The limit of Separation of variables

In chapter 2 we introduce the Separation of variables to solve the problem. But in the processes of solving the problem we can find the limit of Separation of variables.

1. The differential operator L must be separable.

Example:

).

( ) 0 , (

, 0 ) , ( ) , 0 (

0 , 0

, 0 2

x f x u

t u t u

t x u

u u_{tt xt xx}

=

=

=

>

<

<

= + +

π

π

(3.1)

We can not use the Separation of variables to solve the problem (3.1). Because if )

( ) ( ) ,

(x t X x T t

u = .

We substitute u into the differential equation, and divide u , this gives

" 0 ' 2 '

"

= +

+ X

X XT

T X T

T . (3.2)

By (3.2) we can know the equation in (3.1) is not sepsrable.

Then if the differential equation contains u_xt, then the problem can not be solved by Separation of variables.

2. All boundary conditions must be on lines x=constant. That is, the range of x must be bounded.

3. The linear operators defining the boundary conditions at x=constant must involve no partial derivatives of u with respect to t , and their coefficients must be independent of

t .

3.2 Sine- and cosine-transform v.s Fourier transform

In general we use sine- or cosine- transform to solve the half-infinite slab heat conduction problem. But we can also use Fourier transform to solve this problem if we extend f(x) to

∞

<

<

∞

− x as an odd or even function. Recalling the problem (2.12), we extend f(x) to

∞

By the solution (2.11) and (3.3)

. is the same as the solution (2.13).

3.3 Fourier Transform and Laplace Transform

In chapter 2, we use the Fourier transform and Laplace transform to solve the infinite-slab heat conduction problem and we gain two solutions (2.11) and (2.17). We must identify that tow solutions are the same.

With the Fourier transform

∫

Then we have

is the same as the solution (2.17)

For Fourier transform we need to integrate the function from -∞ to ∞, then we usually take Fourier transform into PDE with respect to x for fixed t because of x∈ . R

Similar to Laplace transform we need to integrate the function from 0 to ∞, then we take Laplace transform into PDE with respect to t for fixed x because of t>0.

III. Develop the function ∏

=

−

=

n

j

zj

z z

f

1

) ( )

(

to solve linear

parabolic PDE

We know that there are some differential equations whose solution space is in the Riemann surface. In this chapter, we want to compute the integrals

∫

_{γ f(}¹_z)^{dz, where γ} is in the Riemann surface of algebraic curve

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( . We will develop an algorithm such

that we can compute the integrals

∫

∏

=

−

γ dz

z z

n

j

j 1

) (

1 by Mathematica®5.

Before computing integrals, it is necessary to discuss the Riemann surface of

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( .

4.1. Fundamental introduction

For simplicity, we take f(z)= z to define a single-value and analytic function on the Riemann surface.

Now we let z∈C, and use polar form for z. That is,

(4.2) .

(4.1) ,

) 2 (θ π θ

= +

=

i i

re re z

Then by (4.1)

2 2

1 θ

ei

r

z = ,

and by (4.2)

2 2 ) 1

(2 2 ) 1 2 ( 2 2

1 θ

θ π π

θ i i

i

e r e

r e

r

z = = ⁺ = −

+

.

Therefore f(z)= z is a multi-valued function at each z∈C and is not analytic on C .

How to make f(z)= z to be a single-valued and analytic at every point on C ? Consider two cuts from 0 to −∞ (i.e.the negative real axis) and

Let

P₁=

{

C＼(−∞,0]｜ 1 ^{arg [ , )}

}

−

− +

∈

= π π

θ z

and

{

^C

P₂ = ＼(−∞,0]｜ 2 ^{arg [ ,3 )}

}

−

∈ +

= π π

θ z

as Fig. 4-1 shows.

Fig. 4-1 P1, P2 plane

Define

z z

f₁( )= , z∈P₁.

z z

f₂( )= , z∈P₂. Then

2 2 1 1

1

) (

iθ

e z z z

f = = is single-valued at each z∈P₁ and analytic on P , ₁

) ( )

( ^{2 2} ₁

1 2 2

1 2

2 2 1 2 2

1 2

1 1

1 2

z f e

z e e z e

z e z z z

f = = ⁱ = ⁱ = ^{i i} =− ⁱ =−

+ θ

π θ π

θ θ

is also single-valued at each z∈P₂ and analytic on P₂.

