COMPUTATION OF SIMPLEST NORMAL FORMS OF DIFFERENTIAL EQUATIONS ASSOCIATED WITH A DOUBLE-ZERO EIGENVALUE

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1 International Journal of Bifurcation and Chaos, Vol, No 5 (2) c World Scientific Publishing Company COMPUTATION OF SIMPLEST NORMAL FORMS OF DIFFERENTIAL EQUATIONS ASSOCIATED WITH A DOUBLE-ZERO EIGENVALUE Y YUAN and P YU Department of Applied Mathematics, University of Western Ontario, London, Ontario, Canada N6A 5B7 Received June 8, 2; Revised August 6, 2 In this paper a method is presented for computing the simplest normal form of differential equations associated with the singularity of a double zero eigenvalue Based on a conventional normal form of the system, explicit formulae for both generic and nongeneric cases are derived, which can be used to compute the coefficients of the simplest normal form and the associated nonlinear transformation The recursive algebraic formulae have been implemented on computer systems using Maple The user-friendly programs can be executed without any interaction Examples are given to demonstrate the computational efficiency of the method and computer programs Introduction Normal form theory is one of the useful tools in analyzing complex dynamical behavior of a nonlinear system such as bifurcation and instability The method, based on successive coordinate transformations, can be used to systematically construct a simple form of the original differential equation, and thus greatly simplify the analysis for the original system [Guckenheimer & Holmes, 983; Elphick et al, 987; Nayfeh, 993; Chow et al, 994] The basic idea of the conventional (classical) normal form theory is using the linear singularity of a system at an equilibrim to form a Lie bracket operator and then repeated employing the operator to remove higher order nonlinear terms, as many as possible However, it was later found that conventional normal forms are not simplest forms and actually could be further simplified (eg see [Cushman & Sanders, 986, 988; Chua & Kokubu, 988a, 988b] Roughly speaking, in conventional normal form theory kth-order nonlinear transformation terms are used to eliminate kth-order nonlinear terms of the system, while in simplest normal form theory kth-order nonlinear transformation terms are not only used to simplify kth-order nonlinear terms but are also used to elminate higher order terms Since computation of simplest normal forms is much more complicated than that of conventional normal forms, computer algebra systems such as Maple, Mathematics, Reduce, etc have been used (eg see [Algaba, 998; Yu, 999]) The automatic Maple program developed by Yu [999] can be executed to find the simplest normal forms for Hopf and generalized Hopf bifurcations on various computer systems with very little preparation for an input file The conventional normal form for a double zero singularity has been considered by many researchers Author for correspondence pyu@pyuapmathsuwoca 37

2 38 Y Yuan & P Yu and obtained using different approaches (eg see [Guckenheimer & Holmes, 983; Yu & Huseyin, 986; Chow et al, 994; Broer et al, 995; Bi & Yu, 998]) Recently, simplest normal forms for the double zero singularity have been considered using different methodology [Kokubu et al, 988; Baider & Sanders, 992; Wang et al, 2], and symbolic computations have been discussed (eg see [Yu & Yuan, 999; Algaba et al, 2]) In this paper, we present a method which is based on matrix theory and linear algebra to compute the simplest normal form for the double zero singularity This approach provides a direct guideline for developing computer software Indeed, automatic Maple programs have been implemented on computer systems which can be executed without any interaction up to, in principle, any order To understand the idea which will be used in the further reduction of normal forms, let us consider a simple example associated with single zero singularity Suppose a general system is given by ẋ = Jx + f(x), x R N, () where function f and its first derivative vanishes at the origin an equilibrium of system (); and the Jacobian J evaluated at the origin is given by [ ] J =, A R (N ) (N ), (2) A in which A is hyperbolic (ie all eigenvalues of A have nonzero real parts) The conventional normal form of system () can be obtained by first applying the center manifold theory and then the conventional normal form theory However, it may be derived by using a combined approach [Bi & Yu, 998] without the aid of center manifold theory The conventional normal form (CNF) is given by n ẏ = a j y j (3) j=2 where constant coefficients a j are expressed explicitly in terms of the coefficients of function f Note that the expression of right-hand side is actually a polynomial of y This CNF can be further simplified by a nonlinear transformation to the following simplest normal form (SNF): (i) If a 2, then the SNF is u = a 2 u 2 + a 3 u 3 (4) (ii) If a 2 = = a m = but a m (m 3), then the SNF is u = a m u m + b 2m u 2m (m 3), (5) where the coefficient b 2m is expressed explicitly in terms of a j s The above results can be proved by using the method of mathematical induction In the next section, the SNF for the generic case of a double zero singularity is discussed in detail, while Sec 3 discusses the SNF of a nongeneric case Symbolic computation is outlined in Sec 4 Examples are given in Sec 5 to show the applicability of the method, and conclusions are drawn in Sec 6 Maple source codes and sample input data are listed in appendices for reference 2 SNF Generic Case Finding the SNF for a double zero singularity is much more complicated than the single zero case Consider the system given by ẋ = Jx + f(x), x R N, (6) where function f and its first derivative vanishes at the origin, andj is given by J =, A R (N 2) (N 2), (7) A in which A is hyperbolic With the CNF theory, one may use the nonlinear transformation (NT) x = y + x 2 = y 2 + x i = n n j j=2 k= n j=2 k= j j=2 k= j h k(j k) y k yj k 2, h 2k(j k) y k yj k 2, h ik(j k) y k yj k 2 (i =3, 4,, N) (8) to transform Eq (6) into a CNF, up to nth-order, written in either of the two ways [Guckenheimer & Holmes, 983]: n ẏ = y 2 + a j y j, j=2 (9) n ẏ 2 = a j2 y j ; j=2

3 Computation of Simplest Normal Forms 39 or ẏ = y 2, n ẏ 2 = (a j y j + a j2y j y 2 ) j=2 () where a j s and a j2 s are constant coefficients which can be determined from the coefficients of function f(x) It is noted from Eqs (9) and () that the second form () may be considered simpler than the first form (9) However, in this paper, we do not compare such kind of different normal forms, but consider how to further simplify conventional normal forms, and therefore we will use Eq () as the CNF throughout the paper The results are summarized in the following theorem Theorem Assume that by the normal form theory, the CNF of system (6) is given by Eq (), then the SNF of system (6) up to an arbitrary order can be found as u = u 2, u 2 = a 2 u 2 + a 22u u 2 + a 3 u 3 n ) + (b (3j+) + b (3j+2)2 u u 3j+ j= () if a 2 a 22, where the coefficients b (3j+) and b (3j+2)2 are expressed explicitly in terms of a ij s Proof The first step in the proof is straightforward: By using any methods of conventional normal forms (eg using the Maple programs developed by Bi and Yu [998]), one can obtain the CNF () for system (6) The major step (second step) of the proof is to show that the CNF () can be further simplified to form () by using a near-identity NT The idea is to eliminate the terms in Eq (), as many as possible, order by order, and thus Eq () indeed gives the simplest normal form of system (6) We shall employ the method of mathematical induction to prove the theorem To start the proof, one may assume a general nonlinear transformation described by y i = u i + n j j=2 k= c ik(j k) u k uj k 2 (i =, 2) (2) Then the second step can be proved by the method of mathematical induction It is easy to verify that the theorem is true when n =2, 3, 4 First consider n = 2: We want to see if any of the second-order terms in Eq () can be eliminated by using the following general second-order NT: y i = u i +c i2 u 2 +c i u u 2 +c i2 u 2 2 (i =, 2) (3) It should be noted that under any transformations, a new normal form should be at least as simple as the CNF () Thus, we may assume that upon applying the NT (3), the normal form up to second-order is in the form of u = u 2, u 2 = b 2 u 2 + b (4) 22u u 2 which is actually Eq () truncated at the thirdorder terms The basic idea is to find whether we can set b 2, and/or b 22 zero by choosing suitable coefficients c ijk s in Eq (3) To find the equations for determining the c ijk s, substituting Eqs (3) and (4) into Eq (), and then balancing the secondorder terms on both sides of the resulting equations yields the following six linear algebraic equations for the six unknown c ijk s, which can be grouped as two sets of decoupled equations, written in the matrix from: 2 2 c 2 c c 2 c 22 c 2 c 22 = (5) a 2 b 2 a 22 b 22 It is obvious to see from Eq (5) that c = c 22, c 22 = c 2 = c 2 =, b 2 = a 2, b 22 = a 22 Therefore, the NT (3) can be written as (6) y = u + c 2 u c 22u u 2, y 2 = u 2 + c 22 u 2 2, (7)

4 3 Y Yuan & P Yu and the SNF up to second-order then becomes u = u 2, u 2 = a 2 u a 22u u 2 (8) Equation (8) suggests that no matter what NT one may choose, and the second-order terms in the CNF () cannot be eliminated, and therefore, the normal form up to second-order cannot be further simplified It is noted from Eq (5) that the coefficients c 22 and c 2 (in fact, c 2 does not appear in these equations) can be chosen arbitrarily Of course, the simplest choice for the two coefficients is c 22 = c 2 =, which leads to the transformation y i = u i (i =, 2) implying that nothing has been performed However, as it will be seen, the two coefficients can be used to simplify higher order normal forms This is the key idea used in further simplifying a normal form obtained through the CNF theory When n = 3, similarly we want to eliminate the third-order terms in the CNF () by using the following third-order NT: y = u + c 2 u c 22u u 2 + c 3 u c 2 u u c 2u 2 u 2 + c 3 u 3, y 2 = u 2 + c 22 u c 23u c 22u u c 22 u 2 u 2 + c 23 u 3 (9) Assume that the third-order terms in the normal form are the same as that given, in general, by the corresponding part of the CNF, and then add these terms to the previously obtained (second-order) SNF to construct a form up to third-order terms as follows: u = u 2, u 2 = a 2 u 2 + a 22u u 2 + b 3 u 3 + b 32u 2 u 2 (2) Following the procedure used to obtain the secondorder SNF, one can find eight linear algebraic equations for solving the eight unknown third-order coefficients Again, the eight equations can be divided into two decoupled groups as listed below: c 3 3 c 2 2 c 2 c 3 = c 23 3 c 22 2 c 22 c 23 a 2 c 22 a 22 c 22 2a 2 c 2 2a 22 c 2 a 3 b 3 a 32 b 32 2a 2 c 2 a 22 c 2 = + a 3 b 3 a 32 b 32 a 2 a 22 2a 2 2a 22 c 22 + c 2 (2) 2a 2 a 22 from which we can easily see that b 3 = a 3 In order to find a normal form at this order as simple as possible, we may set b 32 =, and thus c 22 can be uniquely determined Consequently, we obtain b 3 = a 3, b 32 =, c 22 = a 32 3a 2 (22)

5 Computation of Simplest Normal Forms 3 and c 23 = a 32 3, c 2 = c 23, 3 c 3 a 2 2 c 2 2 c = a 22 a a 2 c 22 2a 2 2a 22 2a 2 a 22 c 2, (23) and the SNF up to third-order terms is thus given by u = u 2, u 2 = a 3 u 3 + a 2u 2 + a 22u u 2 (24) When n = 4, we can also find a group of equations, which may be called key equations and can be rearranged in two parts: the first part is given by [ ] a4 b 4 2a2 a 2 + a 42 b 42 a 22 2a 2 2a 2 c 3 c 2 c = 3 a 22a a 3 a 32 (25) c a 2 22 It is noted from Eq (23) that the vector, consisting of the four coefficients c 3, c 2, c 22 and c 22, can be uniquely solved from this equation in terms of c 2 and known coefficients a ij s since the matrix is nonsingular Then substitute this vector into Eq (25) to obtain two linear equations involving three unknowns b 4, b 42 and c 2 Now we want to find the possibility of eliminating b 4 and b 42 Itis easy to see that at most only one of them can be removed by choosing c 2 If let b 42 =,thenc 2 can be solved from a decoupled equation and thus b 4 can be determined from another equation So, the results are b 42 =, c 2 = 2a 2 2 a (9a 42 a 2 a 2 22 a 32 9a 3 a 32 ), 22 b 4 = 4a 22 (4a 4 a 22 + a 2 22a 32 5a 42 a 2 +5a 3 a 32 ) The second part consists of the equations of (26) c 4 c 3 c 22 = c 23 c 222 c 23 2a 2 2a 22 a 2 a 22 3a 2 3a 22 c 23 + c 3 + D (27) 2a 2 a 22 and c 3 = c 24 (28) In Eq (27) D represents the known expressions which have been obtained from the previous steps The D i, i =2, 3, 4, 5usedinthelatter equations have the same meaning Next, suppose that the theorem is true up to (n )th-order (n 3), we then show that it is also true up to nth-order and therefore, the theorem is true up to any order To achieve this, substituting Eq (2) into Eq (), with the aid of Eq (), and then balancing the nth-order coefficients in the resulting equation yields the following linear algebraic equations, written in the matrix form:

6 32 Y Yuan & P Yu c n n n c (n ) c (n 2)2 c n = v (29) n c 2n n c 2(n ) n 2 c 2(n 2)2 c 2n where the 2(n+)-dimensional vector v contains the expressions which are functions of the coefficients, some of which have been obtained in the previous steps, while one or two are determined in the current order, and others will be found in higher order equations It is noted that there exist two particular equations in Eq (29) which contain the coefficients b n and b n2 These two coefficients play a major role in determining the SNF of this order There are three cases: (i) When n =3k +(k ), there are two groups of key equations: the first group is given by an b n + F a n2 b C + F 2 C F k C k = D 2, n2 (3) where n 4 n 4 F = 2a 2 a 2, a 22 (n 2)a 2 2a 2 n i 3 n i 3 F i = b (i+) (i =2, 3,, k), 2b (i+) c (n i) c 2(n i 2) C i = (i =, 2,, k) c 2(n i ) c 2(n i ) (3) in which F i s are 2 2(n i ) matrices and C i s are 2(n i ) vectors The second group of equations are written as C = A n {B (n )C 2 + B 2(n ) C B (k )(n ) C k + α c 2k + β c 22k+ + + α (k ) c 3k 2 + β (k ) c 23k + α k c 3k }, C 2 = A n 2 {B (n 2)C B (k )(n 2) C k + α 2 c 2k + β 2 c 22k+ + + α 2(k ) c 3k 2 + β 2(k ) c 23k + α 2k c 3k }, C k = A n k {α kc 2k + β k c 22k+ + + α k(k ) c 3k 2 + β k(k ) c 23k + α kk c 3k } (32)

7 Computation of Simplest Normal Forms 33 Here, it should be noted that only vector C is directly generated at this order, while the remaining vectors C 2, C 3,, C k are actually obtained from the previous order equations through a recursive procedure α ij and β ij (i =, 2,, k; j =, 2,, k ) are 2(n i ) vectors, which are expressed explicitly in terms of the known coefficients a ij s and b ij s, including many zero components For example, 2k+j 3 2k+j 3 α j =(,,, (2k + j )b (k+2 j),,,,,,,, ) T (j =,, k ), 2k+j 3 2k+j 3 β j =(,,, (2k + j )b (k+ j),,,,,,,, ) T (j =,, k ), 3k 4 β (k ) =(,,, (3k 2)a 2, (3k 2)a 22,,,, 3(k )a 2, 3(k )a 22, ) T, 3k 4 (33) 3(k ) 3(k ) α k =(,,, (3k )a 2, (3k )a 22,,, 2a 2,a 22 ) T A n is a 2(n 2) 2(n 2) matrix, given by n n 2 2 A n =, (34) n 2 n 3 and A n represents the inverse matrix of A n B i(n ) s are given as follows: [ ] M O B (n ) =, (35) M 2 M 3 in which a 22 2a 2 a 22 2a 2 2a 22 3a 2 a 22 2a 2 M =, M 2 =, (n 4)a 22 a 22 a 22 3a 2 2a 22 4a 2 M 3 =, (n 4)a 22 (36)

8 34 Y Yuan & P Yu where O and M i (i =, 2, 3) are (n 2) (n 3) matrices, whereas B i(n ) (i =2, 3,, k ) are 2(n 2) 2(n 2 i) matrices and have the form: B i(n ) = (37) Next, substituting Eqs (3) (37) into Eq (3) results in where and an b n + a n2 b n2 + + ( 2 ( k ) c 2k + Y i v i c 22k+ + i= i= ( k i= Y i V i ) c 2k+ + ( k 2 i= Y i V 2 i ) c 22k+2 ) Y i V (k 2) i c 3k 2 +(Y V (k 2)2 )c 23k +(Y V (k ) )c 3k = D 3, (38) V (k 2)2 β (k ), V (k ) α k, Y = F A n, Z = F 2, Y i+ =(Y i B (n i) + Z i )A n i i Z i+ = Y j B (i j+2)(n j) + F i+2, j= (i =, 2,, k 2), (39) 2(k ) 2(k ) v i =(,,, 2kb (k i+),,,,,,,, ) T, 2(k ) 2(k ) v k =(,,, 2ka 2, 2ka 22,,,, (2k )a 2, (2k )a 22, ) T,

