ONE DIMENESIONAL WAVE EQUATIONS
M.Sc. Graduate Seminar
Gemeda Tolessa
November, 2012
Haramaya University
ONE DIMENESIONAL WAVE EQUATIONS
A Graduate Seminar Submitted to the Department of Mathematics,
School of Graduate Studies
HARAMAYA UNIVERSITY
In partial Fulfillment of the Requirement for the Degree of
MASTER OF SCIENCE IN MATHEMTICS
(DIFFERENTIAL EQUATIONS)
By
Gemeda Tolessa
Advisor
Tsegaye Gedif (PhD)
November, 2012
Haramaya University
ii
SCHOOL OF GRADUATE STUDIES
HARAMAYA UNIVERSITY
As member of the Examination Board of the Final M.Sc. Open Defense, we certify that we have
read and evaluated this Graduate Seminar prepared by Gemeda Tolessa Entitled: OneDimensional Wave Equations and recommended that it be accepted as fulfilling the Graduate
Seminar requirements for the Degree of Master of Science in Mathematics.
___________________
Name of Chair person
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Signature
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Name of External Examiner
Signature
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Name of Internal Examiner
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Date
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Date
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Signature
Date
Final approval and acceptance of the Graduate Seminar is contingent upon the submission of the
final copy of the Graduate Seminar to the Council of Graduate Studies (CGS) through the
Department Graduate Committee (DGC) of the candidate’s department.
I hereby certify that I have read this Graduate Seminar prepared under my direction and
recommend that it be accepted as fulfilling the Graduate Seminar requirement.
Tsegaye Gedif (Ph.D.)
Name of Advisor
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Date
PREFACE
The objective of this seminar is to discuss the basic solution techniques associated with the one
Dimensional Wave Equations. It insights the general concepts of One Dimensional Wave
Equations. This seminar deals with a wave equation in one dimensional with methods of
solutions namely method of separation and d’Alembert’s method which may be successfully
applied to solve different types of partial differential equations.
The one-dimensional wave equations can be exactly solved by d'Alembert's solution and
separation of variables. Here, on this seminar paper considerable attention is given to the
techniques of separation of variables and d'Alembert's solution. However, additional methods
such as Laplace transforms, Fourier transforms and integral representations are also possible.
i
ACKNOWLEDGEMENTS
Different people have assisted me in various ways while doing this seminar paper. First and
foremost I would like to express my deepest gratitude to my advisor Dr. Tsegaye Gedif for his
unreserved guidance and immediate feedback. He critically read my draft and gave me critical
comments and constructive suggestions. His comments are insightful.
My special indebtedness goes to my wife, Aberash Gunja and her sympathy, love, morale, and
hospitality they provided me during the study of the whole courses.
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LIST OF ABBREVIATIONS
BC : Boundary Conditions
BVP: Boundary Value Problem
NBC: Neumann Boundary Condition
CW: Cauchy Problem
IC : Initial Conditions
ICW: Inhomogeneous Cauchy Problem
IVP: Initial Value Problem
ODE :Ordinary Differential Equation
PDE : Partial Differential Equation
MDW: Mixed Cauchy Problem
𝑐1 : Continuously Differentiable
𝑐 2 : Continuously Differentiable twice
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TABLE OF CONTENTS
PREFACE
I
ACKNOWLEDGEMENTS
II
LIST OF ABBREVIATIONS
III
1. 1 INTRODUCTION
1
1.2 Preliminary Concepts
1.2.1 Classification of partial differential equations
1.2.2 The canonical forms of second order partial differential equations
1.2.3 Reduction partial differential equation into a canonical form
1.2.4 Method and Techniques for solving Partial differential equations
2
2
4
4
6
ONE DIMENSIONAL HOMOGENEOUS WAVE EQUATIONS
9
2.1 Derivation of One Dimensional Wave equations
9
2.2 General Solution of wave equations
11
2.3 Initial Value Condition: Cauchy Problem
2.3.1 Domain of Dependence
2.3.2 Domain of Influence
2.3.3 Well-posedness
14
16
17
18
2.4 Separation of Variable
19
2.5 The Wave Equation on the Half line: Reflection Method
21
2.6
23
Mixed Problems for the Wave Equations
ONE - DIMENSIONAL NON-HOMOGENEOUS WAVE EQUATIONS
26
3.1. Initial Condition: Cauchy problem for the Non-homogeneous wave equation 26
3.2. D’Alembert’s Solution for Non-homogeneous Wave Equations
26
3.3 SUMMARY
31
REFERENCES
32
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CHAPTER ONE
1. 1 Introduction
Investigations in the theory of wave propagation lead quite naturally to the consideration of a
linear, hyperbolic partial differential equation the wave equation. In 1727, John Bernoulli (16671748) treated the vibrating string problems by imagining the string to be a thread having a
number of equally placed weight along it. However, because his governing equation was time
independent. It was not truly partial differential equation. The French Mathematician Jean Le
Rond Almebrt (1717-1783) derived the One Dimensional Wave equations as we know it today
by letting the number of weight in the Bernoulli ‘s model become infinite while at the same time
allowing the place between them to go to zero . His famous solution widely known as d’ Almbert
solution, appeared around 1746, six years before the separation of variable technique was
introduced by Daniel Bernoulli (1700-1782), Leonard Euler (1707-1783), Bernoulli and Josef
Lous Lagrange (1736-1813) all solved the wave equations in the mid 1700’s by Method of
separation of variables .
Wave equations are examples of hyperbolic partial differential equations. In its simplest form,
the wave equation concerns a time variable t, one spatial variables 𝑥 and a scalar function
𝑢(𝑥, 𝑡)whose values could model the height of a wave. The one dimensional wave equation has
form
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0
where 𝑥 is the (spatial) and c is a fixed constant. Solutions of this equation that are initially zero
outside some restricted region propagate out from the region at a fixed speed in one spatial
direction, as do physical waves from a localized disturbance; the constant c is identified with the
propagation speed of the wave. This equation is linear, as the sum of any two solutions is again a
solution: in physics this property is called the superposition principle.
1
1.2 Preliminary Concepts
This chapter introduces basic concepts and definitions for partial differential equations (PDE)
and solutions to a variety of PDEs. Applications of the method of separation of variables a
represented for the solution of second-order PDEs.
Definition1.1: A partial differential equation (PDE) is an equation that1has an unknown function
depending on at least two variables, contains some partial derivatives of the unknown function.
Equations involving one or more partial derivatives of a function of two or more independent
variables are called partial differential equations (PDEs).Well known examples of PDEs are
following.
Notations:
i) 𝑡, 𝑥, 𝑦, 𝑧 ( 𝑜𝑟 𝑒. 𝑔. 𝑟, 𝜃, ∅) ) – the independent variables (here, t
represents time while the other variables are space coordinates),
ii) 𝑢 = 𝑢(𝑡, 𝑥, , , , ) - the independent variables (the unknown function),
iii) the partial derivatives will be denoted as follows (e.g.)
𝜕𝑢
𝜕𝑢
𝜕2 𝑢
𝑢𝑥 = 𝜕𝑥 , 𝑢𝑥𝑦 = 𝜕𝑦𝜕𝑥, 𝑢𝑥𝑥 = 𝜕𝑥 2 , etc. is used
1.2.1 Classification of partial differential equations
The partial differential equations may be classified by the number of their independent variables,
that is, the number of variables that the unknown function depends on.
Definition 1.2: A PDE is linear if the dependent variable and all its derivatives appear in a linear
form.
Example 1.1:
PDE in two variables: 𝑢𝑡 = 𝑢𝑥𝑥 , (𝑢 = 𝑢(𝑡, 𝑥))
1
1
PDE in three variables: 𝑢𝑡 = 𝑢𝑟𝑟 + 𝑟 𝑢𝑟 + 𝑟 2 𝑢𝜃𝜃 , 𝑢(𝑢 = 𝑢(𝑡, 𝑟, 𝜃))
PDE in four variables: 𝑢𝑡 = 𝑢𝑥𝑥 + 𝑢𝑦𝑧 , (𝑢 = 𝑢(𝑡, 𝑥, 𝑦, 𝑧))
2
Definition 1.3: The order of a partial differential equation is the order of the highest partial
derivative in the equation.
Example 1.2:
First order: 𝑢𝑡 = 𝑢𝑥 ,
Second order: 𝑢𝑡 = 𝑢𝑥𝑥 , 𝑢𝑥𝑦 = 0
Third order: 𝑢𝑡 + 𝑢𝑢𝑥𝑥𝑥 = sin(𝑥),
Fourth order: 𝑢𝑥𝑥𝑥 = 𝑢𝑡𝑡
A second –partial differential linear PDE into variables can be written in the following general
form:
𝐴𝑢𝑥𝑥 + 𝐵𝑢𝑥𝑦 + 𝐶𝑢𝑦𝑦 + 𝐷𝑢𝑥 + 𝐸𝑢𝑦 + 𝐹𝑢 = 𝐺
Where A, B, C, D, E, and F are coefficients, and G is non-homogeneous term. All quantities are
constants or functions of x and y. The second –order linear is either
Hyperbolic: if 𝐵 2 − 4𝐴𝐶 > 0 e.g., heat flow and diffusion-type problems.
Parabolic: if 𝐵 2 − 4𝐴𝐶 = 0 e.g. vibrating systems and wave motion problems.
Elliptic:
if 𝐵 2 − 4𝐴𝐶 < 0 ., steady-state, potential-type problems
Example 1.3: 𝒚𝒖𝒙𝒙 + 𝒖𝒚𝒚
> 0 𝑓𝑜𝑟 𝑦 < 0 (ℎ𝑦𝑝𝑒𝑟𝑏𝑜𝑙𝑖𝑐)
= 𝟎 → 𝐵 − 4𝐴𝐶 = −4𝑦 { = 0 𝑓𝑜𝑟 𝑦 = 0 (𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑖𝑐)
< 0 𝑓𝑜𝑟 𝑦 > 0 (𝑒𝑙𝑙𝑖𝑝𝑡𝑖𝑐)
2
Definition 1.4: A second-order PDE in two variables (x and t) is linear and homogeneous if it
can be written in the following form
𝐴𝑢𝑥𝑡 + 𝐵𝑢𝑥𝑡 + 𝐶𝑢𝑡𝑡 + 𝐷𝑢𝑥 + 𝐸𝑢𝑡 + 𝐹𝑢 = 0
where the coefficients A, B, C, D, E, and F do not depend on the dependent variable 𝑢 = 𝑢(𝑥, 𝑡)
any of its derivatives though can be functions of independent variables (𝑥, 𝑡).
3
1.2.2 The canonical forms of second order partial differential equations
Any second-order linear PDE (in two variables)
𝐴𝑢𝑥𝑥 + 𝐵𝑢𝑥𝑦 + 𝐶𝑢𝑦𝑦 + 𝐷𝑢𝑥 + 𝐸𝑢𝑦 + 𝐹𝑢 = 𝐺
(where A, B, C, D, E, F, and G are constants or functions of x and y)can be transformed into the
so-called canonical form. This can be achieved by introducing new coordinates
𝜉 = 𝜉(𝑥, 𝑦) , 𝜂 = 𝜂(𝑥, 𝑦)
(in place of x and y) that simplify the equation to its canonical form.
For the hyperbolic PDE (i.e. if 𝐵 2 − 4𝐴𝐶 > 0) ,there are two possibilities
𝒖𝝃𝝃 − 𝒖𝜼𝜼 = 𝒇(𝝃, 𝜼, 𝒖, 𝒖𝝃 , 𝒖𝜼 ) , or
𝑢𝜉𝜂 = 𝑓(𝜉, 𝜂, 𝑢, 𝑢𝜉 , 𝑢𝜂 )
For the parabolic PDE (i.e. if 𝐵 2 − 4𝐴𝐶 = 0)
𝒖𝝃𝝃 = 𝒇(𝝃, 𝜼, 𝒖, 𝒖𝝃 , 𝒖𝜼 )
For the elliptic PDE (i.e. if 𝐵 2 − 4𝐴𝐶 < 0)
𝒖𝝃𝝃 + 𝒖𝜼𝜼 = 𝒇(𝝃, 𝜼, 𝒖, 𝒖𝝃 , 𝒖𝜼 )
1.2.3 Reduction partial differential equation into a canonical form
Introduce the new coordinates 𝜉 = 𝜉(𝑥, 𝑦) and 𝜂 = 𝜂(𝑥, 𝑦)
𝑢𝑥 = 𝑢𝜉 𝜉𝑥 + 𝑢𝜂 𝜂𝑥 , 𝑢𝑦 = 𝑢𝜉 𝜉𝑦 + 𝑢𝜂 𝜂𝑦
𝑢𝑥𝑥 = 𝑢𝜉𝜉 𝜉 2 𝑥 + 2𝑢𝜉𝜂 𝜉𝑥 𝜂𝑥𝑥 + 𝑢𝜂𝜂 𝜂2 𝑥 + 𝑢𝜉 𝑢𝑥𝑥 + 𝑢𝜂 𝑢𝑥𝑥 ,
𝑢𝑦𝑦 = 𝑢𝜉𝜉 𝜉 2 𝑦 + 2𝑢𝜉𝜂 𝜉𝑦 𝜂𝑦𝑦 + 𝑢𝜂𝜂 𝜂2 𝑦 + 𝑢𝜉 𝑢𝑦𝑦 + 𝑢𝜂 𝑢𝑦𝑦 ,
𝑢𝑥𝑦 = 𝑢𝜉𝜉 𝜉𝑥 𝜉𝑦 + 𝑢𝜉𝜂 (𝜉𝑥 𝜂𝑦 + 𝜉𝑦 𝜂𝑥 ) + 𝑢𝜂𝜂 𝜂𝑥 𝜂𝑦 + 𝑢𝜉 𝜉𝑥𝑦 + 𝑢𝜂 𝜂𝑥𝑦 .
Substituting these values into the original equation to obtain a new form
4
̃ 𝑢𝜉 + 𝐸̃ 𝑢𝜂 + 𝐹𝑢 = 𝐺
𝐴̃𝑢𝜉𝜉 + 𝐵̃ 𝑢𝜉𝜂 + 𝐶̃ 𝑢𝜂𝜂 + 𝐷
Where the new coefficient are as follows:
𝐴̃ = 𝐴𝜉𝑥 2 + 𝐵𝜉𝑥 𝜉𝑦 + 𝐶𝜉𝑦 2 ,
𝐵̃ = 2𝐴𝜉𝑥 𝜂𝑥 + 𝐵(𝜉𝑥 𝜂𝑦 + 𝜉𝑦 𝜂𝑥 ) + 2𝐶𝜉𝑦 𝜂𝑦
̃ = 𝐴𝜉𝑥𝑥 + 𝐵𝜉𝑥𝑦 + 𝐶𝜉𝑦𝑦 + 𝐷𝜉𝑥 + 𝐸𝜉𝑦,
𝐶̃ = 𝐴𝜂𝑥 2 + 𝐵𝜂𝑥 𝜂𝑦 + 𝐶𝜂𝑦 2 , 𝐷
𝐸̃ = 𝐴𝜂𝑥𝑥 + 𝐵𝜂𝑥𝑦 + 𝐶𝜂𝑦𝑦 + 𝐷𝜂𝑥 + 𝐸𝜂𝑦,
𝜉 = 𝜉(𝑥, 𝑦) , 𝜂 = 𝜂(𝑥, 𝑦).
Introducing the new coordinates
̃ 𝑢𝜉 + 𝐸̃ 𝑢𝜂 + 𝐹𝑢 = 𝐺
𝐴̃𝑢𝜉𝜉 + 𝐵̃ 𝑢𝜉𝜂 + 𝐶̃ 𝑢𝜂𝜂 + 𝐷
To find the coefficients 𝐴̃, 𝐵̃, 𝐶̃ , and solve for 𝜉 and 𝜂.
Using the new coordinate for the coefficients and homogenous part (to replace 𝑥 = 𝑥(𝜉, 𝜂) and
𝑦 = 𝑥(𝜉, 𝜂)) in the new canonical form
For the hyperbolic equation the canonical form
𝑢𝜉𝜂 = 𝑓(𝜉, 𝜂, 𝑢, 𝑢𝜉 , 𝑢𝜂 )
is achieved by setting 𝐴̃,= 𝐵̃ = 𝐶̃ =0 ,
𝐴̃ = 𝐴𝜉𝑥 2 + 𝐵𝜉𝑥 𝜉𝑦 + 𝐶 = 0 , ̃𝐶 = 𝐴𝜂𝑥 2 + 𝐵𝜂𝑥 𝜂𝑦 + 𝐶𝜂𝑦 2 ,
This can be written as
𝜉
2
𝜉
𝑦
𝜉𝑥
𝜉𝑦
𝑦
and
𝜂𝑥
𝜂𝑦
𝜉𝑦
=
−𝐵+√𝐵2 −4𝐴𝐶
2𝐴
,
𝜂𝑥
𝜂𝑦
=
5
𝑦
, we find the so called characteristic
equations:
𝜉𝑥
𝜂
𝐴 (𝜂𝑥 ) + 𝐵 (𝜂𝑥 ) + 𝐶 = 0,
𝑦
Solving these two quadratic equations for
2
𝜂
𝐴 (𝜉𝑥 ) + 𝐵 (𝜉𝑥 ) + 𝐶 = 0,
−𝐵−√𝐵2 −4𝐴𝐶
2𝐴
The new coordinates equated to constant values define the parametric lines of the new
coordinates .That means that the total derivative s are zero, i.e.,
𝜉(𝑥, 𝑦) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
→ 𝑑𝜉 = 𝜉𝑥 𝑑𝑥 + 𝜉𝑦 𝑑𝑦 = 0 →
𝜂(𝑥, 𝑦) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
→ 𝑑𝜂 = 𝜂𝑥 𝑑𝑥 + 𝜂𝑦 𝑑𝑦 = 0 →
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝜉 ,
= − 𝜉𝑥
𝑦
𝜂 ,
= − 𝜂𝑥
𝑦
and can be easily integrated to find the implicit solution , 𝜉(𝑥, 𝑦) = 𝑐𝑜𝑛𝑠𝑡. and
𝜂(𝑥, 𝑦) = 𝑐𝑜𝑛𝑠𝑡. that is, the new coordinates ensuring the simple canonical form of the PDE.
1.2.4 Method and Techniques for solving Partial differential equations
In developing a solution to a partial differential equation by separation of variables, one assume
that it is possible to separate the contributions of the independent variables into separate
functions that each involve only one independent variable. For example, consider the dependent
variable u that depends on independent variables t and x as specified in the following partial
differential equation and restricting conditions (i.e., one initial and two boundary conditions)
𝜕𝑢
𝜕𝑡 2
=
𝜕2 𝑢
𝑡 > 0 𝑎𝑛𝑑 0 < 𝑥 < 𝐿
𝜕𝑡 2
𝑢(𝑥, 0) = 0
I.C
B.C.1:
B.C.2:
𝜕𝑢
𝜕𝑡
𝑎𝑡 𝑡 = 0 for
(𝑜, 𝑡) = 0
(.1.1)
0<𝑥<𝐿
𝑎𝑡 𝑥 = 0 for 𝑡 > 0
𝑢(𝑙, 𝑡) = 0
𝑎𝑡 𝑥 = 𝐿 for 𝑡 > 0
To solve this problem by applying separation of variables we have assume that:
𝑢(𝑥, 𝑡) = 𝑋(𝑥)𝑇(𝑡)
(1.2)
Differentiating (1.2) twice with respect to x and t and substiituting in the equation (1.1) we
obatian:
𝑋(𝑥)𝑇̈(𝑡) = 𝑋̈(𝑥)𝑇(𝑡),
6
Or
𝑇̈ (𝑡)
𝑋̈ (𝑥)
𝑐 2 𝑇(𝑡)
= 𝑋(𝑥) , thus the equality is one of the functions of different variables,so both
𝑇̈ (𝑡)
quotients have to be constant.Say
𝑇(𝑡)
𝑋̈ (𝑥)
2
= 𝑋(𝑥) =  ,then we can solve each ordinarydifferential
2
2
equation separatly.We have the followwing three cases: −  ,  , 𝑎𝑛𝑑   0 .
Case 1. When the constant is , −  ,then the solution for
2
𝑋̈ (𝑥)
𝑋(𝑥)
= −  , 𝑎𝑟𝑒:
2
𝑋(𝑥) = 𝑐1 𝑠𝑖𝑛(𝜆𝑥) + 𝑐2 𝑐𝑜𝑠(𝜆𝑥)𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑜𝑟
𝑇̈(𝑡)
= −𝜆2 , 𝑎𝑟𝑒:
𝑇(𝑡)
𝑇(𝑡) = 𝑑1 𝑠𝑖𝑛(𝜆𝑡) + 𝑑2 𝑐𝑜𝑠(𝜆𝑡).Then
Then , 𝑢(𝑥, 𝑡) = (𝑑1 𝑠𝑖𝑛(𝜆𝑡) + 𝑑2 𝑐𝑜𝑠(𝜆𝑡))(𝑐1 𝑠𝑖𝑛(𝜆𝑥) + 𝑐2 𝑐𝑜𝑠(𝜆𝑥)).
Case 2. When the constant is 𝜆2 ,then the solution for
𝑋̈(𝑥)
𝑋(𝑥)
=
 2 , 𝑎𝑟𝑒:
𝑋(𝑥) = 𝑐1 𝑒 𝜆𝑥 + 𝑐2 𝑒 −𝜆𝑥 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑓𝑜𝑟
𝑇̈(𝑡)
= 𝜆2 , 𝑎𝑟𝑒:
𝑇(𝑡)
𝑇(𝑡) = 𝑑1 𝑒 𝜆𝑡 + 𝑑2 𝑒 −𝜆𝑡 .Then𝑢(𝑥, 𝑡) = (𝑑1 𝑒 𝜆𝑡 + 𝑑2 𝑒 −𝜆𝑡 )(𝑐1 𝑒 𝜆𝑥 + 𝑐2 𝑒 −𝜆𝑥 )
Case 3.When the constant is 0,thenthe equations become𝑥̈ (𝑥) = 𝑇̈(𝑡) = 0, and
𝑋(𝑥) = 𝑐1 𝑥 + 𝑐2 ,and 𝑇(𝑡) = 𝑑1 𝑡 + 𝑑2 .
Then, 𝑢(𝑥, 𝑡) = (𝑑1 𝑡 + 𝑑2 )(𝑐1 𝑥 + 𝑐2 )
Definition 1.9 A function𝑓(𝑡) is piecewise continuous on an interval[𝑎, 𝑏] if the interval can be
partitioned by a finite number of points 𝑎 = 𝑡0 < 𝑡 < 𝑡1 < 𝑡𝑛 = 𝑏 so that
a. 𝑓(𝑡) is continuous on each subinterval(𝑡𝑖−1 , 𝑡𝑖 )
b. 𝑓(𝑡) approaches a finite limit as the endpoints of each subinterval are approached
from both side within the interval.
Here is the graph of piecewise continuous function 𝑡
𝑓𝑖𝑔. 1.1
7
`
Here is example of a piecewise continuous functions:
𝑥 + 4 ,𝑥 ≥ 0
𝑓(𝑥)={𝑥 2 , 0 < 𝑥 < 5
7,
𝑥≥5
In other words, a piecewise continuous function is a function that has a finite number of breaks
in it and doesn’t blow up to infinity anywhere.
Definition 1.10: A function is f(x) is "even" when:
f(x) = f(-x) for all x
In other words there is symmetry about the y-axis
Fig 1.2
Definition 1.11 A function f(x) is "odd" when:
f(-x) = -f(-x) for all x
And we get origin symmetry
Fig 1.3
8
CHAPTER TWO
One Dimensional Homogeneous Wave equations
2.1 Derivation of One Dimensional Wave equations
When deriving a differential equation corresponding to a given physical problems, we usually
have to make simplified assumptions to ensure that the resulting equation does not become too
complicated. We know these important facts from the ordinary and partial differential equations.
To derive one – dimensional wave equations, we assume the following assumptions. Consider a
flexible string of length 𝐿 tightly stretched between two points 𝑥 = 0 and 𝑥 = 𝐿 on the 𝑥–axis,
with its ends. To determine its deflections (displacement from 𝑥–axis) at any point 𝑥 and at any
time 𝑡, let us take the following assumptions:
1. The string is uniform. ( i.e. its mass per length is constant)
2. The string is perfectly elastic and so offers no resistance to any bending.
3. The tension is too large that the action of gravitational force on the string is negligible.
4. The motion of the string is a small transverse in a vertical.( i.e. each particle of the
string moves strictly in vertical plane)
5. No other forces acting on the string.
Under this assumptions we may expect the solution 𝑢(𝑥, 𝑡) of the differential equation to be
obtained and will reasonably well describe small vibration of the physical ‘’ no idealized ‘’ string
of small homogenous under a large tension.
To obtain the differential equation we consider the force acting on small portion of the string
(Fig.2.1). Since the string does not offer resisting to bending, tension is tangential to the curve of
the string at each point. Let 𝑇1 and 𝑇2 be the tensions at the points 𝑃 and 𝑄 of the portions.
9
Since there is no motion in the horizontal direction, the horizontal component of the tension must
be constant.
𝑢(𝑥, 𝑡)
Q
𝛼
𝛽
𝑇2
p
𝑇1
𝐹𝑖𝑔. 2.1
0
𝑥
∆𝑥
𝐿
Using the notations in fig 2.1, we obtain
𝑇1 𝑐𝑜𝑠𝛼 = 𝑇2 𝑐𝑜𝑠𝛽 = 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
(2.1)
By Newton’s second law of motion, the resultant of these two forces is equal the mass 𝜌∆𝑥 of
the portion times the acceleration(𝑎 =
𝜕2 𝑢
𝜕𝑡 2
),evaluated at some point between 𝑥 and 𝑥 + ∆𝑥; here
𝜌 is the mass of undeflected string per length, ∆𝑥 is the length of the portion the undeleted
string. Hence
𝜕2 𝑢
𝑇1 𝑠𝑖𝑛𝛼 − 𝑇2 𝑠𝑖𝑛𝛽 = 𝜌∆𝑥 𝜕𝑡 2
(2.2)
Using equation (2.1) we can divide this by 𝑇1 𝑐𝑜𝑠𝛼 = 𝑇2 𝑐𝑜𝑠𝛽 = 𝑇 , obtaining
𝑇1 𝑠𝑖𝑛𝛼
𝑇1𝐶𝑂𝑆𝛼
𝜕2 𝑢
𝑇 𝑠𝑖𝑛𝛽
− 𝑇2𝑐𝑜𝑠𝛽 = 𝑡𝑎𝑛𝛼 − 𝑡𝑎𝑛𝛽 = 𝜌∆𝑥 𝜕𝑡 2
(2.3)
2
Now, 𝑡𝑎𝑛𝛼 and 𝑡𝑎𝑛𝛽 are slope of the string at 𝑥and 𝑥 + ∆𝑥 respectively.
𝜕𝑢
𝜕𝑢
𝑡𝑎𝑛𝛼 = (𝜕𝑥 ) , 𝑡𝑎𝑛𝛽 = (𝜕𝑥 )
𝑥
𝑥+∆𝑥
Here, we have to write partial derivatives because 𝑢 also depends on 𝑡 .
Dividing equation (2.3)
by∆𝑥 , thus we have
1 𝜕𝑢
𝜕𝑢
𝜌 𝜕 2𝑢
[( ) − ( )
]=
∆𝑥 𝜕𝑥 𝑥
𝜕𝑥 𝑥+∆𝑥
𝜏 𝜕𝑡 2
10
𝜕 2𝑢 𝜌 𝜕 2𝑢
=
𝜕𝑥 2 𝑇 𝜕𝑡 2
𝜕 2𝑢 𝑇 𝜕 2𝑢
=
𝜕𝑡 2 𝜌 𝜕𝑥 2
If we let ∆𝑥 approaches to zero, we obtain the linear partial differential equation.
𝜕 2𝑢
𝜕 2𝑢
2
=𝑐
,
𝜕𝑡 2
𝜕𝑥 2
𝑤ℎ𝑒𝑟𝑒 𝑐 2 =
𝑇
,
𝜌
𝑇 𝑖𝑠 𝑏𝑒𝑖𝑛𝑔 𝑡ℎ𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑛𝑑 ρ 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒𝑎𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦,
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0
(2.4)
This is the so called One Dimensional Wave Equation, which governs our problems we see that
it is homogenous and of the second order. The notation 𝑐 2 (in stead of 𝑐) for the physical
constant
𝑇
𝜌
has been chosen to indicate that this constant is positive. “One Dimensional’’
indicates that the equation involves only one space variable 𝑥.
2.2 General Solution of wave equations
The partial differential equation
𝜕2 𝑢
𝜕𝑡 2
𝜕2 𝑢
= 𝑐 2 𝜕𝑥 2
(2.5)
Definition 2.1. The solutions of (2.5) are called the two families of characteristics of the
equation.
The equation (2.5) can be written as
(
𝜕
𝜕
𝜕
𝜕
𝜕
𝜕
−𝑐 )( + 𝑐 )𝑢 = 0
𝜕𝑡
𝜕𝑥 𝜕𝑡
𝜕𝑥
𝜕
𝜕
(𝜕𝑡 + 𝑐 𝜕𝑥) = 0or (𝜕𝑡 − 𝑐 𝜕𝑥) = 0
11
(2.6)
The integrals are straight lines
𝑥 + 𝑐𝑡 = 𝑐1, 𝑥 − 𝑐𝑡 = 𝑐2
Introducing the characteristic coordinates
𝑥 + 𝑐𝑡 = 𝜉 ,𝑥 − 𝑐𝑡 = 𝜂
𝜕𝑢 𝜕𝑢 𝜕𝜉 𝜕𝑢 𝜕𝜂 𝜕𝑢
𝜕𝑢
𝜕𝑢 𝜕𝑢
(1) +
(1) =
=
+
=
+
𝜕𝑥 𝜕𝜉 ∂x 𝜕𝜂 ∂x 𝜕𝜉
𝜕𝜂
𝜕𝜉 𝜕𝜂
𝜕𝑢 𝜕𝑢 𝜕𝜉 𝜕𝑢 𝜕𝜂 𝜕𝑢
𝜕𝑢
𝜕𝑢
𝜕𝑢
𝜕𝑢 𝜕𝑢
=
+
= (1) + (1) = (𝑐) + (−𝑐) = 𝑐 ( − )
𝜕𝑡 𝜕𝜉 ∂t 𝜕𝜂 ∂t 𝜕𝜉
𝜕𝜂
𝜕𝜉
𝜕𝜂
𝜕𝜉 𝜕𝜂
 2u  2u
 2u  2u

2

x 2  2
nd n 2
(2.7)
2
 2u
 2u  2u 
2  u

 c  2  2

t 2
nd n 2 
 
2.8
Substituting in eq. (2.5)
  2u
  2u
 2u  2u 
 2u  2u 

c 2  2  2
 2   c 2  2  2

nd n 
nd n 2 
 
 
 2u
4c
0

 2u
 0, c  0

2
( 2.9)
(2.9) is called the canonical form of one dimensional wave equation
Integrating (2.9) with respect to 𝜂
𝜕
𝜕𝑢
∫ 𝜕𝜂 ( 𝜕𝜉 ) 𝑑𝜂=∫ 0𝑑𝜂
u
 g ( )

where g is a function
12
Further integrating with respect to 𝜉
u   g ( )d  c( )
Relabeling in more conventional notation gives
u ( , )  F ( )  G ( )
where 𝐹 is an arbitrary function and 𝐺(𝜂) = ∫ 𝑔(𝜂)𝑑 𝜂. Hence the general solution of (2.5) has
the form
𝑢(𝑥, 𝑡) = 𝐹(𝑥 + 𝑐𝑡) + 𝐺(𝑥 − 𝑐𝑡)
(2.10)
The function 𝐹(𝑥 + 𝑐𝑡)is a wave traveling to the left with the same speed, and it is called a
backward wave. Indeed c can be called the wave speed. Equation (2.10) demonstrates that any
solution of the wave equation is the sum of two such traveling waves. This observation will
enable us to obtain graphical representations of the solutions (the graphical method)
Let us further discuss the general solution (2.12) Consider the (x, t) plane. The following two
families of lines are called the characteristics of the wave equation. For the wave equation, the
1
characteristics are straight lines in the (𝑥, 𝑡)plane with slopes ± 𝑐 .
t
x  ct  
x  ct  
t