-π⁺ π⁺

π⁻ 3π⁻

P1 P2

(b) (a)

Let

{

1 =

D (−∞,0]｜argz=π

}

, as shown in Fig. 4-2.

Fig. 4-2 D₁ =

{

(−∞,0]｜argz=π

}

If z∈P₁ and arg tends to z π⁻，then ²

1 2 2

1 2 arg 2 1

z i e z e

z

z ⁱ

i z

=

≈

=

π

,

If z∈P₂ and argz tends to π⁺，then ²

1 2 2

1 2 arg 2 1

z i e z e

z

z ⁱ

i z

=

≈

=

π

, So z is continuous cross the cut (−∞,0] for z∈D₁.

We define

z z

f₃( )= ，z∈D₁, then

2 1 2 2

1

3(z) z z e iz

f = = ⁱ =

π

for z∈D₁ and analytic on D . ₁

Let

{

2 =

D (−∞,0]｜argz=3π

}

, as shown in Fig. 4-3.

π⁻ D1

π⁺ π

= z arg

Fig. 4-3 D₂ =

{

(−∞,0]｜argz=3π

}

If z∈P₂ and arg tends to z 3π⁻, then ²

1 2

3 2 1 2 arg 2 1

z i e

z e

z

z ⁱ

i z

−

=

≈

=

π

,

If z∈P₁ and arg tends to z −π⁺, then ²

) 1 ( 2 2 1 2 arg 2 1

z i e

z e

z

z ⁱ

i z

−

=

≈

= ⁻

π

, So z is continuous cross the cut (−∞,0] for z∈D₂.

We define

z z

f₄( )= , z∈D₂, then

) ( )

( ² ₃

1

4 z iz f z

f =− =− for z∈D₂ and analytic on D . ₂ According the discuss above, we can construct a single-valued function for z. We have the conclusion as the following:

Let R₂ =P₁∪P₂ ∪(−∞,0] and a function F:R₂ →C，define











∈

∈

∈

∈

=

2 4

1 3

2 2

1 1

, ) (

, ) (

, ) (

, ) ( )

(

D z z f

D z z f

P z z f

P z z f z F

then F(z) is single-valued and analytic at every point z∈R₂. Note that f₁(z)=−f₂(z) and f₃(z)=−f₄(z).

Moreover, F(z) is defined on a Riemann surface R₂ which is a generalization of the complex plane to a surface of more than one sheet such that a multi-valued function has only one value corresponding to each point on the surface.

3π⁻ D2

-π⁺ π

3 argz=

4.2. Riemann surface of the algebraic curve ∏

=

−

=

n

j

zj

z z

f

1

) ( )

(

with z

_j

∈ R

Consider

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( ，z_j∈R and z ＞₁ z ＞₂ z ＞…＞₃ z_n with n distance

branch points.

4.2.1 The cut structure of f(z)

Since f(z) is a two-valued function, we need branch cuts to define a single-valued and analytic function. But how can we construct branch cuts？

In this paper we by face the left direction to do cut explained. For convenience, let n=2 and n=3 to see what is going on？

First we check if there is any cut, for n=2 and z₂ =1 , z₁ =2, as shown in Fig. 4-4.

Fig. 4-4 The branch points are z₂ =1 and z₁=2

Consider −1∈(−∞,1), then we have



=−

−

=

−

− π

) π 2 arg(

) 1 1

arg( .



=−

−

=

−

− π

) π 3 arg(

) 2 1

arg( .

Taking − : π ²

) 1 2 ( 2 2 1 2 1

6 3

2 3

2⋅ − = =−

−

−π i

e . (4.3)

Taking π : ²

) 1 2 (2 2 1 2 1

6 3

2 3

2⋅ − = =−

−

i π

e . (4.4) Since (4.3) = (4.4), there is no cut in (−∞,1)

1 2

+ -

+ -

Consider (1,2) 2

3∈ , then we have

0 2) arg(1 ) 2 1

arg(3− = = ,



=−

−

=

− π

) π 2 arg( 1 ) 2 2

arg(3 .

