Mathematic aspect

Consider the Sine-Gordon equation is

U_xx (x, y)-U_yy(x, y) s i n [ U ( x ,y ) ] 0 (s-G) (3.0.2) Let tx-y

Then

U( t )

x ( t ) t U y) (x,

U_x  



 

- U( t )

y ( t ) t U y) (x,

U_y   



 

U_xx (x, y) ²U( t )

^U_yy^{(x, y)} ^{2U( t )}

So we can rewrite equation (3.0.2) ²U( t )-²U( t )s i n [ U ( t ) ]0

Let h² -w² 1

We get U(t) sin[U(t)] 0 (3.0.3) Compare with equation (3.0.1). This is a equation describing ideal pendulum

Multiple U(t)

U(t)U(t) U(t)sin[U(t )]0 (3.0.4) Integrated by t

U²( t )-c o s U ( t )E₂ 2

1   where E₂  -1 is a constant Add 1

U²(t) [1 -c o s U ( t ) ]1 E₂ E₁ 2

1      (3.0.5)

We can see U (t) 2

1  ₂ as kinetic energy and [1-cosU(t)] as potential energy and this system has total energy E₁







2



 2



^u

0.5 1.0 1.5 2.0

P .E

Figure 3.3

The relation between potential energy P.E and angle u

From (3.0.5)

U²(t) [1-c o s U ( t ) ]E₁ 2

1   

U(t)  2E₁-2 [ 1-c o s U ( t ) ] (3.0.6)

1

c o s U ( t ) ]

-2 [ 1 -2E

(t) U

1

 

Integrated We get

^



0^{U ( t )} 1

d ] c o s -2 [ 1 -2E

t 1 

 (3.0.7)

3.1.0 Ideal Pendulum:

A system of pendulum with no friction is called ideal pendulum.

When no friction is present, the coefficient b vanish. We get the equation

0 ) ( gsin (t)

U  U t 







For convenience we suppose g 1

 

We can rewrite this equation U(t) sin U(t) 0 as first-order system in the usual manner by letting the variable v represent the angular velocity U(t) .The corresponding system is

Equilibrium points of this system are (nπ,0) for nZ

3.1.1 Apply Linearzied to analysis Ideal pendulum:

For system

V

3.1.2 Apply the Hamiltonian system to analysis Ideal pendulum:

Consider U(t)  sin[U(t)]  0

Let H(U,V)= -cosU 2

1

v

²

Because

V H V dt dU



 



U - H -cosU dt

dV



 



it is Hamilonian system with Hamiltonian function H(U,V)= -cosU 2

1

v

²

Figure 3.4 Level curve of H(U,V)

3.1.3 Apply the Jacobian elliptic function to solve the Ideal pendulum motion:

We want to solve (3.0.5) by Jacobian elliptic functions Consider ^



0^U(t)

1

d ] cos -2[1 -2E

t 1 

 with 1 E₂  E₁ i.e. ^



0^U(t) _

2

d 2cos 2E

t 1 

 (3.1.1)

a. If 0  E₁  2 , i.e. -1 E₂ 1

=



0^U(t)

^



²

U ( t )

0 2 2

dx x sin k -1

k

Let ysin x

^



²

sin U(t)

0 2 2 2 dy

y -1

1 y k -1 k 1

So

2 U ( t ) s i n k) , k

sn( t 

i.e. ,k) k sn( t sin 2

U(t)  1 ^-1 where

2 2E k 2

2 

 (3.1.5)

3.1.4 The graph of the Ideal pendulum motion:

1.



