Greg s Notes

8/10/2019 Greg s Notes

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1. Introduction To perturbation Theory & Asymptotic Expansions

Example 1.0.1. Considerx 2 = cosh x (1.1)

For = 0 we cannot solve this in closed form. (Note: = 0 x= 2)The equation defines a function x : (, ) R (some range of either side of 0)We might look for a solution of the form x = x0 +x1 +

2x2 + and by subsititungthis into equation (1.1) we have

x0+x1+2x2+ 2 = cosh(x0+x1+ )

Now for = 0, x0= 2 and so for a suitably small

2 +x1 2 cosh(2 +x1+ ) x1 cosh(2) x() = 2 + cosh() +

For example, if we set = 102 we getx= 2.037622...where the exact solution isx= 2.039068...

1.1. Landau or Order Notation.

Definition 1.1.1.Letfandg be real functions defined on some open set containingzero, ie: 0 D R. We say:

(1) f(x) =O(g(x)) asx (Big O)ifK > 0 and > 0 such that(, ) D andx (, )|f(x)| 0 versions ofo, O with one sided limits limo+

.

We use a series 1, 12 , ,

32 or in general: 0(), 1() . . .

12

+1() =o( 12

()) is what is needed.

Example 1.4.2 (Singular Model Equation).Considerx+x= 1, x(0) = 0, and suppose x = x0+x1O(2)Then(x0+x1) +x0+x1= 1 +O(2)

0 : x0= 1 but x(0) = 0 cannot solve the initial condition.

We can rescale time, ie: t= which gives= t

and dt

d=,

dx

d =

dx

dt

dt

d =x

dx

d +x= 1, x(0) = 0

Now usex = x0+x1O(2)0 :

dx0d

+x0= 1, x0(0) = 0 x0= 1 e = 1 e t1 :

dx1d

+x1= 1, x1(0) = 0 x1= 0 (similarly for2, 3 etc...)

Therefore the solution is:x= 1 et

Example 1.4.3(Singular In The Domain).Consider x+x2 = 1, x(0) =o, t >0 and assume x = x0+x1+

2x2O(2)

0 :x0= 1, x0(0) = 0 x0= t

1 :

(t+x1) +(t+x1)2 = 1 +O(2) 1 +x1+(t2 + 2tx1) = 1 +O(2)

x1+t2 =O(2) x1+t2 = 0 x1 = t3

3(If we carry on we find the 2 term

t5) The solution is:

x= t t33

+O(3)

This is not regular for t [0,)Example 1.4.4(Damped Harmonic Motion with Small Damping ( >0)).Consider x + x + x= 0, x(0) = 0, x(0) = 1 and assumex = x0+ x1+

2x2O(3)

0 :x0+x0 = 0, x0(0) = 0, x0= 1 x0= sin t

1 :x1+ x0+x1= 0 x1+x1= cos t, x1(0) = 0, x1(0) = 0

and we should get: x1= t2sin t whereby the solution is:

x= sin t 1

2 t sin t


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This again is not uniform for t [0,)The exact solution is: x = e

2

t sin(

1 = 2

4)t, the expansion above is only good

for small t.

1.5. Asymptotic Expansions.

Definition 1.5.1. A sequence of functions{n}, n= 0, 1, 2, . . . is called anasymptotic series asx x0 if

limxx0

n+1(x)

n(x) = 0

(ie: n+1(x) =o(n(x)))

Note: We could have x0 = or a one-sided limit x x+0Examples:

(i) x12 , 1, x

12 , x , x

32 , . . . x 0+

(ii) 1, 1x

, 1x ln(x) ,

1

x32 ln(x)

, x (iii) tan x, (x )2, (sin x)3, x

Taylors Theorem:

f(x) =N

n=0

f(n)(x)(x x0)n

n!

+

remainder

RN(x) , f CN+1[x0

r, x0+r], r >0

There are remainder formulas to bound RN(x), for example,

Cauchy:|RN(x)| MNrN+1

(n+ 1)! , MN >0, RN =O((x x0)N+1) asx x0.

It is important to remember that Taylor RN 0 as N In fact we knowplenty of power series that do not converge, eg:

Nn=0

(1)nn!xndiverges for all x.Another famous non-convergent series is Stirlings formula for n!:

ln n! = n ln n+1

2ln(2n) +

1

2231n 1

233251n3+O(

1

n4)

Definition 1.5.2. Let

{n

} be an asymptotic sequence asx

x0.

The sumN

n=0

ann(x) is called an asymptotic expansion of f with N terms if

f(x) N

n=0

ann(x) = o(N(x)).

The coefficientsanare called thecoefficientsof the asymptotic expansion.

n=0

ann(x)

is called an asymptotic series.

