11.4: D- Center Manifolds Depending on Parameters
In this appendix we describe the situation of center manifolds that depend on a parameter. The theoretical framework plays an important role in bifurcation theory.
As when we developed the theory earlier, we begin by describing the set-up. As before, it is important to realize that when applying these results to a vector field, it must be in the following form.
\(\dot{x} = Ax+f(x, y, \mu)\),
\[\dot{y} = By+g(x, y, \mu), (x, y, \mu) \in \mathbb{R}^{c} \times \mathbb{R}^{s} \times \mathbb{R}^{p}, \label{D.1}\]
where \(\mu \in \mathbb{R}^p\) is a vector of parameters and the matrices A and B are have the following properties:
-
A \(c \times c\) matrix of real numbers
having eigenvalues with zero real parts, -
B \(s \times s\) matrix of real numbers
having eigenvalues with negative real parts,
and f and g are nonlinear functions. That is, they are of order two or higher in x, y and \(\mu\), which is expressed in the following properties:
\[\begin{array} {cc} {f(0, 0, 0) = 0,} & {Df(0, 0, 0) = 0,} \end{array} \nonumber\]
\[\begin{array} {cc} {g(0, 0, 0) = 0,} & {Dg(0, 0, 0) = 0,} \end{array} \label{D.2}\]
and they are \(C^r\), r as large as is required to compute an adequate approximation the center manifold. With this set-up \((x, y, \mu) = (0, 0, 0)\) is a fixed point for (D.1) and we are interested in its stability properties.
The conceptual "trick" that reveals the nature of the parameter dependence of center manifolds is to include the parameter \(\mu\) as a new dependent variable:
\(\dot{x} = Ax+f(x, y, \mu)\),
\(\dot{\mu} = 0\),
\[\dot{y} = By+g(x, y, μ), (x, y, μ) \in \mathbb{R}^c \times \mathbb{R}^s \times \mathbb{R}^p, \label{D.3}\]
The linearization of (D.3) about the fixed point is given by:
\(\dot{x} = Ax\),
\(\dot{\mu} = 0\),
\[\dot{y} = By, (x, y, μ) \in \mathbb{R}^c \times \mathbb{R}^s \times \mathbb{R}^p. \label{D.4}\]
Even after increasing the dimension of the phase space by p dimensions by including the parameters as new dependent variables, the fixed point \((x, y, \mu) = (0, 0, 0)\) remains a nonhyperbolic fixed point. It has a c + p dimensional invariant center subspace and a s dimensional invariant stable subspace given by:
\[E^c = \{(x, y, \mu) \in \mathbb{R}^c \times \mathbb{R}^s \times \mathbb{R}^p, |y = 0\}, \label{D.5}\]
\[E^s = \{(x, y, \mu) \in \mathbb{R}^c \times \mathbb{R}^s \times \mathbb{R}^p,| x = 0, \mu = 0\} , \label{D.6}\]
respectively.
It should be clear that center manifold theory, as we have already developed, applies to (D.3). Including the parameters, \(\mu\) as additional dependent variables has the effect of increasing the dimension of the "center variables", but there is also an important consequence. Since \(\mu\) are now dependent variables they enter in to the determination of the nonlinear terms in the equations. In particular, terms of the form
\(x_{i}^{\ell} μ_{j}^{m} y_{k}^{n}\),
now are interpreted as nonlinear terms, when \(\ell + m + n > 1\), for nonnegative integers \(\ell\), m, n. We will see this in the example below.
Now we consider the information that center manifold theory provides us near the origin of (D.3).
- In a neighborhood of the origin there exists a \(C^r\) center manifold that is represented as the graph of a function over the center variables, \(h(x, \mu)\), it passes through the origin (h(0, 0) = 0) and is tangent to the center subspace at the origin (Dh(0, 0) = 0)
- All solutions sufficiently close to the origin are attracted to a trajectory in the center manifold at an exponential rate.
- The center manifold can be approximated by a power series expansion.
It is significant that the center manifold is defined in a neighborhood of the origin in both the x and \(\mu\) coordinates since \(\mu = 0\) is a bifurcation value. This means that all bifurcating solutions are contained in the center manifold. This is why, for example, that without loss of generality bifurcations from a single zero eigenvalue can be described by a parametrized family of one dimensional vector fields.
Example \(\PageIndex{40}\)
We now consider an example which was exercise 1b from Problem Set 8.
\(\dot{x} = \mu x+10x^2\),
\(\dot{μ} = 0\),
\[\dot{y} = x-2y, (x, y) \in \mathbb{R}^2, \mu \in \mathbb{R}. \label{D.7}\]
The Jacobian associated with the linearization about \((x, \mu, y) = (0, 0, 0)\) is given by:
\[\begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {1}&{0}&{-2} \end{pmatrix}. \label{D.8}\]
It is easy to check that the eigenvalues of this matrix are 0, 0, and 2 (as we would have expected). Each of these eigenvalues has an eigenvector. It is easily checked that two eigenvectors corresponding to the eigenvalue 0 are given by:
\[\begin{pmatrix} {2}\\ {0}\\ {1} \end{pmatrix}. \label{D.9}\]
\[\begin{pmatrix} {0}\\ {1}\\ {0} \end{pmatrix}. \label{D.10}\]
and an eigenvector corresponding to the eigenvalue \(-2\) is given by:
\[\begin{pmatrix} {0}\\ {0}\\ {1} \end{pmatrix}. \label{D.11}\]
From these eigenvectors we form the transformation matrix
\[T = \begin{pmatrix} {2}&{0}&{0}\\ {0}&{1}&{0}\\ {1}&{0}&{1} \end{pmatrix}. \label{D.12}\]
with inverse
\[T^{-1} = \begin{pmatrix} {\frac{1}{2}}&{0}&{0}\\ {0}&{1}&{0}\\ {-\frac{1}{2}}&{0}&{1} \end{pmatrix}. \label{D.13}\]
The transformation matrix, T, defines the following transformation of the dependent variables of (D.7):
\[\begin{pmatrix} {x}\\ {\mu}\\ {y} \end{pmatrix} = T \begin{pmatrix} {u}\\ {\mu}\\ {v} \end{pmatrix} = \begin{pmatrix} {2}&{0}&{0}\\ {0}&{1}&{0}\\ {1}&{0}&{1} \end{pmatrix} \begin{pmatrix} {u}\\ {\mu}\\ {v} \end{pmatrix} = \begin{pmatrix} {2u}\\ {\mu}\\ {u+v} \end{pmatrix} \label{D.14}\]
It then follows that the transformed vector field has the form:
\[\begin{pmatrix} {\dot{u}}\\ {\dot{\mu}}\\ {\dot{v}} \end{pmatrix} = \begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{-2} \end{pmatrix} \begin{pmatrix} {u}\\ {\mu}\\ {v} \end{pmatrix} + T^{-1} \begin{pmatrix} {\mu (2u)+10(2u)^2}\\ {0}\\ {0} \end{pmatrix} \label{D.15}\]
or
\[\begin{pmatrix} {\dot{u}}\\ {\dot{\mu}}\\ {\dot{v}} \end{pmatrix} = \begin{pmatrix} {0}&{0}&{0}\\ {0}&{0}&{0}\\ {0}&{0}&{-2} \end{pmatrix} \begin{pmatrix} {u}\\ {\mu}\\ {v} \end{pmatrix} + \begin{pmatrix} {\frac{1}{2}}&{0}&{0}\\ {0}&{1}&{0}\\ {-\frac{1}{2}}&{0}&{1} \end{pmatrix} \begin{pmatrix} {2\mu u+40u^2}\\ {0}\\ {0} \end{pmatrix} \label{D.16}\]
or
\(\dot{u} = \mu u+20u^2\),
\(\dot{μ} = 0\),
\[\dot{v} = -2v- \mu u-20u^2. \label{D.17}\]