Computing the first 2 terms of the Taylor series exp. for the center manifold & find the reduced equation on the center manifold.

Question

I am having difficulty solving the problem below. It is from Meiss Dynamics book. Can I please receive help solving the following system? Thank you

Consider the system $$x' = y$$ $$y'=-y+ax^2 + bxy.$$ Compute the first two terms of the Taylor series expansion for the center manifold and find the reduced equation on the center manifold. For what values of $a$ and $b$ is the origin stable? Unstable? Semi-stable? Note the linearization at the origin is not in Jordan Canonical Form.

Answer 1

The equilibrium is attained at the solutions for

$$ \cases{ y=0\\ -y+a x^2+b x y = 0 } $$

so $(0,0)$ is the equilibrium point. To qualify it we compute the jacobian at this point giving

$$ J = \left( \begin{array}{cc} 0 & 1 \\ 0 & -1 \\ \end{array} \right) $$

with eigenvalues $(1,\ 0)$ so the equilibrium manifold is one-dimensional. To find this manifold we proceed as follows.

For the dynamical system

$$ \cases{ \dot x=f(x,y)\\ \dot y=g(x,y) } $$

Proposing the solution

$$ y=h(x) = \sum_{k=1}^n a_k x^k $$

we have

$$ \dot y=h_x(x)\dot x = h_x(x)f(x,h(x))=g(x,h(x)) $$

assuming $n=4$ equating the $x$ powers we arrive at

$$ \left\{ \begin{array}{rcl} a_1&=&0 \\ a_2 &=& a \\ a_3 &=& a b-2 a^2\\ \end{array} \right. $$

and solving we have

$$ h(x) = a x^2+a(b-2a) x^3+ O(x^4) $$

as a near origin approximation.

Follows a plot showing the stream plot for $a = -\frac 12, b = 1$ showing in thick blue a near origin center manifold segment and in red dashed, a path beginning at $(0.5,0.5)$

NOTE

The central manifold approximate flow for $n=4$ is given by

$$ \dot x = h(x) = a x^2+a (b-2a) x^3+ O(x^4) $$

Answer 2

I am going to assume that

$a \ne 0, \tag 1$

for reasons which will become clear in what follows. Under this assumption, the system

$\dot x = y, \tag 2$

$\dot y = -y + ax^2 + bxy ,\tag 3$

has a single equilibrium point at $(0, 0)$, for setting

$\dot x = \dot y = 0, \tag 4$

we see from (2) that

$y = 0, \tag 5$

and then from (3) that

$ax^2 = 0, \tag 6$

whence

$x = 0. \tag 7$

Here we have used assumption (1), for without it we cannot conclude that the critical set of the system (2)-(3) is a single point.

The Jacobian matrix of this system at $(x, y)$ is given by

$J(x, y) = \begin{bmatrix} 0 & 1 \\ 2ax + by & bx - 1 \end{bmatrix}; \tag 8$

at $(0, 0)$ this becomes

$J(0, 0) = \begin{bmatrix} 0 & 1 \\ 0 & -1 \end{bmatrix}; \tag 9$

this matrix has characteristic polynomial

$\det \left( \begin{bmatrix} -\lambda & 1 \\ 0 & -1 - \lambda \end{bmatrix} \right) = \lambda(\lambda + 1); \tag{10}$

the roots of this polynomial are

$\lambda = 0, -1, \tag{11}$

both of which are real. Thus $(0, 0)$ is not a center.

There has been some discussion in the comments suggesting that perhaps the system (2)-(3) should be replaced with

$\dot x = y, \tag{12}$

$\dot y = -x + ax^2 + bxy; \tag{13}$

this system has to zeroes; from (12) we have

$y = 0, \tag{12}$

and then (13) becomes

$0 = -x + ax^2 = x(ax - 1), \tag{13}$

whence

$x = 0, a^{-1}; \tag{14}$

therefore we need inspect the two zeroes $(0, 0)$ and $(a^{-1}, 0)$.

In this case the Jacobian matrix takes the form

$J(x, y) = \begin{bmatrix} 0 & 1 \\ -1 + 2ax + by & bx \end{bmatrix}, \tag{15}$

and we have

$J(0, 0) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}; \tag{16}$

the characteristic polynomial is now

$\det \left( \begin{bmatrix} -\lambda & 1 \\ -1 & -\lambda \end{bmatrix} \right) = \lambda^2 + 1, \tag{17}$

and the roots are

$\lambda = \pm i, \tag{18}$

so that $(0, 0)$ is in fact a center.

The Jacobian at $(a^{-1}, 0)$ is

$J(a^{-1}, 0) = \begin{bmatrix} 0 & 1 \\ 1 & ba^{-1} \end{bmatrix}, \tag{19}$

with characteristic polynomial

$\det \left ( \begin{bmatrix} -\lambda & 1 \\ 1 & ba^{-1} - \lambda \end{bmatrix} \right ) = \lambda^2 - a^{-1}b\lambda - 1, \tag{20}$

the roots of which are

$\lambda = \dfrac{a^{-1}b \pm \sqrt{b^2 a^{-2} + 4}}{2}. \tag{21}$

More to follow. Stay tuned.