Note: Some people use the stronger definition: O(N(x))

Notation 1.5.3. f(x)

n=0

ann(x) asx

x0


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Clearly any Taylor series is an asymptotic series.

Example: Find an asymptotic expansion for f(x) =

0

et

1 +xtdt

(1 +xt)1 = 1 xt+x2t2 + f(x) =

0

(1 xt+x2t2 + )et dt

Now, it can be shown that

0

tnet dt= n! and hence

f(x) = 1 x+ 2!x2 3!x3 +

which diverges by the ratio test n!xn

(n 1)!xn1= n|x| x = 0.

It could still be be an asymptotic expansion however; wed need to check

f(x) N

n=0

(1)nn!xn =o(xN)

This is a special case of:

Lemma 1.5.4(Watsons Lemma). Letfbe a function with convergent power seriesand radius of convergenceR, andf(t) =O(et) ast

(for some >0) then:

0

eatf(t) dt

n=0

fn(o)

an+1

In last example we had f(x) =

0

et

1 +xdt as x 0+ letxt = u, a= 1/x, then

0

eau

u+u2du (looks like Watsons)

Example 1.5.5(Incomplete Gamma Function). .

(a, x) =

x0

ta1eet

dt=

x0

ta1 dtN

n=0

(t)nn!

=

Nn=0

(1)nn!

x0

tn+a1 dt=N

n=0

(1)nn!(n+a)

xn+a

Note: Power series under integral is convergent, hence uniformly convergent. Wehave a convergent power series for (a, x)inx

Example 1.5.6. .

Ei(x) =

x

et

t

dt (exponential integral)


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E1(x) =

1

etx

t dt (Cauchy principal value), it turns out that E1(x) = Ei(x)

Ei(x) =+ ln(x) +x+ x2

2 2!+ x3

3 3!+ O(x4)

where=

0

exln(x) dx= limn

Nk=0

1

k ln(n)

0.5772

2. ODEs In The Plane

We consider systems of ODEs of the form:x= u(x, y)y = v(x, y)

wherex(0) =x0, y(0) =y0.

(Note: A 2nd order ODE can be expressed as a coupled system of 1 st order ODEssince if x+f(x)x+g(x) = 0 then if we say x= y it follows that y= x so we get

x= yy = f(x) g(x)

Where in this case it turns out thatu(x, y) =y, v(x, y) = f(x) g(x))

2.1. Linear Plane autonomous Systems.

x= ax+byy = cx+dy the 1D case is easy: x= ax, x(t) =x(0)eatExample 2.1.1. .

Consider x= y

3x= 2yIf we sayx =

xy

then we can write x=

3 00 2

x

We solve these two ODEs to get

x(t) =x(0)e3t, y(t) = y(0)e2t (which may be written as)

x(t) =

e3t 0

0 e2t

x(0)y(0)

We can construct a Phase plot with solutions beingcurves in the plane called trajectories or orbits.

We could eliminate t as follows:

y = y(0) x

x(0)

23

, x(0) = 0

x

y

Figure 2.1.1:


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Example 2.1.2.

Similar to the above example consider x= x

y = yWe solve in both cases to get x(t) = x(0)et, y(t) = y(0)et and eliminate t suchthat:

y= y(0) x

x(0)

1

Noting that x =

1 00 1

x we end up with a phase plot which looks like this

Figure 2.1.2:

Example 2.1.3(Simple Harmonic Motion: x+ x= 0).

x= yy= x x=

0 11 0

x

x(t) =A cos t+Bsin t, x(0) =Ax(t) = A sin t+Bcos t, x(0) =B

x=

x(0)cos t+y(0)sin tx(0)sin t+y(0)cos t

=

cos t sin t sin t cos t

rotation matrix

x(0)y(0)

The orbits are circles.

x

y

Figure 2.1.3:

Example 2.1.4(Damped Harmonic Motion: x+bx+x= 0). x= yy= x by x=

0 11 b

x

Try x(t) =et giving the characteristic polynomial

2

+b+ 1 =b

2 b2

4 1


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Ifbis small then b2

4 1< 0 such that

=b

2 i

1 b

2

4 =+i

and so

x(t) =et(A cos(t) +Bsin(t))

xx

yy

Figure 2.1.4:

Theorem 2.1.5. LetA R22 be a real matrix with eigenvalues1, 2 then:(i) If1=2 are real then there exists an invertible matrixP such that

P1AP=

1 00 2

(ii) If1= 2 then eitherA is diagonal, A= I orA is not diagonal and thereis aP such that

P1AP= 10 (iii) If1= + i, 2= i, = 0 then there is aP such that

P1AP=

How does this help? Put x= Ax and let y= P1xthenx= Py, y= P1x, Px= PAxand so

y = P1APy

This allows us to generalise the work we did above.