Taking − : π ²

1 2)

( 2 1 2 1

4 1 2

1 2 1 2 1 2

1 ⋅ − = eⁱ =i

−π

. (4.5)

Taking π : ²

1 2)

( 2 1 2 1

4 1 2

1 2 1 2 1 2

1 ⋅ − = eⁱ =−i

π

. (4.6)

Since (4.5) ≠ (4.6), there is a cut in (1,2)

Hence we have the branch cut in [1,2], as shown in Fig. 4-5.

Fig. 4-5 The cut structure for n=2 branch points in horizontal

But we can use the simpler way to get branch cut. Recall Fig. 4-4. When crossing the cut even times in each line section, it will not change sign. When crossing the cut odd times in each line section will change sign, this implies the line section will form a branch cut. Hence we have the branch cut in [z₂,z₁]. The cut structure is shown in Fig.4-6.

Fig.4-6 The cut structure for four branch points in horizontal

Now given n branch points, If n is even, then the branch cuts are [z_n,z_n−1]、

1 2

Z2 Z1

] ,

[z_n−2 z_n−3 …… and [z₂,z₁]. If n is odd ， then the branch cuts are (−∞,z_n]、 ]

,

[z_n−1 z_n−2 …… and [z₂,z₁]. Show as Fig.4-7.

with n ∈ even

with n ∈ odd

Fig.4-7 branch cuts with n ∈ even and n ∈ odd

4.2.2 The algebraic and geometric structure for Riemann surface of horizontal cut For simplicity, we use n=3 to discuss the structure for Riemann surface of

∏

=

−

=

3

1

) ( )

(

j

zj

z z

f in horizontal cut.

(Ⅰ) Algebraic structure

As Fig.4-8 shows, (−∞,z₃]、[z₂,z₁] represent the cuts in this Riemann surface and 〝+〞,

〝–〞are defined as following(the initial edge with +, the terminal edge with –) :

Fig.4-8 The algebraic structure for three branch points in horizontal Z1

Z2 Z3 Z4

Zu-2 Zu-1

Zu

Zu-3 Z4

Zu-1 Zu-2 Z3 Z2 Z1

Zu

Z1

+ - -

+

Z Z2

Z3

Case 1 : If z∈ I⁺( + edge of sheet Ⅰ )

(Ⅱ) Geometric structure

After knowing the algebraic structure, we will discuss about how to construct a geometric

structure for Riemann surface of

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( . According to algebraic structure for

Riemann surface, we know that if n is even, then the branch cuts are [z_n,z_n−1]、 ]

,

[z_n−2 z_n−3 …… and [z₂,z₁]. It implies we have 1 2 −

n holes. If n is odd, then the branch

cuts are (−∞,z_n]、[z_n−1,z_n−2]…… and [z₂,z₁]. It implies we have 2

−1

n holes. And we

obtain one sheet with two edges in each cut by taken of counterclockwise which labeled the edge of lower- cut with + and the edge of upper- cut with –. Since there are two surface, one is, say sheet Ⅰwith arg f(z)∈[−π,π); another is, say sheet Ⅱ with argf(z)∈[π,3π).

By definition, the – edge of sheet Ⅰis joined to the + edge of sheet Ⅱ, and the + edge of sheet Ⅰ is joined to the – edge of sheet Ⅱ. Whenever crossing the cut, we pass from one sheet to the other sheet and the value is continuous which from our construction.

Note that f(z)∣_II = − f(z)∣_I and for f(z), supra-half-ball represents sheet Ⅰ, and infra-half-ball represents sheet Ⅱ .

We take n=3 to discuss the geometric structure for Riemann surface of

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( in horizontal cuts, as shown in Fig.4-9.

Fig.4-9 The geometric structure for Riemann surface with n=3 in horizontal cut

- -

-∞ + Z3 Z2 + Z1

- -

+ +

-∞ Z3 Z2 Z1

-∞ + Z3

-∞ Z3

-∞ - Z3 Z2 - Z1

Z2 + Z1

Z2 Z1

Ⅱ

Ⅰ

Ⅰ

Ⅱ

+ +

- -

-∞

Z3 Z2

Z1

Ⅱ

Ⅰ

(Ⅰ,+)=(Ⅱ,-)

(Ⅰ,-)=(Ⅱ,+)

(Ⅲ) Algebraic structure v.s Geometric structure

We also use n=3 to discuss. Before talking about the relation between algebraic structure and geometric structure, we need to denote something as the following :

(a) If the curve is drawn by solid line :

In algebraic structure, it means the curve is in sheet Ⅰ ;

In geometric structure, it means the curve is in the overhead Riemann surface.

