1.2 The domain is a circular - 線性橢圓偏微分方程

ⅡⅡ

Ⅱ-1.2 The domain is a circular

We consider a solution u of Laplac ′es equation in the unit circle x² + y² <1 equation in these coordinates is

1 1 0 and satisfies

u(1,θ)= f(θ) . (2-5)

The function f(θ) is a given continuously differentiable function which is periodic of period 2 . The solution π u(r,θ) must also be periodic of period 2 in π θ.

We apply separation of variables to Laplac ′ equation by seeking solutions of the es form R(r)θ(θ).

Substituting，we have

It is easy to see that has solutions of period 2π if and only if λ=n² with

n ，corresponding to these eigenvalues n² we have the eigenfunctions cos(nθ) and sin(nθ). Phere are two eigenfunctions corresponding to each eigenvalue except

which is a full Fourier series .

Hence，we deduce that

We examine the function

∑

^∞

differentiable for r<1，and its derivatives may be formed by term-by term differentiation of its series. Then

In below，we give a example to illustrate above statement.

Example 2

By equation (2-6)，we have

∑

^∞

Ⅱ

Ⅱ----2222 Finite Fourier transform to construct solution of system ofLaplac ′esequation

We shall now treat the corresponding non-homogeneous problem

u_xx+u_yy = F(x,y) for 0< x<π and 0< y<1 , ( 2-7)

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

To solve the above non-homogeneous problem，we expand the solution in a Fourier

∂ is continuous，its finite sine transform is given by

∫

^π ⁼ simpler operation of multiplying its finite sine transform by (−n²).

If ₂

y u

∂

∂ is continuous，we can interchange integration and differentiation to show that

∫

^π ⁼_π

∫

^π_∂^∂

Taking the finite sine transform of both sides of (2-7) therefore leads to the equation

) , (x y F u u_xx+ _yy =

⇒ _π²

∫

₀^π^u^xx⁽^x^,^y⁾^sin^nxdx⁺_π²

∫

₀^π^u^yy⁽^x^,^y⁾^sin^nxdx⁼ _π²

∫

₀^π^F⁽^x^,^y⁾^sin^nxdx

⇒ ( ) ( ) ₂ ( ) ( )

2 b y B y

dy y d b

n _n + _n = _n

− for n=1,2,3,...

⇒ b_n″(y)−n²b_n(y)=B_n(y) .

The condition u(x,0)=0 means that

0 ) 0

( =

bn .

Taking sine transform has reduced the problem (2-7) for a partial differential to the problem for an ordinary differential equation，that is







″ − 0 ) 0 (

) ( ) ( )

( ²

n n

y B y b n y

b .

Solving this by a method，we can use Gree ′ function to solves it，and the solution has ns Fourier sine series form. By Schwar ′ inequality for sums and zs Parseva ′ ls equation ，we have proved the series

∑

^bn(^y)sin^nx converges uniformly for 0≤ x≤π , 0≤ y≤1. Under this condition，we get

∑

^∞

sin ) ( )

, (

n y nx

b y

u .

In below，we give a example to illustrate above statement.

Example

Given u_xx+u_yy = y(1−y)sin³x

⇒ _π²

∫

₀^π ^u^xx ^sin ^nxdx ⁺_π²

∫

₀^π ^u^yy ^sin ^nxdx ⁼ _π²

∫

₀^π ^y⁽¹⁻ ^y⁾^sin ³ ^x^⋅^sin ^nxdx

⇒ ⁽⁻ⁿ²⁾^bⁿ⁽^y⁾⁺^bⁿ^″⁽^y⁾⁼_π² ^y⁽¹⁻^y⁾

∫

₀^π^sin³^x^⋅^sin^nxdx^for ⁿ ⁼ ¹^,²^,...

⇒ ^bⁿ^″⁽^y⁾⁻ⁿ²^bⁿ⁽^y⁾⁼ _π² ^y⁽¹⁻ ^y⁾

∫

₀^π^sin³ ^x^sin^nxdx ^.

In this case，



Let v=e^ry ⇒ v₁(x)=e^ny and v₂(x)=e⁻^ny

⇒ ^bⁿ⁽^y⁾⁼

∫

₀^y^G⁽^y^,^ξ⁾^f⁽^ξ⁾^d^ξ ⁺

∫

_y¹^G⁽^y^,^ξ⁾^f⁽^ξ⁾^d^ξ

Therefore，the solution is

ⅡⅡ-3 Fourier Transform to construct solution of system of Laplac ′es equation

Just as problems on the finite intervals lead to Fourier series，problems on the whole line (−∞,∞) lead to Fourier transform. To understand this relationship，consider a

where the coefficients are

^cⁿ ⁼ ₂¹_l

∫

₋^l_l ^f⁽^y⁾^e⁻ⁱⁿ^l^π^y^dy^.

The coefficients c_n define the function f(x) uniquely in the interval (−l,l). The Fourier integral comes from letting l→∞. However，this limit is one of the trickiest in all mathematics because the interval grows simultaneously as the terms change. If we write

l k nπ

= ，and substitute the coefficients into the series，we get

e l the limit we should expect k to become a continuous variable，and the sum to become an integral. The distance between two successive k ′ is s

k πl

∆ ，which we may think of as becoming dk in the limit. Therefore，we expect the result

f x [ f(y)e ^ikydy]e^ikxdk 2

) 1

( ⁼ _π

∫

₋^∞_∞

∫

₋^∞_∞ ⁻ . ( 2-8 ) Another way to state the above identity (2-8) is

^f⁽^x⁾⁼ ₂¹_π

∫

₋^∞_∞^F⁽^w⁾^e^ikx ₂^dw_π where ^F⁽^w⁾⁼

∫

₋^∞_∞ ^f⁽^x⁾^e^iwx^dx^.

Let ^f^{^}⁽^w⁾⁼

∫

₋^∞_∞ ^f⁽^x⁾^e^iwx^dx , ( 2-9 ) then

1 ^

( ) lim ( )

L iwx

L L

f x f w e dw

−

→∞ −

∫

If the integral in (2-9) converges，it is called the Fourier transform of f(x). It is sometimes denoted by F[ f]. The integral converges if

∫

₋^∞_∞ ^f⁽^x⁾^dx^does.

The Fourier transform of f(x) is

∫

₋^∞_∞

= f w f x e dx

w f

F[ ]( ) ( ) ( ) ^iwx

and the inverse Fourier transform is

∫

₋^∞_∞ ⁻

− =

=F f f w e dw

f ( ) ^iwx

2 ] 1 [ ) (

1 ^

π ^.

For functions of two variables，say u(x,y)，and we define

∫

₋^∞_∞

≡u w y u x y e dx

y w u

F[ ]( , ) ( , ) ( , ) ^iwx

A basic property of the Fourier transform is that the k th derivative u^{( k}⁾ with ,...

2 ,

k transforms to an algebraic expression，that is

) , ( ) ( ) , ](

[

y w u iw y

w u

F ^k = − ^k ,

confirming our comment that derivatives are transformed to multiplication. This formula is easily proved by integration by parts.

One of the many important formulae which is used in this field is given in the convolution theorem. The convolution f ∗ of two functions f and g is defined by g

∫

₋^∞_∞ ⁻ ⁼

∫

₋^∞_∞ ⁻

∗g x f u g x u du f x u g u du

f )( ) ( ) ( ) ( ) ( )

( .

Now

∫

₋^∞_∞ ₋^∞_∞ ⁻

∗g e f u g x u dudx

F[ ] ^iwx ( ) ( )

∫

₋^∞_∞ ^f⁽^u⁾

∫

₋^∞_∞^g⁽^x⁻^u⁾^e^iwx^dxdu^.

After applying this change of variables in above equation，we deduce the convolution theorem which states that

∫

₋^∞_∞ ₋^∞_∞ ⁺

∗g f x g ve dvdx

F[ ] ( ) ( ) ^iw⁽^x ^v⁾

∫

₋^∞_∞ f(x)e^iwxdx⋅

∫

₋^∞_∞g(v)e^iwvdv= F[f]F[g]= ^{^}f⋅g^{^} ^, and

∫

₋^∞_∞

− f⋅g = f ∗g = f u g x−u du

F [ ] ( ) ( )

^ 1 ^

. This is useful relationship in solving differential equations.

Following is a table of some important basic properties of transforms

) (x

f ( )

w f

1 f ′ iw^{^}f

2 xf(x) _^′

f i 3 f(x−a) e⁻^iaw f^{^}

4 e^iaxf(x) ( )

a w

f −

5 af(x)+bg(x) a ^{^}f+bb^{^}

6 f(ax) 1 [ ]

a F w a

Table 2-1. Basic properties of transforms In below，we give a example to illustrate above statement.

Example 2 Example 2 Example 2

Example 2----4 4 4 :::: 4 ( Using Fourier Transform to solve Laplac ′es equation )

Consider u_xx+u_yy =0 in the half plane y ≥0 subject to the boundary condition

) ( ) 0 ,

(x x

u =δ with x∈R and the condition u(x,y)→0 as x² +y² →∞.

: Solution

Using Fourier transform with respect to x ，

∫

₋^∞_∞

=u w y u x y e dx

y x u

F[ ( , )] ( , ) ( , ) ^iwx

and

^ 2 2

]

[ u_yy

F u =

∂

∂ ，

2 ^ 2

) ( ]

[ iw u

F u = −

∂

∂ .

Which implies

u satisfies the ODE

2 ^

−w u

u_yy for y >0 , F[(w,0)]=1.

The solutions of the ODE are e^±^wy. We must reject a positive exponent since

u would grow exponentially as w →∞ and would not have Fourier transform.

So u(w,y)=e⁻^w^y

. Therefore，

^u ^x ^y ⁼ _π

∫

₋^∞_∞^e⁻^y^w^e^iwx^dw

2 ) 1 ,

( ，w∈R and y≥0 .

This improper integral clearly converges for y >0. It is split into to parts and integrated directly as

eiwx

y y ix

u ⁻

= −

) ( 2 ) 1 ,

( π ^︱

∞

0+ e^iwx ^wy

y ix

+ ) ( 2

π ^︱

∞

−

1 1 ) 2 (

ix y ix

y + +

= −

π ^.

) (x² y²

= +

π ^.

Ⅱ

Ⅱ-4 Finite Difference to construct solution of system of Laplac ′es equation

One scheme for solving all kinds of partial differential equations is to replace the derivatives by difference quotients，converting the equation to a difference equation. We

then write the difference equation corresponding to each point at the intersections of a gridwork that subdivides the region of interest at which the function values are unknown.

Solving these equations simultaneously gives values for the function at each node that approximate the true values. We begin with the two-dimensional case.

Let h= x∆ = equal spacing of gridwork in the x−direction，see Figure 2-1. We assume that the function f(x) has a continuous fourth derivative. By Taylor series，

⁴ It follows that

Figure 2-1. Taking five interior points

A subscript notation is convenient :

2 ( ₂)

In above equation，the subscripts on f indicate the x−values at which it is evaluated.

∂ ，by holding y constant and evaluating the function at three points

where x equals x ，_n x_n + and h x_n− . The partial derivative h 2 2

y u∂

∂ is similarly computed，holding x constant. We require that fourth derivatives with respect to both variables exist.

To solve the Laplac ′ equation on a region in the es xy− plane，we subdivide the

Replacing the Laplac ′ equation by the finite difference equation，we get es

We call the points (i+1,j)、(i−1,j)、(i, j+1) and (i, j−1) the nearest neighbors of the mesh point ( ji, ). If ( ji, ) and all its nearest neighbors lie in D+C，we call ( ji, ) an interior point.

It is common to take ∆x=∆y=h，resulting in considerable simplification，so that

1 [ 4 ] 0

, 1 , 1 , , 1 , 2 1 ,

2 = + + + − =

∇ _i _j v_i₊ _j v_i₋ _j v_i_j₊ v_i _j₋ v_i_j

v h . ( 2-9 )

Note that five points are involved in the relationship of equation (2-9)，points to the right、left、above and below the central point(x_i,y_j). The approximation has O(h²) error，

provided that u is sufficiently smooth. This formula is referred to as the five-point star formula.

The system we get in this way has exactly one solution. To prove this，suppose that there were two solutions，

{ }

ui_,j and

{ }

vi_,j of (2-9) in D with identical boundary values.

Their difference

{

ui_,j −vi_,j

}

also satisfies (2-9) in D but with zero boundary values. By the maximum principle，u_i_,_j −v_i_,_j ≤0，hence u_i_,_j =v_i_,_j. So there is at most one solution.

Now，if we define the error function w=u−v .

The boundary value problem for u is therefore properly posed. As h→0 the error v

w= − approaches zero. That is， v converges to u . In below，we give a example to illustrate above statement.

Example 2 Example 2 Example 2

Example 2----5 : 5 : 5 : ( Using Finite Difference to solve 5 : Laplac ′es equation )

Find u(x,y) such that

=0 + _yy

xx u

u ,

u(x,o)=u(x,10)=u(o,y)=0 , u(20,y)=100 .

Solution

We replace the differential equation by a difference equation:

0 ] 4 1 [

, 1 , 1 , , 1 ,

2 u_i₊1_j +u_i₋ _j +u_i _j₊ +u_i_j₋ − u_i _j = h

⇒ u_i₊₁_,_j +u_i₋₁_,_j +u_i_,_j₊₁+u_i_,_j₋₁−4u_i_,_j =0

⇒ 4u_i_,_j −u_i₊₁_,_j −u_i₋₁_,_j −u_i_,_j₋₁−u_i_,_j₊₁ =0 . Suppose we choose h=5，the system of equations is

(0 0 0 4 ) 0 5

2 + +u2+ − u = ,

( 0 0 4 ) 0 5

2 3

2 u1+ +u + − u = ,

( 0 100 0 4 ) 0 5

2 u2+ + + − u = .

We can write equations as matrix form and usung 〝Metlab〞to solve.

The solution to the set of equations is easy when there are only three of them :

u₁ =1.786 、 u₂ =7.143 、 u₃ =26.786 .

Ⅲ

Ⅲ Ⅲ.... The limit of the methods of solving Ellpitic PDE

In this chapter，we want to analysis the limit of four methods of solving Laplac ′es equation in chapter Ⅱ.

u₁ u₂ u₃

0 ﾟ 100 ﾟ

0 ﾟ 0 ﾟ 0 ﾟ

1、The limit of Separation of variables

The standard technique for solving PDE on bounded ( rectangular) domains is called _s separation of variables. The idea is to assume that the unknown function u =u(x,y) in an initial boundary value problem can be written as a product of a funvtion of x and a function of

y ，that is，u(x,y)= X(x)Y(y). Thus，the variables separate. If the method is to be successful，when this product is substituted into the PDE ，the PDE separates into two

ODE ，one for s X(x) and one for Y( y). Therefore，we are left with an ODE boundary value problem for X(x) and an ODE for Y( y). When we solve for X(x) and

) ( y

Y ，we will have a product solution u(x,y) of the PDE that satisfies the boundary conditions.

Whether or not the method of separation of variables can be applied to a particular

problem depends not only on the differential equation but also on the shape of the boundary and on the form of the boundary conditions.

Three things are needed to apply the method to a problem in two variables x and y :

(a) The differential operator L must be separable. For example，this elliptic equation

=0 + + _xy _yy

xx u u

u ，it can not use Separation of variables to find solution.

(b) All initial and boundary conditions must be on lines x≡constant and y≡constant. (c) The linear operators defining the boundary conditions at x≡constant must involve no

partial derivatives of u with respect to y ，and their coefficients must be independent of y . Those at y≡constant must involve no partial derivatives of u with respect to x ，and their coefficients must be independent of x .

That the method of separation of variables can only be applied to a special class of problems.

2、、、、Finite Fourier transform

To solve the nonhomogenous problem，we expand the solution in a Fourier sine series.

The Finite Fourier transforms，are simply Fourier coefficients. Whenever a homogeneous problem can be solved by separation of variables in the form of a Fourier series，the Finite

Fourier transform reduces the partial differential equation to an infinite system of ordinary differential equations. These equations can then be solved by the methods of one-sided

s n

Gree ′ function or Gree ′ function . The Finite Fourier transform is using half - ns space domain.

3、、、、Fourier transform

The Fourier transforms are first encountered in elementary differential equations courses as a technique for solving linear ， constant-coefficient ordinary differential equations; Fourier transforms convert an ODE into an algebra problem. The ideas easily extend to PDE ，where the operation of Fourier transformation converts _s PDE_s into ODEs. Thus the Fourier transforms is useful as a computational tool in solving differential equations. In PDE_s the Fourier transform is usually applied to the spatial variable when it varies over whole line. That is，the Fourier transform is using whole space domain.

4、、、、Finite Difference

The finite difference method is using the domain of rechangular domain or irregular shape. This methos solution form is discrete solution and it is the approximate solution (value).

All we need to do is to continue to make h smaller. However，this procedure runs into severe difficulties. It is apparent that the number of equations increases inordinately fast. With h=1.25

，we would have 105 discrete interior points; with h=0.625，we have 465 discrete interior points and so on. Storing a matrix with 105 rows and 105 columns would require 105 of ² computer memory. Few computer systems allow us such a generous partition，and overlaying memory space from disk storage would be extremely time-consuming. Along with memory requirements，we worry about execution times.

Compared with four methods Compared with four methods Compared with four methods Compared with four methods :

(a) The homogeneous problem can be solved by Separation of variables 、Fourier Transform 、Finite Difference. But to solve the nonhomogeneous problem，we can use Finite Fourier Transform.

(b) Separation of variables、Finite Fourier Transform and Fourier Transform reduces the partial differential equation to ordinary differential equations and facilitates us to solve.

Transform is continuous，whereas the solution caused by Finite Difference is discrete type and it is the approximate solution.

(d) The Separation of variables method can be applied to rectangle domain; the Finite Fourier Transform method can be applied to half-space domain; the Fourier Transform method can be applied to whole space domain ( whole line ); the Finite Difference methid can be applied to rectangular domain or irregular shape domain.

Ⅳ

Ⅳ Ⅳ. . . . Integral evaluations on three-sheeted Riemann surface of genus N

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

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

∏

在文檔中線性橢圓偏微分方程 (頁 26-45)