THE SOLUTION STABILITY OF     

VAN DER POL?S  EQUATION .  

                                                                                                                  by CO.   H . TRAN . 

   

                                                                                                           HUI -  NCU   HCMC   

                                                                                                                  Vietnam                                                                                     coth123@math.com   &  cohtran@math.com                                                                                                                   Copyright  2004                                                                                                         November  06   2004  ------------------------------------------------------------------------------------------------------------------------------------------------------------------------ **  Abstract  : The Van der Pol  differential quation  is solved  by  averaging  method  . ** Subjects:  Vibration Mechanics , The Differential equations  . ------------------------------------------------------------------------------------------------------------------------------------------------------------------------Introduction This worksheet demonstrates Maple's  capabilities in finding  the graphical solution and  dealing with the stability of the steady state solution of Van der Pol 's differential equation  .                                     All rights reserved.  Copying or transmitting of this material without the permission of the authors is not allowed . We consider the  Van Der Pol differential equation  :      

 

 

                                                                                                       [0] Two topics that we will be in discussion are -  Finding the steady state solution of this equation by averaging method .-  Estimating the stability of solution obtained . 1 .Define the model of  problem :  We examine the effect of non-linear system under external force caused by the AC generator ( see fig 1. )         ( fig 1. ) The differential equation of this model is given in the form                                                                              +                                              [1] After simplifying we obtain  :                                                            [2] 

2 . Construct the algorithm .   

Generally the  Van Der Pol differential equation can be expressed by   :                                                                                                                     [3]In the special case let        and       the  equation will be rewritten as :          By using the transform                                                                         [4]Then       ω        and substitute these relations to   ( 2 )  it follows :           ω Or                                                                                                          [5] Thus we begin with  :                                                                                        [6]    To normalize this equation we find the solution which is expressed in the form                                   x = acos ( t + γ )It is advantageous to write   ?  =  t + γ  then the solution will be   x =  acos?                                                                            [7]From the transform      x =  acos?  ,   x? =   asin ? .  We have    x?  =  dx?/dt  =  a?sin?   acos? . ?? By substituting to ( 6 ) , it gives             a'sin? a cos?.? '+a cos? μ (1   cos? ).(a sin? ) = 0                                             [8]In the other hand        dx/dt  =  x? =  a?cos?   asin? .?? =  asin?                                                                          [9] Obviously we reach to the following system of differential equations                                  [10]       

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Warning, the name GramSchmidt has been rebound
Warning, the protected names norm and trace have been redefined and unprotected

 

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(1)

 

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(2)

 

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(3)

 

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	(4)

 

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By using the symbols   a(tt)  =  a?(t)   ,  ?(tt)  =  ??(t)                                                                                      [11] Execute the  averaging method for  a?(t) and  ??(t) , we have  

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(5)

 

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(6)

 

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The expressions of   a? and  ??  are calculated in the forms :   a?  =  μ <   >  =  [12] 

??  =   μ <  >   =                                        [13]  

 

The steady state solution occurs when   a0  =  0  or  a0  =  2     If   a0  =  0  then  x =   x?  =  0   this is a  trivial solution ( equilibrium   ) .  If   a0  =  2   then x =   2cos?  and  x?  = -2sin? .                                                                                                          [14] The necessary and sufficient condition for solution stability includes a?  = da/dt  =  Ψ(a)   with   Ψ(ao)  =  0  and   Ψ?(ao)  <  0                                 

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(7)

 

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	(8)

 

>	subs(a=2,DaohamcuaPsi(a));

 

>	print("Gia tri cua a''(t) tai a = 2 la : a''(2) = ",subs(a=2,DaohamcuaPsi(a)));

 

>	subs(a=0,DaohamcuaPsi(a));

 

>	print("Gia tri cua a''(t) tai a = 0 la : a''(0) = ",subs(a=0,DaohamcuaPsi(a)));

 

 

 

 

 

 

 

	(9)

 

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Thus  if   ao   =  0  ,    a?(0) =      then  the solution is not stable  .  

If   ao   =  0  ,    a?(0) =    then  the solution is stable asymtotically. 

 

Note  :  By solving  the equation ( 12 ) for the vibration  amplitude a(t) ( ? slowly varying  coefficients ? ) then finding the solution expression x(t)  of  Van Der Pol differential equation , we get   

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a0(tt);

 

>	diff_eq:= diff(a(t),t)=-mu*a(t)*(-4+(a(t)^2))/8;

 

>	init_con:=a(0)=ao;biendo:=[dsolve({diff_eq,init_con}, {a(t)})];

 

>	x:=biendo[1]*cos(phi);

 

 

 

 

 

 

 

 

 

	(10)

 

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                             with     ?  =  t  +  γ  .                                                                                                                 [15]                                    

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(11)

 

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Warning, the previous bindings of the names RationalCanonicalForm and adjoint have been removed and they now have an assigned value

 

>	with(plots):a:=0.5;y:=t->2*cos(t + gamma)/sqrt((a^2-a^2*exp(-0.1*t)+4*exp(-0.1*t))/a^2);

 

>	plot(y(t),t=0..4*Pi);

 

>	animate(plot,[2*cos(x*t + gamma)/sqrt((a^2-a^2*exp(-0.1*t)+4*exp(-0.1*t))/a^2),x=0..4*t],t=0..Pi,frames=100 );

 

 

 

 

 

 

 

 

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Use graphical method to consider the solution stability of  Van Der Pol's  differential equation : 

>	restart;with(DEtools):mu:=0.5;

 

>	DEplot({(D@@2)(x)(t)+x(t)-mu*(1-x(t)^2)*D(x)(t)=0},{x(t)},t=-10..50,[[x(0)=1,D(x)(0)=1]],stepsize=0.05,title=`Nghiem on dinh cua pt Van Der Pol`);

 

 

 

 

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3 . Conclusion .  From  graphical results , we reach to conclusion that  the steady state solution stability of Van Der Pol's  diffential equation must be  precise  and estimating the property of solution obtained is very necessary . As presented  above , we might also use the normalization to ( 2 )  by determining the non-trivial solution in the form  

                                                                                        [16]                                                                  

 

with    k  =  1 Otherwise     Substitute   x ?  to  ( 6 )  after simplifying it follows :
 

 

 

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>	f:=vector([[0], [mu*F]]);

 

>	V:=linsolve(A,f);

 

 

 

 

 

 

 

 

 

Warning, the name GramSchmidt has been rebound
Warning, the protected names norm and trace have been redefined and unprotected
	(12)

 

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We rewrite the expressions  of                  M?     =                                                                           N?      =                                                                                                        [17]      

With    

 

Use averaging method for  ( 17 )  we get  :  

 

M ?  =  μ <- Fsint >  =    

N?  =   μ <  Fcost >  =                                                                             [18]    

 

The steady state solution exists when  M = 0   and    N = 0 ( trivial solution )                                        Or            M =  2   or  M = -2    and   N =  0 

Thus    x  =   2cost   , x  = 2cost                                           If   M =  0  and   N  = 0    or     N  =  4      Then     x  =   4sint   The steady state solutions can be formed generally   :  

       

 

But these forms are equivalent to   x =   2cos?       [ see  ( 14 )  ] 

 

Next we begin with   Krylov ? Bogoliubov approximate method for the Van der Pol's  differential equation   

 

 

 

with   ) x '       ,  and  μ  is a small constant  .    

 

The solution will be estimated  by     x  =  r(t) cos?(t)       

 

               By taking the first order approximate terms  ( neglecting the second and third order errors  of  constant   μ )  we can find the amplitude r(t) and the global phase function ?(t) from the following system  

                                               [19]     

 

If the initial condition   r(0) = ro  is satisfied  then the solution of  ( 15 )  will be equivalent to the solution of second order linear differential equation   

 

[20] 

 

with the error of estimation based on the order of     

 

The first order approximate is noticeable in the case of periodic vibration , because  the equivalent linear differential equation  gives us the accumulation and dissipation of energy based on vibration period which we might obtain from the given non-linear differential equation .  Therefore it is useful to apply the equivalent second linear differential equation to observe the non-linear resonant phenomenon  .REFERENCES  

[1] Menh .C. Nguyen   , Dao dong phi tuyen  , Manuscript  , Vien Co hoc , Hanoi 2002 . [2] Korn . G , Korn . T ., So tay toan hoc , Trans Vietnamese , NXB DH-THCN , Hanoi 1978 

 

 

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