[go: up one dir, main page]
More Web Proxy on the site http://driver.im/ Skip to content
Licensed Unlicensed Requires Authentication Published by De Gruyter April 1, 2023

Implicit Runge–Kutta Schemes for Optimal Control Problems with Evolution Equations

  • Thomas G. Flaig EMAIL logo

Abstract

In this paper we discuss the use of implicit Runge–Kutta schemes for the time discretization of optimal control problems with evolution equations. The specialty of the considered discretizations is that the discretizations schemes for the state and adjoint state are chosen such that discretization and optimization commute. It is well known that for Runge–Kutta schemes with this property additional order conditions are necessary. We give sufficient conditions for which class of schemes these additional order condition are automatically fulfilled. The focus is especially on implicit Runge–Kutta schemes of Gauss, Radau IA, Radau IIA, Lobatto IIIA, Lobatto IIIB and Lobatto IIIC collocation type up to order six. Furthermore, we also use a SDIRK (singly diagonally implicit Runge–Kutta) method to demonstrate, that for general implicit Runge–Kutta methods the additional order conditions are not automatically fulfilled. Numerical examples illustrate the predicted convergence rates.

Award Identifier / Grant number: SPP 1253

Funding statement: The work was partially supported by the DFG priority program SPP 1253.

A Order Conditions

We repeat these order conditions up to order four in Table 2, the conditions of order five in Table 3 and the conditions of order six in Tables 47.

Table 2.

The order conditions for Runge–Kutta discretization for the state equation and optimal control problems, see also [4, Tables 2–4] and [8, Tables 1 and 2]. All summations go from 1 to the number of stages s:

  1. Abbreviations:

    c i = a i j , d j = b i a i j .

  2. Order conditions for the state equation without control:

    Order Conditions
    (O1) 1 b i = 1 .
    (O2) 2 d i = 1 2 ,
    (O3) 3 c i d i = 1 6 , b i c i 2 = 1 3 ,
    (O4) 4 b i c i 3 = 1 4 , b i c i a i j c j = 1 8 , d i c i 2 = 1 12 , d i a i j c j = 1 24 .

  3. Additional order conditions for optimal control problems:

    Order Additional  conditions
    (A3) 3 d i 2 b i = 1 3 ,
    (A4) 4 c i d i 2 b i = 1 12 , d i 3 b i 2 = 1 4 , b i b j c i a i j d j = 5 24 , d i b j a i j d j = 1 8 .

Table 3.

The order conditions of order five for Runge–Kutta discretization for the state equation and optimal control problems, see also [4, Table 6]. All summations go from 1 to the number of stages s:

  1. Order conditions of order five for the state equation without control (computed with Mathematica):

    (O5-1) b i a i k a k j c i c j = 1 30 , a j k c j d j c k = 1 40 , b i a i j c i c j 2 = 1 15 ,
    (O5-2) c j 3 d j = 1 20 , b i b k a l k a i l c i d k = 11 120 , b i a i j c i 2 c j = 1 10 ,
    (O5-3) a k j c j 2 d k = 1 60 , b i c i 4 = 1 5 , b i ( a i j c j ) 2 = 1 20 .

  2. Additional order conditions of order five for optimal control problems (see also [4, Table 6]):

    (A5-1) 1 b k a l k c k d k d l = 1 40 , 1 b k c k 2 d k 2 = 1 30 , 1 b l 2 c l d l 3 = 1 20 ,
    (A5-2) 1 b k a k l d k 2 c l = 1 60 , 1 b m 3 d m 4 = 1 5 , b i a i k a i j c j c k = 1 20 ,
    (A5-3) a l k a k j c j d l = 1 120 , 1 b k a l k d k c l d l = 7 120 , b i b j b k a j k a i k c i c j = 2 15 ,
    (A5-4) b i b k a i k c i c k d k = 7 120 , b i b l 2 a i l c i d l 2 = 3 20 , 1 b k a m k a l k d l d m = 1 20 ,
    (A5-5) 1 b l 2 a m l d l 2 d m = 1 10 , 1 b k a m l a l k d k d m = 1 30 , b i b k a l k a i k c i d l = 3 40 ,
    (A5-6) b i b k a i k a i l d k c l = 3 40 , b i b l b m a i m a i l d l d m = 2 15 , b i b k a i k c i 2 d k = 3 20 ,
    (A5-7) 1 b l b m a l m d l 2 d m = 1 15 .

Table 4.

The order conditions of order six for Runge–Kutta discretization for the state equation without control (computed with Mathematica). All summations go from 1 to the number of stages s:

(O6-1) c j 4 d j = 1 30 , a l m a k l a j k d j c m = 1 720 , b i a i j c i 2 c j 2 = 1 18 ,
(O6-2) a j k c j d j c k 2 = 1 90 , b i a i j c i c j 3 = 1 24 , a k j c j 3 d k = 1 120 ,
(O6-3) b i a i j c i 3 c j = 1 12 , a j k c j 2 d j c k = 1 60 , b i a i j a j k c i c j c k = 1 48 ,
(O6-4) a l j a j k c j c k d l = 1 240 , a k l a j k c j d j c l = 1 180 , b i a j k a i j c i 2 c k = 1 36 ,
(O6-5) b i a i k a k j c i c j 2 = 1 72 , a l k a k j c j 2 d l = 1 360 , b i a i k a i j c j 2 c k = 1 36 ,
(O6-6) b i a i l a i k a k j c j c l = 1 72 , b i a k l a j k a i j c i c l = 1 144 , b i c i 5 = 1 6 ,
(O6-7) b i c i ( a i j c j ) 2 = 1 24 , b i a i j ( a j k c k ) 2 = 1 120 .

Table 5.

Part 1 of the additional order conditions of order six for Runge–Kutta discretization for optimal control problems, see also [4, Table 6]. All summations go from 1 to the number of stages s:

(A6-1) 1 b n 4 d n 5 = 1 6 , 1 b m 3 c m d m 4 = 1 30 , 1 b l 2 c l 2 d l 3 = 1 60 ,
(A6-2) 1 b k c k 3 d k 2 = 1 60 , b i b k a i k c i 2 c k d k = 2 45 , 1 b k a l k c k d k c l d l = 1 72 ,
(A6-3) b i b l 2 a i l c i 2 d l 2 = 19 180 , 1 b l 2 a m l d l 2 c m d m = 2 45 , 1 b m 2 b n a n m d m 2 d n 2 = 1 18 ,
(A6-4) 1 b k a k l d k 2 c l 2 = 1 180 , a l m d l 2 c m d m b l b m = 1 90 , b i b k a i k c i 3 d k = 7 60 ,
(A6-5) b i b k a i k c i c k 2 d k = 1 40 , 1 b k a l k c k 2 d k d l = 1 120 , 1 b k a l k d k c l 2 d l = 1 30 ,
(A6-6) b i b l 2 a i l c i c l d l 2 = 1 30 , 1 b l 2 a m l c l d l 2 d m = 1 60 , 1 b k a k l c k d k 2 c l = 1 120 ,
(A6-7) a l m c l d l 2 d m b l b m = 1 40 , b i b m 3 a i m c i d m 3 = 7 60 , 1 b m 3 a n m d m 3 d n = 1 12 .

Table 6.

Part 2 of the additional order conditions of order six for Runge–Kutta discretization for optimal control problems, see also [4, Table 6]. All summations go from 1 to the number of stages s:

(A6-8) a l m d l 3 c m b l 2 = 1 120 , a n l a m l d l d m d n b l 2 = 1 24 , a m k a l k c k d l d m b k = 1 120
(A6-9) a m n d m 3 d n b m 2 b n = 1 24 , a m l a l k d k c l d m b k = 1 80 , b i a i m a i l c i d l d m b l b m = 11 120 ,
(A6-10) a n l a l m d l d m d n b l b m = 1 48 , a n m a n l d l d m d n b l b m = 1 24 , b i b j a j k a i k c i c j c k b k = 1 24 ,
(A6-11) b i b j a j l a i l c i c j d l b l 2 = 11 120 , b i a l k a i l c i d k c l b k = 11 240 , b i a l k a i k c i c k d l b k = 1 60 ,
(A6-12) b i a m l a i l c i d l d m b l 2 = 7 120 , b i a l m a i l c i d l d m b l b m = 11 240 , b i a i k a i j c i c j c k = 1 24 ,
(A6-13) b i a i k a i l c i d k c l b k = 7 120 , a m k a k l d k c l d m b k = 1 240 , b i a i k a k l c i d k c l b k = 1 80 ,
(A6-14) b i a i m a l m c i d l 2 b l b m = 7 180 , a j l a j k d j c k c l = 1 120 , a l k a l m d k d l c m b k = 1 60 ,
(A6-15) b i a l k a i k c i 2 d l b k = 19 360 , a m k a l k c l d l d m b k = 1 45 , a n m a l m d l 2 d n b l b m = 1 36 ,
(A6-16) a l m a k l d k 2 c m b k = 1 360 , b i a i m a m l c i d l 2 b l 2 = 13 180 , a n m a l n d l 2 d m b l b m = 1 72 ,
(A6-17) a n m a m l d l 2 d n b l 2 = 1 36 , b i a i m a i l d l 2 c m b l 2 = 19 360 , b i a i n a i m d m 2 d n b m 2 b n = 7 72 ,
(A6-18) b i a i k a l k c i c l d l b k = 13 360 , a m k a l m d k c l d l b k = 7 360 , b i a i l a l k c i c k d k b k = 7 360 ,
(A6-19) a m l a l k c k d k d m b k = 1 180 , b i a i l a i k c k d k c l b k = 1 45 , b i a i m a i l c l d l d m b l b m = 13 360 ,
(A6-20) b i b j a j k a i k c i 2 c j b k = 7 72 , b i a l k a i l c i 2 d k b k = 13 180 , b i a i k a i l d k c l 2 b k = 7 180 .

