I am a biologist trying to understand the ODE of growth tumour:
Let $N(t)$ denote the number of cancerous cells at time $t$, which are proliferating at a rate form
$$\frac{dN}{dt}=\frac{\lambda N}{\alpha}\left[1-\left(\frac{N}{\theta}\right)^{\alpha}\right] ... (1)$$
where $\alpha(\geq 0 )$ and $\theta(\geq 0 )$ are determined from the growth characteristics of the tumour.
and then it says that the solution to this equation is:
$$ N(t) = N_0 \{ \left[ \frac{N_0}{\theta} \right]^{\alpha} + e^{\lambda t} \left( 1-\left[ \frac{N_0}{\theta}\right]^{\alpha}\right) \}^{-1/\alpha} ...(2)$$
where $N_0 = N(t=0)$.
I still don`t understand how you get this answer, I tried to do the simple form, i.e, solving the Gompertzian growth rate, $$\frac{dN}{dt}= -\lambda N \ ln \left( \frac{N}{\theta}\right) $$, I solved this using the method of separable variables and with some help hehe
My first question is: How can you generalised Gompertzian growth into (1) ? It is not clear for me how you can get (2)?
And then the paper at some point consider now the effect of exposing a tumour cell population to a cycle-nonspecific drug at concentration $C(t)$, thus it gets a new generalised equation:
$$ \frac{1}{N} \ \frac{dN}{dt} = \begin{cases} \lambda - \mu C(t), \ N \leq N_c,\\ \lambda + \frac{\lambda_1}{\alpha} \left[1- \left(\frac{N}{N_c}\right)^{\alpha}\right]- \mu C(t) , \ N \geq N_c \end{cases}$$
Here my question is why is the meaning of the cases in the paper, for example, biologically talking, what does it mean considering the case $N_0 \geq N_c \lambda - \mu C_0 \geq 0$?
This is the paper: https://core.ac.uk/download/pdf/82695656.pdf
Really, I would appreciate your hints, help, comments to understand this paper. Thank you for your time and help in advance.
Consider the function $u(t)=N(t)^{-α}$. Then the derivative of this is $$ u'(t)=-αN(t)^{-α-1}N'(t)=-λ[N(t)^{-α}−θ^{-α}]=-λ[u(t)-θ^{-α}].\tag1 $$ This now is a linear equation, and additionally separable, giving the solution $$ u(t)-θ^{-α}=[u(0)-θ^{-α}]e^{-λt}.\tag2 $$ Now replace $u$ for $N$ to get \begin{align} (N(t)/θ)^{-α} &= 1+[(N(0)/θ)^{-α}-1]e^{-λt}=(N(0)/θ)^{-α}\left((N(0)/θ)^{α}+[1-(N(0)/θ)^{α}]e^{-λt}\right),\tag3 \\~\\ N(t) &= \frac{N(0)}{[(N(0)/θ)^{α}(1-e^{-λt})+e^{-λt}]^{1/α}}.\tag4 \end{align}