Poincaré–Lindstedt method

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]

All efforts of geometers in the second half of this century have had as main objective the elimination of secular terms.

— Henri Poincaré, Les Méthodes Nouvelles de la Mécanique Céleste, preface to volume I

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

${\displaystyle {\ddot {x}}+x+\varepsilon \,x^{3}=0\,}$

for t > 0, with 0 < ε ≪ 1.^[6]

Consider initial conditions

${\displaystyle x(0)=1,\,}$   ${\displaystyle {\dot {x}}(0)=0.\,}$

A perturbation-series solution of the form x(t) = x₀(t) + ε x₁(t) + ... is sought. The first two terms of the series are

${\displaystyle x(t)=\cos(t)+\varepsilon \left[{\tfrac {1}{32}}\,\left(\cos(3t)-\cos(t)\right)-{\tfrac {3}{8}}\,t\,\sin(t)\right]+\cdots .\,}$

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 ${\displaystyle t\sin(t)}$. The Poincaré–Lindstedt method allows for the creation of an approximation that is accurate for all time, as follows.

In addition to expressing the solution itself as an asymptotic series, form another series with which to scale time t:

${\displaystyle \tau =\omega t,\,}$   where   ${\displaystyle \omega =\omega _{0}+\varepsilon \omega _{1}+\cdots .\,}$

We have the leading order ω₀ = 1, because when ${\displaystyle \epsilon =0}$, the equation has solution ${\displaystyle x=\cos(t)}$. Then the original problem becomes

${\displaystyle \omega ^{2}\,x''(\tau )+x(\tau )+\varepsilon \,x^{3}(\tau )=0\,}$

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:

${\displaystyle {\begin{aligned}x_{0}&=\cos(\tau )\\{\text{and }}x_{1}&={\tfrac {1}{32}}\,\left(\cos(3\tau )-\cos(\tau )\right)+\left(\omega _{1}-{\tfrac {3}{8}}\right)\,\tau \,\sin(\tau ).\end{aligned}}}$

So the secular term can be removed through the choice: ω₁ = ⁠3/8⁠. 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

${\displaystyle x(t)\approx \cos {\Bigl (}\left(1+{\tfrac {3}{8}}\,\varepsilon \right)\,t{\Bigr )}+{\tfrac {1}{32}}\,\varepsilon \,\left[\cos {\Bigl (}3\left(1+{\tfrac {3}{8}}\,\varepsilon \,\right)\,t{\Bigr )}-\cos {\Bigl (}\left(1+{\tfrac {3}{8}}\,\varepsilon \,\right)\,t{\Bigr )}\right].\,}$

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 ${\displaystyle \epsilon ^{n}x_{n}}$ consists of a harmonic term ${\displaystyle a_{n}\cos(t)+b_{n}\cos(t)}$, plus some super-harmonic terms ${\displaystyle a_{n,2}\cos(2t)+b_{n,2}\cos(2t)+\cdots }$. The coefficients of the super-harmonic terms are solved directly, and the coefficients of the harmonic term are determined by expanding down to order-(n+1), and eliminating its secular term.

See chapter 10 of ^[5] for a derivation up to order 3, and ^[8] for a computer derivation up to order 164.

Consider the van der Pol oscillator with equation$${\displaystyle {\ddot {x}}+\epsilon (x^{2}-1){\dot {x}}+x=0}$$where ${\displaystyle \epsilon }$ is a small positive number. Perform substitution to the second order:

${\displaystyle \tau =\omega t,\,}$   where   ${\displaystyle \omega =1+\epsilon \omega _{1}+\epsilon ^{2}\omega _{2}+O(\epsilon ^{3})}$

which yields the equation$${\displaystyle \omega ^{2}{\ddot {x}}+\omega \epsilon (x^{2}-1){\dot {x}}+x=0}$$Now plug in ${\displaystyle x=x_{0}+\epsilon x_{1}+\epsilon ^{2}x_{2}+O(\epsilon ^{3})}$, and we have three equations, for the orders ${\displaystyle 1,\epsilon ,\epsilon ^{2}}$ respectively:$${\displaystyle {\begin{cases}{\ddot {x}}_{0}+x_{0}=0\\{\ddot {x}}_{1}+x_{1}+2\omega _{1}{\ddot {x}}_{0}+(x_{0}^{2}-1){\dot {x}}_{0}=0\\{\ddot {x}}_{2}+x_{2}+(\omega _{1}^{2}+2\omega _{2}){\ddot {x}}_{0}+2\omega _{1}{\ddot {x}}_{1}+2x_{0}x_{1}{\dot {x}}_{0}+\omega _{1}(x_{0}^{2}-1){\dot {x}}_{0}+{\dot {x}}_{1}(x_{0}^{2}-1)=0\end{cases}}}$$The first equation has general solution ${\displaystyle x_{0}=A\cos(\tau +\phi )}$. Pick origin of time such that ${\displaystyle \phi =0}$. Then plug it into the second equation to obtain (after some trigonometric identities)$${\displaystyle {\ddot {x}}_{1}+x_{1}+(A-A^{3}/4)\sin \tau -2\omega _{1}A\cos \tau -(A^{3}/4)\sin(3\tau )=0}$$To eliminate the secular term, we must set both ${\displaystyle \sin \tau ,\cos \tau }$ coefficients to zero, thus we have $${\displaystyle {\begin{cases}A=A^{3}/4\\2\omega _{1}A=0\end{cases}}}$$yielding ${\displaystyle A=2,\omega _{1}=0}$. In particular, we found that when ${\displaystyle \epsilon }$ increases from zero to a small positive constant, all circular orbits in phase space are destroyed, except the one at radius 2. Now solving ${\displaystyle {\ddot {x}}_{1}+x_{1}=2\sin(3\tau )}$ yields ${\displaystyle x_{1}=B\cos(\tau +\phi )-{\frac {1}{4}}\sin(3\tau )}$. We can always absorb ${\displaystyle \epsilon B\cos(\tau +\phi )}$ term into ${\displaystyle x_{0}}$, so we can WLOG have just ${\displaystyle x_{1}=-{\frac {1}{4}}\sin(3\tau )}$.

Now plug into the second equation to obtain$${\displaystyle {\ddot {x}}_{2}+x_{2}-(4\omega _{2}+1/4)\cos \tau -{\frac {3}{4}}\cos 3\tau -{\frac {5}{4}}\cos 5\tau =0}$$To eliminate the secular term, we set ${\displaystyle \omega _{2}=-{\frac {1}{16}}}$.

Thus we find that ${\displaystyle \omega =1-{\frac {1}{16}}\epsilon ^{2}+O(\epsilon ^{3})}$.

This is an example of parametric resonance.

Consider the Mathieu equation ${\displaystyle {\ddot {x}}+(1+b\epsilon ^{2}+\epsilon \cos(t))x=0}$, where ${\displaystyle b}$ is a constant, and ${\displaystyle \epsilon }$ is small. The equation's solution would have two time-scales, one fast-varying on the order of ${\displaystyle t}$, and another slow-varying on the order of ${\displaystyle T=\epsilon ^{2}t}$. So expand the solution as $${\displaystyle x(t)=x_{0}(t,T)+\epsilon x_{1}(t,T)+\epsilon ^{2}x_{2}(t,T)+O(\epsilon ^{3})}$$Now plug into the Mathieu equation and expand to obtain$${\displaystyle {\begin{cases}\partial _{t}^{2}x_{0}+x_{0}=0\\\partial _{t}^{2}x_{1}+x_{1}=-\cos(t)x_{0}\\\partial _{t}^{2}x_{2}+x_{2}=-bx_{0}-2\partial _{tT}x_{0}-\cos(t)x_{1}\end{cases}}}$$As before, we have the solutions$${\displaystyle {\begin{cases}x_{0}=A\cos(t)+B\sin(t)\\x_{1}=-{\frac {A}{2}}+{\frac {A}{6}}\cos(2t)+{\frac {B}{6}}\sin(2t)\end{cases}}}$$The secular term coefficients in the third equation are $${\displaystyle {\begin{cases}{\frac {1}{12}}\left(-12bA+5A-24B'\right)\\{\frac {1}{12}}\left(24A'-12bB-B\right)\end{cases}}}$$Setting them to zero, we find the equations of motion:

${\displaystyle {\frac {d}{dT}}{\begin{bmatrix}A\\B\end{bmatrix}}={\begin{bmatrix}0&{\frac {1}{2}}({\frac {1}{12}}+b)\\{\frac {1}{2}}({\frac {5}{12}}-b)&0\\\end{bmatrix}}{\begin{bmatrix}A\\B\end{bmatrix}}}$

Its determinant is ${\displaystyle {\frac {1}{4}}(b-5/12)(b+1/12)}$, and so when ${\displaystyle b\in (-1/12,5/12)}$, the origin is a saddle point, so the amplitude of oscillation ${\displaystyle {\sqrt {A^{2}+B^{2}}}}$ grows unboundedly.

In other words, when the angular frequency (in this case, ${\displaystyle 1}$) in the parameter is sufficiently close to the angular frequency (in this case, ${\displaystyle {\sqrt {1+b\epsilon ^{2}}}}$) of the original oscillator, the oscillation grows unboundedly, like a child swinging on a swing pumping all the way to the moon.

^[9]

For the van der Pol oscillator, we have ${\displaystyle \omega \sim 1/\epsilon }$ for large ${\displaystyle \epsilon }$, so as ${\displaystyle \epsilon }$ becomes large, the serial expansion of ${\displaystyle \omega }$ in terms of ${\displaystyle \epsilon }$ diverges and we would need to keep more and more terms of it to keep ${\displaystyle \omega }$ bounded. This suggests to us a parametrization that is bounded:$${\displaystyle r:={\frac {\epsilon }{1+\epsilon }}}$$Then, using serial expansions ${\displaystyle \epsilon \omega =r+c_{2}r^{2}+c_{3}r^{3}+c_{4}r^{4}+\cdots }$ and ${\displaystyle x=x_{0}+rx_{1}+r^{2}x_{2}+\cdots }$, and using the same method of eliminating the secular terms, we find ${\displaystyle c_{2}=1,c_{3}={\frac {15}{16}},c_{4}={\frac {13}{16}}}$.

Because ${\displaystyle \lim _{\epsilon \to \infty }r=1}$, the expansion ${\displaystyle \epsilon \omega =r+c_{2}r^{2}+c_{3}r^{3}+c_{4}r^{4}+\cdots }$ allows us to take a finite number of terms for the series on the right, and it would converge to a finite value at ${\displaystyle \epsilon \to \infty }$ limit. Then we would have ${\displaystyle \omega \sim 1/\epsilon }$, which is exactly the desired asymptotic behavior. This is the idea behind Shohat expansion.

The exact asymptotic constant is ${\displaystyle \epsilon \omega \to {\frac {2\pi }{3-2\ln 2}}=3.8936\cdots }$, which as we can see is approached by ${\displaystyle 1+c_{2}+c_{3}+c_{4}=3.75}$.