Table 7.

Part 3 of the additional order conditions of order six for Runge–Kutta discretization for optimal control problems, see also [4, Table 6]. All summations go from 1 to the number of stages s:

(A6-21) b i b j a j l a l k a i k c i c j b k = 1 18 , b i b j a j l a j k a i k c i c l b k = 7 144 ,
(A6-22) b i b j a j m a i m a i l c j d l b l b m = 61 720 , b i a l m a i l a i k d k c m b k = 7 360 ,
(A6-23) b i a i m a m l a l k c i d k b k = 19 720 , b i a i m a i l a l k d k c m b k = 13 360 ,
(A6-24) b i a i m a i n a n l d l d m b l b m = 1 18 , b i a i k a m k a l m c i d l b k = 7 360 ,
(A6-25) a n k a m n a l m d k d l b k = 1 144 , b i a i m a m k a l k c i d l b k = 13 360 ,
(A6-26) a n m a m k a l k d l d n b k = 1 72 , b i a i m a i k a l k d l c m b k = 19 720 ,
(A6-27) b i a i m a i l a n l d m d n b l b m = 7 144 .

B Coefficients of Runge–Kutta Schemes

We repeat the Butcher tableaux for selected Runge–Kutta schemes of order two and three in Table 8a, order four in Table 9, order five in Table 10 and order six in Table 11.

Table 8.

Coefficients of Runge–Kutta schemes of order two and three (see also [9, 10, 11]).

  1. Coefficients of the Störmer–Verlet discretization for the state (cf. [9, Table 2.1]):

    0 0 0 1 1 2 1 2 1 2 1 2

  2. Coefficients for the Radau IA method of order three (cf. [11, Table IV.5.3.]):

    0 1 4 - 1 4 2 3 1 4 5 12 1 4 3 4

  3. Coefficients for the Radau IIA method of order three (cf. [11, Table IV.5.5.]):

    1 3 5 12 - 1 12 1 3 4 1 4 3 4 1 4

Table 9.

Coefficients of Runge–Kutta schemes of order four (see also [9, 10, 11]).

  1. Coefficients for the Gauss scheme of order four (cf. [10, Table II.7.3],[11, Table IV.5.1.]):

    1 2 - 3 6 1 4 1 4 - 3 6 1 2 + 3 6 1 4 + 3 6 1 4 1 2 1 2

  2. Coefficients for an L-stable SDIRK method of order four (cf. [11, Formula (6.16)]):

    1 4 1 4 3 4 1 2 1 4 11 20 17 50 - 1 25 1 4 1 2 371 1360 - 137 2720 15 544 1 4 1 25 24 - 49 48 125 16 - 85 12 1 4 25 24 - 49 48 125 16 - 85 12 1 4

  3. Coefficients for the Lobatto IIIA method of order four (cf. [11, Table IV.5.7.]):

    0 0 0 0 1 2 5 24 1 3 - 1 24 1 1 6 2 3 1 6 1 6 2 3 1 6

  4. Coefficients for the Lobatto IIIB method of order four (cf. [11, Table IV.5.9.]):

    0 1 6 - 1 6 0 1 2 1 6 1 3 0 1 1 6 5 6 0 1 6 2 3 1 6

  5. Coefficients for the Lobatto IIIC method of order four (cf. [11, Table IV.5.11.]):

    0 1 6 - 1 3 1 6 1 2 1 6 5 12 - 1 12 1 1 6 2 3 1 6 1 6 2 3 1 6

Table 10.

Coefficients of Runge–Kutta schemes of order five (see also [9, 10, 11]).

  1. Coefficients for the Radau IA method of order five (cf. [11, Table IV.5.3.]):

    0 1 9 - 1 - 6 18 - 1 + 6 18 6 - 6 10 1 9 88 + 7 6 360 88 - 43 6 360 6 + 6 10 1 9 88 + 43 6 360 88 - 7 6 360 1 9 16 + 6 36 16 - 6 36

  2. Coefficients for the Radau IIA method of order five (cf. [11, Table IV.5.5.]):

    4 - 6 10 88 - 7 6 360 296 - 169 6 1800 - 2 + 3 6 225 4 + 6 10 296 + 169 6 1800 88 + 7 6 360 - 2 - 3 6 225 1 16 - 6 36 16 + 6 36 1 9 16 - 6 36 16 + 6 36 1 9

Table 11.

Coefficients of Runge–Kutta schemes of order six (see also [9, 10, 11]).

  1. Coefficients for the Gauss scheme of order six (cf. [11, Table IV.5.2.]):

    1 2 - 15 10 5 36 2 9 - 15 15 5 36 - 15 30 1 2 5 36 + 15 24 2 9 5 36 - 15 24 1 2 + 15 10 5 36 + 15 30 2 9 + 15 15 5 36 5 18 4 9 5 18

  2. Coefficients for the Lobatto IIIC method of order six(cf. [11, Table IV.5.11.]):

    0 1 12 - 5 12 5 12 - 1 12 5 - 5 10 1 12 1 4 10 - 7 5 60 5 60 5 + 5 10 1 12 10 + 7 5 60 1 4 - 5 60 1 1 12 5 12 5 12 1 12 1 12 5 12 5 12 1 12

  3. Coefficients for the Lobatto IIIA method of order six (cf. [11, Table IV.5.7.]):

    0 0 0 0 0 5 - 5 10 11 + 5 120 25 - 5 120 25 - 13 5 120 - 1 + 5 120 5 + 5 10 11 - 5 120 25 + 13 5 120 25 - + 5 120 - 1 - 5 120 1 1 12 5 12 5 12 1 12 1 12 5 12 5 12 1 12

  4. Coefficients for the Lobatto IIIB method of order six (cf. [11, Table IV.5.9.]):

    0 1 12 - 1 - 5 24 - 1 + 5 24 0 5 - 5 10 1 12 25 + 5 120 25 - 13 5 120 0 5 + 5 10 1 12 25 + 13 5 120 25 - 5 120 0 1 1 12 11 - 5 24 11 + 5 24 0 1 12 5 12 5 12 1 12

C Proof of Theorem 3.7

Full proof of Theorem 3.7.

The idea of the proof is to use the simplifying assumptions (B${(p)}$), ((C${(\eta)}$)), ((D${(\zeta)}$)) to reduce the additional order conditions to the classic order conditions or order conditions of lower order, which have already been reduced to the order conditions of the uncontrolled system. As all the numerical schemes fulfill the order conditions for the uncontrolled systems, these conditions can be used to calculate the value of the reduced expression.

Surely the way of the application of the simplifying assumptions is not unique, here one possibility is presented. A first goal in the reduction of order conditions with a fraction b i is to use ((D${(\zeta)}$)) to produce an additional b i which cancels out. In the following we discuss the reduction of all the additional order conditions.

  1. For the first additional order condition of (A5-1) we use the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 , the last condition of (O4) and the first condition of (O5-3). This yields

    k l 1 b k a l k c k d k d l = k l d l a l k c k - k l a l k c k 2 d l = 1 24 - 1 60 = 1 40 .

  2. For the second additional order condition of (A5-1) we use the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 and the third condition of (O4) and the second condition of (O5-3) to get

    k 1 b k c k 2 d k 2 = k c k 2 d k - k c k 3 d k = 1 12 - 1 20 = 1 30 .

  3. For the last order condition of (A5-1) we use again the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 , the first condition of (O3) the third condition of (O4) and the first condition of (O5-2). This gives

    l 1 b l 2 c l d l 3 = c l d l - 2 c l 2 d l + l c l 3 d l = 1 6 - 2 12 - 1 20 = 1 20 .

  4. For the first condition of (A5-2) we apply the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 and use the last condition of (O4) and the second condition of (O5-1), which gives

    k l 1 b k a k l c l d k 2 = k l a k l c l d k - k l a k l c l d k c k = 1 24 - 1 40 = 1 60 .

  5. For the second condition of (A5-2) we use the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 , the condition (O2), the first condition of (O3), the third condition of (O4) and the first condition of (O5-2) to end with

    m 1 b m 3 d m 4 = m d m - m 3 d m c m + m 3 d m c m 2 - m d m c m 3 = 1 5 .

  6. For the third condition of (A5-2) we apply the simplifying assumption ((C${(\eta)}$)) for η = 2 twice and get with the second condition of (O5-3) the result

    i j k b i a i k a i j c j c k = i b i ( k a i k c k ) ( j a i j c j ) = 1 4 i b i c i 4 = 1 20 .

  7. For the first condition of (A5-3) we apply again the simplifying assumption ((C${(\eta)}$)) for η = 2 and the use of the first condition of (O5-3) yields

    j k l a l k a k j c j d l = k l a l k d l ( j a k j c j ) = 1 2 k l a l k d l c k 2 = 1 120 .

  8. For the second condition of (A5-3) we apply first the simplifying assumptions ((D${(\zeta)}$)) for η = 1 and then the definition of c l and the simplifying assumption ((C${(\eta)}$)) for η = 2 . Together with the third condition of (O4) and the first condition of (O5-2) this gives

    k l 1 b k a l k d k c l d l = l c l d l ( k a l k ) - l c l d l ( k a l k c k ) = l c l 2 d l - 1 2 l c l 3 d l = 7 120 .

  9. For the last condition of (A5-3) we apply the simplifying assumption ((D${(\zeta)}$)) for η = 2 twice and get with (O1), the second condition of (O2) and the second condition of (O5-3) the result

    i j k b i b j b k a j k a i k c i c j = j k b j b k a j k c j ( i b i a i k c i ) = 1 2 k ( 1 - c k ) ( j b j a j k c j )
    = 1 4 k b k ( 1 - c k ) ( 1 - c k 2 ) = 1 4 k ( b k - 2 b k c k 2 + b k c k 4 ) = 2 15 .

