I'm having trouble reviewing for a calc test and I'm stumbling on this particular question. If I could get a step by step or solution to study then that would be more than helpful!
Studying a growth rate for a cucumber using the following the equation:
a) Von Bertalanffy model of tumor growth
b) $y'=\alpha (1-\frac{ln(y)}{ln(C)})y$,$y(0)=y_0$ – this is a Gompertz model
1) Find the equilibria of the equation
2) Find whether each equilibrium is stable or unstable.
And my question also is how to write this two model with parameters.
Best Answer
The classical Gompertz model for tumor growth is given by the equations
$$\frac {dN}{ dt }=G (t) N(t)$$ $$\frac {dG}{ dt }=- \alpha G (t)$$
with initial conditions $N(0)=N_0$ and $G(0)=G_0$. Here $ N(t)$ is the number of cells at time $t $, $ G(t)$ is the "per capita" growth rate (that is to say, birth rate minus death rate at time $t$), and $\alpha $ is a positive constant. The model assumes that the tumor growth rate decays exponentially over time at rate $\alpha $, as a result of increasing death rates, decreasing reproduction rates, or both. Integrating the second equation leads to $G (t)=G_0 e^{-\alpha t} $, so that the model is usually written as
$$\frac {dN}{ dt }=G_0 \, e^{-\alpha t} N (t) $$
where the solution of this nonautonomous ODE is
$$ N (t)= N_0 \, e^{\alpha^{-1}(G_0-G(t))}$$
For practical purposes, however, the initial model is often simplified by setting $G(t) = (\beta− \alpha \log {N})$, where $\beta$ is an additional positive constant. This allows to write the following common formulation of the Gompertz model:
$$\frac {dN}{ dt }=N(\beta− \alpha \log {N})$$
An advantage of this simplification is that we have an autonomous equation, where the equilibrium points corresponds to the zeroes of the RHS. Thus, we have to solve
$$N(\beta− \alpha \log {N})=0$$
The solutions are $N=0$ and $N=e^{\beta/\alpha} $, which are then the equilibrium points (note that the equilibrium at $N=0$ is only theoretical, since in this point the equation is not defined). To assess the stability of these equilibrium points, considering that by definition $N \geq 0$, we can note that
$$N(\beta− \alpha \log {N}) >0 \,\,\,\text{for}\,\,\, 0<N <e^{\beta/\alpha}$$
$$N(\beta− \alpha \log {N}) <0 \,\,\,\text{for}\,\,\, N >e^{\beta/\alpha}$$
So, if $$ 0<N <e^{\beta/\alpha}$$ then $N' >0$, which means that the solutions increase towards the equilibrium point $N=e^{\beta/\alpha} $. Also, if $$N >e^{\beta/\alpha}$$ then $N'<0$, which means that the solutions decrease towards the equilibrium point $N=e^{\beta/\alpha} $. Because all solutions near $N=0$ increase and then move away from this point, and because all solutions near $N <e^{\beta/\alpha}$ converge towards this point, we can conclude that the theoretical equilibrium at $N=0$ is unstable, whereas the equilibrium at $N=e^{\beta/\alpha} $ is stable.
Regarding the simpler von Bertalanffy model, its most classic formulation is
$$\frac {dM}{dt}= \alpha M^{2/3}−\beta M$$
where $M $ is the tumoral mass, and $\alpha $ and $\beta $ can be interpreted as per-capita birth and death rates, respectively. The model assumes that the birth rate is proportional to the surface of the tumor, while the death rate is proportional to the mass of the tumor. Typically, this model is simplified assuming that the growth is homogeneous in the three dimensions, so that calling $L $ a linear dimension of the tumor we can write $$M (t)=k L (t)^3$$ where $k $ is a constant. From this, it follows $$dM=3k L^2 dL $$ and substituting in the original equation we get
$$\frac {dL}{dt}= \frac {1}{3}(\tilde {\alpha} -\beta L)$$
where $ \tilde {\alpha}= \alpha/k^{1/3} $. This shows that
$$L = \frac {\tilde {\alpha}}{ \beta} $$
is the only equilibrium point of the von Bertalanffy model, and that it is asymptotically stable.