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[SOLVED] How was u_1 and u_2 star determined

dwsmith

Well-known member
Feb 1, 2012
1,673
$$\begin{array}{lcl}
\frac{du_1}{d\tau} & = & (1-u_1-a_{12}u_2)\\
\frac{du_2}{d\tau} & = & (1-u_2-a_{21}u_1)
\end{array}
$$

The steady states for this system are $(0,0)$, $(1,0)$, and $(0,1)$.

But then the book says without explanation

$\displaystyle u_1^* = \frac{1-a_{12}}{1-a_{12}a_{21}}$

and

$\displaystyle u_2^* = \frac{1-a_{21}}{1-a_{12}a_{21}}$

Never mind solve the problem
 
Last edited:

CaptainBlack

Well-known member
Jan 26, 2012
890
$$\begin{array}{lcl}
\frac{du_1}{d\tau} & = & (1-u_1-a_{12}u_2)\\
\frac{du_2}{d\tau} & = & (1-u_2-a_{21}u_1)
\end{array}
$$

The steady states for this system are $(0,0)$, $(1,0)$, and $(0,1)$.

But then the book says without explanation

$\displaystyle u_1^* = \frac{1-a_{12}}{1-a_{12}a_{21}}$

and

$\displaystyle u_2^* = \frac{1-a_{21}}{1-a_{12}a_{21}}$

Never mind solve the problem
The are the solutions to the pair of equations for a steady state:

$$\begin{array}{lcl}
\frac{du_1}{d\tau} & = & 0\\
\frac{du_2}{d\tau} & = & 0
\end{array}
$$

Which is the system of simultaneous equations:

$$\begin{array}{lcl}
u_1+a_{12}u_2 & = & 1\\
u_2+a_{21}u_1 & = & 1
\end{array}
$$

Also, $(0,0)$, $(1,0)$, and $(0,1)$ are not in general the steady states of the system.

CB
 
Last edited: