3
$\begingroup$

For homework, I have to solve the following problem: consider the system of ODE's \begin{equation} x'=-x^2+a \\ y'=-y \end{equation} where the parameter $a$ is a real number. I have to characterize the equilibrium points and draw a bifurcation diagram. I am asked to linearize the system first. I have missed a couple of classes and I am struggling to understand how I am supposed to work here.

If $a$ is positive, then there are two equilibrium points $(\sqrt{a},0)$ and $(-\sqrt{a},0)$, if $a$ is equal to zero, then there is one equilibrium point $ \left ( 0,0 \right )$ and finally if $a$ is negative, then no equilibrium points exist. The Jacobi matrix of the system is

\begin{pmatrix} - 2x & 0\\ 0 & -1 \end{pmatrix}

which is diagonal, hence its eigenvalues are $-2x$ and $-1$. Then, I calculate this matrix at the equilibrium points:

If $a>0$, then the two Jacobi matrices are \begin{pmatrix} -2\sqrt{a} & 0\\ 0 & -1 \end{pmatrix} which has two real and negative eigenvalues, so $(\sqrt{a},0)$ is a stable point and \begin{pmatrix} 2\sqrt{a} & 0\\ 0 & -1 \end{pmatrix} which has one negative and one positive eigenvalue, so $(-\sqrt{a},0)$ is a saddle point.

If $a=0$ the Jacobi matrix is \begin{pmatrix} 0 & 0\\ 0 & -1 \end{pmatrix} which has one zero eigenvalue so the linearization method does not provide any information about the stability of $\left ( 0,0 \right )$. As my notes suggest, for this point we find the eigenvectors of this matrix and the orbits of the solutions. But I do not understand why and what information we obtain from this. What can I say about the stability of $\left ( 0,0 \right )$ then? What about the saddle point, do I use a similar method? I understand that there is a bifurcation for $a=0$ but I don't know how am I supposed to draw the diagram without a computer. Another thing I find in the notes in another example is that we divide the equations and find a relationship between the solutions which does not involve time, for some reason.

Thanks in advance, any insight on this will be appreciated, right now I am lost.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Hint:

By integration, when $a=0$, either

$$x=0$$ or

$$\frac1x=t+C.$$

$\endgroup$
1
  • $\begingroup$ I know that the system can be solved, my problem is that I do not understand the reason for which we find eigenvalues. The orbits of the solutions tell me whether the solutions approach zero or they move away from it, I presume, right? $\endgroup$ Commented Nov 13, 2018 at 14:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .