Hartman-Grobman Theorem in n DimensionsDefinition: An equilibrium point of is said to be hyperbolic if all eigenvalues of the Jacobian matrix have non-zero real parts.

Theorem If is a hyperbolic equilibrium of

=�� � �x f(x)

� �*D f(x )

�*x

�*x

=�*=�� � �x f(x),

�

then in a neighborhood of , is topologically equivalent to the linearized vector field(this implies linear stability ⇔⇔⇔⇔ nonlinear stability)

=�� � � �

*x D f (x ) x

� =�� � �x f(x)

�*x

�� � �

We now want to talk about ways to detect limit cycles � Simulation of van der Pol oscillator� Bendixson criterion� Poincare Bendixson theorem

Recall the van der Pol oscillator

µ > 0

= = − − µ −� � 2The equation of motion is : x y, y x y(x 1)

�

for

µ > 0�H 0

-1 1

stablelimit cycle

>x 1

Bendixson’s (Negative ) Criterion The objective is toprove conditions under which no limit cycles can exist.

Theorem: Ifthen there are no closed phase paths in a simply connected (no holes!) region of phase space in whichthe divergence of the vector field ( ) is of one sign.� The proof is by contradiction!!

= =� �x f(x,y), y g(x,y) is the system,

+,x ,yf g

�

� The proof is by contradiction!!Proof: Let ℑℑℑℑ be such a closed path. Inside the region,

the divergence is of one sign

x

y

ℑℑℑℑ

nt

2 2( , ) /( , )

Here

f x yt f g

g x y� �

= +� �� �

By Divergence Theorem

However, the area integral on left cannot vanish ���� we have a contradiction ���� the enclosing curve ℑℑℑℑ is not a closed path.Ex: Consider the system

ℑ

+ = • =ℑ

�� � ��,x ,yinterior of(f g )dx dy t n ds 0

+ + =�� � 3x (x) x 0

�

Ex: Consider the system We want to prove that no limit cycles are possible any where in phase space.Soln: The system in vector form is

= = = − − =� � 3x y f(x,y), y y x g(x,y)

�+ = − − = −2 2

,x ,yThen, f g 0 3y 0 3ynon positive

+ + =�� � 3x (x) x 0

never changes sign in the entire phase space! i.e., no limit cycles can exist! (Bendixson’s criterion)There are generalizations: Bendixson-Dulac criterionPoincare Bendixson Theorem (only valid in 2D)- Let ℜℜℜℜ be a closed bounded region in the phase plane;- Let be continuously differentiable;- Let ℜℜℜℜ not contain any fixed points;

=�� � �x f (x )

+,x ,yf g

�

- Let ℜℜℜℜ not contain any fixed points;- Suppose that there is a trajectory ℑℑℑℑ which is confined to ℜℜℜℜ (starts and stays always in ℜℜℜℜ);

Then either ℑℑℑℑ is a closed orbit (limit cycle) or it spirals towards it as → ∞t .

Notes:- conditions (1) – (3) are easily satisfied; - for condition (4)

Trapping region or positively invariant region

�

no equilibrium pointsin this region

Ex 1:

When (a) Show that a closed orbit (limit cycle) exists at r = 1.(b) Show that the limit cycle continues to exist for � > 0 (� is small)(a)

= − + µ θ� 2consider the system r r(1 r ) r cosθ =� 1

µ = 0 :

µ = = − θ =�� 2For 0, r r(1 r ), 1

�

Since r and θθθθ equations are decoupled, it is enough to consider the r equation: the velocity function is

µ = = − θ =��For 0, r r(1 r ), 1

2( ) (1 ). ,' ' 0,1.

lim ( , )

v r r r So the equilibria

for r are at r These define

it cycle in x y plane

= −=

r=1θθθθ=t

When � ≠≠≠≠ 0, we need to find a trapping region in which the limit cycle is contained:

xrmin

y

�

We then need to determine<

>�

�

�max

min

r such that r 0r such that r 0

> � − + µ θ >� 2Now r 0 r(1 r ) (r)cos 0

� − > −µ θ � − > µ2 2r(1 r ) (r)cos r(1 r ) r

rmax

rmin

� − < −µ θ � − < −µ2 2r(1 r ) r cos r(1 r ) r< � − + µ θ + µmaxr 1

� < − µminr 1

Combining the two results gives

Thus, we have constructed a trapping region which does not contain a fixed point: from Poincare Bendixson Theorem, a limit cycle must exist in this annular region.

