NONLINEAR MECHANICAL SYSTEM We always assumed that the force F(x) exerted by the spring on

the mass is a linear function of x:F(x)= -kx (Hooke’s law).

However every spring in nature is nonlinear ( even if only slightly so).

Moreover, springs in some automobile suspension system deliberately are designed to be linear.

So now we allow the force function F(x) to be nonlinear. Because F(0)=0 at the equilibrium position x=0, we may assume that F has a power series expansion of the form

......)( 32 bxaxkxxF

We take k>0 so that the reaction of the spring is directed opposite of the displacement when x is sufficiently small.

Suppose we know that the mass on a spring is connected also to a dashpot that provides a force of resistance proportional to the velocity

of the mass. If the spring is still assumed nonlinear, then the equation of motion of the mass is

(1)

Where c >0 is the dashpot constant. If b >0, then the equivalent first order system

(2)

(3) Has critical point (0,0) and if we take m=1 for convinience,

then at (0,0) the characteristic of equation of the corresponding linear system is

(4)

dxdyy /

)0,( k

DUMPED NONLINEAR VIBRATIONS

3''' bxkxcxmx

3bxcykxdtdym

ydtdx

0)( 2 kckc

if we follow that (0,0) is a stable node if , a stable spiral point if

EXAMPLE:Suppose that m=1, k=5, and c=2. Then the non linear system in () is

In this example no explicit solution for the phase plane trajectories is available, so we proceed to investigate the critical points (0,0) ,(2,0) and (-2,0)

At (0,0) the linearized system

Has characteristic equation with roots .Hence, (0,0) is a stable spiral point of (18) and the liniearized position function of the mass is of the form

45

kc 42 kc 42

3

4525 xyx

dtdy

ydydx

yydtdy

ydtdx

25

0522 i21

)2sin2cos()( tBtAetx t

An exponential damped oscillation about x=0 at (2,0) the substitution u=x-2, v=y in (18) yields the system

with corresponding critical point (0,0). The liniearized sytem

Has characteristics equation with roots and

. It follows that (2,0) is an unstable saddle point of the original system in (18). A similar analysis shows that (-2,0) is also unstable point.The phase

32

45

215210 uuvu

dtdv

vdydx

vudtdv

vdtdu

210

01022 01111

01112

THE NON LINEAR PENDULUM

In section 3.4 we derived the equation

(20)

For the undamped oscillations of the simple pendulum shown in Fig 6.4.6. there we used the approximation for small to replace Eq 20 with the linear model

(21)

Where . The general solution (22)

Of Eq 21 describes oscillations about the equilibrium position with circular frequency

and amplitude The linear model does not adequately describe the possible motions of the pendulum

for large values of . For instance, the equilibrium solution

sin

0sin2

2

Lg

dtd

022

2

dtd

Lg2

tBtAt sincos)(

0 22 BAC

We also want to include the possibility of resistance proportional to velocity, so we consider the general non linear pendulum equation

(23)

We examine first the undamped case, in which c=0. with and

the equivelent first order system is

(24)

Upon substuting the Taylor series for sin x we get

(25)

Thus (24) is an almost linear system of the form

0sin22

2

dtdc

dtd

)()( ttx

)(')( tty

xdtdy

ydtdx

sin2

.......!5!3

52322

xxxdtdy

ydtdx

),(

),(

yxgdycxdtdy

yxfbyaxdtdx

With and the critical points of the system in (24) are the points

with n as integer. The nature of the critical point depends upon whether n is even or odd.

If is even then, because

The substitution in (24) yields the system

(26)

Having (0,0) as the corresponding critical point. Just as in (25), the liniearization of (26) is the system

(27)

0),( yxf ......!531/(),( 532 xxyxg

)0,( n)0,( n

mn 2

umu sin)2sin( yvmxu ,2

udtdv

vdtdu

sin2

udtdv

vdtdu

2

For which (0,0) is the familiar stable center with elliptical trajectories

If n= 2m+1 is odd then, because

The substitution in (24) yields the system

(28)

Having (0,0) as the corresponding critical point. The liniearization

(29)

umu sin)]12[sin(

yvmxu ,)12(

udtdv

vdtdu

sin2

udtdv

vdtdu

2

Of (28) has general solution

So we see that (0,0)is the unstable saddle point

In the undamped case we can see how these centers and saddle points fit together by solving explicity for the trajectories of (24). If we write

and separate the variables,

then the integration from x=0 to x=x yields (30)

We write E for the arbitary constant of integration because,if physical units are so chosen that m=L=1, then the first term on the left is the kinetic energy and

tAtBtvtBtAtu

sinhcosh)(

sinhcosh)(

yx

dtdxdtdy

dtdy sin2

0sin2 xdxydy

Exy )cos1(21 22

The second term the potential energgy of the masss on the end of the pendulum. Then Eis the total mechanical energy;Eq(300 thus expresses conservation of mechanical energy for the undamped pendulum.

If we solve Eq 30 for y and use a half identity, we get the equatition (31)

For the trajectories. THE SEPARTICES

xEy21sin42 22

If the pendulum is released from rest with initial conditions (32)

The Eq 30 with t=0 reduces to

(33)

Hence if , so a periodic oscillations of the pendulum ensues. To determine the period of this oscillation, we substract Eq 33 from Eq 30 and write the result (with and ) in the form

(34)

the period T of time required for one complete oscillation is four times the amount of time required for to decrease from to one-fourth of an oscillation. Hence we solve (34) for and integrate to get

(35)

)0()0(x 0)0(')0( y

E )cos1(2

22E 0

x dtdy

)cos(cos)(21 22

dtd

0ddt

0 coscos24 dT

To attempt to evaluate this integral we first use the identityAnd get (36)

Where (37)

Next the substitution yields

(38)

Finally the substitution

(39)

The integral in (39) is the eliptic integral of the first kind that is often denoted by

. It can be evaluated numerically as follows. First we use the binonial series

(40)

)2(sin21cos 2

0

22 )2(sin

4

k

dT

2sin

k

)2sin()1( ku

1

0222 1)(1(

4

uku

duT

)2,( kF

n

n

xn

nx

1 )2(42)12(311

11

2

022 sin1

4

k

dT

sinu

With to expand the integrand in (39). Then we integrate termwise using the tabulated integral formula

(41)

The final result is the formula

(42)for the period T of the nonlinear pendulum released from rest with

initial angle in terms of the liniearized period and

1sin 22 kx

)2(42)12(31

2sin

2

0

2

nndn

]))2(42)12(31(1[2 22

1

n

n

kn

nT

.......])642531()

4231()

21(1[ 624222

0

kkkT

)0(

20 T )2sin(k

In this example, we investigate what happens when the populations x and y still have to compete for the same finite resources, but each grows according to a logistic growth law so that

Where a, b, c, d and k are positive constants. We are concerned only and This system of equations (10.18) is nonlinear. In order to avoid unnecessary algebra,

we will consider the case where k=5, a=b, and d=c, so that (10.18) becomes

Where c>0. this means that in the absence of competition, the carrying capacity for both x and y is 5/c. why?

Our first step is to find the equilibrium points, which in the case of (10.19) occur when

That is when x=0 or y=5-cx, and y=0 or y=(5-x)/c. if , we have four equilibrium points, namely (0,0), (0,5/c),(5/c,0) and (5/(c+1), 5/(c+1). If c=1, we have an infinite number of

POPULATION MODEL

),('),('

bxdykyyaycxkxx

0x 0y

)5(')5('

xcyyyycxxx

0)5(0)5(

xcyyycxx

1c

equilibrium points, namely (0,0) and all points on the line y=5-x. in the two following examples we will consider c<1 and c>1.

Example of Population ModelWe illustrate the case c<1 by considering c

= 5/8, so that (10.19) becomes

10.20 with equilibrium points at (0,0), (0,8) and

(40/13,40/13).

If we use a nullcline analysis on (10.20), we find that there are four lines to consider: x = 0 and y = 5- 5x/8 (x-nullclines, which have vertical arrows in the phase plane), and y = 0 and y = 8(5-x)/5 (y nullclines, which have horizontal arrows).

The directions of the arrows are determined from (10.20) and are shown in Figure 10.21

We notice that the equilibrium points occur only where a nullcline with vertical arrows intersects one with horizontal arrows; namely, at (0,0), (0,8), (8,0) and (40/13,40,13). The points (5,0) and (0,5) are not equilibrium points.

In order to facilitate the discussion, we have labeled the regions I, II, III and IV. We notice that if an orbit enters region I, it cannot escape and appears to be attracted to the equilibrium point (0,8) as t increases.

Similarly, orbits entering region III cannot escape and appear to be attracted to the equilibrium point (8,0) as t increases.

Orbits in region II are either drawn to the equilibrium point (40/13, 40/13) or cross a nullcline into regions I or III. Thus, the equilibrium point (0,0) behaves like an unstable node, the equlibrium points (0,8) and (8,0) behaves like stable nodes, and the equilibrium point (40/13, 40/13) behaves like a saddle point.

This behavior can be confirm by linearizing the nonlinear system about each of its equilibrium points. The Jacobian matrix is

At the equilibrium point (0,0) this gives

which has determined D = 25 and trace T = 10 so D = This is a rare case , so the Linearization Theorem guarantees only that (0,0) is unstable because T > 0.

graph

We now turn to the equilibrium point (40/13, 40/13), where

Here the determinant is -65. and the trace is -50/13, so (40/13,40/13) behaves like a saddle

point.

We now turn to the equilibrium point (0,8), where

Here the determinant is 15 and the trace is -8,

so the equilibrium point (0,8) behaves like a stable node. By symmetry, a similar analysis applies to the equilibrium point (8,0).

We can also evaluate those orbits numerically, and these are shown in Figure 10.22 along with the direction field.

We notice that there appear to be straight-line orbits approaching (40/13,40/13) from both the left and the right. If we substitute into (10.20), we find m = 1, so y = x is a straight-line solution. The curve y = x is a separatrix.

So, what is the ultimate fate of these populations?

If initially the x population exceeds the y population, then the y population become extinct, and as t increases. If initially the y population, then the x population becomes extinct, and as t increases.

graph

• Method of variation of parameter which- in principle (that is, if the integrals that appear can be evaluated)-can always be used to find a particular solution of the nonhomogeneous linear differential equation• (18)provided that we already know the general equation (19)of the associated homogeneous equation (20)

Variation of parameters

Suppose that we replace the constant, or parameters, in the complementary function in (19) variables: functions of . We ask whether it is possible to choose these function in such a way that the combination (21)

is a particular solution of the nonhomogeneous Eq.(18). It turns out that this is always possible.

THEOREMLet and linearly independent solutions of . If the functions and are chosen so that their derivatives satisfy

Then will be the solution of the non homogeneous equations

Proof. Preparing to force to satisfy the differential equation into the differential equation, we compute the derivative

Since we have two functions to determine, we can impose a second condition on and namely, that

So that simplifies to Now compute from this simplified expression for . We find

Substitution into and regrouping terms gives

Since and satisfy the associated homogeneous equation, the first two terms equal zero. Thus for our second requirement.

Before illustrating the variation of parameter method with a more compelling example we’ll outline in general terms the steps that follow the application of Theorem 3.5. The equations

are the result of applying the theorem. We can solve this system of equations for and by elimination or by Cramer’s rule. Either way we find

 

The expression in the denominators is the same in both formulas and we can write it as the two-by-two determinant

Called the Wronskian determinant of Since we’re dealing with second order constant-coefficient equations, we can check directly that is never zero:(i) If if then (ii) If then

If the solutions and not independent, one of them would be a constant multiple order, so would be zero for all .

To complete the solution, integrate the formula for , to find and ,and then combine with , to get a particular solution

Example:

Use variation of parameters to find a particular solution for each of the following constant coefficient equations.a)

The equation has homogeneous solutions and Since and the two conditions of theorem 3.5 translate into

Adding these two equations gives so We integrate and take .Subtracting the second

equation from the first gives so

We integrate and take According to the

Theorem 3.5, we’ll gets a solution of the nonhomogeneous differential equation by writing

We recognize the term as a part of the general homogeneous solution, so we write the general solution of the nonhomogeneous equation as

b)

The complementary function is , so

Hence the equations are

we can easily solve these equations for

Hence we take

and

Thus our particular solution is

that is,

Integral Transform Formulas MethodAssume is continuous on an interval containing , and let be the solution of the

normalized constant coefficient equation

Satisfying = = 0. If , are the roots of the characteristic equation By that :

i) if , are real and unequal,

ii) If = ,

iii) If , = α ± iβ, β ≠ 0,

In each cases, we use an integral operator defined by

Where the variable is held fixed during the integration. The operator acts as an inverse integral operator to the differential operator with initial condition .

The function g has one of the three forms

Those forms are example of an integral transform of a type called a convolution integral that convert into .

Since the variable x that pccur in the upper limit and in the difference must be treated as a constant with respect to integration in the other variable, some care is necessary in doing the integration.

We can separate the x variable from the integration process. For example, the addition formula for the sine function enables us to write

EXAMPLE 1 OF INTEGRAL TRANSFORM FORMULASHere is an equation in which the right-hand side is itself a

solution of the associated homogeneous equation :

The homogenous solution is corresponding to a double root of the

equation The function .

Since the differential equation is normalized, we can compute the solution satisfying

from

Note that integration is with respect to t, so x is temporarily fixed. We find

This is the solution with .By Corollary 3.2 the general solution is

To satisfy initial conditions of the form we use the flexibility inherent in the

general homogeneous part of the solution .

Since the particular solution given by the integral transform formula satisfies , ,

we just need to determine values for and in so that the independent solutions

and satisfy and For this example, we have , , and .

Hence we solve andTo get and . Thus our solution is

EXAMPLE 2 OF INTEGRAL TRANSFORM FORMULASHere is an example involving trigonometric functions

The characteristic roots of the associated homogenous equation are , . The function is

then so the integrand in the integral transform formula is then .

The integration with respect to t follows from using formula

or more directly using the identity that proves formula

above :

Thus the particular solution satisfying is