  2

 2 ^u

0.5 1.0 1.5 2.0 2.5 3.0

P .E

Figure 3.5

The relation between potential energy P.E and angle u with total energy

E₁ 3



  2

 2 ^u

 2

 1 1 2 v

Figure 3.6

The relation between vector v and angle u with E₁ 3

^ 2 ^   2 ^u

0.5 1.0 1.5 2.0

P .E

Figure 3.7

The relation between potential energy P.E and angle u with total energy

E₁ 2



  2

 2 ^u

 2

 1 1 2 v

Figure 3.8

The relation between vector v and angle u with E₁  2







2



 2



^u

0.5 1.0 1.5 2.0

P .E

Figure 3.9

The relation between potential energy P.E and angle u with total energy

E₁  0.5







2



 2



^u

 1.0

 0.5 0.5 1.0 v

Figure 3.10

The relation between vector v and angle u with E₁ 0.5

2. From (3.0.2)

U( t )s i n [ U ( t ) ]0 with t  hx  wy and h² -w² 1

Let h 2 w  3

a. Graph of the ideal pendulum motion with E₁ 1 (0  E₁  2)

Figure 3.11

b . Graph of the ideal pendulum motion with E₁ 2

Figure 3.12

c .Graph of the ideal pendulum motion with E₁  3 (2  E₁)

Figure 3.13

3.2.0 Pendulum motion with friction:

Recall that the second-order equation governing the motion of the pendulum is

0 sin[U(t)]

l (t) g U m (t) b

U    

Where b is the coefficient of damping m is the mass of the pendulum bob, g is the acceleration of gravity, and l is the length of the pendulum arm.

For convenies we let B m

b  and 1

l

g  .And rewrite this equation as first-order system

in the usual manner by letting the variable v represent the angular velocity U^(). The corresponding system is

V dt dU 

 dt

dV -BV- sinU

3.2.1 Apply Linearzied to analysis pendulum with friction:

For system

V dt dU 

-BV- sinU

1. The linearized system at the equilibrium point



n,0



for n is odd integer

_

































v u B 1

1 0

dt dv dt du

The Jacobian matrix of the system is J = _







 B 1

1

0 and its eigenvalues are

2 4 B B

-  ² 

 

There are saddle points at the equilibrium point



n,0



for n is odd integer

2. The linearized system at the equilibrium point



n,0



for n is even integer

_

































v u B 1

-1 0

dt dv dt du

The Jacobian matrix of the system is J = _







 B 1

-1

0 and its eigenvalues are 1

There are center points at the equilibrium point



ⁿ^,0



for n is even integer.

3.2.2 Apply the nullclines to analysis pendulum with friction:

For system

V dt dU 

 dt

dV -BV- sinU

U-nullcline is  

U,V



;V  0

 If V>0 then

0

dt

dU  This means vector filed “right”

If V<0 then

0 dt

dU  This means vector filed “left”

V-nullcline is  

U, V



;BV sinU 0

 If

BV sinU 0

then

0

dt

dV  This means vector filed “up”

If

BV sinU  0

then

0 dt

dV  This means vector filed “down”

Combine this two graphs

It obvious

1. There are saddle points at the equilibrium point



n,0



for n is odd integer.

2. There are center or spiral sink or spiral source points at the equilibrium point



ⁿ^,0



for n is even integer.

Chapter 4 Physical Applications of Elliptic functions

In the paragraphs below, we study five physical models‟ differential systems. An ideal whirling chain, and Duffing‟s Equation. The other three describe the motions of orbit planets. These parts mainly follow the reference [4].

4.1 Whirling Chain:

Consider a uniform length l of rope or chain, whose ends are fixed at point 0 and A and which is set rotating about the axis OA with constant angular velocity .

For ideal case

1. Gravity will be neglected

2. It will be assumed that the chain always lies in a plane through the axis of rotation.

We shall take O to be the origin of axes Ox, Oy, the x-axis lying along OA, and y-axis lying in the plane of the chain at same instant t (Fig.). Consider the motion of an element ds  PQ of the chain, where P and Q have coordinates (x, y), (x+dx, y+dy) respectively.

The forces acting on this element are tensions T and T+dT at its ends P and Q, and their lines of action are the tangents to the chain at these points; let these tangents make angles ,  d respectively with the x-axis.

Figure 4.1

Resolving the forces tangentially and normally, we obtain components (dT, Td ) respectively. The element moves around a circle of radius y with angular velocity  and its acceleration is accordingly ²y directed in the negative sense parallel to the y-axis.

We can now write down the tangential and normal components of the equation of motion thus;

) sin(

y ds

-dT   ²  , Td - ds²y cos() (4.1.1) Where is the mass per unit length of the chain.

Dividing these equations, we find that Which integrates to give the equation

) sec(

T

T  ₀  , (4.1.3)

T0 being the tension at the point B where   0.

Substituting for T in the second equation (4.1.1), we now deduce that

) This equation integrates to

)

and, after integration from , this leads to the equation

 

Reference to the standard form (1.3.79) now shows that

(y/b) We conclude the equation of chain is

)

y  . Clearly, therefore, it is necessary that

2K

ac/2   (4.1.10) and the equation of the chain can be written

(2Kx/a) sn b y

By eliminating and between equations (4.1.8), (4.1.9 and (4.1.10), we arrive at the equation

K ) k -b(1

ak

2  (4.1.11) Since K is a known function of k, this equation determine k and K when a, b are given.

 can then be found from equation (4.1.9) .

For example, if k=0.5, then K=1.6858 and, thus, a/b=2.53 and b² 4

 3

 .

Instead of b being specified, the length l of the chain may be given. This can be related to the other parameters, thus:

l ^



0^{a }

2]dx (dy/dx) [1

^



0^{a }

2 2

2cn (2Kx/a)dn (2Kx/a)] dx (2bK/a)

[1

^



0^{a  }

2 2

2/k )cn (2Kx/a)dn (2Kx/a)] dx (4k

[1

^



0^{a } 2

2) d n( 2 K x / a )-1 ] d x k

[ ( 2 /

dn ( u ) d u-a K

k

a ^2K

0 2

2



 

-a K k

2 a E

2

 (4.1.12) Where ^



0^K

2(u)du dn

E .With k=0.5, K=1.6858 as before, we read from the table E=1.4675 and hence, l=1.321a.

Remark11:

E (u, k) is a function defined by



 ^u

0

2(v, k)dv dn

k) E(u,

And ^{ }



0^K

2(v, k)dv dn

k) E(K,

E .

It can be saw a function of k.

4.2 Duffing’s Equation:

This is the equation governing the oscillations of mass attached to the end of a

spring whose tension (or compression) T is related to its extension x by an equation of the form

x3

x

T   (4.2.1)

 is always positive.

1. If  0, the spring obeys Hooke‟s law and the oscillations are simple harmonic.

2. If  0, the tension increases with the extension more rapidly than required by Hooke‟s law and the spring is said to be hard.

3. If  0, the tension increases less rapidly than required by the law and the spring is said to be soft.

By a suitable time choice of the unit of time, the equation of motion of the mass can be put into the form (Remark12)

0 x x

x  ³ 

 (4.2.2) which is the canonical form of Duffing’s Equation.

We want to solve this system.

1. The case of a hard spring, for which   0. Integrating to obtain t, we find



having referred to the standard form (1.3.80). Inversion now shows that

t}

with modulus given by indicating that, as the amplitude of the oscillation increases, the period decreases and the frequency therefore increases.

2. The case of a soft spring, for which   0. oscillation (x=0) and x is increasing. Thus, x=0 at t=0, and the equation for the time is



after reference to the standard form (1.3.79) , (to apply this result, we must assume

2 last equation, we obtain

t} where the modulus is determined by

2

Thus, the period of oscillation is given by

)

) a 8 (1 1 2

T     ² (4.2.14) to O( ), in agreement with (4.2.10). For a soft spring, therefore, the frequency of oscillation decreases as the amplitude increases.

Remark12:

By Newton‟s law

-ma T 

Where m is the mass and a is the acceleration of the mass So we get the equation

x -m -ma T x

x ³    

 Divide m

0 x m x m

x    ³ 





Let m and





    m

0 x x

x  ³ 



We note that   0 if   0 the spring is hard and   0 if  0the spring is soft.

