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
– 𝑐2𝛻2𝑢 = 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 + 𝛥𝑢2
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 = 𝑢𝑡𝑡
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 = 𝑢𝑡𝑡
(1)
Boundry Conditions:
u(0, t) = 0 for all t > 0
u(l, t) = 0 for all t > 0
(2)
(3)
Initial Conditions
u(x, 0) = f(x)
ut(x, 0) = g(x)
for all 0 < x <l
for all 0 < x <l
(4)
(5)
http://logosfoundation.org/kursus/wave.pdf
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 𝑢(𝑥, 𝑡) =
𝑋(𝑥)𝑇(𝑡)for some function 𝑋(𝑥) that depends on x but not t and
some function 𝑇 (𝑡) 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
𝑋′′(𝑥)
1 𝑇′′(𝑡)
(𝑡) ⇐⇒
= 2
𝑋(𝑥)
𝑐 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
1 𝑇′′(𝑡)
= 𝜎 2
= 𝜎
𝑋(𝑥)
𝑐 𝑇(𝑡)
⇐⇒
𝑋 ′′ 𝑥 − 𝜎𝑋 𝑥 = 0
𝑇 ′′ 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 𝑑1 , 𝑑2 , 𝑑3 , 𝑑4
For arbitrary 𝑑1 , 𝑑2 , 𝑑3 , 𝑑4
http://logosfoundation.org/kursus/wave.pdf
Step 2 – Imposition of Boundaries
𝑋(0) = 𝑋(𝑙) = 0
For 𝜎 = 0
𝑑1 = 𝑑2 = 𝑋(𝑥) = 0
Thus, this solution is discarded.
http://logosfoundation.org/kursus/wave.pdf
Step 2 – Imposition of Boundaries
𝑋(0) = 𝑋(𝑙) = 0
For 𝜎 ≠ 0, when 𝑋 0
𝑑1 = −𝑑2 = 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,
𝛼𝑘 = 2𝚤𝑑1 𝑑3 + 𝑑4 and 𝛽𝑘 = −2𝑑1 (𝑑3 − 𝑑4 )
𝑑1 , 𝑑3 , 𝑑4 are allowed to be any complex numbers
𝛼𝑘 and 𝛽𝑘 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 𝛼𝑘 and 𝛽𝑘
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<x<l, has a unique representation based on its Fourier Series:
(7)
Which can also be written as:
http://logosfoundation.org/kursus/wave.pdf
Step 3 – Imposition of the Initial Condition
For the coefficients. We can make (7) match (4′) by
choosing ℎ(𝑥) = 𝑓(𝑥) and 𝑏𝑘 = 𝛼𝑘 .
Thus 𝛼𝑘 =
2
𝑙
𝑙
𝑘𝜋𝑥
𝑓(𝑥) sin
0
𝑙
𝑑𝑥.
Similarly, we can make (7) match (5′) by choosing
𝑐𝑘𝜋
ℎ(𝑥) = 𝑔(𝑥) and 𝑏𝑘 = 𝛽𝑘
𝑙
Thus
𝑐𝑘𝜋
𝑙
𝛽𝑘 =
2 𝑙
𝑔(𝑥)
𝑙 0
𝑘𝜋𝑥
sin
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
𝑘𝜋
is just a constant times sin( 𝑥) . As x runs from 0 to l,
𝑙
𝑘𝜋
the argument of sin( 𝑥) runs from 0 to 𝑘𝜋, which is k
𝑙
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
𝑘𝑐𝜋
cos( 𝑡)
𝑙
𝑘𝑐𝜋
sin( 𝑡).
𝑘𝑐𝜋𝑙
is just a constant times
plus a constant times
As t
𝑘𝑐𝜋
increases by one second, the argument,
𝑡, of both cos( 𝑡) and
𝑙
𝑙
𝑘𝑐𝜋
𝑘𝑐
𝑘𝑐𝜋
sin( 𝑡) increases by 𝑙 , which is cycles (i.e. periods). So the
𝑙
2𝑙
𝑐
𝑐
fundamental oscillates at cps, the first harmonic oscillates at 2 cps,
2𝑙
2𝑙
𝑐
the second harmonic oscillates at 3 cps and so on. If the string has
2𝑙
𝑇
density ρ and tension T , then we have seen that 𝑐 =
. So to
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
Example
Problem:
Example
Let l = 1, therefore,
It is very inefficient to use the integral formulae
to evaluate 𝛼𝑘 and 𝛽𝑘 . 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, 8th Edition, Erwin
Kreyszig
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
Neglecting horizontal motion
 we have vertical
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
Fres = T∆y(tan β – tan α)

Fres = TΔy[ux(x+ Δx,y1)-ux(x,y2)]
Similarly Fres on other two sides is given by
Fres = TΔx[uy(x1, y+ Δy)-uy(x2,y)]
Using Newtons Second Law we get


 2u
xy 2  Tyu x x  x, y1   u x x, y2   Tx u y x1 , y  y   u y x2 , y 
t
Which gives us the wave equation:
2
2

 2u

u

u
2


c

2
 x 2
t 2

y





…..(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) u  g ( x, y)
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