  10. For the first condition of (A5-4) we use the simplifying assumption ((D${(\zeta)}$)) for η = 1 and η = 2 , the second condition of (O4), the second condition of (O3) and the second condition of (O5-3) to get

    i k b i b k a i k c i c k d k = i k b i a i k c i c k - c k 2 ( i b i c i a i k ) = 1 8 - 1 2 c k 2 b k ( 1 - c k 2 ) = 7 120 .

  11. For the second condition of (A5-4) we use again the simplifying assumptions ((D${(\zeta)}$)) for η = 1 and η = 2 . The remaining expressions are treated with (O1), the simplifying condition (B${(p)}$) for p = 2 , the second condition of (O3), the first condition of (O4) and the second condition of (O5-3). This gives

    i l b i b l 2 a i l c i d l 2 = i l b i a i l c i ( 1 - c l ) 2 = l ( 1 - c l ) 2 ( i b i c i a i l ) = 1 2 l b l ( 1 - c l ) 2 ( 1 - c l 2 )
    = 1 2 l ( b l - 2 b l c l + 2 b l c l 3 - b l c l 4 ) = 3 20 .

  12. For the last condition of (A5-4) we use first the simplifying assumptions ((D${(\zeta)}$)) for η = 1 we get due to symmetry properties

    l m k 1 b k a m k a l k d l d m = l m k b l b m b k a m k a l k ( 1 - c l ) ( 1 - c m )
    (C.1) = l m k b l b m b k a m k a l k - 2 l m k b l b m b k a m k a l k c l + l m k b l b m b k a m k a l k c l c m .

    The last term is the third condition of (A5-3) and therefore we already know how to tread this term. On the first term of (C.1) we apply the simplifying assumptions ((D${(\zeta)}$)) for η = 1 twice and get with (O1), (B${(p)}$) for p = 2 and the second condition of (O3)

    l m k b l b m b k a m k a l k = k 1 b k ( m b m a m k ) ( l b l a l k ) = k b k ( 1 - c k ) 2 = 1 3 .

    For the remaining term of (C.1) the use of ((D${(\zeta)}$)) for η = 1 and η = 2 and (O1), (B${(p)}$) for p = 2 , the second condition of (O2) and the first condition of (O2) yields

    l m k b l b m b k a m k a l k c l = k 1 b k ( m b m a m k ) ( l b l a l k c l ) = 1 2 k b k ( 1 - c k ) ( 1 - c k 2 ) = 5 24 .

    Altogether we have

    l m k 1 b k a m k a l k d l d m = 1 3 - 5 12 + 2 15 = 1 20 .

  13. For the first condition of (A5-5) we start with the use of the simplifying assumption ((D${(\zeta)}$)) for η = 1 and the definition of c m . The last condition of (O4), the first condition of (O5-3) and the first condition of (O3) give

    l m 1 b l 2 a m l d l 2 d m = l m a m l ( 1 - c l ) 2 d m = l m a m l d m - 2 l m a m l c l d m + l m a m l c l 2 d m
    = m c m d m - 1 12 + 1 60 = 1 6 - 1 15 = 1 10 .

  14. For the second condition of (A5-5) the use of ((D${(\zeta)}$)) for ζ = 1 , the definition of c i , the last condition of (O4), the simplifying assumption ((C${(\eta)}$)) for η = 2 and the first condition of (O5-3) yields

    k m l 1 b k a m l a l k d k d m = k m l a m l a l k d m - k m l a m l a l k d m c k
    = m l a m l c l d m - m l a m l d m ( k a l k c k ) = 1 24 - 1 2 m l a m l d m c l 2
    = 1 24 - 1 120 = 1 30 .

  15. For the last condition of (A5-5) we apply the simplifying assumption ((D${(\zeta)}$)) for ζ = 2 , the definition of c l , first condition of (O5-3) and the first condition of (O3) to get

    i l k b i b k a l k a i k c i d l = l k 1 b k a l k d l ( i b i a i k c i ) = 1 2 l k a l k d l - 1 2 l k a l k d l c k 2
    = 1 2 l c l d l - 1 120 = 3 40 .

  16. For the first condition of (A5-6) we use the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 and the simplifying assumption ((C${(\eta)}$)) for ζ = 2 three times. With the definition of c i , the first condition of (O4) and the second condition of (O5-3) we get

    i k l b i b k a i k a i l d k c l = i k b i a i k ( l a i l c l ) - i b i ( k a i k c k ) ( l a i l c l )
    = 1 2 i b i c i 2 ( k a i k ) - 1 4 i b i c i 4 = 1 2 i b i c i 3 - 1 20 = 1 8 - 1 20 = 3 40 .

  17. To the second condition of (A5-6) we apply the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 once, use the definition of c i , the third condition of (A4) and the first condition of (A5-6). This yields

    i l m b i b l b m a i m a i l d l d m = i l m b i b m a i m a i l d m - i l m b i b m a i m a i l c l d m
    = i l m b i b m a i m c l d m - i l m b i b m a i m a i l c l d m = 2 15 .

  18. The application of the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 to the last condition of (A5-6) together with the last condition of (O5-2), the definition of c i and the first condition of (O4) gives

    i k b i b k a i k c i 2 d k = i k b i a i k c i 2 - i k b i a i k c i 2 c k = i k b i c i 3 - 1 10 = 3 20 .

  19. For the condition (A5-7) we use the simplifying assumption ((D${(\zeta)}$)) for ζ = 1 two times, the definition of c l , the first condition of (O3), the third and fourth conditions of (O4) and the second condition of (O5-1) and get

    l m 1 b l b m a l m d l 2 d m = l m a l m d l - l m a l m d l c l - l m a l m d l c m + l m a l m d l c l c m
    = l m c l d l - l m c l 2 d l - l m a l m d l c m + l m a l m d l c l c m = 1 15

Altogether we have derived the additional order conditions with the use of the simplifying assumptions and (B${(p)}$) for p 2 , ((C${(\eta)}$)) for η = 1 , 2 , ((D${(\zeta)}$)) for ζ = 1 , 2 , and the order conditions (O1)–(O4) and (O5-1)–(O5-3). ∎

D Proof of Theorem 3.9

Full proof of Theorem 3.9.

We follow the same ideas as in Appendix C. Note that the simplifying condition (B${(p)}$) up to order p = 4 are part of the order condition.

  1. For the first additional order condition of (A6-1) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 and the order conditions (O2), the first condition of (O3), the third condition of (O4), the first condition of (O5-2) and the first condition of (O6-1)

    n 1 b n 4 d n 5 = n d n b n 4 b n 4 ( 1 - c n ) 4 = n d n - 4 c n d n + 6 d n c n 2 - 4 d n c n 3 + d n c n 4
    = 1 2 - 4 6 + 6 12 - 4 20 + 1 30 = 1 6 .

  2. For the second order condition of (A6-1) we use the definition of d j = b i a i j together with((D${(\zeta)}$)) for ζ = 1 , the first condition of (O3), the third condition of (O4), the first condition of (O5-2) and the first condition of (O6-1)

    m 1 b m 3 c m d m 4 = m 1 b m 3 c m d m b m 3 ( 1 - c m ) 3 = m d m c m - 3 d m c m 2 + 3 d m c m 3 - d m c m 4
    = 1 6 - 3 12 + 3 20 - 1 30 = 1 30 .

  3. For the third condition of (A6-1) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 , the third condition of (O4), the first condition of (O5-2) and the first condition of (O6-1)

    l 1 b l 2 c l 2 d l 3 = l 1 b l 2 c l 2 d l b l 2 ( 1 - c l ) 2 = l d l c l 2 - 2 d l c l 3 + d l c l 4
    = 1 12 - 2 20 + 1 30 = 1 6 .

  4. For the first condition of (A6-2) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 , the first condition of (O5-2) and the first condition of (O6-1)

    k 1 b k c k 3 d k 2 = k 1 b k c k 3 d k b k ( 1 - c k ) = k c k 3 d k - c k 4 d k
    = 1 20 - 1 30 = 1 60 .

  5. For the second condition of (A6-2) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 , the third condition of (O5-2) and the third condition of (O6-1)

    i k b i b k a i k c i 2 c k d k = i k b i b k a i k c i 2 c k b k ( 1 - c k ) = i k b i a i k c i 2 c k - i k b i a i k c i 2 c k 2
    = 1 10 - 1 18 = 2 45 .

  6. For the third condition of (A6-2) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 , second condition of (O5-2) and the first condition of (O6-2)

    l k 1 b k a l k c k d k c l d l = l k 1 b k a l k c k b k ( 1 - c k ) c l d l = l k a l k c k c l d l - l k a l k c k 2 c l d l
    = 1 40 - 1 90 = 1 72 .

  7. For the first condition of (A6-3) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , the third condition of (O6-1), the first condition of (O4), ((C${(\eta)}$)) for η = 2 and the second condition of (O5-3)

    i l b i b l 2 a i l c i 2 d l 2 = i l b i b l 2 a i l c i 2 b l 2 ( 1 - c l ) 2 = i l b i a i l c i 2 - 2 i l b i a i l c i 2 c l + i l b i a i l c i 2 c l 2
    = i b i c i 3 - 2 i b i c i 2 l a i l c l + 1 18 = 1 4 - i b i c i 4 + 1 18
    = 1 4 - 1 5 + 1 18 = 19 180 .

  8. For the second condition of (A6-3) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , ((C${(\eta)}$)) for η = 2 , the first condition of (O6-2), the third condition of (O4) and the first condition of (O5-2)

    m l 1 b l 2 a m l d l 2 c m d m = m l 1 b l 2 a m l b l 2 ( 1 - c l ) 2 c m d m
    = m c m d m l a m l - 2 m c m d m l a m l c l + m l a m l c m c l 2 d m
    = m c m 2 d m - m c m 3 d m + 1 90 = 1 12 - 1 20 + 1 90 = 2 45 .

  9. For the third condition of (A6-3) we use the definition of d j = b i a i j together with ((D${(\zeta)}$)) for ζ = 1 twice for d m and once for d n ,((C${(\eta)}$)) for η = 1 twice, ((C${(\eta)}$)) for η = 2 , the first condition of (O3), the forth condition of (O4), the first condition of (O5-3), the third condition of (O4), the first condition of (O5-2) and the first condition of (O6-2)

    m n 1 b m 2 b n a n m d m 2 d n 2 = m n 1 b m 2 b n a n m b m 2 ( 1 - c m ) 2 d n b n ( 1 - c n )
    = n d n m a n m - 2 m n a n m d n c m + m n a n m d n c m 2 - n d n c n m a n m
    + 2 n d n c n m a n m c m - m n a n m d n c m 2 c n
    = n d n c n - 2 m n a n m d n c m + m n a n m d n c m 2 - n d n c n 2
    + 2 n d n c n 3 - m n a n m d n c m 2 c n
    = 1 6 - 2 24 + 1 60 - 1 12 + 1 20 - 1 90 = 1 18 .

  10. For the first condition of (A6-4) we use ((D${(\zeta)}$)) for ζ = 1 , the first condition of (O5-3) and the first condition of (O6-2)

    k l 1 b k a k l d k 2 c l 2 = k l 1 b k a k l d k b k ( 1 - c k ) c l 2
    = k l a k l d k c l 2 - k l a k l d k c k c l 2 = 1 60 - 1 90 = 1 180 .

  11. For the second condition of (A6-4) we use ((D${(\zeta)}$)) for ζ = 1 for d l and d m , the forth condition of (O4), ((C${(\eta)}$)) for η = 2 , the first condition of (O5-3), the first condition of (O6-2) and the first condition of (O5-2)

    l m 1 b l b m a l m d l 2 c m d m = l m 1 b l b m a l m d l b l ( 1 - c l ) c m b m ( 1 - c m )
    = l m a l m c m d l - l c l d l m a l m c m - l m a l m c m 2 d l + l m a l m c l c m 2 d l
    = 1 24 - 1 2 l c l 3 d l + 1 60 + 1 90 = 1 24 - 1 2 1 20 + 1 60 + 1 90 = 1 90 .

  12. For the third condition of (A6-4) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , the second condition of (O5-3) and the first condition of (O6-3)

    i k b i b k a i k c i 3 d k = i k b i b k a i k c i 3 b k ( 1 - c k ) = i b i c i 3 k a i k - i k b i a i k c i 3 c k
    = i b i c i 4 - i k b i a i k c i 3 c k = 1 5 - 1 12 = 7 60 .

  13. For the first condition of (A6-5) we use ((D${(\zeta)}$)) for ζ = 1 , the third condition of (O5-1) and the second condition of (O6-2)

    i k b i b k a i k c i c k 2 d k = i k b i b k a i k c i c k 2 b k ( 1 - c k ) = i k b i a i k c i c k 2 - i k b i a i k c i c k 3
    = 1 15 - 1 24 = 1 40 .

  14. For the second condition of (A6-5) we use ((D${(\zeta)}$)) for ζ = 1 , the first condition of (O5-3) and the third condition of (O6-2)

    l k 1 b k a l k c k 2 d k d l = l k 1 b k a l k c k 2 b k ( 1 - c k ) d l
    = l k a l k c k 2 d l - l k a l k c k 3 d l = 1 60 - 1 120 = 1 120 .

  15. For the third condition of (A6-5) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the first condition of (O5-2) and the first condition of (O6-1)

    k l 1 b k a l k d k c l 2 d l = k l 1 b k a l k b k ( 1 - c k ) c l 2 d l
    = l c l 2 d l k a l k - l c l 2 d l k a l k c k
    = l c l 3 d l - l 1 2 c l 4 d l = 1 20 - 1 2 1 30 = 1 30 .

    Instead of ((C${(\eta)}$)) for η = 2 and the first condition of (O6-1) one could use the second condition of (O6-3) directly.

  16. For the first condition of (A6-6) we use ((D${(\zeta)}$)) for ζ = 1 , the second condition of (O4), the third condition of (O5-1) and the second condition of (O6-2)

    i l b i b l 2 a i l c i c l d l 2 = i l b i b l 2 a i l c i c l b l 2 ( 1 - c l ) 2
    = i l b i a a i l c i c l - 2 i l b i a a i l c i c l 2 + i l b i a a i l c i c l 3
    = 1 8 - 2 1 15 + 1 24 = 1 30 .

  17. For the second condition of (A6-6) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 2 , the first condition of (O5-3), the third condition of (O6-2) and the third condition of (O4)

    m l 1 b l 2 a m l c l d l 2 d m = m l 1 b l 2 a m l c l b l 2 ( 1 - c l ) 2 d m
    = m d m l a m l c l - 2 m l a m l c l 2 d m + m l a m l c l 3 d m
    = 1 2 m d m c m 2 - 2 60 + 1 120 = 1 2 1 12 - 2 60 + 1 120 = 1 60 .

  18. For the third condition of (A6-6) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 2 twice, the first condition of (O5-2) and the first condition of (O6-1)

    k l 1 b k a k l c k d k 2 c l = k l 1 b k a k l c k d k b k ( 1 - c k ) c l
    = k c k d k l a a k l c l - k c k 2 d k l a a k l c l = 1 2 k c k 3 d k - 1 2 k c k 4 d k
    = 1 2 ( 1 20 - 1 30 ) = 1 120 .

  19. For the first condition of (A6-7) we use ((D${(\zeta)}$)) for ζ = 1 for d l and d m , ((C${(\eta)}$)) for η = 1 twice and η = 2 twice, the third condition of (O4), the first condition of (O5-3) twice and the first condition of (O6-1)

    l m 1 b l b m a l m c l d l 2 d m = l m 1 b l b m a l m c l d l b l ( 1 - c l ) b m ( 1 - c m )
    = l c l d l m a l m - l c l 2 d l m a l m - l c l d l m a l m c m + l c l 2 d l m a l m c m
    = l c l 2 d l - l c l 3 d l - 1 2 l c l 3 d l + 1 2 l c l 4 d l
    = 1 12 - 1 20 - 1 2 1 20 + 1 2 1 30 = 1 40 .

  20. For the second condition of (A6-7) we use ((D${(\zeta)}$)) for ζ = 1 and ζ = 2 , the third condition of (O6-6), the second condition of (O5-3), the first condition of (O4), the second condition of (O3), (O2) (together with the definition of d j and ((C${(\eta)}$)) for η = 1 ) and the condition of (O1)

    i m b i b m 3 a i m c i d m 3 = m 1 b m 3 b m 3 ( 1 - c m ) 3 i b i a i m c i = 1 2 m ( 1 - c m ) 3 b m ( 1 - c m 2 )
    = 1 2 m b m c m 5 - 3 b m c m 4 + 2 b m c m 3 + 2 b m c m 2 - 3 b m c m + b m
    = 1 2 ( 1 6 - 3 5 + 2 4 + 2 3 - 3 2 + 1 ) = 7 60 .

  21. For the third condition of (A6-7) we use ((D${(\zeta)}$)) for ζ = 1 , the first condition of (O3) (together with ((C${(\eta)}$)) for η = 1 ), the last condition of (O4), the first condition of (O5-3) and the third condition of (O6-2)

    m n 1 b m 3 a n m d m 3 d n = m n 1 b m 3 a n m b m 3 ( 1 - c m ) 3 d n
    = n d n m a n m - 3 m n a n m d n c m + 3 m n a n m d n c m 2 - m n a n m d n c m 3
    = 1 6 - 3 24 + 3 60 - 1 120 = 1 12 .

  22. For the first condition of (A6-8) we use ((D${(\zeta)}$)) for ζ = 1 , the last condition of (O4), the second condition of (O5-1) and the second condition of (O6-3)

    l m 1 b l 2 a l m d l 3 c m = l m 1 b l 2 a l m d l b l 2 ( 1 - c l ) 2 c m
    = l m a l m d l c m - 2 l m a l m d l c l c m + l m a l m d l c l 2 c m = 1 24 - 2 40 + 1 60 = 1 120 .

  23. For the second condition of (A6-8) we use ((D${(\zeta)}$)) for ζ = 1 to get

    m n l 1 b l 2 a n l a m l d l d m d n = m n l 1 b l 2 a n l a m l b l ( 1 - c l ) d m d n
    = m n l 1 b l a n l a m l d m d n - m n l 1 b l a n l a m l c l d m d n .

    As we have no additional d l for producing a b l ( 1 - c m ) to cancel, we need to follow an other strategy for simplification. For the first sum we use ((D${(\zeta)}$)) for ζ = 1 for d m and then ((D${(\zeta)}$)) for ζ = 1 and ζ = 2 , the first condition (O3) (as l a n l = c n ), the last condition of (O4) and the first condition of (O5-3)

    m n l 1 b l a n l a m l d m d n = n l 1 b l a n l d n m a m l b m - n l 1 b l a n l d n m a m l b m c m
    = n l 1 b l a n l d n b l ( 1 - c l ) - 1 2 n l 1 b l a n l d n b l ( 1 - c l 2 )
    = 1 2 n l 1 b l a n l d n - n l a n l d n c l + 1 2 n l a n l d n c l 2
    = 1 2 1 6 - 1 24 + 1 2 1 60 = 1 20 .