= − µ = + µmin maxChoose r 0.99 1 ; r 1.01 1

Ex 2: Glycolyis (model for break down of sugar). In some yeast extracts, this process proceeds in an oscillatory fashion.The equations of motion for the system are:

Here, a and b are reaction constants, a, b > , 0

= − + +� 2x x ay x y

= − −� 2y b ay x y

�

Here, a and b are reaction constants, a, b > , 0

x: concentration of ADP

y: concentration of fructose

Ex

x: concentration of ADP y: concentration of fructose, a, b > 0;

Want to construct a trapping region; andUnderstand under what conditions (a, b) do we get limit cylces?

2 2x x y(a x ), y b y(a x )= − + + = − +� �

��

cylces?Consider the caseof a=0.15, b=1.0Equilibrium point is asymptotically stable

�

Now, consider the caseof a=0.1, b=0.8Equilibrium point is unstable, there is

��

is unstable, there isa stable limit cycle

To plot the vectorfield, first consider the null-clines, i.e., lines of zero x or y velocities:

= − + + = � = +� 2 2x x y(a x ) 0 y x /(a x ) and= − + = � = +� 2 2y b y(a x ) 0 y b /(a x )

= + =�2y x /(a x ) or x 0

yb/a

> >� �x 0,y 0> � �x 0,y 0

This says that For x > b,

The above shows that i.e., we have a trapping region!However, we still cannot apply Poincare Bendixson!

We have a real trapping region only if the equilibrium is unstable.Linearize about

< −dy 1dx

= = +* * 2x b , y b /(a b )2� �− + +

��

�

Jacobian =

We can write the characteristic equation and determine the stability of the equilibrium

* *

2

2(x,y) x ,y

1 2xy a x

2xy (a x ) =

� �− + +� �� �− − +� �

τ < 0 !

� �λ = τ ± τ − ∆� �� �

21,2

1Thus, the eigenvalues are 4

2

∆ = = + >2det (Jac) a b 0− + − + +τ = =

+

4 2 2

2b (2a 1) b (a a )

and trace (Jac)a b

τ > 0

Now

The equilibrium is a stable node or spiral ifIt is unstable if

��

τ > 0It is unstable if

1.2

0.6

1.0

0.140.120 a

b equil. stableequil. unstablestable limit cycle

Perturbation TechniquesAlgebraic Equations:

Ex 1: Consider the quadratic equation (or, a polynomial of any degree)

(1)When εεεε=0, reduces to the solvable quadratic equation

(2)

2 (3 2 ) 2 0, 0 1x xε ε ε− + + + = <

Thus, there exist a Taylor series expansion in powers of εεεε which is convergent for every value of εεεε within some radius of convergence εεεε0. Therefore, let

(4)

The aim is now to find the coefficients

� � � � �4

20 3

2 30 1 2 3

1sec ( )( )

( )( ) ( )

( ) . . . . . i iifirst order ondzeroth third O terms

O orderorder orderOO O

x x x x x xε

εεε ε

ε ε ε ε ε∞

=

= + + + + =�

, 0,1, 2,3,4,......ix i =

��

Substituting eq (4) into eq (1) and expanding the resulting expression as an infinite series in εεεε:

Now,

i

2 3 2 2 30 1 2 3 0 1 2 3( ...) (3 2 )( ...) 2 0.x x x x x x x xε ε ε ε ε ε ε ε+ + + + − + + + + + + + =

2 3 2 2 2 30 1 2 3 0 0 1 2 3

2 3 21 2 3

( ...) 2 ( ...)

( ...)

x x x x x x x x x

x x x

ε ε ε ε ε εε ε ε

+ + + + = + + + +

+ + + +

2 2 2 30 0 1 0 2 1 0 3 1 22 (2 ) (2 2 ) . . . . . .x x x x x x x x x xε ε ε= + + + + + +

and

Eq (1) becomes

(5)Setting coefficient of each power of εεεε to zero seperately gives

2 3 20 1 2 3 0 1 0 2 1(3 2 )( . . . ) 3 (3 2 ) (3 2 ) . . .x x x x x x x x xε ε ε ε ε ε+ + + + + = + + + + +

2 2 20 0 0 1 1 0 0 2 1 2 1( 3 2) (2 3 2 1) (2 3 2 ) . . . . 0x x x x x x x x x x xε ε− + + − − + + + − − + =

��

seperately gives(6)

(7)

(8)Similar equations obtained at higher orders in εεεε.

0 20 03 2 0x xε − + =

10 1 1 02 3 2 1 0x x x xε − − + =

2 20 2 1 2 12 3 2 0x x x x xε + − − =

Consider eq (6). It has the two zeros (or, roots)(9)

(Note that eq (6) is the same as the “reduced equation”). Now, consider eq (7). It can be written as

(10)This is a linear nonhomogeneous equation in the unknown variable with the right hand side being

01 021, 2.x x= =

1 0 0(2 3) 2 1x x x− = −

�

unknown variable with the right hand side being known due to the roots in eq (9). For each of the roots in eq (9), eq (10) can be solved for provided

, which is clearly the case. Thus, the two solutions of eq (10) are

(11)

1x0(2 3) 0x − ≠

11 01

12 02

1 1,

3 2.

x when x

and

x when x

= − =

= =

Next, consider eq (8) at written in a slightly different form:

(12)This again is a linear equation for the unknown with nonhomogeneous known term on the right hand side. This equation can also be solved since . The solutions corresponding to the two roots are

2( )O ε

22 0 1 1(2 3) 2x x x x− = −

2x

0(2 3) 0x − ≠

01 02,x x3 1x when x= = −

�

are

(13)

Combining the results so far, the perturbation expansions for the two roots of the quadratic polynomial are given by

21 11

22 12

3 1

3 3.

x when x

and

x when x

= = −

= − =

(14)

The correctness of these expansions can be verified by directly writing the two roots of the quadratic eq (1),

and and then expanding them in a Taylor series for small values of εεεε.

2 31

2 32

1 3 ( ),

2 3 3 ( ).

x O

x O

ε ε εε ε ε

= − + +

= + − +

2[(3 2 ) (3 2 ) 4(2 )] / 2x ε ε ε= + ± + − +

��

small values of εεεε.Remarks:

1. The power series expansion in eq (4) is also called a straightforward expansion.2. Using this expansion, expanding the nonlinear function in powers of the small parameter εεεε, and collecting terms of different order in the

( , )f x ε

parameter εεεε gives an infinite sequence of problemswhich need to be solved to find the zeros of the function.3. The first problem in the infinite sequence is the reduced problem in the limit of εεεε →→→→ 0. All subsequent problems are linear equations with nonhomogeneous terms dependent on the solutions obtained at an earlier level. The

��

solutions obtained at an earlier level. The coefficient matrices (a scalar quantity in ex 1) are the Jacobian evaluated at the solutions of the reduced problem. Thus, if the linear equation for first order correction x1 is solvable (i.e., the coefficient matrix is nonsingular), all equations for higher order corrections are also solvable.

4. In the ex considered, only three terms have been explicitly evaluated. One can show that the power series is convergent, i.e., for a fixed εεεε within the radius of convergence, the partial sum has a finite limit as n →→→→ ∞∞∞∞. Also, the limit is the actual zero of the function. This is a consequence of “implicit function theorem”.

5. Usually, one does not find all the terms in the

��

5. Usually, one does not find all the terms in the power series, i.e., truncate the series with some finite number of terms (say, N). Clearly, cannot then talk about the convergence of the series as no limit in N →→→→ ∞∞∞∞ can be considered. We are then satisfied if the answer given by the solution with N terms is close to the true solution, atleast for sufficiently small εεεε.

Thus, one says that the difference between the true solution and the N term solution is small, and in fact the error in truncation at N terms is atmost of

for small εεεε. This is really the essence of an asymptotic expansion as opposed to a convergent expansion.

Ex 2: Consider the quadratic equation

( )NO ε

��

Ex 2: Consider the quadratic equation(15)

The reduced eq. (in the limit as εεεε →→→→ 0) is(16)

The roots are:(17)

2 (1 ) 0, 0 1,x xτ ε τ ε τ− + − + = <

Now assume the solutions (roots) of eq (15) as power series

(18)

Substituting this expansion into eq (15), expanding the resulting expression in powers of εεεε gives:

� � � � �4

20 3

2 30 1 2 3

1sec ( )( )

( )( ) ( )

( ) . . . . . i iifirst order ondzeroth third O terms

O orderorder orderOO O

x x x x x xε

εεε ε

ε ε ε ε ε∞

=

= + + + + =�

��

the resulting expression in powers of εεεε gives:

(19)Setting coefficient of each power of εεεε identically zero gives an infinite sequence of problems:

(20)(21)

0 0 0 1 0

2 20 2 1 1

( 1)( ) [(2 1 ) ]

[(2 1 ) ] . . . . 0

x x x x x

x x x x

τ ε τε τ

− − + − − +

+ − − + + + =

0 0( 1)( ) 0x x τ− − =

0 1 0(2 1 )x x xτ− − = −

(22)Eq (20) has the two roots

(23)

Now consider eq (21) for the correction to the roots of the zeroth order problem. Eq (21) is linear in and can be solved provided . Assuming this, the solutions to eq (21) are

20 2 1 1(2 1 ) ( )x x x xτ− − = − +

01 021, .x x τ= =

0 , 1,2ix i =

1x 0(2 1 ) 0x τ− − ≠

τ= − − =

��

(24)

Similarly, eq (22) has the solutions

(25)

11 011/(1 ) 1x when xτ= − − =

321 01/(1 ) 1x when xτ τ= − − =

12 02/(1 )and x when xτ τ τ= − =

322 02/(1 )and x when xτ τ τ= − =

Combining the terms in the solutions obtained so far,we get the three terms expansions for the two rootseq (15) as

(26)Remarks:

1. The series solutions in eqs (26) can be shown to be

2 3 31

2 3 32

1 /(1 ) /(1 ) ( ),

/(1 ) /(1 ) ( )

x O

x O

ε τ ε τ τ ετ ετ τ ε τ τ ε

= − − − − +

= + − + − +

��

convergent to the true zeros of the quadratic eq (15).If the convergence properties of a series areindependent of the parameter ττττ, it is said to beuniform. However, here the radius of convergence inεεεε, εεεε0, is a function of the parameter ττττ. Thus, this seriesexpansion is non-uniform in ττττ.

Eqs (26) show that solutions breakdown (become “non-uniform”) as ττττ →→→→ 1 since higher order terms become unbounded.

2. We first try to find the “region of non-uniformity” of the solution, i.e., the values of the parameter ττττ for which the expansion in eqs (26) is not useful (i.e., the series with small number of terms does not approximate the true solution reasonably well for atleast small values of the parameter εεεε).

��

atleast small values of the parameter εεεε). Note that the non-uniformity arises when (ττττ-1) is small and is for some constant αααα.To see exactly where this happens, consider the conditions under which successive terms in the expansion become of the same order (size).

( )O αε

The zeroth and first order terms in the expansion are of the same order when

(27)The first order and the second order terms are of the same order when

(28)

[ /( 1)] (1) (1 ) ~ ( )O or Oε τ τ ε− = −

2 3 2

1/ 2

[ /( 1)] [ /(1 ) ] (1 ) ~ ( )

(1 ) ~ ( )

O or O

or O

ε τ ε τ τ ετ ε

− = − −−

�

Thus, for small εεεε, the regions of non-uniformity in ττττare dependent on scaling relative to the small parameter εεεε . There is a region of width of and there is a region of width of . Both regions are centered at ττττ=1 and the region of width of is wider than the region of width of . Thus, we need to consider the non-uniformity of .

( )O ε1/ 2( )O ε

1/ 2( )O ε( )O ε

1/ 2( )O ε

Now we construct a solution which is valid in the region of width of around ττττ=1. This is called an inner solution and the region around ττττ=1 is called an inner layer. For this solution, introduce a scaling of ττττ by

(29)where σσσσ is independent of εεεε. Then, eqs (15) and (29) give

1/ 2( )O ε

1/ 2(1 )τ ε σ− =

1/ 2( 1)( 1 )x x ε σ ε σ− − + = −

�

give(30)

When εεεε=0, eq (30) reduces to(31)

that has the double root x=1. With this in mind, assume the solution of eq (30) in a power series in the form

1/ 2( 1)( 1 )x x ε σ ε σ− − + = −

2( 1) 0x − =

1/ 2 3/ 2 / 21 2 3

1

1 . . . . 1 n nn

x x x x xε ε ε ε∞

=

= + + + + = +� (32) Substituting this soln in eq (30) gives

(33)

At the lowest order in εεεε, the reduced equation is(34)

1/ 2 1/ 2 1/ 2 1/ 21 1 1( . . )( . . ) (1 . . )x x xε ε ε σ ε ε+ + + = − + +

2 3/ 21 1 ( ) 0or x x Oε ε σ ε ε+ + + =

21 1 1 0x xσ+ + =

��

(34)with roots

(35)Thus, the solutions of eq (30), upto the second order terms, are

(36)

1 1 1 0x xσ+ + =2

1 [ 4] / 2x σ σ= − ± −

1/ 2 2

1/ 2 2

1 ( 4) / 2 ( )

1 ( 4) / 2 ( )

x O

x O

ε σ σ ε

ε σ σ ε

= − + − +

= − − − +

Clearly, these solutions are regular at ττττ=1 or σσσσ=0, that is, they do not exhibit any singularity. In terms of the original parameter ττττ, these solutions are given by

(37)

This form of the solutions is useful in the neighborhood of ττττ=1. Far away from ττττ=1, it is more

1/ 2 1/ 2 21

1/ 2 1/ 2 22

1 [ (1 ) (1 ) / 4)] / 2 ( )

1 [ (1 ) (1 ) / 4)] / 2 ( )

x O

x O

ε ε τ τ ε ε

ε ε τ τ ε ε

−

−

= − − + − − +

= − − − − − +

��

neighborhood of ττττ=1. Far away from ττττ=1, it is more appropriate to use the straightforward expansions given in eqs (26).

Differential EquationsWe now consider differential equations. We start with straightforward expansions, and then see if they fail (become nonuniform for some parameter values.

Consider the Duffing eqn: Let m=1, F(t)=0, c=0,

k=1+ εεεεx2.x

��

εεεε

Then, the eqn of motion isWe do know some characteristics of the system:

mk

xck F(t)

+ + ε = ε�� �3u u u 0, 1

The force-deflection curve is:

The potential function is:

u

F(u)

V(u)

��

V(u)=u2/2+εεεε u4/4

The total energy function in the (u,du/dt) phase plane is given by:

H(u,du/dt)=(du/dt)2/2+V(u)Thus, all solutions are periodic orbits around (0,0).

u

(1)

We seek an approximate solution in the form of an expansion in εεεε (asymptotic expansion)

(2)

Substituting (2) into (1)

�+ + ε =�

= = ��

��

�

3Consider the eqn u u u 0with initial conditions u(0) a,u(0) 0

∞

=ε = ε = + ε + ε� �m 2m 0 1 2

m 0u(t; ) u (t) u (t) u u

��

Substituting (2) into (1)

Now, collect terms of like powers in εεεε !

+ ε + ε + + + ε + ε �2 20 1 2 0 1 2(u u u ....) (u u u )

+ ε + ε + ε =�2 30 1 2(u u u ) 0

ε + = = =�� �0 0 0 0 0: u u 0, u (0) a, u (0) 0

ε + + = = =�� �1 31 1 1 10: u u u 0, u (0) 0, u (0) 0

�0

01

Solution to eqn : u a cos(t)

Now, plug into equation :

ε =

ε+ = −�� 31 1u u (a cos t)

��

+ = −��1 1u u (a cos t)−+ = +��

3

1 1a

or u u (cos3t 3cos t)4

as t → ∞

−= +−

3 3

1a 3a

The solution is u (t) cos(3t) t sin(t)4(9 1) 8

Note the [tsin(t)] term in the solution for u1(t). It arisesdue to the resonant excitation term on the right handside.

� is no longer a “correction” term and the straight forward expansion is not uniformly valid in t, ie., for all times.

1u

ε = + ε + = +

ε − + +

�0 13 3

The total solution with two terms is :u(t; ) u u acos(t)

[ a cos(3t) / 32 3a t sin(t) / 8] ....

��

As becomes of same order in magnitude as

1t ~ 1/ , uε ε0u

Straight forward expansion breaks down at multiple levels:- when the two term expansion

is not uniformly valid

- It predicts non-periodic motions due to the secular terms

( )t O 1/ ,= ε0 1u u u= + ε +�

��

- there is no indication of frequency dependence on the amplitude of response ‘a’

- main downfall is the implicit assumption that depend on the same time scale! 0 1 2u ,u , u ,.... etc

Method of Multiple Scales- More used in the physics literature- Applies equally to conservative and nonconservativeproblems, and non-autonomous problems

Basic idea:Introduce multiple time scales in the problem:

= ε = �nnT t n 0,1, 2

�

areindependent variables and

As now ‘u’ is a function of many independent variables ���� total derivative changes to partial.

= ε = �nT t n 0,1, 2

= = ε = ε �20 1 2 nThen T t , T t, T t, and T

−

== ε + ε� �

m 1m m

m 0 1 mm 0

u(t) u (T ,T , ,T ) O( )

∂ ⋅ ∂ ⋅⋅ = + +∂ ∂

�0 1

0 1

dTd ( ) ( ) dTThen, ( )

dt T dt T dt

≡ + ε + ε �20 1 2D D D∂ ∂ ∂≡ ≡ ≡

∂ ∂ ∂0 1 20 1 2Here D ; D ; D ;......

T T T

� �= ⋅ = + ε + ε +� �2

0 1 22d d d d

Also ( ) [D D D ...]

�

⋅ = + ε + ε + +�2

2 2 20 0 1 1 0 22

dThus, ( ) D 2 D D (D 2D D )

dt

= ⋅ = + ε + ε +� �� �

0 1 22Also ( ) [D D D ...]

dt dt dtdt� � � �∂ ∂ ∂ ∂= + ε + + ε + ε + +� � � �∂ ∂ ∂ ∂� � � �

�0 0 0 1 1

0 1 0 1

D dT D D D.... .....

T dt T T T

Getting back to Duffing’s Oscillator:

Let(2)

Note that this is a 3-term expansion of the solution.Substituting (2) into (1), and using

+ ω + ε =�� 2 30u u u 0 (1)= + ε

+ ε + ε0 0 1 2 1 0 1 2

2 32 0 1 2

u u ,T ,T ,T ..) u (T ,T ,T ..)

u (T ,T ,T ..) O( )

⋅ = + ε + ε + +�2 2 2 2 20 0 1 1 0 2d ( ) / dt D 2 D D (D 2D D )

��

gives(3)

(4)

ε + ω =0 2 20 0 0 0: D u u 0

⋅ = + ε + ε + +�0 0 1 1 0 2d ( ) / dt D 2 D D (D 2D D )

ε + ω = − −1 2 2 30 1 0 1 0 1 0 0: D u u 2D D u u

(5)

( )ω − ω= +0 0 0 0i T i T0 1 2 1 2Solution of (3) is

u A(T ,T )e A(T ,T )e (6)

Substituting this into equation (4) givesωi T2 2 2

ε + ω = − −2 2 20 2 0 2 0 1 1 0 2 0: D u u 2D D u 2D D u

− −2 21 o 0 1D u 3u u

��

ω+ ω = − ω + 0 0i T2 2 20 1 0 1 0 1D u u [2 i D A 3A A] eω− +0 03i T3A e C.C

+ = + +3 3Note that we have used her[ :

(X X) X 3 X X(X

e

X)= + + +3 2 3X 3X X X C.C.]

In order for the ratio to be bounded for all the secular terms in the solution must vanish ����

(7)This gives the equation that defines the amplitude ‘A’of the zeroth-order term in the asymptotic solution.Then,

1 0u /u 0T ,

ω + =20 12 i D A 3 A A 0

ω ω= + +ω�����

o 0 0 03

i T 3i T1 1 2 2

Au B(T ,T )e e C.C.

��

(8)

This equation gives the solution for the first correction term to the zeroth-order solution. Here the amplitudes A and B are as yet undefined.

= + +ω�����

�����

1 1 2 20homogenous

particular

u B(T ,T )e e C.C8

.

Before moving on to equation (5), consider the requirement that secular terms vanish, i.e., equation (7)

Let

Then,

ω + =20 12 i D A 3 A A 0φ= 1 2i (T ,T )1 2 1 2

1A(T ,T ) a(T ,T )e

2φ= + φ 1 2i (T ,T )1 1 1 2 1 1 2

1D A [{D a(T ,T )} i{D (T ,T )}]e

2

��

Now, separating real and imaginary parts, and using

gives ∂ ∂φ= − ω + =∂ ∂

2

01 1

a 3a0, and 0

T T 8

φ = φ + φie cos isin

= + φ1 1 1 2 1 1 2D A [{D a(T ,T )} i{D (T ,T )}]e2

Consider briefly that only, i.e., we have a 2-term expansion. In that case,

Then

1A(T )

= = φ = + φω

20 0 1 0

0

3a const a and a T

8φ ω� =

20 0 1 0

1 0i i3a T /81A(T ) a e e

2

ω − ωi T i T

��

Then

Noting the expression for A(T1), we get

ω − ω= +0 0 0 0i T i T0 1 1u A(T ) e A(T )e

ω + ε ω +φ� = +2

0 0 0 0 00 0

i{( 3 a /8 )T }1u a e C.C2

.

One can write this solution as

where

is the frequency of oscillation. The amplitude and

ω = ω + ε + εω

220

00

3aO( )

8

ω +φ� = +)0 00 0i( T1u a e C.C

2.

��

is the frequency of oscillation. The amplitude and phase can be decided by initial conditions!φ0 0a and