Center Manifold for PDE Dynamics: A Concrete Example

Center Manifold for PDE Dynamics: A Concrete Example

Consider the 1-dimensional parabolic problem

ut=uxx+(μ+1)uu2,u(x,t)Hper1(0,2π)×0.

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

Linear Analysis

Write the equation as

ut=Lu+N(u;μ),

where L=xx+1 and N(u;μ)=μuu2. We can find the eigenvalues through the linearization at (u,μ)=(0,0),

Leikx=λkeikx,

yielding the eigenvalues: λk=1k2,k.

Center Subspace (Ec) The center subspace corresponds to eigenvalues with zero real part (λk=0).

1k2=0k=±1.

Thus, the center subspace is spanned by the modes eix and eix.

Ec=span{eix,eix}.

Hyperbolic Subspace (Eh) The hyperbolic subspace corresponds to eigenvalues with non-zero real parts (k±1). Therefore, the hyperbolic subspace is spanned by all the other modes.

Eh=span{eikx|k,k±1}.

1. Projections

We define the projection operators Pc and Ph onto the center and hyperbolic subspaces, respectively:

The projection operator Pc onto the center subspace Ec for a function u(x), using the standard L2 inner product f,g=12π02πf(x)g(x)dx, is given by

Pcu=u,eixeix+u,eixeix

The projection onto the hyperbolic subspace is the complement: Phu=(IPc)u

For convenience, we use the projection acting on the Fourier series of a function f(x)=fkeikx throughout this article.

Projection onto Center Subspace (Pc)

Pcf:=f1eix+f1eix

Projection onto Hyperbolic Subspace (Ph)

Phf:=fPcf=k±1fkeikx

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 vEc and wEh, and then substitute it into the PDE:

vt+wt=L(v+w)+μ(v+w)(v+w)2=Lw+μv+μw(v2+2vw+w2)

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

v(x,t)=A(t)eix+A(t)eix,w(x,t)=k±1wk(A(t),A(t),μ)eikx

We project this equation onto Ec and Eh.

Projection onto Ec:

vt=Pc[μvv22vww2]

because the t-derivative does not change the modes of w on the LHS (so Lwt=0), and Pc(Lw)=L(Pcw)=0, Pc(μw)=0 on the RHS. Extracting the coefficient of eix gives the reduced equation with respect to A:

A˙=μAv2+2vw+w2,eix

Note: This projection produces another reduced equation that involves A˙ by collecting the eix terms. We are omitting it because it is simply a conjugate copy of the A˙ equation. In general, a center subspace spanned by two generic basis functions would result in a 2-dimensional reduced system.

Projection onto Eh:

wt=Lw+Ph[μv+μwv22vww2]

Since Ph(μv)=0 (because vEc), this simplifies to:

wt=Lw+μwPh[v2+2vw+w2]

To summarize, we have now computed the reduced equation on Ec up to the first order

A˙=μA+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 2vw and w2 terms in the w-equation as they are of order 3 and 4, respectively. Recall that the reduced Eh equation for w is:

wtLwμw=Ph(v2)

We expand w in the hyperbolic modes:

w(x)=w0(μ,A,A)+w2(μ,A,A)e2ix+w2(μ,A,A)e2ix+

Now compute each term of the reduced Eh equation.

RHS term Pc(v2):

v2=(Aeix+Aeix)2=A2e2ix+2AA+A2e2ix

All terms of v2 are in Eh (modes 0 and ±2). So Pc(v2)=v2.

LHS term wt: Using the chain rule and the fact that w depends on time only through A and A:

wt=wAA˙+wAA˙

At leading order, A˙μA. (we only need the linear term from the previous section here because w is already second order; higher order corrections to A˙ would push the result to O(|A|4)). Thus,

wtμAwA+μAwA

To be very, very precise, we can assume that wk take the form

wk(μ,A,A)=α,β,γwk,α,β,γμαAβAγ

in which case the derivatives wA and wA 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 w0=C0AA, w2=C2A2, and w2=C2A2 for some C0,±2(μ). How do I know that these are all the terms for each wk? In other words, what happens if we include another term in one of the wk’s, say, w0=C0AA+C0A2? In this case, the e0ix (constant) terms of the equation becomes

C0(A˙A+AA˙)+2C0AA˙Substitite A˙μA(1+μ)(C0AA+C0A2The only A2 term)=2AA

Note that this is the only equation that involves C0 and C0 because the linear operator (tLμ) on the LHS does not change the mode of the input function. The constant term on the RHS only has an AA term in its coefficient and hence forces C0=0. This observation enables us to eliminate most of the terms with zero coefficients.

We solve for C0,±2 by matching terms with the same frequencies. Substitute the ansatz into the reduced Eh equation:

(tLμ)(C2A2e2ix+C0AA+C2A2e2ix)=A2e2ix+2AA+A2e2ix

Solve for the Mode k=0:

Substitute w0=C0AA:

  1. LHS wt term: Using A˙μA, w0A=C0A, and w0A=C0A, we have wt(μA)(C0A)+(μA)(C0A)=2μC0AA=2μw0
  2. LHS Lw term: Lw0=1w0=w0
  3. RHS: The constant term in v2 is 2AA.

Assemble the equation for w0:

2μw0w0μw0=2AA (μ1)w0=2AA w0=21μAA=21μ|A|2

Solve for the Mode k=2:

Substitute w2=C2A2:

  1. LHS wt term: w2A=2C2A,w2A=0 wt(μA)(2C2A)=2μC2A2=2μw2
  2. LHS L0w term: Lw2=3w2
  3. RHS: The e2ix term in v2 is A2.

Assemble the equation for w2:

2μw2(3w2)μw2=A2 (μ+3)w2=A2 w2=1μ+3A2

By symmetry (since u is real), w2=w2=1μ+3A2.


4. Assemble the Reduced Equation

Return to the equation for A˙:

A˙=μAv2+2vw+w2,eix

We analyze the terms inside the bracket that generate the eix mode.

  1. v2: Contains only 0,±2 modes. No resonance.

  2. w2: Although contains resonating terms, it is of order A4. Neglect.

  3. 2vw: This is the interaction term.

    2vw=2(Aeix+Aeix)(w0+w2e2ix+w2e2ix)

    The following terms are proportional to eix:

    • 2(Aeix)w0=2Aw0eix
    • 2(Aeix)(w2e2ix)=2Aw2eix

    Substituting the expressions for w0 and w2 found in the previous step:

    • 2Aw0=2A(21μAA)=41μA|A|2
    • 2Aw2=2A(1μ+3A2)=2μ+3A|A|2

Substitute these back into the A˙ equation:

A˙=μA(41μ2μ+3)A|A|2+O(|A|4)

If we set μ=0, the equation becomes

A˙=μA103A|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),A(t)) is the coordinate in the center subspace spanned by e±ix. 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 smooth 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)=h(0)=0+Dh(0)=0v+12D2h(0)v2+
  • 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.

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