Finance

∂
‧ 2

is also called a convection-diffusion or reaction-diffusion equations depending whether

=0

f or v
=0

respectively.

1.2.2 Finance

Suppose we consider an option, which is contract giving its holder the right (but not the obligation) to buy (or sell) some asset, such as a number of stock-market shares, at some specified time, say T, when the exercise price, a previously agreed sum of money E, is paid for the asset. Suppose the asset is a share which is expected to gain in value in 0<t<T, but whose price is subject to unpredictable factor. If we hold an options, we can set up a

“portfolio” of the option to protect ourselves against unpredictability. To do this we need to assess the value V( ts, ) of the option to buy a share at time T as a function of current time t and the asset value S. We suppose we have a cash balance M, and we hold a number ∆ , which may vary in time, of the assets. Thus, the portfolio value is P =M +S∆+V . The cash balance accrues interest at a rate r ; it also changes when we buy or sell assets, in a short time dt, we receive rMdt in interest and spend − Sd on assets. In the same time, t he asset ∆ price changes by dS and the option value by dV , so the overall change is

dV dS rMdt dV

dS Sd

Sd rMdt

dP= − ∆+ ∆+∆ + = +∆ + .

Now we suppose the instantaneous “rate-of-return” on the asset varies randomly.

dx S dt

dS =µ +σ , (1.2)

where µ is a deterministic “growth rate” for the asset; dx is a small normal random variable of mean zero and variance dt , and σ is a parameter which measures how “volatile” the share price is.

By Taylor’ Expansion series for dV , than the risk-free interest rate r for a risk-free portfolio, so

rPdt

Hence, we derive the Black-Scholes equation

)

I. Methods of solving the linear parabolic PDE

In this chapter we introduce six methods to solve the heat conduction problem.

2.1 Separation of Variables

We first consider the heat conduction problem

2

2 0, 0 , 0 (0, ) ( , ) 0, 0

( , 0) ( ). 0

u u

x t

t x

u t u t t

u x f x x

π π

π

∂ ∂

− = < < >

∂ ∂

= = ≥

= ≤ ≤

(2.1)

We look for a specific type of solution; namely, a product of a function of x only and a function of t only

) ( ) ( ) ,

(x t X x T t

u = .

We substitute the function u into the differential equation, and divide u . This gives 0

'' '

=

− X

k X T

T ,

or

X kX T

T^{' ''}

= . (2.2) The left-hand side of this equation depends only upon t . The right-hand side is independent of t . We say that heat equation (in 0<x<l) is separable.

If we take the partial derivative with respect to t of both sides of the separated equation, we find that

0 ] [

'

T = T dt

d .

It follows that

λ

− T = T^'

, where λ is a constant. Then by (2.2) we have

λ

− X = k X

''

.

Thus u(x,t)= X(x)T(t) is a solution of heat equation if and only if X and T satisfy the two

ordinary equations

, 0

, 0

' ''

= +

= +

T T

X kX

λ

λ (2.3)

for some constant λ.

We can solve the two ordinary differential equations (2.3) to obtain particular solutions of the partial differential equation (2.1). For each value of λ the equation kX ^''+ Xλ =0 has two linearly independent solutions. The families of solutions of (2.1) are given by

cos , sin , 0

1, , 0 , . 0

t t

t x t x

k k

e x e x for

k k

x for

e e for

λ λ

λ λ

λ λ

λ λ

λ λ λ

− −

− −

− + − −

>

=

<

Now, consider the boundary conditions. Since we wish to have u=0 for x=0 and x=1, we only consider those solutions of the first equation (2.3) which also satisfy these conditions.

We must have

'' 0, 0 (0) ( ) 0.

kX X x

X X

λ π

π

+ = < <

= =

This homogeneous problem always has the trivial solution X=0, but we are interested in cases where this is not the only solution.

Case1. λ> 0,

kx b k x a x

X λ λ

cos sin

)

( = + .

X(0)= 0 tells us that

kx a x

X λ

sin )

( = ,

and X(π) = 0 gives

sin 0

a k

λπ = .

Then X need not be identically zero if and only if

sin 0 k

λπ = ,

ie.

, n 1, 2, 3 ...

k n

λπ = π =

That is,

2 , n 1, 2, 3 ...

n n k

λ = =

These value λ_n are called the eigenvalues of the problem, and the functions ( ) sin , n 1, 2, 3,

Xn x = nx = …

are the corresponding eigenfunctions.

Case2. λ= 0,

X(x) = a + b x, We know

X(0) = a + b*0 = 0 ⇒ a = 0 ⇒ X(x) = bx, and

X(π) = b*π= 0 ⇒ b= 0.

This gives the trivial solution X= 0.

Case3. λ< 0,

x

k x b

e a x

X ^k

-

-e )

(

λ λ

+

=

−

. We have

0 1

* 1

* ) 0

( =a +b =

X ,

and

( ) ^{k k} 0

X a e b e

λ λ

π π

π

− −

= + − = ,

but these can not find a or b.

We know

sinh , cosh

2 2

x x x x

k k k k

e e e e

x x

k k

λ λ λ λ

λ λ

− − − −

− −

− − − +

= = ,

so that a general solution

k x B

k x A

x

X λ −λ

− +

= sinh cosh

)

( .

Therefore

k x A

x X B

B A B

A

X −λ

⇒ =

⇒ =

= +

= +

= sinh0 cosh0 *0 *1 0 0 ( ) sinh )

0

( .

And

( ) sinh 0 0 0

2

k k

e e

X A A A

k

λ λ

π π

π λπ

− −

− − −

= = = = ⇒ = .

This also gives the trivial solution X= 0.

Having found a sequence of values of λ, we can look at the corresponding functions T(t).

These are easily seen to be multiples of e^{−n kt2} We have constructed the particular solutions

- 2

( , ) sin e ^{n kt} u x tn = nx .

Which satisfy all the homogeneous conditions of the problem (2.1). The same is true of any finite linear combination. We attempt to represent the solution of (2.1) as an infinite series in the functions u_n :

2

1

( , ) _n sin ^{n kt}

n

u x t a nx e

∞

−

=

=

∑

. (2.4) We need to determine the coefficients a_n such that u( tx, ) satisfies the initial condition

) ( ) 0 ,

(x f x

u = . Thus we require

1

( , 0) ( ) _n sin

n

u x f x a nx

∞

=

= =

∑

^.

But the right side is just the Fourier sine series of the function f(x) on the interval ( l 0, ).

Therefore the coefficients a are the Fourier coefficients given by _n

0

2 ( ) sin , 1, 2,...

an f x nx dx n

π

=π

∫

= (2.5)

Therefore we have obtained a solution to (2.1) given by the infinite series (2.4) where the coefficients a are given by (2.5). _n

Substituting the expression for the a_n into the solution formula (2.4) allows us to write the solution as

2

2

1 0

0 1

0

( , ) (2 ( ) sin ) sin

(2 sin sin ) ( )

( , , ) ( ) .

n kt

n

n kt

n

u x t f n d e nx

e n nx f d

K x t f d

π

π

π

ξ ξ ξ

π

ξ ξ ξ

π

ξ ξ ξ

∞ −

=

∞

−

=

=

=

=

∑ ∫

∫ ∑

∫

Remark:

If is defined and integrable on the interval [−π,π], then its Fourier series of the form

∑

^∞

=

+ +

1

0 cos sin

2 _n aⁿ nx bⁿ nx

a ,

where





, 2 , 1 , sin ) 1 (

, 2 , 1 , 0 , cos ) 1 (

=

=

=

=

∫

∫

−

−

n dx nx x f b

n dx nx x f a

n n

π

π π

π

π π

Example 2.1 :(Using Separation of Variables)

Then the solution is

∑

^∞

2.2 Finite Fourier Transformation and nonhomogeneous problem

Recall the heat conduction problem (2.6). The solution of the problem is given by

∑

^∞

We shall now treat the corresponding nonhomogeneous problem

2

by expanding the solution in a Fourier series in terms of the same set of functions.

To solve the above nonhomogeneous problem, we expand the solution in a Fourier sine

The set of sine coefficients

∫

which is called undetermined coefficient and is also called the finite sine transform of u( tx, ). If u_xx is continuous, its finite sine transform is given by

Taking the finite sine transform of both sides of (2.7) leads to the equation )

0 differential problem (2.8), (2.9). We now solve this problem

. sine series

∑ ∫

^∞ definition (2.9) of B_n(t), and formally interchange integration and summation. This gives

,

(2.7), we must simply replace (2.10) by

Example 2.2 : ( Using Finite Fourier transformation)

sin3 , 0 , 0

Then the solution is

.

2.3 Fourier transforms

First let us begin with functions of one variable. The Fourier transform of a function

If f is absolutely integrable, ie.

∫

^∞

∞

of the Fourier transform is the simple form of the inversion formula, or inverse transform. It is

∫

formulas for Fourier transform:

1. F[f'](w)=−iwF[f](w)

[ ( ) ( )

The convolution theorem for Fourier transform

If f(x) and g(x) are both absolutely integrable and square integrable, then problem.

.

∂ are continuous in x and t, and absolutely integrable in x, uniformly in t. Then

u and

. problem

)

Then we solve the ordinary differential problem

,

whose solution is

t Then the solution formula is

∫

Example 2.3 : (Using Fourier transformation to solve the infinite-slab heat conduction problem)

.

4

Then the solution is

.

2.4 Sine and Cosine transforms

If f(x) is given for 0< x<∞, its sine transform is defined as

Hence, the inverse theorem becomes

∫

The sine transform is clearly an odd function of w. Hence the integral on the right becomes

∫

^{L wxFs f dw}

0

] [ sin

4 . Thus the inversion theorem is

∫

( ) 2 _s[ [ ]( )]( )_s

f x F F f w x

=π ,

for a function f(x) defined for 0< x<∞. Similarly, we can define the cosine transform

∫

∞

≡

0

cos ) ( ]

[f f x wxdx

F_c ,

for a function f(x) defined for 0< x<∞.

If we extend f(x) to −∞< x<∞ as an even function, ie. f(−x)= f(x), we have ]

[ 2 )

(w F f

f^ = _c .

The function F_c[ f] is even in w. Hence the inversion theorem becomes

∫

∞

=

0

] [ 2 cos

)

(x wxF f dw

f _c

π ^,

or

( ) 2 _c[ [ ]( )]( )_c

f x F F f w x

=π ,

for a function f(x) defined for 0< x<∞.

Sine and cosine transform are often useful in treating problems with boundary conditions only at x=0. And we can note that

], [ )

0 ( ' ] '' [

], [ )

0 ( ] '' [

2 2

f F w f

f F

f F w w f f F

c c

s s

−

−

=

−

=

provided f(x) and f'(x)→0 as x→∞. Thus sine transform is particularly useful when )

0 (

f is given, while the cosine transform is useful when f'(0) is known.

Example 2.4 : (Using sine or cosine transformation to solve the heat conduction problem in a half-infinite slab)

.

continuous and absolutely integrable in x for each fixed t. we have

∫

The problem coincides with the solution of the heat conduction in an infinite slab, provided we extend f(x) to −∞< x<∞ as an odd function. The corresponding solution u( tx, ) of the problem (2.12) is then also odd at x=0, and hence u(0,t)=0

2.5 Laplace transforms

We consider a function f(x) which vanishes for negative values of x:

0 0 )

(x = for x<

f .

Then if e^−s1xf(x) is absolutely integrable, so is e^−sxf(x) for s≥s₁. It follows that the

Fourier transform f(ξ)



is analytic in a half-plane lnξ >s₁. We define the Laplace transform

∫

∞

≡ − 0

) ( )

](

[f s e f x dx

L ^sx ,

or

) ](

[ ) ](

[f s F f is

L ≡ .

By integration by parts we find that

).

0 ( ' ) 0 ( ] [ ] '' [

), 0 ( ] [ )

( '

] ' [

2 0

f sf f L s f L

f f sL dx x f e f

L ^sx

−

−

=

−

=

=

∫

^{∞ −}

The convolution theorem

] [ ] [ ]

[f g L f L g

L ∗ = ⋅

follows from that for the Fourier transform.

By inversion theorem for the Fourier transform, we can find that the inverse formula for the Laplace transform is

+

∫

∞ −

→

=

iL s

iL s

x

L

d e f i L

x

f σ σ

π

) σ

](

2 [ ) 1

(

lim

where s>s₁ so that L[f](σ) is analytic for Reσ ≥s₁, and the path is vertical.

We consider the problem of heat conduction in an infinite slab, as mentioned in Section 2.3.

In Section 2.3 we use Fourier transform to solve it.

bounded

We suppose that x

Thus, this gives

0 function of x. We now solve the equation (2.14) by means of the Fourier transform.

Let

The solution of the problem (2.15) is

By the convolution theorem for the Fourier transform and

( )

is multiple-valued, and we want to choose a particular branch cut of it. We choose that branch cut of

Figure 2.1 Contour C

Since g(σ) is analytic inside this contour,

∫

⁼

∫

⁺

∑∫

⁼

= +

−

C n C

t iL

s

iL s

t t

n

d g e d

g e g

e ( ) ( ) ( ) 0

6

2

σ σ σ

σ

σ ^{σ σ}

σ .

We let L→∞ and ε →0, we can find that

Contour C2:

θ σ

π π θ

σ =s+Le^iθ from to , d =Lie^iθd 2

, .

∫

∫

^{= π + + ⋅ ⋅}

π

θ θ

σ σ σ ^θ θ

2

)

( ( )

) (

2

d e i L Le s g e

d g

e ^{s Le t i i}

C

t ⁱ

.

Since

L Le s

Le s Le s

s Le

s g

i i i

i

≤ − +

= +

= +

=

+ 1 1 1 1

) (

12 θ θ

θ θ

approach zero as L→∞. By Jordan’s lemma the integrals over this contour C2 approach zero.

C²

C¹

C³ C⁴

C⁵

C⁶

s s+il

s-il

0 ε Re z

Im z

Similarly to contour C6 , π σ θ

over the contour C6 approach zero.

Contour C4:

.

Hence by the inversion formula

∫

2.6 Finite Difference method

Suppose that u =u( tx, ) is a function of two independent variables. The first partial derivatives of u are defined as limits of difference quotients.

h

We can use the respective difference quotients to approximate these partial derivatives, To analyze the truncation errors (TE) associated with the approximation (2.18) and (2.19), we require Taylor’s theorem in two variables.

Assume that u is twice continuously differentiable and that h is positive. From Taylor’s theorem, we have

h

We also require finite-difference formulas for the higher order derivatives of u. To obtain a difference formula for u_xx, we use Taylor’s theorem to write

)24

where TE be simplified by the intermediate-value theorem.

Example 2.5 : (Use Finite Difference method)

).

We replace the boundary conditions and the initial conditions by

).

)

If the constants A and B are bounds for u_xxxx 12

The inequality (2.27) can be written

.

Therefore

k

) stability condition for the problem.

Example 2.6 : (Use Finite Difference method)

1

Using separation of variables as in Section 2.1, the exact solution is x

t x

u( , )=100^−π2tsinπ . By equation (2.20) and (2.23)

) 0

In this problem we must suppose that

2 0

T=0.5 Numerical Exact

x=0.0 0.000000 0.000000

x=0.1 0.204463 0.222241

x=0.2 0.388912 0.422728

x=0.3 0.535291 0.581836

x=0.4 0.629273 0.683989

x=0.5 0.661656 0.719188

x=0.6 0.629273 0.683989

x=0.7 0.535291 0.581836

x=0.8 0.388912 0.4227528

x=0.9 0.204463 0.222241

x=1.0 0.000000 0.000000

Table 2.1 Comparison of the numerical solution and exact solution k=0.005, h=0.1

(2) 6

=1

r Choosing k =0.001667, h=0.1, N =10 gives the numerical solution shown in Table 2.2.

T=0.5 Numerical Exact

x=0.0 0.000000 0.000000

x=0.1 0.222040 0.222241

x=0.2 0.422346 0.422728

x=0.3 0.581309 0.581836

x=0.4 0.683370 0.683989

x=0.5 0.718538 0.719188

x=0.6 0.682270 0.683989

x=0.7 0.581309 0.581836

x=0.8 0.422346 0.4227528

x=0.9 0.222040 0.222241

x=1.0 0.000000 0.000000

Table 2.1 Comparison of the numerical solution and exact solution k=0.001667, h=0.1

II. Comparison with various solving methods

3.1 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).

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

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