How to convert this system of non-linear diff equations into vector form

[Stenle3777]

dx1/dt = 0.25*x1 - 0.001 * x1 * x2 - 2.5 * (10^-5) * x1^2

dx2/dt = -0.5*x2 + 0.001 * x1 * x2 - 5 * (10^-4) * x2^2

Initial conditions are x1(0) = 500; x2(0)=50, if that's any help.

I need to transform this system into the form: dx/dt = f(x), where dx and f(x) are vectors.

I've read that I need to do Jacobian matrix, but I don't know what to do with it after I compute it.

Any help and ideas will be much appreciated. Thank you.

 

[Deleted member 4993]

dx1/dt = 0.25*x1 - 0.001 * x1 * x2 - 2.5 * (10^-5) * x1^2

dx2/dt = -0.5*x2 + 0.001 * x1 * x2 - 5 * (10^-4) * x2^2

Initial conditions are x1(0) = 500; x2(0)=50, if that's any help.

I need to transform this system into the form: dx/dt = f(x), where dx and f(x) are vectors.

I've read that I need to do Jacobian matrix, but I don't know what to do with it after I compute it.

Any help and ideas will be much appreciated. Thank you.

What did you get for the Jacobian?

Please share your work.

 

[Stenle3777]

This is the matrix I got and its Determinant.

J =

0.25 - 0.0001 * x2	-0.001 * x1
0.001 * x2	-0.5+0.001 * x1

D = (x1 + 2 * x2 - 500 ) / 4000

And I don't know where to go from here.

 

[HallsofIvy]

First, you can't "convert a non-linear differential equations into vector form" because vector spaces are necessarily linear. The best you can do is approximate the non-linear differential equations by linear equations then convert them to vector form.

For these equations the best approximation by linear equations is simply dropping the "non-linear" parts, x1*x2, x1^2, and x2^2 to get
dx1/dt= 0.25 x1 and
dx2/dt= 0.50 x2.

Now let \(\displaystyle \vec{x}= \begin{pmatrix}x1(t) \\ x2(t) \end{pmatrix}\) so that
\(\displaystyle \frac{d\vec{x}}{dt}= \begin{pmatrix}\frac{dx1}{dt} \\ \frac{dx2}{dt}\end{pmatrix}= \begin{pmatrix}0.25 x1 \\ 0.50 x2\end{pmatrix}\).

You could also write that as

\(\displaystyle \frac{d\vec{x}}{dt}= \begin{pmatrix}0.25 & 0 \\ 0 & 0.50 \end{pmatrix}\vec{x}\).