x
𝑭𝒊𝒈2.2
13
x
2.3 Initial Value Condition: Cauchy Problem
The approach to the solution of PDE (One dimensional wave equations) has some similarity to
the case of an ODE. That is, after we solve for the general solution of the PDE itself, we apply
the prescribed auxiliary conditions in order to find the particular solution. Specifically, after the
general solution is found in terms of the arbitrary functions F and G , we proceed to find the
particular solution by finding the specific form of the arbitrary functions F and G that also satisfy
the prescribed auxiliary conditions.
Definition 2.2 Finding those solutions that satisfy the initial conditions (initial data) is called the
Cauchy problem.
To describe the motion of the string completely it is necessary also to specify suitable initial and
boundary conditions for the displacement of 𝑢(𝑥, 𝑡). The ends are assumed to be fixed, and
therefore the boundary conditions are:
𝑢(𝑥, 0) = 0,
(2.11)
𝑢(𝐿, 𝑡) = 0, 𝑡 ≥ 0
(2.12)
Since the differential equation (2.1) is of the second order with respect to 𝑡 , it is plausible to
prescribe two initial conditions .
𝑢(𝑥, 0) = 𝑓(𝑥), 0 ≤ 𝑥 ≤ 𝐿
(2.13)
The Cauchy initial value problem for the wave equation is to find a 𝐶 2 solution of
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0
𝑢(𝑥, 0) = 𝛼(𝑥), 𝑢𝑡 (𝑥, 0) = 𝛽(𝑥)
𝛼, 𝛽 ∈ 𝐶 2 (−∞, ∞)are given
THEOREM 2.1 There exists a unique solution of the Cauchy initial value problem and this
solution is given by d’Alembert’s formula
1
1
𝑥+𝑐𝑡
𝑢(𝑥, 𝑡) = 2 [𝛼(𝑥 − 𝑐𝑡) + 𝛼(𝑥 + 𝑐𝑡)] + 2𝑐 ∫𝑥−𝑐𝑡 𝛽(𝑠)𝑑𝑠
where 𝛼(𝑥) ∈ 𝐶 2 , 𝛽(𝑥) ∈ 𝐶 1
𝑢(𝑥, 0) = 𝛼(𝑥)
𝑢𝑡 (𝑥, 0) = 𝛽(𝑥)
14
(2.14)
Proof. Assume there is a solution 𝑢(𝑥, 𝑡)of the Cauchy initial value problem, then it
follows from (2.14) that
𝑢(𝑥, 0) = 𝑓(𝑥) + 𝑔(𝑥) = 𝛼(𝑥)
(2.15)
𝑢𝑡 (𝑥, 0) = 𝑐𝑓 ′ (𝑥) − 𝑐𝑔′ (𝑥) = 𝛽(𝑥)
(2.16)
From (2.15 ) we obtain
𝑓 ′ (𝑥) + 𝑔′ (𝑥) = 𝛼 ′ (𝑥).
Which implies, together with (2.16 )
𝑐𝛼′ (𝑥)+𝛽(𝑥)
𝑐
𝑓 ′ (𝑥) =
2
𝑐𝛼′ (𝑥)−𝛽(𝑥)
𝑐
𝑔′ (𝑥) =
2
Then
𝑓(𝑥) =
𝛼(𝑥) 1 𝑥
+ ∫ 𝛽(𝑠) 𝑑𝑠 + 𝐶1
2
2𝑐 0
𝑔(𝑥) =
𝛼(𝑥) 1 𝑥
− ∫ 𝛽(𝑠) 𝑑𝑠 + 𝐶2.
2
2𝑐 0
The constants 𝐶1 ,𝐶2 satisfy
𝐶1 + 𝐶2 = 𝑓(𝑥) + 𝑔(𝑥) − 𝛼(𝑥) = 0,
𝑢(𝑥, 𝑡) =
1
1 𝑥+𝑐𝑡
[𝛼(𝑥 − 𝑐𝑡) + 𝛼(𝑥 + 𝑐𝑡)] + ∫
𝛽(𝑠) 𝑑𝑠
2
2𝑐 𝑥−𝑐𝑡
This completes the proof.
Thus each solution of the Cauchy initial value problem is given by d’Alembert’s formula. On the
other hand, the function 𝑢(𝑥, 𝑡) defined by the right hand side of (2.14) is a solution of the initial
value problem.
Example 2.1 Solve the following Cauchy Problem
𝑢𝑡𝑡 − 𝑢𝑥𝑥 = 0, (𝑥, 𝑡)𝜖ℛ × (0, 𝑡)
𝑢(𝑥, 0) = 𝑐𝑜𝑠𝑥
𝑢𝑡 (𝑥, 0) = 𝑥
Solution:
By (2.14) Putting 𝑐 = 1 and 𝛼(𝑥) = 𝑐𝑜𝑠𝑥
𝑐𝑜𝑠(𝑥 + 𝑡) + 𝑐𝑜𝑠(𝑥 − 𝑡) 1 𝑥+𝑡
𝑢(𝑥, 𝑡) =
+ ∫ 𝑥𝑑𝑠
2
2 𝑥−𝑡
= 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑡 + 𝑥𝑡
15
Remark. 𝑐𝑜𝑠(𝐴 + 𝐵) = 𝑐𝑜𝑠𝐴𝑐𝑜𝑠𝐵 − 𝑠𝑖𝑛𝐴𝑠𝑖𝑛𝐵
Example: 2.2 Let u ( x, t ) be a solution of the wave equation
u tt  c 2 u xx  0 ,
This is defined in the whole plane. Assume that u is constant x  2  ct.
Prove that u t  u x  0 .
Proof. The solution u ( x, t ) has the form
Since u2  ct , t   constant, it follows that f (2  2ct )  g (2) =constant
Set s  2  2ct , we have f (s ) =constant. Consequently
u ( x, t )  g ( x  ct ) Computing now the
expression u t  cu x , we obtain
u t  cu x = cf ' x  ct   cg ' x  ct   0
Example: 2.3 Find the solution ux, t  of the Cauchy problem
u xx  utt  0 u x,0  x 2 , u t x,0   4 x 2 .
Solution: Since f ( x)  x 2 , g ( x)  4 x 2 , the D’Almebert formula yields
u  x, t  
x  t 2  x  t 2  x  t 4  x  t 4
2
2.3.1 Domain of Dependence
Definition 2.4 The value of u at any point (x0,t0) depends only on the initial data f and g (t=0) in
a certain interval called the domain of dependence of u at (x0,t0) , which is given by [x0-ct0,
x0+ct0].The value 𝑢 at (𝑥0 , 𝑡0 )is determined by the restriction of initial functions 𝑓 and 𝑔in the
interval [𝑥0 − 𝑐𝑡0 , 𝑥0 + 𝑐𝑡0 ]on the 𝑥 −axis, whose end points are cut out by the
characteristics: 𝑥 − 𝑥0 = ±(𝑡 − 𝑡0 ),
16
through the point(𝑥0, 𝑡0 ). The characteristic triangle ∆(𝑥0 , 𝑡0 )is defined as the triangle ℜ × ℜ+
in with vertices
𝐿(𝑥0 − 𝑐𝑡0 , 0),𝐵(𝑥0 + 𝑐𝑡0 , 0) , 𝑅(𝑥0 , 𝑡0 )
For every (𝑥1 , 𝑡1 ) ∈ ∆(𝑥0, 𝑡0 )
[𝑥1 − 𝑐𝑡1 , 𝑥1 + 𝑐𝑡1 ] c [𝑥0 − 𝑐𝑡0 , 𝑥0 + 𝑐𝑡0 ], ∆(𝑥1, 𝑡1 ) c ∆(𝑥0, 𝑡0 )
and (𝑥1 , 𝑡1 ) is determined by the values 𝑓 and 𝑔 on [𝑥1 − 𝑐𝑡1 , 𝑥1 + 𝑐𝑡1 ]
𝐹𝑖𝑔. 2.3
Example 2.4 For the Cauchy probblem
𝑢𝑥𝑥 − 𝑢𝑡𝑡 = 0 , 𝑢(𝑥, 0) = 𝑥 2 , 𝑢𝑡 (𝑥, 0) = 4𝑥 3
The domain of dependence with respect to the point (−3,5) is −8 ≤ 𝑥 ≤ 2.
2.3.2 Domain of Influence
Definition 2.5 Given an interval [a , b] of x at the initial time, we also find that the initial data in this
interval can only affect the value of u in a limited region of the domain called the domain of influence of
the interval [a , b] .
The point(𝑥0 , 0)on the 𝑥 − axis influence the value of 𝑢 at (𝑥, 𝑡) in the wedge –shaped region.
𝐼(𝑥0 ) = {(𝑥, 𝑡): 𝑥0 − 𝑐𝑡 ≤ 𝑥 ≤ 𝑥0 + 𝑐𝑡, 𝑡 ≥ 0}.
17
For any
𝑃1 (𝑥1, 𝑡1 ) ∈ 𝐼(𝑥0 ), ∆(𝑥1, 𝑡1 ) ∩ 𝐼(𝑥0 ) ≠ ∅
𝑃1 (𝑥1, 𝑡1 )  𝐼(𝑥0 ), ∆(𝑥1, 𝑡1 ) ∩ 𝐼(𝑥0 ) = ∅
These are the points inside the forward (truncated) characteristic cone that is defined by the
base[𝑎, 𝑏] and the edges 𝑥 + 𝑐𝑡 = 𝑎, 𝑥 − 𝑐𝑡 = 𝑏(it is the union of the regions I–IV of 𝐹𝑖𝑔. 2.4)
𝑭𝒊𝒈. 𝟐. 𝟒
2.3.3 Well-posedness
Definition 2.6 The problems are well- posed in the sense of Hadamard if the following three
conditions are satisfied:
(i)
There exists a solution;
(ii)
The solution is unique;
(iii)
The solution is stable.
Statement (iii) means that small variations of the initial data yield small variations on the
corresponding solutions. This is also referred to as continuous dependence upon the initial data.
The meaning of small variation is made precise in terms of the topology suggested by the
problem. A problem that does not satisfy any one of these conditions is called ill-posed.
Example 2.5: For second order hyperbolic PDE problems ,the vibrating string is the most
frequently use as an example of a well- posed problem. If we think of 𝑢(𝑥, 𝑡) as a vibrating as
representing the vertical displacement at position 𝑥 and time 𝑡 of an ideal string which is static
equilibrium occupies the horizontal line joining 𝑥 = 0and 𝑥 = 𝐿.Then the following IBVP
models the movement of the string subject to an initial displacement given 𝑓(𝑥)and initial
velocity given by 𝑔(𝑥).
18
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0,
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
Example2.6 An example of ill-posed, second order hyperbolic problem.
A dramatic example of an ill-posed, second order hyperbolic PDE is given by the following BVP
for the one dimensional wave equation. It can be shown that if 𝑇 is irrational, then the only
solution of this BVP for the wave equation is identically zero; whereas if 𝑇 is rational, the
problem has infinitely many non-trivial solution. This solution fails to depend continuously on
the data mainly on the size of the region on which the problem is stated.
2.4 Separation of Variable
If we we tie the string at both ends we can have the following boundary conditions:
𝑢(0, 𝑡) = 𝐴(𝑡), 𝑢(𝐿, 𝑡) = 𝐵(𝑡),Where 𝐴 ,𝐵 are 𝐶 1 piecewice functions .When the
boundary values 𝐴 and 𝐵 are 0 we obtian the Dirichlet Problem for wave equation:
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0,0 < 𝑥 < 𝐿, 𝑡 > 0 Differential Equation(DE)
𝑢(0, 𝑡) = 0, 𝑢(𝐿, 𝑡) =0
𝑡 > 0 Boundary Condtion(BC)
u(x, 0) = f(x), ut (x, 0) = g(x) ,0 < 𝑥 < 𝐿 Initial Condition (IC)
THEOEREM 2.2A soluton of the wave problem:
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0,0 < 𝑥 < 𝐿, 𝑡 > 0
𝑢(0, 𝑡) = 0, 𝑢(𝐿, 𝑡) =0
(𝐷𝐸)
𝑡 > 0( 𝐵𝐶 )
𝑢(𝑥, 0) = 𝑓(𝑥), 𝑢𝑡 (𝑥, 0) = 𝑔(𝑥) ,0 < 𝑥 < 𝐿
𝐿(𝐼𝐶 )
is given by:
∞
𝑢(𝑥, 𝑡) = ∑ [𝐴𝑛 𝑠𝑖𝑛 (
𝑛=1
𝑤ℎ𝑒𝑟𝑒: 𝑓(𝑥) =
𝑛𝜋𝑐𝑡
∑∞
𝑛=1 𝐵𝑛 𝑠𝑖𝑛 (
∞
𝐿
𝑛𝜋𝑐𝑡
𝑛𝜋𝑐𝑡
𝑛𝜋𝑐
) + 𝐵𝑛 𝑐𝑜𝑠 (
)] 𝑠𝑖𝑛 (
),
𝐿
𝐿
𝐿
) , 𝑎𝑛𝑑
𝜋𝑛𝑥
̇ 𝐵𝑛 𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 ∶
) . 𝑇ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝐴𝑛 𝑎𝑛𝑑
𝐿
𝑔(𝑥) = ∑ 𝐴𝑛 𝑠𝑖𝑛 (
𝑛=1
19
2 𝐿
𝑛𝜋𝑥
2 𝐿
𝑛𝜋𝑥
𝐵𝑛 = ∫ 𝑓(𝑥) 𝑠𝑖𝑛 (
) , 𝐴𝑛 = ∫ 𝑔(𝑥) 𝑠𝑖𝑛 (
).
𝐿 0
𝐿
𝐿 0
𝐿
Proof. Let’s take a look at the boundary conditions:
𝑢(0, 𝑡) = 0, 𝑢(𝐿, 𝑡) =0
The only solution for 𝑢(𝑥, 𝑡) that satisfy them is
𝑢(𝑥, 𝑡) = (𝑑1 𝑠𝑖𝑛(𝜆𝑐𝑡) + 𝑑2 𝑐𝑜𝑠(𝜆𝑐𝑡))(𝑐1 𝑠𝑖𝑛(𝜆𝑥) + 𝑐2 𝐶𝑂𝑆(𝜆𝑥)), and the boundary
conditions translated into:
(𝑑1 𝑠𝑖𝑛(𝜆𝑐𝑡) + 𝑑2 𝑐𝑜𝑠(𝜆𝑐𝑡))(𝑐1 𝑠𝑖𝑛(0) + 𝑐2 𝐶𝑂𝑆(0)) = 0
(𝑑1 𝑠𝑖𝑛(𝜆𝑐𝑡) + 𝑑2 𝑐𝑜𝑠(𝜆𝑐𝑡))(𝑐1 𝑠𝑖𝑛(𝜆𝐿) + 𝑐2 𝐶𝑂𝑆(𝜆𝐿)) = 0, ∀𝑡 > 0
namely :
𝑐2 = 0
( 2.17)
𝑐1 𝑠𝑖𝑛(𝜆𝐿) = 0
( 2.18)
From equation ( 2.18 ) we obtain 𝜆 =
𝜋𝑛
𝐿
. and


 n 
 n 
 n 
u x, t    d1n sin  ct   d 2 n cos ct cn sin  x .
 L 
 L 
 L 
n 1 
The only condition left to check are the initial conditions:

𝑢(𝑥, 0) = 𝑓(𝑥) =

𝑛𝜋𝑥
𝐵𝑛 𝑠𝑖𝑛 (
n 1
𝐿
),

𝑢(𝑥, 0) = 𝑔(𝑥) =
𝐿
Then, 𝑢(𝑥, 𝑡) = ∑∞
𝑛=1 [𝑛𝜋𝑐 𝐴𝑛 𝑠𝑖𝑛 (
𝑛𝜋𝑐𝑡
𝐿

𝐴𝑛 𝑠𝑖𝑛 (
n 1
𝑛𝜋𝑐𝑡
) + 𝐵𝑛 𝑐𝑜𝑠 (
𝐿
𝑛𝜋𝑥
)
𝐿
𝑛𝜋𝑐
)] 𝑠𝑖𝑛 (
𝐿
).
THEOREM 2.3 Let𝑢1 (𝑥, 𝑡) and 𝑢2 (𝑥, 𝑡) be two solutions of the problem:
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0,0 < 𝑥 < 𝐿, 𝑡 > 0
(DE)
𝑢(0, 𝑡) = 𝐴(𝑥), 𝑢(𝐿, 𝑡) =𝐵(𝑡), 𝑡 > 0( BC)
𝑢(𝑥, 0) = 𝑓(𝑥), 𝑢𝑡 (𝑥, 0) = 𝑔(𝑥) ,0 < 𝑥 < 𝐿
(IC)
where 𝐴, 𝐵, 𝑓, 𝑔 are piecewise continuous functions. Then 𝑢1 (𝑥, 𝑡) = 𝑢2 (𝑥, 𝑡) for all points in
the domain.
20
Proof. Let 𝑢(𝑥, 𝑡) = 𝑢1 (𝑥, 𝑡) − 𝑢2 (𝑥, 𝑡),t hen 𝑣 satisfies the wave equation with initial
conditions: 𝑢(𝑥, 0) = 𝑢𝑡 (𝑥, 𝑡) = 0, and boundary conditions 𝑢(0, 𝑡) = 0, 𝑢(𝐿, 𝑡) = 0.Our goal is
to prove that 𝑢(𝑥, 𝑡) = 0
∀𝑥, 𝑡.
In order to accomplish this, define:
𝐿
𝐻(𝑡) = ∫0 [𝑐 2 𝑢𝑥 (𝑥, 𝑡)2 + 𝑢𝑡 (𝑥, 𝑡)2 ]𝑑𝑥
We will prove that 𝐻(𝑡) = 0 first. Differentiating with respect to 𝑡we obtain:
̇ = ∫𝐿[𝑐 2 2𝑢𝑥 𝑢𝑥𝑡 + 2𝑢𝑡 𝑢𝑡𝑡 ]𝑑𝑥
𝐻(𝑡)
0
𝐿
= 2𝑐 2 ∫0 [𝑢𝑥 𝑢𝑥𝑡 + 𝑢𝑡 𝑢𝑥𝑥 ]𝑑𝑥 (from eq. (2.4))
𝐿
2
= 2𝑐 ∫
0
𝛿
(𝑢𝑥 𝑢𝑡 )𝑑𝑥 = 2𝑐 2 [𝑢𝑥 (𝑥, 𝑡)𝑢𝑡 (𝑥, 𝑡)]𝐿
0
𝛿𝑥
2
= 2𝑐 (𝑢𝑥 (𝐿, 𝑡)𝑢𝑡 (𝐿, 𝑡) − 𝑢𝑥 (0, 𝑡)𝑢(0, 𝑡)) = 0.
Since 𝐻̇ (𝑡) = 0, 𝐻(𝑡) is constant, and as 𝐻(0) = 0, we conclude that 𝐻(𝑡) = 0
𝐿
Then 𝑢𝑡 (𝑥, 𝑡) = 0, and 𝑢(𝑥, 𝑡) = 𝑢(𝑥, 𝑡) − 𝑢(𝑥, 0) = ∫0 𝑢𝑡 (𝑥, 𝑡)𝑑𝑡 = 0
2.5 The Wave Equation on the Half line: Reflection Method
Definition 2.8(Even and odd extension) Suppose that a function 𝑓(𝑥) is continuous and defined
on the interval [0, 𝜋]. we can extend this function n to the interval[−𝜋, 𝜋]. This can be done in
two ways:
1. 𝑓(𝑥) is an even extension if
𝑓(−𝑥),
𝑓(𝑥),
−𝜋 ≤ 𝑥 < 0,
0 ≤ 𝑥 < 𝜋.
−𝑓(−𝑥),
𝑓(𝑥),
−𝜋 ≤ 𝑥 < 0,
.
0 ≤ 𝑥 < 𝜋.
𝑓𝑒𝑣𝑒𝑛 (𝑥) = {
2. 𝑓(𝑥)is an odd extension if
𝑓𝑜𝑑𝑑 (𝑥) = {
21
To illustrate a further application of d’Alembert formula, let us next consider this initial
/boundary-value problem on the half-lineℝ+ = {𝑥 > 0}:
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0 𝑖𝑛 ℝ+ × (0, ∞)
{𝑢(𝑥, 0) = 𝑓(𝑥), 𝑢𝑡 (𝑥, 0) = 𝑔(𝑥) 𝑜𝑛 ℝ+ × (0, ∞) {𝑡 = 0}
𝑢(𝑥, 0) = 0 𝑜𝑛 {𝑡 = 0} × (0, ∞) ,
(2.19)
Where 𝑓, 𝑔 are given, with 𝑓(𝑥, 0) = 𝑔(𝑥, 0) = 0.
We convert (2.19) into the form (3) by extending 𝑢, 𝑓, 𝑔 to odd reflection. That is, we get
𝑢̃(𝑥, 𝑡) ≔ {
𝑢(𝑥, 𝑡)(𝑥 ≥ 0, 𝑡 ≥ 0)
−𝑢(−𝑥, 𝑡)(𝑥 < 0, 𝑡 > 0),
𝑓(𝑥)(𝑥 ≥ 0)
𝑓̃(𝑥) ≔ {
−𝑓(−𝑥)(𝑥 < 0)
𝑔(𝑥)(𝑥 ≥ 0)
𝑔̃(𝑥) ≔ {
−𝑔(−𝑥)(𝑥 < 0).
Then (2.20) becomes
𝑢̃𝑡𝑡 = 𝑐 2 𝑢̃𝑥𝑥
𝑖𝑛 ℝ × (0, ∞)
{
𝑢̃ = 𝑓̃ , ̃
𝑢𝑡 = 𝑔̃ 𝑜𝑛 ℝ × (𝑡 = 0)
Hence d’Alembert’s formula (2.14) implies
1
1 𝑥+𝑐𝑡
𝑢(𝑥, 𝑡) = 2 [𝑓̃(𝑥 + 𝑐𝑡) + 𝑓̃(𝑥 + 𝑐𝑡)] + 2𝑐 ∫𝑥−𝑐𝑡 𝑔̃(𝑦)𝑑𝑦
Recalling the definitions 𝑢̃, 𝑓̃ , 𝑔̃above, we transform this express to read for𝑥 ≥ 0,𝑡 ≥ 0
1
[𝑓(𝑥 + 𝑐𝑡) + 𝑓(𝑥 + 𝑐𝑡)] +
𝑢(𝑥, 𝑡) = {12
[𝑓(𝑥 + 𝑐𝑡) − 𝑓(𝑐𝑡 − 𝑥)] +
2
1
𝑥+𝑐𝑡
𝑔(𝑦)𝑑𝑦 if x ≥ ct ≥ 0
∫
2𝑐 𝑥−𝑐𝑡
1
𝑥+𝑐𝑡
𝑔(𝑦)𝑑𝑦 if 0 ≤ 𝑥 ≤ 𝑐𝑡.
∫
2𝑐 −𝑥+𝑐𝑡
(2.20)
If 𝑔 = 0, we can understand formula (2.20) as saying that an initial displacement ℎ splits into
two parts, one moving to the right with speed 𝑐 and the other to the left with speed 𝑐. The latter
then affects off 𝑥 = 0, where the vibrating string is held fixed.
22
Note that 𝑢(𝑥, 𝑡) is a continuous function if the compatibility condition𝜑(0)=0 is satisfied.
Otherwise 𝑢(𝑥, 𝑡) is a discontinuous solution and the jump of 𝑢(𝑥, 𝑡) on the characteristic
𝑥 = 𝑐𝑡 is
(𝑐𝑡 + 0, 𝑡) − 𝑢(𝑐𝑡 − 0, 𝑡) = 𝜑(0).
We can do the same for the problem with the Neumann boundary condition, considering even
extensions of initial data. Let us consider the problem
We can do the same for the problem with the Neumann boundary condition, considering even
extensions of initial data. Let us consider the problem
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0
0 < 𝑥 < ∞, 𝑡 > 0
(𝐶𝑁𝑊): {
𝑢(𝑥, 0) = 𝜑(𝑥), 𝑢𝑡 (𝑥, 0) = 𝜓(𝑥) 0 < 𝑥 < ∞
𝑢(𝑥, 𝐿) = 0
𝑡≥0
.
In this case the problem (𝐶𝑁𝑊) to (CW) with initial functions𝜑0(𝑥) 𝑎𝑛𝑑 𝜓0 (𝑥), 𝑤ℎ𝑒𝑟𝑒
𝜑0 (𝑥) ≔ {
𝜑(𝑥) 𝑖𝑓 𝑥 ≥ 0,
𝜑(−𝑥) 𝑖𝑓 𝑥 ≤ 0
As before we can show that the problem (𝐶𝑁𝑊) has a unique
1
1
(𝜑(𝑥 + 𝑐𝑡) + 𝜑(𝑥 − 𝑐𝑡)) + (Ψ(𝑥 + 𝑐𝑡) − Ψ(𝑥 − 𝑐𝑡))
, 𝑥 > 𝑐𝑡
2
2𝑐
𝑢(𝑥, 𝑡) = {
1
1
(𝜑(𝑥 + 𝑐𝑡) + 𝜑(𝑐𝑡 − 𝑥)) + (Ψ(𝑥 + 𝑐𝑡) − Ψ(𝑐𝑡 − 𝑥)), 0 < 𝑥 < 𝑐𝑡,
2
2𝑐
𝑡
where Ψ(𝑡) = ∫0 𝜓(𝑠)𝑑𝑠.
On a semi domain, we often have a slightly more involved process
2.6 Mixed Problems for the Wave Equations
Let us consider the problem (CW) on a finite interval[0, 𝑙]which Dirichlet boundary conditions at
the end- points 𝑥 = 0 and 𝑥 = 𝑙 .This is the problem
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0,
0 < 𝑥 < 𝑙, 𝑡 > 0,
(𝑀𝐷𝑊): { 𝑢(𝑥, 0) = 𝜑(𝑥), 𝑢𝑡 (𝑥, 0) = 𝜓(𝑥),
𝑢(0, 𝑡) = 𝑢(𝑙, 𝑡) = 0,
𝑡 ≥ 0.
23
0 ≤ 𝑥 ≤ 𝑙,
It can be interpreted as vibrations of a string with clamped ends, for instance vibrations of a
guitar string. We can get the solution of the problem (𝑀𝐷𝑊) again using the method of
reflection in this case through both ends. We extend the initial data 𝜑(𝑥)and 𝜓(𝑥) given on the
interval(0, 𝑙) to the whole line using “odd" extensions 𝜑𝑒𝑜 (𝑥)and 𝜓𝑒𝑜 (𝑥) with respect to both
sides 𝑥 = 0 and 𝑥 = 𝑙 , where
𝜑(𝑥),
0<𝑥<𝑙
𝜑𝑒𝑜 (𝑥) ≔ {
−𝜑(−𝑥) − 𝑙 < 𝑥 < 0
𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑜𝑑 2𝑙.
Consider the problem (𝐶𝑊𝑒𝑜 ):
𝑣𝑡𝑡 − 𝑐 2 𝑣𝑥𝑥 = 0, 𝑥 ∈ ℝ , 𝑡 > 0 ,
(𝐶𝑊𝑒𝑜 ): ={
𝑥∈ℝ
𝑣(𝑥, 0) = 𝜑𝑒𝑜 (𝑥),
𝑣𝑡 (𝑥, 0) = 𝜓𝑒𝑜 (𝑥),
𝑥∈ℝ
Section 2.2 it has a solution
1
1 𝑥+𝑐𝑡
𝑣(𝑥, 𝑡) = (𝜑𝑒𝑜 (𝑥 + 𝑐𝑡) + 𝜑𝑒𝑜 (𝑥 − 𝑐𝑡)) + ∫
𝜓 (𝑠) 𝑑𝑠.
2
2𝑐 𝑥−𝑐𝑡 𝑒𝑜
Its restriction
𝑢(𝑥, 𝑡) = 𝑣(𝑥, 𝑡)\0≤𝑥≤𝑙
gives the unique solution of the problem(𝑀𝐷𝑊). Note that the solution formula is characterized
by a number of reflections at each end point.
𝑥 = 0 𝑎𝑛𝑑𝑥 = 𝑙along characteristics through reflecting points. They divide the domain ℝ =
{(𝑥, 𝑡): 0 < 𝑥 < 𝑙, 𝑡 > 0}into diamond-shaped domains with sides parallel to characteristics and
within each diamond the solution𝑢(𝑥, 𝑡) is given by a different formula.
On the data 𝜑and 𝜓we impose the compatibility condition
𝜑(0) = 𝜑(𝑙) = 𝜓(0) = 𝜓(𝑙) = 0
(2.26)
In this case the solution 𝑢(𝑥, 𝑡) is a continuous function on ℝ.
Note that 𝑢(𝑥, 𝑡 )𝜖𝐶 2 (ℝ) if
𝜑(0) = 𝜑(𝑙) = 𝜑 ′′ (0) = 𝜑 ′′ (𝑙) = 𝜓(0) = 𝜓(𝑙) = 0
(2.27)
We can do the same for the problem with the Neumann boundary condition, considering even
extensions of initial data. Namely, let us consider the problem
24
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 0,
0 < 𝑥 < 𝑙, 𝑡 > 0,
(𝑀𝐷𝑊): { 𝑢(𝑥, 0) = 𝜑(𝑥), 𝑢𝑡 (𝑥, 0) = 𝜓(𝑥),
𝑢𝑥 (0, 𝑡) = 𝑢𝑥 (𝑙, 𝑡) = 0,
𝑡 ≥ 0.
0 < 𝑥 < 𝑙,
In this case we reduce the problem (MDW)to (CWee ) with initial functions φee (x)and
ψee (x),where
𝜑(𝑥),
0<𝑥<𝑙
𝜑𝑒𝑒 (𝑥) ≔ {
𝜑(−𝑥) − 𝑙 < 𝑥 < 0
𝑒𝑥𝑡𝑒𝑛𝑑𝑒𝑑 𝑡𝑜 𝑏𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑜𝑑 2𝑙.
As before the problem(𝐶𝐷𝑊) admits the unique solution
𝑢(𝑥, 𝑡) = 𝑤(𝑥, 𝑡) \ 0 xt
where 𝑤(𝑥, 𝑡) is the solution of the problem
𝑤𝑡𝑡 − 𝑐 2 𝑤𝑥𝑥 = 0,
𝑥 ∈ ℝ , 𝑡 > 0,
(𝐶𝑊𝑒𝑒 ): {
𝑤(𝑥, 0) = 𝜑𝑒𝑒 (𝑥), 𝑥 ∈ ℝ ,
𝑤𝑡 (𝑥, 0) = 𝜓𝑒𝑒 (𝑥), 𝑥 ∈ ℝ .
Example 2.8 Solve the problem (CDW) with 𝑐 = 1, 𝜓 = 0 and
3𝜋 5𝜋
𝑐𝑜𝑠 3 𝑥, 𝑥𝜖ℝ ( , ) ,
2 2
𝜑(𝑥) = {
3𝜋
5𝜋
0, 𝑥𝜖ℝ+ ( , ) .
2 2
Solution. The solution of the odd extended problem
𝑢𝑡𝑡 − 𝑢𝑥𝑥 = 0, 𝑥 ∈ ℝ , 𝑡 > 0,
{ 𝑢(𝑥, 0) = 𝜑0 (𝑥), 𝑥 ∈ ℝ,
𝑢𝑡 (𝑥, 0) = 0, 𝑥 ∈ ℝ.
1
is𝑣(𝑥, 𝑡) = 2 (𝜑0 (𝑥 + 𝑡) + 𝜑0 (𝑥 − 𝑡)). The original problem has the solution
1
(𝜑(𝑥 + 𝑡) + 𝜑(𝑥 − 𝑡))
𝑥 > 𝑡,
2
𝑢(𝑥, 𝑡) = {
1
(𝜑(𝑥 + 𝑡) − 𝜑(𝑡 − 𝑥)) 0 < 𝑥 < 𝑡.
2
𝜋 3𝜋
The profile of 𝑢(𝑥, 𝑡) at successive instances are𝑡 = 0, 2 ,
25
2
, 2𝜋,
5𝜋 7𝜋
2
,
2
CHAPTER THREE
One - Dimensional Non-homogeneous Wave Equations
3.1. Initial Condition: Cauchy problem for the Non-homogeneous wave equation
Consider the following non-homogeneous Cauchy problem
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 𝐹(𝑥, 𝑡), −∞ < 𝑥 < ∞, 𝑡 > 0
(3.1)
𝑢(𝑥, 0) = 𝑓(𝑥), −∞ < 𝑥 < ∞,
(3.2)
𝑢𝑡 (𝑥, 0) = 𝑔(𝑥), −∞ < 𝑥 < ∞,
(3.3)
This problem models, for example, the vibration of a very long string in the presence of an
external force 𝐹. As in the homogeneous case 𝑓 and𝑔 are given functions that represent the
shape and vertical velocity of the string at time 𝑡 = 0
3.2. D’Alembert’s Solution for Non-homogeneous Wave Equations
Definition 3.1 The transformation(𝜉, 𝜂) = (𝜉(𝑥, 𝑦), 𝜂(𝑥, 𝑦))is called a change of Coordinates (or
a nonsingular transformation) if the Jacobian 𝐽 ≔ 𝜉𝑥 𝜂𝑦 − 𝜉𝑦 𝜂𝑥 ofthe transformation does not
vanish at any point (x, y).
It can be split into two problems –one homogeneous with nonzero initial data (CW) which we
solve, and one inhomogeneous with zero data
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 =0 𝑥 𝜖ℝ , 𝑡 > 0
𝑢(𝑥, 0) = 0, 𝑥 𝜖ℝ ,
(3.4)
𝑢(𝑥, 0) = 0,𝑥 𝜖ℝ .
26
If 𝑢1 (𝑥, 𝑡) and 𝑢2 (𝑥, 𝑡)are solutions of (CW) and (3.4) respectively then 𝑢(𝑥, 𝑡) = 𝑢1 (𝑥, 𝑡)+
𝑢2 (𝑥, 𝑡) is a solution of (𝐼𝐶𝑊).Let us consider (3.4).Making change of variable 𝜉 = 𝑥 + 𝑐𝑡,
𝜂 = 𝑥 − 𝑐𝑡,
We transform (3.4) into
1
𝑈𝜉𝜂 = − 4 𝐹(𝜉, 𝜂),
(3.5)
𝑈(𝜉, 𝜉) = 𝑈𝜉 (𝜉, 𝜉) = 𝑈𝜂 (𝜉, 𝜉) = 0 (3.6) Where
𝜉+𝜂 𝜉−𝜂
,
),
2
2𝑐
𝜉+𝜂 𝜉−𝜂
𝐹(𝜉, 𝜂) = 𝑓 (
,
).
2
2𝑐
𝑈(𝜉, 𝜂) = 𝑢 (
Integrating (3.5) with respect to 𝜂 we have
𝑈𝜉 (𝜉, 𝜉) − 𝑈𝜉 (𝜉, 𝜂) = −
1 𝜉
∫ 𝐹(𝜉, 𝑠) 𝑑𝑠
4𝑐 2 𝜂
Which, in view of (3.6), yield
1 𝜉
𝑈𝜉 (𝜉, 𝜂) = 2 ∫ 𝐹(𝜉, 𝑠) 𝑑𝜉
4𝑐 𝜂
Integrating the last equation with respect to 𝜉
𝑈(𝜉, 𝜂) − 𝑈(𝜂, 𝜂) =
1
𝜉
1 𝜉 𝑧
∫ ∫ 𝐹(𝜉, 𝑠) 𝑑𝑠
4𝑐 2 𝜂 𝜂
𝑧
𝑈(𝜉, 𝜂) = 2𝑐 2 ∫𝜂 ∫𝜂 𝐹(𝜉, 𝑠) 𝑑𝑠.
Let us make change of variables
𝑠 = 𝜎 − 𝑐𝜏,
𝑧 = 𝜎 + 𝑐𝜏,
Which has Jacobian 𝐽 =
𝜕(𝑠,𝑧)
𝜕(𝜎,𝜏)
= 2𝑐 ≠ 0
The last change transforms the domain of integration
𝐷 = {(𝑠, 𝑧): 𝜂 ≤ 𝑠 ≤ 𝑧, 𝜂 ≤ 𝑧 ≤ 𝜉}
into
𝐷′ = {(𝜎, 𝜏): 𝑥 − 𝑐(𝑡 − 𝜏) ≤ 𝜎 ≤ 𝑥 + 𝑐(𝑡 − 𝜏), 0 ≤ 𝜏 ≤ 𝑡}.
Indeed by, 𝜂 ≤ 𝑠 ≤ 𝑧 ≤ 𝜉, we have
𝑥 − 𝑐𝑡 ≤ 𝜎 − 𝑐𝜏 ≤ 𝜎 + 𝑐𝜎 − 𝑐𝜏 ≤ 𝑥 + 𝑐𝑡.
27
(3.7)
Then it follows
0 ≤ 2𝑐𝜏 ≤ 2𝑐𝑡 ⇔ 0 ≤ 𝜏 ≤ 𝑡,
and
𝑥 − 𝑐(𝑡 − 𝜏) ≤ 𝜎 ≤ 𝑥 + 𝑐(𝑡 − 𝜏).
The solution of (2.1) is
𝑡
1
𝑥+𝑐(𝑡−𝜏)
1
𝑢2 (𝑥, 𝑡) = 2𝑐 ∫0 ∫𝑥−𝑐(𝑡−𝜏) 𝑓(𝜎, 𝜏) 𝑑𝜎𝑑𝜏 = 2𝑐 ∫ ∫∆(𝑥,𝑡) 𝑓(𝜎, 𝜏)𝑑𝜎𝑑𝜏
(3.8)
Where ∆(𝑥, 𝑡)denotes the characteristic triangle. We prove that (ICW) has a solution given the
exact formula. We prove that problem (𝐼𝐶𝑊)has a solution given by the exact formula
1
1
𝑥+𝑐𝑡
𝑥+𝑐(𝑡+𝜏)
𝑢(𝑥, 𝑡) = 2 (𝜑(𝑥 + 𝑐𝑡) + 𝜑(𝑥 − 𝑐𝑡)) + 2𝑐 (∫𝑥−𝑐𝑡 ∫𝑥−𝑐(𝑡−𝜏) 𝑓(𝜎, 𝜏)) 𝑑𝜎𝑑𝜏
Note that from (3.9) it follows that the well-posed of (ICW) , for 0 ≤ 𝑡 ≤ 𝑇,
We have
|𝑢(𝑥, 𝑡)| ≤ |𝜑|∞ +
1
1
|𝜓|∞ 2𝑐𝑇 +
|𝑓| ∬
𝑑𝜎𝑑𝜏
2𝑐
2𝐶 ∞,𝑇 ∆(𝑥,𝑡)
Note that from (3.6) it follows the well-posedness of(𝐶𝑊)
Indeed, , for 0 ≤ 𝑡 ≤ 𝑇 ,we have
𝑢|(𝑥, 𝑡)| ≤ ‖𝜑‖∞ 2𝑐𝑇 +
1
||𝑓||∞, 𝑇 ∬ 𝑑𝛿𝑑𝜏
2𝑐
∆(𝑥,𝑡)
𝑇2
≤ ‖𝜑‖∞ + 𝑇‖𝜓‖∞ + ‖𝑓‖∞, 𝑇
2
because
 d d  S () 
 x ,t
2cT .T
.
2
Then for ∈> 0, there exists 𝛿 ∈ (0,
∈
1+𝑇+
𝑇2
2
) such that if ‖𝜑‖∞ < 𝛿, ‖𝜓‖ < 𝛿 and ‖𝑓‖∞,𝑇 < ∈. It
follows ‖𝑢‖∞,𝑇 < ∈,which proves the continuousdependence.
Example 3.1 Find the value solution 𝑢(𝑥, 𝑡) at the point (1,1) of the Cauchy problem
𝑢𝑡𝑡 − 𝑢𝑥𝑥 = 1, 𝑢(𝑥, 0) = 𝑥 2 , 𝑢𝑡 (𝑥, 0) = 1
Solution:
From (2.1) 𝑐 = 1 ,𝑡 = 0
𝑢(1,1) =
1
1 2
1
[𝑓(1 + 1) + 𝑓(1 − 1)] + ∫ 𝑔(𝑥) 𝑑𝑥 − ∬ 𝐺(𝑥, 𝑡)𝑑𝑅
2
2 0
2
𝑅
28
(3.9)
=
22 + 02 1 2
1 1 2−𝑦
+ ∫ 1𝑑𝑥 − ∫ ∫ 1𝑑𝑥𝑑𝑦
2
2 0
2 0 𝑦
5
=
2
Example 3.2 Find the value solution 𝑢(𝑥, 𝑡) at the point (𝑥, 𝑡) of the Cauchy problem
𝑢𝑡𝑡 − 𝑢𝑥𝑥 = 1
𝑢(𝑥, 0) = 𝑥 2
𝑢𝑡 (𝑥, 0) = 1
Solution: Consider (𝑥1 , 𝑡1 ) to be a point the plane. Then, putting 𝑐 = 1
𝑢(𝑥, 𝑡) =
𝑥1 +𝑡1
(𝑥1 + 𝑡1 )2 + (𝑥1 − 𝑡1 )2
1 𝑡1 −𝑡+𝑥1 +𝑡1
+∫
𝑑𝑥 − ∫ ∫
𝑑𝑥𝑑𝑦
2
2 0 𝑡+𝑥1 −𝑡1
𝑥1 −𝑡1
1
= 𝑥1 2 + 𝑡 2 + 𝑡1
2
Finally, letting (𝑥1, 𝑡1 ) = (𝑥, 𝑡) yields the solution
1
𝑢(𝑥, 𝑡) = 𝑥 2 + 𝑡 2 + 𝑡.
2
Example3.3: Solve the following Cauchy problem
𝑢𝑡𝑡 − 𝑐 2 𝑢𝑥𝑥 = 𝑒 𝑥 − 𝑒 𝑥 , (𝑥, 𝑡) ∈ ℝ × (0, ∞)
𝑢(𝑥, 0) = 𝑥
𝑢𝑡 (𝑥, 0) = 𝑠𝑖𝑛𝑥.
Solution: We use 𝑒 𝑥 − 𝑒 −𝑥 = 2𝑠𝑖𝑛ℎ𝑥 and formula
Remark. 𝑐𝑜𝑠ℎ(𝑎 + 𝑏) = 𝑐𝑜𝑠ℎ𝑎𝑐𝑜𝑠𝑏 + 𝑠𝑖𝑛ℎ 𝑎 𝑠𝑖𝑛ℎ 𝑏.
Then by (3.9)
29
(𝑥 + 3𝑡) + (𝑥 − 3𝑡) 1 𝑥+3𝑡
1 𝑡 𝑥+3(𝑡+𝑠) 𝑦
(𝑒 − 𝑒 −𝑦 )𝑑𝑥𝑑𝑦
𝑢(𝑥, 𝑡) =
+ ∫
sin 𝑦 𝑑𝑦 + ∫ ∫
2
6 𝑥−3𝑡
6 0 𝑥−3(𝑡−𝑠)
1
𝑥 + 3𝑡 1 𝑡 𝑦
𝑥 + 3(𝑡 + 𝑠)
= 𝑥 − 𝑐𝑜𝑠𝑦 ⌊
+ ∫ (𝑒 + 𝑒 −𝑦 ) ⌊
𝑑𝑠
6
𝑥 − 3𝑡 6 0
𝑥 − 3(𝑡 − 𝑠)
1
1 𝑡
𝑥 + 3(𝑡 + 𝑠)
[𝑐𝑜𝑠(𝑥
=𝑥+
− 3𝑡) + 𝑐𝑜𝑠(𝑥 + 3𝑡)] + ∫ cosh 𝑠 |
𝑑𝑠
6
3 0
𝑥 − 3(𝑡 − 𝑠)
1
2
2
= 𝑥 + 𝑠𝑖𝑛𝑥𝑠𝑖𝑛3𝑡 − 𝑠𝑖𝑛ℎ𝑥 + 𝑠𝑖𝑛ℎ𝑥𝑐𝑜𝑠ℎ3𝑡.
3
9
9
30
3.3 SUMMARY
Like ordinary differential equations, partial differential equations are equations to be solved in
which the unknown element is a function, but in PDEs the function is one of several variables,
and so of course the known information relates the function and its partial derivatives with
respect to the several variables. Again, one generally looks for qualitative statements about the
solution. For example, in many cases, solutions exist only if some of the parameters lie in a
specific set (say, initial data). Various broad families of PDE's admit general statements about
the behavior of their solutions. This area has a long-standing close relationship with the physical
sciences, especially physics, thermodynamics, and quantum mechanics: for many of the topics in
the field, the origins of the problem and the qualitative nature of the solutions are best.
A general linear PDE may be viewed as seeking the kernel of a linear map defined between
appropriate function spaces. (Determining which function space is best suited to the problem is
itself a nontrivial problem and requires careful functional analysis as well as a consideration of
the origin of the equation. Indeed, it is the analysis of PDEs which tends to drive the
development of classes of topological vector spaces.) The perspective of differential operators
allows the use of general tools from linear algebra, including eigen space decomposition
(spectral theory) and index theory.
Generally,
1. Wave equations are examples of second order partial differential equations
2. One dimensional wave equation are mathematical models of most physical
phenomena.( Sound Wave ,musical Technology)
3. The solution of one dimensional equation can be successfully solved by separation
variables and d’Alembert’s method
4. Finding a specific solution to a 1D.W Eq. typically requires an initial condition as well
as boundary conditions.
31
References
Andrews, L.C.1986 .Elementary partial differential equations with boundary value problems,
Academic press ,Orlando.
E.L. Ince,1944.Ordinary Differential Equations.Mineda, NY: Dover.
Erwin kreyszig, 2005. Advanced Engineering Mathematics, Wiley john Wiley & Sons, Inc.
U.S.A
G.F. Carrier and C.E. Pearson, 1988 .Partial Differential Equations, Theory and
Technique,second edition. Boston, MA: Academic Press.
Lonis p. Stavroulak and T.A. Stepan. Partial Differential equation –An Introduction with
mathematical maple ,second edition.
N.H. Fletcher and T.D. Rossing, 1998.The Physics of Musical Instruments.New York,
NY:Springer-Verlag,
Pinchover Y. and J. Rubinstein, 2005.An Introduction to Partial Differential equation. U.S.A
R. Courant and D. Hilbert, 1996. Methods of Mathematical Physics, Vols. I, II, New York, NY:
John Wiley & Sons.
Walter A. Strauss, 1992 Partial Differential Equations. Sons, Inc.
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