Multiple scale technique
Second order systems
Lesson 11
Weakly nonlinear oscillator
d2u
dt2 + ω02u = εF(u, du dt ).
it is a general form of Duffing’s equation
d2u
dt2 + u ∓ εu3 = 0 with F = ±u3 or van der Pol’s oscillator
d2u
dt2 + u − ε(1 − u2)du
dt = 0 with F = (1 − u2)du
dt .
For the first equation the method of renormalization
Type of solutions of the second order ODE.
The multiple scale technique permits successfully treat all types of weakly nonlinear oscillators. But
before introducing the method we review the types of the the second order ODE.
We start from the unperturbed equation
d2u
dt2 + ω02u = 0.
Its solution is
1 du
The phase plane
Let us introduce the coordinate system with x = u and
y = dudt . Then
x = u = A cosω0t + B sin ω0t
and
y = du
dt = −ω0A sin ω0t + ω0B cos ω0t.
Eliminating t yields the solution curve on (x, y) plane
x2 + y2
ω02 = A2 + B2.
The phase plane
The curves x2 + y2 = A2 + B2 are ellipses with
The phase plane
The differential equation ddt2u2 + dudt + 2u = 0 has the general solution
u = e−2t
A cos
√7
2 t + B sin
√7 2 t
where
A = u(0) and − A
2 +
√7B
2 = du dt (0)
The phase plane
as we see the solutions are the spiral into the origin
Linear autonomous equations
There are three types of behavior of solutions of autonomous linear differential equations:
1. periodic orbits,
2. decay towards the origin, 3. diverge to the infinity.
Nonlinear autonomous equations
The nonlinear autonomous differential equations has one more special type of solutions limit cycle.
Occurrence of this type depends both on the form of the equation and on the initial conditions. As an
example consider the equation
d2u
dt2 = −u +
1 − u2 − du dt
2du dt .
Nonlinear autonomous equations
Let us write the equation ddt2u2 = −u + 1 − u2 − dudt 2dudt using the phase plane variables
x = u, y = dx
dt = du dt .
Then the equation is equivalent to the system
dx
dt = y | · x
dy
dt = −x + (1 − x2 − y2)y | · y
Nonlinear autonomous equations
d
dt(x2 + y2) = 2(1 − x2 − y2)y2.
Using the polar coordinates
x = r cos θ r = sin θ
we get the
d
dt(r2) = 2(1 − r2)r2 sin2 θ.
Nonlinear autonomous equations
There is three possibility for the equation
d
dt(r2) = 2(1 − r2)r2 sin2 θ
1. If r = 1 then dtd (r2) = 0 and r = const give the periodic solution.
2. If r > 1 then dtd (r2) < 0 and the solution decrease to the periodic solution.
3. If r < 1 then dtd (r2) > 0 and the solution increase to the periodic solution.
This situation called the stable limit cycle.
Limit cycle
Types of solutions for nonlinear ODE
The solutions of nonlinear autonomous second order differential equations belong to the following four
classes
1. periodic solutions, 2. limit cycles,
3. solutions tends towards a fixed value 4. solutions tends to infinity.
The renormalization is success only for the first type of solutions.
Limitation of renormalization
Let us compare two problems
d2u
dt2 + u = εu, u(0) = 1, du
dt (0) = 0 d2v
dt2 + v = −εdv
dt , v(0) = 1, dv
dt (0) = 0
We look for the two term expansions
u(t, ε) = u0(t) + εu1(t)
and
Limitation of renormalization
The second coefficients are following
d2u1
dt2 + u1 = cos t, u1(0) = du1
dt (0) = 0 ⇒ u1 = t
2 sin t, d2v1
dt2 + v1 = − sint, v1(0) = dv1
dt (0) = 0 ⇒ v1 = 1
2(−t cos t + sin t).
Limitation of renormalization
Introducing the strained coordinates t = s + εf1, we get for the first equation
cos t + 1
2εt sin t = cos(s + εf1) + 1
2εs sin s
= cos s − εf1 sin s + 1
2εs sin s
Choosing f1 = 2s we have u = cos s + O(ε) where
t = s + εs
2 + O(ε2)
Limitation of renormalization
Expressing s
s = t
1 + ε
2 + O(ε2)−1
= t
1 − ε
2 + O(ε2) .
Thus
u = cos t
1 − ε
2 + . . . .
The exact solution is
u = cos t√
1 − ε = cos t
1 − ε
2 + . . . .
The renormalization provides a means of evaluating
Limitation of renormalization
Introducing the strained coordinates t = s + εf1 for the second equation
cos t − 1
2ε(t cos t − sin t) = cos(s + εf1) − 1
2ε(s cos s − sin s)
= cos s − εf1 sin s − 1
2εs cos s + 1
2ε sin s
f1 = −s2 cot s is not acceptable since it ia singular for
s = π, 2π, 3π, . . .. Trying to make the expansion of solution uniform the strained expansion becomes nonuniform. Let us compare with the exact solution.
Time changing amplitude
u = e−εt2
cos r
1 − ε2
4 t + ε 2
q
1 − ε42
sin r
1 − ε2 4 t
.
It is not a periodic solution since the frequency of oscillation and the amplitude are changed.
The frequency of oscillation changed from 1 to q1 − ε42
The amplitude e−εt2 decays.
Let us see why the renormalization failed in general case.
Limitation of renormalization
d2u
dt2 + u = εF(u, du
dt ), u(0) = a, du
dt (0) = 0
Starting from two term expansion u0 + εu1 we get
u0 = a cos t and
d2u1
dt2 + u1 = εF(a cos t, −a sin t).
Since F is a periodic function, it can be decomposed into the Fourier series
Limitation of renormalization
The homogeneous solution is A cos t + B sin t, so the term a1 cos t + b1 sint will generate a particular solution
t
2(a1 sin t − b1 cos t).
The two term expansion is
a cos t + ε t
2(a1 sint − b1 cos t) + nonsecular terms Using the renormalization t = s + εf1 we get
s nonsecular terms
Limitation of renormalization
We require f1 = a21as − b1s cot 2sa. We see that iff b1 = 0 we avoid the problems with cot 2sa. In this case the expansion is
u = a cos s + O(ε), t = s + εa1s
2a + O(ε2) ⇒ u = a cos t
1 − εa1
2a + . . . .
The expansion can generate only periodic solutions.
Limitation of renormalization
For Duffing’s equation we had F = −u3, so
F(a cos t, −a sin t) = −a3 cos3 t = a3
4 (3 cos t + cos 3t)
and renormalization gave the positive result.
For van der Pol’s oscillator F = (1 − u2)dudt
F = (1 − a2 cos2 t)(−a sin t) = (−a + a3
4 ) sin t + a3
4 sin 3t.
If a 6= 2 the renormalization fails. The case a = 2
corresponds to the limit cycle and the normalization is