9 where v i, V j i and V j2 i (i, j =, 2,, k ) are 2(n i ) vectors V j i 2(k )+j 2(k )+j (,,,,,,,,, ) T Computation of Simplest Normal Forms 35 and V j2 i have the same form of but they are different vectors Now we want to prove that in Eq (38) except for the first three terms, all other terms are indeed zero First we show that k i= Y iv i = To achieve this, note that Y = F A n n 4 n 4 = 2a 2 a 2 a 22 (n 2)a 2 2a 2 n 4 n 4 =, (4) n 4 n 4 Y B (n ) = [ ] M O M 2 M 3 n 7 n 7 =, (4) then n 5 n 5 Z = F 2 =, (42) Y 2 =(Y B (n ) + Z )A n 2 n 9 n 9 =, 2i 2i Y i =, (43)

10 36 Y Yuan & P Yu and therefore, 2(k ) 2i 2i Y i v i = (44) 2(k ) Consequently, when i k or2i 2(k ), Eq (44) becomes Y i v i =, i =, 2,, k (45) Having proved k i= Y iv i =, it is easy to see that k j i= Y iv j i =and k j i= Y i V j2 i =for j k because these two vectors have more zero components than v i does Finally, Eq (38) becomes an b n + c a n2 b n2 2k = D 3 (46) Equation (46) clearly indicates that we can set b n2 = by explicitly choosing a unique c 2k in terms of the known coefficients, and then b n is also uniquely determined (ii) When n =3k +2(k ), we can find a similar equation like Eq (38) as follows: an b n + c a n2 b n2 2(2k+) ( k ) + Y i u i c (2k+) = D 4, (47) i= where the terms related to c 2(2k+j) and c (2k+j) (j =2, 3,, k) have been neglected since they are all zero, which becomes more clear below Here, 2k 2k u i =(,,, (2k +)b (k i+2),,,,,,,, ) T, 2k 2k u k =(,,, (2k +)a 2, (2k +)a 22,,, 2a 2,a 22 ) T, (48) in which u i (i =, 2,, k)are2(n i ) vectors Then, a simple manipulation shows that 2k 2i 2i Y i u i =, (49) 2k

11 Computation of Simplest Normal Forms 37 and thus, when i k (ie 2i 2k 3and2 2i 2k 2), Eq (49) yields Y i u i = for i =, 2,, k, (5) which implies that the summation in Eq (47) actually has only one term Y k u k Next consider i = k For this case, we have 2k (2k +)a 2 2k 2k (2k +)a 22 Y k u k = s k t k 2k 2a 2 a 22 ( ) =, (2k +)a 2 s k +2a 2 t k (5) where s k =(Y k ) 2,2k =((Y k B (n k+) + Z k )A n k ) 2,2k =(Y k B (n k+) A n k ) 2,2k 2k 3 2k + n 3k + a 2 2k + n 3k + a 22 2k 3 2k 3 =(,,, s k,,,,, t k,, ) 2k 3 2 n 3k + a 2 n 3k + a 22 = 2k n 3k + s 2 k + n 3k + t k a 2, (52)

12 38 Y Yuan & P Yu t k =(Y k ) 2,4k+ =((Y k B (n k+) + Z k )A n k ) 2,4k+ =(Y k B (n k+) A n k ) 2,4k+ 2k 3 2k + (n 3k +)(n 3k) a 2 2k + (n 3k +)(n 3k) a 22 2k 3 2k 3 =(,,, s k,,,,, t k,, ) 2k 3 ( ) n 3k + k 2a2 n 3k n 3k + a 22 [ 2k = (n 3k +)(n 3k) s k + 2 ( ) ] n 3k n 3k + k t k a 2 Here, the subscripts of Y k denote the row and column at which the component is located In general, we can find 2i s i = n 3i + s 2 i + n 3i + t i a 2, [ 2i t i = (n 3i +)(n 3i) s i + 2 ( ) ] (i =2, 3,, k) (53) n 3i n 3i + i t i a 2, A simple calculation for Y yields Y = F A n = 2a 2 n a 22 n n 4 {}}{ a 2 2 (n )(n 2) n 2 a 22 (n )(n 2) 3a 2 n 3 n 4 {}}{, (54) and then comparing the components leads to s = a 2 and t = 3/(n 3)a 2 = 3/(3k )a 2 It follows from Eq (53) that s 2 = t 2 = = ( [ 3 n 5 3 n 3 2 ) a 2 2 = 3 n 3 a2 2 = 3 3k a2 2, n 5 3 (n 5)(n 6) 3 n 3 2 ] n 6 n 5 2 a (n 5)(n 6) a2 2 = 3 5 (3k )(3k 4) a2 2, (55)

13 and then we can use the method of mathematical induction to show, in general, that s i = ( )i 3 5 (2i ) 5 8 (3i ) t i = ( )i 3 5 (2i +) 2 5 (3i ) a i 2, Computation of Simplest Normal Forms 39 a i 2, (i =, 2,, k) (56) It is clear from Eq (56) that t k 2k + = s k 2 Therefore, we have [see Eq (5)] Y i u i = or (2k +)s k +2t k = (57) and finally, Eq (47) becomes an b n + a n2 b n2 for i =, 2,, k, (58) c 22k+ = D 4 (59) Similarly, we may set b n2 = by explicitly solving for a unique c 2(2k+) from the above equation in terms of known coefficients Then b n is uniquely determined (iii) When n =3k +3(k ), similar to case (ii), we may obtain where an b n + a n2 b n2 c 2k+ ( k ) + Y i w i c 22k+2 = D 5 (6) i= 2k 2k w i =(,,, (2k +)b (k i+2),,,,,,,, ) T for i =, 2,, k, (6) 2k 2k w k =(,,, (2k +)a 2, (2k +)a 22,,,, 2ka 2, 2ka 22, ) T in which w i (i =, 2,, k)are2(n i ) vectors Similarly, we can prove that Y i w i = for i =, 2,, k, 2k (2k +)a 2 (2k +)a 22 2k 2k Y k w k = s k t k 2k ( ) = (2k +)a 2 s k 2ka 2 t k 2ka 2 2ka 22 (62)

