MHB Peak in single ODE within a system

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Carla1985
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Hi all, I have the system of nonlinear ODEs:

$$
\begin{align}
\frac{dX}{dt}=&-k_+ A X+k_-Y \\
\frac{dY}{dt}=&\ k_+ A X-k_-Y-\alpha k_+ X Y +\beta Z \\
\frac{dZ}{dt}=&\ \alpha k_+ X Y -\beta Z
\end{align}
$$

I also have a conservation law that says $D=X+Y+2Z$. Obviously it is not possible to find exact solutions to the system as it is nonlinear, but I noticed when running an ODE solver that the time course curve of $Z$ sometimes has a peak and then fall, dependant on the parameters. I was wondering if there is any way to be able to find a condition on whether I get a peak or not?

Regards
Carla
 
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At a base level, you could say that a peak like that means there is an interval $(a,b)$ containing a point $c$ such that $\dfrac{dZ}{dt}>0 \; \forall \, t\in(a,c)$ and $\dfrac{dZ}{dt}<0 \; \forall \, t\in(c,b)$. So, based on your system of ODE's, you could compare $\alpha k_+ X Y -\beta Z$ with $0$. Put another way, when is $\alpha k_+ X Y > \beta Z?$

Do you have any constraints (other than your conservation law, which looks like it could reduce the order of the system by 1. I assume $D$ is a constant?) on the constants $\alpha, \beta, k_+,$ and $k_-?$
 
Does your "conservation law" mean that D= X+ Y+ 2Z is a constant? If so then, taking the derivative of both sides with respect to t, $\frac{dX}{dt}+ \frac{dX}{dt}+ 2\frac{dZ}{dt}= 0$. Using the derivative formulas given, it follows that $\alpha k_+XY- \beta Z= 0$ so you can replace the XY term by $\beta Z$ to get a linear system.
 
Hi,

Yes, sorry, I should have explained better, D is a constant value. The only other constraints are $\alpha, \beta, k_+, k_-\geq 0$. I will work through your suggestions. Thank you very much for the help.

Regards
Carla
 
Carla1985 said:
D is a constant value. The only other constraints are $\alpha, \beta, k_+, k_-\geq 0$.
Is $A$ a constant as well?
 
Krylov said:
Is $A$ a constant as well?

Yes $A$ is constant. The only variables are $X,Y$ and $Z$
 
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