If a nonlinear system depends on a parameter, then the equilibrium points can change as the parameter varies. In other words, as the parameter changes, a bifurcation can occur. Consider the one-parameter system family of systems dx = x2 – a =-y(+2+1), where a is the parameter.(a) Show that the system has no equilibrium points if a < 0. (b) Show that the system has two equilibrium points if a > 0. (c) Show that the system has exactly one equilibrium point if a = 0. (d) Find the linearization of the equilibrium point for a = 0 and compute the eigenvalues of this linear system.

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varshamittal029 varshamittal029 · Answer 1 · 2022-12-22T12:38:37-05:00

If a<0, then there is no x-null lines, the system has no equilibrium points.

If a=0, it has only one equilibrium points.

If a>0, the system has two equilibrium points.

Points are (-√a,0) and (√a,0)

The system changes from no fixed points to two fixed points when it increases through a=0.

a is a bifurcation point.

The Jawbian matrix,

J = [tex]\left[\begin{array}{ccc}2x&0\\-2xy&x^{2}-1\\\end{array}\right][/tex]

Now,

J[0,0] = [tex]\left[\begin{array}{ccc}0&0\\0&-1\\\end{array}\right][/tex] =0

J[-√a,0] = [tex]\left[\begin{array}{ccc}-2\sqrt{a} &0\\0&-a-1\\\end{array}\right][/tex] = 2a√a + 2√a

If a=0, then it will be a live equilibrium, one of the values are 0, For a>0, the point (-√a,0) is sink and (√a,0) is a saddle points.

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