14 32 Y Yuan & P Yu It has been seen from case (ii) that [ (2k +)s k + 2t k ]a 2 =,soifa 2,then (2k +)a 2 s k 2ka 2 t k Hence, k Y i w i = i= (63) and Eq (6) then becomes an b n + c 2k+ + c 22k+2 = D 5 a n2 b n2 (64) Unlike the cases (i) and (ii), for case (iii) we may set b n = b n2 = by uniquely determining c 22(k+) and c (2k+) in terms of the known coefficients explicitly The proof of the Theorem is completed From the above proof, we have seen that the pattern of the coefficients of the SNF is given as follows: When n =3k +or3k +2, b n2 =,b n, while when n =3k +3,b n = b n2 = The recursive formulae derived in this section can be straightforwardly used to code symbolic computer software Maple programs have been developed which can be executed on different computer systems The Maple source code is given in Appendix A 3 SNF Nongeneric Case The SNF for the generic case given in the previous section is found under the condition a 2 a 22 In this section, we shall briefly discuss the nongeneric case when a 2 a 22 = More specifically, if a 2 = a 3 = = a µ =anda 22 = a 32 = = a ν2 =, then what is the SNF? Baider and Sanders [992] have given a detailed study on the SNF which can be divided, in general, into three subcases: (I) µ<2ν, (II)µ>2ν and (III) µ =2ν They developed a tool called grading functions to find the form of the simplest normal forms The first two subcases were proved by Baider and Sanders [992] Later, Kokubu et al [996] and Wang et al [2] proved the subcase (III) and provided a form of the SNF However, it is noted that the form and method presented in these papers are not easy to be applied for computing an explicit SNF for a given system The approach given in the previous section can be straightforwardly used to code symbolic computer programs In this section, we shall discuss the computation method for the subcase (III) For this subcase, Wang et al [2] obtained the following form Suppose the original system is described by the following two-dimensional differential equations: ẏ = y 2 +hot, ẏ 2 = αy νy (65) 2 + βx 2ν+ +hot, where hot represents higher order terms of homogeneous polynomials of y and y 2 Then the SNF is u = u 2, u 2 = αu ν u 2 + βu 2ν+ + b 2ν u 2ν u 2 + a m u m (66) + m=2ν+2 n=ν+ n(mod(ν+)) ν,ν b n u n u 2, where the coefficients a m and b n are uniquely determined from the coefficients of the original system (65) Note that the original system (65) is a twodimensional general center manifold, which is not the general n-dimensional system (6), nor the CNF described by the two-dimensional center manifold described by Eq (9) or Eq () Determining the coefficients a m and b n is probably the most cumbersome part of the computation in applications For this reason, we have paid attention to computing the SNF efficiently Similar to the generic case, we can, based on the CNF (), obtain the following result Assume that by the CNF theory, the normal form of system (6) is given by Eq (), and in addition, a 2 =a 3 = =a 2(q ) =, a (2q ), (q 2), a 22 =a 32 = =a (q )2 =, a q2, (67) then the SNF of system (6) up to an arbitrary order is u = u 2, u 2 = a q2 u q u 2 + a (2q ) u 2q + a (2q )2 u 2q 2 u 2 + m=2q b m u m + n=q+ n(mod(q)) q,q b n2 u n u 2, (68)

15 Computation of Simplest Normal Forms 32 where the coefficients b m and b n2 are uniquely determined from the coefficients of the original system () The proof is similar to that for the generic case and thus the details are omitted In the following, we will discuss how to develop symbolic computation From the computation view point, based on the CNF () and the SNF (68), we want to use the coefficients of the NT c ik(j k) [see Eq (2)], to eliminate certain CNF coefficients a ij such that the CNF () is reduced to the SNF (68) The main procedure is substituting the NT (2) and the SNF (68) back into the CNF (), and then balancing the coefficients for each order This results in a set of linear algebraic equations for determining c ik(j k) and b ij First, it is easy to show that no c ik(j k) can be used to eliminate the first three terms in Eq (68) which have the same coefficients as that in the CNF Next note that the terms u n u 2 in the last summation expression of Eq (68) have been eliminated from the CNF () when n(mod(q)) equals q and q Therefore, the key step in the computation is to find the particular c ik(j k) s which can be used to eliminate those terms The linear algebraic equations corresponding to the missing terms u n u 2 comes from the coefficient of the term u n u 2 of the second balancing equation We call this coefficient as COEF(n, ) for convenience In addition, notice that the first coefficient c ik(j k) to be used is c 22, which is similar to the generic case [see Eq (22)] Then, by defining k =[n/q] (the notation [x] denotes the largest integer which is less than x), we obtain the following results: () When k =,c 22 can be uniquely determined from COEF(2q, ) since it involves c 22 only (2) When k>andn(mod(q)) = q, c k can be uniquely determined from COEF((k +)q 2, ) (3) When k > and n(mod(q)) = q, c 2k can be uniquely determined from COEF((k +)q, ) We have used the above formulae to develop general recursive Maple algorithms for computing the explicit expressions of the coefficients of the SNF and the corresponding NT The Maple programs are outlined in next section 4 Outline of Symbolic Computation The detailed procedure and formulae given in the previous sections can be directly applied to develop symbolic computation programs We have used Maple computer algebra system to develop a software package which can be conveniently used to compute the SNF and associated NT for a double zero singularity The program only requires a simple preparation for an input file from a user The Maple programs are outlined below, and the source codes can be found in Appendix A Since the programs for the generic and nongeneric cases are similar, we will not distinguish the two cases here (a) Read a prepared input file The input file indicates the upper boundary order of the computation of the SNF, Ord, and the conditions for the nongeneric case, ie the integer q For a particular given system, the input file also gives the coefficients of the original differential equations (ie the CNF of the system), a jkl (b) Initial step: For the generic case (i = 2), set b 2 = a 2 and b 22 = a 22 For the nongeneric case (i = q), set b q2 = a q2, b (2q ) = a (2q ) and b (2q )2 = a (2q )2 (c) For a suborder i(p i Ord, p =3forthe generic case, p = q + for the nongeneric case), recursively compute the coefficients of the SNF and the corresponding NT (i) Set the general SNF dy j, j =, 2using the undetermined coefficients b jk (ii) Create the variable H jk which is used to eliminate the terms higher than the given Ord in the nonlinear functions f j so that computation time can be reduced (iii) Set the general nonlinear functions f j, j =, 2, using H jk (iv) Substitute the NT H jk and the SNF dy j into the differential equations, and pick up the expression of the particular suborder i, Res j (v) From Res j obtained above, find the expressions of the coefficients for each binomial y l ym 2 (vi) By balancing the coefficient of each binomial, solve the coefficients of the ith-order NT, c jkl (vii) Solve the coefficients of the ith-order SNF, b jk (d) Write the SNF and NT into the output file Nform

16 322 Y Yuan & P Yu 5 Examples In this section, we shall apply the results presented in previous sections and the Maple programs we developed to compute the SNF for three examples: one is a generic case and two are nongeneric cases 5 Generic case Consider the following six-dimensional system: ẏ = y 2, ẋ = x 2 +(x x 2 ) 2 + x 3, ẋ 2 = x 2 + x x 2 x 3 x 6, ẋ 3 = x 3 + x x 2, ẋ 4 = 2x 4 + x x 2x 5, ẋ 5 = 3x 5 + x 6 + x 2 + x2 5, ẋ 6 = x 5 3x 6 + x 2 2 x 5x 6, (69) where the linearized system evaluated at the equilibrium x i = has a double zero eigenvalue λ = λ 2 =, two real eigenvalues λ 3 = andλ = 2, and a complex conjugate eigenvalue λ 5,6 = 3 ± i The Jacobian of system (69) is in the real Jordan canonical form J = (7) First, to find the CNF of system (69), we may apply the symbolic computer program written in Maple, developed by Bi and Yu [998] Executing the program on a PC gives the following CNF, up to twelveth order: ẏ 2 = y 2 +3y y 2 3y 3 + y2 y y4 7 5 y3 y y y4 y y y5 y y y6 y y y7 y y y8 y y y y 2 (7) y y y y y y 2 The coefficients given in the above equation can be written in the form of a j and a j2 according to formula () By noting that a 2 =anda 22 = 3, we know that this is a generic case Then we use these coefficients to execute the Maple program for the generic case (see Appendix A for the Maple source code) to find the SNF for this example u = u 2, u 2 = u 2 +3u u 2 3u u4 6 5 u u u (72) u u It should be noted that although the original differential equation (69) has integer coefficients only, the Maple program can deal with other real numbers or symbolic notations The examples given in the next subsection have noninteger coefficients and purely symbolic notations

17 Computation of Simplest Normal Forms Nongeneric case The first nongeneric example is described by the following five-dimensional system: ẋ = x x x x5 3, ẋ 2 = 2 3 x x 3 +5x 2 x x2 2, ẋ 3 = 3 x 3 x x x 2x 5, (73) ẋ 4 = 2x 4 + x 5 + x x 2x 3, ẋ 5 = x 4 2x x2 3 + x 2x x x 2 x 4, where the linearized system evaluated at the equilibrium x i = has a double zero eigenvalue λ = λ 2 =, one real eigenvalue λ 3 = (/3) and a complex conjugate eigenvalues λ 4,5 = 2 ± i The Jacobian of system (73) is in the real Jordan canonical form J 2 = (74) Again we may execute the earlier developed Maple program [Bi & Yu, 998] to find the CNF up to, say, twelveth-order: ẏ = y 2, ẏ 2 = 6 25 y2 y y3 y y y4 y y y5 y y y y y y 7 y y 9 y 8 y y y 9 y y 2 y y y y 2 y (75)

18 324 Y Yuan & P Yu Note from Eq (75) that a 2 = a 3 = a 4 =,a 5 = (2/25) and a 22 =,a 32 = (6/25), so this is a nongeneric case for q = 3 [see Eq (67)] Therefore, we may use the coefficients given in Eq (75) to execute the Maple program for computing the SNF of nongeneric case to obtain u = u 2, u 2 = 6 25 u2 u u u4 u u u u 6 u u8 u u u 9 u 2 u 2 u (76) The SNF given by Eqs (72) and (76) indeed shows that a CNF can be further simplified To conclude this section, we present another example which has been considered by Algaba et al [2] Assume we start from the CNF (): ẏ = y 2, ẏ 2 = n j=2 with the assumption (a j y j + a j2y j y 2 ) a 2 =, a 22, a 3, (77) which indicates that this is a nongeneric case with q = 2 The SNF form given by Algaba et al [2], called hypernormal form, is u = u 2, u 2 = a 22 u u 2 + a 3 u 3 + a 32u 2 u 2 + a 4 u 4 + a 5 u 5 + b 52u 4 u 2 + b 62 u 5 u 2 + b 72 u 6 u 2 + b 82 u 7 u 2 + b 92 u 8 u 2 + b 2 u 9 u 2 + b 2 u u 2 + b 22 u u 2 + (78) where 2a 3 a 22 a 32 a 42 +5a 6 a a 3 a 6 a 22 b 52 =a 52 5a 4 a 2 22 a 42+63a 3 a 4 a 42 a 3 (9a 3 +a 2 22 ), (79) However, executing our Maple program for the same system results in the following SNF: u = u 2, u 2 = a 22 u u 2 + a 3 u 3 + a 32u 2 u 2 + a 4 u 4 where + a 5 u 5 + b 6u 6 + b 7u 7 + b 8u 8 + b 9 u 9 + b u + b u + b 2u 2 + b 6 = a 6 7a 3 a 2 22 a 52+5a 4 a 42 a a 3a 32 a 42 a 22 (8) +63a 52 a a 3 a 4 a 42 5a 22 (9a 3 +a 2 22 ), (8) It is noted that the SNF (8) obtained by our Maple program is different from that given by Eq (78)

