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Backward Euler differentiates a stiff scalar decay fit on its exact branch

Example

For u′=−ku with fixed u0>0 and step size h>0, take k∈(0,(2h)−1) and the one-stage backward Euler equation Y=u0−hkY. Its fixed-point map sends the invariant interval [u0/2,u0] into itself and has contraction factor hk<1/2, so the branch selected by Small-step implicit Runge–Kutta stages have a unique branch and controlled residual error is well defined. The observation is zero and the terminal loss is J(k)=12Y(k)2.

Facts & Assumptions

Given: The scalar backward Euler stage equation, the stated parameter interval, and the terminal quadratic objective.

[F1]

The one-stage implicit Runge--Kutta residual has the selected stage root and residual bound under its invariant-ball contraction hypotheses (Small-step implicit Runge–Kutta stages have a unique branch and controlled residual error).

[F2]

On a nonsingular selected stage branch, implicit differentiation gives DθY=−GY−1Gθ and differentiates the Runge--Kutta update (An implicit Runge-Kutta stage has the stated parameter sensitivity on its selected branch).

[F3]

Backward Euler is the one-stage implicit Runge--Kutta method with A=(1) and b=(1) (Backward Euler step equation and algebraic residual).

Verification

1.1F1F2F3givenalgebra

To apply [F1] with the prescribed center u0, use the larger closed ball B=[0,2u0] of radius R=u0. For F(Y)=−kY, take M=2ku0 and L=k; then hM=2hku0<R and q=hk<1/2. The field is C1 on R with derivative bounded by L, so the nonsingularity hypotheses inherited by [F2] hold on this ball. Solving the scalar residual gives (1+hk)Y=u0, hence Y=u0/(1+hk). Since 1+hk>0, this is the unique root in the stated interval and agrees with the branch in [F1].

2.1F2step 1.1algebra∎

Differentiate the residual (1+hk)Y−u0=0: (1+hk)Yk+hY=0, so [F2] gives Yk=−hu0/(1+hk)2. Therefore Jk=YYk=−hu02/(1+hk)3.

Depends on

Used by

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