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Sat 24 Sep, 2005 12:31 am
Hi all! I'm working on the following linearization problem:
Given a non-linear system:
dx/dt = x ( a - 4 + 2y ) - a
dy/dt = y ( a + 1 - x) - 2a
where a is a constant that may be positive or negative.
( 1 , 2 ) is an equilibrium solution as both dx/dt and dy/dt are zero at ( 1 , 2 ).
So, we can linearize this non-linear system about ( 1 , 2 ).
Ater doing that we get a linearized system as follows:
dx_1/dt = a*x_1 + 2*y_1
dy_1/dt = -2*x_1 + a*y_1
which has the general form:
dx_1/dt = a*x_1 + Beta*y_1
dy_1/dt = -Beta*x_1 + a*y_1
and in our particular system Beta equals 2.
Now, here's the part where I'm stuck:
Given the linearization is of the form above, for what values of a and Beta is x = 1, y = 2 stable?
And there's also a hint: Eliminate x_1 or y_1.
Can anyone guide me through this part. Thanks!
The real part of the characteristic roots of the matrix
Code:
a Beta
-Beta a
must be negative.
And for any value of Beta, if a<0 the liniearised system is stable at the equilibrium point (1,2).
Right, but how do I show this using elimination on either x_1 or y_1? Thanks!
You can transform the linearized first order simultaneous equations to a second order equation by elimination:
x'' = 2ax' + (a*a + Beta*Beta)x
(here prime means the derivative with respect to t)
and by assuming the solutions of the form e^(qt), you can derive the characteristic equation
q*q - 2aq - (a*a+Beta*Beta) = 0
where q is the (generic) characteristic root.
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