    For the second sum we use ((D${(\zeta)}$)) for ζ = 1 for d m and then((D${(\zeta)}$)) for ζ = 1 and ζ = 2 , the last condition of (O4), the first condition of (O5-3) and the last condition of (O6-2)

    m n l 1 b l a n l a m l c l d m d n = n l 1 b l a n l c l d n m a m l b m ( 1 - c m )
    = n l 1 b l a n l c l d n b l ( 1 - c l ) - 1 2 n l 1 b l a n l c l d n b l ( 1 - c l 2 )
    = 1 2 n l a n l c l d n - n l a n l c l 2 d n + 1 2 n l a n l c l 3 d n
    = 1 2 1 24 - 1 60 + 1 2 1 120 = 1 120 .

    Together we have shown

    m n l 1 b l 2 a n l a m l d l d m d n = 1 20 - 1120 = 1 24 .

  24. For the third condition of (A6-8) we use ((D${(\zeta)}$)) for ζ = 1 for d l , the last condition of (O4), the first condition (O5-3) and the third condition of (O6-2)

    k l m 1 b k a m k a l k c k d l d m = k m 1 b k a m k c k d m l b l a l k ( 1 - c l )
    = k m 1 b k a m k c k d m ( b k ( 1 - c k ) - 1 2 b k ( 1 - c k 2 ) )
    = 1 2 k m a m k c k d m - k m a m k c k 2 d m + 1 2 k m a m k c k 3 d m
    = 1 2 1 24 - 1 60 + 1 2 1 120 = 1 120 .

  25. For the first condition of (A6-9) we use ((D${(\zeta)}$)) for ζ = 1 for b m twice and for d n once, the definition c i = a i j , the first condition of (O3), the last condition of (O4), the third condition of (O4), the second condition of (O5-1), the first condition of (O5-2), the second condition of (O6-3)

    m n 1 b m 2 b n a m n d m 3 d n = m n b m 2 b n b m 2 b n a m n d m ( 1 - c m ) 2 ( 1 - c n )
    = m c m d m - m n a m n c n d m - 2 m d m c m 2
    + 2 m n a m n d m c m c n + m c m 3 d m - m n a m n d m c m 2 c n
    = 1 6 - 1 24 - 2 12 + 2 40 + 1 20 - 1 60 = 1 24 .

  26. For the second condition of (A6-9) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , the first condition of (O5-3) and the third condition of (O6-2)

    m l k 1 b k a m l a l k d k c l d m = m l k b k b k a m l a l k ( 1 - c k ) c l d m
    = m l a m l c l d m k ( a l k - a l k c k )
    = m l a m l c l 2 d m - 1 2 m l a m l c l 3 d m
    = 1 60 + 1 2 1 120 = 1 80 .

  27. For the third condition of (A6-9) we use ((D${(\zeta)}$)) for ζ = 1 for d l and d m , the simplifying condition ((C${(\eta)}$)) for η = 1 four times and for η = 2 twice, the first condition of (O4), the third condition of (O5-3) and the third condition of (O6-6)

    i l m b i b l b m a i m a i l c i d l d m = i l m b l b m b l b m b i a i m a i l c i ( 1 - c l ) ( 1 - c m )
    = i b i c i m a i m l a i l - i l b i a i l c i c l m a i m
    - i m b i a i m c i c m l a i l + i b i c i m a i m c m l a i l c l
    = i b i c i 3 - 2 i k b i a i k c i 2 c k + 1 4 i b i c i 5
    = 1 4 - 2 10 + 1 4 1 6 = 11 120 .

  28. For the first condition of (A6-10) we use ((D${(\zeta)}$)) for ζ = 1 for d m and d n , the simplifying condition ((C${(\eta)}$)) for η = 1 twice, the last condition of (O4), the first condition of (O5-3), the first condition of (A5-3) and the first condition of (O6-4)

    l m n 1 b l b m a n l a l m d l d m d n = l m n b l b m b l b m a n l a l m d n ( 1 - c l ) ( 1 - c m )
    = l n a n l d n m a l m - n l a n l d n c l m a l m - l m n a n l a l m d n c m + l m n a n l a l m d n c l c m
    = l n a n l d n c l - n l a n l d n c l 2 - l m n a n l a l m d n c m + l m n a n l a l m d n c l c m
    = 1 24 - 1 60 - 1 120 + 1 240 = 1 48 .

  29. For the second condition of (A6-10) we use ((D${(\zeta)}$)) for ζ = 1 for d l and d m , ((C${(\eta)}$)) for η = 1 four times and for η = 2 twice, the third condition of (O4), the second condition of (O5-1) and the first condition of (O6-1)

    l m n 1 b l b m a n m a n l d l d m d n = l m n b l b m b l b m a n m a n l d n ( 1 - c l ) ( 1 - c m )
    = n d n m a n m l a n l - n l a n l d n c l m a n m
    - n m a n m d n c m l a n l + n d n l a n l c l m a n m c m
    = n c n 2 d n - n l a n l d n c l c n - n m a n m d n c m c n + 1 4 n d n c n 4
    = 1 12 - 2 40 + 1 4 1 30 = 1 48 .

  30. For the third condition of (A6-10) we use ((D${(\zeta)}$)) for ζ = 2 twice, the condition (O2), the first condition of (O4) and the third condition of (O6-6)

    i j k b i b j b k a j k a i k c i c j c k = k 1 b k c k i b i a i k j b j a j k c j
    = 1 4 k 1 b k c k b k 2 ( 1 - c k ) 2
    = 1 4 ( k c k b k - 2 k c k 3 b k + k c k 5 b k )
    = 1 4 ( 1 2 + 1 4 + 1 6 ) = 1 24 .

  31. For the first condition of (A6-11) we use ((D${(\zeta)}$)) for ζ = 1 and ((D${(\zeta)}$)) for ζ = 2 three times, (B${(p)}$) for p = 1 (which is (O1)), the second condition of (O2), the second condition of (O5-3), the second condition of (O4) and the second condition of (O6-2)

    i j l b i b j b l 2 a j l a i l c i c j d l = i j l b i b j b l b l 2 a j l a i l c i c j ( 1 - c l )
    = l 1 b l i b i a i l c i j b j a j l c j - j l b j b l a j l c l c j i b i a i l c i
    = 1 4 l 1 b l b l 2 ( 1 - c l 2 ) 2 - 1 2 j l b j b l a j l c l c j b l ( 1 - c l 2 )
    = 1 4 ( l b l - 2 l b l c l 2 + l b l c l 4 ) - 1 2 ( j l b j a j l c j c l - j l b j a j l c j c l 3 )
    = 1 4 ( 1 - 1 3 + 1 5 ) - 1 2 ( 1 8 - 1 24 ) = 11 120 .

  32. For the second condition of (A6-11) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the third condition of (O5-1) and the second condition of (O6-2)

    i k l b i b k a l k a i k c i d k c l = i k l b i b k b k a l k a i k c i ( 1 - c k ) c l
    = i l b i a i k c i c l k a l k - i l b i a i k c i c k k a l k c l
    = i l b i a i k c i c l 2 - 1 2 i l b i a i k c i c k 3 = 1 15 - 1 2 1 24 = 11 240 .

  33. For the third condition of (A6-11) we use ((D${(\zeta)}$)) for ζ = 2 , the last condition of (O4) and the third condition of (O6-2)

    i k l b i b k a l k a i k c i c k d l = k l 1 b k a l k c k d l i b i a i k c i = 1 2 k l 1 b k a l k c k d l b k ( 1 - c k 2 )
    = 1 2 k l a l k c k d l b k - 1 2 k l a l k c k 3 d l b k = 1 2 1 24 - 1 2 1 120 = 1 60 .

  34. For the first condition of (A6-12) we use ((D${(\zeta)}$)) for ζ = 1 and ζ = 2 , the first condition of (O3) (together with ((C${(\eta)}$)) for η = 1 ), the last condition of (O4), the first condition of (O5-3) and the third condition of (O6-2)

    i l m 1 b l 2 b i a m l a i l c i d l d m = i l m b l b l 2 a m l ( 1 - c l ) d m i b i a i l c i
    = 1 2 l m b l 2 b l 2 a m l d m ( 1 - c l ) ( 1 - c l 2 )
    = 1 2 ( m d m l a m l - l m a m l d m c l - l m a m l d m c l 2 + l m a m l d m c l 3 )
    = 1 2 ( 1 6 - 1 24 - 1 60 + 1 120 ) = 7 120 .

  35. For the second condition of (A6-12) we use ((D${(\zeta)}$)) for ζ = 1 twice, the second condition of (O4) (together with ((C${(\eta)}$)) for η = 1 ), the third condition (O5-1) (together with ((C${(\eta)}$)) for η = 1 ), the first condition (O5-1) and the last condition of (O6-3)

    i l m b i b l b m a l m a i l c i d l d m = i l m b l b m b l b m b i a l m a i l c i ( 1 - c l ) ( 1 - c m )
    = i l b i a i l c i m a l m - i l b i a i l c i c l m a l m - i l m b i a l m a i l c i c m + i l m b i a l m a i l c i c l c m
    = 1 8 - 1 15 - 1 30 + 1 48 = 11 240 .

  36. For the third condition of (A6-12), we use ((C${(\eta)}$)) for η = 2 twice and the third condition of (O6-6)

    i j k b i a i k a i j c i c j c k = i b i c i ( k a i k c k ) ( j a i j c j ) = 1 4 i b i c i 5 = 1 4 1 6 = 1 24 .

  37. For the first condition of (A6-13), we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 2 and η = 1 , the second condition of (O5-3) and the first condition of (O6-3)

    i k l b i b k a i k a i l c i d k c l = i k b i b k a i k c i b k ( 1 - c k ) l a i l c l = 1 2 i k b i a i k c i 3 ( 1 - c k )
    = 1 2 i b i c i 3 k a i k - 1 2 i k b i a i k c i 3 c k = 1 2 i b i c i 4 - 1 2 i k b i a i k c i 3 c k
    = 1 2 1 5 - 1 2 1 12 = 7 120 .

  38. For the second condition of (A6-13), we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 2 , the first condition of (O5-3) and the third condition of (O6-2)

    k l m 1 b k a m k a k l d k c l d m = k m b k b k a m k d m ( 1 - c k ) l a k l c l
    = 1 2 ( k m a m k d m c k 2 - k m a m k d m c k 3 ) = 1 2 ( 1 60 - 1 120 ) = 1 240 .

  39. For the third condition of (A6-13), we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 2 , the first condition of (O5-1) and the second condition of (O6-2) and the third condition of (O4)

    i k l b i b k a i k a k l c i d k c l = i k l b k b k b i a i k a k l c i c l ( 1 - c k )
    = i k l b i a i k a k l c i c l - i k b i a i k c i c k l a k l c l
    = i k l b i a i k a k l c i c l - 1 2 i k b i a i k c i c k 3 = 1 30 - 1 2 1 24 = 1 80 .

  40. For the first condition of (A6-14), we use ((D${(\zeta)}$)) for ζ = 1 , the third condition of (A5-5), ((D${(\zeta)}$)) for ζ = 2 , ((C${(\eta)}$)) for η = 1 , the first condition of (O6-2)

    l m b i b l b m a i m a l m c i d l 2 = l m b i b l b l b m a i m a l m c i d l ( 1 - c l )
    = l m b i b m a i m a l m c i d l - l m 1 b m a l m d l c l i b i a i m c i
    = 3 40 - 1 2 ( l m b m b m a l m d l c l ( 1 - c m 2 ) )
    = 3 40 - 1 2 ( l d l c l m a l m - l m a l m d l c l c m 2 )
    = 3 40 - 1 2 ( l d l c l 2 - 1 90 ) = 3 40 - 1 2 ( 1 12 - 1 90 ) = 7 180 .

  41. For the second condition of (A6-14), we use ((C${(\eta)}$)) for η = 2 twice and the first condition of (O6-1)

    j k l a j l a j k d j c k c l = j d j ( l a j l c l ) ( k a j k c k )
    = 1 4 j d j c j 4 = 1 4 1 30 = 1 120 .

  42. For the last condition of (A6-14), we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the second condition of (O5-1) and the second condition of (O6-3)

    k l m 1 b k a l k a l m d k d l c m = k l m b k b k a l k a l m ( 1 - c k ) d l c m
    = l m a l m d l c m k a l k - l m a l m d l c m k a l k c k
    = l m a l m d l c m c l - 1 2 l m a l m d l c m c l 2 = 1 40 - 1 2 1 60 = 1 60 .

  43. For the first condition of (A6-15) we use ((D${(\zeta)}$)) for ζ = 1 ,

    i k l b i b k a l k a i k c i 2 d l = i k l b i b l b k a l k a i k c i 2 ( 1 - c l )
    (D.1) = i k b i b k a i k c i 2 l b l a l k - i k b i b k a i k c i 2 l b l a l k c l .

    For the first term we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , the last condition of (O5-2), and the first condition of (O4) (which is (B${(p)}$) for p = 4 )

    i k b i b k a i k c i 2 l b l a l k = i k b i b k a i k c i 2 b k ( 1 - c k ) = i b i c i 2 k a i k - i k b i a i k c i 2 c k
    = i b i c i 3 - 1 10 = 1 4 - 1 10 = 3 20 .

    For the second term of (D.1), we use ((D${(\zeta)}$)) for ζ = 2 , ((C${(\eta)}$)) for η = 1 , the last condition of (O6-1), and the first condition of (O4)

    i k b i b k a i k c i 2 l b l a l k c l = 1 2 i k b i b k a i k c i 2 b k ( 1 - c k 2 )
    = 1 2 i b i c i 2 k a i k - 1 2 i k b i a i k c i 2 c k 2
    = 1 2 i b i c i 3 - 1 2 1 18 = 1 2 1 4 - 1 36 = 7 72 .

    Altogether we have

    i k l b i b k a l k a i k c i 2 d l = 3 20 - 7 72 = 19 360 .

  44. For the second condition of (A6-15), we use ((D${(\zeta)}$)) for ζ = 1 , the last condition of (A5-5), and the first condition of (A6-15), which has just been proven without use of the second condition of (A6-15))

    k l m 1 b k a m k a l k c l d l d m = k l m b l b k a m k a l k c l ( 1 - c l ) d m
    = k l m b l b k a m k a l k c l d m - k l m b l b k a m k a l k c l 2 d m = 3 40 - 19 360 = 1 45 .

  45. For the last condition of (A6-15), we use ((D${(\zeta)}$)) for ζ = 1 , the last condition of (A5-4) and the second condition of (A6-15), which has just been proven without use of the last condition of (A6-15))

    l m n 1 b l b m a n m a l m d l 2 d n = l m n b l b l b m a n m a l m d l d n ( 1 - c l )
    = l m n 1 b m a n m a l m d l d n - l m n 1 b m a n m a l m c l d l d n = 1 20 - 1 45 = 1 36 .

  46. For the first condition of (A6-16) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 2 , the first condition of (O5-3) and the first condition of (O6-2)

    k l m 1 b k a l m a k l d k 2 c m = k l m b k b k a l m a k l d k ( 1 - c k ) c m = k l a k l d k ( 1 - c k ) m a l m c m
    = 1 2 ( k l a k l d k c l 2 - k l a k l d k c k c l 2 ) = 1 2 ( 1 60 - 1 90 ) = 1 360 .

  47. For the second condition of (A6-16) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , the first condition of (O5-1), the first condition of (O6-4)and the second condition of (O4)

    i l m b i b l 2 a i m a m l c i d l 2 = i l m b l 2 b l 2 b i a i m a m l c i ( 1 - c l ) 2
    = i m b i a i m c i l a m l - 2 i l m b i a i m a m l c i c l + i l m b i a i m a m l c i c l 2
    = i m b i a i m c i c m - 2 30 + 1 72 = 1 8 - 19 360 = 13 180 .

  48. For the last condition of (A6-16) we use ((D${(\zeta)}$)) for ζ = 1 twice, ((C${(\eta)}$)) for η = 1 , the first condition of (A5-3) , the second condition of (O6-4), the last condition of (O4) and the second condition of (O5-1)

    l m n 1 b l b m a n m a l n d l 2 d m = l m n b l b m b l b m a n m a l n d l ( 1 - c l ) ( 1 - c m )
    = l n a l n d l ( 1 - c l ) m a n m - l m n a l n a n m d l c m + l m n a l n a n m d l c l c m
    = l n a l n d l c n - l n a l n d l c l c n - 1 120 + 1 180 = 1 24 - 1 40 - 1 360 = 1 72 .

  49. For the first condition of (A6-17) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the last condition of (O4), the first condition of (O5-3) and the second condition of (O6-5)

    l m n 1 b l 2 a n m a m l d l 2 d n = l m n b l 2 b l 2 a n m a m l d n ( 1 - c l ) 2
    = m n a n m d n l a m l - 2 m n a n m d n l a m l c l + l m n d n a n m a m l c l 2
    = m n a n m d n c m - m n a n m d n c m 2 + l m n d n a n m a m l c l 2
    = 1 24 - 1 60 + 1 360 = 1 36 .

  50. For the second condition of (A6-17) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 twice, the first condition of (O4), the last condition of (O5-2) and the last condition of (O6-1)

    i m l 1 b l 2 b i a i m a i l d l 2 c m = i m l b l 2 b l 2 b i a i m a i l ( 1 - c l ) 2 c m
    = i m b i a i m c m l a i l - 2 i m b i a i m c m l a i l c l + i l b i a i l c l 2 m a i m c m
    = i m b i a i m c i c m - i m b i a i m c m c i 2 + 1 2 i l b i a i l c l 2 c i 2
    = 1 8 - 1 10 + 1 2 1 18 = 19 360 .

  51. For the last condition of (A6-17) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 five times and η = 2 twice, the last condition of (O6-4), the second condition of (O3), the second condition of (O4), the last condition of (O5-1), the second condition of (O4) and the second condition of (O5-3)

    i m n b i b m 2 b n a i n a i m d m 2 d n = i m n b m 2 b n b m 2 b n b i a i n a i m ( 1 - c m ) 2 ( 1 - c n )
    = i b i ( n a i n ) ( m a i m ) - 2 i m b i a i m c m ( n a i n ) + i m b i a i m c m 2 ( n a i n )
    - i m b i a i n c n ( m a i m ) + 2 i b i ( n a i n c n ) ( m a i m c m ) - i m n b i a i n a i m c n c m 2
    = i b i c i 2 - 2 i m b i a i m c i c m + i m b i a i m c i c m 2 - i m b i a i n c n c i + 1 2 i b i c i 4 - 1 36
    = 1 3 - 2 8 + 1 15 - 1 8 + 1 2 1 5 - 1 36 = 7 72 .

  52. For the first condition of (A6-18) we use ((D${(\zeta)}$)) for ζ = 2 , ((C${(\eta)}$)) for η = 2 , the first condition of (O6-2) and the third condition of (O4)

    i k l b i b k a i k a l k c i c l d l = k l 1 b k a l k c l d l ( i b i a i k c i ) = 1 2 k l b k b k a l k c l d l ( 1 - c k 2 )
    = 1 2 l c l d l k a l k - 1 2 k l a l k c l d l c k 2 = 1 2 l c l 2 d l - 1 2 1 90
    = 1 2 1 12 - 1 180 = 13 360 .

