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Author's personal copy3398 S. Blanes, E. Ponsoda / Journal of Computational and Applied Mathematics 236 (2012) 3394–3408Let us consider the linear homogeneous equationY ′ (t) = M(t)Y(t), Y(t 0 ) = I n , (17)with Y(t), M(t) ∈ C n×n , and denote the fundamental matrix solution by Φ(t, t 0 ). If we consider that the solution can bewritten in the exponential form, Φ(t, t 0 ) = exp(Ω(t, t 0 )), where Ω(t, t 0 ) = ∞Ω n=1 n(t, t 0 ), we obtain the Magnus seriesexpansion [28] whereΩ 1 (t, t 0 ) = tΩ 2 (t, t 0 ) = 1 2t 0M(t 1 ) dt 1 , tt 0dt 1 t1t 0dt 2 [M(t 1 ), M(t 2 )] , . . . .Here [A, B] ≡ AB − BA is the matrix commutator of A and B. We denote the truncated expansion by kΨ [k] (t, t 0 ) = exp Ω i (t, t 0 ) . (19)i=1The first order approximation agrees with the first order approximation for most exponential methods like e.g. the Fer [29]or Wilcox [30] expansions, but they differ at higher orders. In general, exponential methods (and most expansions) convergefor tt 0∥M(t 1 )∥ dt 1 < ξ, (20)where ξ is a constant depending on the method. For the Magnus expansion (ME), we have that, for the 2-norm, ξ = π, beingthis a sufficient but not necessary condition [31] (see also [11] and the references therein).(18)3.1. Magnus expansion for the non-homogeneous problemSuppose the game can be influenced by external forcesx ′ (t) = A(t)x(t) + B 1 (t)u 1 (t) + B 2 (t)u 2 (t) + z(t), x(0) = x 0 ,where z(t) ∈ R n is the uncertainty, see [32]. This problem can be solved using most results with minor modifications. Noticethat this problem is equivalent toy ′ (t) = Ā(t)y(t) + ¯B1 (t)u 1 (t) + ¯B2 (t)u 2 (t), y(0) = y 0 ,where y = [x, 1] T ∈ R n+1 and A(t) z(t)Ā(t) =, ¯B1 (t) =0 0 B1 (t), ¯B2 (t) =0 B2 (t).0Then, the methods previously considered can be used.On the other hand, we have seen that a matrix RDE can be written as a linear system. However, coupled RDEs whereR ij ≠ 0 do not admit this linear form. For this reason, it is convenient to consider the generalization of the ME to thenonlinear case.3.2. The Magnus expansion for nonlinear systemsLet us consider the nonlinear and non-autonomous equationz ′ = f (t, z), z(t 0 ) = z 0 ∈ R n , (21)where the evolution operator (z(t) = Φ t (z 0 )) satisfies the operational equationddt Φt = Φ t L f (t,y) , y = z 0 , (22)with solution Φ t (y)| y=z0 . Here, L f = f · ∇ y , with f = [f 1 , f 2 , . . . , f n ] T , is the Lie derivative (or Lie operator) associated withf , acting on differentiable functions F : R n −→ R m asL f F(z) = F ′ (z)f (z),where F ′ (y) denotes the Jacobian matrix of F (see [11] for more details).
Author's personal copyS. Blanes, E. Ponsoda / Journal of Computational and Applied Mathematics 236 (2012) 3394–3408 3399Since L f is a linear operator, we can then use directly the Magnus series expansion to obtain the formal solution of (22)for t ∈ [t 0 , T] as Φ T = exp(L ω(z0 )), with ω = i ω i. The first two terms are nowω 1 (z 0 ) = Tω 2 (z 0 ) = − 1 2t 0f (s, z 0 ) ds, (23) Tt 0ds 1 s1t 0ds 2 (f (s 1 , z 0 ), f (s 2 , z 0 )), (24)where (f , g) denotes the Lie bracket with components (f , g) i = f · ∇ z g i (z) − g · ∇ z f i (z). Observe that for the integrals (23)and (24) z 0 , on the vector field, is frozen (on the Lie derivatives, one has to compute the derivatives with respect to z and toreplace them by z 0 ). If the series converges, the exact solution, z(T) = Φ T (z 0 ) can be approximated by the 1-flow solutionof the autonomous differential equationy ′ = ω [n] (y), y(0) = z 0 , (25)ω [n] = niω i , i.e. y(1) ≃ z(T) (see also [11] and references therein for more details). Notice that the ME can be considered asan average on the explicitly time-dependent functions appearing on the vector field. It still remains to solve the autonomousproblem (25). A simple method (useful when an efficient algorithm is known for the problem (21) when the time is frozen)will be presented.Example 1. As an illustrative example, let us consider the scalar nonlinear and non-autonomous equationz ′ = α(t)z k + β(t)z m , z(t 0 ) = z 0 ,then Eqs. (23) and (24) provideω 1 (z 0 ) = a z k 0 + b zm 0 ,where the constants a, b, c are given bya = Tt 0α(s) ds, b =c = − m − k2 Tt 0ds 1 s1ω 2(z 0 ) = c z k+m−10, Tt 0β(s) ds,t 0ds 2 (α(s 1 )β(s 2 ) − β(s 1 ), α(s 2 )).Then, the solution, y(1), of the autonomous (solvable) equationy ′ = ay k + by m + c y k+m−1 , y(t 0 ) = z 0 ,is the approximation to z(T) by the second order Magnus expansion. If k = 1, m = 2 this is a RDE and the Magnusapproximation can be seen as the exact solution of another autonomous RDE where the time-dependent function α(t) isreplaced by a and β(t) by b + c.3.3. Magnus integratorsIn those cases where it is not feasible to obtain analytical solutions, the numerical integration can be the choice. The timeinterval is divided on a mesh, t 0 < t 1 < · · · < t N−1 < t N = T , and an appropriate scheme is used on each interval. Forsimplicity in the presentation, we consider a constant time step, h = (T − t 0 )/N and t n = t 0 + nh, n = 0, 1, . . . , N. In thefollowing, we consider Magnus integrators. To this purpose, it is useful to notice that for the truncated expansion (19)Ψ [1] (t + h, t) = Φ(t + h, t) + O(h 3 ), Ψ [2] (t + h, t) = Φ(t + h, t) + O(h 5 ).If the integrals in (18) cannot be computed analytically, we can approximate them by a quadrature rule, and this can be doneby computing the matrix M(t) on the points of a single quadrature rule [14] (see also [11,12]).In some cases, the matrix M(t) is explicitly known, but in some other cases it is only known on a mesh. Then, in orderto present methods which can be easily adapted for different quadrature rules we introduce the averaged (or generalizedmomentum) matricesM (i) (h) ≡ 1 tn +h i 1 h/2t − t1/2 M(t) dt = t i M(t + th i t nh i 1/2 ) dt, (26)−h/2for i = 0, 1, . . . with t 1/2 = t n + h/2. Then, it is clear that Ψ [1] (t + h, t) = e M(0) (h) for the second order method, butfourth-order methods can be obtained by takingΨ [2]1 (t + h, t) = exp M (0) (h) + [M (1) (h), M (0) (h)] . (27)
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