F G  c 2 FxxG  FyyG

• Separating variables we get

G
1
2



F

F



xx
yy
2
cG F
• This gives two equations: for the time function
G(t) we have 
2
…..(6)
G  G  0
And for the Amplitude function F(x,y) we have
…..(7)
Fxx  Fyy  2 F  0
which is known as the Helmholtz equation
• Separation of Helmholtz equation:
F(x,y) = H(x)Q(y)
• Substituting this into (7) gives
 d 2Q

d 2H
2
Q   H 2  HQ 
2
dx
 dy

…..(8)
• Separating variables
1 d 2H
1  d 2Q
2 
2






Q


k
2
2


H dx
Q  dy

• Giving two ODE’s
2
d
H
(9)
 k 2H  0
dx2
And (10)
d 2Q
2

p
Q0
2
dy
where
p   k
2
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
Hm(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 function
As p2 = ν2-k2 and λ=cν we have
λ = c(k2+p2)1/2
Hence λmn = cπ(m2/a2+n2/b2)1/2
Therefore
…..(12)
mx
ny
umn x, y, t   Bmn cos mn t  B sin mn t sin
sin
…(13)
a
b

*
mn

Solution of the Entire Problem:
Double Fourier Series


…..(14)
u ( x, y, t )   umn ( x, y, t )
m 1 n 1



  Bmn cosmn t  B
*
mn
m 1 n 1
mx
ny
sin mn t sin
sin
a
b

Using (3)
mx
ny
ux, y,0   Bmn sin
sin
 f ( x, y) (15)
a
b
m 1 n 1


• Using Fourier analysis we get the generalized
Euler formula
4
mx
ny

f ( x, y ) sin
sin
dxdy (16)


ab 0 0
a
b
b a
Bmn
And using (4) we obtain
*
mn
B
4

abmn
mx
ny
0 0 g ( x, y) sin a sin b dxdy (17)
b a
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/ft2, the
initial velocity is zero and the initial
displacement is


f x, y   0.1 4x  x 2 2 y  y 2

ft
Solution
c  T /   12.5 / 2.5  5 [ft / sec ]
2
2
2
*
Bmn
 0 as g(x, y)  0
4
mx
ny
2
2

0.1 4 x  x 2 y  y sin
sin
dxdy


42 0 0
4
2
2 4
Bmn

0 m, n even
 0.426
m, n odd
3 3
mn


Which gives
1
 5
ux, y, t   0.426 3 3 cos
 4
m, n odd m n
mx
ny

m  4n t sin
sin
4
2

2
2
Ref: Advanced Engineering Mathematics, 8th Edition,
Erwin Kreyszig
19.4 Vibrating String Solutions
19.4.1 d’Alembert’s Solution
• Solution for the wave equation
2
 2u

u
2
c
2
t
x 2
(1)
can be obtained by transforming (1) by
introducing independent variables
v  x  ct,
z  x  ct (2)
• u becomes a function of v and z.
• The derivatives in (1) can be expressed as
derivatives with respect to v and z.
ux  uvvx  uz z x  uv  uz
u xx  uv  u z x  uv  u z x vx  uv  u z z z x
 uvv  2uvz  u zz
• We transform the other derivative in (1)
similarly to get
utt  c uvv  2uvz  uzz 
2
• Inserting these two results in (1) we get
 2u
uvz 
0
zv
 (3)
which gives
u   (v )   ( z )
from(2)
u ( x, t )   ( x  ct)  ( x  ct)
(4)
• This is called the d’Alembert’s solution of the
wave equation (1)
D’Alembert’s solution satisfying initial
conditions
ux,0  f x 
ut x,0  g x 
(5)
(6)
ux,0   ( x)  ( x)  f x 
(7)
ut x,0  c ' ( x)  c ' ( x)  g x  (8)
Dividing (8) by c and integrating we get
x
1
 ( x)  ( x)  k ( x0 )   g ( s) ds  (9)
c x0
where k ( x0 )   ( x0 )  ( x0 )
• Solving (9) with (7) gives
x
1
1
1
 ( x)  f ( x)   g ( s) ds  k ( x0 )
2
2c x0
2
x
1
1
1
 ( x)  f ( x)   g ( s) ds  k ( x0 )
2
2c x0
2
• Replacing x by x+ct for φ and x by x-ct for ψ
we get the solution
x  ct
1
1
u ( x, t )   f x  ct   f x  ct  
g ( s) ds

2
2c x ct
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
sin t if 0  t  2
moved by w(0, t )  f (t )  
 0 otherwise
w( x, t )  0 for t  0
(iii) lim
x 
Solution
2
2w
2  w
c
2
t
x 2
• Wave equation:
• With f as given and using initial conditions
w( x,0)  0
w
0
t t 0
• Taking the Laplace transform with respect to t
2
2w 2

w

w
2
L 2   s Lw swx,0 
 c L 2 
t t 0
 t 
 x 
2 
2
  2 w    st  2 w


 st
L 2    e
dt

e
w( x, t )dt  2 Lw( x, t )
2
2 
x
x 0
x
 x  0
•We thus obtain
2

W
2
2
s W c
x 2
 2W s 2
thus
 2W 0
2
x
c
•Which gives
W ( x, s)  A(s)esx / c  B(s)e sx / c
• Using initial condition


limW ( x, s)  lim  e  st w( x, t )dt   e  st lim w( x, t )dt  0
x 
x 
0
0
x 
• 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
x
x
 x
w( x, t )  sin  t   if  t   2
c
c
 c
and zero otherwise
Travelling wave solution
Ref: 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/separa
tion.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