given equation

Concept Introduction:

To check the stability at any point, differentiate the equation and put the value in it.

Quotient Rule for differentiation ${\left(\frac{\text{f}}{\text{g}}\right)}^{\text{'}}=\frac{\text{f}\text{'}\text{g}-\text{g}\text{'}\text{f}}{{\text{g}}^{\text{2}}}$ and other basic rules of differentiation.

For small value of x predation is very weak, thus the budworm population increases exponentially, when $\text{x}\to 0$

Answer to Problem 1E

Solution:

Instability at ${\text{x}}^{\text{*}}\text{ = 0}$ is proved below.

Explanation of Solution

Given information:

$\frac{\text{dx}}{\text{dτ}}\text{ = rx }\left(\text{1}-\frac{\text{x}}{\text{k}}\right)-\frac{{\text{x}}^{\text{2}}}{{\text{1 + x}}^{\text{2}}}$

The given expression is dimensionless for budworm population growth

$\frac{\text{dx}}{\text{dτ}}\text{ = rx }\left(\text{1}-\frac{\text{x}}{\text{k}}\right)-\frac{{\text{x}}^{\text{2}}}{{\text{1 + x}}^{\text{2}}}$

Where $\text{τ}$ is the dimensionless time, r is the growth rate, and k is the carrying capacity

To check the stability of the equation at ${\text{x}}^{\text{*}}\text{ = 0}$, differentiate the equation with respect to x and put $\text{x = 0}$

$\frac{\text{d}}{\text{dx}}\left(\text{rx }\left(\text{1- }\frac{\text{x}}{\text{k}}\right)\text{ - }\frac{{\text{x}}^{\text{2}}}{{\text{1 + x}}^{\text{2}}}\right)$

$\frac{\text{d}}{\text{dx}}\left(\text{rx - }\frac{{\text{rx}}^{\text{2}}}{\text{k}}\text{ - }\frac{{\text{x}}^{\text{2}}}{{\text{1 + x}}^{\text{2}}}\right)=\text{r}-\frac{\text{2rx}}{\text{k}}-\frac{\text{2x}}{{\left({\text{1+x}}^{\text{2}}\right)}^{\text{2}}}$

By substituting $\text{x = 0}$ in the above equation

$\frac{\text{d}}{\text{dx}}\left(\text{rx }\left(\text{1- }\frac{\text{x}}{\text{k}}\right)\text{ - }\frac{{\text{x}}^{\text{2}}}{{\text{1 + x}}^{\text{2}}}\right)=\text{r}$

Given condition is $\text{ r > 0}$, therefore $\text{x = 0}$ is always an unstable fixed point.

This result is also verified by plotting graph for equation

$\dot{\text{x}}\text{ = rx }\left(\text{1}-\frac{\text{x}}{\text{k}}\right)-\frac{{\text{x}}^{\text{2}}}{{\text{1 + x}}^{\text{2}}}$

From the above graphs, it is clear that for $\text{r > 0}$, there is always an unstable point at $\text{x = 0}$

Conclusion

For the given type of system, there is always an unstable fixed point at $\text{x = 0}$