$\begingroup$

What is the solution to the Gompertz differential equation subject to $P(0)=P_0$?

The Gompertz differential equation is $dP/dt=P(a-b\ln(P))$.

Sorry to bother anyone but I looked online and couldn't find it.

$\endgroup$ 3

1 Answer

$\begingroup$

I'll solve the Gomptertz equation

$\dfrac{dP(t)}{dt} = P(t)(a - b \ln P(t)) \tag 1$

with initial condition

$P(t_0) = P_0; \tag 2$

that is, I will allow the initial time to be any $t_0 \in \Bbb R$.

Clearly, we must look for solutions with $P(t) \ne 0$, lest $\ln P(t)$ be undefined. Bearing this in mind, we may divide (1) through by $P(t)$ and obtain

$\dfrac{1}{P(t)}\dfrac{dP(t)}{dt} = a - b \ln P(t), \tag 3$

whence

$\dfrac{1}{P(t)}\dfrac{dP(t)}{dt} + b \ln P(t) = a, \tag 4$

which we may write as

$\dfrac{d}{dt} \ln P(t) + b \ln P(t) = a; \tag 5$

careful scrutiny of (5) motivatetes the substitution

$f(t) = \ln P(t), \tag 6$

and (5) becomes

$\dfrac{df(t)}{dt} + bf(t) = a, \tag 7$

which we immediately recongize as as first-order, linear, time invariant system which may readily be solved in closed form; also, the substitution (6) mandates the introduction of the transformed initial condition

$f(t_0) = \ln P(t_0) = \ln P_0. \tag 8$

The solution of (7) with initial condition (8) is well-known to be

$f(t) = e^{-b(t - t_0)}(f(t_0) + \displaystyle \int_{t_0}^t ae^{b(s - t_0)}ds), \tag 9$

and the integral is easily evaluated

$\displaystyle \int_{t_0}^t ae^{b(s - t_0)}ds = \left(\dfrac{a}{b}e^{b(s - t_0)} \right )_{t_0}^t = \dfrac{a}{b} \left (e^{b(t - t_0)} - 1 \right ); \tag{10}$

we bring it all together by using (10) in (9):

$f(t) = e^{-b(t - t_0)}(f(t_0) + \dfrac{a}{b} \left (e^{b(t - t_0)} - 1 \right )) = f(t_0)e^{-b(t - t_0)} + \dfrac{a}{b} \left (1 - e^{-b(t - t_0)} \right ) , \tag {11}$

or

$f(t) = \left (f(t_0) - \dfrac{a}{b} \right ) e^{-b(t - t_0)} + \dfrac{a}{b}. \tag{12}$

We now invert (6):

$P(t) = e^{f(t)}, \tag{13}$

or, using (8),

$P(t) = \exp \left (\left (\ln P_0 - \dfrac{a}{b} \right ) e^{-b(t - t_0)} + \dfrac{a}{b} \right)$ $= \exp \left (\left (\ln P_0 - \dfrac{a}{b} \right ) \exp \left (-b(t - t_0) \right ) + \dfrac{a}{b} \right). \tag{14}$

as the solution to the Gompertz equation (1).

Note Added in Edit, Saturday 30 September 2017 8:52 AM PST: We can also directly insert (13) into (1); we obtain

$\dfrac{dP(t)}{dt} = e^{f(t)}\dfrac{df(t)}{dt}, \tag{15}$

and

$\ln P(t) = f(t), \tag{16}$

whence

$e^{f(t)}\dfrac{df(t)}{dt} = e^{f(t)}(a - bf(t)); \tag{17}$

the factor of $e^{f(t)}$ cancels out and we are left with (7), from which we may proceed as above. End of Note.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy