Center Manifold for PDE Dynamics: A Concrete Example…

Consider the $1$ -dimensional parabolic problem

\begin{matrix} u_{t} = u_{x x} + (μ + 1) u - u^{2}, \\ u (x, t) \in H_{per}^{1} (0,2 π) \times ℝ_{\geq 0} . \end{matrix}

We study the local dynamics near the trivial solution $u = μ = 0$ .

Linear Analysis

Write the equation as

u_{t} = L u + N (u; μ),

where $L = \partial_{x x} + 1$ and $N (u; μ) = μ u - u^{2}$ . We can find the eigenvalues through the linearization at $(u, μ) = (0,0)$ ,

L e^{i k x} = λ_{k} e^{i k x},

yielding the eigenvalues: $λ_{k} = 1 - k^{2}, k \in ℤ$ .

Center Subspace ( $E^{c}$ ) The center subspace corresponds to eigenvalues with zero real part ( $λ_{k} = 0$ ).

1 - k^{2} = 0 ⟹ k = \pm 1.

Thus, the center subspace is spanned by the modes $e^{i x}$ and $e^{- i x}$ .

E^{c} = span {e^{i x}, e^{- i x}} .

Hyperbolic Subspace ( $E^{h}$ ) The hyperbolic subspace corresponds to eigenvalues with non-zero real parts ( $k \neq \pm 1$ ). Therefore, the hyperbolic subspace is spanned by all the other modes.

E^{h} = span {e^{i k x} | k \in ℤ, k \neq \pm 1} .

1. Projections

We define the projection operators $P_{c}$ and $P_{h}$ onto the center and hyperbolic subspaces, respectively:

The projection operator $P_{c}$ onto the center subspace $E^{c}$ for a function $u (x)$ , using the standard $L^{2}$ inner product $⟨ f, g ⟩ = \frac{1}{2 π} \int_{0}^{2 π} f (x) \overline{g} (x) d x$ , is given by

P_{c} u = ⟨ u, e^{i x} ⟩ e^{i x} + ⟨ u, e^{- i x} ⟩ e^{- i x}

The projection onto the hyperbolic subspace is the complement: $P_{h} u = (I - P_{c}) u$

For convenience, we use the projection acting on the Fourier series of a function $f (x) = \sum f_{k} e^{i k x}$ throughout this article.

Projection onto Center Subspace ( $P_{c}$ )

P_{c} f : = f_{1} e^{i x} + f_{- 1} e^{- i x}

Projection onto Hyperbolic Subspace ( $P_{h}$ )

P_{h} f : = f - P_{c} f = \sum_{k \neq \pm 1} f_{k} e^{i k x}

2. Substitute the Ansatz

We decompose the solution $u$ onto components on the center and hyperbolic subspaces:

u (x, t) = v (x, t) + w (x, t)

where $v \in E^{c}$ and $w \in E^{h}$ , and then substitute it into the PDE:

\begin{matrix} v_{t} + w_{t} & = L (v + w) + μ (v + w) - {(v + w)}^{2} \\ = L w + μ v + μ w - (v^{2} + 2 v w + w^{2}) \end{matrix}

because $L v = 0$ . Here, $v$ and $w$ can be expressed using the amplitude $A (t)$ and the bases of the center and hyperbolic subspaces:

\begin{matrix} v (x, t) & = A (t) e^{i x} + \overline{A} (t) e^{- i x}, \\ w (x, t) & = \sum_{k \neq \pm 1} w_{k} (A (t), \overline{A} (t), μ) e^{i k x} \end{matrix}

We project this equation onto $E^{c}$ and $E^{h}$ .

Projection onto $E^{c}$ :

v_{t} = P_{c} [μ v - v^{2} - 2 v w - w^{2}]

because the $t$ -derivative does not change the modes of $w$ on the LHS (so $L w_{t} = 0$ ), and $P_{c} (L w) = L (P_{c} w) = 0$ , $P_{c} (μ w) = 0$ on the RHS. Extracting the coefficient of $e^{i x}$ gives the reduced equation with respect to $A$ :

\dot{A} = μ A - ⟨ v^{2} + 2 v w + w^{2}, e^{i x} ⟩

Note: This projection produces another reduced equation that involves $\dot{\overline{A}}$ by collecting the $e^{- i x}$ terms. We are omitting it because it is simply a conjugate copy of the $\dot{A}$ equation. In general, a center subspace spanned by two generic basis functions would result in a $2$ -dimensional reduced system.

Projection onto $E^{h}$ :

w_{t} = L w + P_{h} [μ v + μ w - v^{2} - 2 v w - w^{2}]

Since $P_{h} (μ v) = 0$ (because $v \in E^{c}$ ), this simplifies to:

w_{t} = L w + μ w - P_{h} [v^{2} + 2 v w + w^{2}]

To summarize, we have now computed the reduced equation on $E^{c}$ up to the first order

\dot{A} = μ A + O (| A |^{2}) .

Higher Orders

To close the reduced equation at cubic order $O (| A |^{3})$ , we need $w$ at quadratic order $O (| A |^{2})$ . Thus, we can neglect the $2 v w$ and $w^{2}$ terms in the $w$ -equation as they are of order $3$ and $4$ , respectively. Recall that the reduced $E^{h}$ equation for $w$ is:

w_{t} - L w - μ w = - P_{h} (v^{2})

We expand $w$ in the hyperbolic modes:

w (x) = w_{0} (μ, A, \overline{A}) + w_{2} (μ, A, \overline{A}) e^{2 i x} + w_{- 2} (μ, A, \overline{A}) e^{- 2 i x} + \dots

Now compute each term of the reduced $E^{h}$ equation.

RHS term $P_{c} (v^{2})$ :

v^{2} = {(A e^{i x} + \overline{A} e^{- i x})}^{2} = A^{2} e^{2 i x} + 2 A \overline{A} + {\overline{A}}^{2} e^{- 2 i x}

All terms of $v^{2}$ are in $E^{h}$ (modes 0 and $\pm 2$ ). So $P_{c} (v^{2}) = v^{2}$ .

LHS term $w_{t}$ : Using the chain rule and the fact that $w$ depends on time only through $A$ and $\overline{A}$ :

w_{t} = \frac{\partial w}{\partial A} \dot{A} + \frac{\partial w}{\partial \overline{A}} \dot{\overline{A}}

At leading order, $\dot{A} \approx μ A$ . (we only need the linear term from the previous section here because $w$ is already second order; higher order corrections to $\dot{A}$ would push the result to $O (| A |^{4})$ ). Thus,

w_{t} \approx μ A \frac{\partial w}{\partial A} + μ \overline{A} \frac{\partial w}{\partial \overline{A}}

To be very, very precise, we can assume that $w_{k}$ take the form

w_{k} (μ, A, \overline{A}) = \sum_{α, β, γ \in ℤ} w_{k, α, β, γ} μ^{α} A^{β} {\overline{A}}^{γ}

in which case the derivatives $\frac{\partial w}{\partial A}$ and $\frac{\partial w}{\partial \overline{A}}$ becomes explicit. The radius of convergence of this power series is not a concern as the center manifold only makes sense locally.

3: More Substitution and Projections

Assume $w_{0} = C_{0} A \overline{A}$ , $w_{- 2} = C_{- 2} {\overline{A}}^{2}$ , and $w_{2} = C_{2} A^{2}$ for some $C_{0, \pm 2} (μ)$ . How do I know that these are all the terms for each $w_{k}$ ? In other words, what happens if we include another term in one of the $w_{k}$ ’s, say, $w_{0} = C_{0} A \overline{A} + C_{0}^{*} A^{2}$ ? In this case, the $e^{0 i x}$ (constant) terms of the equation becomes

\underset{Substitite \dot{A} \approx μ A}{\underset{⏟}{C_{0} (\dot{A} \overline{A} + A \dot{\overline{A}}) + 2 C_{0}^{*} A \dot{A}}} - (1 + μ) (C_{0} A \overline{A} + \underset{The only A^{2} term}{\underset{⏟}{C_{0}^{*} A^{2}}}) = 2 A \overline{A}

Note that this is the only equation that involves $C_{0}$ and $C_{0}^{*}$ because the linear operator $(\partial_{t} - L - μ)$ on the LHS does not change the mode of the input function. The constant term on the RHS only has an $A \overline{A}$ term in its coefficient and hence forces $C_{0}^{*} = 0$ . This observation enables us to eliminate most of the terms with zero coefficients.

We solve for $C_{0, \pm 2}$ by matching terms with the same frequencies. Substitute the ansatz into the reduced $E^{h}$ equation:

(\partial_{t} - L - μ) (C_{- 2} {\overline{A}}^{2} e^{- 2 i x} + C_{0} A \overline{A} + C_{2} A^{2} e^{2 i x}) = A^{2} e^{2 i x} + 2 A \overline{A} + {\overline{A}}^{2} e^{- 2 i x}

Solve for the Mode k=0:

Substitute $w_{0} = C_{0} A \overline{A}$ :

LHS $w_{t}$ term: Using $\dot{A} \approx μ A$ , $\frac{\partial w_{0}}{\partial A} = C_{0} \overline{A}$ , and $\frac{\partial w_{0}}{\partial \overline{A}} = C_{0} A$ , we have $w_{t} \approx (μ A) (C_{0} \overline{A}) + (μ \overline{A}) (C_{0} A) = 2 μ C_{0} A \overline{A} = 2 μ w_{0}$
LHS $L w$ term: $L w_{0} = 1 \cdot w_{0} = w_{0}$
RHS: The constant term in $- v^{2}$ is $- 2 A \overline{A}$ .