19 Computation of Simplest Normal Forms 325 Starting from the sixth-order term (the second line of the equation u 2 ), Algaba et al used the form u k u 2 while we use the form u k, but the numbers of the terms higher than the fifth-order are the same However, it is noted by comparing the first line of the equation u 2 that Eq (78) has an extra term b 52 This seems to show that our SNF (8) is simpler than the hypernormal form (78) 6 Conclusions A method is presented to compute the simplest normal forms for a double zero singularity Explicit, recursive formulae have been derived which can be directly implemented on a computer algebra system User-friendly symbolic computation programs written in Maple have been developed Examples are given to show the applicability of the methodology and the efficiency and convenience of the Maple programs Acknowledgment This work was supported by the Natural Sciences and Engineering Research Council of Canada References Algaba, A, Freire, E & Gamero, E [998] Hypernormal form for the Hopf-zero bifurcation, Int J Bifurcation and Chaos 8, Algaba, A, Freire, E & Gamero, E [2] Characterizing and computing normal forms using Lie transforms: A survey, Dynamics of Continous, Discrete and Impulsive Systems (DCDIS), to appear Baider, A & Sanders, J A [992] Further reduction of the Takens Bogdanov normal forms, J Diff Eqn 99, Bi, Q & Yu, P [998] Computation of normal forms of differential equations associated with non-semisimple zero eigenvalues, Int J Bifurcation and Chaos 8, Broer, H W, Chow, S-N, Kim, Y I & Vegter, G [995] The hamiltonian double-zero eigenvalues, Fields Inst Commun 4, 29 Chow, S-N, Li, C-Z & Wang, D [994] Normal Forms and Bifurcation of Planar Vector Fields (Cambridge University Press, Cambridge) Chua, L O & Kokubu, H [988a] Normal forms for nonlinear vector fields Part I: Theory and algorithm, IEEE Trans Circuits Syst 35, Chua, L O & Kokubu, H [988b] Normal forms for nonlinear vector fields Part II: Applications, IEEE Trans Circuits Syst 36, 5 7 Cushman, R & Sanders, J A [986] Nilpotent normal forms and representation theory of sl(2,r), Contemporary Mathematics 56, 3 5 Cushman, R & Sanders, J A [988] Normal form for (2;n-nilpotent vector field, using invariant theory, Physica D3, Elphick, C, Tirapegui, E, Brachet, M E, Coullet, P & Iooss, G [987] A simple global characterization for normal forms of singular vector fields, Physica A29, Guckenheimer, J & Holmes, P [983] Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields (Springer-Verlag, NY) Kokubu, H, Oka, H & Wang, D [996] Linear grading function and further reduction of normal forms, J Diff Eqn 32, Nayfeh, A H [993] Methods of Normal Forms (John Wiley, NY) Wang, D, Li, J, Huang, M & Jiang, Y [2] Unique normal forms of Bogdanov Takens singularity, J Diff Eqn 63, Yu, P & Huseyin, K [986] Static and dynamic bifurcations associated with a double zero eigenvalue, Dyn Stab Syst, Yu, P [999] Simplest normal forms of Hopf and generalized Hopf bifurcations, Int J Bifurcation and Chaos 9, Yu, P & Yuan, Y [999] Simplest normal form of a codimension-two system, Proc 7th CANCAM, Hamilton, Canada, May 3 June 3, 999, pp Appendix A Maple Source Codes A Maple code for generic case read input: solution := proc(n) local i: global c, cof: for i from n by - to do

20 326 Y Yuan & P Yu c[2,i,n-i] := solve(cof[2,i-,n-i+], c[2,i,n-i]): c[,i,n-i] := solve(cof[,i-,n-i+], c[,i,n-i]): end: b[2,] := a[2,]: b[2,2] := a[2,2]: dy := y2: dy2 := b[2,]*y^2 + b[2,2]*y*y2: du := u2: du2 := b[2,]*u^2 + b[2,2]*u*u2: H[,] := y: H[2,] := y2: H[,2] := H[,] + c[,,2]*y2^2 + c[2,,2]*y*y2: H[2,2] := H[2,] + c[2,,2]*y2^2: for m from 3 to Ord do H[,m] := H[,m-]: H[2,m] := H[2,m-]: for j from to m do H[,m] := H[,m] + c[,j,m-j]*y^j*y2^(m-j): H[2,m] := H[2,m] + c[2,j,m-j]*y^j*y2^(m-j): dy2 := dy2 + b[m,]*y^m + b[m,2]*y^(m-)*y2: for m from 3 to Ord do f[] := H[2,m]: f[2] := : for j from 2 to m do f[2] := f[2]+a[j,]*h[,m-j+]^j+a[j,2]*h[,m-j+]^(j-)*h[2,m-j+]: for i from to 2 do res[i] := diff(h[i,m],y)*dy + diff(h[i,m-],y2)*dy2 - f[i]: res[i] := subs(y=eps*y, y2=eps*y2, res[i]): res[i] := subs(eps=,diff(res[i],eps$m)/m!): for i from to 2 do for j from to m do cof[i,j,m-j] := diff(res[i],y$j,y2$(m-j))/j!/(m-j)!: cof[i,j,m-j] := simplify(subs(y=,y2=, cof[i,j,m-j])): if m=3 then b[m,2] := : c[2,,2] := solve(cof[2,m-,]+m*cof[,m,], c[2,,2]): b[m,] := solve(cof[2,m,], b[m,]): solution(m): else if modp(m,3)= then k := (m-)/3: b[m,2] := : c[,,2*k] := solve(cof[2,m-,]+m*cof[,m,], c[,,2*k]): b[m,] := solve(cof[2,m,], b[m,]): solution(m):