  53. For the second condition of (A6-18) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the second condition of (O5-1) and the first condition of (O6-2)

    k l m 1 b k a m k a l m d k c l d l = k l m b k b k a m k a l m c l d l ( 1 - c k )
    = l m a l m c l d l k a m k - l m a l m c l d l k a m k c k
    = l m a l m c l d l c m - 1 2 l m a l m c l d l c m 2 = 1 40 - 1 2 1 90 = 7 360 .

  54. For the third condition of (A6-18) we use ((D${(\zeta)}$)) for ζ = 1 , the first condition of (O5-1) and the first condition of (O6-5)

    i l m b i b k a i l a l k c i c k d k = i l m b k b k b i a i l a l k c i c k ( 1 - c k )
    = i l m b i a i l a l k c i c k - i l m b i a i l a l k c i c k 2 = 1 30 - 1 72 = 7 360 .

  55. For the first condition of (A6-19) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 2 , the second condition of (O6-5) and the first condition of (O5-3)

    k l m 1 b k a m l a l k c k d k d m = k l m b k b k a m l a l k c k ( 1 - c k ) d m
    = l m a m l d m k a l k c k - k l m a m l a l k c k 2 d m
    = 1 2 l m a m l d m c l 2 - 1 360 = 1 2 1 60 - 1 360 = 1 180 .

  56. For the second condition of (A6-19) we use ((D${(\zeta)}$)) for ζ = 1 , the first condition of (O5-1) and the third condition of (O6-5)

    i l k b i b k a i l a i k c k d k c l = i l k b k b k b i a i l a i k c k ( 1 - c k ) c l
    = i l k b i a i l a i k c k c l - i l k b i a i l a i k c k 2 c l = 1 20 - 1 36 = 1 45 .

  57. For the third condition of (A6-19) we use ((D${(\zeta)}$)) for ζ = 1 twice, ((C${(\eta)}$)) for η = 1 and η = 2 twice, the third condition of (O6-5), the second condition of (O4), the third condition of (O5-1) and the second condition of (O5-3)

    i l m b i b l b m a i m a i l c l d l d m = i l m b l b m b l b m b i a i m a i l c l ( 1 - c l ) ( 1 - c m )
    = i l b i a i l c l ( 1 - c l ) m a i m - i b i ( l a i l c l ) ( m a i m c m ) + i l m b i a i l a i m c l 2 c m
    = i l b i a i l c i c l - i l b i a i l c i c l 2 - 1 4 i b i c i 4 + 1 36
    = 1 8 - 1 15 - 1 4 1 5 + 1 36 = 13 360 .

  58. For the first condition of (A6-20) we use ((D${(\zeta)}$)) for ζ = 2 , ((C${(\eta)}$)) for η = 1 , the third condition (O6-1) and the first condition (O4)

    i j k b i b j b k a j k a i k c i 2 c j = i k b i b k a i k c i 2 j b j a j k c j = 1 2 i k b i b k a i k c i 2 b k ( 1 - c k 2 )
    = 1 2 ( i b i c i 2 k a i k - i k b i a i k c i 2 c k 2 ) = 1 2 ( i b i c i 3 - 1 18 )
    = 1 2 ( 1 4 - 1 18 ) = 7 72 .

  59. For the second condition of (A6-20) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the third condition of (O5-2) and the third condition of (O6-1)

    i l k b i b k a l k a i l c i 2 d k = i l k b i b k a l k a i l c i 2 b k ( 1 - c k ) = i l b i a i l c i 2 k a l k - i l b i a i l c i 2 k a l k c k
    = i l b i a i l c i 2 c l - 1 2 i l b i a i l c i 2 c l 2 = 1 10 - 1 2 1 18 = 13 180 .

  60. For the last condition of (A6-20) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , the third condition of (O6-5) and the third condition of (O5-1)

    i k l b i b k a i k a i l d k c l 2 = i k l b i b k a i k a i l b k ( 1 - c k ) c l 2
    = i l b i a i l c l 2 k a i k - i k l b i a i k a i l c k c l 2 = i l b i a i l c l 2 c i - 1 36 = 1 15 - 1 36 = 7 180 .

  61. For the first condition of (A6-21) we use ((D${(\zeta)}$)) for ζ = 2 , ((C${(\eta)}$)) for η = 2 , the first condition of (O6-5) and the second condition of (O4)

    i j k l b i b j b k a j l a l k a i k c i c j = j k l b j b k a j l c j a l k ( i b i a i k c i ) = 1 2 j k l b j b k a j l c j a l k b k ( 1 - c k 2 )
    = 1 2 ( j l b j a j l c j k a l k - j k l b j a j l c j a l k c k 2 )
    = 1 2 ( j l b j a j l c j c l - 1 72 ) = 1 2 ( 1 8 - 1 72 ) = 1 18 .

  62. For the second condition of (A6-21) we use ((D${(\zeta)}$)) for ζ = 2 , ((C${(\eta)}$)) for η = 1 and η = 2 , the second condition of (O4) and the last condition of (O6-1)

    i j k l b i b j b k a j l a j k a i k c i c l = i j k l b j b k a j l a j k c l i b i a i k c i = 1 2 i j k l b j b k a j l a j k c l b k ( 1 - c k 2 )
    = 1 2 i j l b j a j l c l k a j k - 1 2 i j k l b j a j k c k 2 l a j l c l
    = 1 2 i j l b j a j l c l c k - 1 4 i j k l b j a j k c k 2 c j 2 = 1 2 1 8 - 1 4 1 18 = 7 144 .

  63. For the first condition of (A6-22) we use ((D${(\zeta)}$)) for ζ = 1 and ζ = 2 , ((C${(\eta)}$)) for η = 1 four times, the second condition of (O3), the last condition of (O5-1), the second condition of (O4) and the last condition of (O6-4)

    i j l m b i b j b l b m a j m a i m a i l c j d l = i l m b i b l b m a i m a i l b l ( 1 - c l ) j b j a j m c j
    = 1 2 i l m b i b m a i m a i l ( 1 - c l ) b m ( 1 - c m 2 )
    = 1 2 i b i ( m a i m ) ( l a i l ) - 1 2 i m b i a i m c m 2 l a i l
    - 1 2 i m b i a i l c l m a i m + 1 2 i m l b i a i m a i l c m 2 c l
    = 1 2 ( i b i c i 2 - i m b i a i m c m 2 c i - i m b i a i l c l c i + i m l b i a i m a i l c m 2 c l )
    = 1 2 ( 1 3 - 1 15 - 1 8 + 1 36 ) = 61 720 .

  64. For the second condition of (A6-22) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the first condition of (O5-1) and the last condition of (O6-4)

    i k l m b i b k a l m a i l a i k d k c m = i k l m b i b k a l m a i l a i k b k ( 1 - c k ) c m
    = i l m b i a l m a i l c m k a i k - i l m b i a l m a i l c m k a i k c k
    = i l m b i a l m a i l c m c i - 1 2 i l m b i a l m a i l c m c i 2 = 1 30 - 1 2 1 36 = 7 360 .

  65. For the first condition of (A6-23) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 , the second condition of (O6-6) and the first condition of (O5-1)

    i k l m b i b k a i m a m l a l k c i d k = i k l m b i b k a i m a m l a l k c i b k ( 1 - c k )
    = i l m b i a i m a m l c i k a l k - i k l m b i a i m a m l a l k c i c k
    = i l m b i a i m a m l c i c l - 1 144 = 1 30 - 1 144 = 19 720 .

  66. For the second condition of (A6-23) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 once and for η = 2 four times, the second condition of (O5-3) and the last condition of (O6-1)

    i k l m b i b k a i m a i l a l k c m d k = i k l m b i b k a i m a i l a l k c m b k ( 1 - c k )
    = i b i ( l a i l ( k a l k ) ) ( m a i m c m ) - i l b i a i l ( m a i m c m ) ( k a l k c m c k )
    = 1 4 ( i b i c i 4 - i l b i a i l c i 2 c l 2 ) = 1 4 ( 1 5 - 1 18 ) = 13 360 .

  67. For the first condition of (A6-24) we use ((D${(\zeta)}$)) for ζ = 1 twice, ((C${(\eta)}$)) for η = 1 four times and for η = 2 four times, the second condition of (O4), the last condition of (O5-1), the last condition of (O5-2) and the last condition of (O6-1)

    i l m n b i b l b m a i m a i n a n l d l d m = i l m n b i b l b m a i m a i n a n l b l b m ( 1 - c l ) ( 1 - c m )
    = i n b i a i n ( m a i m ) ( l a n l ) - i n b i a i n ( m a i m ) ( l a n l c l )
    - i n b i a i n ( l a n l ) ( m a i m c m ) + i n b i a i n ( m a i m c m ) ( l a n l c l )
    = i n b i a i n c i c n - 1 2 i n b i a i n c i c n 2 - 1 2 i n b i a i n c n c i 2 + 1 4 i n b i a i n c i 2 c n 2
    = 1 4 - 1 2 1 15 - 1 2 1 10 + 1 18 = 1 18 .

  68. For the second condition of (A6-24) we use ((D${(\zeta)}$)) for ζ = 2 , ((C${(\eta)}$)) for η = 1 , the second condition of (O6-5), ((C${(\eta)}$)) for η = 2 and the third condition of (O4)

    i k l m b i b k a i k a m k a l m c i d l = k l m 1 b k a m k a l m d l i b i a i k c i = 1 2 k l m 1 b k a m k a l m d l b k ( 1 - c k 2 )
    = 1 2 ( l m a l m d l k a m k - k l m a m k a l m d l c k 2 )
    = 1 2 ( l d l m a l m c m - 1 360 ) = 1 2 ( l d l c l 2 - 1 360 ) = 1 2 ( 1 12 - 1 360 ) = 7 360 .

