Continuous growth

[gadelke]

Continuous growth - law of motion

I think this might be an exponential growth question but I'm not sure how to go about solving it?

Suppose I have some continuous time law of motion of

, which is the value of

at time

.

The growth rate of

with respect to some generic time

is as follows:

So the rate of growth at any time depends on the current value of

at that time.

If I know what the value of

at time

, how would I find out the value of

at time

(in terms of

)?

My intuition would be that the solution is in the form:

I think I cannot just integrate the growth rate because only the value of

is known, but any

will have to be derived iteratively through the growth rate equation?

How would I solve this problem? Any help would be greatly appreciated.

 

[HallsofIvy]

From \(\displaystyle \frac{d\mu_t}{dt}= \lambda(1+\mu_t)\) (since there is, in this problem, only one independent variable, t, we don't need the \(\displaystyle \partial\)), we have \(\displaystyle \frac{d\mu_t}{1+ \mu_t}= \lambda dt\). To integrate on the left, let \(\displaystyle x= 1+ \mu_t\) so that \(\displaystyle dx= d\mu_t\). The equation becomes \(\displaystyle \frac{dx}{x}= \lambda dt\). Integrating both sides, \(\displaystyle ln(|x|)= \lambda t+ C\). Taking the exponential of both sides we have \(\displaystyle x(t)= C'e^{\lambda t}\) where \(\displaystyle C'= e^C\).

Since \(\displaystyle x= 1+ \mu_t\), \(\displaystyle 1+ \mu_t= C'e^{\lambda t}\) so that \(\displaystyle \mu_t= C'e^{\lambda t}- 1\). Knowing that \(\displaystyle \mu_a= C_a\), we have \(\displaystyle \mu_a= C'e^{\lambda a}- 1= C_a\) so that \(\displaystyle C'e^{\lambda a}= C_a+ 1\) and \(\displaystyle C'= (C_a+ 1)e^{-\lambda a}\).

That is, \(\displaystyle \mu_t= (C_a+ 1)e^{-\lambda a}e^{\lambda t}= (C_a+ 1)e^{\lambda(t- a)}\).

 

[gadelke]

From \(\displaystyle \frac{d\mu_t}{dt}= \lambda(1+\mu_t)\) (since there is, in this problem, only one independent variable, t, we don't need the \(\displaystyle \partial\)), we have \(\displaystyle \frac{d\mu_t}{1+ \mu_t}= \lambda dt\). To integrate on the left, let \(\displaystyle x= 1+ \mu_t\) so that \(\displaystyle dx= d\mu_t\). The equation becomes \(\displaystyle \frac{dx}{x}= \lambda dt\). Integrating both sides, \(\displaystyle ln(|x|)= \lambda t+ C\). Taking the exponential of both sides we have \(\displaystyle x(t)= C'e^{\lambda t}\) where \(\displaystyle C'= e^C\).

Since \(\displaystyle x= 1+ \mu_t\), \(\displaystyle 1+ \mu_t= C'e^{\lambda t}\) so that \(\displaystyle \mu_t= C'e^{\lambda t}- 1\). Knowing that \(\displaystyle \mu_a= C_a\), we have \(\displaystyle \mu_a= C'e^{\lambda a}- 1= C_a\) so that \(\displaystyle C'e^{\lambda a}= C_a+ 1\) and \(\displaystyle C'= (C_a+ 1)e^{-\lambda a}\).

That is, \(\displaystyle \mu_t= (C_a+ 1)e^{-\lambda a}e^{\lambda t}= (C_a+ 1)e^{\lambda(t- a)}\).

Thank you - this definitely makes sense!

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Also as an aside - is it possible to express this in terms of \(\displaystyle \mu_{t} = \mu_{a} + \text{growth of } \mu \text{ in } (a, t] \) for \(\displaystyle t > a \)? I think this general form might also be useful to have.