Wave Equations

Wave Equations

Institute of Lifelong Learning, University of Delhi pg. 1

Subject: Math

Lesson: Wave Equations

Course Developer: Dr. Preeti Jain

College/Department: A.R.S.D. College, University of Delhi

Wave Equations


Table of Contents:

Chapter : Wave Equations

1: Learning Outcomes

2: Introduction

3: Homogeneous Wave Equation

4: Initial Boundary Value Problem

5: Non- Homogeneous Boundary Conditions

6: Vibration of Finite Strings with Fixed Ends

7: Non- homogeneous Wave Equations

8: Riemann Method

9: Goursat Problem

10: Spherical Wave Equation

Summary

Exercises

Glossary

References/ Further Reading

Wave Equations


1. Learning Outcomes:

After you have read this lesson , you will be able to understand, define

and simplify the homogeneous wave equation, non- homogeneous

wave equation, non- homogeneous boundary conditions, initial

boundary value problem, finite string problem with fixed ends,

Riemann problem, Goursat problem and spherical wave equation. You

should be able to differentiate between homogenous wave equations

and non- homogeneous wave equations. You will also be able to

understand why it is said that the solution of finite vibrating string

problem with fixed ends is more complicated then the problem of

infinite vibrating string. Moreover, you should be able to apply the

knowledge to solve various initial boundary problems which arises in

many practical problems.

2. Introduction:

Differential equations enables us to solve various problems which we

come across in many mathematical problems. Our main concern here

is on boundary value problems. Boundary value problems are

problems in differential equations in which certain conditions (which

we can also call as constraints) are imposed on the equations. After

finding the general solution of these problems, we apply these

additional condition to get a solution which also satisfies the boundary

conditions. We also come across the boundary value problems in many

physical quantities. Generally, these conditions are specified at the

extremes. Here, when we are dealing with wave equations, we will be

considering both the cases homogeneous wave equation and non-

homogeneous wave equation. If we define (initialize) the independent

variable at the lower boundary of the domain, we can often term

homogeneous wave equations as initial boundary value problems.

Wave Equations


3. Homogeneous Wave Equation:

The equation 2 0tt xxu c u is the standard example of hyperbolic

equation. There are two real characteristic slopes at each point ,x t .

This equation is popularly known as wave equation in one dimension

and describes the propagation (bi-directional) of waves with finite

speed c . The characteristics are therefore, curves in the real domain

of the problem. Oscillatory (not always periodic) behaviour in time.

Here time reversible is permissible as it reverses the direction of wave

propagation.

Now, we will obtain the solution of one dimensional wave equation in

free space. For this, consider the Cauchy problem of an infinite string

with the initial conditions

2 0 , , 0tt xxu c u x R t (3.1)

,0 ,u x f x x R (3.2)

,0 ,tu x g x x R (3.3)

To solve equation (3.1), we first reduce it into canonical form. The two

characteristic coordinates

, ,x ct x ct

transforms equation (3.1) into 0.u

After performing two straightforward integration, we get

,u ,

where and are arbitrary functions to be determined, (provided

they are differentiable twice). Thus, the general solution of wave

equation in terms of original variables x and t is

,u x t x ct x ct , (3.4)

,0u x x x f x , (3.5)

Wave Equations


,0tu x c x x g x . (3.6)

Integrating the last equation and then simplifying for and , we get

0

1 1

2 2 2

x

x

Kx f x g d

c , (3.7)

0

1 1

2 2 2

x

x

Kx f x g d

c , (3.8)

0x and K called the arbitrary constants.

The solution of wave equation is, therefore, given by

1 1

,2 2

x ct

x ctu x t f x ct f x ct g d

c

. (3.9)

Solution shown in equation (3.9) is known as DAlembert solution of

the Cauchy problem for one dimensional wave equation.

if double derivative of f and derivative of g exist then by direct

substitution it is evident that ,u x t satisfies the equation (3.1). The

DAlembert solution describes two distinct waves- one moves to right

direction and other on the left both with speed c. Physically,

x ct represents a propagating wave progressing in the negative x-

x

y

0 0,x t

R

O

Figure 1

Range of influence

Domain of dependence

Wave Equations


direction and x ct represents a propagating wave progressing in

the positive x- direction. These two waves do not interact with

themselves nor with each other and hence the solutions are

superposed. Waves do not change their shape as they propagate. The

results depend continuously on the initial data and boundary data, so

the equation is well posed and the solution is unique.

As one can see, the solution of equation(3.1), consists of terms of

f x ct and integral of g. Equation (3.9), which is a solution of

equation (3.1), further suggests that the solution does not depend on

values of function f x which is independent to initial data. And the

end points of the interval are ,x ct x ct which is known as domain of

dependence of ,x t . Further, the solution depends on the integral

value of g between the limits ( x ct ).

Now, the value of u at 0 0,x t does not depend on all the values of u

and tu at 0,t but only on the values in the interval

0 0 0 0,x ct x ct (domain of dependence of 0 0,x t ). If we change f away

from these two points 0 0x ct and 0 0x ct , and g outside this interval,

value of u does not changes at 0 0,x t . Thus, at the point 0 0,x t the

range of influence is the infinite sector R as shown in figure 1.

Value Addition: What we learnt

From this section, we learnt that waves do not change their shape as

they propagate. They only vary in speed and they propagate with

constant speed c in both directions.

Wave Equations


Example 1: Solve the partial differential equation 2 3 0,xx xy yyu u u

with the given initial conditions

,0 sin ,u x x ,0yu x x .

