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Noncommuting two-by-two matrices distinguish Lie and Strang local errors

Example

On R2, let a(x)=Ax and b(x)=Bx, where A=(0100),B=(0010). The fields are smooth and their exact flows exist globally. With the bracket convention in Lie and Strang compositions of exact split flows, the Lie split has a nonzero h2 local defect and the symmetric Strang split has a nonzero h3 local defect at x=e1=(1,0)T.

Facts & Assumptions

Given: The two matrices and the corresponding linear vector fields on R2.

[F1]

Lie and Strang compositions of exact split flows: for linear fields a(x)=Ax, b(x)=Bx, the bracket is [a,b](x)=(BA−AB)x.

[F2]

Lie and Strang splitting errors have commutator expansions: the Lie defect is 12h2[a,b](x)+O(h3) and the Strang defect has leading coefficient −124[a,[a,b]](x)−112[b,[a,b]](x) at order h3.

[F3]

Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes: matrix products are computed by row-by-column multiplication.

[F4]

The exponential series of a bounded operator: every finite matrix M has an everywhere-defined exponential etM with e0=I and derivative MetM, so its linear ODE flow exists for all real t.

Proof

1.1F1F3F4algebra

By [F4], the linear fields have global exact flows. Direct multiplication gives BA=(0001),AB=(1000),C:=BA−AB=(−1001). By [F1], [a,b](x)=Cx.

2.1F1F3step 1.1algebra

Multiplying once more gives CA−AC=(0−200),CB−BC=(0020). These are the matrices of [a,[a,b]] and [b,[a,b]] under the same convention.

2.2F2step 1.1algebra

At e1, Ce1=−e1≠0, so [F2] gives Lie local defect −12h2e1+O(h3), whose leading coefficient is nonzero. Thus its local error has exact order two.

3.1F2step 2.1algebra∎

The Strang leading matrix from step 2.1 is −124(CA−AC)−112(CB−BC)=(01/12−1/60). It sends e1 to (0,−1/6)T≠0, so [F2] gives a nonzero cubic leading local defect and exact local order three.

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