The Growth-Defense Trade-off

Now we will look at a similar problem where the new control variabe, $y(x)$ is now the proportion of resources that are allocated to growth and $1-y(x)$ are the proportion allocated to defense. We will keep the growth form as it was in the last equation. We now define $l(x)$ and $m(x)$ and then solve the problem as before.

The probability that an individual is alive at some time $x$ is now determined by the differential equation

$$l(x)'=-(\mu(x)-\gamma(1-y(x))l(x)$$

where $\gamma$ is the reduction in death rate that would occur if the tree allocated all it resources to defense, $y(x) = 0$. This equation is solvable

$$l(x) = e^{-\int \mu(x)dx + \gamma(x-Y(x))}$$

This checks with our intuition as well, since allocating everything to growth, $Y(x) = x$, means that our death rate is unchanged. Any reduction in growth leads to a our death rate being reduced.

Size is defined as in the previous section to be

$$m(x)=m_{\infty}(1-e^{-a Y(x) })$$

The problem is to choose $y(x) \epsilon [0,1]$ such that

$$J(y(x)) = \int_{0}^{\infty} e^{-\int \mu(x)dx - \gamma(x-Y(x))} \alpha (1-e^{-aY(x)})^{\beta} dx$$

is maximized.

The Euler-Legrange equation simplifies in this instance to

$$0 \leq \frac{\partial f}{\partial Y}$$

When solved for $Y(x)$, the solution is

$$Y(x) \leq \frac{1}{a} ln\left(1+\frac{\beta a}{\gamma}\right)$$

The solution is again piecewise. First the organism should grow, as rapidly as it can, and then at some age, $x_{o} = \frac{1}{a} ln\left(1+\frac{\beta a}{\gamma}\right)$, it should switch completely over to defense. The age of the switch is reduced by increasing $a$ or $\gamma$, and pushed further back in time by increases to reproductivity sensitivity to size, $\beta$.