Example 2.1.6. For case 1 in the theorem, consider u= P

1

APuthen

u1= 1u1, u2= 2u2

(sinceP1APu is just a matrix of eigenvectors acting on u)Thenui(t) = ui(0)eit and so

u(t) =

e1t 0

0 e2t

u1(0)u2(0)

Nowx= Pu so x(t) = P

e1t 0

0 e2t

u1(0)u2(0)

= P

e1t 0

0 e2t

P1

x1(0)x2(0)

u1 and u2 are related by eliminating t:

u1(t) =u1(0)e1t

u1

u1(0) 11

=et

, then u2= u2(0) u1

u1(0)21

.


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x

y

1> 2> 0 1< 2< 0

Figure 2.1.5:

For 1=2, distinct eigenvectorsv1, v2, Avi = iviHalf lines through the origin in eigen-directions are trajectories.

x

yeige

ndirect

ions

e.g. 1> 0 >2

Figure 2.1.6:


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2.2. Phase Space Plots. .

Eigenvalues real,

different, same sign

Node: Source

Eigenvalues real,

different, same sign

Node: Sink

Eigenvalues real,

different, opposite

sign

Saddle

Eigenvalues real,

equal,

> 0, A= I

Node: Source

Eigenvalues real,

equal

< 0, A= I

Node: Sink

Eigenvalues real,

equal

> 0, AI

Degenerate Source

Eigenvalues real,

equal

< 0, AI

Degenerate Sink

Eigenvalues

complex,

1= +i, 2= -i

< 0, 0

Stable Spiral

Eigenvalues

complex,

1= +i, 2= -i

> 0, 0

Unstable Spiral

Eigenvalues

purely imaginary,

1= i, 2-i, 0

Ellipse

Figure 2.2.1:


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Example 2.2.1. consider

x= 3x+yy= 2x 2y

The critical points of this system are at

00

, and the Jacobian is given by

3 12 2

we find the eigenvalues by setting A I=3 1

2 2

= 0, ie:

2 + 5+ 4 = 0

=

4,

1 this corresponds to a node (sink)

As for the associated eigenvectors;

A 4I=

1 12 2

v=

12

is in the null space (hence an eigenvector)

A I=2 1

2 1

v =

12

is the other eigenvector

We can get further information to help in curve sketching by considering isoclines:

y

x=

dy

dx=

2x 2y3x+y

Figure 2.2.2:

2.3. Linear Systems. A linear system for which the eigenvalues are wholly imag-inary (= ) is called called a centre.The characteristic equation, in general ofA R22 =2 (traceA)+ det A.In this case (for imaginary), 2 +2 = 0, traceA = 0, det A > 0.

Consider a simple case: x= yy = cx

eliminatet to gety

x =

dy

dx = cx

y

ydy =

cx dx y2

2 +

cx2

2 = const

This is the equation for an ellipse.To determine the direction of the arrows, set y = 0 then on x axis x= 0, y =cx

and

xy

is a vector in the direction of solutions, and we that for x positive x is

positive.


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The Oscillatory nature of these graphs dont reflect any real life situations.

y(t)

x(t)

Figure 2.3.1:

2.4. Linear Approximations. .

Consider x= u(x, y), y= v(x, y), Critical points occur when u = v = 0.Let (x0, y0) be critical points, put= x x0, = y y0and Taylor expand about(x0, y0) to get (near equilibrium point)

u(x, y) =u(x0, y0) +

u

x

(x0, y0)

+u

y

(x0, y0)

+O(2 +2) as (, ) (0, 0)

v(x, y) =v(x0, y0) +

v

x

(x0, y0)

+v

y

(x0, y0)

+O(2 +2) as (, ) (0, 0)

and so

uv

=

u

x

u

yv

x

v

y

+O(2 +2)

We can now make the approximation

=

u

x

u

yv

x

v

y

(x0, y0)

+O(2 +2)

which is a linear system.

Example 2.4.1(Preditor-Prey). .We wish to model the dynamics between predators and prey. Without consideringexternal & environmental variables, as the number of predators increases, we expect

the population growth rate of the prey should lessen; this then, should result in aslow down in the growth rate of predators (as there is more competition for fewerprey).let x be a population of prey (eg: rabbits), y be a population of predators (eg:foxes) withx, y >0 (note: this model relies on large x and y such that we are ableto talk about derivatives etc... (since x, y are integers!)

A simple model is

x= x(a y)y= y(c+x) (a, c, , >0).

Thenu(x, y) = x(a y), v(x, y) = y(c+x) and so for an equilibrium,u= v = 0, x(a y), y(c+x) = 0

so foru = 0, either x = 0 or a

y= 0 (y =

a

)


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for v = 0, either y = 0 orc+x= 0 (x= c

)

Therefore the criticals are at(0, 0), (

c

,

a

)

More specifically if we put a = 1, = 12 , c= 34 , =

14 then:

x= x(1 y2

)y = y( 3

4+ x

4)

with critical points at (0, 0), (3, 0)

Near (0, 0):

=

1 00

34

The corresponding eigenvalues being:

1= 1, v1=

10

, 2= 3

4, v2=

01

This is a saddle. Near (3, 2):

=

0 3412 0

with eigenvalues:

1,2= i

3

2 ie: 2 +2 = const, so ellipses.

Now

dy

dx =

y( x4

34

)

x(1 y2 ) 3

4ln x+ ln y y

2 x

4 = const.It is possible, albeit tricky to show this is a closed curve.

x(t)

y(t)

Figure 2.4.1:

Example 2.4.2(Circular Pendulum). .

We consider a circular pendulum given by

non-dimensional units x+ sin x= 0 where x is an angle.

m

mg

x

Figure 2.4.2:

Note that for small angles x, x sin x and so we get x+x = 0 (simple harmonic


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The only critical is at (0, 0)

u

x

u

yv

x

v

y

=

0 1

1 3x2 0

=

0 11 0

evaluated at (0, 0)

This is a centre = iExample 2.5.2 (Soft Spring). .We could change the sign before and simulate a soft spring ie: x+x x3 = 0The critical points in this case lie at (0 , 0), ( 1

, 0) and the jacobian is given by

0 11 + 3x2 0

forx close to1

we have a source; which, in physical terms, means we are actually

adding energy to the system.

3. Limit Cycles

Orbits, that is, trajectories of a system of ODEs cannot cross.

x(0)

x(t)

x(0) = x(T)

Figure 3.0.1:

Ifx=

x(t)y(t)

and x(0) =x(T) (for T= 0) then x(0) = x(T).

when this happens we have a periodic orbit, eg a clock, oscillator, cycle in theeconomy, biology etc...Suppposing we have such a cycle what can be said about what happens around it?It could be the case that both outside and inside the orbit we spiral towards it;but then again, something else could happen instead. The equilibrium point at the

centre doesnt give us this information.

Figure 3.0.2:

Example 3.0.3 (Cooked up). Consider a system of the contrived form:

x= y

x(x2 +y2

1) 1

y = x y(x2 +y2 1) 2


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we see that by construction, whenx, ysatisfyx2 +y2 = 1 we obtain simple harmonicmotion. There is only one equilibriums point and that occurs at (0,0). The Jacobianat this point is given by

f

x

f

yg

x

g

y

(x,y)=(0,0)

=

1 11 1

The characteristic equation is 2 2+ 2 = 0 which has roots = 1 i. This isan unstable spiral.

If instead however, we look at this in polar coordinates

r2 =x2 +y2, tan = y

x

Differentiating implicitely we get:

2rr= 2xx+ 2yy, = xy yxx2 +y2

If we multiply (1) and (2) by x & y respectively we get;

xx= x(y x(x2 +y2 1)), yy= y(x y(x2 +y2 1)) xx+yy= (x2 +y2)(r2 1) rr = r2(r2 1)

r= r(r2 1)This reveals that there is also an equilibrium point at r = 1 (ie; for r = 1 we stay

on the circle). Furthermore for r > 1, r < 0 which is a stable spiral, whilst forr 0 which is the unstable spiral we found above.

r < 0

r > 0

Figure 3.0.3:

Now if we multiply (1) and (2) by y & x respectively and subtract we get:

xy yx= x2 xy(r2 1) +y2 +xy(r2 1) = x2 +y2 =r2

=

r2

r2 = 1

The equilibrium point correctly told us that locally we have an unstable spiral butit failed to illuminate the behaviour as we move further out.

We shall establish a couple of results that allow us determine when & wherethere there exist no closed orbits. First recall

Theorem 3.0.4 (Divergence Theorem/Divergence theorem in the plane). Let Cbe a closed curve, letA R2 be the region it encloses, andu, v be functions withcontinuous derivatives. Then

A

u

x+

v

ydx dy=

C

u dy v dy=C

udy

dsds v dx

dsds

where(x(s), y(s)) is a parameterisation ofC


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Theorem 3.0.5(Bendixons Negative criterion). Consider the system

x= u(x, y)y= v(x, y)

withu andv continuously differentiable.

LetA R2 be a region of the plane for which ux

+v

y does not change sign.

Then there is no closed orbit contained withinA.

