Solving a form of the logistic equation to arrive a given solution

142 Views Asked by At

I am writing my bachelor thesis on modelling of city growth and using the book Cities and Complexity by Michael Batty.

On page 394, while modelling the growth as spatial epidemic, he writes:

$$\frac{d N(t)}{dt} = \alpha C N(t) \left(1-\frac{N(t)}{C} \right) $$ [The above equation] is one form of the logistic equation, which on integration yields $$N(t)=\frac{C}{1+\left(\frac{C}{N(0)}-1\ \right) e^{-\alpha C t}}.$$

Now, this is a tad odd. Lets solve this the usual way:

$$\frac{d N(t)}{dt} = \alpha C N(t) \left(1-\frac{N(t)}{C} \right)$$

$$\frac{d N(t)}{N(t) \left(1-\frac{N(t)}{C} \right)} = \alpha C dt$$ $$\int \frac{d N(t)}{N(t) \left(1-\frac{N(t)}{C} \right)} =\int \alpha C dt$$ $$\ln(N(t))-\ln(N(t)-C) =\alpha C t$$ $$\frac{N(t)}{N(t)-C} =e^{\alpha C t}$$ $$N(t) =e^{\alpha C t} (N(t)-C)$$ $$N(t)\left(1-e^{\alpha C t}\right) =-C e^{\alpha C t}$$ $$N(t) =-C \frac{e^{\alpha C t}}{1-e^{\alpha C t}} $$ $$N(t) =\frac{C}{e^{-\alpha C t}-1} $$

They look a bit similar, but where did the $N(0)$ come from? I would say that he takes defined integral over $t$, but I am not sure.

1

There are 1 best solutions below

1
On BEST ANSWER

$N(0)$ is the initial condition, that is, the value of $N$ at the initial instant $t=0$.

In the integration of the equation, you forgot to include an arbitrary constant $K$ in the fourth equation: $$ \ln(N(t))-\ln(N(t)-C) =\alpha C t+K. $$ Imposing the initial condition you can find the value of $K$ and the solution given in the book.