04-the wave equation

7/27/2019 04-The Wave Equation

1/41

The Wave Equation

Analytic Solution using Separation

of Variables


2/41

The Wave Equation 2

Motion of an Elastic String

Consider an elastic string (e.g. a guitar string) of

length L, fixed at both ends.

This is distorted and at some instant, say t = 0, released

to that it vibrates.

The problem is to calculate the transverse (i.e. vertical)deflection of the string u(x,t) at any point x between the

fixed endpoints and for any time t > 0.


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The Wave Equation 3

Elastic String

u

x x+xO

PQ

T1

T2

L


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The Wave Equation 4

Outline of Solution State modelling assumptions and initial conditions

Derive the (Partial) Differential Equation (PDE),

and associated boundary conditions

Convert PDE into Ordinary Differential Equations

(ODEs) using Separation of Variables Solve ODEs that satisfy the boundary conditions

Use Fourier Series to satisfy the initial conditions


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The Wave Equation 5

Modelling Assumptions

We assume the following:

The mass of the string per unit length (the line density) is

constant i.e. a homogeneous string. This string isperfectly elastic and does not offer any resistance tobending.

The tension caused by stretching the string is so largecompared to the deflection caused by gravity that thelatter can be neglected.

The string performs small transverse motions in a

vertical plane; that is every particle in the string movesstrictly vertically so that the deflection and the slope atevery point of the string always remain small in absolutevalue.


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The Wave Equation 6

The Differential Equation

To obtain the differential equation, we consider the forces

acting on a small element of the string.

Since the string does not offer any resistance tobending, the tension is tangential to the curve of the

string at each point.

Let T1 and T2 be the tensions at the endpoints, P and Q, ofthis portion of string.

Since there is no horizontal motion, the horizontal

component of the string must be constant.Using the notation of the previous diagram, we obtain

(1)constantT)cos(T)cos(T 21 ===


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The Wave Equation 7


In the vertical direction, we have two forces:

the vertical components of the tension and

where the minus sign appears because the component at P

is directed downwards.

By Newton's second law, the resultant of these two forces

is equal to the mass, , of the string element, times

the acceleration in the vertical direction , evaluated atsome point between x and

)sin(T1 )sin(T2

x

2

2

t

u

xx +


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The Wave Equation 8


Hence

using eq(1), we can divide this by to obtain

(2)

i.e.

2

2

12 t

ux)sin(T)sin(T

=

2

2

1

1

2

2

t

u

T

x

)cos(T

)sin(T

)cos(T

)sin(T

=

2

2

t

u

T

x)tan()tan(

=


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The Wave Equation 9


Now tan() and tan() are the slopes of the string at x andx + x respectively:

so and

which gives

i.e.

xx

u)tan(

=xxx

u)tan(

+

=

2

2

xxx t

u

T

x

x

u

x

u

=

+

2

2

xxx t

u

Tx

u

x

u

x

1

=

+


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The Wave Equation 10


If we now let x approach zero,

the left hand side of this equation becomes the partial

derivative of with respect to x,

i.e. the second order partial derivative .

The equation therefore can be written as

where

x

u

2

2

x

u

2

2

22

2

t

u

c

1

x

u

=

=

T

c

2


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This is known as the One-Dimensional Wave Equation.

It is a second order partial differential equation whichdescribes how the vibrations evolve in time in one space

dimension.

The notation c2

rather than c for the physical constantT/ has been chosen to indicate that this constant ispositive.

2

2

22

2

t

u

c

1

x

u

=

(3)


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Boundary ConditionsVibrations in an elastic string are governed by the one-dimensional wave equation

The solution of this equation also has to satisfy certainboundary conditions due to the fact that the two ends of thestring are fixed.

The string is fixed for all times, i.e. for all values of t, at x = 0,and is similarly fixed at x = L .

The boundary conditions are therefore

u(0,t) = 0 and u(L,t) = 0 for all t (4)

2

2

22

2

xuc

tu

=


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Initial Conditions

The form of the motion of the string will also depend on its

initial deflection profile (i.e. u(x,0) ) and on the initial velocityprofile (i.e. ).

These will be functions of x only so we write

(5)

The problem will therefore be solved if we can find a solution

of the one-dimensional wave equation (3) which also

satisfies the boundary conditions (4) and the initial

conditions (5).

0tt

u

=

)x(gt

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

0t

=

==


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Method of Solution

Outline

1. The first step is to apply the method ofseparation of

variables to the partial differential equation to obtaintwo ordinary differential equation.

2. The second step is to determine a solution of the two

ordinary differential equations that satisfy the boundaryconditions.

3. Finally, using Fourier Series, we shall compose these

solutions to get a solution which satisfies the initialconditions.


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We first assume that the solution of the wave equation

u(x,t) which is a function which depends on both x and t,

can be written as the product of a function of x only and afunction of t only.

(6)

We then substitute this back into the wave equation.

Clearly we need to differentiate with respect to x and t.

We use the dot notation to indicate differentiation withrespect to t and the dash notation to denote differentiation

with respect to x.