(b) If the curve is drawn by dash line :

In algebraic structure, it means the curve is in sheet Ⅱ ;

In geometric structure, it means the curve is in the ventral Riemann surface.

We give some example to show that the curve in algebraic structure and its corresponding in geometric structure in Fig.4-10 to Fig.4-12.

Fig.4-10 The rule in algebraic structure and geometric structure

Fig.4-11 The rule in algebraic structure and geometric structure

-∞

Z3 Z2

Z1

Ⅱ

Ⅰ

-

+

Z1 Z2

Z3

-

+

Z1

Z3 Z2

Ⅱ

Ⅰ

-

+

Z1 Z2

Z3

-

+

-∞

Fig.4-12 The rule in algebraic structure and geometric structure

4.3. Riemann surface of the algebraic curve

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( withz_j∈C

In this section, we discuss the vertical cut structure. We define that (z, f(z)) belong to sheet Ⅰ if and only if

∏

=

−

∈

−

n

j

zj

z

1

2) 2 , [ 3 ) (

arg π π

, i.e. )

,2 2 [ 3 )

arg( π π

−

∈

−z_j

z for each j .

And f(z)∣_II = − f(z)∣_I . 4.3.1 The vertical cut structure

We consider

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( with z_j∈C for j=1,2,3,⋅⋅⋅⋅⋅,n and we by face the

up direction to do cut explained. The method of analyzing the vertical cut structure is the same as horizontal cut structure.

Then we can use the simpler way to get branch cut. We take n=4 with z₁ =i、z₂ =2i、 i

z₃ =3 and z₄ =4i, that is, z₁<z₂<z₃<…<z_n, as shown in Fig.4-13.

Fig.4-13 The cut appears at z ＜z_j for each z_j

Z1

Z3 Z2

Ⅱ

Ⅰ

-

+

Z1 Z2

Z3

-

+

-∞

Z₁ Z₂ Z₃

Z₄ 4i 3i 2i

ReZ ImZ

i +

+ +

+ -

- -

-

When crossing the cut even times in each line section, it will not change sign. When crossing the cut odd times in each line section will change sign，this implies the line section will form a branch cut. Hence we have the branch cuts in [z₄,z₃] and [z₂,z₁]. The cut structure is showed in Fig.4-14.

Fig.4-14 The cut structure for four branch points in vertical

4.3.2 The algebraic and geometric structure for Riemann surface of vertical cut

For simplicity, we use n=4 to discuss the structure for Riemann surface of

∏

=

−

=

4

1

) ( )

(

j

zj

z z

f in vertical cut. In the cut structure, we still depend on the

countclockwise to take〝+〞、〝–〞 sign. The definition of solid-line and dash-line are the same as horizontal cut case.

(a) (b)

Fig.4-15 The algebraic structure for four branch points in vertical + - + -

Z₂

Z₁ Z₃ Z₄

+ - + - Z₁ Z₃

Z₂ Z₄

Z Z1 i Z2

Z3

Z4 4i 3i 2i ImZ

ReZ

(Ⅰ) Algebraic structure

(Ⅱ) Geometric structure

The construct a geometric structure for Riemann surface of

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( is the

same as horizontal cuts.

By above example and illustration, we discusses the geometric structure for Riemann surface in vertical cuts. Show as Fig.4-16 (page 52).

4.4. The integrals over

^a

,

b

cycles

We want to evaluate

∫

_af(¹_z)^{dz and}

∫

_bf(¹_z)^dz for n branch points where a , b represent the a , b cycles over the Riemann surface of

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( with z_j∈C,

and develop an algorithm such that the integrals can be easily computed.

4.4.1 The a, b cycles over the Riemann surface of

∏

=

−

=

n

j

zj

z z

f

1

) ( )

(

(A) In horizontal cut :

Let z₁,z₂,⋅⋅⋅⋅⋅,z_n be the n branch points in x−axis with z_j∈C, then

∏

=

−

=

n

j

zj

z z

f

1

) ( )

( forms a N−holes Riemann surface where N∈Z⁺ ∪

{ }

0 and







= −

= −

even n n foe

N

odd n n for

N

2 2 2

1

Fig.4-16 The geometric structure for Riemann surface with n=4 in vertical cuts

4ί 3ί 2 ί ί

- -

+ +

4 ί 3 ί 2 ί ί

4 ί + 3 ί

4 ί 3 ί

4 ί - 3 ί 2 ί - ί 2 ί + ί

2 ί ί

Ⅱ

Ⅰ

Ⅰ

Ⅱ

+ +

- -

4 ί

3 ί 2 ί

ί

Ⅱ

Ⅰ

(Ⅰ,+)=(Ⅱ,-)

(Ⅰ,-)=(Ⅱ,+)

+ +

- -

So there are N a, b cycles. The Fig.4-17 represents the a, b cycles in the Riemann surface for n is even and the Fig.4-18 is the case for n is odd.

Fig.4-17 a , b cycles for horizontal cuts of even branch points

Fig.4-18 a , b cycles for horizontal cuts of odd branch points

(B) In vertical cut :

Let z₁,z₂,⋅⋅⋅⋅⋅,z_n∈C be the n branch points where n is even and z2k = z2k−1，

,2 , 2 ,

1 n

k = ⋅⋅⋅⋅ . There are 2

−2

n a, b cycles in the Riemann surface showed in Fig.4-19.

For a_k cycle, it encloses the cut z_2k−1z_2k , b_k cycle is passed through the cut z_2k−1z_2k from one sheet to the other.

Z_u

a3 a₂ a₁

b3

b₂

b₁ -

+

- +

- +

- Z_u−3 Z⁴ Z₃ Z2 + Z₁ Z_u−1 Z_u−2

Z_u

a3

a₂ a₁

b3

b₂

b₁ -

+

-

+ -

+

-

Z₄ Z₃ Z2 + Z₁

Z_u−1 Z_u−2

Fig.4-19 a , b cycles for vertical cuts

Let z₁,z₂,⋅⋅⋅⋅⋅,z_n∈C be the n branch points where n is even and z2k = z2k−1,

,2 , 2 ,

1 n

k = ⋅⋅⋅⋅ . There are 2

−2

n a , b cycles in the Riemann surface shown in Fig.4-20.

Fig.4-20 a, b cycles for vertical cuts + -

+ -

+ - + -

Z₁ Z_n−1

Z_2k Z_2k−1

Z₄

Z₃ Z₂ Z_n

a_k

a₂

a₁ b₁

b₂ b_k

Z_n−1

Z_n + -

Z₃ Z₁

+ -

+ -

+ -

b₁ b₂

b_k

Z_2k Z_2k−1

a₁ a₂

a_k

Z₂ Z₄

4.4.2 About 〝〝〝〝 Mathematica 〞〞〞〞 and how to modify

All programs in this paper are run by Mathematica®5. But we can not compute directly, before computing we need to give some adjustments. Since Mathematica®5 reads argument of any complex number in (−π,π] only, then it just gives right answer in sheet Ⅰ in horizontal cuts ( expect at the argument −π ).

Consider the branch points z_j, j=1,2,…,n . In horizontal cut structure we define

) , [ )

arg(z−z_j ∈ −π π , for z∈C. In vertical cut structure we define ) ,2 2 [ 3 )

arg( π π

−

∈

−z_j

z ,

for z∈C. But in Mathematica®5, it defines arg(z−z_j)∈(−π,π], for z∈C. Then before

computing the integral we must modify the function so that we can get the correct value.

In horizontal cut structure the value of z−z_j in our Theory and in Mathematica®5 are

different at a point z with arg(z−z_j)=−π , so we must modify the function z−z_j at π

−

=

− )

arg(z z_j . But if point z with arg(z−z_j)=−π is only a point on the contour r, then

it can not influence the value of

∫

_r ^z−zjdz so that we need not modify the function zj

z− .

In vertical cut structure the value of z−z_j in our Theory and in Mathematica®5 are

different at some points z with , )

2 [ 3 )

arg( π π

−

−

∈

−z_j

z so that we must modify the function

zj

z− at , )

2 [ 3 )

arg( π π

−

−

∈

−z_j

z .

Besides, the askew cut structure is the same as horizontal cut and vertical cut structure. So

we define )

4 ,9 [4 )

arg( π π

∈

−z_j

z , for z∈C. It implies that we must modify the function

zj

z− at )

4 ,9 ( )

arg( π

π

∈

−z_j

z .

4.4.3 Evaluation of

∫

_af(¹_z)^{dz and}

∫

_bf(¹_z)^dz

In this section we give three examples about the horizontal cut, vertical cut and askew cut.

In the three examples we try to modify the function f(z).

Example 4.1 :

Let n=6, and z₁=4、z₂ =3、z₃ =2、z₄ =1、z₅ =−1 and z₆ =−2 are six branch points , as shown in Fig.4-21.

If

∏

=

−

=

6

1

2 1

) ( ) (

j

zj

z z

f ，then ?

) (

1 =

∫

_{rf z} ^{dz where r=a,b cycles.}

We use Mathematica®5 to compute the integral.

Fig.4-21 a, cycles for six branch points in horizontal cut b

(i) For a cycle :

Fig.4-22 a a, ^∗ cycles for six branch points in horizontal cut -

+

-

+ -

+

-2 O

-1 1 2 3 4

a a*

b -

+

- + -

-2 O +

-1 1 2 3 4

a

We know that

∫

_a ^f(z)dz=

∫

_a*^f(z)dz, we only compute the integral along a^* path.

For the equivalent path a^* :

Since arg(z−z_j)=−π is not the valid range in Mathematica®5, we must modify f(z), as shown in Table 4.1. ( M means the value of f(z) in Mathematica®5.)

Value of f(z)

Branch points (1,0) to (2,0) (2,0) to (1,0)

z 1 −M +M

z 2 −M +M

z 3 −M +M

z4 +M +M

z 5 +M +M

z 6 +M +M

Sheet Ⅰ or sheet Ⅱ ( Sheet Ⅰ) +M ( Sheet Ⅰ) +M

Total −M +M

Table 4.1 we must modify the valve of f(z) for a^* cycle in Mathematica®5.

By Mathematica®5,

2 1 2

49

1 2 1

1 1 1

2 3.3819 10 1.13022

( )dz ( )dz ( )dz i

f z f z f z

−

∫

+

∫

= −

∫

= × − − ^.

Therefore the integral over a cycle is ₁

1 1 49

3.3819 10 1.13022

( ) ( )

a dz a dz i

f z ^∗ f z

= = × − −

∫ ∫



^.

(ii) For b cycle :

Fig.4-23 ,b b^∗ cycles for six branch points in horizontal cut

For the equivalent path b^* :

Since the interval (−1,1) and (2,3) have no cut, so solid line in sheet Ⅰimplies + sign and dash line in sheet Ⅱ implies – sign . since arg(z−z_j)=−π is not the valid range in Mathematica®5, we get the Table 4.2.

Value of f(z) Branch points

(1,0) to (2,0) (2,0) to (1,0)

z1 −M +M

z2 −M +M

z 3 −M +M

z4 +M +M

z 5 +M +M

z 6 +M +M

Sheet Ⅰ or sheet Ⅱ (Sheet Ⅰ) +M (Sheet Ⅱ) −M

Total −M −M

Table 4.2 we must modify the valve of f(z) for b^* cycle in Mathematica®5.

- +

- + -

-2 O +

-1 1 2 3 4

b b*

By Mathematica®5,

1 3 1 2 2 1

1 2 1 3 1 2

1 1 1 1 1 1

( )dz ( )dz ( )dz ( )dz ( )dz ( )dz

f z f z f z f z f z f z

−

− + − − − −

∫ ∫ ∫ ∫ ∫ ∫

49i 10 77621 . 3 0760776 .

0 + × ⁻

−

= .

Therefore the integral over b cycle is

1 1 49

0.0760776 3.77621 10

( ) ( )

b dz b dz i

f z ^∗ f z

= = − + × −

∫ ∫



^.

Example 4.2 :

Let n=6，and z₁ =1+2i、z₂ =1、z₃ =3i、z₄ =i、z₅ =−1+3i and z₆ = 1− +i are six branch points, as shown in Fig.4.24

If

∏

=

−

=

6

1

2 1

) ( ) (

j

zj

z z

f , then ?

) (

1 =

∫

_r ^dz

z

f ^{where r=a,b cycles.}

Fig.4.24 a, cycles for six branch points in vertical cut b

We use Mathematica®5 to compute the integral. Note that in vertical cut structure we must

modify the function z−z_j at , )

2 [ 3 )

arg( π π

−

−

∈

−z_j

z .

(i) For a cycle:

For the equivalent path a^* : shown in Fig. 4.25 b

+ - + -

+ -

- O 1

1+2ί a

ί -1+3 3

-1+ί

Fig.4.25 a* cycle for example 4.2

We can get the Table 4.4. For example, , ) 2 [ 3 )

arg( ₁ π π

−

−

∈

− z

z for z along the path (0,3i)

to (0.2i) so that we must to modify z−z₁ .

Value of f(z)

Branch points (0,3i) to (0.2i) (0,2i) to (0,i) (0,2i) to (0,3i) (0,i) to (0,2i)

z 1 −M +M −M +M

z 2 −M −M −M −M

z 3 +M +M +M +M

z4 _{−M −M +M +M}

z 5 +M +M +M +M

z 6 +M +M +M +M

Sheet Ⅰ or

sheet Ⅱ ^{+M +M +M +M}

Total −M +M +M −M

Table 4.4 we must modify the valve of f(z) for a^* cycle in Mathematica®5.

By Mathematica®5,

2 2 3

3 2 2

1 1 1 1 1

( ) ( ) ( ) ( ) ( )

i i i i

a dz i dz i dz i dz i dz

f z f z f z f z f z

∗ = − + − +

∫ ∫ ∫ ∫ ∫

=1.38321−2.33762i Therefore the integral over a cycle is

1 1

1.38321 2.33762

( ) ( )

a dz a dz i

f z = ^∗ f z = −

∫ ∫



^.

1 -1+3ί

O

+ - + -

+ -

1+2ί

ί 3ί

-1+ί

a*

(ii) For b cycle:

For the equivalent path b^* :as shown in Fig.4.26

Fig.4.26 b^* cycle for example 4.2

We get the Table 4.6. For example, , )

2 [ 3 )

arg( ₁ π π

−

−

∈

− z

z for z along -1+ί to 1 so that we

must to modify z−z₁ .

Value of f(z)

Branch points -1+i to 1 1 to -1+i

z1 +M +M

z2 −M −M

z 3 +M +M

z 4 +M +M

z 5 +M +M

z 6 +M +M

Sheet Ⅰ or sheet Ⅱ +M −M

Total −M +M

Table 4.6 we must modify the valve of f(z) for b^* cycle in Mathematica®5.

By Mathematica®5 ,

1 1 1

1 1 1

1 1 1 1

2 0.590344 1.16143

( ) ( ) ( ) ( )

i i

b dz i dz dz dz i

f z f z f z f z

∗

− + − +

= − − + + = = −

∫ ∫ ∫ ∫

^.

Therefore the integral over b cycle is

0

+ - + - + -

1+2ί

ί 3ί

-1+ί -1+3ί

1

1 1

0.590344 1.16143

( ) ( )

b dz b dz i

f z = ^∗ f z = −

∫ ∫



^.

Example 3 :

Let n=4, z₁ =1、z₂ =i、z₃ =−i and z₄ =−1 are four branch points form a askew cut as shown in Fig.4.27.

If

∏

=

−

=

4

1

2 1

) ( ) (

j

zj

z z

f ，then ?

) (

1 =

∫

_r ^dz

z

f where r=a,b cycles

Note that in askew cut structure we must modify the function z−z_j at

4 ) ,9 ( )

arg( π

π

∈

−z_j

z .

Fig.4.27 a, cycles for four branch points in askew cut b

(i) For a cycle :

For the equivalent path a : as shown in Fig.4.28 ^*

Fig.4.28 a ^* cycle for example 4.3 -ί

O a^*

-1

1 +

-

+ - ί -ί

O

a ί

-1

b + -

+ -

We get the Table 4.8. For example, ) 4 ,9 ( )

arg( ₁ π

π

∉

− z

z for z along -1 to i and i to -1 so that

we must not modify z−z₁.

Value of f(z)

Branch points -1 to i i to -1

z =1 1 +M +M

z =i 2 −M −M

z =-i 3 +M +M

z4=-1 −M +M

Sheet Ⅰ or sheet Ⅱ +M +M

Toatl +M −M

Table 4.8 we must modify the valve of f(z) for a cycle in Mathematica®5. ^*

By Mathematica®5,

*

1

1 1

1 1 1 1

2 2.62206 2.62206

( ) ( ) ( ) ( )

i i

a dz dz i dz dz i

f z f z f z f z

−

− −

= − = = −

∫ ∫ ∫ ∫



^.

Therefore the integral over a cycle is

1 1

2.62206 2.62206

( ) ( )

a dz a dz i

f z = ^∗ f z = −

∫ ∫



^.

(ii) For b cycle :

For the equivalent path b : as shown in Fig. 4.29 ^*

Fig.4.29 b cycle for example 4.3 ^* -ί

-1 O

1 +

-

+ - ί

b^*

We get the Table 4.10. For example, ) 4 ,9 ( )

arg( ₁ π

π

∈

− z

z for z along the path -1 to 1 so that

we must modify z−z₁ .

Value of f(z)

Branch points -1 to 1 1 to -1

z =1 1 −M +M

z =i 2 −M −M

z =-i 3 +M +M

z4=-1 −M −M

Sheet Ⅰ or sheet Ⅱ +M −M

Toatl −M −M

Table 4.10 we must modify the valve of f(z) for b ^* cycle in Mathematica®5.

By Mathematica®5,

*

1 1

1 1

1 1 1

( ) ( ) ( ) 0

b dz dz dz

f z f z f z

−

= − − − =

∫ ∫ ∫



^.

Therefore the integral over b cycle is

*

1 1

( ) ( ) 0

b dz b dz

f z = f z =

∫ ∫

 

^.

4.5 Application for Riemann integral

Recalling the heat conduction problem in section 2.5 we use Laplace and Fourier transformation to solve it. But we want to solve the integral

. ) ( 1 )

4 ( ) 1 , (

-lim

^{e e d f y dy}

t i x

u ^{x y t}

iL s

iL L s

σ σ π

σ − σ

− +

−

∞

∞ →∞

∫

∫

=

Since the path is from s− to il s+il, we must not modify the integrate in Mathematica®5.

Let f y( )=y², and by Mathematica®5 we can get

. 1 )

4 ( ) 1 1 , (

-

lim

^{e e d ydy x}

x i

u ^{x y}

iL s

iL L s

=

= ^{− −}

+

−

∞

∞ →∞

∫

∫

^σ

π σ

σ σ

(4.11)

We also compute the integrate in (2.17), and we can get that

x ydy e

x u

y x

=

=

∫

^∞

∞

−

−

−

2 ) 1 1 ,

( ^{( ) 4}

2

π ^,

is the same as (4.11). As shown in Table. 4.11, u

∼

is the value of u which is computed by Mathematica®5.

t 1 2 3 4 5 6 7 8 9 10

u

∼ 2

2+x 4+x² 6+x² 8+x² 10+x² 12+x² 14+x² 16+x² 18+x² 20+x² u 2+x² 4+x² 6+x² 8+x² 10+x² 12+x² 14+x² 16+x² 18+x² 20+x²

Table 4.11 compare u with u

∼

References

1. H. F. Weinberger, A First Course in Partial Differential Equations with Complex Variables and Transform Methods, 1919.

2. Kevorkian J., Partial Differential Equations : analytical solution techniques, 1989.

3. Tayler, Alan B., Mathematical models in applied mechanics, 2001.

4. DuChateau, Paul./Zachmann, David W., Applied partial differential equations, 2002.

5. Ockendon, J. R., Applied partial differential equations, 1999.

6. Su-Chuan Yao, NCTU, Master thesis, Integral Evaluation on Three-sheeted Riemann Surface of Genus N of Type I, Taiwan, 2000.

7. Hsiao, Yu-Wei, NCTU, Master thesis, Integral Evaluations on Three-sheeted Riemann Surfaces of Genus N of Type II, Taiwan, 2000.

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