Paper VI: Refinement defects and the Duhamel identity

3 Refinement defects and the Duhamel identity

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How to read the result

Read the generator defect as a mismatch between two declared paths: evolve and then coarsen, or coarsen and then evolve. The Duhamel identity transports that mismatch through time. Its estimate depends on the stated operator norms and growth bounds.

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Let \(X_f\) and \(X_c\) be fine and coarse spaces, with generators \(A_f\) and \(A_c\), and let

\[ C:X_f\to X_c \]

be a coarsening map. Define the generator-level refinement defect

\[ \boxed{\Delta=CA_f-A_cC.} \]

The corresponding semigroup defect is

\[ \mathfrak D(t)=Ce^{tA_f}-e^{tA_c}C. \]

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Theorem 3.1: Exact Duhamel refinement identity

For every \(t\ge0\),

\[ \boxed{ Ce^{tA_f}-e^{tA_c}C = \int_0^t e^{(t-r)A_c}\Delta e^{rA_f}\,dr. } \]

In particular, \(\Delta=0\) implies exact semigroup intertwining at every time.

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Proof of Theorem 3.1

Set

\[ F(r)=e^{(t-r)A_c}Ce^{rA_f}. \]

Then

\[ F'(r)=e^{(t-r)A_c}(CA_f-A_cC)e^{rA_f}. \]

Integrating from \(0\) to \(t\) gives \(F(t)-F(0)=Ce^{tA_f}-e^{tA_c}C\) and proves the formula.

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Corollary 3.2: Norm bound

Suppose

\[ {\left\lVert e^{tA_f}\right\rVert}\le M_f e^{\omega_f t}, \qquad {\left\lVert e^{tA_c}\right\rVert}\le M_c e^{\omega_c t}. \]

Then

\[ {\left\lVert\mathfrak D(t)\right\rVert} \le M_fM_c{\left\lVert\Delta\right\rVert} \int_0^t e^{\omega_c(t-r)}e^{\omega_fr}\,dr. \]

For contraction semigroups,

\[ \boxed{{\left\lVert\mathfrak D(t)\right\rVert}\le t{\left\lVert\Delta\right\rVert}.} \]

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Exact generator intertwining gives a vanishing semigroup defect. A controlled perturbation produces a nonzero defect propagated by the Duhamel identity.

3.1 Resolvent form

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For \(z\) in the common resolvent set, let

\[ R_f(z)=(zI-A_f)^{-1}, \qquad R_c(z)=(zI-A_c)^{-1}. \]

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Theorem 3.3: Resolvent refinement identity

\[ \boxed{ CR_f(z)-R_c(z)C =R_c(z)\Delta R_f(z). } \]

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Proof of Theorem 3.3

Multiply the left side by \(zI-A_c\) on the left and \(zI-A_f\) on the right. The result is

\[ (zI-A_c)C-C(zI-A_f)=CA_f-A_cC=\Delta. \]

Multiplication by the two resolvents recovers the stated identity.

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