group 2 bhadouria , arjun singh glave , theodore dean han, zhe

Group 2

Bhadouria, Arjun SinghGlave, Theodore Dean

Han, Zhe

Chapter 5. Laplace TransformChapter 19. Wave Equation

Wave Equation

Chapter 19

Overview

• 19.1 – Introduction– Derivation– Examples

• 19.2 – Separation of Variables / Vibrating String– 19.2.1 – Solution by Separation of Variables– 19.2.2 – Travelling Wave Interpretation

• 19.3 – Separation of Variables/ Vibrating Membrane• 19.4 – Solution of wave equation

– 19.4.1 – d’Alembert’s solution– 19.4.2 – Solution by integral transforms

19.1 - Introduction

• Wave Equation

– Uses:• Electromagnetic Waves• Pulsatile blood flow• Acoustic Waves in Solids• Vibrating Strings• Vibrating Membranes

http://www.math.ubc.ca/~feldman/m267/separation.pdf

Derivation

u(x, t) = vertical displacement of the string from the x axis at position x and time tθ(x, t) = angle between the string and a horizontal line at position x and time tT(x, t) = tension in the string at position x and time t ρ(x) = mass density of the string at position x


Derivation

• Forces:

• Tension pulling to the right, which has a magnitude T(x+Δx, t) and acts at an angle θ(x+Δx, t) above horizontal

• Tension pulling to the left, which has magnitude T(x, t) and acts at an angle θ(x, t) below horizontal

• The net magnitude of the external forces acting vertically F(x, t)Δx

• Mass Distribution:


DerivationVertical Component of Motion

Divide by Δx and taking the limit as Δx → 0.


Derivation


Derivation

For small vibrations:

Therefore,


Derivation

Substitute into (2) into (1)


DerivationHorizontal Component of the Motion

Divide by Δx and taking the limit as Δx → 0.


Derivation

• For small vibrations: and

Therefore,


Solution

For a constant string density ρ, independent of xThe string tension T(t) is a constant, andNo external forces, F


Separation of Variables; Vibrating String

19.2.1 - Solution by Separation of Variables

Scenario

u(x, t) = vertical displacement of a string from the x axis at position x and time tl = string length

Recall:(1)

Boundry Conditions:u(0, t) = 0 for all t > 0 (2)u(l, t) = 0 for all t > 0 (3)

Initial Conditions u(x, 0) = f(x) for all 0 < x <l (4)ut(x, 0) = g(x) for all 0 < x <l (5)

http://logosfoundation.org/kursus/wave.pdf

ProcedureThere are three steps to consider in order to solve this problem: Step 1:• Find all solutions of (1) that are of the special form for some

function that depends on x but not t and some function that depends on t but not x.

Step 2:• We impose the boundary conditions (2) and (3).

Step 3: • We impose the initial conditions (4) and (5).


Step 1 – Finding Factorized Solutions

Let:Since the left hand side is independent of t the right hand side must also be independent of t. The same goes for the right hand side being independent of x. Therefore, both sides must be constant (σ).



(6)



Solve the differential equations in (6)



If , there are two independent solutions for (6)

If ,



Solutions to the Wave EquationFor arbitrary and arbitrary

For arbitrary


Step 2 – Imposition of Boundaries

For

Thus, this solution is discarded.



For , when When

Therefore,



Since , in order to satisfy An integer k must be introduced such that:

Therefore,



Where, and are allowed to be any complex numbers and are allowed to be any complex numbers


Step 3 – Imposition of the Initial Condition

From the preceding:

which obeys the wave equation (1) and the boundary conditions (2) and (3), for any choice of and



The previous expression must also satisfy the initial conditions (4) and (5):

(4’)

(5’)



For any (reasonably smooth) function, h(x) defined on the interval 0<x<l, has a unique representation based on its Fourier Series:

(7)Which can also be written as:



For the coefficients. We can make (7) match (4 ) by ′choosing and .

Thus .

Similarly, we can make (7) match (5 ) by choosing and ′

Thus



Therefore,

(8)Where,



The sum (8) can be very complicated, each term, called a “mode”, is quite simple. For each fixed t, the mode

is just a constant times . As x runs from 0 to l, the argument of runs from 0 to , which is k half–periods of sin. Here are graphs, at fixed t, of the first three modes, called the fundamental tone, the first harmonic and the second harmonic.



The first 3 modes at fixed t’s.



For each fixed x, the mode

is just a constant times plus a constant times . As t increases by one second, the argument, , of both and increases by , which is cycles (i.e. periods). So the fundamental oscillates at cps, the first harmonic oscillates at 2cps, the second harmonic oscillates at 3cps and so on. If the string has density ρ and tension T , then we have seen that . So to increase the frequency of oscillation of a string you increase the tension and/or decrease the density and/or shorten the string.


Example

Problem:

Example

Let l = 1, therefore,

It is very inefficient to use the integral formulae to evaluate and . It is easier to observe directly, just by matching coefficients.

Example

Separation of Variables; Vibrating String

19.2.2 - Travelling Wave Interpretation

Travelling WaveStart with the Transport Equation:

where, u(t, x) – functionc – non-zero constant (wave speed)x – spatial variable

Initial Conditions

http://www.math.umn.edu/~olver/pd_/pdw.pdf

Travelling WaveLet x represents the position of an object in a fixed coordinate frame. The characteristic equation:

Represents the object’s position relative to an observer who is uniformly moving with velocity c.

Next, replace the stationary space-time coordinates (t, x) by the moving coordinates (t, ξ).


Travelling WaveRe-express the Transport Equation:

Express the derivatives of u in terms of those of v:


Travelling Wave

Using this coordinate system allows the conversion of a wave moving with velocity c to a stationary wave. That is,


Travelling Wave

For simplicity, we assume that v(t, ξ) has an appropriate domain of definition, such that,

Therefore, the transport equation must be a function of the characteristic variable only.


The Travelling Wave Interpretation


Travelling Wave

Revisiting the transport equation,

Also recall that:


Travelling Wave

At t = 0, the wave has the initial profile

• When c > 0, the wave translates to the right.

• When c < 0, the wave translates to the left.

• While c = 0 corresponds to a stationary wave form that

remains fixed at its original location.


Travelling WaveAs it only depends on the characteristic variable, every solution to the transport equation is constant on the characteristic lines of slope c, that is:

where k is an arbitrary constant. At any given time t, the value of the solution at position x only depends on its original value on the characteristic line passing through (t, x).


Travelling Wave


19.3 Separation of VariablesVibrating Membranes

• Let us consider the motion of a stretched membrane

• This is the two dimensional analog of the vibrating string problem

• To solve this problem we have to make some assumptions

Physical Assumptions

1. The mass of the membrane per unit area is constant. The membrane is perfectly flexible and offers no resistance to bending

2. The membrane is stretched and then fixed along its entire boundary in the xy plane. The tension per unit length T is the same at all points and does not change

3. The deflection u(x,y,t) of the membrane during the motion is small compared to the size of the membrane

Vibrating MembraneRef: Advanced Engineering Mathematics, 8th Edition, Erwin

Kreyszig

Derivation of differential equation

We consider the forces acting on the membraneTension T is force per unit lengthFor a small portion ∆x, ∆y forces are approximately T∆x and T∆yNeglecting horizontal motion we have vertical components on right and left side as T ∆y sin β and -T ∆y sin αHence resultant is T∆y(sin β – sin α)As angles are small sin can be replaced with tangents Fres = T∆y(tan β – tan α)

Fres = TΔy[ux(x+ Δx,y1)-ux(x,y2)]Similarly Fres on other two sides is given byFres = TΔx[uy(x1, y+ Δy)-uy(x2,y)]

Using Newtons Second Law we get

Which gives us the wave equation:

…..(1)

yxuyyxuxTyxuyxxuyTtuyx yyxx ,,,, 21212

2

2

2

2

22

2

2

yu

xuc

tu

Vibrating Membrane: Use of double Fourier series

• The two-dimensional wave equation satisfies the boundary condition(2) u = 0 for all t ≥ 0 (on the boundary of membrane)

• And the two initial conditions(3) u(x,y,0) = f(x,y) (given initial displacement f(x,y)And (4) ),(

0

yxgtu

t

Separation of Variables

• Letu(x,y,t) = F(x,y)G(t) …..(5)

• Using this in the wave equation we have

• Separating variables we get

GFGFcGF yyxx

2

22

1

yyxx FFFGc

G

• This gives two equations: for the time function G(t) we have

…..(6)And for the Amplitude function F(x,y) we have

…..(7)which is known as the Helmholtz equation

• Separation of Helmholtz equation:F(x,y) = H(x)Q(y) …..(8)

• Substituting this into (7) gives

02

GG

02 FFF yyxx

HQ

dyQdHQ

dxHd 2

2

2

2

2

• Separating variables

• Giving two ODE’s(9)

And (10)

where

222

2

2

2 11 kQdyQd

QdxHd

H

022

2

HkdxHd

022

2

QpdyQd

222 kp

Satisfying boundary conditions

• The general solution of (9) and (10) areH(x) = Acos(kx)+Bsin(kx) and Q(y) = Ccos(py)

+Dsin(py)Using boundary condition we get H(0) = H(a) = Q(0) = Q(b) = 0which in turn givesA = 0; k = mπ/a; C = 0; p = nπ/bm,n Ε integer

• We thus obtain the solutionHm(x) = sin (mπx/a) and Qn(y) = sin(nπy/b)

• Hence the functions(11)Fmn(x) = Hm(x)Qn(y) = sin(mπx/a)sin (nπy/b)

Turning to time functionAs p2 = ν2-k2 and λ=cν we haveλ = c(k2+p2)1/2

Hence λmn = cπ(m2/a2+n2/b2)1/2 …..(12)Therefore

…(13) byn

axmtBtBtyxu mnmnmnmnmn

sinsinsincos,, *

Solution of the Entire Problem:Double Fourier Series

…..(14)

Using (3)

