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A bounded approximate Jacobian can destroy Rosenbrock stiff damping

Statement

For the scalar family y′=−λy with λ>0 and y0=1, take h=4/λ, J~=0, and T~=0 in the exact-arithmetic stage equations of the two-stage method. Then W=1, its amplification factor is 5, and this approximate-J method amplifies the mode even though the exact-J method at z=−4 has modulus less than one. The zero approximation is bounded. This calculation demonstrates loss of stiff damping and makes no order or uniform-convergence claim.

Facts & Assumptions

Given: The scalar equation, λ>0, h=4/λ, and the stated approximate derivatives.

[F1]

Two-stage second-order Rosenbrock W-method with shared factorization: the stages satisfy Wk1=f+αhT~, F1=f(y+hk1/2) for this autonomous equation, Wq2=F1−k1, k2=q2+k1, and y+=y+hk2.

[F2]

The selected exact-J two-stage Rosenbrock method is scalar L-stable: for the exact Jacobian, R(z)=[1+(1−2α)z]/(1−αz)2 and it is L-stable.

Proof

Given: The data in the Statement.

1.1givenF1algebra

Here W=1 and k1=f(y)=−λy. Since hk1/2=−2y, the midpoint state is −y and F1=f(−y)=λy.

2.1step 1.1F1algebra

The second right-hand side is F1−k1=2λy, so q2=2λy and k2=q2+k1=λy. Therefore y+=y+hλy=5y, giving amplification factor 5.

3.1givenF2step 2.1algebra

For the exact-J branch, [F2] gives R(−4)=[1−4(1−2α)]/(1+4α)2=(5−42)/(5−22)2, because α=1−1/2. Since 25<32<36, we have 5<42<6, so ∣5−42∣<1; also 22<4, so 5−22>1. Thus ∣R(−4)∣<1. This differs from step 2.1 and proves that boundedness of J~ alone does not preserve stiff damping.

4.1step 1.1step 2.1algebra∎

The approximate-J calculation in step 2.1 and the exact-J factor comparison establish the stated damping failure. No order or convergence conclusion follows from this stiff-scaled calculation.

Finite check

With λ=4 and h=1, the approximate-J output is 5y, while the exact-J factor is (5−42)/(5−22)2≈−0.1393.

Depends on

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