Solution: The given equation is hyperbolic in nature. The two

characteristic coordinates takes the form

3

yx , x y .

Therefore,

2xxu u u u ,

1 2

9 3yyu u u u ,

1 2

3 3xyu u u u ,

The canonical form (we get after substitution) is 0u .

Therefore,

,3

yu x y x x y

, (3.10)

,0 sinu x x x x , (3.11)

1

,03

y y yu x x x x ,

Integrating, the above eq. we get

0

0

21

3 2

xx

xx

x x d

(3.12)

where 0x is an arbitrary point.

Solving equation (3.11) and (3.12), we get

0

23 3sin

4 4 2

x

x

x x

,

0

21 3sin

4 4 2

x

x

x x

,

Wave Equations


21 3

, sin sin .4 4 3 3

y yu x y x y x xy

Value Addition: Activity 1

Find the solution of the following partial differential equations with the

given initial condition:

1. 0, , 0,tt xxu u x R t ,0 sin3 , ,0 cos3 .tu x x u x x

2. 9 0, , 0,tt xxu u x R t 3,0 , ,0 cos .tu x x u x x x

4. Initial Boundary Value Problem:

Two initial conditions where the wave starts from and two boundary

conditions (how the wave and boundary interact, the wave might be

scattered or absorbed are required for the IBVP). We mention some

conditions on the function constituting a solution of the BVP.

1. The solution must be continuous in the region for which the

problem is posed, up to the boundary of the region.

2. The solution must have a continuous second derivatives within

the region and satisfy the given equation and also satisfy the

given boundary conditions of the region.

3. The solution (if the region is three- dimensional and infinite)

approaches to zero as we display a given point to an infinte

distance along an arbitrary ray C.

The purpose of this section is to make you familiar with the effect of

boundary conditions on the solution.

Value Addition: Did you know

In an initial boundary value problem, an initial condition is imposed

over the problem while in a boundary value problem two conditions

are imposed one at initial point and other at the end point.

Wave Equations


4.1. Semi- Infinite String with Fixed End:

the purpose of this segment is to make you aware, how to solve the

problem of semi- infinite vibrating string with a fixed end, i.e.,

2 0, 0 , 0tt xxu c u x t ,

,0 ,u x f x 0 ,x

,0tu x g x ,0 ,x

0, 0,u t 0 .t (4.1)

Now, x ct represents a straight line passing through the origin which

divides the region into two halves. Here, we obtain two waves which

propagates in both directions. One in the positive x- direction with the

velocity c which we obtain from x ct and one in the negative x-

direction with the velocity c which we obtain from x ct . Therefore,

for x ct solution is determined by equation (3.9). For x ct as shown

in figure 2, some part of the solution still lies in the first quadrant for

the interval ,x ct x ct as from D Alembert formula

,u x t x ct x ct . (4.2)

Applying boundary conditions, we get

x

y

0 0,x t

x ct

O

Figure 2

x ct

x ct

0 0 ,0x ct 0 0 ,0x ct

Wave Equations


0, 0u t ct ct .

Hence,

ct ct

Replacing ct by x- ct, we obtain

0

1 1

2 2 2

ct x Kx ct f ct x g d

c

.

The solution is therefore, divided into two parts and is given by

1 1

,2 2

x ct


c

for ,x ct (4.3)

1 1

,2 2

x ct

ct xu x t f x ct f ct x g d

c

for .x ct (4.4)

The solution will exist if f and g also satisfies these additional

conditions, 0 0 0 0f f g , along with the conditions of

differentiability on them (i.e., f should be twice continuously

differentiable and g should be continuously differentiable).

Example 2: Solve the initial boundary-value problem

4 0, 0 , 0tt xxu u x t

4,0 ,u x x 0 ,x

,0 0,tu x 0 ,x

0, 0,u t 0.t

Solution: The given problem is the problem of Semi- Infinite String

with a Fixed End.

Here 4x is twice continuously differentiable, 0g x and

0 0 0 0f f g . Thus, all the conditions of the problem are

satisfied. Hence, solution is given by equation(4.3) and (4.4), i.e.,

For 2 ,x t 4 41

, 2 22

u x t x t x t

,

For 2 ,x t 4 41

, 2 22

u x t x t t x

.

Wave Equations


Also, 0, 0,u t is satisfied by ,u x t for 2 ,x t ( 0t ).


Solve the following initial boundary-value problems

4 0, 0 , 0, 0, 0, 0,tt xxu u x t u t t

( ,0) sin , ,0 0, 0 .tu x x u x x

4.2. Semi-Infinite String with a Free End:

Now, our purpose is to makes you familiar with the problem of Semi-

Infinite String with a Free End. The differential equation for the

problem is as follows:

2 0, 0 , 0tt xxu c u x t ,

,0 ,u x f x 0 ,x

,0tu x g x ,0 ,x

0, 0xu t ,0 ,t (4.5)

As explained above here, the solution for x ct is given by

equation(3.9). To compute solution for x ct (the portion which lies in

first quadrant 0t ), consider equation(3.4) from the DAlembert

solution

,u x t x ct x ct ,

Applying boundary conditions on it, we get

0, 0.xu t ct ct

After integrating we get

ct ct K ,

where K is a constant of integration

Thus,

Wave Equations


0

1 1

2 2 2

ct x Kx ct f ct x g d

c

.

The solution is given by

1 1

,2 2

x ct


c

for ,x ct (4.6)

0 0

1 1,

2 2

x ct ct x

u x t f x ct f ct x g d g dc

for x ct .

(4.7)

The solution will exist if f and g also satisfies these additional

conditions, 0 0 0f g , along with the conditions of differentiability

on them.