Proof. Suppose for contradiction there exists a closed orbit in A. Then this orbitforms a closed curveC in A.

A

A

x(0)= x(T)

Figure 3.0.4:

Let A A be the region enclosed by C (ie A = C). Then by the divergencetheorem

T0

(xy yx) dt=T

0

udy

dtdt v dx

dtdt=

C

u dy v dy=

A

u

x+

v

ydx dy = 0

but

u

x +

v

y is either > 0 or < 0x, y A hence A

= 0

Example 3.0.6 (returned to cooked example). . x= u = y x(x2 +y2 1)y = v = x y(x2 +y2 1)

u

x= 3x2 y2 + 1, v

y = x2 3y2 + 1 and so

u

x+

v

y = 4x2 4y2 + 2 = 4(x2 +y2) + 2

For values ofx, y such that x2 +y2 > 12 this fails to be always positive or always

negative; and so we cannot say there exists no closed orbit. (Though we can beconfident there is no such orbit for

x2 +y2 =r < 12

)

Bendixons Criterion is not much use for answering the question

Is there a closed orbit between r = 1 orr = 12

?

Example 3.0.7 (Damped Harmonic Motion). .For x+ 2x+x = 0 we have x= u = y, y= v = 2y xThis is a stable spiral at (0,0) and

u

x+

v

y = 0 2which is constant

Hence there is no sign change for u

x+

v

yand so no closed orbits.


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Example 3.0.8 (General Damped Oscillator). .This system is characterised by x + f(x)x

dampingf(x)>0

+g(x) = 0 which with the usual substitu-

tion x= y yields x= yy= f(x)y g(x)

Now u

x+

v

y = 0 f(x) is always negative and so general damped systems of this

form have no closed orbits.

3.1.

Theorem 3.1.1 (Poincare -Bendixon Theorem). Given a region A R2 and anorbit of a system of ODEs C which remains in At 0 then C must approacheither a limit cycle or equilibrium.

Remark 3.1.2.

(1) orbit = trajectory = solution curve(2) limit cycle = closed orbit that nearby orbits approach(3) We can use this result with time running backwards, ie: orbits come

from unstable equilibrium or closed orbits.

3.2. energy (brief). .

consider an oscillator of the form

x= yy = f(x) characterising x+f(x) = 0

Since there is no damping we expect energy to be conserved.

Consider=x2

2 +F(x) =

y2

2 +F(x) where

x2

2 can be considered kinetic energy,

F(x) the potential energy. Then

d(x, y)

dt =yy+f(x) = (x+F(x))x

SetF= f then d

dt= 0 along solution curves.

So is constant on a solution (x(t), y(t)). we call this a first integral

Example 3.2.1 (Duffings Equation). .This is the hard spring system we met earlier x+2x+x3 = 0

f(x) = 2x+ x3 F(x) = 2x2

2 +

x4

4 + some constant we need not worry

about in this context ofconstant solution curves. Then(x, y) = y2

2 +

2x2

2 +

x4

4for > 0. As (x, y) is constant then the solutions are bounded for all t (closedcurves).

We can say that x2 +y2 max(2, 22

) ( y2

2 +

2x2

2 +

x4

4 ) = max(2,

2

2). Ie;

solutions stay in the circle.For constant we can check that for y

= 0 there are two solutions for x.


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4. Lindstedts Method

Example 4.0.2 (Duffings Equation). x+2x+x3 = 0, 0<


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0 = ; actually we could choose what we want for 0 and depending on howdifficult we wnat to make things, this choice determines 1 later. It makes senseto keep things simple and choose 0 such that for = 0 we have x() =x(t))

Now d

dt= so

dx

dt =

dx

d

d

dt =

dx

d (so x= x) and

d2x

dt2 = 2

d2x

d2and so returning to the ODE we have

2x x3 = 0

(=01+1+ )2(x0 + x1+ ) + (x0+x1+ ) (x0+x1+ )3 = 0Terms in 0:

x0+x0= 0

As before we will assume initial conditions x(0) = a, x(0) = 0 to pick a criticalpoint on some curve. We know there exists a solution with these properties byassuming ellipsoidal shaped orbits; and by varyinga we get all the curves.Since there is no in initial conditions we get

x0(0) =a, x1(0) = 0, x0(0) =x

1(0) = 0

Then the solution for x0 is

x0= a cos

Terms in 1:

x1 + 21x0+ x1

x30 = 0

x1+ x1 = 21(a cos()) + (a cos())3 = 0

x1+ x1 = (21a+3

4a3)cos() +

1

4a3 cos(3)

The whole point of doing this was to eliminate the cos() term which would haveintroduced a secular term and so we set 21a+

34

a2 = 0

1= 38

a3

so it now remains to solve

x1+x1=1

4a3 cos(3)

We try x1= A cos(3) 9A cos(3) +A cos(3) = 14 a3 cos(3) A= a3

32The general solution for x1 before applying boundary conditions is

x1 = cos() +sin() a3

32cos(3)

Nowx1(0) = 0 = a332 and x1(0) = 0 = 0 giving

x1= a3

32cos() a

3

32cos(3)

Thus

x(t) =a cos(t) +a3

32(cos(t) cos(3t))

Where = 1 38 a

2

+


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5. Method of Multiple scales

In the last section we used Lindstedts method to take account of varying fre-quencies; now we develop a more general method for situations with two timescales.An example of where this is relevant is the damped circular pendulem

x+ 2x+ sin x= 0

There are two things going on here; there is an oscillation which is captured by onetime scale; and a slow loss of energy due to the damping term captured by anothertime scale. By considering two scales we are able to capture different features of

the system.

t T

Figure 5.0.2:

Consider small oscillations and small energy

x+x+ sin x= 0 (D.H.M)

If we did a standard expansion wed end up with a solution

x(t, ) = sin t+ 2

t sin t+ and this is not a uniform expansion for the solution. The exact solution for thissystem is:

x= 1

1 24

et2 sin

t

1 2

4

where

x(0) = 1x(0) = 0

5.1. The Method. .

(1) Introduce a new variable T = t, and think of t as fast time, and T asslow time.(2) Treatt andT as independent variables for the function x(T , t , ). Using

the chain rule we have

dx

dt =

x

t

t

t+

x

T

T

t +

x

=0

t

=x

t +

x

Td2x

dt2 =

d

dt

xt

+x

T

=

t

xt

+x

T

1 +

T

xt

+x

T

=2x

t2 + 2

2x

tT +2

x

T

(3) Try an expansionx = x0(t, T) +x1(t, T) + where of course T =t


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(4) Use the extra freedom ofx depending on Tto kill off any secular terms.

5.2. Application of Multiple scales to D.H.M.

Example 5.2.1. we consider x+ x+ x = 0 with boundary conditions x(0) =0, x(0) = 1, this becomes

2x

t2 + 2

2x

tT +2

2x

T2+

xt

+x

T

+x= 0

Now let X=x0+x1+2x2+ to get

2

t2+2

2

tT+ 2

2

T2(x0 +x1 +2x2 +

)+

t+

T(x0 +x1 +2x2 +

)

+x0+x1+2x2+ = 0

Terms in 0:2x0t2

+x0= 0

this is a PDE with solution

x0=A0(T)cos t+B0(T)sin t

whereA0(T), B0(T) are some functions ofT (as opposed to just constants) Usingboundary conditions x(0) = 0, x(0) = 1 then

x0(0) =A0(T)cos t+B0(T)sin t= 0 A0(0) = 0

x0(0) = A0(T)sin t+B0(T)cos t= 1 B0(0) = 1Terms in 1:

2x1t2

+ 22x0tT

+ x0

t +x1= 0

Now x0

t = A0(T)sin t+ B0(T)cos t and

2x0tT

= dA0(T)dT

sin t+dB0(T)

dT cos t

and so it remains to solve

2x1t2

+x1= 2dA0(T)

dT sin t+

dB0(T)

dT cos t

+A0(T)sin t B0(T)cos t

The RHS contains terms in sin(t), cos(t) which will induce secular terms. wetherefore choose A0(T) and B0(T) such that they go away. In other words:

2dA0

dT +A0= 0

A0 = A0(0)e

12

T = 0

2dB0dT

+B0= 0 B0=B0(0)e12 T =e 12T

and so

x0(t) =e1

2T sin t= e

t2 sin t

to get thex1term we would need higher order terms to fully specify it. Notice thatwith the exception of constants, the first order part captures most of the featuresof the exact solution.

In general we would have multiple time scales: T0= t, T1= t, . . . , T n= nt

If we consider a series x0(T0, T1, . . . , T n) +x1(T0, T1, . . . , T n) +. . . then

d

dt =

t =T0+

T1+2

T2+


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Example 5.2.2(Van Der Pols Equation). Consider

x+(x2 1)x+x= 0Immediately we see that if x2 > 1 we have damping, x2 < 1 we have negativedamping; so it would seem there is a tendency to head towards x = 1. Perhaps wecould use energy methods; anyhow...2

t2+ 2

2

tT +

(x0+x1+ )

+((x20+ 1)

+

T(x0+x1+ ) +x0+x1+ = 0

Terms in 0:2x0t2

+x0 = 0 x0 = A(T)cos t+B(T)sin tWe can write this in complex form

x0= A0(T)eit +A0(T)e

it (noting thatz +z = 2Re(z))

We can justify this step since in general, ifz = a+ib then

(a+ib)eit + (a ib)eit =a(eit +eit) +ib(eit eit) = 2a cos t 2b sin tTerms in 1:

2x1t2

+x1+ 22x0tT

+ (x20 1)x0t

= 0

2x1t2

+x1+ 2(Aieit Aieit) + (Aeit +Aeit) 1(iAeit iAeit) = 0

whereA= A(T) = A

TIf we wish to find A(T) as opposed to x1 we need only consider secular terms. Sowe need to equate coefficients ofeit (andet) to zero and kill them off.considering terms in eit we have:

2iA iA iA2A+ 2iA2A= 0 2A A+A2A= 0

Similarly, if we consider eit we get the conjugate of this expression; ie; whateverA kills off theeit terms also kills off the eit terms.

We should now use the polar form ofA, as |A2

A| = |A3

|, and if we write A = |A|ei

with = arg(A) and|A| = 12

a then

x0= a cos(t+) and we now wish to find a

Now dA

dT = 12

da

dTei + 12 aie

i ddT

and so it remains to solve

aei +iaei 12

aei + a2

4e2ia

2ei = 0

which dividing through byei gives

a=1

2a+

1

8a3 ia= 0

We now equate real and imaginary parts:

= 0 is constant


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Which leaves us the following ODE

a=1

2a 1

8a3

There are a number of methods to solve this; one of which being making the sub-stitution cos = 1

2aand using the fact that

1

sin =

1

sin

csc + cot

csc + cot =

csc2 + csc cot

csc + cot =

dd

(csc + cot )

csc + cot

1

sin d= log(csc() + cot())

This would still require a bit of work and so we shall instead treat a2 as a variableand use partial fractions.

da2

dT = 2a

da

dT =a2 a

4

4

4

a2(4 a2)da2 =

1a2

+ 1

4 a2

da2 = log|a2| log|4 a2| =T+T0

a2

4 a2 =keT a2(1 +keT) = 4keT a2 = 4k

eT +k

a(T) = 21 +k1eT

Hence

x0(t, T) = 21 +k1eT

cos(t+)

Example 5.2.3 (Duffings Equation). .We return again to Duffings equation x+ 2x+ x3 in the context of multiplescales. Let T =x then

x=dx

dt =

+

, x=

2x

t2 =

2x

t2 + 2

2x

tT +

22x

T2

and so putting this into the ODE above gives2t2

+ 2 2

tT

(x0+x1+ ) +2(x0+x1+ ) +(x0+x1+ )3

Terms in 0:

2

x0t2

+2x0= 0 x0 = Aeit +AeitTerms in 1:

2x1t2

+ 22x0tT

+2x1+x30= 0

Now

22x0tT

= 2(Aieit iAeit)whilst

x30= A3e3it + 3A2Aeit + 3AA2eit +A3e3it

and so2x1t2

+2x1+ 2(Aieit iAeit)

+A3e3it + 3A2Aeit + 3AA2eit +A3e3it = 0


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we now wish to kill off the secular terms eit and eit

terms in eit:2iA+ 3A2A= 0

Again setA = 12

aei such thatA= 12

aei + 12

aiei. Then

2i( 12 aei + 12 aie

i) + 34 a2e2i12 ae

i = 0

ia a+ 38 a3 = 0The imaginary term ia= 0 a= a0 (a constant)This leaves us with a simple ODE to solve

a= 38

a30

with solution

(T) = 3

8

a20

t (sinceT =T)

Thus the solution we get working as far as the first term x0 is

x= x0+O() =a0cos((+ 38

a20

)t) +O()

which as far as the first term goes is identical (except for the sign of 38

a20

(westarted with +x3 in this example)) to that which we got for Lindstedts method.

6. The WKB Method

In this section we focus on eigenvalue problems for 2nd order ODEs, specifically

d2ydx2

+2r(x)y = 0, y(0) =y(1) = 0 (6.1)

ie;y is a function ofx.

6.1. Motivation. stationary waves, for example, a string on a musical instrumentsatisfy

T2y

x2 =

2y

t2

Where T is the tension, the density of string, y the displacement, and x is thedistance along the string.

T

(x)

x

y

Figure 6.1.1:

y(0, t), y(1, t) = 0 the string is pinned down at both ends.Standing wave solutions have the form

y(x, t) = y(x)cos t

and so we reduce the wave equation to

d2y

dx2 +

(x)

T 2y= 0

this is in the form of (6.1) where = , r = (x)

t


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6.1.1. Constant density. Consider the case (x) = constant, then

d2y

dx2+

2

c2y = 0 (wherec2 = T

)

This gives

y(x) =A cos(x) +Bsin(x) (where2 = 2

c2)

Using the boundary conditions gives

y(0) = 0 A= 0, y(1) = 0 Bsin = 0 = n (B= 0, n Z)thus solutions have the form

y = b sin(nx) (6.2)

Solutions only exist for Z; ie, special or eigen values.In general we only get closed solutions for a simple r(x).Note: we are using boundary conditions y(0), y(1) rather than initial conditionsy(0), y(0). This means solutions do not exist for all (ie; they have to fit).Like eigenvectors the eigenfunctions y can be scaled by any B .

6.2. The WKB Approximation. This method stands for Wentzel, Kramer,Brillouin(though the work done by Jeffreys 2 years earlier had not beenrecognised by these three, and so he is often neglected credit for it).

if or are very large, the zeros of (6.2) are very close together. we intend to usean asymptotic expansion for large.ifr was constant in (6.1) we would have a solution of the form

y(x) =Aex

r

So we will try solutions of the form

y= eg(x,)

Then

dy

dx =g (, x)eg(x,) and

d2y

dx2 =2(g(, x))2eg(x,) +g(, x)eg(x,)

we substitute into (6.1) and divide through by 2eg(,x) to get

1

g+ (g)2 +r = 0

This is a 1st order ODE in g , ie:

1

h+h2 +r= 0

Noting that 1

behaves like (since we are considering large), we look for anexpression of the form

h(x) =h0(x) + 1

h1(x) + 12

h2(x) +O( 13

)

and sinceg = h= h0(x) + 1 h1(x) +

12

h2(x) +O( 13 ) then the originalexpression fory becomes

y = eg0+g1+1

g2+O(

1

2)


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Example 6.2.1 (Airys Equation). .Consider

y+2xy= 0

usingy = eg0+g1+1g2+O(

1

2) then

y= eg0+g1+1 g2+(g0+g1+ )

y= eg0+g1+1 g2+(g0+g1+ )2 +eg0+g1+ 1 g2+(g0+g1+ )

we substitute this into the original expression to get

(g0+g1+ )2 +g0+g1+ +2x= 0

Terms in 2:g02 +x = 0 g0= ix 12

g0= 23 ix32 +A

we ignore the constant because as it merely factorises out ofy as a constantmultiple of the solution and is of no real value to us.

Terms in 1 using g0= 23 ix

32 :

g0+ 2g0g1= 0 12 ix

12 + 2ix

12 g1

= 0 14x

+ g1= 0

g1= 14log(x) +Bwe didnt need the absolute value signs for log in this case since 0 < x, 1. Also, if

we were to consider the other solution for g0 then since g0 inherits the same signwe still get the same solution for g1.The general solution is

y = Aei23

x3/2 14

log(x) +Bei23

x3/2 14

log(x)

= 1

x14

Ccos( 2

3x

32 ) +D sin( 2

3x

32 )

+O(1

2) as

In general, ify = y +2r(x)y = 0 then

y(x, ) = 1

r(x)14

A cos(

xa

r(t)dt) +Bsin(

xa

r(t)dt)

+O( 1

2) as

Returning to the particular problem, we still havent found or used boundaryconditions.

to find approximate eigenvalues for large:

y(0) = 0 C= 0y(1) = 0 sin( 2

3) = 0 2

3= n

32

n (for large n Z)Example 6.2.2. .this time consider y +2(1 + 3 sin2 x)2y = 0, y(0) =y(1) = 0, again usingy= eg0+g1+ such that

(g0+g1+ )2 + (g0+g1+ ) +2(1 + 3 sin2 x)2 = 0

terms in 2:

g02 + (1 + 3 sin2 x)2 = 0 g0 = i(1 + 3 sin2 x)2


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we now use the angle sum trig identity sin2 x= 12

(1 cos2x) to getg0= i( 5

2+ 3

2(cos2x))

g0= i( 52 x 34sin 2x)terms in 1 (taking positive g0):

2g0g1+g0= 0 2(1 + 3 sin2 x)g1+ 3sin 2x= 0 g1= 3sin2x

2(1 + 3 sin2 x)

g1= 12log(1 + 3 sin2 x)Againin this examplewed gain no new information if we used the other form of

g0. Putting it all together we havey =

1

(1 + 3 sin2 x)12

Ccos(( 5

2x 3

4sin 2x)) +D sin(( 5

2x 3

4sin2x))

Now we use the boundary conditions

y(0) = 0 C= 0, y(1) = 0 ( 52 34sin2) =n n n5

2 3

4sin2

For example, plugging in some numbers, say n = 8 then our approximation yields8= 13.824 whilst the exact solution is 13.820

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