## Linearization

For this system of differential equations

$$\frac{dx}{dt} = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000})$$
$$\frac{dy}{dt} = y(-0.5 + \frac{x}{1000+x})$$
for all $$x,y \geq 0$$
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
$$\frac{dx}{dt} = (1.054u +2.55*10^{-4} u^2 -v) \frac{u+1000}{u+2000}$$
$$\frac{dy}{dt} = \frac{0.5uv - 544u}{2000 + u}$$
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
$$\left(\begin{array}{cc}0.527&-0.5\\0.272&0\end{array}\right)$$
is this right so far?? Any help would be appreciated, greatly!

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 Quote by stunner5000pt For this system of differential equations $$\frac{dx}{dt} = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000})$$ $$\frac{dy}{dt} = y(-0.5 + \frac{x}{1000+x})$$ for all $$x,y \geq 0$$ 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 $$\frac{dx}{dt} = (1.054u +2.55*10^{-4} u^2 -v) \frac{u+1000}{u+2000}$$ $$\frac{dy}{dt} = \frac{0.5uv - 544u}{2000 + u}$$ 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 $$\left(\begin{array}{cc}0.527&-0.5\\0.272&0\end{array}\right)$$ is this right so far?? Any help would be appreciated, greatly!
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):

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

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

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

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:

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

Taking Partials of the above f(x,y) and g(x,y), we get:
fx(x,y) = (0.8) - (1.6)*x/(3129) - (y)*(1000)/{(x + 1000)^2}
fy(x,y) = (-x)/(x + 1000)
gx(x,y) = (y)*(1000)/{(x + 1000)^2}
gy(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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