# Solving linear system, finding equilibrium and bifurcation points

For homework, I have to solve the following problem: consider the system of ODE's $$$$x'=-x^2+a \\ y'=-y$$$$ where the parameter $$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.

## 1 Answer

Hint:

By integration, when $$a=0$$, either

$$x=0$$ or

$$\frac1x=t+C.$$

• 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? Commented Nov 13, 2018 at 14:19