1 1

*

1 1

sinsinsincos

),,(),,(

m nmnmnmnmn

m nmn

byn

axmtBtB

tyxutyxu

1 1

)15(),(sinsin0,,m n

mn yxfbyn

axmByxu

• Using Fourier analysis we get the generalized Euler formula

And using (4) we obtain

)16(sinsin),(4

0 0

b a

mn dxdybyn

axmyxf

abB

)17(sinsin),(4

0 0

* b a

mnmn dxdy

byn

axmyxg

abB

Example

• Vibrations of a rectangular membraneFind the vibrations of a rectangular membrane of sides a = 4 ft and b = 2 ft if the Tension T is 12.5 lb/ft, the density is 2.5 slugs/ft2, the initial velocity is zero and the initial displacement is

ft241.0, 22 yyxxyxf

Solution]sec/ft[55.2/5.12/ 222 Tc

0 y)g(x, as0* mnB

oddnmnm

evennm

yxynxmyyxxBmn

,426.0,0

dd2

sin4

sin241.024

4

33

2

0

4

0

22

Which gives

nm odd

ynxmtnmnm

tyxu,

2233 2

sin4

sin44

5cos1426.0,,

Ref: Advanced Engineering Mathematics, 8th Edition, Erwin Kreyszig

19.4 Vibrating String Solutions19.4.1 d’Alembert’s Solution

• Solution for the wave equation

can be obtained by transforming (1) by introducing independent variables

)1(2

22

2

2

xuc

tu

)2(, ctxzctxv

• u becomes a function of v and z.• The derivatives in (1) can be expressed as

derivatives with respect to v and z.

• We transform the other derivative in (1) similarly to get

zvxzxvx uuzuvuu

zzvzvv

xzzvxxzvxzvxx

uuuzuuvuuuuu

2

zzvzvvtt uuucu 22

• Inserting these two results in (1) we get

which gives

• This is called the d’Alembert’s solution of the wave equation (1)

)3(02

vzuuvz

)4()()(),((2) from

)()(

ctxctxtxu

zvu

D’Alembert’s solution satisfying initial conditions

)6(0,

)5(0,

xgxuxfxu

t

)8()(')('0,

)7()()(0,

xgxcxcxu

xfxxxu

t

Dividing (8) by c and integrating we get

)()()( where

)9()(1)()()(

000

0

0

xxxk

dssgc

xkxxx

x

• Solving (9) with (7) gives

• Replacing x by x+ct for φ and x by x-ct for ψ we get the solution

x

x

x

x

xkdssgc

xfx

xkdssgc

xfx

0

0

)(21)(

21)(

21)(

)(21)(

21)(

21)(

0

0

ctx

ctx

dssgc

ctxfctxftxu )(21

21),(

19.4.2 Solution by integral transforms

Laplace TransformSemi Infinite string

Find the displacement w(x,t) of an elastic string subject to:

(i) The string is initially at rest on the x axis(ii) For time t>0 the left end of the string is

moved by(iii)

otherwise0

2t0 ifsin)(),0(

ttftw

0for t0),(lim

txwx

Solution

• Wave equation:• With f as given and using initial conditions

• Taking the Laplace transform with respect to t

2

22

2

2

xwc

tw

0

0)0,(

0

ttwxw

2

22

0

22

2

0,xwLc

twxswwLs

twL

t

),(),( 2

2

02

2

02

2

2

2

txwLx

dttxwex

dtxwe

xwL stst

•We thus obtain

0 thus 2

2

2

2

2

222

Wcs

xWxWcWs

•Which gives csxcsx esBesAsxW // )()(),(

• Using initial condition

• This implies A(s) = 0 because c>0 so esx/c

increases as x increases.• So we have W(0,s) = B(s)=F(s)• So W(x,s)=F(s)e-sx/c

• Using inverse Laplace we get

0),(lim),(lim),(lim00

dttxwedttxwesxW

x

stst

xx

otherwise zero and

2 ifsin),(

cxt

cx

cxttxw

Travelling wave solutionRef: Advanced Engineering Mathematics, 8th Edition, Erwin

Kreyszig

References

• H. Brezis. Functional Analysis, Sobolev Spaces and Partial Differential Equations. 1st Edition., 2011, XIV, 600 p. 9 illus. 10.3

• R. Baber. The Language of Mathematics: Utilizing Math in Practice. Appendix F

• Poromechanics III - Biot Centennial (1905-2005)• http://www.math.ubc.ca/~feldman/m267/

separation.pdf• http://logosfoundation.org/kursus/wave.pdf• http://www.math.umn.edu/~olver/pd_/pdw.pdf

• Advanced engineering mathematics, 2nd edition, M. D. Greenberg

• Advanced engineering mathematics, 8th edition, E. Kreyszig

• Partial differential equations in Mechanics, 1st edition, A.P.S. Selvadurai

• Partial differential equations, Graduate studies in mathematics, Volume 19, L. C. Evans

• Advanced engineering mathematics, 2nd edition, A.C. Bajpai, L.R. Mustoe, D. Walker

group 2 bhadouria , arjun singh glave , theodore dean han, zhe

Documents

imposition of boundaries

introduction http

factorized solutions

initial condition http

time t x

x axis

imposition of boundarieshttp

initial conditions