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RKC stability polynomials and certified negative-real intervals

Statement

On the scalar test equation y′=λy, write z=hλ and let R(z) be the factor by which a successful method step multiplies y.

For first-order RKC with s≥1 stages, R1,s(z)=Ts(1+z/s2). It has degree s, satisfies R1,s(0)=R1,s′(0)=1, and its real absolute stability set {z∈R:∣R1,s(z)∣≤1} is exactly [−2s2,0].

For second-order RKC with s≥2 stages, both the linear algorithm and the scalar-linear specialization of the nonlinear algorithm have R2,s(z)=as+bsTs(1+w1z),w1=3s2−1,bs=s2−13s2,as=1−bs. This polynomial has degree s and satisfies R2,s(0)=R2,s′(0)=R2,s′′(0)=1. It is absolutely stable on the real interval [−2(s2−1)/3,0]; this is a certified interval and is not claimed to be the maximal real stability set.

Let d≥1, let A be a self-adjoint endomorphism of Rd with its Euclidean inner product, and assume ⟨Av,v⟩≤0 for every v. Let ρ=max⁡i(−λi), where λi are the eigenvalues of A. If hρ≤2s2, first-order RKC is nonexpansive in the Euclidean norm; if hρ≤2(s2−1)/3, linear second-order RKC is nonexpansive. The same linear contraction statement holds for the nonlinear RKC method when applied to f(y)=Ay. No corresponding claim is made for nonnormal matrices.

Facts & Assumptions

Given: The stated RKC recurrences, s≥1 for first-order RKC, s≥2 for second-order RKC, and the self-adjoint nonpositive operator hypotheses in the matrix clause.

[F1]

First-order stabilized explicit Runge–Kutta–Chebyshev step: Y0=y, Y1=(I+hA/s2)Y0, and Yj=2(I+hA/s2)Yj−1−Yj−2 for j≥2 when f(y)=Ay.

[F2]

Second-order linear stabilized Runge–Kutta–Chebyshev step: V0=y, V1=(I+w1hA)V0, Vj=2(I+w1hA)Vj−1−Vj−2, and the output is asV0+bsVs.

[F3]

Second-order nonlinear stabilized Runge–Kutta–Chebyshev step: the nonlinear stage recurrence uses b0=b1=b2=1/4, aj=1−bj, and the displayed μj,νj,mj,gj coefficients.

[F4]

Chebyshev polynomials of the first and second kinds by their three-term recurrences: Tn+2(x)=2xTn+1(x)−Tn(x) and Un+2(x)=2xUn+1(x)−Un(x).

[F5]

Chebyshev bounds and endpoint derivatives used by stabilized Runge–Kutta methods: ∣Tn(x)∣≤1 on [−1,1], ∣Tn(x)∣>1 for ∣x∣>1 when n≥1, the degrees/leading coefficients are n,2n−1 for Tn, and Ts′(1)=s2, Ts′′(1)=s2(s2−1)/3.

[F6]

Polynomial evaluation at an endomorphism: p(T)=∑kakTk: for a finite polynomial p(x)=∑kakxk, p(A)=∑kakAk with A0=I.

[F7]

Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space: self-adjoint means A∗=A on the stated real inner-product space.

[F8]

Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis: a self-adjoint endomorphism on a finite-dimensional real inner-product space has an orthonormal eigenbasis.

[F9]

Real and complex inner-product spaces and their induced length, Orthonormal families, complete orthonormal systems and Hilbert bases, and The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn: the Euclidean inner product is a real inner product; in an orthonormal basis, squared length is the sum of squared coordinates and ∥v∥22=∑ivi2.

[A1]

A is self-adjoint and ⟨Av,v⟩≤0 for every v; h>0 and the stated spectral-radius inequality holds in the applicable branch.

Proof

1.1F1F4F5F6inductionalgebra

For f(y)=Ay, put X=I+hA/s2 in [F1]. The Chebyshev recurrence [F4] and the definition of polynomial evaluation [F6] give Yj=Tj(X)y by induction from T0(X)y=y and T1(X)y=Xy. Hence the output factor is R1,s(z)=Ts(1+z/s2). Its degree is s because [F5] gives nonzero leading coefficient 2s−1 and the affine argument has nonzero slope.

1.2F2F3F4F5inductionalgebra

In [F3], specialize to f(y)=λy. Define the scalar polynomial Pj by Yj=Pj(z)y for every initial scalar y; the linear recurrence makes this definition valid even when y=0. Then P0=1, P1=1+b1w1z, and for j≥2, Pj=(1−μj−νj)+μjPj−1+νjPj−2+mjzPj−1+gjz. Set x=1+w1z. The base values satisfy P0=a0+b0T0(x) and P1=a1+b1T1(x). If Pj−1=aj−1+bj−1Tj−1(x) and Pj−2=aj−2+bj−2Tj−2(x), the coefficient of Tj−1 is μjbj−1+mjzbj−1=2bj(1+w1z)=2bjx, and the coefficient of Tj−2 is νjbj−2=−bj. The constant part is 1−μj−νj+μjaj−1+νjaj−2+mjzaj−1+gjz=1−μjbj−1−νjbj−2+z(mjaj−1+gj)=1−bj=aj, because gj=−aj−1mj, μjbj−1=2bj, and νjbj−2=−bj. Thus Pj=aj+bjTj(x) by [F4]. The same coefficient induction with z replaced by hA proves that the nonlinear recurrence specialized to f(v)=Av has update asI+bsTs(I+w1hA). The linear algorithm [F2] has that same shifted-Chebyshev recurrence and final weights, so its final factor is the same R2,s. The degree is s, since bs>0, w1>0, and [F5] gives degree s with nonzero leading coefficient.

2.1F5F10step 1.1step 1.2algebra

By [F5], Ts(1)=1, Ts′(1)=s2, and Ts′′(1)=s2(s2−1)/3. The scalar chain rule and product rule [F10] applied to the polynomials in steps 1.1–1.2 therefore give R1,s(0)=1 and R1,s′(0)=Ts′(1)/s2=1. For second-order RKC, as+bs=1, bsw1s2=1, and bsw12Ts′′(1)=s2−13s29(s2−1)2s2(s2−1)3=1. Differentiating the polynomial in step 1.2 gives the stated values of R2,s(0), R2,s′(0), and R2,s′′(0).

2.2F5step 1.1step 1.2algebra

For first-order RKC, ∣R1,s(z)∣≤1 exactly when x=1+z/s2 lies in [−1,1], by [F5]; solving these two inequalities gives −2s2≤z≤0. For second-order RKC, bs>0 and as=1−bs>0 for s≥2. If −2/w1≤z≤0, then x=1+w1z∈[−1,1], so [F5] gives ∣as+bsTs(x)∣≤as+bs∣Ts(x)∣≤as+bs=1. Since 2/w1=2(s2−1)/3, the claimed interval follows.

3.1A1F2F6F7F8F9step 1.2step 2.2algebra∎

By [F8], choose an orthonormal eigenbasis ei with Aei=λiei. The nonpositive premise [A1] gives λi≤0 by applying it to ei. For any vector v=∑iαiei, polynomial evaluation gives R(hA)v=∑iαiR(hλi)ei. Under the first-order spectral bound, every hλi∈[−2s2,0], so step 2.2 gives ∣R1,s(hλi)∣≤1; orthonormality then yields ∥R1,s(hA)v∥2=∑iαi2R1,s(hλi)2≤∑iαi2=∥v∥2. Under the second-order spectral bound, the same calculation uses the interval from step 2.2 for R2,s. The scalar specialization in step 1.2 shows it also applies to nonlinear RKC on this linear field.

Remarks

The matrix conclusion uses an orthonormal eigenbasis and therefore does not follow from eigenvalue inclusion alone for a nonnormal matrix. For RKC2, the interval proved above is a guaranteed stability interval; the statement does not claim that it is the largest real interval in the absolute stability set.

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