The Growth-Reproduction Trade-off

The Euler equation is useful in solving a number of applied problems involving tradeoffs over the lifetime of an individual organism. For example, plants must establish an investment plan for incoming resources. These resources can be used for reproduction, growth, or defense. In the simple problem that I work out here, resources are traded off between their allocation to growth or reproduction. The control parameter $y(x)$ will control the proportion of resources allocated to growth and the remainder, $1-y(x)$, will be used for reproduction.

The most difficult part of many problems in the calculus of variations is in their formulation. Here we must come up with a legimitate model that describes the lifetime of a typical plant. We are given at least a head start by the general equation of life history theory

$$R_{o} = \int_{0}^{\infty} l(x)m(x)dx$$

This equation says that the net reproductive value of an individual $R_{o}$ is equivalent to the integral of the product of the probability that the individual is alive at age $x$, $l(x)$, with the fecundity at age $x$, $m(x)$, taken from time $0$ to $\infty$.

Using this equation, we can now let the product of $1-y(x)$ and $m(x)$ be the amount of reproduction that is realized at a given age $t$. In this way, if all resources are being used strictly for growth, then the value of $m(x)$ grows while actual reproductive output stays at nil.

From data on tree size and reproduction, it is a feasible and often made assumption that seed production is allometric with tree size. Therefore, we adjust $(1-y(x))m(x)$ to $(1-y(x)) \alpha m(x)^{\beta}$, where $\alpha$ and $\beta$ are allometric constants. Now we will define $l(x)$ and $m(x)$ as follows

$$l(x)' = -\mu l(x)$$

where $\mu$ is the death rate at age $x$, such that

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

.

$$m(x)'=a(m_{\infty}-m(x))y(x)$$

where $a$ is the growth rate and $m_{\infty}$ is the maximum size, such that

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

.

We can now define a new term $Y(x)$ to clean up the above equation, letting

$$Y(x)=\int y(x) dx$$

such that

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

.

Now we essentially have the problem in the calculus of variations form. All that remains to be done is to solve it, which we can do using the Euler-Legrange equation that we derived in section 1. However, just in cased you missed it, I will now restate the problem.

Given a function of the form

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

find $y(x) \epsilon [0,1]$ such that $J(y(x))$ is maximized.

Now we apply the Euler-Legrange equation, taking careful note of the fact that we have relabeled our control parameters such that we now have a functional of the form $f(Y(x),y(x);x)$. Also, realize that $Y(x)'=y(x)$ from the fundamental theorem of calculus. Now the Euler-Legrange equation turns the entire problem into puddy in our hands. First, we recognize the functional as

$$f(Y(x),y(x);x) = e^{-\int \mu dx} \alpha (1-e^{-a Y(x)})^{\beta}(1- y(x))$$

Then we write the Euler-Legrange equation:

$$0 = \frac{\partial f}{\partial Y}-\frac{d}{dx} \frac{\partial f}{\partial y}$$

There is but one more thing we must realize in order for our answer to make any sense. It is, to my mind, the particle of genius in the Engen and Saether (1994) paper from which the problem originates. Without it, our constraints on $y(x)$ are meaningless.

If $J(y(x))'>0$ then we must continue to increase $y(x)$. It is little more than replacing the equals sign of the Euler-Legrange equation with a greater than or equal to sign. Our solution can now be piecewise, which it in fact turns out to be. Now

$$0 \leq \frac{\partial f}{\partial Y}-\frac{d}{dx} \frac{\partial f}{\partial y}$$

.

Now we simply do the math.

$$\frac{\partial f}{\partial Y} = e^{-\int \mu dx} \alpha (1-e^{-a Y(x)})^{\beta-1} (1-y(x)) a e^{-a Y(x)} \beta$$

$$\frac{\partial f}{\partial y} = - e^{-\int \mu dx} \alpha (1-e^{-a Y(x)})^{\beta}$$

$$\frac{d}{dx} \frac{\partial f}{\partial y} = \mu e^{-\int \mu... ...u dx} \alpha (1-e^{-a Y(x)})^{\beta-1} a y(x) e^{-a Y(x)} \beta$$

Then

$$\frac{\partial f}{\partial Y}-\frac{d}{dx} \frac{\partial f}{\partial y} =$$

$$e^{-\int \mu dx} \alpha (1-e^{-a Y(x)})^{\beta-1} (1-y(x)) a ... ...u dx} \alpha (1-e^{-a Y(x)})^{\beta-1} a y(x) e^{-a Y(x)} \beta$$

$$\geq 0$$

This simplifies to

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

which is the solution to our problem. It says that $Y(x)$ is to be at its maximal value, i.e. $y(x) = 1$, until

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

.

In this way, the solution is piecewise, and looks more familiar in the form

$$Y(x) = \left\{ \begin{array}{ll} x & x < x_{o} \\ \frac{1... ...ac{\beta a}{\mu(x)}\right) & x \geq x_{o} \end{array} \right.$$

where

$$x_{o} = \frac{1}{a} ln\left(1+\frac{\beta a}{\mu(x)}\right)$$

This solution tells us a number of things. First, the solution is of the bang-bang form unless $\mu(x)$ is decreasing with age. This makes intuitive sense because we might expect trees to be indeterminate growers in situations where increased size further prolongs the reproductive life of the tree. All constant or increasing death rates should give rise to determinate growers. Increasing growth rate, $a$, reduces the age at first reproduction, as does reducing the reproductive sensitivity to size, $\beta$, and increasing the death rate over the lifespan. The age at first reproduction is $x_{o}$.