# General question about differential equations

I have never studied differential equations and have just run across some notes that involve one. I have a discrete function P that maps points in time (e.g. day1, day2, day3....) onto some real number on the interval [0, 1].

$P(t+1) = (1-3a)P(t) + a(1-P(t))$

the notes then define the change in P to be:

$\Delta P(t) = P(t+1) - P(t) = (1-3a)P(t) + a(1-P(t)) - P(t) = a - 4aP(t)$

Ok, fine. The notes then go on to say let's instead treat time as continuous. As such the rate of change of this continuous function is:

$dP(t)/dt = a - 4aP(t)$

The notes then state that the solution to this differential equation is:

$P(t) = 1 / 4 + (P(0) - 1 / 4)e^{-4at}$

My question is for what values of $t$ does the above equation hold? It appears to me from the broader context of this problem that this equation does not in fact hold for any specific $t$ but only for an unfixed $t$ as $t$ increases without bound. As such, if I wanted to know the value of $P(500)$ in the continuous case I can't actually use this equation. Is this understanding correct?

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Please allow me to use the symbol $n$ for discrete times and the symbol $t$ for continuous times. You consider in fact two different dynamics, one, in discrete time, of a sequence $(P_n)$ such that $$P_{n+1}=a+(1-4a)P_n$$ for every nonnegative integer $n$ and the other, in continuous time, of a function $(Q(t))$ such that $$Q'(t)=a-4aQ(t)$$ for every nonnegative real number $t$. (Yes, I use the letter $Q$ for the solution in continuous time.)

In the case at hand, it happens that both dynamics may be solved explicitly: one has $$P_n=(1-4a)^n(P_0-1/4)+1/4,$$ and $$Q(t)=e^{-4at}(Q(0)-1/4)+1/4.$$ The similarity is striking (and for good reasons) but one should keep in mind that $(P_n)$ and $(Q(t))$ are different objects and given by different formulas.

For example it is not true that the sequence $(Q(n))$ of the values of the function $Q$ considered at integer times is $(P_n)$, even if one chooses identical initial conditions $P_0=Q(0)$ (the exception being the common fixed point $P_0=Q(0)=1/4$, yielding $Q(t)=P_n=1/4$ for every nonnegative integer time $n$ and every nonnegative real time $t$).

In the end, if you are interested in $P_{500}$ (the result of $500$ iterations of the dynamics in discrete time), you should use the formula valid in discrete time (the one with a power of $1-4a$) but if you are interested in $Q(500)$ (the result of following the continuous path $t\mapsto Q(t)$ from time $t=0$ up to time $t=500$), you should use the formula valid in continuous time (the one with an exponential).

To solve the discrete time dynamics, a two-steps general technique is first, to look for fixed points, that is, value(s) $b$ such that if $P_n=b$ then $P_{n+1}=b$ (here $b=1/4$), and second, to use a fixed point as the new origin. Here, one considers the dynamics of the sequence $R_n=P_n-1/4$. This reads $R_{n+1}=(1-4a)R_n$ with $R_0=P_0-1/4$. From this formula one should be able to deduce the value of $R_n$ and then of $P_n$.
The same centering-around-a-fixed-point idea works in continuous time: $Q'(t)=0$ means that $Q(t)=1/4$ hence one considers $S(t)=Q(t)-1/4$, which, o miracle, is a solution of the easy-to-solve equation $S'(t)=-4aS(t)$...
Some kind of cautionary tale should probably be added here, stressing that the analysis of discrete time dynamics through their (simpler) continuous time counterparts may be misleading. Consider for example the discrete dynamics $P_{n+1}=1-2P_n$ and its continuous translation $Q'(t)=1-3Q(t)$. Then $Q(t)\to\frac13$ for every $Q(0)$ although $P_n$... Well, you may want to check this for yourself. – Did Aug 26 '11 at 20:42
What general technique did you use for finding an equation to express $P_n$? – Dejas Aug 26 '11 at 20:44
See edit.   – Did Aug 26 '11 at 20:49
I'm having problems reproducing $R_{n+1} = (1-4a)R_n$. I get $R_{n+1} = P_{n+1} - 1/4 = a + P_n(1-4a) - 1/4 = a - 4aP_n + P_n - 1/4 = a - 4aP_n + R_n = a(1-4P_n)R_n$. In any case, this math feels powerful. I've either seen it and forgotten it or never learned it. Where would these ideas be found in a collegiate curriculum? – Dejas Aug 26 '11 at 22:59
$...=a−4aP_n+P_n−1/4=P_n(1-4a)+a-1/4=P_n(1-4a)-(1/4)(1-4a)=...$ – Did Aug 27 '11 at 0:11