General solution to a Growth equation

Question

I'd like to compute a formula that describes a population growth. The population starts with $N(t=0)$ individuals. At each time step there are births and deaths. The number of births at time $t$ is given by $N(t-1)f$, where $f$ being the fecundity. The number of deaths is given by $N(t-ls)$, where $ls$ being the lifespan. Therefore the recursive equation is: $$ N(t)=N(t-1)+N(t-1) f-N(t-ls) $$ which also equals: $$ N(t)=N(t-1) (f+1)-N(t-ls) $$ What is the general solution? I mean an equation that gives the value $N(t)$ in function of $ls$, $f$ and $N(0)$ only.

Step-by-step solution is more than welcome.

UPDATE:

$ls$ and $t$ are integers.

Answer 1

Generating functions are your friend. I will show how to get your generating function from the recurrence.

I changed your notation a little.

You have $n(t) = a n(t-1)-n(t-m)$, or $n(t+m) = a n(t+m-1)-n(t)$ .

Let $f(x) =\sum_{t=0}^\infty n(t) x^t $.

$x^m f(x) =\sum_{t=0}^\infty n(t) x^{t+m} =\sum_{t=m}^\infty n(t-m) x^{t} $ and $x f(x) =\sum_{t=0}^\infty n(t) x^{t+1} =\sum_{t=1}^\infty n(t-1) x^{t} $, so

$\begin{align} f(x)-ax f(x)+x^m f(x) &=\sum_{t=0}^\infty n(t) x^t -a\sum_{t=1}^\infty n(t-1) x^{t} +\sum_{t=m}^\infty n(t-m) x^{t}\\ &=\sum_{t=0}^{m-1} n(t) x^t -a\sum_{t=1}^{m-1} n(t-1) x^{t} +\sum_{t=m}^\infty (n(t)-an(t-1)+n(t-m)) x^{t}\\ &=\sum_{t=0}^{m-1} n(t) x^t -a\sum_{t=1}^{m-1} n(t-1) x^{t}\\ &= g(t)\\ \end{align} $

where $g(t) =\sum_{nt0}^{m-1} n(t) x^t-a\sum_{t=1}^{m-1} n(t-1) x^{t} =n(0)+\sum_{t=1}^{m-1} (n(t)-an(t-1)) x^t $ is a polynomial of degree $m-1$ that incorporates the initial conditions of the $n(i)$.

Therefore $f(x) =\dfrac{g(x)}{h(x)} $ where $h(x) = 1-ax+x^m$.

The properties of the $n(i)$ thus depend on the initial conditions (incorporated in $g(x)$) and the roots of $h(x)$. In particular, if $h(x)$ has no real roots, the $n(i)$ will oscillate. If $h(x)$ has positive real roots, the $n(i)$ will grow (unless there are special initial conditions) like $c/r^i$, where $r$ is the root of $h$ of smallest value.

I will now try to determine what I can about the roots of $h$.

I will assume that $a > 0$.

$h(0) = 1$.

Since $h'(x) =-a+mx^{m-1} $, $h'(x)=0$ for $x=x_0 =(a/m)^{1/(m-1)} $, so $h'(x) < 0$ for $0 \le x < x_0$ and $h'(x) > 0$ for $x > x_0$. Therefore $h$ has a minimum at $x_0$.

To see if $h$ has any real root, we need to see if $h(x_0) < 0$. If not, $h$ has no real roots.

$\begin{align} h(x_0) &=1-ax_0+x_0^m\\ &=1-x_0(a-x_0^{m-1})\\ &=1-x_0(a-a/m)\\ &=1-(a/m)^{1/(m-1)}a(1-1/m)\\ &=1-(a/m)^{1+1/(m-1)}(m-1)\\ &=1-(a/m)^{m/(m-1)}(m-1)\\ \end{align} $.

Therefore $h(x_0) < 0$ if $(a/m)^{m/(m-1)}(m-1) > 1$ or $a > m/(m-1)^{(m-1)/m}$.

Therefore, for any $m$, if $a$ is large enough, the $n(i)$ will grow exponentially.

Note that if $a = m/(m-1)^{(m-1)/m}$, $h$ has a double root at $x_0$, and the $n(i)$ will grow differently.

I'll take this a little further and leave it at that.

Let's see what happens for moderately large $m$.

$m/(m-1)^{(m-1)/m} =\dfrac{m}{m-1}(m-1)^{1/m} $.

$(m-1)^{1/m} =e^{\ln(m-1)/m} \approx 1+\ln(m-1)/m $ and $\dfrac{m}{m-1} =\dfrac{1}{1-1/m} \approx 1+1/m $, so $m/(m-1)^{(m-1)/m} \approx (1+\ln(m-1)/m)(1+1/m) \approx 1+(1+\ln(m-1))/m $.

Therefore $a$ does not have to be too large for $h$ to have real roots.

That's enough for now.

Answer 2

In principle, this is a simple constant-coefficient difference equation. The problem is that $ls$ can be any positive integer, and this makes the equation difficult to solve specifically. To wit, assume that the solution $N(t) = A r^t$, where $A$ is some constant and $r$ a growth rate. Then we may find $r$:

$$r^t-(f+1) r^{t-1} +r^{t-ls}=0$$

or

$$r^{ls}-(f+1) r^{ls-1}+1=0$$

In general, this equation may be solved numerically for $r$, and produces $ls$ distinct values of $r$ i.e., $r_1, r_2, \ldots r_{ls}$. Then the solution is

$$N(t) = \sum_{k=1}^{ls} A_k r_k^t$$

where the $A_k$ are found from, e.g., initial conditions.

Answer 3

Since your equation is a recurrence relation of order $ls$ (unless $ls=0$, in which case it's of order $0$, but I'll assume that that's not the case), the general solution is going to have $ls$ arbitrary constants (including $N_0$, but not including $f$). I kind of doubt you'll be able to get a "nice", explicit solution, but what I can get you is a "nice" ordinary generating function.

To make notation easier, I'll write $n$ instead of $t$, $N_n$ instead of $N(t)$, $r$ instead of $f+1$, and $a$ instead of $ls$. Then the recurrence relation is $$N_n-rN_{n-1}-N_{n-a}=0.$$ Define $F(x):=\sum_{n=0}^\infty N_nx^n$ to be our ordinary generating function for $N_n$. Multiplying everything in the above equation by $x^n$ and summing from $n=a$ to $\infty$, we get $$\begin{aligned} 0 &=\sum_{n=a}^\infty N_nx^n - r\sum_{n=a}^\infty N_{n-1}x^n + \sum_{n=a}^\infty N_{n-a}x^n\\ &= F(x)-(N_0+N_1x+\cdots + N_{a-1}x^{a-1})\\ &\quad{}-rx(F(x)-(N_0+N_1x+\cdots + N_{a-1}x^{a-2}))\\ &\quad{}+x^aF(x)\\ &= (1-rx+x^a)F(x) -(N_0+c_1x+c_2x^2+\cdots+c_{a-1}x^{a-1}), \end{aligned}$$ where $c_1,\cdots,c_{a-1}$ are arbitrary constants. Then $$ F(x)=\frac{N_0+c_1x+c_2x^2+\cdots+c_{a-1}x^{a-1}}{1-rx+x^a}.$$

Returning to the original notation, we get $$ \sum_{t=0}^\infty N(t)\,x^t = \frac{N(0)+c_1x+c_2x^2+\cdots+c_{ls-1}x^{ls-1}}{1-(f+1)x+x^{ls}}.$$