How Does the Gompertz Equation Model Tumor Growth?

[Jamin2112]

Homework Statement

As long as N isn't too small, the growth of cancerous tumors can be modeled by the Gompertz equation,

dN/dt=-rN*ln(N/K),

where N(t) is proportional to the number of cells in the tumor, and r, K > 0 are parameters.

(a) Sketch the graph f(N)=-rn*ln(N/k) verses N, find the critical points, and determine whether each is asymptotically stable or unstable.

(b) For the initial condition N(0)=N₀ (N₀>0), solve the Gompertz equation for N(t). Does this agree with your results from (a) in the limit as t -->∞?

(c) Sketch N(t) vs. t for initial conditions (i) 0 < N₀ < K, (ii) N₀=K, and (iii) N₀ > K.

Homework Equations

The Attempt at a Solution

(a)

Below is my sketch. I figured, since r,K,N>0 (can't have negative size of tumor), this is just a natural log function upside down and stretched out a little bit.

I'm not sure how to tell if it's asymptotically stable. I know that on the graph of N(t) versus t, it will start out with a positive slope; decrease in slope until the slope is zero at N(t)=K; then the slope will gradually become more negative.

(b)

dN/dt=-rN*ln(N/K)
==> dN / (N*ln(N/K) = -rdt
==> d/dN (ln(ln(N/K))) = -rdt
==> ln(ln(N/K) = -rt + C
==> ln(N/K) = e^-rt+C = ©e^-rt (©=e^c. I'm going to merge my constant shizzle while I solve)
==>N/K = e^©ert
==>N=N(t)= Ke^©ert = ®e(-rt) (®=ke^©)


... This is where I'm stuck. I've always been told that I can merge constants as I go along; but doing so in this problem screws everything up because I need K to stay with me!

 

[Eynstone]

Check your derivation : ln(ln(N/K) ) doesn't make sense when N <K. It's better to handle the 3 cases separately.

 

[epenguin]

Sketch the graph f(N)=-rn*ln(N/k) that should be f(N)=-rN*ln(N/k) verses N, find the critical points, and determine whether each is asymptotically stable or unstable.


I'm not sure how to tell if it's asymptotically stable. I know that on the graph of N(t) versus t, it will start out with a positive slope; decrease in slope until the slope is zero at N(t)=K; then the slope will gradually become more negative.

(b)

dN/dt=-rN*ln(N/K)
==> dN / (N*ln(N/K) = -rdt
==> d/dN (ln(ln(N/K))) = -rdt
==> ln(ln(N/K) = -rt + C
==> ln(N/K) = e^-rt+C = ©e^-rt (©=e^c. I'm going to merge my constant shizzle while I solve)
==>N/K = e^©ert
==>N=N(t)= Ke^©ert = ®e(-rt) (®=ke^©)


... This is where I'm stuck. I've always been told that I can merge constants as I go along; but doing so in this problem screws everything up because I need K to stay with me!

Your question suggests to me having been copied to here not quite accurately. At any rate a natural reading of the first sentence above seems to be asking for critical points of f. There is one maximum of f but that is an inflection point of N against t. It doesn't come to me to think stability is something you characterise inflection points by.

Confusion between f and N has made your red sentence all wrong but I am sure you will get it right when you realize that.

For the asymptotic stability you do not need to solve the d.e., you just need to look at the sign of f, i.e. of dN/dt, near N=0, and near the top limit N=K (both sides of that one).

For your problem with the arbitrary constant it is a bit fiddly.

You got to ln(N/K) = Ce^-rt (I agree with your integration) and you are probably thinking (N/K) = e^Ce-rt, ah that's N = Ke^Ce^-rt my K seems to have been swallowed into a lumped arbitrary constant. But it hasn't - e^Ce-rt is not e^Ce^-rt but (e^C)^e-rt, not the same as multiplying K by e^C, think about it.