國
立
交
通
大
學
應 用 數 學 系
碩
士
論
文
線性雙曲型偏微分方程之研究
Topics on linear hyperbolic equations
研 究 生：陳美如
指導教授：李榮耀 博士
線性雙曲型偏微分方程之研究
Topics on linear hyperbolic equations
研 究 生：陳美如 Student: Mei-Zu Chen
指導教授：李榮耀 博士 Advisor: Dr. Jong-Eao Lee
國 立 交 通 大 學
應 用 數 學 系
碩 士 論 文
A Thesis
Submitted to Department of Applied Mathematics College of Science
National Chiao Tung University in Partial Fulfillment of the Requirements
for the Degree of Master in
Applied Mathematics June 2007
線性雙曲型偏微分方程之研究
學生：陳美如 指導教授：李榮耀博士
國 立 交 通 大 學 應 用 數 學 系
摘 要
本論文主要研究線性的雙曲型偏微分方程(線性雙曲 PDEs)。首先，我們舉 給幾個屬於此類型的實際例子。再來，使用幾種典型的方法來解線性雙曲 PDEs。 同時並以不同形式來表示解，並且確定解的一致性。 當我們對 PDEs 使用積分轉換時(對於變數是整條實數線使用 Fourier 轉換； 變數是半射線使用 Laplace 轉換)，再藉由逆積分轉換(inversion Fouier transform or inversion Laplce transform)來得到 PDEs 的解是必要的。但是執行逆積分轉換時， 經常那些被積分會出現平方根。然而，平方根在複數平面上是多值的。為了能正 確地進行逆轉換，我們利用適當的代數分析來建構多值函數的黎曼曲面，使其變 成單值函數，以致於我們能正確地在分析上和數值上完成逆積分轉換。最後由一 些例子說明整個架構。Topics on linear hyperbolic equations
Student: Mei-Zu Chen Advisor: Dr. Jong-Eao Lee
Department of Applied Mathematics National Chiao Tung University
Abstract
We study the linear hyperbolic partial differential equations (linear hyperbolic PDEs). First, we give some practical examples and show that they are governed by such type of the equations. Next, we apply several classical methods to solve the linear hyperbolic PDEs with the solutions being expressed in various forms. We then identify those solutions.
When we apply Fourier and Laplace transformations to the whole- and half-line PDEs,it is necessary to perform the inverse Fourier and Laplace transformations to derive the PDE solutions, and it is quite often that those integrals involve the square root operator which is multi-valued in the complex plane. In order to perform the inverse transformations correctly, we develop the Riemann surfaces from the complex plane with the proper algebraic structures to assure that the square root is now a single-valued function on the surfaces, and we are able to accomplish the inverse transformations analytically and numerically. Some examples are given to illustrate the entire scheme.
致 謝
這篇論文能夠順利完成，首先要感謝我的指導教授李榮耀老師的
指導與鼓勵，不厭其煩的幫我潤飾修辭，以及讓我學習到做學問和做
事應有的態度，使學生受益良多；同時，也感謝口試委員邵錦昌教授、
莊重教授和郎正廉教授所給予學生的建議和指教，使論文能完整的呈
現。
在這兩年來，感謝同學佳樺、文雯、筱凡、育慈和雁婷陪伴我在
新竹的日子。無論在學業或生活上，都給我許多寶貴的意見。另外，
感謝健勳、練爸爸和練媽媽的關心和鼓勵，讓我度過了最美好的研究
所生活。
最後感謝辛苦栽培我的父母，謝謝你們一路支持我的選擇，給予
我最大的鼓勵與支持，讓我完成無憂無慮的完成學業，心中充滿無限
的感激，僅以這篇論文獻給我親愛的家人。
Contents
1 Introduction 1
1.1 The Advection Equation . . . 1
1.2 The Wave Equation in One Dimension . . . 2
1.3 The Telegraph Equation . . . 4
2 Solutions of Linear Hyperbolic Equations 7 2.1 Solution of the Advection Equation . . . 7
2.2 D’Alembert’s Solution to the Wave Equation . . . 8
2.3 The method of Separation of Variables to the Wave Equation . . . 11
2.4 The Fourier Transforms to the Wave Equation . . . 13
2.5 The Laplace Transforms to the Wave Equation . . . 15
2.6 The method of characteristics to the Wave Equation . . . 17
2.7 The method of Finite Difference to the Wave Equation . . . 20
3 The Wave equation in different domains 22 3.1 Solution of the finite string problem . . . 22
3.2 Solution of the infinite problem . . . 24
4 Riemann surface of genus N 25 4.1 Introduction . . . 25
4.2 The algebraic structure on Riemann surface . . . 27
4.3 The geometric structure on Riemann surface . . . 31
4.4 The integrals over a, b cycles on Riemann surface . . . 33
1
Introduction
We begin our study of linear hyperbolic equations by showing classical examples. First, we present a simple transport for first order partial differential equations, and then we extend our discussion to system of first order equations in electronics. Later, we will show the wave equation for second order partial differential equations. Under several hypothesis of physical phenomenon, a vibrating string problem is changed into one dimensional wave equation. Conversely, we can give a proper approximation to physics by discussing the solution of this mathematical model. Hence, we analyze the partial differential equations to observe physical problems.
1.1
The Advection Equation
Definition 1.1. Let u(x, t),F(x, t) be m × 1 vector and A(x, t), B(x, t) be m × m matrix .The
system of first order equations
ut(x, t) + A(x, t)ux(x, t) + B(x, t)u(x, t) = F(x, t) (1.1)
is said to be hyperbolic if A(x, t) is real diagonalized. Obviously, a single real equation
ut(x, t) + cux(x, t) = F(x, t)
is a hyperbolic equation. Let the particles of pollutant be transported from left to right with a constant speed c in a river. Denote u(x, t) the density of particles at the position x and time t in the river. Suppose this river is so narrow that no particles get scattered.
First, we consider there no particles get lost or added. At time t, the amount of the particles of pollutant in an interval [a,b], where 0 < a < b, is
M =
Z b
a
u(x, s)dx.
Let s, h > 0. The particles of time s + h are of time s transported to right with distance ch centimeters. Hence, the amount of the particles of pollutant at time s + h is equal to the
amount at time s, i.e. M = Z b a u(x, s)dx = Z b+ch a+ch u(x, s + h)dx. (1.2)
By the First Fundamental of Calculus, we differentiate the equation (1.2) for b, then it becomes
u(b, s) = u(b + ch, s + h). (1.3)
Again we differentiate the equation (1.3) for h, hence
0 = cux(b + ch, s + h) + ut(b + ch, s + h).
Let h=0, we derive a homogeneous transport equation.
ut(b, s) + cux(b, s) = 0,
where b is arbitrary, hence we get the advection equation.
If the particles get lost or added in the river, then we obtain the nonhomogeneous ad-vection equation
ut+ cux = F,
where F is the amount of particles which get loss or added per length at position x and time
t.
1.2
The Wave Equation in One Dimension
Definition 1.2. The linear partial differential equation for second order
A(x, t)uxx+ B(x, t)uxt+ C(x, t)utt+ D(x, t)ux + E(x, t)ut+ F(x, t)u = 0
is said to be hyperbolic, parabolic, or elliptic at (x0, t0) if B2(x0, t0) − 4A(x0, t0)C(x0, t0) is
positive, zero, or negative, respectively.It is hyperbolic, parabolic, or elliptic in the domain
For the most part in this paper, we discuss the type of linear hyperbolic equations. The wave equation in one dimension is
utt(x, t) − c2uxx(x, t) = F(x, t),
where 0 < x < l, t > 0. It is hyperbolic in its region. Because A(x, t) = −c2, B(x, t) = 0, C(x, t) = 1, B2(x, t) − 4A(x, t)C(x, t) = 0 − 4 × (−c)2× 1 = 4c2 > 0,for all 0 < x < l, t > 0.
Now, we want to show how the motion of a string as a mathematical equation under several assumptions.
(1) The string with length l is flexible and elastic. It is so flexible such that it offers no resistance to bending. Hence, the tension is in the direction of tangent to the profile of the string. In an elastic uniformly string, the density is a constant (mass per unit length). (2) There is no elongation of a single element of the string. By Hooke’s law the tension is constant.
(3)The string has small transverse vibration.
(4)The weight of the string is small compared with the tension in the string.
Denote u(x, t) the displacement from equilibrium position at time t and position x . Let ρ and T be a constant density and tension at time t and position x. For any two closed points
x and x + ∆x in the string at time t as shown in Figure 1
E D O t x x+'x x T T
From Newton’s Law F = ma, we get a equation for equivalent vertical force
T sin β − T sin α = ρ∆xutt. (1.4)
Since small transverse vibration of a string, sin α ≈ tan α and sin β ≈ tan β. So, the equation (1.4) becomes
tan β − tan α = ρ∆x
T utt. (1.5)
At time t, tan α = (ux)x and tan β = (ux)x+∆x, so equation (1.5) obtains
1
∆x[(ux)x+∆x− (ux)x] =
ρ
Tutt. (1.6)
Let ∆x be sufficiently small, then we derive the homogenous wave equation
utt(x, t) = c2uxx(x, t) = 0,
where c2 = Tρ.
Let there be an external force to a string. Hence, it appears an nonhomogeneous term.
utt(x, t) − c2uxx(x, t) = f (x, t),
where c2 = T
ρ and f (x, t) the external force per unit length at position x and time t.
Remark 1.3. The value c is a wave speed. It is clear the unit of a tension T is kg · m/s2and
of a density ρ is kg/m. Then the unit of c =
q T
ρ is m/s . Here, the unit of c is indeed the
unit of the speed.
Finally, we successfully transform a physical phenomenon into a mathematical equation under several properly hypotheses. By this mathematical equation, we can get a lot of information for the motion of the string vibrating. Thus, we will discuss the wave equation in detail in a later chapter.
1.3
The Telegraph Equation
The wave equation which we have discussed in section 1.2 can be replaced by the system of first order equations. Now, we want to present how the wave equation changes into the
system of first order equations in electronics. Suppose a pair of transmission lines has a voltage V(x, t) across them and a current I(x, t) at position x and time t. The part of it is an interconnection of elements: capacitance, resistance, leakage resistance and inductance. Denote C the capacitance per unit length, R the resistance per unit length, G the leakage resistance per unit length and L the inductance per unit length.
Conductance G is the ability of an element to conduct electric current per unit length, and then conductance is the reciprocal of resistance
G = 1
R =
I
V.
Let µ, N, A be the permeability of core current, number of turns and cross section area of the inductor, respectively. When the current passes through an inductor, it is found that the voltage across the inductor is directly proportional to the time rate of change
V = LdI dt, where L = N 2µA l .
By Kirchhoff’s Current Law and Kirchhoff’s Voltage Law, we have system (1.7)
V(x, t) − RI(x, t)∆x − L∂I(x, t)
∂t ∆x = V(x + ∆x, t),
I(x, t) − GV(x, t)∆x − C∂V(x, t)
∂t ∆x = I(x + ∆x, t).
(1.7)
Let ∆x be small enough, then system (1.7) becomes the following system (1.8)
Vx(x, t) = −RI(x, t) − LIt(x, t), Ix(x, t) = −GV(x, t) − CVt(x, t). (1.8) Let u = I V ,
then system (1.8) can be expressed the standard form of (1.1). Here,
A = 0 1L 1 C 0 ,
possesses two distinct real eigenvalues, A is real diagonalizable. Clearly, this system is hyperbolic. Differentiate the first and second of system (1.8) for x and t, respectively. Then we obtain
Vxx(x, t) = −RIx(x, t) − LItx(x, t),
Ixt(x, t) = −GVt(x, t) − CVtt(x, t).
(1.9)
Subtracting L times of second equation from first equation the system (1.9), hence we get the partial order equation for second order
CLVtt(x, t) − Vxx = −(GL + CR)Vt(x, t) − GRV(x, t). (1.10)
The equation
Vtt(x, t) − c2Vxx = −aVt(x, t) − bV(x, t). (1.11)
is called the telegraph equation, where c = CL1 , a = GL + CR, b = GR.
Let this transmission line have no energy lost, then R = 0, G = 0. Hence, above equation (1.11) becomes the homogenous wave equation
Vtt(x, t) −
1
CLVxx = 0.
Similarly, we can derive the homogenous wave equation of I
Itt(x, t) −
1
2
Solutions of Linear Hyperbolic Equations
2.1
Solution of the Advection Equation
Example 2.1. Using the method of characteristic to solving a I.V.P. of the advection equa-tion. ut(x, t) + 2tux(x, t) = 0, −∞ < x < ∞, t > 0, u(x, 0) = e−x2, −∞ < x < ∞. The characteristic is dx dt = 2t, x(0) = ξ.
Along this characteristic
x(t) = t2+ ξ,
the solution u(x, t) is a constant. Because
d dtu(x(t), t) = ut+ ux dx dt = ut+ 2tux = 0
Hence, the solution
2.2
D’Alembert’s Solution to the Wave Equation
Consider a finite string problem with two fixed ends
utt(x, t) = c2uxx(x, t), 0 < x < l, t > 0, u(x, 0) = f (x), 0 ≤ x ≤ l, ut(x, 0) = g(x), 0 ≤ x ≤ l, u(0, t) = 0, t ≥ 0, u(l, t) = 0, t ≥ 0. (2.1)
Chosen a new coordinate transformation (ξ, η)
ξ = x + ct, η = x − ct.
Hence, the wave equation (2.1) becomes
−4c2u
ξη = 0.
So, the solution is
u(x, t) = p(ξ) + q(η), (2.2)
where p(ξ), q(η) are arbitrary functions of ξ, η, respectively. First, we discuss a solution only depends on p(ξ). Fixed x + ct = ξ, then the solution is
u(x, t) = p(ξ),
Since,
dx
dt = −c,
Hence, along this characteristic x + ct = ξ, the wave is move to left with velocity c. Next, we consider a solution only depends on q(η). First, we
Fixed x − ct = η, then the solution is constant of
Since
dx
dt = c.
Hence, along this characteristic x − ct = η, the wave is move to right with velocity c. Finally, we combine p(ξ) and q(η) together. One wave for p(ξ) propagates to left along the line x + ct = ξ with speed c; another q(η) for propagates to right along the line x − ct = η with speed c.
Applying initial conditions of (2.1) to a general solution (2.2), hence we get
p(ξ) = 1 2f (ξ) + 1 2c Z ξ 0 g( ¯x)d ¯x + p(0), ξ ∈ [0, l], q(η) = 1 2f (η) − 1 2c Z η 0 g( ¯x)d ¯x − p(0), η ∈ [0, l],
Thus, the D’Alembert’s solution is
u(x, t) = 1 2[ f (x + ct) + f (x − ct)] + 1 2 Z x+ct x−ct g( ¯x)d ¯x,
where 0 ≤ x − ct ≤ x + ct ≤ l. The above solution of the D’Alembert’s solution form is only valid on the region shown as Figure 2.
2 l c l 0
x-ct=0
x+ct=l
t xConsider an nonhomogeneous finite string problem with two fixed ends. After a corrdi-nate transformation (ξ, η) as above, then the wave equation with extra force term becomes
uξη( ξ + η 2 , ξ − η 2c ) = − 1 4c2F( ξ + η 2 , ξ − η 2c ), Integrating two sides of above equation for ξ from η to ξ, we get
uη( ξ + η 2 , ξ − η 2c ) − 1 2f 0(η) − 1 2cg(η) = − 1 4c2 Z ξ η F(¯ξ + η 2 , ¯ξ − η 2c )d ¯ξ,
And we integrate above equation for η from η to ξ, then it yields the nonhomogeneous D’Alembert’s solution u(x, t) = 1 2[ f (x + ct) + f (x − ct)] + 1 2c Z x+ct x−ct g( ¯x)d ¯x + 1 2c Z t 0 Z x+c(t−¯t) x−c(t−¯t) F( ¯x, ¯t)d ¯xd ¯t,
Applying the boundary condition, we try to extend the valid domain of the D’Alembert’s solution to D = {(x, t) | −∞ < x < ∞, t ≥ 0} by extending functions f , g and F. The follow-ing table is that extends the domain of two functions f , g and F correspondfollow-ing to different boundary conditions from [0, l] to (−∞, ∞).
x = 0 extend functions f , g and F x = l
u(0, t) = 0 odd at points 0, l u(l, t) = 0
u(0, t) = 0 odd at points 0,even at l ux(l, t) = 0
ux(0, t) = 0 even at points 0,odd at l u(l, t) = 0
ux(0, t) = 0 even at points 0, l ux(l, t) = 0
Table 1: Extend functions f , g and F.
Example 2.2. Find the D’Alembert’s solution for the following problem. utt(x, t) = uxx(x, t), 0 < x < π, t > 0,
u(x, 0) = sin x, 0 ≤ x ≤ π, ut(x, 0) = cos x, 0 ≤ x ≤ π,
u(0, t) = 0, t ≥ 0, u(π, t) = 0, t ≥ 0.
Using D’Alembert’s solution formula, we get u(x, t) = 1 2[sin(x + ct) + sin(x − ct)] + 1 2c Z x+ct x−ct cos( ¯x)d ¯x, (2.3)
where 0 ≤ x − ct ≤ x + ct ≤ π. According to boundary conditions of this problem, f and
g are odd at the point x = 0 and even at the point x = π. Then f (x) = sin x, g(x) = cos x, x ∈ R. Hence,
u(x, t) = sin x cos t + cos x sin t, x ∈ R, t > 0 (2.4)
2.3
The method of Separation of Variables to the Wave Equation
In this section, we introduce a common method to solve the initial boundary value problem. The strategy of this method is separate independent variables for the function.
Example 2.3. Using the method of separation of variables to solve the forced vibration of rectangular membrane problem.
utt(x, y, t) = uxx(x, y, t) + uyy(x, y, t) + xy sin t, 0 < x < π, 0 < y < π, t > 0, u(x, y, 0) = 0, 0 ≤ x ≤ π, 0 ≤ y ≤ π, ut(x, y, 0) = 0, 0 ≤ x ≤ π, 0 ≤ y ≤ π, u(0, y, t) = 0, 0 ≤ y ≤ π, t ≥ 0, u(π, y, t) = 0, 0 ≤ y ≤ π, t ≥ 0, u(x, 0, t) = 0, 0 ≤ x ≤ π, t ≥ 0, u(x, π, t) = 0, 0 ≤ x ≤ π, t ≥ 0. (2.5) Let
u(x, y, t) = U(x, y)T (t). (2.6)
Substituting equation (2.6) into the wave equation of problem (2.5), then
UT ” = c2∆UT. Let T ” c2T = ∆U U = −λ.
Then
T ” + λT = 0, ∆U + λU = 0.
Let λ = α2, then we have
T = A cos αt + B sin αt,
where A and B are constant. Again separating the variables of U(x, t), let U(x, t) =
X(x)Y(y). Then it yield two problems
X” − µX = 0, X(0) = 0, X(π) = 0. and Y” + (λ + µ)Y = 0, Y(0) = 0, Y0(0) = 0.
Let µ = −β2and γ2 = (λ + µ) = α2− β2. Hence, the solutions of above problems are
Xm(x) = sin mx,
Yn(y) = sin ny,
where β = m and γ = n. So,
u(x, y, t) = ∞ X m=1 ∞ X n=1
(amncos αmnt + bmnsin αmnt) sin mx sin ny,
where amn= 4 ab Z π 0 Z π 0
u(x, y, 0) sin mx sin nydxdy = 0,
bmn= 4 αmnab Z π 0 Z π 0
ut(x, y, 0) sin mx sin nydxdy = 0.
Assume the solution
u(x, y, t) = ∞ X m=1 ∞ X n=1
and external forcing function F(x, y, t) = ∞ X m=1 ∞ X n=1
Fmn(t) sin αmnt sin mx sin ny,
here Fmn(t) = 4 ab Z π 0 Z π 0
F(x, y, t) sin mx sin nydxdy
= 4(mπ cos mπ − sin mπ)(nπ cos nπ − sinnπ) sin t π2m2n2
Taking u and F into the wave equation in problem (2.5) Hence, we get the following equa-tion
u”mn+ (m2+ n2)umn = Fmn,
where u is twice continuously differentiable with respect to t. Thus,
u(x, y, t) = ∞ X m=1 ∞ X n=1
umn(t) sin αmnt sin mx sin ny,
where umn(t) = 1 αmn Z t 0 Fmnsin(αmn)(t − τ)dτ = 4(−1) m+n+1 mnαmn {sin αmnt[ cos(1 − αmn)t − 1 2(1 − αmn) + cos(1 + αmn)t − 1 2(1 + αmn) ] + cos αmnt[ sin(1 − αmn)t − 1 2(1 − αmn) + sin 1 + αmn)t − 1 2(1 + αmn) ]}, where αmn= √ m2+ n2
2.4
The Fourier Transforms to the Wave Equation
We often use the method of integral transform to solve the problem for initial value prob-lems of the infinite or semi-infinite region. First, we introduce the method of Fourier Trans-form for a variable of all full real line. In general, a variable transTrans-formed is the spatial variable. And we will discuss the solution by Fourier Transform.
Definition 2.4. If f (x) is absolutely integrable, then the Fourier Transform is
F [ f (x)](ω) = ˆf(ω) =
Z ∞
−∞
Problems of partial differential equation can be reduced by problems of ordinary differ-ential equation for the Fourier Transform ˆu(x, t) of u(x, t) by property (2.5). After solving the problem of ˆu(x, t), there is an inversion theorem for Fourier Transform to help we trans-form ˆu(x, t) back u(x, t).
Property 2.5. If f (x) is absolutely integrable, approaches zero as x → ±∞ and has a first
derivative, then
F [ f0(x)](ω) = −iωF [ f (x)](ω)
Property 2.6. If f (x) is absolutely integrable, then
F [eicx
f (x)](ω) = F [ f (x)](ω + c).
Theorem 2.7. If f (x) is absolutely integrable, then the Fourier Transform is
f (x) = 1
2π
Z ∞
−∞
F [ f (x)](ω).
Example 2.8. Using the Fourier Transform to solve a long string problem. utt(x, t) = c2uxx(x, t), −∞ < x < ∞, t > 0,
u(x, 0) = e−|x|, −∞ < x < ∞,
ut(x, 0) = 0, −∞ < x < ∞,
(2.7)
Let the Fourier Transform of u for a variable x is
F [u(x, t)](ω) = ˆu(ω, t) =
Z ∞
−∞
eiωxu(x, t)dx.
Then problem (2.7) is reduced to the following problem (2.8) for second order ordinary differential equation corresponding initial value conditions by property 2.5.
d dt2ˆu(ω, t) = −c 2ω2ˆu(ω, t), −∞ < ω < ∞, t > 0, ˆu(ω, 0) = 2 1 + ω2, −∞ < ω < ∞, d dωˆu(ω, 0) = 0, −∞ < ω < ∞. (2.8)
Then ˆu(ω, t) = 2 1 + ω2 cos ωct = 2 1 + ω2 eiωct− e−iωct 2 By property 2.6, it yields the solution
u(x, t) = 1
2[e
−|x+ct|+ e−|x−ct|].
2.5
The Laplace Transforms to the Wave Equation
In section (2.4), we have introduced a method of Fourier Transform in order to solving problem of infinite line region. Now, we present a method of Laplace Transform for solving a half-line extent. In general, a variable transformed is the time variable. The property of Laplace Transform is similar to Fourier Transform.
Definition 2.9. If f (t) is absolutely integrable, then the Laplace Transform is
L [ f (t)](s) = F(s) =
Z ∞
0
e−stf (t)dt. (2.9)
The properties of Laplace Transform are similar to Fourier Transform.
Property 2.10. If f (t) is absolutely integrable, approaches zero as t → ∞ and has a first
derivative for t > 0, then
L [ f0(t)](s) = sL [ f (t)](s) − f (0).
In general, if f (t) and f(m)(t) are absolutely integrable, m = 1, 2, n − 1, approaches zero as
t → ∞ L [ f(n)(t)](s) = snL [ f (t)](s) − sn−1f (0) − ... − f(n−1)(0). Property 2.11. If L−1[F(s)] = f (t), then L−1[e−asF(s)] = f (t − a) if t > a, 0 otherwise.
Theorem 2.12. If F(s) is the Laplace transform of a real function f (t), with the complex transformed variables s, then the inversion integral is
f (t) = lim L→∞ Z γ+iL γ−iL L [ f (t)](s)est ds. (2.10)
Example 2.13. Using the Laplace Transform to solve one fixed end string problem. utt(x, t) = 4uxx(x, t), 0 < x < ∞, t > 0,
u(x, 0) = 0, 0 < x < ∞, ut(x, 0) = 0, 0 < x < ∞,
u(0, t) = 3 sin t, t ≥ 0.
(2.11)
Let U(x, s) = L [u(x, t)], u(x, t) is bounded. Then problem (2.11) is reduced to the follow-ing problem (2.12) for second order ordinary differential equation by property 2.10.
s2U(x, s) = Uxx(x, s), 0 < x < ∞, t > 0, U(x, 0) = 0, 0 < x < ∞, Ut(x, 0) = 0, 0 < x < ∞, U(0, s) = 3 s2+ 1, s ≥ 0. (2.12) Then we have U(x, s) = aesx+ be−sx, (2.13)
where a and b are constant. Applying a boundary condition of problem (2.12) and U(x, s) is bounded, then
U(x, s) = 3
s2+ 1e
−sx (2.14)
By property 2.11, the solution is
u(x, t) = 3 sin(t − x) if t > x, 0 otherwise.
2.6
The method of characteristics to the Wave Equation
In this section, we will discuss how we approximate the solution for hyperbolic linear equation of second order by two characteristics. The following linear hyperbolic equation for second order is
A(x, t)uxxx, t + B(x, t)ux,t(x, t) + C(x, t)utt(x, t) + e(x, t, ut, ux) = 0, (2.15)
where B2(x, t) − 4A(x, t)C(x, t) > 0 for all (x, t) in its region. Let
∂u(x, t) ∂x = S (x, t), ∂u(x, t) ∂t = T (x, t). (2.16) Then dS = uxxdx + uxtdt, dT = utxdx + uttdt. (2.17)
Substitute equations (2.17) into equation (2.15), then we obtain
A(x, t)(dS − uxtdt
dx ) + B(x, t)uxt(x, t) + C(x, t)(
dT − uxtdx
dt ) + e(x, t, ut, ux) = 0. (2.18)
Multiplying two sides of above equation (2.18) by dxdt, then we obtain [A(dt dx) 2− B(dt dx) + C]uxt− [A dS dx( dt dx) + C dT dx + e( dt dx)] = 0. (2.19)
Let a tangent slope at every point on the curve C satisfy a root of
A(dt dx) 2− B(dt dx) + C = 0. (2.20) Since B2(x, t) − 4A(x, t)C(x, t) > 0,
there are two curve. These are called characteristics. On two characteristics, the equation (2.19) is simplified to solve AdS dx( dt dx) + C dT dx + e( dt dx)] = 0.
The procession for solving this problem is shown as follows.
First, we have to know what the values m of the tangent slope at initial points P and
Q on the curve C are. Let m+ and m− be the right characteristic and the left characteristic
at one point, respectively. As shown in Figure 3. This value is easy to get from equation (2.20). mQ -mP + + R Q P
Figure 3: Two characteristics.
Second, we use lines to approximate these curves and we have to find the point R. An line equation of the right characteristic at the point P is
L1 : t − tP = m+p(xR− xP)
and a line equation of the left characteristic at the point Q is
L2 : t − tQ= m−Q(xR− xQ)
The intersection of above two lines L1 and L2 is the point R. Hence, we can find a pint R
from two lines L1 and L2.
Third, we want to find the value of SR and TR. Denote mav the average of the tangent
slope of any two points in the line. From P to R of L1, we derive the equation
From Q to R of L2, we obtain the equation
AQ(x, t)(SR− SQ)mav+ CQ(TR− TQ) + eQ(tR− tQ) = 0,
where an index of variables A, C, S , T, G and t represents the values of A, C, S , T, G and t at that point. Finally, we have value of u at the point R from following equation (2.21)
uR− uP =
1
2(SP+ SR)(xR− xP) + 1
2(TP+ TR)(tR− tP) (2.21)
Example 2.14. Using the method of characteristics to calculate displacement u at time t = 0.3 of the point on the stringx = 0.4.
utt(x, t) = uxx, 0 < x < l, t > 0, u(x, 0) = 1 2x(1 − x), 0 ≤ x ≤ l, ut(x, 0) = 0, 0 ≤ x ≤ l, u(0, t) = u(l, t) = 0, t ≥ 0. (2.22)
A(x, t) = −1, B(x, t) = 0, C(x, t) = 1 and e(x, t) = 0 in the partial differential equation of
problem (2.15). Since m+P = 1 and m−Q = −1, the right characteristic through P and the left characteristic through Q are L1 : x − t = 0.1, L2 : x + t = 0.7, respectively. Here, S (x, t) =
and T (x, t) = 0, 0 ≤ x ≤ l, t > 0. Then SP = 0.4, SQ = −0.2, TP = 0, TQ = 0.
From P to R in L1, we have
(−1)(SR− 0.4) + TR = 0.
From Q to R in L2, we get
(SR+ 0.2) + TR = 0.
Those equations imply SR = 0.1, TR = −0.3. Finally, the value of u at the point R from
equation (2.21) is
0.045 +0.4 + 0.1
2 (0.4 − 0.1) +
0 − 0.3
2 (0.3 − 0) = 0.075. Thus, a displacement u at time t = 0.3 of the point on the string x = 0.4 is 0.075.
2.7
The method of Finite Difference to the Wave Equation
The methods of integral transforms (like as Fourier, Laplace transform) and separation of variables are only valid on special problems. And we obtain an infinite integral form or a sum of infinite series form of solutions. If we can not compute them exactly, we want to approximate them by numerical methods. Suppose a function u(x, t) and its derivatives are continuous and finite. We give v(x, t) a good approximation to u(x, t).
A stretch string fixed two points problem (2.14) has done by the methods of character-istic. Now, we try to solve it by the method of finite difference for l = 1. Denote mesh parameter h = Nl and k = h such that x = 0, h, 2h, . . . , Nh = 1, t = 0, k, 2k, . . . , N = 0.1. Let a mash function ν(nh, mk) satisfies
Λh[ν] = − ν((n + 1)h, mk) − 2ν(nh, mk) + ν((n − 1)h, mk) h2 + ν(nh, (m + 1)k) − 2ν(nh, mk) + ν(nh, (m − 1)k) k2 = 0, u(x, 0) = 1 2x(1 − x), 0 ≤ x ≤ 1, ut(x, 0) = 0, 0 ≤ x ≤ 1, u(0, t) = u(1, t) = 0, t ≥ 0. (2.23)
Then we get a recursion formula
ν(nh, (m + 1)k) = ν((n + 1)h, mk) − ν((n − 1)h, mk) − ν(nh, (m − 1)k), (2.24)
m ≥ 1. But there does not have the value ν at time −1 of point nh in above recursion
formula (2.24). Initial conditions of problem (2.23) provides some information
ν(nh, −k) = ν(nh, k). (2.25)
Put (2.25) into a recursion formula (2.24), then it becomes
ν(nh, k) = 1
2[ν((n + 1)h, 0) − ν((n − 1)h, 0)]. (2.26) Thus, the following table and Figure 4 shown the discrete values
t 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0 0.045 0.08 0.105 0.12 0.125 0.12 0.105 0.08 0.045 0 0.1 0 0.04 0.075 0.1 0.115 0.1225 0.115 0.1 0.075 0.04 0 0.2 0 0.03 0.06 0.085 0.1025 0.105 0.1025 0.085 0.06 0.03 0 0.3 0 0.02 0.04 0.0625 0.075 0.0825 0.075 0.0625 0.04 0.02 0
Table 2:Discrete data to problem (2.14) with l = 1.
0 0.2 0.4 0.6 0.8 1
x
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14u
t=0 t=0.1 t=0.2 t=0.3)3
The Wave equation in different domains
3.1
Solution of the finite string problem
Consider a stretch string fixed two end points problem (2.1). The D’Alembert’s solution of a finite string problem is
u(x, t) = 1 2[ f (x + ct) + f (x − ct)] + 1 2 Z x+ct x−ct g( ¯x)d ¯x,
where f (x) and g(x) odd extend about points x = 0 and x = l. They extend the domain
D = {(x, t) | 0 ≤ x ≤ l, t > 0} to {(x, t) | x ∈ R, t > 0}.
Standing wave solutions of a finite string is
u(x, t) = ∞ X n=1 [ancos( nπc l )t + bnsin (nπcl)t] sin( nπ l x), where an= 2 l Z l 0 f (x) sin(nπ l )xdx, bn = 2 nπc Z l 0 g(x) sin(nπ l )xdx.
Proposition 3.1. The integral form of D’Alembert’s solution is the same as the infinite series form from the method of separation of variables.
u(x, t) = 1 2[ f (x + ct) + f (x − ct)] + 1 2 Z x+ct x−ct g( ¯x)d ¯x = 1 2[ ∞ X n=1 ansin( nπ l )(x + ct) + ∞ X n=1 ansin( nπ l )(x − ct)] + 1 2c Z x+ct x−ct nπc l ∞ X n=1 bnsin( nπ l ¯xd ¯x) = ∞ X n=1 ansin( nπ l x) cos( nπc l t) + nπ 2l ∞ X n=1 Z x+ct x−ct bnsin( nπ l ) ¯xd ¯x = ∞ X n=1 ansin( nπ l x) cos( nπc l t) − 1 2[ ∞ X n=1 bncos( nπ l )(x + ct) − ∞ X n=1 bncos( nπ l )(x − ct)] = ∞ X n=1 ansin( nπ l x) cos( nπc l )t + bnsin( nπ l x) sin (nπcl)t = ∞ X n=1 [ancos( nπc l )t + bnsin (nπcl)t] sin( nπ l x).
The integral form of D’Alembert’s solution is the same as the series form of separation of variables.
There are differences between D’Alembert’s formula and Separation of variables. The method of separation of variables also solves the special problem of the wave equation with nonconstant c. This method is restricted to on a finite boundary condition. Like as rectangle,circle or cylinder,and so on. In fact, it does not work for all equations. For instance, a partial differential equation with the variable coefficients. It is hard to separate the equation such that one side of only involving in x and the other in y. Even if it with constant coefficients, it may be not apply to.For example, uxx+ uxy+ uyy= 0 is not separable
in rectangular coordinates. But it can be in polar coordinates.
There is the comparison between an analytic solution and numerical solutions in Ex-ample (2.14) with l = 1 in following Table.
x 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Analytic solution 0 0.02 0.04 0.06 0.075 0.08 0.075 0.06 0.04 0.02 0 Finite Difference 0 0.02 0.04 0.0625 0.075 0.0825 0.075 0.0625 0.04 0.02 0 Characteristic 0 0.02 0.04 0.06 0.075 0.08 0.075 0.06 0.04 0.02 0 Table 3: Discrete solutions by different methods to Example (2.14) with l = 1 at t = 0.3.
From above Table 3, the solution by characteristics is an exact solution. Because charac-teristics are just lines in this example for constant speed c, there is no error in it. Hence, we can get a correct solution. The solution by the finite difference method is approximate to the accurate solution. It is a stable solution with an error of second order. Comparing to analytic solution, the numerical solution is the discrete data. We must evaluate the value one by one. If it is not easy to find the analytic solution, then it is a good choice to evaluate the discrete solution by the computer and give a good approximation for analytic solution.
3.2
Solution of the infinite problem
utt(x, t) = c2uxx(x, t), −∞ < x < ∞, t > 0, u(x, 0) = f (x), −∞ < x < ∞, ut(x, 0) = g(x), −∞ < x < ∞, lim x→∞u(x, t) = limx→∞ ∂u(x, t) ∂t = 0, t > 0.Fourier Transform of this initial-fixed boundary value problem on x is
d2 dt2ˆu(ω, t) + c 2ω2ˆu(ω, t) = 0, −∞ < ω < ∞, t > 0, ˆu(ω, 0) = F(ω), −∞ < ω < ∞, d dt ˆ u(ω, 0) = G(ω), −∞ < ω < ∞. (3.1)
Problem (3.2) becomes ordinary differential equation of ω for second order, then the solution is ˆu(ω, t) = F(ω) cos ωct + G(ω) ct sin ωct = F(ω)e iωct+ e−iωct 2 + G(ω) cω eiωct− e−iωct 2i . (3.2)
By Inversion of Fourier Transform Theorem, we have
u(x, t) = 1 2[ f (x + ct) + f (x − ct)] + 1 2 Z x+ct x−ct g( ¯x)d ¯x,
4
Riemann surface of genus N
The procession of solving the general initial value problem of linear hyperbolic equations by Fourier transform or Laplace transform may meet the integral of the square root func-tion. Here, we present how to deal with this multi-valued functions such that the integral is meaning and correct evaluated by Mathematica.
4.1
Introduction
We know the square root function is a multi-valued function in the complex plane. Because
z = reiθ= ei(θ+2π), Then √ z = √ rei(2θ) √ rei(θ+2π2 ) = −√rei(θ2)
It is clear that a square root function is a two-valued function defined on the complex plane C. So, it is not continuous on C. Moreover, it is not analytic on C. Hence, it is meaningless for evaluate the integral of multi-valued function. The only way to change multi-valued functions into single value function is redefined the domain. Denote two the square root functions
f1(z) = |z| 1 2e iarg(z) 2 and f2(z) = |z| 1 2e iarg(z) 2 .
Let D1 = {C \ (−∞, 0] | arg(z) ∈ [−π, π)} and D2 = {C \ (−∞, 0] | arg(z) ∈ [π, 3π)}
be the domains of square root functions f1(z) and f2(z), respectively. Here, functions f1(z)
and f2(z) are single valued function in each domain. Since f1(z) and f2(z) are both
dis-continuous on the negative real line, denote branch cuts C1 = {(−∞, 0] | arg(z) = π} and
y x -ʌ ʌ y x 3ʌ ʌ
Figure 5: The domain D1 and D2.
y x ʌ + ʌ -arg z S 3ʌ - S 3 argz -ʌ+ y x
Figure 6: The branch cut C1and C2.
Let functions defined on above two cuts be
f3(z) = i|z| 1 2, z ∈ C 1. and f4(z) = −i|z| 1 2, z ∈ C 2.
And functions f3(z) and f4(z) are single valued function on each cuts. These domains are
on Riemann surface as following: √ z = f1(z), z ∈ D1 f2(z), z ∈ D2 f3(z), z ∈ C1 f4(z), z ∈ C2
Remark 4.1. The square root function is analytic in the domain D1∪ D2∪ C1∪ C2. Since
functions f1(z), f2(z), f3(z) and f4(z) are analytic in each domains. And any path pass from
D1 to D2through the cut C1 is also continuous. Because z ∈ D1 and argz tends to π, then
f1(z)tends to i|z|
1
2. It is equal to the function f2(z), where argz = π. It is similar to any path from D2 to D1 through the cut C2 is still continuous. Hence, the function is analytic on
Riemann surface. Remark 4.2. f1(z) = − f2(z) and f3(z) = − f4(z). Since z ∈ D2, arg(z) ∈ [π, 3π). f2(z) = |z| 1 2eiθ+2π2 = |z|12ei θ 2eiπ = −|z|12ei θ 2 = − f1(z), where θ ∈ [−π, π).
We will develop an algorithm such that we can evaluate the integrals of a square root function by Mathematica.
4.2
The algebraic structure on Riemann surface
Since f (z) = √z is a two-valued function, we need construct branch cuts to cut our plane.
For every point in complex plane, we have arg(z − 0) ∈ [−π, π). In following Figure 7, + edge is the initial edge and − edge is terminal edge. The argument of + edge is −π and of
0 Ё
Ѐ
Figure 7: The cut plan of √z.
We define (z, f (z)) belong to sheet I if arg(z − 0) ∈ [−π, π); (z, f (z)) belong to sheet II if arg(z − 0) ∈ [π, 3π). Remark 4.3. z ∈ I+: f (z) = |z|12ei −π 2 = −i|z|12 z ∈ I−: f (z) = |z|12ei 2π 2 = i|z|12
Hence, f (z)|I− = − f (z)|I+. Since - edge in I is + edge in II, it means f (z)|II = − f (z)|I.
After presenting the algebraic structure of f (z) = √z on Riemann surface, we have to
show the cut structure for the general horizontal and vertical cuts. Let p(z) = √(z − z1)(z − z2)
For
Case1.z1, z2 ∈ R(Horizontal cut)
z1, z2are branch points of p(z). It is discontinuous on this interval [z1, z2]. Hence, there is a
branch cut between two points.
For example, p(z) = √z − 1√z − 2, chosen a point 0 ∈ (−∞, 1).
z2 Ё z1
+
Figure 8: The algebraic structure of p(z) for two branch points in horizontal.
(ii) For 0 in - edge: Then arg(z − 1) = arg(z − 2) = π
arg(0 − 1) = arg(−1) = −π π arg(0 − 2) = arg(−2) = −π π
Taken the principal argument of a negative number is −π, hence
√ −1 · √−2 = |1|12ei( −π 2)· |2| 1 2ei( −π 2) = |2| 1 2ei(−π) = −|2| 1 2 (4.1)
On the other hand, taken the principal argument of a negative number is π , then
√ −1 · √−2 = |1|12ei( π 2)· |2| 1 2ei( π 2) = |2| 1 2ei(π) = −|2| 1 2 (4.2)
Here, the value in equation (4.1) is equal to equation (4.2). So, p(z) is continuous on this interval (−∞, 1). Pick up one point 32 ∈ (1, 2).
(i) For 32 in + edge: Then arg(z − 1) = arg(z − 2) = −π (ii) For 32 in − edge: Then arg(z − 1) = arg(z − 2) = π
arg(3 2 − 1) = arg( 1 2) = 0 arg(3 2 − 2) = arg( −1 2 ) = −π π
By the same way as above, chosen the principal argument of a negative number is −π,
hence r 1 2· r −1 2 = | 1 2| 1 2ei(0)· |1 2| 1 2ei( −π 2) = |1 2|e i(−π2) = −|1 2|i (4.3)
Chosen taken the principal argument of a negative number is π, then r 1 2 · r −1 2 = | 1 2| 1 2ei(0)· |1 2| 1 2ei( π 2) = 1 2e i(π2)= |1 2|e i(π2) = |1 2|i (4.4)
Since the value in equation (4.3) is unequal to equation (4.4). Then the function is dis-continuous on this interval (1, 2) . So, there is a branch cut in it. Of course, a square root function is continuous for positive real number. Hence, there is a branch cut between two points z1, z2 ∈ N.
Case2.z1, z2 ∈ iR(vertical cut)
The branch cut for the vertical cut structure is the same as horizontal. Denote the interval [z1, z2] be the branch cut. The algebraic structure of p(z) for two branch points in vertical as
shown in Figure 9 We define the sheet is I = {z | arg(z − zj) ∈ [−32π,12π)} and another sheet
Ё z1
z2 +
Figure 9: The algebraic structure of p(z) for two branch points in vertical.
is II = {z | Arg(z − zj) ∈ [12π,52π)}, where zj is the branch point of p(z), for each positive
integer j = 1, 2. + edge is the initial edge and − edge is terminal edge. The argument of + edge is −32π and of − edge is 12π.
Remark 4.4. As we know, a curve crosses the cut from one sheet to another sheet. Hence,
if a curve goes through 2N − 1 cuts for N ∈ N, then a curve will cross to another sheet. So, there is a branch cut at that line segment.
Remark 4.5. By the same way as horizontal case, thus the value in sheet II is minus of
in sheet I. If z ∈ I+(+ edge of sheet I) and z ∈ [z1, z2], then arg(z − z1) = −32π and
arg(z − z1) = 12π. So, p(z) = p(z − z1)(z − z2) = 2 Y j=1 p (z − zj) = |z − z1| 1 2ei(− 3π 4)· |z − z2| 1 2ei(− π 4) = − 2 Y j=1 |z − zj| 1 2
If z ∈ I−(− edge of sheet I), then arg(z − z1) = 12π and arg(z − z1) = −12π. So,
p(z) = p(z − z1)(z − z2) = 2 Y j=1 p (z − zj) = |z − z1| 1 2ei( π 4)· |z − z 2| 1 2ei(− π 4) = 2 Y j=1 |z − zj| 1 2 So, f (z)|I− = − f (z)|I+. It means f (z)|II = − f (z)|I.
4.3
The geometric structure on Riemann surface
After presenting the algebraic structure, we want to know how does it look like in geomet-ric. For the horizontal branch cut case, by above definition, sheet I and sheet II is denoted by I = {z | arg(z − zj) ∈ [−π, π)} and II = {z | arg(z − zj) ∈ [π, 3π)},respectively.We glue
the + edge of sheet I to the − edge of sheet II. When the integral path crossing the branch cut, then it pass from one sheet to the other sheet. If we want to evaluate the value in sheet
II, then we only evaluate the negative value of the value in sheet I. Now, we discuss the
geometric structure on Riemann surface of f (z) = √z. Using stereo projection, we have
complex plane is a point and is projected to a north pole of the surface. Hence, a geometric structure of f as Figure 10 shown. And the geometric structure of vertical cuts on Riemann surface of p(z) is the same as horizontal cuts.
f 0 0 f 0 + Ё ҇ ҈ Ё + f 0 f 0 f Ё Ё + + 0 f + Ё Ё + Ё +
Figure 10: The geometric structure on Riemann surface of p(z) in horizontal cuts.
Consider a general function f (z) =
s n Q j=1
(z − zj). Since z1, z2, ..., zn are the n branch
points, then there are dn2e − 1 holes on Riemann surface, where dn2e =
n 2 n : even n+1 2 n : odd
4.4
The integrals over a, b cycles on Riemann surface
After analyzing the algebraic structure and geometric structure of p(z) = √(z − z1)(z − z2)
on Riemann surface, we want to evaluate the integral of p(z)1 . Since every simple closed curves can be written as a linear combination of a, b cycles. Hence, we discuss the integral of canonical cycles a, b. Consider a general function f (z) =
s n Q j=1 (z − zj). We want to evaluateH a 1 f (z) and H b 1
f (z) by another equivalent paths of a, b cycles such that the integrals
are easier computed.
For horizontal branch cut:
The following Figure 11 is a, b cycles on Riemann surface. For n is odd:
For n is even:
Figure 11: The a, b cycles.
Now, we want to evaluate the integral of f (z) = √(z + 2)(z + 1)(z − 1)(z − 2)(z − 3)(z − 4) over a, b cycles as Figure 12. The algebraic and geometric structure:
Figure 12: a1, b1 cycle.
Figure 13: Geometric structure.
We want to evaluate the integral by Mathematica, but there is something wrong. The argument evaluated by Mathematica is (−π, π], and the argument in our theory is [−π, π). In Mathematica, it regards the argument −π in our theory as π. Hence, we have to modify the value evaluated from Mathematica.
For f (z) = √z
In Theory:
z
Figure 14: The argument in theory.
In Mathematica:
z
Figure 15: The argument in Mathematica.
Therefore, f (z) = √z = −M arg(z) = −π M otherwise
By the same, we only modify the value for any f (z) = √z − zjwhen arg(z − zj) = −π.
For a1cycle:
sheet I, by Cauchy integral Theorem.
Let z = 32 + eiθ,then dz = ieiθdθ. Since arg(z − z
j) ∈ [−π, π) for 1 ≤ j ≤ 6. So,
a1
1
f (z)dz is
same as the value evaluated by Mathematica. Hence,
a1 1 f (z)dz = Z π −π ieiθ q 3 2 + e iθ+ 1 r 3 2+ e iθ+ 2 r 3 2 + e iθ− 1 r 3 2 + e iθ− 2 r 3 2+ e iθ− 3 r 3 2 + e iθ− 4 · dθ = −1.13022i
For the equivalent path a∗1:
a1 Ё + Ё + Ё + 3 2 0 -2 -1 1 a*1 4 Figure 16: a1cycle.
The argument of Mathamatica evaluated is (−π, π], then the argument of − edge is regarded as π. Hence we must modify the integral on − edge by multiple (−1).
1 to 2 2 to 1
Integral path angle value angle value
z − 4 −π −M π +M z − 3 −π −M π +M z − 2 −π −M π +M z − 1 0 +M 0 +M z + 1 0 +M 0 +M z + 2 0 +M 0 +M Sheet I +M I +M Total −M +M
Table 4: Angles and values for z − zj along integral path a∗1.
Hence, a∗1 1 f (z)dz = Z 2 1 1 √ z + 1√z + 2√z − 1√z − 2√z − 3√z − 4dz = −1.13022i
Therefore, the integral over a∗1cycle is equal to over a1cycle
a∗1 1 f (z)dz = I a∗1 1 f (z)dz = −1.13022i For b1cycle:
The integral over b1cycle is same over the circle of radius 52at center 1 and the circle in
the sheet I, by Cauchy integral Theorem. A dotted line is in sheet II. Let z = 1 +52eiθ,then
dz = 52ieiθdθ.Since arg(z − z
j) ∈ [−π, π) for 1 ≤ j ≤ 6. So, H b1 1 f (z) is evaluated correct by Mathematica. Hence,
Figure 17: b1cycle. b1 1 f (z)dz = Z π −π 5 2ie iθ q 5 2+ e iθ+ 1 q 5 2 + e iθ+ 2 q 5 2+ e iθ− 1 q 5 2 + e iθ− 2 q 5 2+ e iθ− 3 q 5 2 + e iθ− 4 dθ = −0.0760776
1 to 2 2 to 1
Integral path angle value angle value
z − 4 −π −M π +M z − 3 −π −M π +M z − 2 −π −M π +M z − 1 0 +M 0 +M z + 1 0 +M 0 +M z + 2 0 +M 0 +M Sheet I +M II −M Total −M −M
Table 5: Angles and values for z − zj along integral path b∗1. b∗1 1 f (z)dz = 2[ Z −1 3 1 √ z + 1√z + 2√z − 1√z − 2√z − 3√z − 4dz] = −0.0760776
For vertical branch cut : The argument evaluated by Mathematica is (−π, π], but the argu-ment in our theory is [−32π, −12π). In Mathematica, it regards the argument belong [−32π, −π] in our theory as [−12π, π]. Hence, we have to modify the value evaluated from Mathematica.
For f (z) = √z − i
In Theory:
i i
z
Figure 18: The argument in theory.
In Mathematica:
z
i i
Figure 19: The argument in Mathematica.
Therefore, f (z) = √z = −M arg(z − i) ∈ [−32π, −12π] M otherwise
By the same, we only modify the value for any f (z) = √z − zj when arg(z − zj) =
[−32π, −12π]. For n is even:
The following Figure 20 is a, b cycles on Riemann surface.
z4 a1 z1 ak z 1 n zn z3 + - + - + - z2 b1 b2 bk z2k z2 1 k a2 Figure 20: a, b cycle.
Now, we want to evaluate the integral of
f (z) = s 1
6
Q j=1
(z − zj)
where z1 = 1 + 2i, z2 = 1, z3= 3i, z4 = i, z5 = −1 + 3i and z6 = −1 + i over a, b cycles as in
Figure 21.
Since the argument of Mathematica is (−π, π] , we must modify the value on the argu-ment [32π, −π] in vertical cut. We regard the branch point as the origin of the rectangular coordinate system in the plane. Hence, we modify the second sign of every rectangular coordinate system of center for each branch point.
1 + Ё 1+2ȓ 3ȓ + Ё -1 O a1 -1+3ȓ -1+ȓ + Ё Figure 21: a1cycle.
The integral over a1 cycle is the same as over the enclosed rectangle in Figure 22.
Integral path (1) (2) (3) (4) (5) (6) (7) z − (1 + 2i) +M +M −M −M −M −M +M z − 1 −M −M −M −M −M −M −M z − 3i +M +M +M +M −M +M +M z − i +M +M +M +M −M −M −M z − (−1 + 3i) +M +M +M +M +M +M +M z − (−1 + i) +M +M +M +M +M +M +M Sheet +M +M +M +M +M +M +M Total −M −M +M +M +M −M +M
Table 6: Angles and values for z − zj along integral path a1.
a1 1 f (z)dz = − Z 1 3+2i −1 3+i 1 f (z)dz + Z −1 3+3i 1 3+2i 1 f (z)dz − Z −1 3+2i −1 3+3i 1 f (z)dz + Z −1 3+i −1 3+2i 1 f (z)dz = 1.38321 − 2.33762i
For the equivalent path a∗1in Figure 23.
ȓ O 1 + Ё + Ё 1+2ȓ 3ȓ -1+ȓ -1+3ȓ * 1 a Figure 23: a∗1cycle.
Integral path 3i to 2i 2i to i i to 2i 2i to 3i z − (1 + 2i) −M +M −M +M z − 1 −M −M −M −M z − 3i +M +M +M +M z − i −M −M +M +M z − (−1 + 3i) +M +M +M +M z − (−1 + i) +M +M +M +M Sheet I I I I Total −M +M +M −M
Table 7: Angles and values for z − zj along integral path a∗1.
a∗1 1 f (z)dz = − Z 2i 3i 1 f (z)dz + Z i 2i 1 f (z)dz − Z 2i i 1 f (z)dz + Z 3i 2i 1 f (z)dz = 1.38321 − 2.33762i For b1 cycle: Figure 24: b1cycle.
Figure 25: b1cycle. Integral path (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) z − (1 + 2i) +M +M +M +M +M +M +M +M −M −M −M +M z − 1 −M −M −M −M −M +M +M +M −M −M −M −M z − 3i +M +M +M +M +M +M +M +M +M −M +M +M z − i −M +M +M +M +M +M +M +M +M +M −M −M z − (−1 + 3i) +M +M +M +M +M +M +M +M +M −M −M −M z − (−1 + i) −M +M +M +M +M +M +M +M +M +M +M +M Sheet I I I I I II II II II II II II Total −M −M −M −M −M −M −M +M −M −M +M −M
Table 8: Angles and values for z − zj along integral path b1.
b1 1 f (z)dz = − Z −1 2+3i −1+2i 1 f (z)dz + Z −1 2+2i −1 2+3i 1 f (z)dz − Z −1+2i −1 2+2i 1 f (z)dz = 0.590344 − 1.16143i
Figure 26: a1cycle.
−1 + i to 1 1 to −1 + i
Integral path value value
z − (1 + 2i) +M +M z − 1 −M −M z − 3i +M +M z − i +M +M z − (−1 + 3i) +M +M z − (−1 + i) +M +M Sheet I II Total −M +M
Table 9: Angles and values for z − zj along integral path b∗1.
b∗1 1 f (z)dz = 2 Z −1+i 1 1 f (z)dz = 0.590344 − 1.16143i
4.5
Solutions to Liner Hyperbolic Equations by Mathematica
Now, we want to solve the infinite problem (4.5) with the source term only involving time-independent.
Example 4.6. Using Laplace transform, and then Fourier transform to solve the following I.V.P. utt(x, t) = uxx(x, t) + sin2 √ t, −∞ < x < ∞, t > 0, u(x, 0) = 0, −∞ < x < ∞, ut(x, 0) = 0, −∞ < x < ∞. (4.5)
First, using the method of Laplace Transform with respect to t, we have
s2U(x, s) − Uxx(x, s) = √ π s√se −1 s, s > 0. Note. L [sin 2√t] = √ π s√se −1 s. Since sin 2√t = 2√t −(2 √ t)3 3! + (2√t)5 5! − (2√t)7 7! + ... Then L [sin 2√t] = 2Γ( 3 2) s32 − 8Γ( 5 2) 3!s52 + 32Γ( 7 2) 5!s72 − 128Γ( 9 2) 7!s92 + ... = √ π s32 [1 − 1 s + 1 2s2 − 1 3s3 + ...] = √ π s√se −1 s.
Using the method of Fourier Transform with respect to x, it becomes
s2U(ω, s) + ωˆ 2U(x, s) =ˆ √ π s√se −1 s2πδ(ω). This implies ˆ U(ω, s) = 1 s s s2+ ω2 √ π s√se −1 s2πδ(ω).
By Convolution Theorem, we derives U(x, t) = 1 s √ π s√se −1 s{F−1[ s s2+ ω2]F −1[2πδ(ω)]} = 1 s √ π s√se −1 s[ ∞ Z ∞ 1 2e −s|x−y|· 1dy] = √ π s3√se −1 s.
By Inversion Theorem of Laplace Transform, we have
u(x, t) = 1
2πiL→∞lim
Z s+iL
s−iL
U(x, τ)eτtdτ, s > 0.
Let G(x, τ) = U(x, τ)eτt and we apply Cauchy’s theorem to the integral of G(x, s) over the contour shown as following Figure 27. Since G(x, s) is analytic inside this contour C,
Figure 27: The integral contour C of G(x, τ).
(1)Along the path C1 of contour C: Let τ = s + Leiθ,π2 ≤ θ ≤ π, dτ = iLeiθdθ
Z C1 G(x, τ)dτ = Z π π 2
e(s+Leiθ)tU(x, s + Leiθ) · iLeiθdθ
Since |U(x, s + Leiθ)| = | √ π (s + Leiθ)3√s + Leiθe − 1 s+Leiθ| ≤ | √ π (L − s)72 |
approach zero as L → ∞. By Jordan’s lemma the integrals over this contour C1approach
zero.
(2)Along the path C2of contour C: Let τ = εeiθ, where −π ≤ θ ≤ π, dτ = iεeiθ. Since
|U(x, εeiθ)| = |
√ π (εeiθ)| 7 2|e− 1 εeiθ| ≤ √ π ε e −1 ε, (4.6)
By L’Hospital Rule twice times, we evaluate the value of the right hand side for above inequality (4.6) as ε → ∞ is lim ε→0 ε−7 2 e1ε = lim ε→0 7 2ε −1 2 e1ε = lim ε→0 7 4ε 5 2 e1ε = 0.
By Jordan’s lemma, the integral over this contour C2approach zero as ε → 0.
(3)Along the path C3of contour C:
Similar as contour C1, let τ = s + Leiθ, π ≤ θ ≤ 3π2, dτ = iLeiθdθ
Then integrals over the contour C3approach zero.Hence, we have
u(x, t) = −1 2πi[ Z 0 ∞ G(x, τ)dτ + Z −∞ 0 G(x, τ)dτ] −∞ to 0 0 to −∞
Integral path angle value angle value
s π +M −π −M
Sheet I +M I +M
Total +M −M
Table 10: Angles and values for s along the equivalent integral path.
We must modify the value on the integral path from a point 0 to −∞ by Mathematica, so
u(x, t) = 1 2πi[−M{ Z 0 ∞ G(x, τ)dτ} + M{ Z −∞ 0 G(x, τ)dτ}] = 1 2πi[2M{ Z −∞ 0 G(x, τ)dτ}], where M{R
CG(x, τ)dτ} is represented by the integral value G evaluated by Mathematica on
source term only involving the time-variable t. Fixed x = 1
At the time t = 1, the value
u(x, t) = 0.396896.
At the time t = 2, the value
u(x, t) = 1.63306.
This is the same as the value evaluated by D’Alembert’s solution for t = 1, 2. Here, D’Alembert’s solution of problem (4.5) is
u(x, t) = 1 2 Z t 0 Z x+(t−τ) x−(t−τ) sin 2√¯td ¯xd¯t = −3 2 √ t cos 2√t + 1 4(3 − 4t) sin 2 √ t.
For fixed x = 1, the displacement in a long string as shown in the Figure 28.
20 40 60 80 100 t -200 -150 -100 -50 50 100 150 u+1,t/
References
[1] Weinberger H.F., A First Course in Partial Differential Equations with Complex Variables and Transform Methods, Chelsea,1965.
[2] Roger Knobel, An introduction to the Mathematical Theory of Waves, 2000. [3] G.D.SMITH, Numerical Solution of Partial Differential Equations,1964. [4] Walter A. Strauss, Partial Differential Equations, 1992.
[5] Tyn Myint-U, Partial Differential Equations of Mathematical Physics,1973.
[6] Yu-Wei Hsiao, NCTU, Master thesis, Integral Evaluations on Three-sheeted Riemann surfaces of Genus N of type II, Taiwan, 2001.
