AEROACOUSTICS
-Lecture 3Using the equations of Fluid motion
to derive the wave equation
Derivation of the Wave Equation:

derived from the linearized Euler equations are:
Mass
Momentum
Temporal derivative of Mass Eq’n
Spatial derivative of Momentum Eq’n
ui '
 '
 o
0
t
xi
o
ui '  p '

0
t
xi
ui '  
  '
 o

   0
t  t
xi  t
  ui '  p'  

 0
 o

xi 
t
xi  xi
Derivation of the Wave Equation
Temporal derivative of Mass Eq’n
Spatial derivative of Momentum Eq’n
ui '  
  '



   0
o
t  t
xi  t
  ui '  p'  

 0
 o

xi 
t
xi  xi
ui ' 
 '  
  o
0
t t
t 
xi 
  ui '    p'
o

0


xi 
t  xi xi
ui ' 
 

 o
0
2
t 
xi 
t
  ui '   2 p'
o

0


xi 
t  xi xi
 2 '
“take derivative”

Simplify
Wave Equations:

derived from the linearized Euler equations are:
 2 '
t 2
ui ' 
 
  o
0
t 
xi 
  ui '   p'
o

0
xi 
t  xi xi
2
Same set of Eq’ns shown on previous page
Subtract
  2 '  
ui '    


 2
 o
   
t 
xi    xi
 t
?
ui ' 

 o
 =
t 
xi 
  ui ' 
o
xi 
t 
ui '   2 p'

 o t   x x


i i

  0  0

 ui '
 ui '

=
o
xi t
t xi
Continuous Functions
o
Wave Equations:

derived from the linearized Euler equations are:
  2 '  
ui '    
 2   o
   
t 
xi    xi
 t
 2'
t 2
ui '


 o t

2
  p'
  x x

i i

  0  0

 2 p'

0
xi xi

The above expression contains both the acoustic density and pressure; we need a
relationship between density and pressure!
need

equation with only one variable
use the equation of state (assumptions)
p  RT
 
1
c
2
p
Need to Derive This Relationship
This is where the assumption of constant speed of sound
is derived, in applications where c varies, it is important
to understand the following derivation
Pressure – density dependency p():

Process: Isentropic (Reversible, Adiabatic)
pVol   cons tan t  C1
m

m  cons tan t
Vol 
p

p


 m  * cons tan t  C2

 po  p 
 o   

 po  p 
 
o   1  
o 


 C2
Small

2
1     1           1     ...

  


o 
o
2 !  o 


Pressure – density dependency p():

Process: Isentropic (Reversible, Adiabatic)
p

Simplifications

po  p 


 o   
p



o

 
o   1  
o 


 C2
Small

2
1     1           1     ...

  


o 
o
2 !  o 


po
o
po

 po  p 
 po  p 

 
o   1    


o




 
po 1    
    po  p 

o 


 
po  po   
  po  p

o

 
po   
  p

o

p
  p
  o
o 

Pressure – density dependency p():

Process: Isentropic (Reversible, Adiabatic)
p
  p
  o
o 

po
 RTo
o
po  o RTo
RTo  p
c 2  p
or
 
1
2
p
c = speed of sound
c
This means no entropy is being produced during the wave propagation
Isentropic Wave Eq’n
 '
2
t 2

 p'
0
xi xi
2
 2 '
t 2
 2 p'
t 2
 2 '
c
0
xi xi
2
 2 p'
c
0
xi xi
2
The wave equation in
terms of acoustic density
fluctuation (’)
The wave equation
The most fundamental equation in acoustics:
the 1D version of the wave equation was first derived in 1747 by D’ Alambert for the
case of the vibrating string (never published because of his strong reservations about
the physical admissibility of its solutions)


Euler (1747-1750) and Lagrange (1759) developed the early theory of sound based
on fluid-dynamics principles

the first derivation of the wave equation in 1D for sound appeared in a paper by Euler
(1759)

it describes the properties of a sound field in space and time

it describes how these properties evolve in space and time
Observations about the wave motion:

waves are disturbances or deviations from a quiet pre-existing condition (such as a
static equilibrium). (Example: ripple on a quiet pond)

TIME plays a vital role. (Example: static displacement of a string or a rubber bond is
a disturbance, but not a wave)

a wave travels with finite speed (c)

all mechanical waves travel in a material medium. Example: sound cannot travel
through vacuum, it needs matter (fluid: air/water; solid) to propagate. There are also
waves (electromagnetic waves, light, X Rays) that require no medium to propagate.

