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I have a system of three ODEs and I have computed the Jacobian matrix.
One of the steady states is (0,0,0) and I am trying to linearize the system around this steady state. In the Jacobian matrix two of the terms are of the form $cx^2\over cx+y$ and $z\over y$.Here, $x,y,z$ are the variables of the system and $c$ is a known constant.

But when substituting x=0,y=0 and z=0 to the above terms it will be $0\over 0$.

So, in this case, can I use L'hospital rule for 2 variables to compute the Jacobian around that steady state?

That is I want to find
$\lim_{(x,y)\to (0,0)}{ cx^2\over cx+y}$ and $\lim_{(z,y)\to (0,0)} {z\over y}$.

I referred to the article on l'Hopital's rule for multi variable functions and with what it says in the article I could not find the limit of
$\lim_{(x,y)\to (0,0)}{ cx^2\over cx+y}$ and the limit of $\lim_{(z,y)\to (0,0)} {z\over y}$ does not exist.

Can someone please let me know a method to evaluate the two terms around the steady state.

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  • $\begingroup$ Just checking: Are you saying the Jacobian matrix contains three terms of the two forms or that it has three eigenvalues of the two forms? (Which could potentially be the same thing, depending on details you haven't shared.) $\endgroup$ – Eric Towers Mar 24 '18 at 8:02
  • $\begingroup$ @clarkson If the partial derivatives are of such a kind as you describe, then clearly the RHS is not $C^1$ at the origin. The linearization method can be applied under some conditions weaker than $C^1$, but you have (at least) to check if the RHS is differentiable at the origin (by definition, I think). What is your system? $\endgroup$ – user539887 Mar 24 '18 at 8:06
  • $\begingroup$ @EricTowers The Jacobian is a 3*3 matrix and J(1,1) term is $cx^2\over cx+y$ and J(2,3) term is $z\over y$ $\endgroup$ – clarkson Mar 24 '18 at 8:08
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With any limitaton in the domain the two limit doesn't exist, indeed simply note that for $t\to 0$

  • $x=0 \quad y=t \implies {cx^2\over cx+y}=0$
  • $x=t \quad y=-ct+t^2 \implies {cx^2\over cx+y}=\frac{ct^2}{ct-ct+t^2}=c$

and

  • $z=0 \quad y=t \implies \frac z y = 0$
  • $z=t \quad y=t \implies \frac z y = 1$
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  • $\begingroup$ Thank you for the answer. Can you please explain what this $t$ is $\endgroup$ – clarkson Mar 24 '18 at 8:09
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    $\begingroup$ It' a parametr which tends to zero, it represents a parametrization of $(x,y,z)\to (0,0,0)$ once we find different paths with different limits then we say that the limit doesn't exist. $\endgroup$ – user Mar 24 '18 at 8:11
  • $\begingroup$ So, in this case isn't there any other way that I can use to linearize the system around (0,0,0) $\endgroup$ – clarkson Mar 24 '18 at 8:16
  • $\begingroup$ This expression have no limit at 0 and are also unbounded, indeed in some trajectories they tends to $\infty$ let try with $z=t$ $y=t^2$ in the second and $x=t$ $y=-ct+t^3$ in the first. $\endgroup$ – user Mar 24 '18 at 8:20
  • $\begingroup$ @clarkson I repeat: Please show your system! $\endgroup$ – user539887 Mar 24 '18 at 8:27

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