Assemble the equation for $w_{0}$ :

2 μ w_{0} - w_{0} - μ w_{0} = - 2 A \overline{A}

(μ - 1) w_{0} = - 2 A \overline{A}

w_{0} = \frac{2}{1 - μ} A \overline{A} = \frac{2}{1 - μ} | A |^{2}

Solve for the Mode k=2:

Substitute $w_{2} = C_{2} A^{2}$ :

LHS $w_{t}$ term: $\frac{\partial w_{2}}{\partial A} = 2 C_{2} A, \frac{\partial w_{2}}{\partial \overline{A}} = 0$ $w_{t} \approx (μ A) (2 C_{2} A) = 2 μ C_{2} A^{2} = 2 μ w_{2}$
LHS $L_{0} w$ term: $L w_{2} = - 3 w_{2}$
RHS: The $e^{2 i x}$ term in $- v^{2}$ is $- A^{2}$ .

Assemble the equation for $w_{2}$ :

2 μ w_{2} - (- 3 w_{2}) - μ w_{2} = - A^{2}

(μ + 3) w_{2} = - A^{2}

w_{2} = - \frac{1}{μ + 3} A^{2}

By symmetry (since $u$ is real), $w_{- 2} = {\overline{w}}_{2} = - \frac{1}{μ + 3} {\overline{A}}^{2}$ .

4. Assemble the Reduced Equation

Return to the equation for $\dot{A}$ :

\dot{A} = μ A - ⟨ v^{2} + 2 v w + w^{2}, e^{i x} ⟩

We analyze the terms inside the bracket that generate the $e^{i x}$ mode.

$v^{2}$ : Contains only $0, \pm 2$ modes. No resonance.
$w^{2}$ : Although contains resonating terms, it is of order $A^{4}$ . Neglect.
$2 v w$ : This is the interaction term.
$2 v w = 2 (A e^{i x} + \overline{A} e^{- i x}) (w_{0} + w_{2} e^{2 i x} + w_{- 2} e^{- 2 i x})$
The following terms are proportional to $e^{i x}$ :
- $2 (A e^{i x}) \cdot w_{0} = 2 A w_{0} e^{i x}$
- $2 (\overline{A} e^{- i x}) \cdot (w_{2} e^{2 i x}) = 2 \overline{A} w_{2} e^{i x}$
Substituting the expressions for $w_{0}$ and $w_{2}$ found in the previous step:
- $2 A w_{0} = 2 A (\frac{2}{1 - μ} A \overline{A}) = \frac{4}{1 - μ} A | A |^{2}$
- $2 \overline{A} w_{2} = 2 \overline{A} (- \frac{1}{μ + 3} A^{2}) = - \frac{2}{μ + 3} A | A |^{2}$

Substitute these back into the $\dot{A}$ equation:

\dot{A} = μ A - (\frac{4}{1 - μ} - \frac{2}{μ + 3}) A | A |^{2} + O (| A |^{4})

If we set $μ = 0$ , the equation becomes

\dot{A} = μ A - \frac{10}{3} A | A |^{2} + O (| A |^{4})

For even higher order expansion of this reduced equation, one can repeat the procedure (substitution, projection, and matching coefficients) based on this order 3 expansion.

As a final reminder, $(A (t), \overline{A} (t))$ is the coordinate in the center subspace spanned by $e^{\pm i x}$ . So this reduced equation describes the dynamics near the trivial solution $(u, μ) = (0,0)$ . Another note is that $w$ should be thought of as the image of $v$ under a smooth1 The regularity of the center manifold is whole another topic. Refer to the book above for more details. map $h$ , i.e., $w = h (v; μ)$ . By inspecting the Taylor expansion of $h$ (suppressing $μ$ for simplicity):

w = h (v) = \underset{= 0}{\underset{⏟}{h (0)}} + \underset{= 0}{\underset{⏟}{D h (0)}} v + \frac{1}{2} D^{2} h (0) v^{2} + \dots

The first term vanishes because the center manifold passes through the point $u = 0$ .
The second term is zero because the center manifold is tangent to the center subspace at $u = 0$ . This is why we can assume $w$ is of order $O (| A |^{2})$ in the calculation.

Reference

Local Bifurcations, Center Manifolds, and Normal Forms in Infinite-Dimensional Dynamical Systems by Mariana Haragus and Gérard Iooss.

The regularity of the center manifold is whole another topic. Refer to the book above for more details.