Asigment Ode
Transcript of Asigment Ode
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