Solution of the ODE $\frac{dx}{dt} = kP$

Question

I am trying to model a system where the rate of change of a variable $x\in [0,1]$ is proportional to a difference, say $P$, where $P$ lies between $-1$ and $1$. I have modeled it as a simple ODE, $\frac{dx}{dt} = kP$, where $P$ lies between $-1$ and $1$, and $k$ is a proportionality constant. Clearly the solution of the problem is: $x =kPt + x_0$. But the problem is that the value of $x$ has to be between $0$ and $1$. How can I ensure that? Any suggestions regarding this would be appreciated. Thanks.

Answer 1

I'm not sure if this is what you are going for, but you seem to have correctly solved the ODE. However, let me walk through what you have done and then I'll provide an answer to your question.

First, you defined an ODE

$$\frac{dx}{dt}=kP$$

where neither $k$ nor $P$ are functions of $x$ or $t$. You then correctly found that this implies

$$x(t)=kPt+x_0$$

where $x_0=x(0)$. Having got this far, you are confused because $x(t)$ is only allowed to range from $[0,1]$.

Now, my solution to this is that $t$ can not range over an infinite range. In fact, solving

$$0=kPt+x_0\Rightarrow t=\frac{-x_0}{k P}$$

$$1=kPt+x_0\Rightarrow t=\frac{1-x_0}{k P}$$

That is, the range of $t$ will depend on $x_0,k,$ and $P$. Otherwise, it looks like your solution to the ODE is correct and I'm not sure why you would restrict the range of your answer to $[0,1]$.