# Linearization with Jacobian Matrix

Assuming I have

$\frac{dx}{dt}=5x^2+2xy+x$

$\frac{dy}{dt}=xy-y$

which leads to a jacobian matrix

$$\begin{pmatrix} 10x+2y & 2y \\ y & x-1 \end{pmatrix}$$

one of the fixed points is $(0,0)$, how do I find the form of the linearized system at that fixed point so that it is at the form of example: $\frac{dx}{dt}=5 \cdot x$

First off: the correct Jacobian would be: $$J(x,y)=\left(\matrix{f_x(x,y) & f_y(x,y) \\ g_x(x,y) & g_y(x,y)}\right),$$ where your system is: $$\frac{d}{dt}\left(\matrix{x\\y}\right)=\left(\matrix{f(x,y)\\g(x,y)}\right).$$ then you have to find a point $P(x^*,y^*)$ such that $f(P)=g(P)=0$. $(0,0)$ satisfies this condition.
Moreover, now you can proxy, in a "small" interval cointaining $P$, the dynamic in the following way: $$\frac{d}{dt}\left(\matrix{x\\y}\right)=\left(\matrix{f(x,y)\\g(x,y)}\right)\approx\left(\matrix{f(P)\\g(P)}\right)+\left(\matrix{f_x(P) & f_y(P) \\ g_x(P) & g_y(P)}\right)\cdot\left(\matrix{x\\y}\right),$$ so, reminding that $\left(\matrix{f(P)\\g(P)}\right)=\left(\matrix{0\\0}\right)$ you then have: $$\frac{d}{dt}\left(\matrix{x\\y}\right)=\approx\left(\matrix{f_x(P) & f_y(P) \\ g_x(P) & g_y(P)}\right)\cdot\left(\matrix{x\\y}\right)=J(P)\cdot\left(\matrix{x\\y}\right).$$ In your case: $$J(x,y)=\left(\matrix{10x+2y+1 & 2x \\ y & x-1}\right);\ P(0,0).$$ Therefore: $$J(P)=\left(\matrix{1 & 0 \\ 0 & -1}\right),$$ which leads to: $$\frac{dx}{dt}=x,$$ $$\frac{dy}{dt}=-y.$$ Or, rather: $$x(t)=x_0e^t,$$ $$y(t)=y_0e^{-t},$$ But be careful that this ODE is ill-posed, as one eigenvalue has real part greater than 0.
• The general form is $dX/dt=J(P)\cdot X$, as I wrote in the answer. Anyway, you should always provide the answer in $dx/dt$ form. – 7raiden7 Mar 28 '14 at 10:45
• In your example, another fixed point id $Q(-1/6,0)$. So $J(Q)=(-2/3,-1/3,0,-7/6)$ leading to $dx/dt=-2/3x-1/3y;\ dy/dt=-7/6y$ – 7raiden7 Mar 28 '14 at 10:56