How to Linearize a System of Differential Equations with Equilibrium Points?

[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 you 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!

 

[xanthym]

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 you 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):

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

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

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

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:
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.


~~

 

[thepatientmental]

Yes, you are on the right track. To linearize this system, we can use the Taylor series expansion around the equilibrium point. In this case, we have three equilibrium points, so we need to linearize the system at each of these points separately.

For the first equilibrium point (0,0), we have x=u and y=v, so the linearized system becomes:

\frac{du}{dt} = 0.8u - \frac{0.8u^2}{3129} - \frac{v}{1000}
\frac{dv}{dt} = -0.5v + \frac{u}{1000}

To simplify this, we can assume that the non-linear terms (u^2 and uv) are small when u and v are close to zero, and thus can be neglected. This gives us the linearized system:

\frac{du}{dt} = 0.8u - \frac{v}{1000}
\frac{dv}{dt} = -0.5v + \frac{u}{1000}

We can now write this in matrix form as:

\left(\begin{array}{cc}0.8&-\frac{1}{1000}\\ \frac{1}{1000}&-0.5\end{array}\right) \left(\begin{array}{c}u\\v\end{array}\right)

This matrix has eigenvalues (0.85, -0.65), indicating that the equilibrium point (0,0) is a saddle point.

Similarly, for the second equilibrium point (1000,1088), we have x=u+1000 and y=v+1088, and the linearized system becomes:

\frac{du}{dt} = -u - \frac{0.8u^2}{3129} - \frac{v}{1000}
\frac{dv}{dt} = -0.5v + \frac{u}{1000}

Again, we can assume that the non-linear terms are small and neglect them, giving us the linearized system:

\frac{du}{dt} = -u - \frac{v}{1000}
\frac{dv}{dt} = -0.5v + \frac{u}{1000}

In matrix form, this becomes:

\left(\begin{array}{cc}-1&-\