sound waves are longitudinal waves where the disturbance moves parallel to the
direction of propagation. Waves on a string or a membrane, shear waves (in
geophysics) and electromagnetic waves are transverse (i.e. the disturbance moves
perpendicular to the direction of propagation)
We should keep in mind that this same wave equation occurs in
a variety of other contexts:

electromagnetic theory

gravity waves in shallow waters

dilatational and shear elastic waves in solids

Magnetohydrodynamics

pressure surges in liquid-filled tubes (blood vessels)
The general form of the wave equation:
2
 2u
2  u
c
0
t 2
xi xi
or
2
 2u
 2u  2u 
2  u
 c  2  2  2   0
2
t
y
z 
 x
where “c” is the speed at which the wave travels
and “u” a physical property associated with the disturbance.
For sound waves “u” is an acoustic pressure or an acoustic density.
The general form of the wave equation:
Vector
 2 p
Tensor
 p
2
t
2
 c2
 p
0
xi xi
t 2
2
or
 c 22 p  0
 2
2
2
  2  2  2
 x
y
z

2



The general form of the wave equation:
  2 p 
c 
0
 x 2 


1D
 2 p 

0
2 

y 
2D
  2 p  2 p  2 p 
c 


0
2
2 
 x 2

y
z 

3D
 2 p
t
 2 p
t
 2 p
t
2
2
2
2
2
2
 2 p
c 
 x 2

Mathematical classification of the wave equation:

2nd order Partial Differential Equation (PDE)

it represents a Marching or a Propagation problem

it is a transient-like problem where the solution at the initial condition and a set of
Boundary Conditions are required in order to solve the problem

it is a hyperbolic in character equation since admits wave like solutions
Classification of the PDEs:

it’s based on the mathematical concept of characteristics

hyperbolic, parabolic, and elliptic

a hyperbolic PDE is characterized by that two distinct families of real characteristic
curves exist.
D’Alembert’s solution of the 1D wave equation
The general form of the 3D wave equation is:
2
 2u
 2u  2u 
2  u
 c  2  2  2   0
2
t
y
z 
 x
equivalent with
2
 2u
2  u
c
0
2
t
xi xi
where “c” is the speed at which the wave travels
and “u” a physical property associated with the disturbance.
For sound waves “u” is an acoustic pressure or an acoustic density.
Want a solution and ideally some kind of similarity solution
D’Alembert’s solution of the 1D wave equation
Let’s consider the 1D wave equation:

it is obtained by dropping the y and z dependence in the 3D wave equation

it is called also the equation for plane waves
2
 2u
2  u
2

c

u

c
u xx
tt
2
2
t
x
It is unusual for a PDE in that a very simple general solution may be found:

in general, waves have sinusoidal shape.

Therefore, let’s assume that u = A sin (a∙x) represents a solution of the wave equation.

However, a solution of the wave equation has to be a function of both time (t) and
distance (x).
Since the simplest combination between “x” and “t” is a linear combination, let’s
assume that our solution is:

u  A  sin( ax  bt )
Solutions of the wave equation
It can be easily proved that for b = ±ac:
u  A  sin( ax  bt ) is a solution of the wave equation.
This implies:
u  A  sin a ( x  ct ) is a solution of the wave equation.
It can be also demonstrated that:
u  A  cos a( x  ct )
u  A  e  j a ( x ct )
(a complex exponential function)
u  A  ln a( x  ct ) (a non-oscillatory function)
u  A  ( x  ct ) n
(a simple algebraic term)
are all solutions of the wave equation!!!
Moreover, due to its linearity, the summation of the solutions of the wave equation
is a solution of the wave equation :

We are led to believe that the solutions of the wave equations are of this form:
u  f ( x  ct )
u  g ( x  ct )
The method of D’Alembert:

provides a solution to the 1D wave equation by introducing the new variables:
  ( x  ct )
  ( x  ct )
Thus: u ( x, t )  u ( , )  u ( ( x, t ); ( x, t ))
2
 2u
2  u
2

c

u

c
u xx
Since the wave equation is:
tt
2
2
t
x
We need expressions for the second order derivatives!
Using the chain rule one can have:

 
 




x  x  x
but:
   x  ct  x   ct 



 1 0  1
x
x
x
x
   x  ct  x   ct 



 1 0  1
x
x
x
x





x  
Derivation of 2nd Derivatives (Good Exercise)

 
 




x  x  x





x  
             









x x   x  x    x  x 
                                 

















x x   x    x    x    x    x    x    x    x 
Previously Shown
   x  ct 

1
x
x
   x  ct 

1
x
x
                     

1 
1  
1 
1  
1 
1  
 1 
 1
x x                        
2
2
 
 
2
 2


2




x

2
In a similar fashion, using the chain rule one can obtain the first time derivative:

 
 




t  t  t
   x  ct  x   ct 



 0  c  c
t
t
t
t
   x  ct  x   ct 



 0  c  c
t
t
t
t
The second derivatives are obtained from:
 2u
  u 
 2u
  u 


and




x 2 x  x 
t 2
t  t 



 c
c
t


Derivation of 2nd Time Derivative:

   




t  t  t



 c
c
t


             









t t   t  t    t  t 
                                 

















t t   t    t    t    t    t    t    t    t 
Previously Shown
   x  ct 

 c
t
t
   x  ct 

c
t
t
 
  
 
  
 
  
 

   

  c   
  c    
  c   
c  
c
  c    
 c
c
t t  
  
  
        
