Linearization

stunner5000pt

For this system of differential equations

[tex] \frac{dx}{dt} = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000}) [/tex]
[tex] \frac{dy}{dt} = y(-0.5 + \frac{x}{1000+x})[/tex]
for all [tex] x,y \geq 0 [/tex]
now the equilibrium points are
(x,y) = (0,0), (1000,1088), (3129,0)

i need to linearize this system such taht i can figure out the eigenvalues of this system and figure out whether this system is a spiral sink, source or whatever

if i use x = u -1000 and y = v-1088
i get answers like this
[tex] \frac{dx}{dt} = (1.054u +2.55*10^{-4} u^2 -v) \frac{u+1000}{u+2000} [/tex]
[tex] \frac{dy}{dt} = \frac{0.5uv - 544u}{2000 + u} [/tex]
but hte denominator throws things off, doesn't it??
but if i just accept it like it is and assumethat as u,v approach zero the non linear terms get insignificant (ya ya not proper math language)
and the matrix becomes
[tex] \left(\begin{array}{cc}0.527&-0.5\\0.272&0\end{array}\right) [/tex]
is this right so far?? Any help would be appreciated, greatly!

xanthym

The following briefly reviews key features of diff-eq Linearization technique. Steps 1 - 4 at bottom of review section provide guidance for OP to obtain problem solution.

Remember, linearization is the process of APPROXIMATING the original non-linear diff-eq system with a "Linearized System" about an Equilibrium Point (x₀,y₀). Thus, let the original non-linear system be given below (assuming no explicit function of "t" on the RHS):

[tex]
\left (
\begin{array}{c}
(dx/dt) \\
(dy/dt)
\end{array} \right )
=
\left (
\begin{array}{c}
f(x,y) \\
g(x,y)
\end{array} \right )
[/tex]

Then the Linearized System will utilize the Jacobian matrix (in red below) evaluated at an Equilibrium Point (x₀,y₀):

[tex]
\left (
\begin{array}{c}
(dx/dt) \\
(dy/dt)
\end{array} \right )
=
\color{red} \left (
\begin{array}{c c}
f_{x}(x_{0},y_{0}) & f_{y}(x_{0},y_{0}) \\
g_{x}(x_{0},y_{0}) & g_{y}(x_{0},y_{0})
\end{array} \right ) \color{black} \cdot
\left (
\begin{array}{c}
(x - x_{0}) \\
(y - y_{0})
\end{array} \right )
[/tex]

By evaluating the Jacobian Eigenvalues, the Linearized System solutions and their characteristics (sink, spiral, etc.) can then be determined. Hence, the first objective is determining the Jacobian from appropriate Partials:

[tex] \frac{dx}{dt} = f(x,y) = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000}) [/tex]
[tex] \frac{dy}{dt} = g(x,y) = y(-0.5 + \frac{x}{1000+x})[/tex]

Taking Partials of the above f(x,y) and g(x,y), we get:
f_x(x,y) = (0.8) - (1.6)*x/(3129) - (y)*(1000)/{(x + 1000)^2}
f_y(x,y) = (-x)/(x + 1000)
g_x(x,y) = (y)*(1000)/{(x + 1000)^2}
g_y(x,y) = (-0.5) + (x)/(x + 1000)

The steps are now the following:
1) Form Jacobian matrix with above Partials.
2) Evaluate Jacobian at an Equilibrium Point (x₀,y₀). ::: <---(by def: f(x₀,y₀) = g(x₀,y₀) = 0)
3) Determine Eigenvalues for the evaluated Jacobian.
4) Determine and interpret Linearized System Solutions from Eigenvalues.


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