7. The Poincaré-Lindstedt method

-1-

7. The Poincaré-Lindstedt method

For  autonomous conservative systems (e.g., motion in a velocity  independent  potential),  a  simple

method exists for updating the frequency of motion of the solution.  Consider the system

˙ ˙ 

x 

+ x = ε f x;ε

( )

     (7. 1)

We search for the natural time scale of the oscillations.  That is, the natural frequency, 

ω

0

 (equals

unity here) is to be updated due to the nonlinear perturbation into a new frequency, 

ω:

  ω =1 + ε ω

1

+ ε

2

ω

2

+ 

     (7. 2)

A new time variable is introduced

τ = ω t

⇒

d

dt

= ω

d

d

τ

(7. 3)

  

d

2

dt

2

= ω

2

d

2

d

τ

2

= 1 + 2ε ω

1

+ ε

2

ω

1

2

+ 2ω

2

(

)

+

(

)

d

2

d

τ

2

If we now expand x and f(x;

ε) in powers of ε

x

=

ε

n

x

n

τ

( )

n

≥0

∑

(7. 4)

  f x;ε

( )

= f

1

x

( )

+ ε f

2

x

( )

+ 

Inserting the expansions in Eq. (7.1) and comparing coefficients of powers of 

ε, order by order, we

obtain through second order

d

2

x

0

d

τ

2

+ x

0

= 0

  (7. 5a)

d

2

x

1

d

τ

2

+ x

1

= f

1

x

0

( )

− 2 ω

1

d

2

x

0

d

τ

2

(7. 5b)

d

2

x

2

d

τ

2

+ x

2

= f

2

x

0

( )

+

df

1

x

0

( )

dx

0

x

1

− 2 ω

1

d

2

x

1

d

τ

2

− ω

1

2

+ 2ω

2

(

)

d

2

x

0

d

τ

2

(7. 5c)

The  lowest  order  is  simply  a  harmonic  oscillation,  however,  not  in  ordinary  time,  but  in  time

measured in units of the modified frequency.  The free parameters 

ω

1

, 

ω

2

, .... are selected so  that

no secular terms develop up to the required order in the perturbation.  For  instance, in the equation

for  x

1

,  the  f

1

(x

0

)  may  have  a  simple  harmonic  term  in  its  Fourier  expansion  with  the  same

frequency as x

0

.  

ω

1

 is chosen so  as to eliminate the simple harmonic altogether, since it leads to

secular behavior.  The solution is obtained as follows

-2-

x

0

= acos τ + ϕ

0

(

)

(7. 6)

d

2

x

1

d

τ

2

+ x

1

= f

1

x

0

( )

+ 2ω

1

acos

τ + ϕ

0

(

)

Here 

ϕ

0

 is a constant phase.  We now expand f

1

 in a Fourier series

f

1

x

0

( )

=

a

n

cosn

τ + ϕ

0

(

)

n

≠ 0

∑

(7. 7)

(Note  that  f

1

(x

0

)  depends  only  on  cos(

τ+ϕ

0

)  and  not  on  sin(

τ+ϕ

0

).    As  a  result,  its  Fourier

expansion  includes  cosines  only).    Now  isolate  the  Fourier  component  that  resonates  with  the

additional cosine term in Eq. (7.6):

a

1

=

1

π

f

1

a cos

χ

(

)

cos

χ dχ

0

2

π

∫

(7. 8)

and require that the two terms cancel each other.  That is,

ω

1

= −

1

2

a

1

a

(

)

    (7. 9)

This can be now carried on  to  any  desired  degree  of  accuracy.    It  is  important  to  note  that  the

frequency must be updated as we go up in the order of the expansion, since, most probably, the true

(but unknown) frequency, 

ω, is affected by ε to all orders.  If we neglect to update the frequency at

a given order, then there will be a mismatch between the true frequency, and the approximate one.

As a result, the approximate solution will describe the exact one only over a limited range in time:

the higher orders will develop secular terms.

The  problem  exists  even  in  the  trivial  case  of  the  harmonic  oscillator  with  a  slightly  perturbed

frequency of Section 5

˙ ˙ 

x 

+ 1 + 2ε

(

)

x

= 0

     (7.10)

We, of course, know how to solve this linear problem,  but  pretend  that  it  requires  a  perturbative

treatment.  Here f(x;

ε) of Eq. (7.1) is simply

f x;

ε

( )

= f

1

x

( )

= −2 x

 (7.11)

Inserting Eq. (7.11) in Eq. (7.8) we find that a

1

=-2a.  Hence, Eq. (7.9) yields

ω

1

=1

   (7.12)

When inserted in Eq. (7.5b), this yields

x

1

= bcos τ + ψ

0

(

)

      (7.13)

-3-

Here, again, 

ψ

0

 is a constant phase.  Eq. (7.5c) now becomes

d

2

x

2

dt

2

+ x

2

= 1 + 2ω

2

(

)

a cos

τ + ϕ

0

(

)

      (7.14)

In order not to have secular terms here, we must have

ω

2

= −

1

2

     (7.15)

yielding

x

2

= ccos τ + ξ

0

(

)

     (7.16)

with 

ξ

0

 a constant phase.  As a result, through second order,

 

τ = 1 + ε +

1

2

ε

2

+ O ε

3

( )

(

)

t

(7.17)

x

= a cos τ + ϕ

0

(

)

+ ε bcos τ + ψ

0

(

)

+ c cos τ + ξ

0

(

)

+ O ε

3

( )

t

= O ε

0

( )

An O(

ε

2

) error is obtained for t≤O(1/

ε) if the ε

2

 term is dropped.  Now, as an example, impose the

initial condition x(0)=1, dx/dt(0)=0.  [Remember  that  dx/dt=

ω(dx/dt)!].    This  yields  a=1,  b=c=0,

ϕ

0

=0, giving

x

= cos τ = cos 1 + ε +

1

2

ε

2

+ O ε

3

( )

(

)

t

which, for t≤O(1/

ε), yields:

x

= cos 1 + ε +

1

2

ε

2

(

)

t

+ O ε

2

( )

t

≤ O 1 ε

( )

(7.18)

It is easy to see that this is simply the beginning of the expansion of the exact known solution

x

= cos 1 + 2 ε t

and that, indeed the error incurred is O(

ε

2

) for t≤O(1/

ε).

As another example, consider, again, the Duffing oscillator:

˙ ˙ 

x 

+ x = −ε x

3

x 0

( )

=1, ˙ 

x  0

( )

= 0

  (7.19)

Here

f x;

ε

( )

= f

1

x

( )

= −x

3

 (7.20)

-4-

and

f

1

x

0

( )

= −a

3

cos

3

τ + ϕ

0

(

)

= −

1

4

a

3

cos3

τ + ϕ

0

(

)

+ 3cos τ + ϕ

0

(

)

(

)

 (7.21)

The coefficient of the resonant term is

a

1

= −

3

4

a

3

(7.22)

yielding

ω

1

=

3

8

a

2

      (7.23)

This is precisely the O(

ε) correction to the frequency found in Section 6.5 (Eq.  (6.51)).    Higher

orders may be computed in a similar manner.

Finally, we point out why the method fails when the function f of Eq. (7.1) depends on the velocity,

dx/dt.  Consider the lowest order contribution, f

1

.  Regardless of whether  the  solution  tends  to  a

periodic limit cycle, or is not be periodic, f

1

(x

0

) is now replaced by f

1

(x

0

,dx

0

/d

τ), with

f

1

x

0

,

dx

0

d

τ

⎛ 

⎝ 

⎜ 

⎞ 

⎠ 

⎟  = f

1

acos

τ + ϕ

0

(

)

,

− asin τ + ϕ

0

(

)

(

)

The corresponding Fourier expansion now is

f x

0

,

dx

0

d

τ

⎛ 

⎝ 

⎜ 

⎞ 

⎠ 

⎟  =

a

n

cosn

τ + ϕ

0

(

)

+ b

n

sin n

τ + ϕ

0

(

)

{

}

n

≥0

∑

Now, in Eq. (7.6), f

1

 will contribute two resonant terms: a cosine term which can be eliminated by

an appropriate choice of 

ω

1

 and a sine term that one cannot eliminate.   The  latter  will  produce  a

secular term.  Hence the method works only for t=O(1).

Exercises

7.1 Prove Eqs. (7.5a-c).

7.2 Use the Poincaré-Lindstedt method to find the first order correction to the frequency in

˙ ˙ 

x 

+ x + ε sin x = 0