4.3 Orbit motion:

4.3.1 Orbits under aμ /r

⁴

Law of Attraction:

Suppose a particle of unit mass is attracted towards a center O by a force  r⁴ , r and  being its polar coordinates at time t in the plane of motion. Then, since the particle‟s energy E and angular momentum h about O will be conserved, we can write down the equations of motion (Remark13)

E 3r -) r r ( 2 1

3 2 2

2   

^

 , r²^ h (4.3.1)

Putting r 1 u and eliminating t between these equations, we arrive at the equation

(u) f u

-u ) d

(du ²  ³  ²  

  , (4.3.2) where  3h²/2   3E 

This equation determines the polar equation of the orbit. Clearly,  0 (we ignore the case of rectilinear motion), but may take any real value. We shall always assume the sense of the motion to be such that increases (i.e., h>0)

Before solve this equation we should verify that there are five cases to consider:

Figure 4.2

1. If ,  0,then f (u) has one real zero at greater than and two complex zeros whose real parts are negative (since the sum of the zeros is  )

For   2  -1

10  5 5 10

u

1000

500 500

f



^u



1 1 2 3 4 u

5 10 15 20 25 30

f



^u



Figure 4.3a Figure 4.3b

For   2  -1

2. If ,  0,f (u) has a double zero at u=0 and a simple zero at u  For   2   0

 4  2 2 4

u

 150

 100

 50 50

f



^u



 1 1 2 3

u

 3

 2

 1 1 2 3 4 f



^u



Figure 4.4a Figure 4.4b

3. If,0 4³ 27 , f (u) has three real zeros u₁,u₂,u₃, satisfying



  





 _{2 3}

1 0 u 2 3 u

u .

1 1 2 3

u

 2

 1 1 2 3 4 5 f



^u



Figure 4.5 For

  2  1

4. If 4³ 27 , f (u) has a pair of coincident zeros at u  2 3 and a simple zero at

3 -u   .

For   2  32 27

1 1 2 3

u

1 1 2 3 4 5 f



^u



1.32 1.34 1.36 1.38 1.40

u 0.002

0.004 0.006 0.008

f



^u



Figure 4.6a Figure 4.6b

5. If  4³ 27 , f (u) has a real zero with negative u and two complex zeros with positive real parts.

 3  2  1 1 2 3 4

u

 40

 20 20

f



^u



Figure 4.7

For   2   2

Since (du d)² 0, by consideration of the sign of f(u) over the range of all positives of u, it is possible to establish the character of each of the possible orbits in these cases without further integration.

Case 1.

For   0, f (u) has one real zero at greater than and two complex zeros whose real parts are negative.

We get f (u) (u-a){(u b)²c²}

a, b, c being all positive and a . Clearly, we need u a to make f (u) positive.

Apply some method

2 1S S (u)

f  (4.3.3a) where (p q) [(u p) -(u -q) ]

2

S₁  1  ^-1  ^{2 2} (4.3.3b) S₂ (pq)^-1[(q b)(u p)²(p-b)(u -q)²] (4.3.3c) p and q are positive numbers given by

a -} c b) {(a

p  ²  ² q {(a b)² c²}a (4.3.3d) Then, integrating equation (4.3.2) we find



   





] q) -b)(u -(p p) b)(u ][(q q) -(u -p) [(u q) du (p 2

2 2

2 2

2 1

- 

 , (4.3.4)

We now make the substitution

p u

q -x u

  (4.3.5)





We have now arrived at a standard from and can make use of the result (1.3.80) to show that the modulus being given by

q The polar equation of the orbit now follows in the form

)

We deduce that, as  increases form 0, the trajectory spirals outward from the center, the mass being at its maximum distance2 (q -p) 1 afrom O when   2K  .

Thereafter, the orbit spirals inward and reaches the center again when  4K  . As before, negative values of  yield the mirror image trajectory, which is identical with the original.

Case 2.

(Use the substitution^u ^{1 v}.)We have ignored the constant of integration, since this can always be eliminated by suitable choice of the line  0. The polar equation of the orbit is now found to be Which is a cardioid. Thus, the particle recedes to a maximum distance 1 2 from the pole and then falls into the center of attraction.

Case 3.

Integrating equation (4.3.2). We get



Changing the variable by the transformation

0)

we reduce the integral to standard form, thus



 10  5 5 10 

0.1 0.2 0.3 0.4 0.5 r

Figure 4.8

2 u 1, u -1,

u₁  ₂  ₃ 

This implies, as  increases from 0, the trajectory spirals outward from the center, achieving maximum distance 1 u₃ , when  K  ; it then spirals back into the pole, arriving there when   2K  .

b. If 0 u  u₂ then xa and the standard form (1.3.84) is used to yield

x}

) u -(u { ns ) u -(u

- 2 ^-1 _{2 1}

1 3 2

1

-  

 (4.3.20)

whence

) ( )sn u -(u u u r

1 ₂

1 2

1  



 (4.3.21) The constant k and  take the same values as before.

 6  4  2 2 4 6 

 4

 2 2 4 r

Figure 4.9

2 u 1, u -1,

u₁  ₂  ₃ 

For this orbit equation (4.3.21),  must equal or exceed   ^{sn-1[ {-u}₁ ^(u₂ ^-u₁^)}]

to give positive values for u and r. When  has this limit value, r is infinite and further increase in  causes r to decrease to a minimum of 1 u₂ when   K  . If

 is increased again, r approaches infinity as   (2K -)/ . Thus, the trajectory

first approaches the center of attraction from infinity and later recedes again to an infinite distance

Case 4.

If  4³ 27 , then

3 2 -u

3 4 cosh u

-) 3 (u ) 3 2 -(u

du _{-12 -1}

2 1

-

 



 

 

 





. (4.3.25)

(Put u -2 3 1 v if u  2 3and 2 3-u 1 v if u  2 3) The equation of the orbit is accordingly

 10 5 5 10



0.5 0.6 0.7 r

Figure 4.10

2 cosh

1 -cosh 2 r 3

 





 , if u 2 3, (4.3.26a)

2 -cosh

1 cosh 2

3







  , if ^u  ² ³, (4.3.26b) In (4.3.26a)

For positive values of  , the orbit spirals outward from the center of attraction, approaching the circle r 3 2 asymptotically from inside circle.

For   2  32 27

In (4.3.26b)

For   cosh ^-12, the orbit spirals inward from infinity, approaching the circler

 2 3

r  asymptotically from outside the circle.

Reflecting these orbits in line   0, we obtain the orbits for negative values of  ,

which represent similar trajectories, traversed in the opposite sense, i.e., diverging inward and outward from he circular motion.

 6  4  2 2 4 6 

 2

 1 1 2 r

Figure 4.11

Case 5.

In case,

} c b) -a){(u (u

f(u)   ²  ² (4.3.27) Where a, b, c are all positive; all positive values of u are now admissible. The analysis proceeds as for Case1. , the signs of a and b being reversed. Thus

a } c b) {(a

p  ²  ²  q {(a b)² c²}-a (4.3.28) andp q.

The orbital equation is as given at (4.3.11)

) ( cn p q

) ( cn -r 1





  (4.3.29) and, as  increases from 0, r increases from 0 and the orbit spirals outward from the center. However, the equation cn() -q p now has a real root  for which r becomes infinite and the trajectory does not return to the center of attraction. Negative values of  provide a mirror image orbit along which the mass can fall into the center of attraction from an infinite distance.

Remark13:

Total energy E = kinetic energy + potential energy  K  U

^ ₂^{1m v2 }



^{F ds}

dr

r ) r r ( 1 2

1 ^r

4 2 2 2

2  





 





^{2 2 2} ₃

3r -) r r ( 2

1 

^

 



Angular momentum h r^m v^ (r,)1(r^,r^) The value of h is

r h  ²

4.3.2 Orbits under aμ /r

⁵

Law of Attraction:

If r⁵is the attraction per unit mass, the particle‟s potential energy in the field is

4r4

- and the equations of energy and angular momentum are

E 4r -) r r ( 2 1

4 2 2

2   

^

 , r²^h. (4.3.30) These equations lead to the equation

g(v) ) (u g u

-u ) d

(du ²  ⁴  ²   ² 

  (4.3.31)

determining the orbits, where

0

2h² 

 

 ,   4E , v u².

) u (v

g(v)  ² is a quadratic and its zeros distinguish five cases

1.   0, both zeros v₁, v₂ are real and v₁  0,v₂  . v  v₂ for g to be positive.

2

 10

 5 5 10

For   2   -1

2.  0, zeros are 0 and  . v on the orbit.

 2  1 1 2 3 4 5

v

 4

 2 2 4 g



^v



   

3. ²

4

0   1 , both zeros are real and satisfy 0 v₁  v₂  . v  v₁or v  v₂ on the orbit.

 2  1 1 2 3 4 5

v

 4

 2 2 4 g



^v



For   2

2

 1



4. ²

4 1

  , coincident zeros at  2

v 1 ; all positive values of v are admissible.

4 2 2 4

v

10

 5 5 10

g



^v



0.8 1.0 1.2 1.4

v

0.10

0.05 0.05 0.10

g



^v



For   2  1

5. ²

4 1

  , zeros are complex with positive real parts; all positive values of v are admissible.

 4  2 2 4

v

 10

 5 5 10

g



^v



For   2   2

Suppose increases with t (h>0) Case 1.

) v -)(u v -(u ) d

(du ²  ² ₁ ² ₂

  , (4.3.32) where

0 ) -4 (1 -2

v₁  1 ²   ,     -)  4

(1 2

v₂ 1 ² .

We must have u  v₂ . Integration leads to the orbital equation

) v u ( nc ) v

-(v_{2 1} ^{-12 -1} ₂

2 1

-  

 , (4.3.33) using the standard integral (1.3.87), the modulus being given by

1 2 2 1

v -v - v

k  . (4.3.34) We deduce that

) ( cn v r 1

2



 , (4.3.35) where



 ² (v₂-v₁) . (4.3.36) Thus, as  increases from-K  toK  , the particle spirals out from the center of attraction to a maximum distance ^{1 v}₂ and then falls back into the center along the mirror image spiral.

Case 2. which is a circle through the center of attraction. Thus, the particle first recedes from O along one semicircle and then falls back into O along the remaining semicircle.

Case 3. and then apply the standard integral (1.3.79) to give

)

attraction from infinity to a minimum distance ^{1 v}₂ and then recedes again to infinity. and apply the standard integral (1.3.84) . This show that

) and recedes from it to a maximum distance 1 v₂ ; thereafter, it falls back into pole

O . This is similar to Case 1.

Solving for u, we derive

) as possible equations for the orbit. The first equation corresponds to a trajectory which spirals outward from O, approaching the circle r (2 ) asymptotically. The alternative orbit spirals inward from infinity and approaches the same circle asymptotically.

Case 5.

For convenience in later calculations, we shall take the zeros of g(v) to be

, Transforming by

t the last equation becomes

)

We shall take p, q to be positive and, clearly, p q. Integration, using the standard form (1.3.89) , now yields

where the modulus is given by

2b a

(8ab)

k   (4.3.50) The polar equation of the orbit now follows in form

)

(Since a b  2, can be found in the interval (0, K).) We conclude that the particle spirals into the center of attraction form infinity as increases from- to.

4.3.3 Relativistic Planetary Orbits:

According to the general theory of relativity, a planet falling freely in the gravitational field of a spherically sun behaves as if it were governed by the Newtonian laws and were attracted to the sun by a non-Newtonian gravitational force of

) per unit mass, where h is the angular momentum per unit mass of the planet about the center of the sun and c is the velocity of light. Thus, its equations of motion are

E As in the previous sections, putting , we now arrive at the equation

2 determining the orbit.

For all planets in the solar system, the term 2u³ c²is always very small by comparison with the remaining terms in equation (4.3.55)

For convenience let

h2

v

u  (4.3.56) and then write equation (4.3.55) in the form

f(v) This circumstance that a planet‟s energy is insufficient to permit its escape from the sun‟s field requires that 0. Also 1, for otherwise(dv d)² 0 in the absence of the relativistic term. is very small and positive for all planets in the solar system, taking its largest value of 5.09 10^-8 for Mercury.

By graphing the function2v -v² v³, it is easy to establish that the zeros of f(v) (in equation(4.3.57)) are all real and satisfy the inequalities0v₁ 1v₂ 2v₃, v₃

being very large. Thus

v)

0.5 1.0 1.5 2.0

Integrating equation (4.3.57), we deduce that





Use of (1.3.87) now yields the result

} is the equation of the orbit.

Substituting the expansions (4.3.60), we calculate that

The modulus id determined by

) where lh²  . This represents the classical elliptical orbit with semi-latusrectum l and eccentricity e.

On the relativistic orbit given by equation (4.3.69), perihelion occurs when   K 

and, on the next occasion, when  3K  . Thus,  increases by 2K  between two passages through perihelion, instead of the increase of 2 expected from the classical theory. The advance of perihelion per revolution is accordingly



For Mercury,   5.09 10^-8 and its period is 88 days. Thus, the advance of perihelion per century predicted by the theory is 43; this is exactly the residual

advance remaining to be explained at the time the new theory was proposed by Einstein.

Remark14:

v3

v is excluded since this would lead to vas   ; i.e., the planet would fall into the sun.

Chapter5 Conclusion

The goal that people research differential equations is to describe some phenomena in the real world. However, most of the phenomena are hard to describe and trying to get the solutions of them is another hard work.

In this paper, we introduce one kind of function defined in complex number, which has some „good‟ properties, and these functions have powerful usage for getting the

solutions of integral equations.

To show the effects of these functions, we choose seven physical examples. By Newton‟s mechanics, general theory of relativity and some laws of motions, we use differential systems to describe these physical phenomena. And then, we use elliptic Jacobian functions to get the solutions of them. In the end, we compare these solutions with physical phenomena.

The last three examples (Orbits under aμ /r⁴ Law of Attraction, Orbits under aμ /r⁵ Law of Attraction, Relativistic Planetary Orbits) explain the evolution of the modern theories in Planetary Orbits. The first two examples obey Newton‟s law, and the last one is the model basing on the general theory of relativity proposed by Einstein. That is the reason for the difference of the estimation of Mercury‟s period.

.

Reference:

1. Paul Blanchard, Robert L. Devaney, Glen R. Hall, Jong-Eao Lee, Differential Equations: A Contemporary Approach, Thomson, 2007.

2. Tom M.Apostol, Modular functions and Dirichlet Series in Number Theory, (2nd ed. ), Springer-Verlag, New York, 1990.

3. K. Chandrasekharan, Elliptic Functions, Springer-Verlag, 1985.

4. Derek F. Lawden, Elliptic Functions and Applications, Springer-Verlag, 1989.

5. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, Cambride University Press, 1927.

6. L. M. Milne-Thomson, Jacobian Elliptic Function Tables, Dover Publications, Inc., New York, 1950.

7. Chih-Chien Hwang, NCTU, Master thesis, A Study of The Elliptic Functions and Their Application, 1988.

8. Muh-Jyh Tzeng, NCTU, Master thesis, Study on Sine-Gordon and Nonlinear Schrodinger equations.

9. Chin-Hui Chen, NCTU, Master thesis, Topic in force Oscillations.

10.Wen-Yu Chien, NCTU, Master thesis, The Exact Theory and Perturbation of the Pendulum Motions, 2009.

在文檔中 橢圓函數之理論與運用 (頁 54-0)