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I have a system \begin{align} \dfrac{dx}{dt}&=-x^2 + 4 y^2, \\ \dfrac{dy}{dt}&=-8 - 4 y + 2 x y. \end{align}

There two singular points $A_1(-2;-1), A_2(4,2)$. I need to know the type of these points.

To do it, in case of linear system, I need to find lambdas, which depend on a,b,c,d koefs. Now, in case of nonlinear system, I'm getting problems:

1) To find lamdas I need to make a linearization of the equations. I don't know how to do that. All that I got in mind is the regression, exactly linear regression.

2)Instead of linear system, there are 2 singular points, hence I think, lambda equation should be changed respectively to this fact. How to find lambdas for two or more singular points?

Lutz Lehmann
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Stdugnd4ikbd
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1 Answers1

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  1. ) You need to compute the Jacobi matrix in the stationary points. The linearization is just the first order Taylor expansion of the right side. You have $$\dot x=f(x,y),\\\dot y=g(x,y).$$ In the linearization you replace them with $$\dot x=f(x_0,y_0)+∂_xf(x_0,y_0)(x-x_0)+∂_yf(x_0,y_0)(y-y_0),\\ \dot y=g(x_0,y_0)+∂_xg(x_0,y_0)(x-x_0)+∂_yg(x_0,y_0)(y-y_0).$$

  2. ) You compute the eigenvalues separately. The linearization gives insight into the local behavior very close to the stationary points, as long as the stationary points are isolated, there is no need to consider any possible interaction (like connecting trajectories) between them.

Lutz Lehmann
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  • As for Taylor series, I don't understand what I need to do. The problem is that I have not $y=some-dependence-on-x$ equation, but $\dfrac{dx}{dt}=x-and-y$. What I have to do? Just get the right expression, make the $y= dep on x$ form or what? – Stdugnd4ikbd Feb 16 '19 at 21:31
  • $t$, not $x$, is the independent variable, $(x(t),y(t))$ is the desired solution curve. – Lutz Lehmann Feb 16 '19 at 21:44
  • Didn't understand what exactly need to do. $x(t), y(t)$ are two equations instead of a single original one. How does the expression, you wrote relates to curve, and why curve at all, if I need a line? – Stdugnd4ikbd Feb 17 '19 at 11:04
  • Ok, in Taylor's series I need to take a derivative of the function. But respectively to which variable? I have at least two - x and y(and t). I need to make a t(x,y) equation? – Stdugnd4ikbd Feb 17 '19 at 11:07
  • Ok, I've linearized the second to $4x-8y$ near $(-2;-1)$ point and $-6x+16y-8$ for $(4,2)$ point. I'll linearize the first one, but I have the last question. To find lamdas, there is an equation with $a,b,c,d$ coefficients, which should be the coefficients of $ax+by$ equation form, but I already have $-6x+16y-8$. What to do with the 8?(sorry for my stupidness) – Stdugnd4ikbd Feb 17 '19 at 19:58
  • The equation for lambdas, I've found is $\lambda_{1,2}=\dfrac{a+d}{2}+-\sqrt{\dfrac{(a+d)^2-4(ad-bc)}{4}}$ – Stdugnd4ikbd Feb 17 '19 at 20:13
  • You do like with any linear system of ODE, you compute a homogeneous solution and add a particular solution. Or you introduce the differences to the stationary point as new variables, then the linearized system will be homogeneous. – Lutz Lehmann Feb 17 '19 at 20:31
  • Unfortunately, I didn't understand, what You wanted to say, but, however, thank You very much, for Your help – Stdugnd4ikbd Feb 17 '19 at 20:52