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consider following nonlinear equation system how solve it? $$x'=|y|$$ $$y'=x$$ and whats the matrix that associated to this system

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I haven't worked out the details, but here's a thought: at first glance, your system appears to be nonlinear (in fact, the vector field isn't even C^1, which makes me nervous) but I suspect (and this will be easy to check) that an initial condition in a given quadrant (e.g. x>0, y>0) will remain in that quadrant for all times. So my gut now tells me that the fundamental solution matrix is going to have a piecewise definition based on the quadrant your initial conditions lie in (this makes sense, because your solutions should not be C^1). –  A Blumenthal Jan 12 '13 at 20:35
    
all none linear system change to this form AX+G(t)=0 by Jacobian matrix i mean whats this matrix and this question is exercise of book and doesn't define initial condition –  Maisam Hedyelloo Jan 12 '13 at 20:56
    
What's the book? On what pages does it discuss your $AX+G(t)=0$? On what page is the exercise? –  Gerry Myerson Jan 25 '13 at 5:49

1 Answer 1

up vote 3 down vote accepted

It seems like you would have to be given some ICs that keep you in a particular quadrant or that you would require that with your solution.

In order to solve, we would have to break this up into two systems as:

$$x'=y$$ $$y'=x$$

This would give us eigenvalues $\lambda_{1,2} = \pm 1$, which yields a solution:

$x(t) = c_1 e^{-t} (e^{2 t}+1)+c_2 e^{-t} (e^{2 t}-1)$, and

$y(t) = c_1 e^{-t} (e^{2 t}-1)+c_2 e^{-t} (e^{2 t}+1)$

$$x'=-y$$ $$y'=x$$

This would give us eigenvalues $\lambda_{1,2} = \pm i$, which yields a solution:

$x(t) = c_2 \sin(t)+c_1 \cos(t)$, and

$y(t) = c_1 \sin(t)+c_2 \cos(t).$

Of course, there would have to be restrictions placed on the ICs and I suppose you can just plot them with different values of x and y and see if they are consistent for all time. It is possible that you would have to glue together the solutions for both systems based on the the values of $x$ and $y$.

Regards

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$+1\;$You look good in green $\;\ddot\smile\;$ –  amWhy Apr 22 '13 at 0:11
    
@amWhy: Thank you, I never feel like I get enough of those and some of you seem to be in the stratosphere with the number of answers and accepts (green)! :-) Yourself included! regards –  Amzoti Apr 22 '13 at 0:16

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