Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe

Transcript Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe

Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe

Chapter 5. Laplace Transform Chapter 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 – – 𝑐 2 𝛻 2 𝑢 = 𝑢𝑡𝑡 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 t θ(x, t) = angle between the string and a horizontal line at position x and time t T(x, t) = tension in the string at position x and time t ρ(x) = mass density of the string at position x http://www.math.ubc.ca/~feldman/m267/separation.pdf

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: • 𝜌(𝑥) 𝛥𝑥 2 + 𝛥𝑢 2 http://www.math.ubc.ca/~feldman/m267/separation.pdf

Derivation

Vertical Component of Motion Divide by Δx and taking the limit as Δx → 0. http://www.math.ubc.ca/~feldman/m267/separation.pdf

Derivation

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

Derivation

For small vibrations: Therefore, http://www.math.ubc.ca/~feldman/m267/separation.pdf

Derivation

Substitute into (2) into (1) http://www.math.ubc.ca/~feldman/m267/separation.pdf

Derivation

Horizontal Component of the Motion Divide by Δx and taking the limit as Δx → 0. http://www.math.ubc.ca/~feldman/m267/separation.pdf

Derivation

• For small vibrations: 𝑐𝑜𝑠θ~1 and 𝑑𝑇/𝑑𝑥(𝑥, 𝑡) ~ 0 Therefore, http://www.math.ubc.ca/~feldman/m267/separation.pdf

Solution

For a constant string density ρ, independent of x The string tension T(t) is a constant, and No external forces, F 𝑐 2 𝛻 2 𝑢 = 𝑢 𝑡𝑡 𝑐 = √(𝑇/𝜌) http://www.math.ubc.ca/~feldman/m267/separation.pdf

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 t l = string length Recall: 𝑐 2 𝛻 2 𝑢 = 𝑢 𝑡𝑡 Boundry Conditions: u(0, t) = 0 for all t > 0 u(l, t) = 0 for all t > 0 Initial Conditions u(x, 0) = f(x) ut(x, 0) = g(x) for all 0 < x

Procedure

There are three steps to consider in order to solve this problem: • Step 1: Find all solutions of (1) that are of the special form 𝑋(𝑥)𝑇(𝑡) for some function 𝑋(𝑥) some function 𝑇 (𝑡) 𝑢(𝑥, 𝑡) = that depends on x but not t and 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).

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

Step 1 – Finding Factorized Solutions

𝑢(𝑥, 𝑡) = 𝑋(𝑥)𝑇(𝑡) Let: 𝑋(𝑥)𝑇 ′′(𝑡) = 𝑐 2 𝑋′′(𝑥)𝑇 (𝑡) ⇐⇒ 𝑋′′(𝑥) 𝑋(𝑥) = 1 𝑐 2 𝑇′′(𝑡) 𝑇(𝑡) 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).

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

Step 1 – Finding Factorized Solutions

𝑋 ′′ 𝑥 𝑋(𝑥) 1 = 𝜎 𝑐 2 𝑇′′(𝑡) 𝑇(𝑡) = 𝜎 ⇐⇒ 𝑋 ′′ 𝑇 ′′ 𝑥 − 𝜎𝑋 𝑥 = 0 𝑡 − 𝑐 2 𝜎𝑇 𝑡 = 0 (6) http://logosfoundation.org/kursus/wave.pdf

Step 1 – Finding Factorized Solutions

Solve the differential equations in (6) 𝑋 𝑥 = 𝑒 𝑟𝑥 𝑇(𝑡) = 𝑒 𝑠𝑡 http://logosfoundation.org/kursus/wave.pdf

Step 1 – Finding Factorized Solutions

If 𝜎 ≠ 0 , there are two independent solutions for (6) If 𝜎 = 0 , http://logosfoundation.org/kursus/wave.pdf

Step 1 – Finding Factorized Solutions

Solutions to the Wave Equation For arbitrary 𝜎 ≠ 0 and arbitrary 𝑑 1 , 𝑑 2 , 𝑑 3 , 𝑑 4 For arbitrary 𝑑 1 , 𝑑 2 , 𝑑 3 , 𝑑 4 http://logosfoundation.org/kursus/wave.pdf

Step 2 – Imposition of Boundaries

𝑋(0) = 𝑋(𝑙) = 0 For 𝜎 = 0 𝑑 1 = 𝑑 2 = 𝑋(𝑥) = 0 Thus, this solution is discarded. http://logosfoundation.org/kursus/wave.pdf

Step 2 – Imposition of Boundaries

For 𝜎 ≠ 0, when 𝑋 0 𝑋(0) = 𝑋(𝑙) = 0 𝑑 1 = −𝑑 2 = 0 When 𝑋(𝑙) = 0 Therefore, http://logosfoundation.org/kursus/wave.pdf

Step 2 – Imposition of Boundaries

Since 𝜎 ≠ 0, in order to satisfy 𝑒 2√𝜎𝑙 = 1 An integer k must be introduced such that: Therefore, http://logosfoundation.org/kursus/wave.pdf

Step 2 – Imposition of Boundaries

Where, 𝛼 𝑘 = 2𝚤𝑑 1 𝑑 3 + 𝑑 4 and 𝛽 𝑘 = −2𝑑 1 (𝑑 3 − 𝑑 4 ) 𝑑 1 , 𝑑 3 , 𝑑 4 are allowed to be any complex numbers 𝛼 𝑘 and 𝛽 𝑘 are allowed to be any complex numbers http://logosfoundation.org/kursus/wave.pdf

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 𝛼 𝑘 and 𝛽 𝑘 http://logosfoundation.org/kursus/wave.pdf

Step 3 – Imposition of the Initial Condition

The previous expression must also satisfy the initial conditions (4) and (5): (4 ’ ) (5 ’ ) http://logosfoundation.org/kursus/wave.pdf

Step 3 – Imposition of the Initial Condition

For any (reasonably smooth) function, h(x) defined on the interval 0

Step 3 – Imposition of the Initial Condition

For the coefficients. We can make (7) match (4′) by choosing ℎ(𝑥) = 𝑓(𝑥) and 𝑏 𝑘 = 𝛼 𝑘 . Thus 𝛼 𝑘 = 2 𝑙 0 𝑙 𝑓(𝑥) sin 𝑘𝜋𝑥 𝑙 𝑑𝑥 . Similarly, we can make (7) match (5′) by choosing 𝑐𝑘𝜋 ℎ(𝑥) = 𝑔(𝑥) and 𝑏 𝑘 = 𝛽 𝑘 𝑙 Thus 𝑐𝑘𝜋 𝑙 𝛽 𝑘 = 𝑙 2 0 𝑙 𝑔(𝑥) sin 𝑘𝜋𝑥 𝑙 𝑑𝑥 http://logosfoundation.org/kursus/wave.pdf

Step 3 – Imposition of the Initial Condition

Therefore, (8) Where, http://logosfoundation.org/kursus/wave.pdf

Step 3 – Imposition of the Initial Condition

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 sin( 𝑘𝜋 𝑙 the argument of sin( 𝑘𝜋 𝑙 𝑥) 𝑥) . As x runs from 0 to l, runs from 0 to 𝑘𝜋 , 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.

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

Step 3 – Imposition of the Initial Condition

The first 3 modes at fixed t’s.

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

Step 3 – Imposition of the Initial Condition

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

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

Problem:

Example

Let l = 1, therefore,

Example

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

Example

Separation of Variables; Vibrating String

19.2.2 - Travelling Wave Interpretation

Travelling Wave

Start with the Transport Equation: where, u(t, x) – function c – non-zero constant (wave speed) x – spatial variable Initial Conditions http://www.math.umn.edu/~olver/pd_/pdw.pdf

Travelling Wave

Let 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).

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

Travelling Wave

Re-express the Transport Equation: Express the derivatives of u in terms of those of v: http://www.math.umn.edu/~olver/pd_/pdw.pdf

Travelling Wave

Using this coordinate system allows the conversion of a wave moving with velocity c to a stationary wave. That is, http://www.math.umn.edu/~olver/pd_/pdw.pdf

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

The Travelling Wave Interpretation

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

Travelling Wave

Revisiting the transport equation, Also recall that: http://www.math.umn.edu/~olver/pd_/pdw.pdf

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.

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

Travelling Wave

As 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).

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

Travelling Wave

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

19.3 Separation of Variables Vibrating 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 Membrane

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

Derivation of differential equation

We consider the forces acting on the membrane Tension T is force per unit length For a small portion ∆x, ∆y forces are approximately T∆x and T∆y



components on right and left side as T ∆y sin β and -T ∆y sin α Hence resultant is T∆y(sin β – sin α) As angles are small sin can be replaced with tangents  F res = T∆y(tan β – tan α)

F res = TΔy[u x (x+ Δx,y 1 )-u x (x,y 2 )] Similarly F res on other two sides is given by F res = TΔx[u y (x 1, y+ Δy)-u y (x 2 ,y)] Using Newtons Second Law we get  



 2



2 





u x



 

1 

2  





u y



1 ,

 

  

y x

2 ,  Which gives us the wave equation:  2



2 

2  2



2   2



2 …..(1)

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) 



t t

 0 

(

)

Separation of Variables

• • Let u(x,y,t) = F(x,y)G(t) …..(5) Using this in the wave equation we have

 



2 

F xx G



F yy G

 • Separating variables we get  

G c

 1



F xx



F yy

    2

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

   2

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

F xx



  2

 0 …..(7)

which is known as the Helmholtz equation Separation of Helmholtz equation: F(x,y) = H(x)Q(y) Substituting this into (7) gives

H dx

   

H d

Q dy

2   2

  …..(8)

• • Separating variables 1

H d

H dx

2   1

 

Q dy

2   2

   

2 Giving two ODE’s (9)

H dx

2 

 0 And (10)



 0 where

2   2 

Satisfying boundary conditions

• The general solution of (9) and (10) are H(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) = 0 which in turn gives A = 0; k = mπ/a; C = 0; p = nπ/b m,n Ε integer

• • We thus obtain the solution H m (x) = sin (mπx/a) and Q n (y) = sin(nπy/b) Hence the functions (11)F mn (x) = H m (x)Q n (y) = sin(mπx/a)sin (nπy/b) Turning to time function As p 2 = ν 2 -k 2 and λ=cν we have λ = c(k 2 +p 2 ) 1/2 Hence λ mn

u mn

Therefore 

  

B mn

= cπ(m 2 /a 2 +n 2 /b 2 ) 1/2 cos 

mn t

 *

B mn

sin 

mn t

 sin …..(12)



sin

a n



…(13)

Solution of the Entire Problem: Double Fourier Series

(

, 

)    

 1

 1 

B mn

  

 1

u mn

(

, cos 

mn t y

)  *

B mn

sin 

mn t

 sin …..(14)



sin

a n



y b

Using (3)



, 0  

    1  1

n B mn

sin



sin

a n





b f

(

)  ( 15 )

• Using Fourier analysis we get the generalized Euler formula

B mn

 4

ab b a

 0 0

(

) sin



sin

a n



y dxdy b

 ( 16 ) And using (4) we obtain *

B mn

 4



mn b a

 0 0

(

) sin



sin

a n



y dxdy b

 ( 17 )

Example

•

Vibrations of a rectangular membrane

Find 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/ft 2 , the initial velocity is zero and the initial displacement is

 0 .

1  4



2  2



2  ft

Solution

2 

/   12 .

5 / 2 .

5  5 [ ft 2 / sec 2 ] *

B mn

 0 as g(x, y)  0

B mn

 4 4  2 0 2  4 0 0 .

1  4



2  2



2  sin



sin 4



2  0 0 .

426

n even m

n odd m

Which gives



  0 .

426 

odd m

3 cos 5  4

2  4

2 

sin



sin 4



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

19.4 Vibrating String Solutions 19.4.1 d’Alembert’s Solution

• Solution for the wave equation  2



2 

2  2



2   ( 1 ) can be obtained by transforming (1) by introducing independent variables









 ( 2 )

• • • u becomes a function of v and z.

The derivatives in (1) can be expressed as derivatives with respect to v and z.

u x u xx



u vv





u v u v v



x u z



x u z

 

z x u v

 2

u vz



u zz





u u v z





v x u





u v



u z



z z x

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

u tt



2 

u vv



u vz



u zz



• Inserting these two results in (1) we get

u vz

  2



 0   ( 3 ) which gives

  (

)   (

) from (2)

(

)   (



)   (



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

D’Alembert’s solution satisfying initial conditions

u u t

     

f g

   

u u t

       (



' ) ( 

)   (

)

  ' (

 ( 5 )  ( 6 )

) 

   (  7 ) ( 8 ) Dividing (8) by c and integrating we get  (

)   (

) 

(

0 )  1

c x



(

)

where

(

0 )   (

0 )   (

0 )  ( 9 )

• • Solving (9) with (7) gives  (

)  1 2

(

)  1 2

c x

0 

x g

(

)

 1 2

(

0 )  (

)  1 2

(

)  1 2



x g

(

)

 1 2

(

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

(

)  1 2







ct x







 1 2

c x x

  

ct ct g

(

)

19.4.2 Solution by integral transforms

Laplace Transform

Semi 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 (iii) moved by

lim  

(

) 

( 0 , 0

)  for t

(

 )  0   sin 0

0   otherwise 2 

Solution

• • • Wave equation:  2



2 

2  2



2 With f as given and using initial conditions

(

, 0 )  0 

 0 

t t

 0 Taking the Laplace transform with respect to t

   

2   

  

   



t t

 0 

    2

w x

2  

    2



2      0 



 2



  2 

2  0 



st w

(

)

  2 



(

)  • We thus obtain



2 thus  2 

x W

2  2



2 

 0 • Which gives

(

) 

(

)

e sx



(

)



• Using initial condition

lim  

(

)  lim

   0 



st w

(

)

  0 



st x

lim  

(

)

 0 • • • • This implies A(s) = 0 because c>0 so e sx/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

(

)  sin

x c



 2 

and zero otherwise

Travelling wave solution

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

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/separa tion.pdf

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

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

• • • • • Advanced engineering mathematics, 2 nd M. D. Greenberg edition, Advanced engineering mathematics, 8 th E. Kreyszig edition, Partial differential equations in Mechanics, 1 st edition, A.P.S. Selvadurai Partial differential equations, Graduate studies in mathematics, Volume 19, L. C. Evans Advanced engineering mathematics, 2 nd A.C. Bajpai, L.R. Mustoe, D. Walker edition,

Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe

Transcript Group 2 Bhadouria, Arjun Singh Glave, Theodore Dean Han, Zhe