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Rosenbrock–Euler has an explicit one-step defect and solve-residual allowance

Statement

Let D⊆R×Rd be open and f:D→Rd be C2 (componentwise, as in Ck Euclidean maps and diffeomorphisms). Let y(s) solve y′=f(s,y) on [t,t+h], h>0, and set J=Dyf(t,y(t)). Suppose the trajectory lies in a region where ∥f∥2≤F, ∥Dyf∥2≤L, and ∥ft+Dyf f∥2≤M1. Fix γ>0 and a real matrix J~. Suppose A=I−γhJ and A~=I−γhJ~ are invertible with ∥A−1∥2≤κ0 and ∥A~−1∥2≤κ1. Write δ=∥J~−J∥2.

For the exact-arithmetic Rosenbrock stage map Φγ,hJ~, and a computed stage k^ whose update residual obeys ∥A~k^−f(t,y(t))∥2≤ρ, the update y^+=y(t)+hk^ satisfies ∥y^+−y(t+h)∥2≤CEh2+κ1hρ,CE:=γκ0LF+M12+γκ0κ1δF. The last term is the separate Jacobian-approximation allowance. The residual is an equation residual in the units of Implicit residuals and their update and derivative units, not a forward-error estimate by itself.

Facts & Assumptions

Given: The stated exact solution, derivative bounds, matrices, inverse bounds, and residual bound.

[F1]

The Rosenbrock–Euler step map with an exact or approximate Jacobian: Φγ,hJ(t,y)=y+h(I−γhJ)−1f(t,y).

[F2]

The one-stage update has an explicit Jacobian-approximation defect: ∥Φγ,hJ~(t,y)−Φγ,hJ(t,y)∥2≤γh2κ0κ1δ∥f(t,y)∥2.

[F3]

Induced matrix norms are compatible with matrix-vector multiplication, submultiplicative, and satisfy ||I|| = 1: the induced Euclidean norm satisfies ∥A−1z∥2≤∥A−1∥2∥z∥2.

[F4]

If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz: a continuously differentiable vector curve has endpoint difference equal to the integral of its derivative, and hence a derivative bound M gives endpoint variation at most M times the interval length.

[F5]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): derivatives of a composition are obtained by composing the total derivatives.

[F6]

Finite-precision one-stage Rosenbrock step with residual status: the stage equation is (I−γhJ~)k=f(t,y) and the update is y+hk.

[F7]

A computed shifted-system residual has a certified upper bound: a successful solve can supply a certified upper bound for the exact linear-equation residual.

[F8]

If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative: since f is componentwise C2, all its first partial derivatives exist and are continuous; applying this theorem at each point makes f totally differentiable there.

[F9]

A total derivative computes every directional derivative, and its matrix is the Jacobian: in the joint variable (t,y), the nonzero standard time and state basis directions identify the corresponding columns of Df with ft and Dyf.

Proof

Given: The data in the Statement.

1.1givenF4F5F8F9F10F11algebra

The componentwise C2 hypothesis and [F8] make f totally differentiable along the solution; [F9] identifies the time and state derivative blocks used in the chain rule [F5], giving y′′(s)=ft(s,y(s))+Dyf(s,y(s))f(s,y(s)). Thus y′ and y′′ are continuous and, by [F11], integrable on the compact interval. For each 0≤r≤h, the vector FTC [F4] applied to y′ gives ∥y′(t+r)−y′(t)∥2≤M1r (the case r=0 is equality). Applying [F4] to y then yields y(t+h)−y(t)−hy′(t)=∫0h(y′(t+r)−y′(t)) dr; [F10] bounds its norm by ∫0hM1r dr=M1h2/2.

1.2givenF1F3F6F7algebra

If r=A~k^−f0, subtraction of the exact stage equation gives k^−k=A~−1r. Thus [F3] and the residual bound yield ∥k^−k∥2≤κ1ρ, and multiplication by h bounds the update difference by κ1hρ.

2.1givenF1F3step 1.1algebra

Put f0=f(t,y(t)). From [F1], Φγ,hJ(t,y(t))−y(t)−hf0=γh2A−1Jf0. By [F3] and the bounds on A−1,J,f0, its norm is at most γκ0LFh2. Combining with step 1.1 bounds the exact-J map's local defect by (γκ0LF+M1/2)h2.

3.1step 2.1F2givenalgebra

Apply [F2] and ∥f0∥2≤F to compare the approximate-J map with the exact-J map. The resulting difference is at most γh2κ0κ1δF. Adding the step 2.1 defect gives ∥Φγ,hJ~(t,y(t))−y(t+h)∥2≤CEh2.

4.1step 3.1step 1.2algebra∎

Add the two bounds from steps 3.1 and 1.2 by the triangle inequality. This is the claimed estimate, with no conversion of ρ to forward error beyond the stated inverse bound.

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