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Exercise.nb                                                                                          1




Double Rayleigh-Duffing oscillator in nearly 1:2 resonance. MSM.

..      
q1   q1  2 q1  b0                            
                                   q2  q1 2  b1 q3  c q3  0
             1                                        1   1
..
q q    2 q  b                  q 2  b q3  c q3  0
                                   q             
  2           2      2     2   0    2    1       2   2      2


2  2 1




Time scales and definitions

  Utility (don't modify)


              OffGeneral::spell1



               Notation`


Time scales

         SymbolizeT0 ; SymbolizeT1 ; SymbolizeT2 ;

         timeScales  T0 , T1 , T2 ;


  Differentiation (don't modify)


              dt1expr_ : Sumi Dexpr, timeScalesi  1, i, 0, maxOrder;
              dt2expr_ : dt1dt1expr  Expand . i_;imaxOrder  0;



  Some rules (don' t modify)



              conjugateRule  A  A, A  A,   ,   , Complex0, n_  Complex0,  n;
Exercise.nb                                                                                                2




              displayRule  q_i_,j_ a__ __  RowTimes  MapIndexedD1
                                                                            211 &, a, qi,j ,

                    A_i_ a__ __  RowTimes  MapIndexedD1
                                                                21 &, a, Ai ,

                    q_i_,j_ __  qi,j , A_i_ __  Ai ;




Equations of motions and resonance conditions
Equations of motion

         EOM  1 2 Subscriptq, 1t  c Subscriptq, 1t3 
                 b1 Subscriptq, 1 't3  b0 Subscriptq, 2 't  Subscriptq, 1 't2 
                  Subscriptq, 1 't  Subscriptq, 1 ''t  0,
                2
              2 Subscriptq, 2t  c Subscriptq, 2t3  b2 Subscriptq, 2 't3 
                 b0 Subscriptq, 2 't  Subscriptq, 1 't2 
                  Subscriptq, 2 't  Subscriptq, 2 ''t  0; EOM  TableForm

         2 q1 t  c q1 t3   q1  t  b1 q1  t3  b0  q1  t  q2  t2  q1  t  0
         2 q2 t  c q2 t3   q2  t  b2 q2  t3  b0  q1  t  q2  t2  q2  t  0
          1
          2



Ordering of the dampings

         smorzrule     ,    ;

Definition of 2 (introduction of the detuning)

         ombrule  2  2   

         2     2 

Definition of the expansion of qi

         solRule  qi_  Sumj1 qi,j 1, 2, 3, j, 3 &;


  Some rules (don' t modify)


              multiScales  qi_ t  qi  timeScales,
                    Derivativen_q_t  dtnq  timeScales, t  T0 ;


Max order of the procedure

         maxOrder  2;

Scaling of the variables
Exercise.nb                                                                                                                         3



         scaling  Subscriptq, 1t   Subscriptq, 1t,
           Subscriptq, 2t   Subscriptq, 2t,
           Subscriptq, 1 't   Subscriptq, 1 't,
           Subscriptq, 2 't   Subscriptq, 2 't, Subscriptq, 1 ''t 
             Subscriptq, 1 ''t, Subscriptq, 2 ''t   Subscriptq, 2 ''t

         q1 t   q1 t, q2 t   q2 t, q1  t   q1  t,
          q2  t   q2  t, q1  t   q1  t, q2  t   q2  t

Modification of the equations of motion: substitution of the rules. Representation.

         EOMa  EOM . scaling . multiScales . smorzrule . ombrule . solRule  TrigToExp 
               ExpandAll . n_;n3  0; EOMa . displayRule

          2  D0 q1,1   3  D0 q1,2    D2 q1,1   2 D2 q1,2   3 D2 q1,3   3  D1 q1,1  
              2 2 D0 D1 q1,1   2 3 D0 D1 q1,2   3 D2 q1,1   2 3 D0 D2 q1,1   2 D0 q1,1 2 b0 
                                                    0               0               0



              2 3 D0 q1,1  D0 q1,2  b0  2 2 D0 q1,1  D0 q2,1  b0  2 3 D0 q1,2  D0 q2,1  b0 
                                                             1



              2 D0 q2,1 2 b0  2 3 D0 q1,1  D0 q2,2  b0  2 3 D0 q2,1  D0 q2,2  b0 
              2 3 D0 q1,1  D1 q1,1  b0  2 3 D0 q2,1  D1 q1,1  b0  2 3 D0 q1,1  D1 q2,1  b0 
              2 3 D0 q2,1  D1 q2,1  b0  3 D0 q1,1 3 b1   2 q1,1  c 3 q3  2 2 q1,2  3 2 q1,3  0,
            2  D0 q2,1   3  D0 q2,2    D2 q2,1   2 D2 q2,2   3 D2 q2,3   3  D1 q2,1  
                                                                     1              1,1     1            1



              2 2 D0 D1 q2,1   2 3 D0 D1 q2,2   3 D2 q2,1   2 3 D0 D2 q2,1   2 D0 q1,1 2 b0 
                                                     0               0               0



              2 3 D0 q1,1  D0 q1,2  b0  2 2 D0 q1,1  D0 q2,1  b0  2 3 D0 q1,2  D0 q2,1  b0  2 D0 q2,1 2 b0 
                                                             1



              2 3 D0 q1,1  D0 q2,2  b0  2 3 D0 q2,1  D0 q2,2  b0  2 3 D0 q1,1  D1 q1,1  b0 
              2 3 D0 q2,1  D1 q1,1  b0  2 3 D0 q1,1  D1 q2,1  b0  2 3 D0 q2,1  D1 q2,1  b0  3 D0 q2,1 3 b2 
              3 2 q2,1  2 2  2 q2,1   2 q2,1  c 3 q3  2 3  2 q2,2  2 2 q2,2  3 2 q2,3  0
                                               2              2,1                      2            2


Separation of the coefficients of the powers of 

         eqEps  RestThreadCoefficientListSubtract   ,   0 &  EOMa  Transpose;

Definition of the equations at orders of  and representation

         eqOrderi_ :  1 &  eqEps1 . q_k_,1  qk,i  

               1 &  eqEps1 . q_k_,1  qk,i    1 &  eqEpsi  Thread
Exercise.nb                                                                                                                          4



         eqOrder1 . displayRule
         eqOrder2 . displayRule
         eqOrder3 . displayRule

         D2 q1,1  2 q1,1  0, D2 q2,1  2 q2,1  0
           0         1            0         2


         D2 q1,2  2 q1,2   D0 q1,1   2 D0 D1 q1,1   D0 q1,1 2 b0  2 D0 q1,1  D0 q2,1  b0  D0 q2,1 2 b0 ,
           0         1

          D2 q2,2  2 q2,2 
               D0 q2,1   2 D0 D1 q2,1   D0 q1,1 2 b0  2 D0 q1,1  D0 q2,1  b0  D0 q2,1 2 b0  2  2 q2,1 
           0         2




         D2 q1,3  2 q1,3 
               D0 q1,2    D1 q1,1   2 D0 D1 q1,2   D2 q1,1  2 D0 D2 q1,1   2 D0 q1,1  D0 q1,2  b0 
           0         1


               2 D0 q1,2  D0 q2,1  b0  2 D0 q1,1  D0 q2,2  b0  2 D0 q2,1  D0 q2,2  b0  2 D0 q1,1  D1 q1,1  b0 
                                                               1


                2 D0 q2,1  D1 q1,1  b0  2 D0 q1,1  D1 q2,1  b0  2 D0 q2,1  D1 q2,1  b0  D0 q1,1 3 b1  c q3 ,
          D2 q2,3  2 q2,3   D0 q2,2    D1 q2,1   2 D0 D1 q2,2   D2 q2,1  2 D0 D2 q2,1  
                                                                                                                           1,1



             2 D0 q1,1  D0 q1,2  b0  2 D0 q1,2  D0 q2,1  b0  2 D0 q1,1  D0 q2,2  b0 
           0         2                                                          1


             2 D0 q2,1  D0 q2,2  b0  2 D0 q1,1  D1 q1,1  b0  2 D0 q2,1  D1 q1,1  b0 
                2 D0 q1,1  D1 q2,1  b0  2 D0 q2,1  D1 q2,1  b0  D0 q2,1 3 b2  2 q2,1  c q3  2  2 q2,2 
                                                                                                        2,1


Resonance condition


         ResonanceCond  1                  2 ;
                                           1
                                           2

Involved frequencies

         omgList  1 , 2 ;


  Utility (don't modify)


              omgRule  SolveResonanceCond,  ,    Flatten1 &  omgList  Reverse



              2  2 1 , 1          
                                      2
                                      2




First-Order Problem
Equations (=0)

         linearSys   1 &  eqOrder1;
         linearSys . displayRule  TableForm

         D2 q1,1  2 q1,1
          0         1

         D2 q2,1  2 q2,1
          0         2
Exercise.nb                                                                                                                        5



Formal solution of the First-Order Problem

         sol1  q1,1  FunctionT0 , T1 , T2 , A1 T1 , T2  ExpI 1 T0   A 1 T1 , T2  Exp I 1 T0 ,
              q2,1  FunctionT0 , T1 , T2 , A2 T1 , T2  ExpI 2 T0   A 2 T1 , T2  Exp I 2 T0 

         q1,1  FunctionT0 , T1 , T2 , A1 T1 , T2   1 T0  A1 T1 , T2   1 T0 ,
          q2,1  FunctionT0 , T1 , T2 , A2 T1 , T2   2 T0  A2 T1 , T2   2 T0 




Second-Order Problem
Substitution of the solution on the Second-Order Problem and representation

         order2Eq  eqOrder2 . sol1  ExpandAll;
         order2Eq . displayRule

         D2 q1,2  2 q1,2   2   T0 1 D1 A1  1  2   T0 1 D1 A1  1 
           0         1

                 T0 1  A1 1  2  T0 1 A2 b0 2  2  T0 1  T0 2 A1 A2 b0 1 2  2  T0 2 A2 b0 2 
                                                1     1                                                    2     2
                                                                                                               2
                 T0 1  1 A1  2 A1 b0 2 A1  2  T0 1  T0 2 A2 b0 1 2 A1  2  T0 1 b0 2 A1 
                                              1                                                            1
                                                                                                                             2
               2  T0 1  T0 2 A1 b0 1 2 A2  2 A2 b0 2 A2  2  T0 1  T0 2 b0 1 2 A1 A2  2  T0 2 b0 2 A2 ,
          D2 q2,2  2 q2,2  2  T0 1 A2 b0 2  2   T0 2 D1 A2  2  2   T0 2 D1 A2  2    T0 2  A2 2 
                                                             2                                                            2

           0         2                    1     1
               2  T0 2  A2 2  2  T0 1  T0 2 A1 A2 b0 1 2  2  T0 2 A2 b0 2  2 A1 b0 2 A1 
                                                                                     2     2            1
                                                                       2
               2  T0 1  T0 2 A2 b0 1 2 A1  2  T0 1 b0 2 A1    T0 2  2 A2  2  T0 2  2 A2 
               2  T0 1  T0 2 A1 b0 1 2 A2  2 A2 b0 2 A2  2  T0 1  T0 2 b0 1 2 A1 A2  2  T0 2 b0 2 A2 
                                                                     1
                                                                                                                             2
                                                             2                                                            2




  Utility (don't modify)


              expRule1i_ : Expa_ : ExpExpanda . omgRulei .  T0  T1 


Terms of type ei 1 T0 in the first equation of the Second-Order Problem and representation

         ST11  Coefficientorder2Eq , 2 . expRule11, ExpI 1 T0  &  1;
         ST11 . displayRule

          2  D1 A1  1    A1 1  2 A2 b0 1 2 A1 

Terms of type ei 2 T0 in the first equation of the Second-Order Problem and representation

         ST12  Coefficientorder2Eq , 2 . expRule12, ExpI 2 T0  &  1;
         ST12 . displayRule

          A2 b0 2 
             1     1


Terms of type ei 1 T0 in the second equation of the Second-Order Problem and representation
Exercise.nb                                                                                                                                                          6



         ST21  Coefficientorder2Eq , 2 . expRule11, ExpI 1 T0  &  2;
         ST21 . displayRule

         2 A2 b0 1 2 A1 

Terms of type ei 2 T0 in the second equation of the Second-Order Problem and representation

         ST22  Coefficientorder2Eq , 2 . expRule12, ExpI 2 T0  &  2;
         ST22 . displayRule

         A2 b0 2  2  D1 A2  2    A2 2  2  A2 2 
           1     1


Scalar product with the left eigenvectors: First-Order AME; representation

         SCond1  1, 0.ST11, ST21  0, 0, 1.ST12, ST22  0;
         SCond1 . displayRule

          2  D1 A1  1    A1 1  2 A2 b0 1 2 A1   0, A2 b0 2  2  D1 A2  2    A2 2  2  A2 2   0
                                                                     1     1


Algebraic manioulation to obtain D1 A1 and D1 A2

         SCond1Rule1  SolveSCond1, A1                                    T1 , T2 , A2            T1 , T2 1  ExpandAll;
                                                                    1,0                     1,0


         SCond1Rule1 . displayRule  TableForm
                     A1
         D1 A1                A2 b0 2 A1
                       2
                     A2                    A1 b0 2
                                              2
         D1 A2                 A2              1
                       2                       2 2


Substitution of the First - Order AME to the Second Order Equations

         order2Eqm 
          order2Eq  . SCond1Rule1 . SCond1Rule1 . conjugateRule . expRule1  & 
              1, 2  ExpandAll; order2Eqm . displayRule

         D2 q1,2  2 q1,2   2  T0 1 A2 b0 2  2 3  T0 1 A1 A2 b0 1 2  4  T0 1 A2 b0 2 
           0         1                      1     1                                             2     2
                                                                2                                                                                     2
               2 A1 b0 2 A1  2  T0 1 b0 2 A1  2 A2 b0 2 A2  2 3  T0 1 b0 1 2 A1 A2  4  T0 1 b0 2 A2 ,
                        1                      1               2                                                     2
                                               3                                                                                    1
                                                    T0 2                                                                               T0 2
          D2 q2,2  2 q2,2   2  2
           0         2                                       A1 A2 b0 1 2  2  T0 2 A2 b0 2  2 A1 b0 2 A1  2  2
                                                                                          2     2            1                                    A2 b0 1 2 A1 

                                                                                                      b0 1 2 A1 A2  2  T0 2 b0 2 A2 
                       1                                                                3
                           T0 2                                                          T0 2                                          2
               2      2             A1 b0 1 2 A2  2 A2 b0 2 A2  2 
                                                               2
                                                                                        2
                                                                                                                                       2


Assumpion on the coefficients

         $Assumptions  1 , 2 , b0 , c, , , b1 , b2   Reals

         1    2         b0    c            b1      b2   Reals

Construction of the particular solution of the first equation of the Second-Order Problem and representation
Exercise.nb                                                                                                                                                                        7



         solOrder2Eqm1 
          SimplifyTableq1,2 T0 , T1 , T2  . TrigToExpFlattenDSolveorder2Eqm1, 1 
                                      order2Eqm1, 2, i, q1,2 T0 , T1 , T2 , T0  . C1  0, C2  0,
                  i, 1, Lengthorder2Eqm1, 2; solOrder2Eqm1 . displayRule


         
             1                                       3  T0 1 A1 A2 b0 2                     4  T0 1 A2 b0 2
                                                                                                            2     2
                  2  T0 1 A2 b0 , 
                              1                                                         ,                                    , 2 A1 b0 A1 ,
             3                                                      4 1                                    15 2
                                                                                                                1



                                                                                                                                                                   
                                                                                                                                                               2
             1                           2          2 A2 b0 2 A2
                                                             2                     3  T0 1 b0 2 A1 A2                   4  T0 1 b0 2 A2
                                                                                                                                             2
                   2  T0 1
                                    b0 A1 ,                               ,                                            ,
             3                                                2
                                                               1
                                                                                                        4 1                                     15 2
                                                                                                                                                     1


         solq21  ExpandTotalsolOrder2Eqm1;
         solq21 . displayRule

         1                                      3  T0 1 A1 A2 b0 2                  4  T0 1 A2 b0 2
                                                                                                    2     2
              2  T0 1 A2 b0 
                          1                                                                                              2 A1 b0 A1 
         3                                                    4 1                                      15 2
                                                                                                            1
                                                                                                                                                              2
             1                              2        2 A2 b0 2 A2
                                                              2                 3  T0 1 b0 2 A1 A2                      4  T0 1 b0 2 A2
                                                                                                                                             2
                  2  T0 1 b0 A1                                                                                    
             3                                                2
                                                               1
                                                                                                    4 1                                     15 2
                                                                                                                                                 1


Construction of the particular solution of the second equation of the Second-Order Problem

         solOrder2Eqm2  Tableq2,2 T0 , T1 , T2  .
                  SimplifyTrigToExpFlattenDSolveorder2Eqm2, 1  order2Eqm2, 2, i,
                                q2,2 T0 , T1 , T2 , T0  . C1  0, C2  0,
              i, 1, Lengthorder2Eqm2, 2; solOrder2Eqm2 . displayRule


         
                   3                                                                                                                         1
                        T0 2                                                                                                                    T0 2
             8 2                 A1 A2 b0 1                   1                                        2 A1 b0 2 A1
                                                                                                                  1              8 2                      A2 b0 1 A1
                                                                       2  T0 2
                                                        ,                        A2
                                                                                    2   b0 ,                                ,                                                 ,
                               5 2                             3                                                   2
                                                                                                                     2
                                                                                                                                                       3 2


                                                                                                                                                     2  T0 2 b0 A2 
                       1                                                                            3
                           T0 2                                                                      T0 2
             8        2             A1 b0 1 A2                                        8          2             b0 1 A1 A2                    1                         2
                                                            ,  2 A2 b0 A2 ,                                                     ,
                               3 2                                                                         5 2                                 3

         solq22  ExpandTotalsolOrder2Eqm2;
         solq22 . displayRule
                                                        3                                                                            1
                                                              T0 2                                                                      T0 2
              1                                     8 2               A1 A2 b0 1                  2 A1 b0 2 A1
                                                                                                             1               8 2                      A2 b0 1 A1
                      2  T0 2
                                   A2
                                      2   b0                                                                                                                            
              3                                                     5 2                                       2
                                                                                                                2
                                                                                                                                                     3 2
                                               1                                                3
                                                   T0 2                                          T0 2
                                      8       2             A1 b0 1 A2            8          2              b0 1 A1 A2               1                             2
             2 A2 b0 A2                                                                                                                       2  T0 2 b0 A2
                                                       3 2                                                5 2                          3

Formal representation of the solution
Exercise.nb                                                                                                                                                              8



         sol2  q1,2  FunctionT0 , T1 , T2 , Evaluatesolq21 ,
               q2,2  FunctionT0 , T1 , T2 , Evaluatesolq22;
         sol2 . displayRule


         q1,2  FunctionT0 , T1 , T2 ,
                                                                      1                                 3  T0 1 b0 2 A1 A2            4  T0 1 b0 2 A2
                                                                                                                                                         2 2
                                                                          2  T0 1 b0 A2 
                                                                                         1                                                                         
                                                                      3                                                  4 1                   15 2
                                                                                                                                                    1



                                                                                                                                                                    ,
                                                                                                                                                                2
                                  1                               2       2 b0 2 A2 A2
                                                                                2                      3  T0 1 b0 2 A1 A2            4  T0 1 b0 2 A2
                                                                                                                                                          2
                   2 b0 A1 A1        2  T0 1 b0 A1                                                                             
                                  3                                                 2
                                                                                     1
                                                                                                                     4 1                       15 2
                                                                                                                                                    1



             q2,2  FunctionT0 , T1 , T2 ,
                                                                           3
                                                                                T0 2
                                                                      8 2               b0 1 A1 A2                 1
                                                                                                                        2  T0 2 b0 A2 
                                                                                                                                        2
                                                                                     5 2                            3
                                                   1                                              1
                                                        T0 2                                        T0 2
                   2 b0 2 A1 A1
                         1                8 2                   b0 1 A2 A1             8       2             b0 1 A1 A2
                                                                                                                               
                         2
                          2
                                                            3 2                                          3 2


                                                                                         2  T0 2 b0 A2 
                                              3
                                                  T0 2
                                  8          2             b0 1 A1 A2             1                            2
                   2 b0 A2 A2                                                  
                                                        5 2                        3




Third-Order Problem
Substitution in the Third-Order Equations

         order3Eq  eqOrder3 . sol1 . sol2  ExpandAll;

Simplifications

         order3Eqpr1, 2  Simplifyorder3Eq1, 2;

         order3Eqpr2, 2  Simplifyorder3Eq2, 2;

Terms of type ei 1 T0 in the first equation of the Third-Order Problem and representation

         ST311  Coefficientorder3Eqpr , 2 . expRule1 , ExpI 1 T0  &  1;
         ST311 . displayRule

         
               1
             60 1
              60  D1 A1  1  60 D2 A1  1  120  D2 A1  2  120  D1 A1  A2 b0 1 2  180 c A2 1 A1  120 
                     D1 A2  b0 2 A1  80 A2 b2 3 A1  180  A2 b1 4 A1  448 A1 A2 b2 2 2 A2  90 A1 A2 b2 1 2 A2 
                                       1                           1                                       1

                                  1          1 0 1               1     1                 0 1                    0     2


Terms of type ei 2 T0 in the first equation of the Third-Order Problem and representation

         ST312  Coefficientorder3Eqpr , 2 . expRule1 , ExpI 2 T0  &  1;
         ST312 . displayRule

         0
Exercise.nb                                                                                                                        9



Terms of type ei 1 T0 in the second equation of the Third-Order Problem and representation

         ST321  Coefficientorder3Eqpr , 2 . expRule11, ExpI 1 T0  &  2;
         ST321 . displayRule


         
                   1
             30 1 2
                  80  D1 A1  A2 b0 2 2  60  D1 A1  A2 b0 1 2  140  D1 A2  b0 2 2 A1  40   A2 b0 2 2 A1 
                   160  A2 b0 2 2 A1  40 A2 b2 3 2 A1  224 A1 A2 b2 2 2 A2  45 A1 A2 b2 1 3 A2 
                                         1                              2                      1                       1

                                1             1 0 1                      0 1 2                  0     2


Terms of type ei 2 T0 in the second equation of the Third-Order Problem and representation

         ST322  Coefficientorder3Eqpr , 2 . expRule12, ExpI 2 T0  &  2;
         ST322 . displayRule

                            30  D1 A2  1 2  30 D2 A2  1 2  30 2 A2 1 2 
                   1
                                                         1
             30 1 2
                   60  D1 A1  A1 b0 2 2  60  D2 A2  1 2  64 A1 A2 b2 3 2 A1  45 A1 A2 b2 2 2 A1 
                   90 c A2 1 2 A2  40 A2 b2 1 3 A2  16 A2 b2 4 A2  90  A2 b2 1 4 A2 
                                        1                        2             0 1                    0 1 2

                         2                2 0      2          2 0 2              2        2


Scalar product with the left eigenvectors: Second-Order AME; representation

         SCond2  1, 0.ST311, ST321  0, 0, 1.ST312, ST322  0;
         SCond2 . displayRule

                          60  D1 A1  1  60 D2 A1  1  120  D2 A1  2 
                   1
                                                     1                           1
                 60 1
                           120  D1 A1  A2 b0 1 2  180 c A2 1 A1  120  D1 A2  b0 2 A1  80 A2 b2 3 A1 
                           180  A2 b1 4 A1  448 A1 A2 b2 2 2 A2  90 A1 A2 b2 1 2 A2   0,
                                                               1                            1          1 0 1

                                  1     1                 0 1                    0     2


                              30  D1 A2  1 2  30 D2 A2  1 2  30 2 A2 1 2  60  D1 A1  A1 b0 2 2 
                       1
                                                           1                                                    1
                 30 1 2
                           60  D2 A2  1 2  64 A1 A2 b2 3 2 A1  45 A1 A2 b2 2 2 A1 
                           90 c A2 1 2 A2  40 A2 b2 1 3 A2  16 A2 b2 4 A2  90  A2 b2 1 4 A2   0
                                             2             0 1                    0 1 2

                                 2                2 0      2          2 0 2              2        2


Algebraic manipulation to obtain D2 A1 and D2 A2
Exercise.nb                                                                                                                                                               10



          SCond2Rule1 
              SolveSCond2, A1                          T1 , T2 , A2             T1 , T2 1 . A1 2,0 T1 , T2   T1 SCond1Rule1
                                                 0,1                     0,1


                                        1, 2 . A2 2,0 T1 , T2   T1 SCond1Rule12, 2 . SCond1Rule1 .
                 SCond1Rule1 . conjugateRule  ExpandAll  Simplify  Expand;
          SCond2Rule1 . displayRule

          D2 A1 

                   2 A1       1                                                3  c A2 A1
                                                                                        1              5                          3                    A2 b2 2 A1
                                                                                                                                                         1 0 1
                                   A2 b0 A1    A2 b0 A1                                              A2 b2 1 A1 
                                                                                                               1 0                    A2 b1 2 A1 
                                                                                                                                       1     1                        
                   8 1         2                                                      2 1            12                         2                        2 2

                   A2 b0 2 A1              A2 b0 2 A1               A2 b0 2 A1             56                             3  A1 A2 b2 2 A2
                                                                                                                                            0 2
                                                                                                     A1 A2 b2 2 A2 
                                                                                                                0                                      ,
                         2 1                      4 1                        2 1               15                                     4 1
                          A2 b0 2
                            1     1              A2 b0 2
                                                   1     1           A2 b0 2
                                                                        1     1               2 A2         A2 b0 1
                                                                                                               1             7
           D2 A2                                                                                                            A1 A2 b2 1 A1 
                                                                                                                                          0
                            4 2
                               2                   8 2
                                                      2                 4 2
                                                                           2
                                                                                              8 2             2 2          4


                                                                                                                                                
                  17  A1 A2 b2 2 A1
                              0 1                     3  c A2 A2
                                                             2             2                           3                     4  A2 b2 2 A2
                                                                                                                                  2 0 2
                                                                              A2 b2 2 A2 
                                                                                  2 0                       A2 b2 2 A2 
                                                                                                             2     2
                           30 2                           2 2            3                           2                              15 1




Reconstitution of the AME and of the solution
dA1
 dt
       ame1

          ame1  A1 1,0 T1 , T2   A1 0,1 T1 , T2  .
              JoinSCond1Rule1, SCond2Rule1 . omgRule1; ame1 . displayRule

           A1         2 A1                              3  c A2 A1
                                                                  1
                                 A2 b0 A1                               
            2            8 1                                  2 1
            2                            3                                                  67
                  A2 b2 1 A1 
                    1 0                         A2 b1 2 A1   A2 b0 2 A1 
                                                 1     1                                           A1 A2 b2 1 A2
                                                                                                           0
            3                            2                                                  15

dA1
 dt
       ame2

          ame2  A2 1,0 T1 , T2   A2 0,1 T1 , T2  .
              JoinSCond1Rule1, SCond2Rule1 . omgRule1; ame2 . displayRule

           A2                      3                      1                    1                        2 A2
                     A2               A2 b0 
                                            1                   A2 b0 
                                                                  1                     A2 b0 
                                                                                           1                      
            2                   16                        32                   16                      16 1
             A2 b0 2
               1     1          61                                    3  c A2 A2
                                                                             2              12
                                        A1 A2 b2 1 A1 
                                                 0                                                A2 b2 1 A2  6 A2 b2 2 A2
                                                                                                     2 0             2     1
                  2 2          30                                         4 1             5

Better representation
Exercise.nb                                                                                                                                                    11



            Collectame1, A1 T1 , T2 , A1 T1 , T2 2 A 1 T1 , T2 ,
                     A1 T1 , T2  A2 T1 , T2  A 2 T1 , T2 , A2 T1 , T2  A 1 T1 , T2  . displayRule


                                                                                      b1 2 A1  A2   b0   b0 2  A1 
                             2                  3c           2               3                                                     67
            A1                          A2
                                           1                         b2 1 
                                                                        0                 1                                                  A1 A2 b2 1 A2
                                                                                                                                                     0
                     2       8 1                  2 1          3               2                                                     15

            Collectame2, A2 T1 , T2 , A1 T1 , T2 2 , A1 T1 , T2 2 A 1 T1 , T2 ,
                     A1 T1 , T2  A2 T1 , T2  A 2 T1 , T2 , A2 T1 , T2 2 A 2 T1 , T2  . displayRule

                         3  b0          b0        1                   b0 2
                                                                             1
            A2 
             1                                            b0                     
                             16         32          16                   2 2

                                         2           61                                 3c            12
                A2                                       A1 b2 1 A1  A2
                                                                   0          2                                b2 1  6 b2 2 A2
                                                                                                                  0            1
                         2              16 1           30                                     4 1       5




Polar form of the AME

  Utility


                traspRule 
                    A1 T1 , T2   A1 t, A 1 T1 , T2   A 1 t, A2 T1 , T2   A2 t, A 2 T1 , T2   A 2 t



                A1 T1 , T2   A1 t, A1 T1 , T2   A1 t, A2 T1 , T2   A2 t, A2 T1 , T2   A2 t


Rewriting of the AME

            amem1  ame1 . traspRule  A1 't

                                     2 A1 t              3  c A1 t2 A1 t
                  A1 t                                                                          b2 1 A1 t2 A1 t           b1 2 A1 t2 A1 t 
            1                                                                                  2                                 3
                                                                                                    0                                  1
            2                            8 1                           2 1                   3                                 2

                 b0 A2 t A1 t   b0 2 A2 t A1 t                                 b2 1 A1 t A2 t A2 t  A1  t
                                                                                      67
                                                                                              0
                                                                                      15

            amem2  ame2 . traspRule  A2 't

                                                                                                            b0 2 A1 t2
                       b0 A1 t2                      b0 A1 t2              b0 A1 t2 
                 3                              1                           1                                    1
                                                                                                                            
                16                             32                          16                                     2 2
                                                              2 A2 t
                      A2 t    A2 t                                            b2 1 A1 t A2 t A1 t 
                1                                                                61
                                                                                        0
                2                                                16 1           30
                3  c A2 t2 A2 t
                                                               b2 1 A2 t2 A2 t  6 b2 2 A2 t2 A2 t  A2  t
                                                        12
                                                                0                           1
                              4 1                       5

Polar form of the complex amplitudes
Exercise.nb                                                                                                                                               12




         polRule  A1 t 
                                              1
                                                  a1t  1t ,
                                              2

              A2 t             a2t  2t , A 1 t               a1t  1t , A 2 t             a2t  2t 
                              1                                       1                                      1
                              2                                       2                                      2

         A1 t               1t a1t, A2 t 
                          1                                       1
                                                                 2t a2t,
                          2                                   2

          A1 t                             a1t, A2 t 
                          1                                    1
                                1t                          2t a2t
                          2                                    2

         polRuleD  JoinpolRule, DpolRule, t


         A1 t               1t a1t, A2 t                 2t a2t, A1 t 
                          1                                       1                                      1
                                                                                                              1t a1t,
                          2                                       2                                      2

          A2 t               2t a2t, A1  t                  1t a1 t           1t a1t 1 t,
                          1                                           1                         1
                          2                                           2                         2

          A2  t                2t a2 t         2t a2t 2 t,
                           1                             1
                           2                           2

          A1 t                             a1 t    1t a1t 1 t,
                          1                            1
                                   1t
                           2                            2

          A2 t                             a2 t    2t a2t 2 t
                          1                            1
                                   2t
                           2                            2

Substitution of the complex amplitudes with the polar form

         eq1  amem1 . polRuleD

         1                                 1                                                     1t 2 a1t            3  c  1t a1t3
               1t  a1t                 1t 2t  a1t a2t b0                                                                   
         4                                 4                                                             16 1                       16 1
              1                                           67                                                 3
                       1t a1t3 b2 1 
                                        0                        1t a1t a2t2 b2 1 
                                                                                  1t a1t3 b1 2 
                                                                                        0            1
             12                            120                               16

                 1t 2t a1t a2t b0 2   1t a1 t    1t a1t 1 t
             1                                        1                  1
             4                                        2                  2

         eq2  amem2 . polRuleD

         1                                 1                                3
               2t  a2t                  2t  a2t              2  1t  a1t2 b0 
         4                                 2                               64
              1                                          1                                       2t 2 a2t
                      2  1t  a1t2 b0                  2  1t  a1t2 b0                                   
             128                                         64                                              32 1
             3  c  2t a2t3                 61                                                3
                                                          2t a1t2 a2t b2 1 
                                                                                  0                        2t a2t3 b2 1 
                                                                                                                            0
                         32 1                     240                                              10

                                                                                              2t a2 t            2t a2t 2 t
             3                                       2  1t a1t2 b0 2
                                                                            1            1                          1
                   2t a2t3 b2 2 
                                      1                                              
             4                                                   8 2                    2                          2

Autonomous equations and separations of the real and imaginary parts
Exercise.nb                                                                                                                                                 13



         eqa1  ComplexExpandExpandeq1  1t 

         1                         1
               a1t                  a1t a2t Cos2 1t  2t b0 

                                                                                                                 a1 t
         4                         4
              3                               1
                     a1t3 b1 2 
                                1                 a1t a2t Sin2 1t  2t b0 2                                   
             16                               4                                                                     2
                     1                                                              2 a1t            3 c a1t3              1
                         a1t a2t Sin2 1t  2t b0                                                                    a1t3 b2 1 
                                                                                                                                             0
                     4                                                                16 1                     16 1           12

                                                                                                                                         a1t 1 t
                         67                                 1                                                                        1
                               a1t a2t2 b2 1 
                                             0                  a1t a2t Cos2 1t  2t b0 2 
                     120                                    4                                                                        2

         eqa2  ComplexExpandExpandeq2  2t 

         1                         3
               a2t                   a1t2 Cos2 1t  2t b0 
         4                         64
                 1                                                            1
                          a1t2 Cos2 1t  2t b0                         a1t2 Sin2 1t  2t b0 

                                                                                                a2 t
             128                                                          64
             3                            a1t2 Sin2 1t  2t b0 2
                                                                          1
                  a2t3 b2 2 
                             1                                                                                 
             4                                                  8 2                                    2
                     1                   1                                                          3
                         a2t              a1t2 Cos2 1t  2t b0                          a1t2 Sin2 1t  2t b0 
                     2                   64                                                     64

                         1                                                        2 a2t          3 c a2t3              61
                                a1t2 Sin2 1t  2t b0                                                                  a1t2 a2t b2 1 
                                                                                                                                                  0
                     128                                                            32 1                   32 1           240

                                                                                                                a2t 2 t
                         3                         a1t2 Cos2 1t  2t b0 2
                                                                                   1                        1
                              a2t3 b2 1 
                                      0                                                                 
                     10                                                8 2                                 2

Definition of 

         phiRule  2 1t  2t  t

         2 1t  2t  t

         phiRuleD  2 't  2 1 't   't

         2 t  2 1 t   t

Polar Amplitude Modulation Equations

         amep1  CollectComplexExpand2 eqa1 .   0, a1t, a2t

              3
                 a1t3 b1 2 
                             1
              8

                                                                                                                                                 a1 t
                                                 1                                         1
             a1t                 a2t             Cos2 1t  2t b0                Sin2 1t  2t b0 2
                              2                   2                                         2
Exercise.nb                                                                                                                                       14



         amep2  CollectComplexExpand2 eqa2 .   0, a1t, a2t

         1                      3
               a2t                a2t3 b2 2 
                                                 1
         2                      2
                              3                                              1
             a1t2                   Cos2 1t  2t b0                   Cos2 1t  2t b0 
                              32                                            64

                                                                                                            a2 t
                      1                                           Sin2 1t  2t b0 2
                                                                                           1
                            Sin2 1t  2t b0 
                   32                                                               4 2

         amep3  CollectComplexExpand  2 eqa1 .   0, a1t, a2t

                           3c          1                        2         67
         a1t3                           b2 1  a1t 
                                            0                                   a2t2 b2 1 
                                                                                         0
                           8 1        6                        8 1       60

                                                                                 Cos2 1t  2t b0 2  1 t
                                  1                                          1
                   a2t               Sin2 1t  2t b0 
                                  2                                          2

         amep4  CollectComplexExpand  2 eqa2 .   0, a1t, a2t

                            3c            3
         a2t3                              b2 1  a1t2
                                               0
                           16 1          5
                  1                                              3                                           1
                           Cos2 1t  2t b0                    Sin2 1t  2t b0                  Sin2 1t  2t b0 
                  32                                            32                                          64

                                                                                                                   2 t
                      61                          Cos2 1t  2t b0 2
                                                                           1                               2
                            a2t b2 1 
                                   0                                                        a2t  
                   120                                          4 2                                      16 1

Reduced Amplitude Modulation Equations

         rame1  Collectamep1 . phiRule, a1t, a2t, t


                                                                                                                               a1 t
              3                                                       1                          1
                 a1t3 b1 2  a1t
                             1                           a2t            Cost b0            Sint b0 2
              8                                     2                  2                          2

         rame2  Collectamep2 . phiRule, a1t, a2t, t

         1                      3
               a2t                a2t3 b2 2 
                                                 1
         2                      2

                                                                                                                                           a2 t
                              3                             1                               1                        Sint b0 2
                                                                                                                                   1
             a1t2                   Cost b0            Cost b0                  Sint b0 
                              32                           64                              32                                 4 2
Exercise.nb                                                                                                                                                  15



         rame3  Expand2 a2t amep3  a1t amep4 . phiRule . phiRuleD

                                         1                                                3
           a1t a2t                      a1t3 Cost b0                            a1t3 Sint b0 
                                         32                                               32
              1                                                                                           2 a1t a2t                2 a1t a2t
                   a1t3 Sint b0   a1t a2t2 Sint b0                                                                                  
           64                                                                                                       4 1                    16 1
           3 c a1t3 a2t                   3 c a1t a2t3               7                                     49
                                                                                a1t3 a2t b2 1 
                                                                                                0                       a1t a2t3 b2 1 
                                                                                                                                      0
                       4 1                           16 1                  40                                    30

                                                   a1t a2t2 Cost b0 2  a1t a2t  t
           a1t3 Cost b0 2
                                1

                             4 2

         fixRule  a1t  a1, a2t  a2, t  

         a1t  a1, a2t  a2, t  

Equations to find Fixed Points

         fix1  CollectExpandSimplifyrame1 . a1 't  0, a2 't  0,  't  0,
                  a1t, a2t, a1t a2t, a1t3 , a2t3 , a1t2 a2t, a1t a2t2  . fixRule

         a1           3                               1                              1
                          a13 b1 2  a1 a2 
                                   1                        Cos b0                   Sin b0 2
              2        8                               2                              2

         fix2 
          CollectExpandSimplifyrame2 . a1 't  0, a2 't  0,  't  0, a1t, a2t,
                   a1t3 , a2t3 , a1t2 a2t, a1t a2t2 , b0 , Cost, Sint . fixRule

         a2           3                                   3                                                     2
                                                                                                                     1
                          a23 b2 2  a12 b0
                                   1                                       Cos  Sin                  
              2        2                                   32       64                                    32       4 2

         fix3  CollectExpandSimplifyrame3 . a1 't  0, a2 't  0,  't  0,
                  a1t, a2t, a1t3 , a2t3 , a1t2 a2t, a1t a2t2 ,
                   a1t3 a2t, a1t a2t3 , b0 , Cost, Sint . fixRule

                              2             2                      3c               49                           3c        7
         a1 a2                                 a1 a23                                  b2 1  a13 a2
                                                                                               0                                 b2 1 
                                                                                                                                   0
                              4 1        16 1                     16 1             30                       4 1          40

                                                                                                   a1 a22 b0  Sin  Cos 2 
                            3                                                          2
                                                                                           1
          a13 b0                             Sin  Cos                    
                            32       64                                      32       4 2

Equilibrium paths in the case a1  0

         ep1  SimplifySolvefix2 . a1  0  0, a2, 1  0


         a2  0, a2                                   , a2                                
                                                                                          

                                              3   b2 1                           3        b2 1

Reconstitution of the solution
Exercise.nb                                                                                                                       16



         qr1 t_  ComplexExpand
              A1 T1 , T2   1 t  A 1 T1 , T2   1 t  solq21 . traspRule . polRule . T0  t

                                          1                 1
         a1t Cost 1  1t             a1t2 b0        a1t2 Cos2 t 1  2 1t b0 
                                          2                 6
          a1t a2t Cos3 t 1  1t  2t b0 2                  a2t2 b0 2
                                                                                    2       a2t2 Cos4 t 1  2 2t b0 2
                                                                                                                             2
                                                                                       
                                   8 1                                      2 2
                                                                                1                             30 2
                                                                                                                  1


         qr2 t_  ComplexExpand
              A2 T1 , T2   2 t  A 2 T1 , T2   2 t  solq22 . traspRule . polRule . T0  t

                                          1                 1                                              a1t2 b0 2
                                                                                                                      1
         a2t Cost 2  2t             a2t2 b0        a2t2 Cos2 t 2  2 2t b0                          
                                          2                 6                                                  2 2
                                                                                                                  2

                                  t 2                                                          3 t 2
          4 a1t a2t Cos              1t  2t b0 1          4 a1t a2t Cos               1t  2t b0 1
                                   2                                                              2
                                                                     
                                   3 2                                                           5 2




Numerical integrations
Numerical values


         1  1, 2  2, b0 
                                  1
                                      , b1  1, b2  1, c  1,   0.05,   0.05,   0, 2  2 1  
                                  2

         1, 2,
                   1
                       , 1, 1, 1, 0.05, 0.05, 0, 2
                   2

Time of integration

         ti  500;

Numerical Intergations of the RAME

         solrame1  NDSolverame1  0, rame2  0, rame3  0,
            a10  0.1, a20  0.1, 0  0.1, a1t, a2t, t, t, 0, ti

         a1t  InterpolatingFunction0., 500., t,
           a2t  InterpolatingFunction0., 500., t,
           t  InterpolatingFunction0., 500., t
Exercise.nb                                                                                                                        17



         GraphicsArrayPlota1t . solrame1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "a1 t",
           Plota2t . solrame1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "a2 t",
           Plott . solrame1, t, 0, ti, PlotStyle  Thick, Frame  True,
            AxesOrigin  0, 0, FrameLabel  "t", "t"

                                                   0.4
               0.5                                                                           2.0
                                                   0.2
                                                                                             1.5
      a1 t




                                                                                      t
                                               a2 t
               0.0                                 0.0
                                                  0.2                                       1.0
              0.5                                0.4                                       0.5
                                                  0.6                                       0.0
                     0   100 200 300 400 500             0   100 200 300 400 500                   0   100 200 300 400 500
                                t                                   t                                         t


Numerical Intergations of the AME

         solramep1  NDSolveamep1  0, amep2  0, amep3  0, amep4  0, a10  0.1,
            a20  0.1, 10  0.1, 20  0.1, a1t, a2t, 1t, 2t, t, 0, ti

         a1t  InterpolatingFunction0., 500., t,
           a2t  InterpolatingFunction0., 500., t,
           1t  InterpolatingFunction0., 500., t,
           2t  InterpolatingFunction0., 500., t

         GraphicsArrayPlota1t . solramep1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "a1 t",
           Plota2t . solramep1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "a2 t",
           Plot1t . solramep1, t, 0, ti, PlotStyle  Thick,
            Frame  True, AxesOrigin  0, 0, FrameLabel  "t", "1 t",
           Plot2t . solramep1, t, 0, ti, PlotStyle  Thick, Frame  True,
            AxesOrigin  0, 0, FrameLabel  "t", "2 t"

                                          0.4                           4                                  6
               0.5                        0.2                                                              5
      a1 t




                                                                  1 t




                                                                                                       2 t




                                                                        3
                                     a2 t




                                          0.0                                                              4
               0.0                                                      2                                  3
                                         0.2                                                              2
              0.5                       0.4                           1                                  1
                                         0.6                           0                                  0
                     0 100200300400500          0 100 300 500
                                                     200 400                0 100200300400500                  0 100200300400500
                            t                          t                           t                                   t


Graphics of the reconstituted solution
Exercise.nb                                                                                                                18



         GraphicsArrayPlotqr1 t . solramep1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "q1 t",
           Plotqr2 t . solramep1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "q2 t"

                                                                   0.4

              0.5
                                                                   0.2
      q1 t




                                                                   0.0




                                                                q2 t
              0.0
                                                                  0.2

              0.5                                                0.4

                                                                  0.6
                     0   100     200       300     400   500             0       100    200       300       400    500
                                       t                                                      t


Numerical Intergations of the original equations

         solorig1  NDSolveJoinEOM, q1 0  0.1, q2 0  0.1, q1 '0  0.1, q2 '0  0.1,
           q1 t, q2 t, t, 0, ti, MaxSteps  1 000 000

         q1 t  InterpolatingFunction0., 500., t,
           q2 t  InterpolatingFunction0., 500., t

         GraphicsArrayPlotq1 t . solorig1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "q1 t",
           Plotq2 t . solorig1, t, 0, ti, PlotStyle  Thick,
            PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "q2 t"

                                                                    0.4

              0.5
                                                                    0.2

                                                                    0.0
      q1 t




                                                                q2 t




              0.0
                                                                   0.2

              0.5                                                 0.4

                                                                   0.6
                     0   100     200       300     400    500                0    100    200          300    400     500
                                       t                                                          t

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Exercise: Multiple Bifurcations of Sample Dynamical Systems

  • 1. Exercise.nb 1 Double Rayleigh-Duffing oscillator in nearly 1:2 resonance. MSM. ..  q1   q1  2 q1  b0    q2  q1 2  b1 q3  c q3  0 1 1 1 .. q q   2 q  b   q 2  b q3  c q3  0 q   2 2 2 2 0 2 1 2 2 2 2  2 1 Time scales and definitions Utility (don't modify) OffGeneral::spell1  Notation` Time scales SymbolizeT0 ; SymbolizeT1 ; SymbolizeT2 ; timeScales  T0 , T1 , T2 ; Differentiation (don't modify) dt1expr_ : Sumi Dexpr, timeScalesi  1, i, 0, maxOrder; dt2expr_ : dt1dt1expr  Expand . i_;imaxOrder  0; Some rules (don' t modify) conjugateRule  A  A, A  A,   ,   , Complex0, n_  Complex0,  n;
  • 2. Exercise.nb 2 displayRule  q_i_,j_ a__ __  RowTimes  MapIndexedD1 211 &, a, qi,j , A_i_ a__ __  RowTimes  MapIndexedD1 21 &, a, Ai , q_i_,j_ __  qi,j , A_i_ __  Ai ; Equations of motions and resonance conditions Equations of motion EOM  1 2 Subscriptq, 1t  c Subscriptq, 1t3  b1 Subscriptq, 1 't3  b0 Subscriptq, 2 't  Subscriptq, 1 't2   Subscriptq, 1 't  Subscriptq, 1 ''t  0, 2 2 Subscriptq, 2t  c Subscriptq, 2t3  b2 Subscriptq, 2 't3  b0 Subscriptq, 2 't  Subscriptq, 1 't2   Subscriptq, 2 't  Subscriptq, 2 ''t  0; EOM  TableForm 2 q1 t  c q1 t3   q1  t  b1 q1  t3  b0  q1  t  q2  t2  q1  t  0 2 q2 t  c q2 t3   q2  t  b2 q2  t3  b0  q1  t  q2  t2  q2  t  0 1 2 Ordering of the dampings smorzrule     ,    ; Definition of 2 (introduction of the detuning) ombrule  2  2    2     2  Definition of the expansion of qi solRule  qi_  Sumj1 qi,j 1, 2, 3, j, 3 &; Some rules (don' t modify) multiScales  qi_ t  qi  timeScales, Derivativen_q_t  dtnq  timeScales, t  T0 ; Max order of the procedure maxOrder  2; Scaling of the variables
  • 3. Exercise.nb 3 scaling  Subscriptq, 1t   Subscriptq, 1t, Subscriptq, 2t   Subscriptq, 2t, Subscriptq, 1 't   Subscriptq, 1 't, Subscriptq, 2 't   Subscriptq, 2 't, Subscriptq, 1 ''t   Subscriptq, 1 ''t, Subscriptq, 2 ''t   Subscriptq, 2 ''t q1 t   q1 t, q2 t   q2 t, q1  t   q1  t, q2  t   q2  t, q1  t   q1  t, q2  t   q2  t Modification of the equations of motion: substitution of the rules. Representation. EOMa  EOM . scaling . multiScales . smorzrule . ombrule . solRule  TrigToExp  ExpandAll . n_;n3  0; EOMa . displayRule  2  D0 q1,1   3  D0 q1,2    D2 q1,1   2 D2 q1,2   3 D2 q1,3   3  D1 q1,1   2 2 D0 D1 q1,1   2 3 D0 D1 q1,2   3 D2 q1,1   2 3 D0 D2 q1,1   2 D0 q1,1 2 b0  0 0 0 2 3 D0 q1,1  D0 q1,2  b0  2 2 D0 q1,1  D0 q2,1  b0  2 3 D0 q1,2  D0 q2,1  b0  1 2 D0 q2,1 2 b0  2 3 D0 q1,1  D0 q2,2  b0  2 3 D0 q2,1  D0 q2,2  b0  2 3 D0 q1,1  D1 q1,1  b0  2 3 D0 q2,1  D1 q1,1  b0  2 3 D0 q1,1  D1 q2,1  b0  2 3 D0 q2,1  D1 q2,1  b0  3 D0 q1,1 3 b1   2 q1,1  c 3 q3  2 2 q1,2  3 2 q1,3  0,  2  D0 q2,1   3  D0 q2,2    D2 q2,1   2 D2 q2,2   3 D2 q2,3   3  D1 q2,1   1 1,1 1 1 2 2 D0 D1 q2,1   2 3 D0 D1 q2,2   3 D2 q2,1   2 3 D0 D2 q2,1   2 D0 q1,1 2 b0  0 0 0 2 3 D0 q1,1  D0 q1,2  b0  2 2 D0 q1,1  D0 q2,1  b0  2 3 D0 q1,2  D0 q2,1  b0  2 D0 q2,1 2 b0  1 2 3 D0 q1,1  D0 q2,2  b0  2 3 D0 q2,1  D0 q2,2  b0  2 3 D0 q1,1  D1 q1,1  b0  2 3 D0 q2,1  D1 q1,1  b0  2 3 D0 q1,1  D1 q2,1  b0  2 3 D0 q2,1  D1 q2,1  b0  3 D0 q2,1 3 b2  3 2 q2,1  2 2  2 q2,1   2 q2,1  c 3 q3  2 3  2 q2,2  2 2 q2,2  3 2 q2,3  0 2 2,1 2 2 Separation of the coefficients of the powers of  eqEps  RestThreadCoefficientListSubtract   ,   0 &  EOMa  Transpose; Definition of the equations at orders of  and representation eqOrderi_ :  1 &  eqEps1 . q_k_,1  qk,i    1 &  eqEps1 . q_k_,1  qk,i    1 &  eqEpsi  Thread
  • 4. Exercise.nb 4 eqOrder1 . displayRule eqOrder2 . displayRule eqOrder3 . displayRule D2 q1,1  2 q1,1  0, D2 q2,1  2 q2,1  0 0 1 0 2 D2 q1,2  2 q1,2   D0 q1,1   2 D0 D1 q1,1   D0 q1,1 2 b0  2 D0 q1,1  D0 q2,1  b0  D0 q2,1 2 b0 , 0 1 D2 q2,2  2 q2,2   D0 q2,1   2 D0 D1 q2,1   D0 q1,1 2 b0  2 D0 q1,1  D0 q2,1  b0  D0 q2,1 2 b0  2  2 q2,1  0 2 D2 q1,3  2 q1,3   D0 q1,2    D1 q1,1   2 D0 D1 q1,2   D2 q1,1  2 D0 D2 q1,1   2 D0 q1,1  D0 q1,2  b0  0 1 2 D0 q1,2  D0 q2,1  b0  2 D0 q1,1  D0 q2,2  b0  2 D0 q2,1  D0 q2,2  b0  2 D0 q1,1  D1 q1,1  b0  1 2 D0 q2,1  D1 q1,1  b0  2 D0 q1,1  D1 q2,1  b0  2 D0 q2,1  D1 q2,1  b0  D0 q1,1 3 b1  c q3 , D2 q2,3  2 q2,3   D0 q2,2    D1 q2,1   2 D0 D1 q2,2   D2 q2,1  2 D0 D2 q2,1   1,1 2 D0 q1,1  D0 q1,2  b0  2 D0 q1,2  D0 q2,1  b0  2 D0 q1,1  D0 q2,2  b0  0 2 1 2 D0 q2,1  D0 q2,2  b0  2 D0 q1,1  D1 q1,1  b0  2 D0 q2,1  D1 q1,1  b0  2 D0 q1,1  D1 q2,1  b0  2 D0 q2,1  D1 q2,1  b0  D0 q2,1 3 b2  2 q2,1  c q3  2  2 q2,2  2,1 Resonance condition ResonanceCond  1  2 ; 1 2 Involved frequencies omgList  1 , 2 ; Utility (don't modify) omgRule  SolveResonanceCond,  ,    Flatten1 &  omgList  Reverse 2  2 1 , 1   2 2 First-Order Problem Equations (=0) linearSys   1 &  eqOrder1; linearSys . displayRule  TableForm D2 q1,1  2 q1,1 0 1 D2 q2,1  2 q2,1 0 2
  • 5. Exercise.nb 5 Formal solution of the First-Order Problem sol1  q1,1  FunctionT0 , T1 , T2 , A1 T1 , T2  ExpI 1 T0   A 1 T1 , T2  Exp I 1 T0 , q2,1  FunctionT0 , T1 , T2 , A2 T1 , T2  ExpI 2 T0   A 2 T1 , T2  Exp I 2 T0  q1,1  FunctionT0 , T1 , T2 , A1 T1 , T2   1 T0  A1 T1 , T2   1 T0 , q2,1  FunctionT0 , T1 , T2 , A2 T1 , T2   2 T0  A2 T1 , T2   2 T0  Second-Order Problem Substitution of the solution on the Second-Order Problem and representation order2Eq  eqOrder2 . sol1  ExpandAll; order2Eq . displayRule D2 q1,2  2 q1,2   2   T0 1 D1 A1  1  2   T0 1 D1 A1  1  0 1   T0 1  A1 1  2  T0 1 A2 b0 2  2  T0 1  T0 2 A1 A2 b0 1 2  2  T0 2 A2 b0 2  1 1 2 2 2   T0 1  1 A1  2 A1 b0 2 A1  2  T0 1  T0 2 A2 b0 1 2 A1  2  T0 1 b0 2 A1  1 1 2 2  T0 1  T0 2 A1 b0 1 2 A2  2 A2 b0 2 A2  2  T0 1  T0 2 b0 1 2 A1 A2  2  T0 2 b0 2 A2 , D2 q2,2  2 q2,2  2  T0 1 A2 b0 2  2   T0 2 D1 A2  2  2   T0 2 D1 A2  2    T0 2  A2 2  2 2 0 2 1 1 2  T0 2  A2 2  2  T0 1  T0 2 A1 A2 b0 1 2  2  T0 2 A2 b0 2  2 A1 b0 2 A1  2 2 1 2 2  T0 1  T0 2 A2 b0 1 2 A1  2  T0 1 b0 2 A1    T0 2  2 A2  2  T0 2  2 A2  2  T0 1  T0 2 A1 b0 1 2 A2  2 A2 b0 2 A2  2  T0 1  T0 2 b0 1 2 A1 A2  2  T0 2 b0 2 A2  1 2 2 2 Utility (don't modify) expRule1i_ : Expa_ : ExpExpanda . omgRulei .  T0  T1  Terms of type ei 1 T0 in the first equation of the Second-Order Problem and representation ST11  Coefficientorder2Eq , 2 . expRule11, ExpI 1 T0  &  1; ST11 . displayRule  2  D1 A1  1    A1 1  2 A2 b0 1 2 A1  Terms of type ei 2 T0 in the first equation of the Second-Order Problem and representation ST12  Coefficientorder2Eq , 2 . expRule12, ExpI 2 T0  &  1; ST12 . displayRule  A2 b0 2  1 1 Terms of type ei 1 T0 in the second equation of the Second-Order Problem and representation
  • 6. Exercise.nb 6 ST21  Coefficientorder2Eq , 2 . expRule11, ExpI 1 T0  &  2; ST21 . displayRule 2 A2 b0 1 2 A1  Terms of type ei 2 T0 in the second equation of the Second-Order Problem and representation ST22  Coefficientorder2Eq , 2 . expRule12, ExpI 2 T0  &  2; ST22 . displayRule A2 b0 2  2  D1 A2  2    A2 2  2  A2 2  1 1 Scalar product with the left eigenvectors: First-Order AME; representation SCond1  1, 0.ST11, ST21  0, 0, 1.ST12, ST22  0; SCond1 . displayRule  2  D1 A1  1    A1 1  2 A2 b0 1 2 A1   0, A2 b0 2  2  D1 A2  2    A2 2  2  A2 2   0 1 1 Algebraic manioulation to obtain D1 A1 and D1 A2 SCond1Rule1  SolveSCond1, A1 T1 , T2 , A2 T1 , T2 1  ExpandAll; 1,0 1,0 SCond1Rule1 . displayRule  TableForm  A1 D1 A1    A2 b0 2 A1 2  A2  A1 b0 2 2 D1 A2     A2  1 2 2 2 Substitution of the First - Order AME to the Second Order Equations order2Eqm  order2Eq  . SCond1Rule1 . SCond1Rule1 . conjugateRule . expRule1  &  1, 2  ExpandAll; order2Eqm . displayRule D2 q1,2  2 q1,2   2  T0 1 A2 b0 2  2 3  T0 1 A1 A2 b0 1 2  4  T0 1 A2 b0 2  0 1 1 1 2 2 2 2 2 A1 b0 2 A1  2  T0 1 b0 2 A1  2 A2 b0 2 A2  2 3  T0 1 b0 1 2 A1 A2  4  T0 1 b0 2 A2 , 1 1 2 2 3 1  T0 2  T0 2 D2 q2,2  2 q2,2   2  2 0 2 A1 A2 b0 1 2  2  T0 2 A2 b0 2  2 A1 b0 2 A1  2  2 2 2 1 A2 b0 1 2 A1  b0 1 2 A1 A2  2  T0 2 b0 2 A2  1 3   T0 2   T0 2 2 2 2 A1 b0 1 2 A2  2 A2 b0 2 A2  2  2 2 2 Assumpion on the coefficients $Assumptions  1 , 2 , b0 , c, , , b1 , b2   Reals 1 2 b0 c   b1 b2   Reals Construction of the particular solution of the first equation of the Second-Order Problem and representation
  • 7. Exercise.nb 7 solOrder2Eqm1  SimplifyTableq1,2 T0 , T1 , T2  . TrigToExpFlattenDSolveorder2Eqm1, 1  order2Eqm1, 2, i, q1,2 T0 , T1 , T2 , T0  . C1  0, C2  0, i, 1, Lengthorder2Eqm1, 2; solOrder2Eqm1 . displayRule  1 3  T0 1 A1 A2 b0 2 4  T0 1 A2 b0 2 2 2 2  T0 1 A2 b0 ,  1 , , 2 A1 b0 A1 , 3 4 1 15 2 1  2 1 2 2 A2 b0 2 A2 2 3  T0 1 b0 2 A1 A2 4  T0 1 b0 2 A2 2 2  T0 1  b0 A1 , , , 3 2 1 4 1 15 2 1 solq21  ExpandTotalsolOrder2Eqm1; solq21 . displayRule 1 3  T0 1 A1 A2 b0 2 4  T0 1 A2 b0 2 2 2 2  T0 1 A2 b0  1   2 A1 b0 A1  3 4 1 15 2 1 2 1 2 2 A2 b0 2 A2 2 3  T0 1 b0 2 A1 A2 4  T0 1 b0 2 A2 2 2  T0 1 b0 A1    3 2 1 4 1 15 2 1 Construction of the particular solution of the second equation of the Second-Order Problem solOrder2Eqm2  Tableq2,2 T0 , T1 , T2  . SimplifyTrigToExpFlattenDSolveorder2Eqm2, 1  order2Eqm2, 2, i, q2,2 T0 , T1 , T2 , T0  . C1  0, C2  0, i, 1, Lengthorder2Eqm2, 2; solOrder2Eqm2 . displayRule  3 1  T0 2  T0 2 8 2 A1 A2 b0 1 1 2 A1 b0 2 A1 1 8 2 A2 b0 1 A1 2  T0 2 ,  A2 2 b0 ,  , , 5 2 3 2 2 3 2 2  T0 2 b0 A2  1 3   T0 2   T0 2 8 2 A1 b0 1 A2 8 2 b0 1 A1 A2 1 2 ,  2 A2 b0 A2 , , 3 2 5 2 3 solq22  ExpandTotalsolOrder2Eqm2; solq22 . displayRule 3 1  T0 2  T0 2 1 8 2 A1 A2 b0 1 2 A1 b0 2 A1 1 8 2 A2 b0 1 A1 2  T0 2   A2 2 b0     3 5 2 2 2 3 2 1 3   T0 2   T0 2 8 2 A1 b0 1 A2 8 2 b0 1 A1 A2 1 2 2 A2 b0 A2    2  T0 2 b0 A2 3 2 5 2 3 Formal representation of the solution
  • 8. Exercise.nb 8 sol2  q1,2  FunctionT0 , T1 , T2 , Evaluatesolq21 , q2,2  FunctionT0 , T1 , T2 , Evaluatesolq22; sol2 . displayRule q1,2  FunctionT0 , T1 , T2 , 1 3  T0 1 b0 2 A1 A2 4  T0 1 b0 2 A2 2 2 2  T0 1 b0 A2  1   3 4 1 15 2 1 , 2 1 2 2 b0 2 A2 A2 2 3  T0 1 b0 2 A1 A2 4  T0 1 b0 2 A2 2 2 b0 A1 A1  2  T0 1 b0 A1    3 2 1 4 1 15 2 1 q2,2  FunctionT0 , T1 , T2 , 3  T0 2 8 2 b0 1 A1 A2 1  2  T0 2 b0 A2  2 5 2 3 1 1  T0 2   T0 2 2 b0 2 A1 A1 1 8 2 b0 1 A2 A1 8 2 b0 1 A1 A2    2 2 3 2 3 2 2  T0 2 b0 A2  3   T0 2 8 2 b0 1 A1 A2 1 2 2 b0 A2 A2   5 2 3 Third-Order Problem Substitution in the Third-Order Equations order3Eq  eqOrder3 . sol1 . sol2  ExpandAll; Simplifications order3Eqpr1, 2  Simplifyorder3Eq1, 2; order3Eqpr2, 2  Simplifyorder3Eq2, 2; Terms of type ei 1 T0 in the first equation of the Third-Order Problem and representation ST311  Coefficientorder3Eqpr , 2 . expRule1 , ExpI 1 T0  &  1; ST311 . displayRule  1 60 1 60  D1 A1  1  60 D2 A1  1  120  D2 A1  2  120  D1 A1  A2 b0 1 2  180 c A2 1 A1  120  D1 A2  b0 2 A1  80 A2 b2 3 A1  180  A2 b1 4 A1  448 A1 A2 b2 2 2 A2  90 A1 A2 b2 1 2 A2  1 1 1 1 1 0 1 1 1 0 1 0 2 Terms of type ei 2 T0 in the first equation of the Third-Order Problem and representation ST312  Coefficientorder3Eqpr , 2 . expRule1 , ExpI 2 T0  &  1; ST312 . displayRule 0
  • 9. Exercise.nb 9 Terms of type ei 1 T0 in the second equation of the Third-Order Problem and representation ST321  Coefficientorder3Eqpr , 2 . expRule11, ExpI 1 T0  &  2; ST321 . displayRule  1 30 1 2  80  D1 A1  A2 b0 2 2  60  D1 A1  A2 b0 1 2  140  D1 A2  b0 2 2 A1  40   A2 b0 2 2 A1  160  A2 b0 2 2 A1  40 A2 b2 3 2 A1  224 A1 A2 b2 2 2 A2  45 A1 A2 b2 1 3 A2  1 2 1 1 1 1 0 1 0 1 2 0 2 Terms of type ei 2 T0 in the second equation of the Third-Order Problem and representation ST322  Coefficientorder3Eqpr , 2 . expRule12, ExpI 2 T0  &  2; ST322 . displayRule  30  D1 A2  1 2  30 D2 A2  1 2  30 2 A2 1 2  1 1 30 1 2 60  D1 A1  A1 b0 2 2  60  D2 A2  1 2  64 A1 A2 b2 3 2 A1  45 A1 A2 b2 2 2 A1  90 c A2 1 2 A2  40 A2 b2 1 3 A2  16 A2 b2 4 A2  90  A2 b2 1 4 A2  1 2 0 1 0 1 2 2 2 0 2 2 0 2 2 2 Scalar product with the left eigenvectors: Second-Order AME; representation SCond2  1, 0.ST311, ST321  0, 0, 1.ST312, ST322  0; SCond2 . displayRule  60  D1 A1  1  60 D2 A1  1  120  D2 A1  2  1 1 1 60 1 120  D1 A1  A2 b0 1 2  180 c A2 1 A1  120  D1 A2  b0 2 A1  80 A2 b2 3 A1  180  A2 b1 4 A1  448 A1 A2 b2 2 2 A2  90 A1 A2 b2 1 2 A2   0, 1 1 1 0 1 1 1 0 1 0 2  30  D1 A2  1 2  30 D2 A2  1 2  30 2 A2 1 2  60  D1 A1  A1 b0 2 2  1 1 1 30 1 2 60  D2 A2  1 2  64 A1 A2 b2 3 2 A1  45 A1 A2 b2 2 2 A1  90 c A2 1 2 A2  40 A2 b2 1 3 A2  16 A2 b2 4 A2  90  A2 b2 1 4 A2   0 2 0 1 0 1 2 2 2 0 2 2 0 2 2 2 Algebraic manipulation to obtain D2 A1 and D2 A2
  • 10. Exercise.nb 10 SCond2Rule1  SolveSCond2, A1 T1 , T2 , A2 T1 , T2 1 . A1 2,0 T1 , T2   T1 SCond1Rule1 0,1 0,1 1, 2 . A2 2,0 T1 , T2   T1 SCond1Rule12, 2 . SCond1Rule1 . SCond1Rule1 . conjugateRule  ExpandAll  Simplify  Expand; SCond2Rule1 . displayRule D2 A1   2 A1 1 3  c A2 A1 1 5 3  A2 b2 2 A1 1 0 1    A2 b0 A1    A2 b0 A1    A2 b2 1 A1  1 0 A2 b1 2 A1  1 1  8 1 2 2 1 12 2 2 2  A2 b0 2 A1  A2 b0 2 A1   A2 b0 2 A1 56 3  A1 A2 b2 2 A2 0 2     A1 A2 b2 2 A2  0 , 2 1 4 1 2 1 15 4 1  A2 b0 2 1 1  A2 b0 2 1 1   A2 b0 2 1 1   2 A2  A2 b0 1 1 7 D2 A2        A1 A2 b2 1 A1  0 4 2 2 8 2 2 4 2 2 8 2 2 2 4  17  A1 A2 b2 2 A1 0 1 3  c A2 A2 2 2 3 4  A2 b2 2 A2 2 0 2    A2 b2 2 A2  2 0 A2 b2 2 A2  2 2 30 2 2 2 3 2 15 1 Reconstitution of the AME and of the solution dA1 dt  ame1 ame1  A1 1,0 T1 , T2   A1 0,1 T1 , T2  . JoinSCond1Rule1, SCond2Rule1 . omgRule1; ame1 . displayRule  A1  2 A1 3  c A2 A1 1    A2 b0 A1   2 8 1 2 1 2 3 67  A2 b2 1 A1  1 0 A2 b1 2 A1   A2 b0 2 A1  1 1  A1 A2 b2 1 A2 0 3 2 15 dA1 dt  ame2 ame2  A2 1,0 T1 , T2   A2 0,1 T1 , T2  . JoinSCond1Rule1, SCond2Rule1 . omgRule1; ame2 . displayRule  A2 3 1 1   2 A2    A2   A2 b0  1  A2 b0  1   A2 b0  1  2 16 32 16 16 1  A2 b0 2 1 1 61 3  c A2 A2 2 12   A1 A2 b2 1 A1  0   A2 b2 1 A2  6 A2 b2 2 A2 2 0 2 1 2 2 30 4 1 5 Better representation
  • 11. Exercise.nb 11 Collectame1, A1 T1 , T2 , A1 T1 , T2 2 A 1 T1 , T2 , A1 T1 , T2  A2 T1 , T2  A 2 T1 , T2 , A2 T1 , T2  A 1 T1 , T2  . displayRule b1 2 A1  A2   b0   b0 2  A1    2 3c 2 3 67 A1   A2 1   b2 1  0 1  A1 A2 b2 1 A2 0 2 8 1 2 1 3 2 15 Collectame2, A2 T1 , T2 , A1 T1 , T2 2 , A1 T1 , T2 2 A 1 T1 , T2 , A1 T1 , T2  A2 T1 , T2  A 2 T1 , T2 , A2 T1 , T2 2 A 2 T1 , T2  . displayRule 3  b0  b0 1  b0 2 1 A2  1     b0   16 32 16 2 2   2 61 3c 12 A2    A1 b2 1 A1  A2 0 2   b2 1  6 b2 2 A2 0 1 2 16 1 30 4 1 5 Polar form of the AME Utility traspRule  A1 T1 , T2   A1 t, A 1 T1 , T2   A 1 t, A2 T1 , T2   A2 t, A 2 T1 , T2   A 2 t A1 T1 , T2   A1 t, A1 T1 , T2   A1 t, A2 T1 , T2   A2 t, A2 T1 , T2   A2 t Rewriting of the AME amem1  ame1 . traspRule  A1 't  2 A1 t 3  c A1 t2 A1 t  A1 t   b2 1 A1 t2 A1 t  b1 2 A1 t2 A1 t  1 2 3   0 1 2 8 1 2 1 3 2  b0 A2 t A1 t   b0 2 A2 t A1 t   b2 1 A1 t A2 t A2 t  A1  t 67 0 15 amem2  ame2 . traspRule  A2 't  b0 2 A1 t2  b0 A1 t2   b0 A1 t2    b0 A1 t2  3 1 1 1   16 32 16 2 2  2 A2 t  A2 t    A2 t   b2 1 A1 t A2 t A1 t  1 61  0 2 16 1 30 3  c A2 t2 A2 t  b2 1 A2 t2 A2 t  6 b2 2 A2 t2 A2 t  A2  t 12  0 1 4 1 5 Polar form of the complex amplitudes
  • 12. Exercise.nb 12 polRule  A1 t  1 a1t  1t , 2 A2 t  a2t  2t , A 1 t  a1t  1t , A 2 t  a2t  2t  1 1 1 2 2 2 A1 t   1t a1t, A2 t  1 1  2t a2t, 2 2 A1 t  a1t, A2 t  1 1  1t  2t a2t 2 2 polRuleD  JoinpolRule, DpolRule, t A1 t   1t a1t, A2 t   2t a2t, A1 t  1 1 1  1t a1t, 2 2 2 A2 t   2t a2t, A1  t   1t a1 t    1t a1t 1 t, 1 1 1 2 2 2 A2  t   2t a2 t    2t a2t 2 t, 1 1 2 2 A1 t  a1 t    1t a1t 1 t,  1 1  1t 2 2 A2 t  a2 t    2t a2t 2 t  1 1  2t 2 2 Substitution of the complex amplitudes with the polar form eq1  amem1 . polRuleD 1 1   1t 2 a1t 3  c  1t a1t3  1t  a1t   1t 2t  a1t a2t b0    4 4 16 1 16 1 1 67 3   1t a1t3 b2 1  0   1t a1t a2t2 b2 1   1t a1t3 b1 2  0 1 12 120 16   1t 2t a1t a2t b0 2   1t a1 t    1t a1t 1 t 1 1 1 4 2 2 eq2  amem2 . polRuleD 1 1 3  2t  a2t    2t  a2t  2  1t  a1t2 b0  4 2 64 1 1   2t 2 a2t 2  1t  a1t2 b0   2  1t  a1t2 b0   128 64 32 1 3  c  2t a2t3 61 3    2t a1t2 a2t b2 1  0   2t a2t3 b2 1  0 32 1 240 10  2t a2 t    2t a2t 2 t 3  2  1t a1t2 b0 2 1 1 1  2t a2t3 b2 2  1  4 8 2 2 2 Autonomous equations and separations of the real and imaginary parts
  • 13. Exercise.nb 13 eqa1  ComplexExpandExpandeq1  1t  1 1  a1t   a1t a2t Cos2 1t  2t b0  a1 t 4 4 3 1 a1t3 b1 2  1 a1t a2t Sin2 1t  2t b0 2   16 4 2 1 2 a1t 3 c a1t3 1   a1t a2t Sin2 1t  2t b0    a1t3 b2 1  0 4 16 1 16 1 12 a1t 1 t 67 1 1 a1t a2t2 b2 1  0 a1t a2t Cos2 1t  2t b0 2  120 4 2 eqa2  ComplexExpandExpandeq2  2t  1 3  a2t   a1t2 Cos2 1t  2t b0  4 64 1 1  a1t2 Cos2 1t  2t b0   a1t2 Sin2 1t  2t b0  a2 t 128 64 3 a1t2 Sin2 1t  2t b0 2 1 a2t3 b2 2  1   4 8 2 2 1 1 3   a2t   a1t2 Cos2 1t  2t b0   a1t2 Sin2 1t  2t b0  2 64 64 1 2 a2t 3 c a2t3 61  a1t2 Sin2 1t  2t b0    a1t2 a2t b2 1  0 128 32 1 32 1 240 a2t 2 t 3 a1t2 Cos2 1t  2t b0 2 1 1 a2t3 b2 1  0  10 8 2 2 Definition of  phiRule  2 1t  2t  t 2 1t  2t  t phiRuleD  2 't  2 1 't   't 2 t  2 1 t   t Polar Amplitude Modulation Equations amep1  CollectComplexExpand2 eqa1 .   0, a1t, a2t 3  a1t3 b1 2  1 8  a1 t  1 1 a1t  a2t   Cos2 1t  2t b0  Sin2 1t  2t b0 2 2 2 2
  • 14. Exercise.nb 14 amep2  CollectComplexExpand2 eqa2 .   0, a1t, a2t 1 3  a2t  a2t3 b2 2  1 2 2 3 1 a1t2   Cos2 1t  2t b0   Cos2 1t  2t b0  32 64  a2 t 1 Sin2 1t  2t b0 2 1  Sin2 1t  2t b0  32 4 2 amep3  CollectComplexExpand  2 eqa1 .   0, a1t, a2t 3c 1 2 67 a1t3  b2 1  a1t  0  a2t2 b2 1  0 8 1 6 8 1 60 Cos2 1t  2t b0 2  1 t 1 1 a2t  Sin2 1t  2t b0  2 2 amep4  CollectComplexExpand  2 eqa2 .   0, a1t, a2t 3c 3 a2t3  b2 1  a1t2 0 16 1 5 1 3 1  Cos2 1t  2t b0   Sin2 1t  2t b0   Sin2 1t  2t b0  32 32 64  2 t 61 Cos2 1t  2t b0 2 1 2 a2t b2 1  0  a2t   120 4 2 16 1 Reduced Amplitude Modulation Equations rame1  Collectamep1 . phiRule, a1t, a2t, t  a1 t 3  1 1  a1t3 b1 2  a1t 1  a2t   Cost b0  Sint b0 2 8 2 2 2 rame2  Collectamep2 . phiRule, a1t, a2t, t 1 3  a2t  a2t3 b2 2  1 2 2  a2 t 3 1 1 Sint b0 2 1 a1t2   Cost b0   Cost b0   Sint b0  32 64 32 4 2
  • 15. Exercise.nb 15 rame3  Expand2 a2t amep3  a1t amep4 . phiRule . phiRuleD 1 3   a1t a2t   a1t3 Cost b0   a1t3 Sint b0  32 32 1 2 a1t a2t 2 a1t a2t  a1t3 Sint b0   a1t a2t2 Sint b0    64 4 1 16 1 3 c a1t3 a2t 3 c a1t a2t3 7 49   a1t3 a2t b2 1  0 a1t a2t3 b2 1  0 4 1 16 1 40 30  a1t a2t2 Cost b0 2  a1t a2t  t a1t3 Cost b0 2 1 4 2 fixRule  a1t  a1, a2t  a2, t   a1t  a1, a2t  a2, t   Equations to find Fixed Points fix1  CollectExpandSimplifyrame1 . a1 't  0, a2 't  0,  't  0, a1t, a2t, a1t a2t, a1t3 , a2t3 , a1t2 a2t, a1t a2t2  . fixRule a1  3 1 1  a13 b1 2  a1 a2  1  Cos b0  Sin b0 2 2 8 2 2 fix2  CollectExpandSimplifyrame2 . a1 't  0, a2 't  0,  't  0, a1t, a2t, a1t3 , a2t3 , a1t2 a2t, a1t a2t2 , b0 , Cost, Sint . fixRule a2  3 3   2 1  a23 b2 2  a12 b0 1   Cos  Sin   2 2 32 64 32 4 2 fix3  CollectExpandSimplifyrame3 . a1 't  0, a2 't  0,  't  0, a1t, a2t, a1t3 , a2t3 , a1t2 a2t, a1t a2t2 , a1t3 a2t, a1t a2t3 , b0 , Cost, Sint . fixRule 2 2 3c 49 3c 7 a1 a2      a1 a23   b2 1  a13 a2 0  b2 1  0 4 1 16 1 16 1 30 4 1 40  a1 a22 b0  Sin  Cos 2  3   2 1 a13 b0  Sin  Cos   32 64 32 4 2 Equilibrium paths in the case a1  0 ep1  SimplifySolvefix2 . a1  0  0, a2, 1  0 a2  0, a2   , a2     3 b2 1 3 b2 1 Reconstitution of the solution
  • 16. Exercise.nb 16 qr1 t_  ComplexExpand A1 T1 , T2   1 t  A 1 T1 , T2   1 t  solq21 . traspRule . polRule . T0  t 1 1 a1t Cost 1  1t  a1t2 b0  a1t2 Cos2 t 1  2 1t b0  2 6 a1t a2t Cos3 t 1  1t  2t b0 2 a2t2 b0 2 2 a2t2 Cos4 t 1  2 2t b0 2 2   8 1 2 2 1 30 2 1 qr2 t_  ComplexExpand A2 T1 , T2   2 t  A 2 T1 , T2   2 t  solq22 . traspRule . polRule . T0  t 1 1 a1t2 b0 2 1 a2t Cost 2  2t  a2t2 b0  a2t2 Cos2 t 2  2 2t b0   2 6 2 2 2 t 2 3 t 2 4 a1t a2t Cos  1t  2t b0 1 4 a1t a2t Cos  1t  2t b0 1 2 2  3 2 5 2 Numerical integrations Numerical values 1  1, 2  2, b0  1 , b1  1, b2  1, c  1,   0.05,   0.05,   0, 2  2 1   2 1, 2, 1 , 1, 1, 1, 0.05, 0.05, 0, 2 2 Time of integration ti  500; Numerical Intergations of the RAME solrame1  NDSolverame1  0, rame2  0, rame3  0, a10  0.1, a20  0.1, 0  0.1, a1t, a2t, t, t, 0, ti a1t  InterpolatingFunction0., 500., t, a2t  InterpolatingFunction0., 500., t, t  InterpolatingFunction0., 500., t
  • 17. Exercise.nb 17 GraphicsArrayPlota1t . solrame1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "a1 t", Plota2t . solrame1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "a2 t", Plott . solrame1, t, 0, ti, PlotStyle  Thick, Frame  True, AxesOrigin  0, 0, FrameLabel  "t", "t" 0.4 0.5 2.0 0.2 1.5 a1 t t a2 t 0.0 0.0 0.2 1.0 0.5 0.4 0.5 0.6 0.0 0 100 200 300 400 500 0 100 200 300 400 500 0 100 200 300 400 500 t t t Numerical Intergations of the AME solramep1  NDSolveamep1  0, amep2  0, amep3  0, amep4  0, a10  0.1, a20  0.1, 10  0.1, 20  0.1, a1t, a2t, 1t, 2t, t, 0, ti a1t  InterpolatingFunction0., 500., t, a2t  InterpolatingFunction0., 500., t, 1t  InterpolatingFunction0., 500., t, 2t  InterpolatingFunction0., 500., t GraphicsArrayPlota1t . solramep1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "a1 t", Plota2t . solramep1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "a2 t", Plot1t . solramep1, t, 0, ti, PlotStyle  Thick, Frame  True, AxesOrigin  0, 0, FrameLabel  "t", "1 t", Plot2t . solramep1, t, 0, ti, PlotStyle  Thick, Frame  True, AxesOrigin  0, 0, FrameLabel  "t", "2 t" 0.4 4 6 0.5 0.2 5 a1 t 1 t 2 t 3 a2 t 0.0 4 0.0 2 3 0.2 2 0.5 0.4 1 1 0.6 0 0 0 100200300400500 0 100 300 500 200 400 0 100200300400500 0 100200300400500 t t t t Graphics of the reconstituted solution
  • 18. Exercise.nb 18 GraphicsArrayPlotqr1 t . solramep1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "q1 t", Plotqr2 t . solramep1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "q2 t" 0.4 0.5 0.2 q1 t 0.0 q2 t 0.0 0.2 0.5 0.4 0.6 0 100 200 300 400 500 0 100 200 300 400 500 t t Numerical Intergations of the original equations solorig1  NDSolveJoinEOM, q1 0  0.1, q2 0  0.1, q1 '0  0.1, q2 '0  0.1, q1 t, q2 t, t, 0, ti, MaxSteps  1 000 000 q1 t  InterpolatingFunction0., 500., t, q2 t  InterpolatingFunction0., 500., t GraphicsArrayPlotq1 t . solorig1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.8, 0.8, Frame  True, FrameLabel  "t", "q1 t", Plotq2 t . solorig1, t, 0, ti, PlotStyle  Thick, PlotRange  Automatic,  0.6, 0.4, Frame  True, FrameLabel  "t", "q2 t" 0.4 0.5 0.2 0.0 q1 t q2 t 0.0 0.2 0.5 0.4 0.6 0 100 200 300 400 500 0 100 200 300 400 500 t t