compartmental model and differential equation

[tioo]

Hello,

I try to write differential equations of the next problem :
Let :
- x1(t) : number of cells of type 1 at time t
- x2(t) : number of cells of type 2 at time t
- r : rate at which a cell of type 1 reproduce
- p : fraction of cells of type 2 that revert of type 1
- k : rate at which a cell of type 2 revert to a type 1 state

I'm only interested in the number of cells of type 1 :

dx1(t)/dt = r*x1(t)+pk*x2(t) + ?

I think that the growth of cells of type 2 that have reverted is missing but I d'ont know how to take into account it in equation.

Thank you
Ps: Sorry for my english (I'm French ).

 

[Ishuda]

Hello,

I try to write differential equations of the next problem :
Let :
- x1(t) : number of cells of type 1 at time t
- x2(t) : number of cells of type 2 at time t
- r : rate at which a cell of type 1 reproduce
- p : fraction of cells of type 2 that revert of type 1
- k : rate at which a cell of type 2 revert to a type 1 state

I'm only interested in the number of cells of type 1 :

dx1(t)/dt = r*x1(t)+pk*x2(t) + ?

I think that the growth of cells of type 2 that have reverted is missing but I d'ont know how to take into account it in equation.

Thank you
Ps: Sorry for my english (I'm French ).

Hey, your English is much better than my French which is almost non-existent

The way I would approach the problem:
(a) x₁(t) = p x₂(t) + f(t) [ p is the fraction of cells of type 2 that revert of type 1. f(t) is unknown at this point]
(b) (p x₂(t))' = k [k is rate at which a cell of type 2 revert to a type 1 state. The ' stands for derivative with respect to t]
(c) (x₁(t))' = r [r is the rate at which a cell of type 1 reproduce]

So, start putting things together:
From (a), (b) and (c) we have
(x₁(t))' = r = [p x₂(t) + f(t)]' = (p x₂(t))' + (f(t))' = k + (f(t))'
or
(f(t))' = r - k
Now integrate (f(t))' to get your (almost) final answer. How many cells of type 1 were there to start with?

Note that there may be some confusion here. Generally when it is said that the rate of change of something is, for example, r, what is meant is that the derivative of that something is r. However, in the case of reproduction, this could (and generally is) interpreted differently and a little more information is needed. That information is the expiration rate. When the expiration rate is not given, it is generally assumed that each cell 'reproduces forever' and the growth is actually a simple exponential. For example suppose each cell reproduces forever [or at least as long as the time frame we are interested in] at the rate s.
Time 0: 1 cell
Time s: 2 cells [the original cell and its prodigy]
Time 2s: 4 cells [the cells at Time s and their 2 prodigy]
Time 3s: 8 cells [the cells at Time 2s and their 4 prodigy]
etc.
The number of cells is then 2^t/s and the rate (derivative) is s^-1 ln(2) 2^t/s where ln(2) is the natural logarithm of 2.

EDIT: This could be handled in this problem by letting r, p and k be functions of t rather than constants as was implied above.

 

[tioo]

Thank you very much for your response ! I think it will help me a lot. I have some problems with my internet connection. I study it quickly and come back. :grin: