To avoid order reduction, the expressions g
(j) in (3) need to vanish in the stiff limit. In line with [9], we define the following criteria:
Definition 2 (Weak Stage Order)
A RK scheme has weak stage order (WSO) \(\tilde {q}\) if there is an A-invariant subspace that is orthogonal to b and that contains the stage order vectors τ
(j) for \(1\le j\le \tilde {q}\).
Theorem 2 (WSO Is the Most General Condition that Ensures g
(j) = 0 for All ζ > 0)
Let coefficients A, b be given. Then g
(j) = 0 for all ζ > 0 and
\(1\leq j \leq \tilde {q}\)
if and only if the corresponding RK scheme has weak stage order
\(\tilde {q}\).
Proof
Let C(G) denote the column space of
$$\displaystyle \begin{aligned} G := \begin{bmatrix} \mathbf{\tau}^{\,(1)}, A \mathbf{\tau}^{\,(1)},A^2 \mathbf{\tau}^{\,(1)}, \ldots, A^{s-1}\mathbf{\tau}^{\,(1)}, \mathbf{\tau}^{\,(2)}, A\mathbf{\tau}^{\,(2)}, \ldots, A^{s-1}\mathbf{\tau}^{\,(\tilde{q})} \end{bmatrix}. \end{aligned}$$
From the Cayley-Hamilton theorem it follows that WSO \(\tilde {q}\) is equivalent to
$$\displaystyle \begin{aligned} {\mathbf{b}}^T A^\ell \mathbf{\tau}^{\,(j)} = 0, \quad \quad 0\leq \ell \leq s-1, \ 1 \leq j \leq \tilde{q}\;. \end{aligned} $$
(4)
⇒ Because C(G) is A-invariant, C(G) is invariant under multiplication by (1 − ζA)−1, i.e. if v ∈ C(G) then for any ζ > 0, the product (1 − ζA)−1
v ∈ C(G). Since b is orthogonal to C(G), we have g
(j) = 0 for all \(1\leq j \leq \tilde {q}\).
⇐= If g
(j) = 0, then ζ
−1
g
(j) = b
T(1 − ζA)−1
τ
(j) = 0 for all ζ > 0. Differentiating both sides of this equation ℓ-times, with respect to ζ, and taking the limit as ζ → 0+, yields the conditions in Eq. (4). □
Definition 3 (Weak Stage Order Eigenvector Criterion)
A RK scheme satisfies the WSO eigenvector criterion of order \(\tilde {q}_e\) if for each \(1\le j\le \tilde {q}_e\), there exists μ
j such that A
τ
(j) = μ
j
τ
(j), and moreover, b
T
τ
(j) = 0.
The WSO eigenvector criterion of order \(\tilde {q}_e\) implies WSO (of at least) \(\tilde {q}_e\). For a given scheme, let p denote the classical order, q the stage order, and \(\tilde {q}\) the weak stage order. Then we have \(\tilde {q} \ge q\) and p ≥ q. Note however that a method with WSO \(\tilde {q}\ge 1\) need not even be consistent; order conditions must be imposed separately.
The WSO eigenvector criterion may serve to avoid OR because it implies that
$$\displaystyle \begin{aligned} g^{(j)} = \zeta{\mathbf{b}}^{\,T}(1-\zeta\mu_j)^{-1}\mathbf{\tau}^{\,(j)} = \frac{\zeta}{1-\zeta\mu_j}{\mathbf{b}}^{\,T}\mathbf{\tau}^{\,(j)}\;, \end{aligned}$$
i.e., it allows one to “push” the stage order residuals past the matrix (1 − ζA)−1, and then use b
T
τ
(j) = 0. Note that the condition b
T
τ
(j) = 0 that is required in Definition 3 is actually automatically satisfied (due to the order conditions) if \(p>\tilde {q}_e\) (or \(p\ge \tilde {q}_e\) for stiffly accurate schemes).
It must be stressed that the concept of WSO (both criteria) is based on the linear test equation (1), hence it is not clear to what extent WSO will remedy OR for nonlinear problems or problems with time-dependent coefficients. In Sect. 4 we numerically investigate some nonlinear test problems.
Finally, we present a limitation theorem on the WSO eigenvector criterion.
Theorem 3
DIRK schemes with invertible A have
\(\tilde {q}_e\le 3\).
Proof
Because the τ
(j) only depend on A, the eigenvector relation in Definition 3 depends only on A, not on b. With A lower triangular, the first k components of τ
(j) depend only on the upper k rows of A; and the same is true for the eigenvector relation as well. Hence, for a scheme to have an A that allows for the WSO eigenvector criterion of order \(\tilde {q}_e\), all upper sub-matrices of A must admit the same, too. We can therefore study A row by row. The first component of τ
(j) equals \((1-\frac {1}{j})a_{11}^j\), which is nonzero for j > 1. Hence, the first row of the equation A
τ
(j) = μ
j
τ
(j) is equivalent to μ
j = a
11. With that, we can move to the second row of the equation, which reads
$$\displaystyle \begin{aligned} (1\!-\!\tfrac{1}{j})a_{11}^j a_{21} + (a_{22}\!-\!a_{11}) \left(a_{11}^{j-1}a_{21} + (a_{21}\!+\!a_{22})^{j-1}a_{22} - \tfrac{1}{j}(a_{21}\!+\!a_{22})^j\right) = 0\;. \end{aligned} $$
(5)
To determine the set of solutions (a
11, a
21, a
22) of (5), we first observe that (5) is homogeneous, i.e., if (a
11, a
21, a
22) solves (5), then (μa
11, μa
21, μa
22) solves (5) as well for any \(\mu \in \mathbb {R}\). It therefore suffices to consider the solutions of (5) in the 2D-plane \((\frac {a_{11}}{a_{21}},\frac {a_{22}}{a_{21}})\). Figure 1 shows the resulting solution curves for j ∈{2, 3, 4}.
One class of solutions lies on the straight line of slope 1 passing through (1, 0). Those schemes are equal-time methods, i.e., RK schemes that have c = ν
e, where \(\nu \in \mathbb {R}\) is a constant. In fact, equal-time schemes satisfy the eigenvector relation for all j. However, they are not particularly useful RK methods, because—among other limitations—they are restricted to second order. This follows because the order 1 and 2 conditions require b
T
e = 1 and \({\mathbf {b}}^{\,T}\mathbf {c} = \frac {1}{2}\). Thus \(\nu = \frac {1}{2}\), and \({\mathbf {b}}^{\,T}{\mathbf {c}}^2 = \nu ^2 = \frac {1}{4}\), which contradicts the order 3 condition \({\mathbf {b}}^{\,T}{\mathbf {c}}^2 = \frac {1}{3}\). Note that the equal-time scenario also covers the points at infinity in Fig. 1, i.e., the schemes with a
21 = 0.
Non-equal-time schemes that satisfy (5) for j = 2 and j = 3 are the following two points in the \((\frac {a_{11}}{a_{21}},\frac {a_{22}}{a_{21}})\) plane: \(P_1 = (-4+3\sqrt {2},\sqrt {2}-1) = (0.2426,0.4142)\) and \(P_2 = (-(\sqrt {2}+1)(\sqrt {2}+2),-(\sqrt {2}+1)) = (-8.2426,-2.4142)\). None of these two points satisfies (5) for j = 4 (green curve in Fig. 1). Therefore \(\tilde {q}_e\le 3\). □
Among the two sets of solutions found in the proof, P
1 implies that a
11, a
21, and a
22 all have the same sign, which is a desirable property. In contrast, P
2 implies that a
21 < 0. Both WSO 3 schemes presented below correspond to the P
1 solution.