  69. For the first condition of (A6-25) we use ((D${(\zeta)}$)) for ζ = 1 , the second condition of (O6-1), ((C${(\eta)}$)) for η = 2 , the first condition of (O5-3)

    k l m n 1 b k a n k a m n a l m d k d l = k l m n 1 b k a n k a m n a l m b k d l ( 1 - c k )
    = l m n a m n a l m d l k a n k - k l m n a n k a m n a l m d l c k
    = l m a l m d l n a m n c n - 1 720 = 1 2 l m a l m d l c m 2 - 1 720
    = 1 2 1 60 - 1 720 = 1 144 .

  70. For the second condition of (A6-25) we use ((D${(\zeta)}$)) for ζ = 2 and ζ = 1 , the first condition of (A6-15), ((C${(\eta)}$)) for η = 2 , the last condition of (O4) and the first condition of (O3)

    i k l m b i b k a i m a m k a l k c i d l = k l m 1 b k a m k a l k d l i a i m b i c i = 1 2 k l m 1 b k a m k a l k d l b m ( 1 - c m 2 )
    = 1 2 k l 1 b k a l k d l m b m a m k - 1 2 k l m b m b k a m k a l k d l c m 2
    = 1 2 k l 1 b k a l k d l b k ( 1 - c k ) - 1 2 19 360
    = 1 2 ( l d l k a l k - k l a l k d l c k ) - 19 720
    = 1 2 ( l d l c l - 1 24 ) - 19 720 = 1 2 ( 1 6 - 1 24 ) - 19 720 = 13 360 .

  71. For the first condition of (A6-26) we use ((D${(\zeta)}$)) for ζ = 1 , ((D${(\zeta)}$)) for ζ = 1 and ζ = 2 , ((C${(\eta)}$)) for η = 1 , the first condition of (A5-3), the second condition of (O6-5) and the last condition of (O4)

    k l m n 1 b k a n m a m k a l k d l d n = k l m n 1 b k a n m a m k a l k d n b l ( 1 - c l )
    = k m n 1 b k a n m a m k d n ( ( l a l k b l ) - ( l a l k b l c l ) )
    = k m n 1 b k a n m a m k d n b k ( ( 1 - c k ) - 1 2 ( 1 - c k 2 ) )
    = ( 1 - 1 2 ) m n a n m d n k a m k - k m n a n m a m k c k d n + 1 2 k m n a n m a m k c k 2 d n
    = 1 2 m n a n m c m d n - 1 120 + 1 2 1 360 = 1 2 1 24 - 1 144 = 1 72 .

  72. For the second condition of (A6-26) we use ((C${(\eta)}$)) for η = 2 , and the first condition of (A6-15)

    i k l m b i b k a i m a i k a l k d l c m = i k l b i b k a i k a l k d l m a i m c m = 1 2 i k l b i b k a i k a l k d l c i 2 = 1 2 19 360 = 19 720 .

  73. For the condition (A6-27) we use ((D${(\zeta)}$)) for ζ = 1 , ((C${(\eta)}$)) for η = 1 and η = 2 , the last condition of (A5-5) and the first condition of (A6-15)

    i l m n b i b l b m a i m a i l a n l d m d n = i l m n b i b l b m a i m a i l a n l b m ( 1 - c m ) d n
    = i l n b i b l a i l a n l d n ( ( m a i m ) - ( m a i m c m ) )
    = i l n b i b l a i l a n l d n c i - 1 2 i l n b i b l a i l a n l c i 2 d n = 3 40 - 1 2 19 360 = 7 144 .

Altogether we have derived the additional order conditions with the use of the simplifying assumptions and (B${(p)}$) for p = 1 , 2 , 3 , 4 , ((C${(\eta)}$)) for η = 1 , 2 , ((D${(\zeta)}$)) for ζ = 1 , 2 , and the order conditions (O1)–(O4), (O5-1)–(O5-3) and (O6-1)–(O6-7). ∎

References

[1] T. Apel and T. G. Flaig, Crank–Nicolson schemes for optimal control problems with evolution equations, SIAM J. Numer. Anal. 50 (2012), no. 3, 1484–1512. 10.1137/100819333Search in Google Scholar

[2] R. Becker, D. Meidner and B. Vexler, Efficient numerical solution of parabolic optimization problems by finite element methods, Optim. Methods Softw. 22 (2007), no. 5, 813–833. 10.1080/10556780701228532Search in Google Scholar

[3] J. F. Bonnans and J. Laurent-Varin, Computation of order conditions for symplectic partitioned Runge–Kutta schemes with application to optimal control, Rapport de recherche RR–5398 2004, http://hal.inria.fr/docs/00/07/06/05/PDF/RR-5398.pdf. Search in Google Scholar

[4] J. F. Bonnans and J. Laurent-Varin, Computation of order conditions for symplectic partitioned Runge–Kutta schemes with application to optimal control, Numer. Math. 103 (2006), no. 1, 1–10. 10.1007/s00211-005-0661-ySearch in Google Scholar

[5] T. Dupont and R. Scott, Polynomial approximation of functions in Sobolev spaces, Math. Comp. 34 (1980), no. 150, 441–463. 10.1090/S0025-5718-1980-0559195-7Search in Google Scholar

[6] T. G. Flaig, Discretization strategies for optimal control problems with parabolic partial differential equations, PhD thesis, Universität der Bundeswehr München, 2013. Search in Google Scholar

[7] W. W. Hager, Rates of convergence for discrete approximations to unconstrained control problems, SIAM J. Numer. Anal. 13 (1976), no. 4, 449–472. 10.1137/0713040Search in Google Scholar

[8] W. W. Hager, Runge–Kutta methods in optimal control and the transformed adjoint system, Numer. Math. 87 (2000), no. 2, 247–282. 10.1007/s002110000178Search in Google Scholar

[9] E. Hairer, C. Lubich and G. Wanner, Geometric Numerical Integration: Structure-Preserving Algorithms for Ordinary Differential Equations, 2nd ed., Springer Ser. Comput. Math. 31, Springer, Berlin, 2006. Search in Google Scholar

[10] E. Hairer, S. P. Nørsett and G. Wanner, Solving Ordinary Differential Equations. I. Nonstiff Problems, Springer Ser. Comput. Math. 8, Springer, Berlin, 1987. 10.1007/978-3-662-12607-3Search in Google Scholar

[11] E. Hairer and G. Wanner, Solving Ordinary Differential Equations. II. Stiff and Differential-Algebraic Problems, 2nd ed., Springer Ser. Comput. Math. 14, Springer, Berlin, 1996. 10.1007/978-3-642-05221-7_1Search in Google Scholar

[12] M. Herty, L. Pareschi and S. Steffensen, Implicit-explicit Runge-Kutta schemes for numerical discretization of optimal control problems, SIAM J. Numer. Anal. 51 (2013), no. 4, 1875–1899. 10.1137/120865045Search in Google Scholar

[13] L. S. Hou, O. Imanuvilov and H.-D. Kwon, Eigen series solutions to terminal-state tracking optimal control problems and exact controllability problems constrained by linear parabolic PDEs, J. Math. Anal. Appl. 313 (2006), no. 1, 284–310. 10.1016/j.jmaa.2005.02.067Search in Google Scholar

[14] J. Lang and J. G. Verwer, W-methods in optimal control, Numer. Math. 124 (2013), no. 2, 337–360. 10.1007/s00211-013-0516-xSearch in Google Scholar

[15] P. Lasaint and P.-A. Raviart, On a finite element method for solving the neutron transport equation, Mathematical Aspects of Finite Elements in Partial Differential Equations, Academic Press, New York (1974), 89–123. 10.1016/B978-0-12-208350-1.50008-XSearch in Google Scholar

[16] J.-L. Lions, Optimal Control of Systems Governed by Partial Differential Equations, Grundlehren Math. Wiss. 170, Springer, New York, 1971. 10.1007/978-3-642-65024-6Search in Google Scholar

[17] D. Meidner and B. Vexler, A priori error estimates for space-time finite element discretization of parabolic optimal control problems. I. Problems without control constraints, SIAM J. Control Optim. 47 (2008), no. 3, 1150–1177. 10.1137/070694016Search in Google Scholar

[18] D. Meidner and B. Vexler, A priori error analysis of the Petrov–Galerkin Crank–Nicolson scheme for parabolic optimal control problems, SIAM J. Control Optim. 49 (2011), no. 5, 2183–2211. 10.1137/100809611Search in Google Scholar

[19] A. Rösch, Error estimates for parabolic optimal control problems with control constraints, Z. Anal. Anwend. 23 (2004), no. 2, 353–376. 10.4171/ZAA/1203Search in Google Scholar

[20] S. Steffensen, M. Herty and L. Pareschi, Numerical methods for the optimal control of scalar conservation laws, System Modeling and Optimization, IFIP Adv. Inf. Commun. Technol. 391, Springer, Heidelberg (2013), 136–144. 10.1007/978-3-642-36062-6_14Search in Google Scholar

[21] F. Tröltzsch, Optimale Steuerung partieller Differentialgleichungen - Theorie, Verfahren und Anwendungen, Vieweg, Wiesbaden, 2005. 10.1007/978-3-322-96844-9Search in Google Scholar

Received: 2022-04-25
Revised: 2022-12-04
Accepted: 2023-01-11
Published Online: 2023-04-01
Published in Print: 2023-10-01

© 2023 Walter de Gruyter GmbH, Berlin/Boston

Downloaded on 21.1.2025 from https://www.degruyter.com/document/doi/10.1515/cmam-2022-0097/html
Scroll to top button