21 Computation of Simplest Normal Forms 327 elif modp(m,3)=2 then k := (m-2)/3: b[m,2] := : c[2,,2*k+]:= solve(cof[2,m-,]+m*cof[,m,], c[2,,2*k+]): b[m,] := solve(cof[2,m,], b[m,]): solution(m): elif modp(m,3)= then k := (m-3)/3: b[m,2] := : b[m,] := : c[,,2*k+]:= solve(cof[2,m,], c[,,2*k+]): c[2,,2*k+2]:= solve(cof[2,m-,]+m*cof[,m,], c[2,,2*k+2]): solution(m): fi: fi: du2 := du2 + b[m,]*u^m: save du, du2, Nform ; A2 Maple code for nongeneric case read input: nu := q-: for i from 2 to 2*nu do a[i,] := : b[i,] := : if nu > then for i from 2 to nu do a[i,2] := : b[i,2] := : fi: solution := proc(n) local i: global c, cof: c[2,n,] := solve(cof[,n,],c[2,n,]): c[,,n-] := solve(cof[,,n],c[,,n-]): for i from n- by - to do c[2,i,n-i] := solve(cof[2,i-,n-i+],c[2,i,n-i]): for i from n by - to 2 do c[,i,n-i] := solve(cof[,i-,n-i+],c[,i,n-i]): end: b[2*nu+,] := a[2*nu+,]: b[nu+,2] := a[nu+,2]: b[2*nu+,2] := a[2*nu+,2]: dy := y2: dy2 := b[2,]*y^2 + b[2,2]*y*y2: du := u2:

22 328 Y Yuan & P Yu du2 := b[2,]*u^2 + b[2,2]*u*u2: H[,] := y: H[2,] := y2: H[,2] := H[,] + c[,,2]*y2^2 + c[2,,2]*y*y2: H[2,2] := H[2,] + c[2,,2]*y2^2: for m from 3 to Ord do H[,m] := H[,m-]: H[2,m] := H[2,m-]: for j from to m do H[,m] := H[,m] + c[,j,m-j]*y^j*y2^(m-j): H[2,m] := H[2,m] + c[2,j,m-j]*y^j*y2^(m-j): dy2 := dy2 + b[m,]*y^m + b[m,2]*y^(m-)*y2: for m from 3 to Ord do f[] := H[2,m]: f[2] := : for j from 2 to m do f[2] := f[2]+a[j,]*h[,m-j+]^j+a[j,2]*h[,m-j+]^(j-)*h[2,m-j+]: for i from to 2 do if res[i] := diff(h[i,m],y)*dy + diff(h[i,m-],y2)*dy2 - f[i]: res[i] := subs(y=eps*y, y2=eps*y2, res[i]): res[i] := subs(eps=,diff(res[i],eps$m)/m!): cof[i,,] := diff(res[i], y): cof[i,,] := diff(res[i], y2): for j from 2 to m do for k from j by - to do cof[i,k,j-k] := diff(cof[i,k-,j-k],y): cof[i,,j] := diff(cof[i,,j-],y2): for j from to m do cof[i,j,m-j] := simplify(subs(y=,y2=,cof[i,j,m-j])/j!/(m-j)!): m<2*nu+2 then solution(m): for k from nu+2 to 2*nu do b[k,2] := a[k,2]: elif nu= then for i from 2*nu+2 to Ord do b[i,2] := : solution(m): if modp(m,2)= then j := m/2: c[2,,j] := solve(cof[2,m-,],c[2,,j]): else j := (m-)/2: c[,,j] := simplify(solve(cof[2,m-,],c[,,j])):

23 else fi: b[m,] := solve(cof[2,m,],b[m,]): Computation of Simplest Normal Forms 329 solution(m): if modp(m-,nu+)=nu- or modp(m-,nu+)=nu then b[m,2] := : k := trunc((m-)/(nu+)): if k= then c[2,,2] := solve(cof[2,2*nu+,],c[2,,2]): elif modp(m-,nu+)=nu- then c[,,k] := solve(cof[2,k*(nu+)+nu-,],c[,,k]): else c[2,,k+] := solve(cof[2,k*(nu+)+nu,],c[2,,k+]): fi: else b[m,2] := solve(cof[2,m-,],b[m,2]): fi: b[m,] := solve(cof[2,m,],b[m,]): fi: du2 := du2 + b[m,]*u^m + b[m,2]*u^(m-)*u2: save Ord, du, du2, Nform ; Appendix B Maple Input Files For convenience, the input data for the SNF of the two numerical examples are listed in this appendix A user may use the Maple source codes listed in Appendix A and the input data given below to execute the programs on one s PC to find the results presented in this paper B The input file for the generic case Ord := 2: a[2,] := : a[2,2] := 3: a[3,] := - 3: a[3,2] := : a[4,] := 7/6: a[4,2] := - 7/5: a[5,] := 7/3: a[5,2] := 937/3: a[6,] := - 427/9: a[6,2] := /375: a[7,] := /25: a[7,2] := /25: a[8,] := /35: a[8,2] := /65625: a[9,] := /325: a[9,2] := /575: a[,]:= /5625: a[,2]:= /4725:

24 33 Y Yuan & P Yu a[,]:= / : a[,2]:= /354375: a[2,]:= / : a[2,2]:= / : B2 The input file for the nongeneric case Ord := 2: q := 3: a[2,] := : a[2,2] := : a[3,] := : a[3,2] := - 6/25: a[4,] := : a[4,2] := - 69/295: a[5,] := - 2/25: a[5,2] := /45325: a[6,] := - 586/32375: a[6,2] := / : a[7,] := /3325: a[7,2] := / : a[8,] := / : a[8,2] := / : a[9,] := / : a[9,2] := / : a[,]:= / : a[,2]:= / : a[,]:= / : a[,2]:= / : a[2,]:= / : a[2,2]:= / :