In perturbation theory, the Poincaré–Lindstedt method or Lindstedt–Poincaré method is a technique for uniformly approximating periodic solutions to ordinary differential equations, when regular perturbation approaches fail. The method removes secular terms—terms growing without bound—arising in the straightforward application of perturbation theory to weakly nonlinear problems with finite oscillatory solutions.[1] [2]
The method is named after Henri Poincaré,[3] and Anders Lindstedt.[4]
The article gives several examples. The theory can be found in Chapter 10 of Nonlinear Differential Equations and Dynamical Systems by Verhulst.[5]
The undamped, unforced Duffing equation is given by
\ddot{x}+x+\varepsilonx3=0
for t > 0, with 0 < ε ≪ 1.[6] Consider initial conditions
x(0)=1,
x(0) |
=0.
A perturbation-series solution of the form x(t) = x0(t) + ε x1(t) + ... is sought. The first two terms of the series are
x(t)=\cos(t)+\varepsilon\left[\tfrac{1}{32}\left(\cos(3t)-\cos(t)\right)-\tfrac{3}{8}t\sin(t)\right]+ … .
This approximation grows without bound in time, which is inconsistent with the physical system that the equation models.[7] The term responsible for this unbounded growth, called the secular term, is
t\sin(t)
In addition to expressing the solution itself as an asymptotic series, form another series with which to scale time t:
\tau=\omegat,
\omega=\omega0+\varepsilon\omega1+ … .
We have the leading order ω0 = 1, because when
\epsilon=0
x=\cos(t)
\omega2x''(\tau)+x(\tau)+\varepsilonx3(\tau)=0
Now search for a solution of the form x(τ) = x0(τ) + ε x1(τ) + ... . The following solutions for the zeroth and first order problem in ε are obtained:
\begin{align} x0&=\cos(\tau)\\ and x1&=\tfrac{1}{32}\left(\cos(3\tau)-\cos(\tau)\right)+\left(\omega1-\tfrac{3}{8}\right)\tau\sin(\tau). \end{align}
So the secular term can be removed through the choice: ω1 = . Higher orders of accuracy can be obtained by continuing the perturbation analysis along this way. As of now, the approximation—correct up to first order in ε—is
x(t) ≈ \cosl(\left(1+\tfrac{3}{8}\varepsilon\right)tr)+\tfrac{1}{32}\varepsilon\left[\cosl(3\left(1+\tfrac{3}{8}\varepsilon\right)tr)-\cosl(\left(1+\tfrac{3}{8}\varepsilon\right)tr)\right].
We solve the van der Pol oscillator only up to order 2. This method can be continued indefinitely in the same way, where the order-n term
\epsilonnxn
an\cos(t)+bn\cos(t)
an,\cos(2t)+bn,\cos(2t)+ …
See chapter 10 of for a derivation up to order 3, and [8] for a computer derivation up to order 164.
Consider the van der Pol oscillator with equationwhere
\epsilon
which yields the equationNow plug inwhere\tau=\omegat,
\omega=1+\epsilon\omega1+\epsilon2\omega2+O(\epsilon3)
x=x0+\epsilonx1+\epsilon2x2+O(\epsilon3)
1,\epsilon,\epsilon2
x0=A\cos(\tau+\phi)
\phi=0
\sin\tau,\cos\tau
A=2,\omega1=0
\epsilon
\ddotx1+x1=2\sin(3\tau)
x1=B\cos(\tau+\phi)-
14 | |
\sin(3\tau) |
\epsilonB\cos(\tau+\phi)
x0
x1=-
14 | |
\sin(3\tau) |
Now plug into the second equation to obtainTo eliminate the secular term, we set
\omega2=-
1 | |
16 |
Thus we find that
\omega=1-
1 | |
16 |
\epsilon2+O(\epsilon3)
This is an example of parametric resonance.
\ddotx+(1+b\epsilon2+\epsilon\cos(t))x=0
b
\epsilon
t
T=\epsilon2t
d | |
dT |
\begin{bmatrix} A\ B \end{bmatrix}=\begin{bmatrix} 0&
12 | |||
|
+b)\\
12 | |||
|
-b)&0\\ \end{bmatrix} \begin{bmatrix} A\\B \end{bmatrix}
Its determinant is
14 | |
(b-5/12) |
(b+1/12)
b\in(-1/12,5/12)
\sqrt{A2+B2}
In other words, when the angular frequency (in this case,
1
\sqrt{1+b\epsilon2}
For the van der Pol oscillator, we have
\omega\sim1/\epsilon
\epsilon
\epsilon
\omega
\epsilon
\omega
\epsilon\omega=r+c2r2+c3r3+c4r4+ …
x=x0+rx1+r2x2+ …
c2=1,c3=
15 | |
16 |
,c4=
13 | |
16 |
Because
\lim\epsilonr=1
\epsilon\omega=r+c2r2+c3r3+c4r4+ …
\epsilon\toinfty
\omega\sim1/\epsilon
The exact asymptotic constant is
\epsilon\omega\to
2\pi | |
3-2ln2 |
=3.8936 …
1+c2+c3+c4=3.75
\scriptstyleE=\tfrac12
x |
2+\tfrac12x2+\tfrac14\varepsilonx4
\scriptstyle
x |