Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If I have this equation:

$p'(t)-p(t)\alpha =0$

I can say that $p$ is a function that represents the size of a population at time t. The rate at which the population grows is constant. The solution will show that the size of the population is proportional to the initial size.

If I have this equation:

$p'(t)-p(t)f(t) =0$

I can say the rate at which the population grows is determined by $f$. The size of the population is still proportional to the initial size.

But if I have:

$p'(t)-p(t)\alpha = h(t)$

It's difficult to determine from the solution what role the initial size of the population, $p(0)$, has. The solution is:

$p(t)=\bigl(\int e^{-\alpha t}h(t)\ dt + c\bigr)\ e^{\alpha t}$

So my question is this: if I'm interpreting these differential equations as growth functions, what does $c$ represent in the last equation? In the previous equations, $c=p(0)$.

share|improve this question
    
$p(0)=c$ still means that $c$ is the initial size of the population if $p(t)$ represents the size of the population at time $t$. –  Matt Sep 12 '12 at 18:20
    
For constant $h(t)$ (breeding plus immigration), there is a fairly natural interpretation as a "virtual" population, since $p(t)$ is a constant plus an exponential term. –  André Nicolas Sep 12 '12 at 19:18
add comment

2 Answers 2

The equation $$p'(t)=f(t)\ p(t)$$ says: The rate $p'$ by which the population $P$ grows is proportional to the current size $p$ of $P$; but the proportionality factor $f$ valid at time $t$ depends on time. If, e.g., the growth rate depends on the seasons, the function $t\mapsto f(t)$ would be a certain periodic function with period 365 days. At any rate, the size of $P$ at some later time $t$ will be proportional to the initial size $p(0)$.

The equation $$p'(t)=\alpha p(t)+h(t)$$ with constant $\alpha\in{\mathbb R}$ says: The population $P$ would rise (or decline) at the constant rate $\alpha$, and therefore would be given by $p(t)=p(0)\ e^{\alpha t}$, if it were not for an additional extraneous influx (or decrease) $h(t)$ dependent only on time $t$, but not on the current size $p(t)$ of the population. In this case it may very well be that for large $t>0$ the initial value $p(0)$ has almost no effect on the actual value $p(t)$. This is the case, e.g., if $\alpha<0$, and $h$ is some periodic function with period $T$. Then for large $t$ there will be a certain "stable" periodic behavior. This "limiting" behavior depends only on $h$, but not on the initial value $p(0)$.

Concerning your last question: When you write the solution in the "more correct" form $$p(t)=\int_0^t e^{\alpha(t-\tau)}\ h(\tau)\ d\tau + p(0)e^{\alpha t}$$ then you can verify by inspection that $p(0)$ plays no rôle for large $t$ when $\alpha<0$.

share|improve this answer
    
I've sloppily figured out that, if $h$ were a constant instead, then $c=p(0)-\frac{h(t)}{\alpha}$, so as $\alpha \to \infty, c \to p(0)$. But h is not a constant, so this is moot (and probably wrong.) –  Korgan Rivera Sep 12 '12 at 23:46
add comment

In your first equation, the solution is $p(t)=ce^{at}$. As you say, you choose $c$ to be $p(0)$ if that is the initial condition you are given. If you are given that $p(2)=10000$, that there were $10000$ individuals at year $2$, you would find $c=10000e^{-2a}$ Similarly, $c$ in the last equation is a constant of integration which you set to match some initial condition you are given.

share|improve this answer
    
In the first and second equation, I don't choose $c$ to be $p(t)$. $p(0)$ is in fact $c$. If I'm given $p(2)= 10000$, $p(0)$ is still $c$. So it's easy to read that the population is proportional to the initial size: the c term in the solutions for the first and second equations I know is p(0). But that's not so for the 3rd equation. –  Korgan Rivera Sep 12 '12 at 18:40
    
@KorganRivera:If you start the integral in the third equation at 0, then $c=p(0)$ again-the integral is zero and the denominator is 1. If you start the integral at a different time, $c$ will be different. This is like any indefinite integral. If you write $\int x dx = x^2+c$, the $c$ will adjust according to the lower limit you use. –  Ross Millikan Sep 12 '12 at 18:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.