Value Addition:

In a semi infinite string we know the initial point from where the

vibration in string starts and waves move only in the positive direction.

whereas in an infinite string we don't know the origin of the vibration

in the string i.e. it moves from to .

Example 3: Solve the initial boundary-value problem

9 0, 0 , 0tt xxu u x t

,0 0,u x 0 ,x

3,0 ,tu x x 0 ,x

0, 0,xu t 0.t

Solution: The given problem is the problem of Semi- Infinite String

with a Free End. Here 3x is continuously differentiable, 0f x and

0 0 0f g . Thus, all the conditions of the problem are satisfied.

Hence, solution is given by equation (4.6) and (4.7), that is

For 3 ,x t 3

43

3

33

1 1,

6 6 4

x tx t

x tx t

u x t d

Wave Equations


4 41 3 324

x t x t ,

For 3 ,x t 3 3

3 3

0 0

1,

6

x t t x

u x t d d

4 41 3 324

x t t x ,

3 31, 4 3 4 324

xu x t x t t x .

We notice here that 0, 0,xu t is satisfied for 3 .x t


Solve the initial boundary- value problem

16 0, 0 , 0tt xxu u x t

,0 cos ,2

xu x

0 ,x

3,0 ,tu x x 0 ,x

0, 0xu t 0 t .

5. Non- Homogeneous Boundary Conditions:

So far we dealt with problems of differential equations with

homogeneous boundary conditions. This section is to make you aware

with equations whose boundary conditions are non- homogeneous.

Consider a initial boundary-value problem with non homogeneous

boundary conditions. We illustrate this by considering two cases:

Case 1:

2 0, 0, 0,tt xxu c u x t

,0 , 0u x f x x ,

,0 , 0tu x g x x ,

Wave Equations


0, , 0u t p t t , (5.1)

Proceeding, as in the case of homogeneous wave equation for x ct .

For 0 x ct by d Alemberts solution for the Cauchy problem we have

,u x t x ct x ct (5.2)

Applying boundary conditions, we get

0,u t ct ct p t ,

ct p t ct .

Replacing ct by x ct , the preceding relation becomes

x

x ct p t ct xc

,

Thus, from equation (21), we get

,x

u x t p t x ct ct xc

1 1

2 2

x ct

ct x

xf x ct f ct x g d p t

c c

(5.3)

Here g must be cont. differentiable, f and p must be twice

continuously differentiable and 20 0 , 0 0 , 0 0 .f p p g p c f

Example 4: Solve the signal problem governed by the wave equation

2 , 0, 0,

,0 ,0 0, 0,

0, , 0.

tt xx

t

u c u x t

u x u x x

u t U t t

(5.4)

Solution: Characteristic curves are given by .dx

cdt

After integration we get

, .x ct x ct

Therefore, general solution of equation (5.4), is

,u x t x ct x ct (5.5)

Applying boundary conditions, for x ct , we have

Wave Equations


,0 0u x x x ,

,0 0tu x c x c x ,

Solving them we get solution for equation (5.5), that is

, 0u x t .

For x ct , we have

0,u t ct ct U t ,

Replacing ct by x-ct , we get

x

x ct U t ct xc

,

Since 0x , we get

x

x ct U tc

.

Therefore,

, 0 for

for

u x t x ct

xU t ct x x ct

c

(5.6)

,x x

u x t U t H tc c

where H is the Heaviside unit step function.

Case 2:

2 0, 0, 0,tt xxu c u x t

,0 , 0u x f x x ,

,0 , 0tu x g x x ,

0, , 0.xu t q t t (5.7)

Applying boundary conditions on equation (5.2), we get

0, .xu t ct ct q t

After integration, we get

Wave Equations


0

.t

ct ct c q d K (5.8)

where K is the constant of integration.

Replacing ct by x-ct in equation (5.8), we get

0

.x

tcx ct ct x c q d K

The solution of the initial boundary- value problem for x ct , is given

by

0 0

0

1 1,

2 2

x ct ct x

xtc

u x t f x ct f ct x g d g dc

c q d

(5.9)

Here g must be continuously differentiable, and f must be twice

continuously differentiable and

0 0 , 0 0 .f q g q

Value Additions:

After reading this we are able to differentiate between homogeneous

and non- homogeneous Boundary conditions of wave equations.

Homogeneous boundary conditions of wave equations are those in

which right hand side is zero (is not a function of time t) i.e. 0, 0u t

etc. whereas non- homogeneous boundary conditions of wave

equations are those in which right hand side is not zero (is a function

of time t) i.e. 0,u t p t etc.


Solve the initial boundary-value problem

2 , 0 , 0,

,0 , 0 ,

,0 , 0 ,

0, , , , 0.

tt xx

t

u c u x l t

u x f x x l

u x g x x l

u t p t u l t q t t

Wave Equations


6. Vibration of Finite Strings with Fixed Ends:

We will now study problem of vibration of string when the length of

string is finite and the string is fixed between two ends. This problem

requires more concern and labour then that of previous problem. It is

due to the repeated reflection of waves from the boundaries which is

not present in the case of infinite string problem. Let the length of

string be l which is fixed at both ends. we require to determine the

solution of

2 , 0 , 0,

,0 , 0 ,

,0 , 0 ,

0, 0, , 0, 0,

tt xx

t

u c u x l t

u x f x x l

u x g x x l

u t u l t t

(6.1)

We already know that the solution of the wave equation is

,u x t x ct x ct .

Now, we apply the initial conditions and get

,0 ,u x f x x x 0 ,x l

,0tu x g x c x c x , 0 .x l

Solving for and , we obtain

0

1 1,

2 2 2

Kf g d

c

0 ,x ct l (6.2)

0

1 1,

2 2 2

Kf g d

c

0 x ct l . (6.3)

Hence,

1 1

, ,2 2

x ct

x ctu x t f x ct f ct x g d

c

(6.4)

The solution is determined uniquely by the initial data in the following

region

Wave Equations


0 , , 0.x l x

x ct t x ct l t tc c

When t is large, the solution depends on the boundary conditions.

Thus, we obtain

0, 0u t ct ct , 0t , (6.5)

, 0u l t l ct l ct , 0t . (6.6)

If we take equation (6.5), we get

, 0 0t ct . (6.7)

Replacing by and then using equation (6.2), we get

0

1 1,

2 2 2

Kf g d

c

0 , 0.l l (6.8)

Thus, we see that the range of is extended to .l l

If we take equation (6.6), (where l ct ) we get

2l , l . (6.9)

Replacing by and then using equation (6.3), we get

2

0

1 12 2 ,

2 2 2

l Kl f l g d

c

0 2 2 .l l l l (6.10)

Thus, the range of is extended to 0 2l . Thus, we can obtain

for every 0 and for every .l

Value Addition:

Here the string under consideration is of finite length and there is

repeated reflection of waves from the boundaries.

Example 5: Find the solution of the initial boundary value problem

Wave Equations


, 0 2, 0,

,0 sin , 0 2,2

,0 0, 0 2,

0, 0, 2, 0, 0.

tt xx

t

u u x t

xu x x

u x x

u t u t t

Solution: Here, the problem is vibration of finite string with fixed

ends

1 1

sin , 0 2,2 2 2 2 2

K Kf

1 1

sin , 0 2,2 2 2 2 2

K Kf

Using boundary conditions, we get

1

sin , 0 2,2 2 2

K

1

sin2 2 2

K

, 2 0,

41

2 sin ,2 2 2

Kl

0 4 2,

1

sin2 2 2

K

, 2 4.

Proceeding in this manner, we determine the solution

1

, sin sin2 2 2

x ct x ctu x t

for all x in 0,2 and for all 0t .

Example 6: Find the solution of the initial boundary value problem

4 , 0 1, 0,

,0 0, 0 1,

,0 1 , 0 1,

0, 0, 1, 0, 0.

tt xx

t

u u x t

u x x

u x x x x

u t u t t

Wave Equations


Solution: Here the problem is again of vibration of finite string with

fixed ends. We shall proceed as follows:

2 3

0 0

1 11 , 0 1,

4 2 4 2 3 2

K Kd

2 3

0

1, 0 1,

4 2 3 2

K

Using boundary conditions, we get

2 3

0

1, 0 1 1 0,

4 2 3 2

K

2

2 3

0

12 ,

4 2 3 2

Kl

0 2 1, 1 2.

Proceeding in this manner, we determine the solution

,u x t

The range of is thus extended to 0 2l . Proceeding in the same

way, we can obtain for all 0 and for all .l Thus, the

solution is determined for all 0 x l and 0t .


Solve the initial boundary value problem

2 , 0 , 0,

,0 cos , 0 1,

,0 0, 0 1,

0, 0, 1, 0, 0.

tt xx

t

u c u x l t

xu x x

l

u x x

u t u t t

Solution: ,u x t for all x in 0, l and for all 0t .

7. Non- homogeneous Wave Equations:

The motive of this section is to make you familiar with the non-

homogeneous wave equation. The partial differential equations were

non- homogeneous either because of the presence of a driving term or

Wave Equations


because of a non- homogeneous boundary condition. The generalized

non- homogeneous wave partial differential equation in one dimension

has the form

2 * , ,tt xxu c u h x t (7.1)

Let us derive a mathematical formula, for a particular case when 1c ,

and then generalize it for any value of c.

Consider the Cauchy Problem for the non- homogeneous wave

equation

, ,xx yyu u h x y (7.2)

with the initial conditions

,0u x f x , (7.3)

,0 ,yu x g x (7.4)

Figure 3

Let 0 0 0,P x y be a point of the plane. Since characteristics

constantx y , are two straight lines drawn through the point 0P with

slopes 1 . They intersect x- axis at 1P and 2P with coordinates

0 0 ,0x y and 0 0 ,0x y respectively. Thus, 0 1 2P PP form a triangle

with sides 0 1 2A A A as shown in figure 3, where the whole closed interior

region of the triangle be denoted by D and boundaries by A .

x

y

0 0 0,P x y

O

0 0x y x y

0 0x y x y

1 0 0 ,0P x y 2 0 0 ,0P x y

1A

0A

2A

Wave Equations


Applying Greens theorem on L.H.S. of equation (7.2), we get

, ,xx yy x yR R

A

u u dR u dy u dx h x y dR (7.5)

Now, along 0 0,A dy x varies from 0 0 0 0tox t x t , x y constant along

1A , therefore, dx dy and along 2A , x y constant, therefore dx dy .

Therefore,

0 1 2

x y y x y x yA A A

A

u dy u dx u dx u dx u dy u dx u dy

0 0 0 0 0 0

0 0 0 0 0 0

, ,0

,0 ,

x y x y x y

yx y x y x y

u dx du du

0 0

0 00 0 0 0 0 0,0 2 , ,0

x y

yx y

u dx u x y u x y u x y

(7.6)

From equation (7.6) and equation (7.5), we get

0 0

0 00 0 0 0 0 0

1 1, ,0 ,0 ,

2 2

x y

yx y R

u x y u dx u x y u x y h x y dR

. (7.7)

The point 0 0,x y which we choose is an arbitrary point it can assume

any value. So, replacing it with any point ,x y , and using initial

conditions in equation (7.7), we get

1 1 1

, ,2 2 2

x y

x y Ru x y g d f x y f x y h x y dR

. (7.8)

Equation (7.8), we obtain when we consider 1c . In general 1c .

Thus, solution of equation (7.1) can be written as ((using

transformation y ct ) here 2*, hh x y c and *gg x c

21 1 1

, * * ,2 2 2

x ct

x ct Ru x t g d f x ct f x ct h x t dR

c

(7.9)

Example 7: Determine the solution of the initial value problem

2 , ,0 , ,0 sin .tt xx tu c u x ct u x x u x x

Solution: The given problem is the problem of non- homogeneous

wave equation. Its solution can be find by substituting

, * and * ,f x g x h x t in equation (7.9), that is,

Wave Equations


2 01 1 1

, sin2 2 2

x ct ct y x ct

x ct y x ctu x t d x ct x ct x ct dxdy

c c

2

2 0

1 1cos

2 2 2

y x ctctx ct

x ct

y x ct

xx ctx dy

c c

2 0sin sin 1

2 22 2

ctx ctx x ct y y ct dy

c c

2

2

2

0

1 1sin sin 2

2 2 2

ct

yx x ct x cty y cty

c

2 21 1 1

sin sin2 2 2

x x ct xt ct .


Solve the following equations

(i) 2 , ,0 0, ,0 3.tt xx tu c u x u x u x

(ii) 1, ,0 sin , ,0 .xx yy yu u u x x u x x

Solution: (i) 21

, 3 .2

u x t t xt

(ii) 21 1

, sin sin .2 2

u x y x y x y xy y

8. Riemann Method:

This section is to make you aware with the method of solving the

linear hyperbolic equation. Now, we shall discuss the Riemann

method of solving second order partial differential equation in two

independent variables 2

, , , ,x yu

f x y u u ux y

. It yields an exact solution

in terms of the adjoint equation to the original equation. The basic idea

is that the solution of a non- characteristic initial value problem in two

Wave Equations


dimensions can be found if the adjoint equation with the specific

boundary conditions can be solved.

It is assumed that the function , , , ,x yf x y u u u is continuous at al points

of a region D* defined by ,x y for all values of

, , , ,x y u p q concerned. The method due to Riemann, represents the

solution in a manner depending explicitly on the prescribed boundary

conditions. Although this often involves the solution of another

boundary value problem for the Greens function.

We shall assume that the equation has already been reduced to

canonical form

,L u f x y (8.1)

L is called the linear operator,

2

a b cx y x y

(8.2)

where ,a x y , ,b x y , ,c x y and ,f x y are differentiable functions in

some region D*.

Now, let ,v x y be a function (to be determined suitably) having

continuous second order partial derivatives. Then, compute

vL u uM v , where M is the operator defined by the relation (known

as adjoint operator to the operator L ).

2 av bvv

M v cvx y x y

, (8.3)

and

xy xyu v

vu uv v uy x x y

,

x x xvau u av vau , y y yvbu u bv vbu ,

so that

Wave Equations


x yvL u uM v U U (8.4)

and

yU auv uv , xV buv u v (8.5)

Now, if M = L, we say the operator L is called self- adjoint. Applying,

Greens theorem to equation (8.4), we have

x yD D C

vL u uM v dxdy U U dxdy Udy Vdx , (8.6)

where is the curve RPQP which bounds the region of integration D

which is in D*.

Let be the continuous curve(as shown in figure 4) which is smooth.

Suppose now, that the values of u and xu or yu are defined along the

curve in the xy plane, and we want to obtain the solution of the

equation (8.1) at the point ,P agreeing with these boundary

conditions. Through P draw lines parallel to x- axis and y-axis as

shown in figure 4, cutting curve in Q and R respectively. Take the

curve to be the closed circuit PQRP . By construction 0dx on PR and

0dy on PQ. Thus,

Q R P

P Q RC

Udy Vdx Vdx Udy Vdx Udy (8.7)

Substituting V from equation(8.5) and integrating by parts, we get

Q Q QQ

xPP P PVdx buvdx uv uv dx . (8.8)

Substituting the values from equation (8.7) and (8.8), in equation

(8.6), we get

Q QQ

xPP PD

vL u uM v dxdy buvdx uv uv dx

R P

Q RUdy Vdx Udy (8.9)

Wave Equations


So far the function v has been arbitrary. Let us choose the function

, ; ,v x y such that it is the solution of the adjoint equation 0M v .

Also the function v should satisfy the following conditions:

1. ,xv b x y v when y (along PQ)

2. ,yv a x y v when x (along PR)

3. 1v when x , y at P.

Such a function if it exists is called Greens function for the problem.

Using, these conditions in equation (8.9), we get

R R

y xP Q Q QD

u uv uv ady bdx uv dy vu dx vf dxdy . (8.10)

Equation (8.10) gives u at the point P when u and xu are given along

the curve . When yu is given along the curve , we use

d uv uv dx uv dyx y

,

Integrating this expression from Q to R ( along the curve ), we get

R

x yR Q Quv uv uv dx uv dy

x

y

,P

R

Q

O

Figure 4: Smooth initial curve

Wave Equations


R R

x y x yQ Qu vdx uv dy uv dx u vdy (8.11)

Substituting Q

uv from equation (8.11) in equation (8.10), we get

R R

x yP R Q QD

u uv uv ady bdx uv dx vu dy vf dxdy (8.12)

Finally, by adding equation (8.12) and (8.10), we obtain the

symmetrical result

1 12 2

R R

x yP R Q Q Qu uv uv uv ady bdx u v dx v dy

1

2

R

x yQ

D

v u dx u dy vf dxdy (8.13)

Thus, the solution of the given equation at any point in terms of the

values of , xu u and yu along a given curve , can be obtained by using

equation (8.10), (8.11) and (8.12), whichever is appropriate. Further,

we observe that the solution at any point , depends on the values

which is popularly known as Cauchy data. Thus, a slight change in the

initial data outside the curve would change the solution only outside

this region.

Value Addition:

Green's Theorem: Let C be a piecewise smooth simple closed curve

in the xy-plane and let D denote the closed region enclosed by C.

Suppose M,N,N

x

and

M

y

are real valued continuous functions in an

open set containing D. Then

D C

N Mdxdy Mdx Ndy

x y

where the line integral is taken around C in the counter clockwise direction.

Wave Equations


Example 8: Prove that for the equation 1 04xy

u u Greens function

is 0, ; ,v x y J x y where 0J denotes Bessels function of

the first kind of order zero.

Solution: Here 10, 0, , & , 0.4a b c f x y

14xyL u u u & 1

4xyM v v v .

The Greens function , ; ,v x y satisfies the following conditions:

1 04xyM v v v

0xv when y

0yv when x

1v when , .x y

To find v , we put 2s x y (8.14)

Thus v v s is a function of one variable

Now 0s at , .x y

Taking logarithm of equation (8.14) and differentiating partially with

respect to x and y, we get

, .2 2

s y s x

x s y s

So,

2

2

1

4xy

d v dvv

ds sds

.

Thus, we obtain 2

2

10

4xy

d v dvv v v

ds sds

Or

2

2 2 2

20 0

d v dvs s s vds ds

(8.15)

Wave Equations


Above equation is Bessels equation of order zero, and so 0v J s is a

solution of equation (8.15)

Hence, 0v J x y .

Also, it satisfies all the properties of the Greens function as

0 0 1 1J v when , ,x y

0 01

2x x

yv J s s J s

x

0xv when y and

0 01

2y y

xv J s s J s

y

0yv when .x


Solve the following problem using Riemann method:

2 2

1 2

0,

, , , .

xx tt

t

x u t u

u x t f x u x t g x

Solution: 3 3

2 2

1 1 1 1,

2 2 4 2

x x

t t

xt xt

f gxu x t f xt tf xt d xt d

t

.

9. Goursat Problem:

This section is to make you aware with the Goursat problem. In

Goursat problem we finds a solution of the hyperbolic partial

differential equation

, , , ,xy x yu F x y u u u (9.1)

which satisfies the following conditions

, ,u x y f x (9.2)

on a curve, say, 0y , (which is characteristic of the given equation)

and

,u x y g x ,

Wave Equations


on a monotonic increasing curve y y x . Further, we assume that

curve intersects the characteristic at the origin.

Here, we are illustrating few examples to make you aware of the

procedure of solving Goursats problem.

Value Additions:

A partial differential equation of second order i.e.,

xx xy yy x yAu Bu Cu Du Eu Fu G

where A, B, C, D, E, F and G are constants, is said to be

(i) Parabolic if 2 4 0B AC

(ii) Hyperbolic if 2 4 0B AC

(iii) Elliptic if 2 4 0B AC .

Example 9: Solve the Goursat problem

3 3 3 3 0xx yy x yxy u x yu y u x u , (9.3)

,u x y f x on 2 2 16y x for 0 4x ,

,u x y g x on 0x for 0 4y ,

where 0 4f g .

Solution: Here the given equation is not in canonical form. So, we

reduce it into canonical form first. In canonical form equation (9.3) can

be written as 0u ,

where,

2 2y x , 2 2y x .

Thus, the general solution is

2 2 2 2,u x y y x y x

Applying the prescribed conditions , we have

2, 16 2 16u x y f x x on 2 2 16y x (9.4)

Wave Equations


2 2,u x y g x y y on 0x (9.5)

We observe that these equations are compatible, since

0 4 16 16f g (9.6)

Now, replacing x by 2 21

162

y x in equation (64) and y by 2 2y x

in equation (9.5), we have

2 2 2 21

16 162

f y x y x

, (9.7)

2 2 2 2 2 2g y x x y x y , (9.8)

but f x on 2 2 16y x .

Therefore, from equation (9.8),

2 2 16 16x y g

0 16f (using (9.6)) (9.9)

Thus,

2 2 2 21, 16 16 0 162

u x y f y x g y x f

2 2 2 21 16 02

f y x g y x f

. (9.10)

Example 10: Solve 2tt xxu c u ,

,u x t f x on t t x ,

,u x t g x on 0x ct ,

where 0 0g f .

Solution: The general solution of the problem, using initial conditions,

is

,u x t x ct x ct , (9.11)

Wave Equations


,u x t f x x ct x x ct x , (9.12)

0 2g x x , (9.13)

Let, s x ct x , the inverse of it is x s . Thus, equation (9.12) may

be written as

f s s s ct s , (9.14)

Replacing 2x by x ct

2

s ct sg s s ct s

. (9.15)

From equation (9.14) and (9.15), we get

02

s ct sf s s g

,

or,

02

x ct ct x ctx ct f x ct g

. (9.16)

Hence, the solution is given by (using equation (9.15) and(9.16) in

(9.11))

, 0 02 2

x ct ct x ctx ctu x t g f x ct g

.


Solve the following problems

3 3 3 3 0,xx yy x yxy u x yu y u x u

2 2, on 8 for 0 2,u x y f x y x x

2 2, on 16 for 2 4,u x y g x y x x with 2 2 .f g

Solution: 2 2 2 28 16

, 22 2

y x y xu x t f g f

.

10. Spherical Wave Equation:

Wave Equations


This section is to make you aware with spherical waves. We are

rewriting wave equation into spherical coordinates in this section.

wave equation is one of the kind of partial differential equations of

second order. In the wave equation the dependent variable

, , ,u u x y z t is a function of three space variable; x, y, z and time

variable t, is

2 0tt xx yy zzu c u u u .

In spherical polar coordinates , ,r , the above equation takes the

form

2 2

2 2 2 2 2

2 1 1 1sin

sin sinr rr

u u uu u

r r r c t

(10.1)

Solution of equation (10.1) is known as spherical symmetric waves if u

depends on r and t. Thus, we have

2

2 2

2 1r rr

uu u

r c t

(10.2)

Introducing a new dependent variable ,U ru r t , equation (10.2)

reduces to

2 .tt rrU c U (10.3)

Equation (10.3) denotes a wave in one dimension.

As obtained earlier, the solution of the equation(10.3) is of the form

1

,u r t r ct r ctr .

As we can see, there are two progressive spherical waves travelling

with constant velocity c in opposite directions (one approaching

towards origin and other going away from origin) .

Here, we are interested only in the solution for outgoing waves

1

,u r t r ctr , (10.4)

Wave Equations


21 1

,ru r t r ct r ct r ctr r .

Here ru is called the radial velocity. In fluid flow, u denotes the

velocity potential so that the total limiting flux through a sphere of

radius r and center at the origin is

20

lim4 , 4rr

Q t r u r t ct

.

1

4

r rr ct c t Q t

c c

(10.5)

Again in fluid flow, if p denotes the pressure at any time t and 0p the

pressure at equilibrium value, the difference between them is given by

0

4t

rp p u Q t

r c

, (10.6)

here we call the density of the fluid.

Now, we shall obtain the solution of equation (10.3), where ,U ru r t

with the boundary conditions

,0u r f r , ,0tu r g r , 0r . (10.7)

As f and g satisfies equation (10.3) they should be continuously

differentiable as given by the dAlembert solution of the Cauchy

problem for the one dimensional wave equation, that is

1 1

2 2

r ct

r ctru U r ct f r ct r ct f r ct g d

c

(10.8)

provided 0.r ct

As for 0, 0t r , which determines the solution ,u r t only up-to the

characteristic r ct in the r- t plane.

However, when r is less than or equal to ct , this solution does not

holds, because in that case f and g are not defined for negative values

of r . If u is finite at 0r (where 0U ) for all 0t , the solution for

r ct exists and can be obtained as follows:

Wave Equations


For 0ct r

1

,2

U r t ct r ct r (10.9)

which for 0r gives

0ct ct for 0ct (10.10)

Now, (using equation (10.9) and (10.10)), we get

1

r tU U ct rc

1

r ct f r ct f r ct r ct g r ctc

(10.11)

In view of the fact 1

r tU Uc

is constant on each characteristic ct r =

constant , it turns out from equation (10.11) for 0r ,

ct ct f ct f ct t g ct .

After integrating we get

0

10

t

t t f t g dc

,

so from equation (10.10), we get

0

10

t

t t t f t g dc

.

Substituting these values in equation (10.9), we get

1 1

,2

ct r

ct ru r t ct r f ct r ct r f ct r g d

r c

(10.12)

Value Addition:

The relation between rectangular coordinated and spherical polar

coordinate is given by

sin cos

sin sin

cos .

x r

y r

z r

Wave Equations


Summary:

So far we have obtained the solution of wave equation which is given

by equation (3.9),

1 1

,2 2

x ct


c

.

The importance of waves lies in the fact that in case of waves, the

signal breaks into two pieces and the waves propagates in opposite

directions at finite speed c (movement is along the characteristic, and

therefore, information is retained). Then we illustrate procedure to find

solution of Semi- infinite string with a fixed end and with a free end

and the solution is given by equations (4.3), (4.4) and (4.6), (4.7)

respectively. Then we discuss non- homogeneous boundary conditions

in differential equations and gave a procedure to simplify differential

equations with these conditions, which is given by equations (5.3) and

(5.9). After that we outline a procedure to solve problem of finite

strings with fixed ends and its solution is given by equation (6.10).

Then we take problem of non- homogeneous wave equation, whose

solution is given by equation (7.9). After that we discuss Riemann

method, which is used for solving Goursat problem and Cauchy

problem for linear hyperbolic partial differential equations given by

equation (8.13). In section 9 we gave procedure to solve Goursat

problem and in the end we gave solution of the spherical wave

equation which is given by equation no (10.8) and (10.12).

Now, we suppose you should be able to answer the following

questions:

True False:

1. Time reversible is not permissible in case of propagation of

waves.

Wave Equations


2. Waves change their shape as they propagate.

3. The partial differential equations were non- homogeneous either

because of the presence of a driving term or because of a non-

homogeneous boundary condition.

4. Riemann method yields an exact solution of the linear hyperbolic

equation in terms of the adjoint equation to the original

equation.

5. Three dimensional wave equation is given by

2 0tt xx yy zzu c u u u where t is time variable and x, y, z are

space variables.

6. The adjoint operator M of operator L is called self-adjoint if M=L.

7. The problem of the finite string is easier than that of the infinite

string.

8. Non- homogeneous one- dimensional wave equation is given by

2 0tt xxu c u .

9. Physically, x ct in D Alembert solution of the Cauchy

problem represents a progressive wave travelling in the positive

x- direction with speed different from c.

10.Bessels Equation of order zero is given by 2

2 2

20

d y dyx x x y

dx dx .

Solutions

1. False ;Time reversible is permissible as it reverses the direction

of wave propagation.

2. False ; Waves do not change their shapes while propagation.

3. True.

4. True.

5. False; Three dimensional wave equation is 2 0tt xx yy zzu c u u u .

6. True.

Wave Equations


7. False; The problem is more complicated due to the repeated

reflection of waves from the boundaries.

8. False; Non- homogeneous wave equation is 2 ,tt xxu c u h x t .

9. False; It represents a progressive wave travelling in the negative

x- direction travelling with speed c without change of shape.

10.True.

Descriptive Questions:

1. Give an expression for the D Alembert solution of the Cauchy

problem.

2. Differentiate between homogeneous and non- homogeneous wave

equation.

Find the solution of each of the following initial value problems

3. 2 20, ,0 sin , ,0 .tt xx tu c u u x x u x x

4. 2 10, ,0 cos , ,0 .tt xx tu c u u x x u x e

5. 2 2, ,0 5, ,0 .xtt xx tu c u e u x u x x

6. 2 sin , ,0 cos , ,0 1 .tt xx tu c u x u x x u x x

7. 2 , ,0 sin , ,0 0.ttt xx tu c u xe u x x u x

8. 2 22, ,0 , ,0 cos .tt xx tu c u u x x u x x

9. 10 9 0, ,0 , ,0 .xx xy yy yu u u u x f x u x g x

10. 2 2 0 0, 0, ,1 , ,1 .xx tt tx u t u x t u x f x u x g x

Determine the solution of the following initial boundary value problem

11. 16 0, 0 , 0, 0, 0, 0,tt xxu u x t u t t

2( ,0) sin , ,0 , 0 .tu x x u x x x

12. 0, 0 , 0, 0, 0, 0,tt xx xu u x t u t t

( ,0) cos , ,0 0, 0 .2

t

xu x u x x

Wave Equations


13. 4 0, 0 1, 0,tt xxu u x t

2( ,0) 0, ,0 1 , 0 1,tu x u x x x x

0, 0, 1, 0, 0.u t u t t

14. Solve the characteristic initial value problem

3 0, 0,xx yy xxu x u u x

2

, on 0 for 0 y 2,2

xu x y f y y

2

, on 4 for 2 y 4,2

xu x y g y y where 2 2 .f g

Solutions:

3. 2 2 31sin cos ,3

x ct x t c t 4. cos cos tx cte

,

5. 2 2 3 21 15 23 2x ct x ct xx t c t e e e

c

6. 21cos cos 2 sin sin cosx ct t x x ctc

7. 2sin cos 2txt ex ct 8. 2 2 2 11 cos sinx t c x ctc

9. 99 9 1, 8 9 8 8y

x

x y

yu x t f x g d f x y

10. 1 1, *2 2xt

x

t

xu x t f xt f g dt

where *g x

g xx

.

11. 3 31, sin cos4 4 4 for 424u x t x t x t x t x t

3 31sin cos4 4 4 for 424x t x t t x x t

12. , cos cos for and .2 2x tu x t x t x t

14. 2 2

, 2 4 02 4 2 4

y x y xu x t f g

Glossary:

Wave Equations


G Goursat Problem: In Goursat problem we finds a solution of the

hyperbolic partial differential equation , , , ,xy x yu F x y u u u which

satisfies the following conditions , ,u x y f x on a curve, say,

0y , (which is characteristic of the given equation) and

,u x y g x , on a monotonic increasing curve y y x .

G Green's Theorem: Let C be a piecewise smooth simple closed

curve in the xy-plane and let D denote the closed region

enclosed by C. Suppose M,N,N

x

and

M

y

are real valued

continuous functions in an open set containing D. Then

D C

N Mdxdy Mdx Ndy

x y

where the line integral is taken around C in the counterclockwise

direction.

H Homogeneous wave equation: A wave equation is said to be

homogeneous if the right hand side of the boundary condition

associated with the wave equation is zero (i.e., it is not a

function of time t) i.e. 0, 0u t .

N Non-Homogeneous wave equation: A wave equation is said to

be non-homogeneous if the right hand side of the boundary

condition associated with the wave equation is not zero (i.e., it

is a function of time t) i.e. 0,u t p t etc.

R Riemann Method: Riemann method of solving second order

partial differential equation in two independent variables

2

, , , ,x yu

f x y u u ux y

yields an exact solution in terms of the

adjoint equation to the original equation.

Wave Equations


References/ Further Reading:

1. Tyn Myint-U and Lokenath Debnath, Linear Partial Differential

Equation for Scientists and Engineers, Springer, Indian reprint,

2006.

2. Ian N. Sneddon, Elements of Partial Differential Equations,

McGraw-Hill, 1957.

3. Bateman H., Partial Differential Equations of Mathematical

Physics, Cambridge University Press, Cambridge (1959).

4. Debnath L., Non-Linear Partial Differential Equation for

Scientists and Engineers, (Second Edition), Birkhauser Verlag,

Boston (2005).

5. Epstein B., Partial Differential Equations, McGraw- Hill, New

York (1962).

6. Hadamard J., Lectures on Cauchys Problem in Linear Partial

Differential Equations, Dover Publications, New York (1952).

Wave Equations

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