1. Separations of Variables

u(x,t) = F(x)G(t)


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and

[ ]

)t(G)x(F

dt)t(Gd)x(F

)t(G)x(Ftt

u

2

2

2

2

2

2

&&

=

=

=

[ ]

)t(G)x(F

)t(Gdx

)x(Fd

)t(G)x(Fxx

u

2

2

2

2

2

2

=

=

=


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Inserting these back into the wave equation we obtain

Dividing both sides by c2F(x)G(t) we obtain

The left hand side of the equation depends only on the

variable t whereas the right hand side depends only on the

variable x. The only way that this can occur is if each sideequals an arbitrary constant, say k.

i.e.

)t(G)x(Fc)t(G)x(F 2 =&&

)x(F

)x(F

)t(Gc

)t(G

2

=

&&

k)x(F

)x(F

)t(Gc

)t(G

2=

=

&&


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Therefore we have two equations

(7)

and

(8)

2

2

G(t)k G(t) c kG(t) 0c G(t) = =

&&&&

0)x(kF)x(Fk)x(F

)x(F==


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2. Satisfying the Boundary Conditions

We shall now determine the functions F(x) and G(t) so that

satisfies the boundary conditions (4), i.e.

and for all t

Considering the first of these equations,

Clearly either F(0) = 0 or G(t) = 0 for all t. The latter solution

means that u(x,t) would be identically equal to zero which is of

no interest, so we must have that F(0) = 0.

Similarly for the second equation, we must have F(L) = 0.

The boundary conditions therefore imply the following

conditions on F(x): (9)

0)t(G)0(F)t,0(u == 0)t(G)L(F)t,L(u ==

0)t(G)0(F)t,0(u ==

0)L(F0)0(F ==


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We now consider the ordinary differential equation (8)

There are three possible possibilities for the value of k.

Firstly k could be zero,

secondly k could be a positive number

thirdly k could be a negative number.

0)x(kF)x(F =


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For k = 0, the equation states that

which has a general solution of the form

where a and b are constants.

If this is the case, then

andand hence F(x) is identically equal to zero.

This means that u(x,t) is also identically equal to zero

which is uninteresting.

0)x(F =

bax)x(F +=

0b0)0(F ==

0a0)L(F ==


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If k > 0, then

where

is some real number.

This means that the general solution of eq(8) is

where A and B are constants.

The boundary condition

whileHence the only solution is the uninteresting solutionu(x,t) = 0.

2k =

xx BeAe)x(F +=

AB0)0(F ==

0A0)L(F ==


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The final possibility is that k < 0.

In this case we can write for some real number p.

Therefore the differential equation for F(x) takes the form

which is the simple harmonic motion equation which has

general solution

(A,B constants)

2pk =

0)x(Fp)x(F 2 =+

)pxsin(B)pxcos(A)x(F +=


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The boundary condition F(0) = 0 implies that A = 0,

while F(L) = 0 implies that Bsin(pL) = 0.

For this to occur either B = 0 or . The former result would again mean that F(x) and hence

u(x,t) is identically equal to zero. Consequently we must

take which implies that pL is an integermultiple of.

Therefore(10)

where n is an integer.

0)pLsin( =

0)pLsin( =

L

n

p

=


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3. Finding the General Solution

We therefore obtain infinitely many solutions for the

function F(x) depending on the value of the integer n

(n = 1, 2, 3, ) (11)

[Note that we could also include the negative integers but

we would obtain essentially the same solutions since

sin(-) = - sin() and so the minus sign is just incorporatedinto the constant Bn.]

== x

L

nsinB)x(F)x(F nn


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We can now solve the second ordinary differential equation

0)t(kGc)t(G 2 =&&

subject to the restriction that 22

L

npk

==

i.e. 0)t(GL

cn)t(G

2

=

+&&

Once again, this is the simple harmonic equation with

an infinite number of solutions depending on the integer n,

n n n

cn cnG(t) G (t) C cos t D sin t

L L

= = +


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The function u(x,t) is the product of the functions F(x)

and G(t)

So we obtain infinitely many solutions of the partialdifferential equation which satisfy the boundary

conditions (4).

n n n n

n n n

cn cn nu (x,t) C cos t D sin t B sin x

L L L

cn cn nu (x,t) C cos t D sin t sin x

L L L

= +

= +

(where in the second line we have simply absorbed the constant Bn

into

the other constants Cn and Dn).

(12)i.e.


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These functions are called eigenfunctions or

characteristic functions

The values are called the characteristic

frequencies of the vibrating string.

The most general solution of the partial differential

equation satisfying the boundary conditions will be a sum

over all possible eigenfunctions, so we obtain:


L

cn

=

=

+

==

1n

nn

1n

n xL

nsint

L

cnsinDt

L

cncosC)t,x(u)t,x(u

(13)


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4. Satisfying the Initial Conditions

The initial conditions that we need to satisfy are given by

equations (5),

i.e

From equation (13) we see that

that is

t 0

uu(x,0) f(x) and g(x)

t =

= =

( ) ( )n n nn 1 n 1

nf(x) u(x,0) u (x,0) C cos 0 D sin 0 sin x

L

= =

= = = +

=

=

1n

n x

L

nsinC)x(f (14)


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The function f(x), which is only defined for , can

however be written as a Fourier sine series given by

These two expressions for the function f(x) are clearly thesame if we identify the constants Cn as the Fourier sine

coefficients, bn given by

0 x L

=

=

1n

n xL

nsinb)x(f (15)

dxxL

nsin)x(f

L

2bC

L

0

nn

== (n = 1, 2, 3 ) (16)


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The second initial condition that needs to be satisfied is

Since u(x,t) is given by

the first partial derivative with respect to the variable t can

be calculated to be

)x(gt

u

0t =

=

=

+

=

1n

nn xL

nsintL

cnsinDtL

cncosC)t,x(u

=

+

=

1n

nn xL

nsint

L

cncos

L

cnDt

L

cnsin

L

cnC

t

u (17)


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Therefore

We must again choose the constants Dn so that the right

hand side coincides with the Fourier sine series for the

function .

This will be the case if

==

=

1n

n

0t

x

L

nsin

L

cnD

t

u(18)

t

u

dxxL

nsin)x(g

L

2

L

cnD

L

0

n

=

(19)


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Hence

Subject to these conditions on the constants Cn and Dn, the

function defined in equation (13) is a complete solution tothe problem.

L

n

0

2 nD g(x)sin x dx

cn L

=

(20)


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For the sake of convenience, we demonstrate it only for

the case when the initial velocity profile g(x) is identically

zero (although the procedure can be used moregenerally however).

Applying the trigonometric product formula for cosines

and sines

Further Simplification

=

= 1n n xLn

sintL

cn

cosC)t,x(u (21)

( ) ( )

+

+

=

ctx

L

nsinctx

L

nsin

2

1x

L

nsint

L

cncos


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We may rewrite the solution (21) as

The expressions in each of the two series are the Fourier sine

components for the initial displacement profile f(x) but with the

variable x replaced by (x ct) in the first case and (x + ct) in

the second.

Thus we may write

where is the odd periodic extension of f(x) with period 2L.

++

=

=

=

)ctx(LnsinC

21)ctx(

LnsinC

21)t,x(u

1n

n

1n

n (22)

[ ])ctx(f~)ctx(f~2

1)t,x(u ++=

f(x)%


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The graph of the function )ctx(f~

is obtained from that of

)x(f by shifting it ct units to the right. This means that )ctx(f

~

represents a wave with profile f(x) travelling to the right with

velocity c units.

Similarly represents a wave with profile f(x)travelling to the left with velocity c units.

)ctx(f

~

+

Equation (22) therefore gives a solution that is a

superposition of these two travelling waves.


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The graph of the function is obtained from that of

f(x) by shifting it ct units to the right.

This means that represents a wave with profilef(x) travelling to the right with velocity c units.

Similarly represents a wave with profile f(x)

travelling to the left with velocity c units. Equation (22) therefore gives a solution that is a

superposition of these two travelling waves.

For completeness, we may write out fully, the solution (13)

)ctx(f~

)ctx(f~

)ctx(f~

+

L L

n 10 0

2 n cn 2 n cn nu(x,t) f(x)sin x dx cos t g(x)sin x dx sin t sin x

L L L cn L L L

=

= +


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ExampleAn elastic string is plucked so that its motion is governed by

the one-dimensional wave equation

Solve the wave equation subject to the following boundary and initial

conditions:

i)

ii)

iii)

2

22

2

2

x

u

ct

u

=

metre1atandorigintheatfixedendpointsi.e.0)t,1(u

0)t,0(u

==

x

0 x 0.510u(x,0) (initial displacement profile)1 x

0.5 x 110


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Solution

The initial displacement profile, f(x), is shown below:

The general solution of the wave equation with fixed boundary

conditions is given by

where in this case, L = 1, and Dn = 0 since the initial velocity profile

is identically zero.

so where

x

u

O0.5 1.0

=

=

+

==

1n

nn

1n

n xL

nsint

L

cnsinDt

L

cncosC)t,x(u)t,x(u

( ) ( )

=

=1n

n xnsintcncosC)t,x(u

=T

c 2


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Solution

To satisfy the initial condition for u(x,t), we have to choose the

constants Cn to be the Fourier sine coefficients of the function f(x).

( )

( )

L

n

0

1

0

0.5 1

0 0.5

10.5

2 2 2 20 0.5

2 2

2C f(x)sin n x dxL

2f(x)sin n x dx

1 1

x 1 x2 sin(n x)dx 2 sin(n x)dx

10 10

1 x cos(n x)x cos(n x) sin(n x) sin(n x)2 2

10n 10n10n 10n

2 nsin

25n

=

=

= +

= + + +

=


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Solution

The complete set of solutions in this case is therefore

2 2n 1

2 2n 1

2

2 nu(x, t) sin cos(cn t)sin(n x)

25n

2 1 n

sin cos(cn t)sin(n x)25 n

cos(3c t)sin(3 x)

cos(c t)sin( x)2 9

cos(5c t)sin(5 x) cos(7c t)sin(7 x)5+

25 49

=

=

=

=

=

K

04-the wave equation

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