- #1
toastermm
- 3
- 0
I'm running into a problem. This is mainly for reading over the summer and I'm working on getting through a dynamical systems book on my own. I've come across a system that I'm not too sure on the procedure.
Consider the following system of differential equations:
[itex]\frac{dX}{dt} = 1 - X - XY - XZ, \\
\frac{dY}{dt} = aXY - Y,\\
\frac{dZ}{dt} = aXZ - Z.
[/itex]
Here, 'a' is a real positive constant and [itex] X,Y,Z \geq 0 . [/itex]
I can find all of the three steady states except for the one where all three exist, i.e. when [itex] \bar{X},\bar{Y},\bar{Z} \neq 0 . [/itex]
When solving for this steady state, I arrive at [itex] \bar{X} = \frac{1}{a} . [/itex]
Then, plugging this back into the first equation, I get the following
[itex] 0 = a -1 - \bar{Y} - \bar{Z}, [/itex]
or,
[itex] a-1 = \bar{Y}+\bar{Z}. [/itex]
Is there a way to solve for [itex]\bar{Y}[/itex] and [itex]\bar{Z}[/itex]?
-as a side note, I've explored this numerically with MATLAB and it looks like it depends on the initial conditions. Could it be a saddle?
Consider the following system of differential equations:
[itex]\frac{dX}{dt} = 1 - X - XY - XZ, \\
\frac{dY}{dt} = aXY - Y,\\
\frac{dZ}{dt} = aXZ - Z.
[/itex]
Here, 'a' is a real positive constant and [itex] X,Y,Z \geq 0 . [/itex]
I can find all of the three steady states except for the one where all three exist, i.e. when [itex] \bar{X},\bar{Y},\bar{Z} \neq 0 . [/itex]
When solving for this steady state, I arrive at [itex] \bar{X} = \frac{1}{a} . [/itex]
Then, plugging this back into the first equation, I get the following
[itex] 0 = a -1 - \bar{Y} - \bar{Z}, [/itex]
or,
[itex] a-1 = \bar{Y}+\bar{Z}. [/itex]
Is there a way to solve for [itex]\bar{Y}[/itex] and [itex]\bar{Z}[/itex]?
-as a side note, I've explored this numerically with MATLAB and it looks like it depends on the initial conditions. Could it be a saddle?