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The continuous ODE adjoint gives the parameter gradient with endpoint terms

Statement

Assume T>0. Let uθ solve the parameter-dependent ODE of Parameter-dependent ODE flows have the variational tangent on a common tube on [0,T]. Let J(θ)=ϕ(uθ(T),θ)+∫0Tℓ(t,uθ(t),θ) dt, where ϕ and ℓ are C1 on a neighborhood of the trajectory tube and their first derivatives are bounded and uniformly continuous there. Let S=Dθuθ be the tangent from that lemma. Let p solve the terminal adjoint problem −p′(t)=Duf(t,uθ(t),θ)Tp(t)+Duℓ(t,uθ(t),θ)T,p(T)=Duϕ(uθ(T),θ)T. Then ∇θJ=Dθϕ(uθ(T),θ)T+Dθu0(θ)Tp(0)+∫0T(Dθℓ(t,uθ,θ)T+Dθf(t,uθ,θ)Tp(t)) dt. This formula separates initial, terminal, running-cost and vector-field parameter terms. It applies only while the common-tube and differentiability hypotheses hold.

Facts & Assumptions

Given: T>0, the common-tube ODE, C1 initial map, objective, and bounded uniformly continuous first derivatives in the Statement.

[F1]

Parameter-dependent ODE flows have the variational tangent on a common tube: the parameter flow is C1 and S′=Duf S+Dθf, S(0)=Dθu0.

[F2]

A tangent and adjoint pairing gives the ODE gradient boundary identity: if S′=AS+B and p′=−ATp−g, then p(T)TS(T)−p(0)TS(0)=∫0T(−gTS+pTB) dt.

[F3]

Linear matrix ODEs have unique global solutions on a fixed interval: a continuous coefficient linear matrix IVP has a unique solution on the full compact interval.

[F4]

The total-derivative chain rule differentiates a composition of C1 Euclidean maps (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

[A1]

All objective derivatives and ODE derivative coefficients are continuous and bounded on the common compact tube, and their first derivatives have the uniform continuity needed to differentiate the integral objective.

Proof

1.1F3givenconstruct

Fix θ and write A(t)=Duf(t,uθ(t),θ) and g(t)=Duℓ(t,uθ(t),θ)T. The coefficient matrix C(t)=(−A(t)T−g(t)00) is continuous. Apply [F3] with initial time T to the homogeneous matrix equation Z′=C(t)Z, choosing an initial matrix whose first column is (Duϕ(uθ(T),θ)T,1)T and whose other columns are zero. Its first column is (p(t),1)T and solves the terminal adjoint equation. Any other solution embeds as a column of such a matrix solution, so uniqueness in [F3] gives uniqueness of p.

1.2F1F4A1given

By [F1] and the chain rule [F4], the terminal derivative is DθϕT+S(T)TDuϕT. For the running term, a parameter increment h gives Δu(t)=S(t)h+o(∥h∥) uniformly in t by [F1]. Uniform continuity of the objective derivatives on the common compact tube gives a uniform first-order remainder o(∥h∥) along the segment from (uθ(t),θ) to (uθ+h(t),θ+h); the common integrable majorant permits integration of this remainder. Applying [F4] pointwise and integrating therefore gives the running derivative ∫0T(DθℓT+STDuℓT) dt. Consequently ∇θJ=DθϕT+S(T)TDuϕT+∫0T(DθℓT+STDuℓT) dt.

1.3F1F2given

In [F2], set A=Duf, B=Dθf, and g=DuℓT. The adjoint equation in the Statement and tangent equation in [F1] satisfy its hypotheses. Transposing that row pairing identity and using p(T)=DuϕT gives S(T)TDuϕT−Dθu0Tp(0)=∫0T(−STDuℓT+DθfTp) dt.

2.1step 1.2step 1.3algebra∎

Substitute step 1.3 into step 1.2; the terms involving STDuℓT cancel, leaving the asserted continuous adjoint gradient formula.

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