      
2
2
 
2  
2 
c
c
c
c
2
2




t

2
2
2
2
D’Alembert’s Eq’n
 2u
 2u
t
2
 c2
 2u
x
2
0
t 2
 2u
2
 2u
2
2
 2u
2  u
c
c
 2c
c
2
2

x

2
2
 2u
2
 2u
2  u
c
 2c
c

2
2
2
2
Plugging in the second derivatives of “u” in space and time into the 1D wave
equation results:
2
2
 2  2u
2  u
2  u
 2c
c
 c
2

2
 
2
2
  2  2u
2  u
2  u
 2c
c
   c
2

2
  

  0

 2u
4c
0

2
 2u
 0  u  0
  
Equation that is simple enough to be solved by direct integration!!!
Finding D’Alembert’s solution

integrating the previously obtained relation with respect to ξ gives:
u
 g ( )


, where g(η) is an arbitrary function of η
then, integrating with respect to η gives:
u ( , )  F ( )  G ( ) , where F(ξ) is an arbitrary function of ξ and
G ( )   g ( )d
Replacing ξ and η by their expressions in terms of x and t,
  ( x  ct )
  ( x  ct )
one can obtain D’Alembert solution of the wave equation:
u ( x, t )  F ( x  ct )  G ( x  ct )
D’Alembert’s solution of the wave equation
u ( x, t )  F ( x  ct )  G ( x  ct )
The particular forms of F and G are determined from the initial data (at t=0):
 (c )
u ( x,0)  f ( x)
u ( x,0)  f ( x)  F ( x)  G ( x)
u
( x ,0 )  g ( x )
t
u
( x,0)  ut ( x,0)  g ( x)  c  F ' ( x)  c  G ' ( x)
t
By multiplying the first expression with “c” and integrating the second one, we have:
c  f ( x)  c  F ( x)  c  G ( x)
 g ( s)ds  c  F ( x)  c  G ( x)

Adding these two expressions from above, gives G(x):
G ( x) 
x
1
1
 f ( x) 
  g ( s )ds
2
2c 0
But: F(x)=f(x)-G(x), therefore:
x
1
1
F ( x)   f ( x) 
 g ( s )ds
2
2  c 0
D’Alembert’s solution of the wave equation
u ( x, t )  F ( x  ct )  G ( x  ct )
With:
x
1
1
F ( x)   f ( x) 
 g ( s )ds
2
2  c 0
x
1
1
G ( x)   f ( x) 
 g ( s )ds
2
2  c 0
The solution of the wave equation with specified initial conditions becomes:
u ( x, t ) 
x  ct
x  ct
1
1
1
1
 f ( x  ct ) 

g ( s ) ds   f ( x  ct ) 

g ( s )ds
0
0
2
2c
2
2c
That’s equivalent with:
u ( x, t ) 
x  ct
f ( x  ct )  f ( x  ct )
1

  g ( s )ds
2
2  c x ct
Characteristics Diagram; Wave equation’s characteristics
The characteristics, if they exist and are real curves
within the solution domain, are related to the
directions in which the “information” can be
transmitted and can be:

lines in 2D

surfaces in 3D
2
 2u
2  u
c
0
2
t
xi xi
 2 ( x  xmax / 2) 
u ( x,0)  sin 
, if x0  x  x1
x1  x0


u ( x,0)  0
, otherwise
y
z
x

the initial conditions

the boundary conditions
u
To identify the characteristics one have to specify
in terms of “u” and its first derivatives:
x
Mihaescu M., “Computational Aeroacoustics based on Large Eddy
Simulation and Acoustic Analogies”, ISBN 91-628-6443-2, 2005
Forward propagating waveform
Let’s consider the function: u  A( x  ct ), where0  x  ct  1
For simplicity let the amplitude “A” being A = 1. Then, the function “u” becomes:
u  x  ct , where0  x  ct  1
ct  x  1  ct
For x = ct results that u = 0
For x = 1 + ct results that u = 1
Let’s represent “u” versus “x” at three different time instances: t = 0, τ, and 2τ
u
for t = 2τ → u = x - 2c τ, with 2cτ < x <1+ 2cτ;
1
for x = 2cτ → u = 0;
t=2τ
0
2cτ
1 + 2cτ
x
for x = 1+ 2cτ → u = 1
u
FORWARD propagating wave
(Outgoing wave)
1
for t = τ → u = x - c τ, with cτ < x <1+ cτ;
for x = cτ → u = 0;
t=τ
0
cτ
u
1 + cτ
x
for x = 1+ cτ → u = 1
for t = 0 → u = x, with 0 < x <1;
1
for x = 0 → u = 0;
t=0
0
1
for x = 1 → u = 1
x
The wave is steadily translated FORWARD, unchanged in shape, along
the x-axis with the speed “c”
Backward propagating waveform
Let’s consider the function: u  A( x  ct ), where0  x  ct  1
Again, for simplicity let the amplitude “A” being A = 1. Then, the function “u”
becomes: u  x  ct , where0  x  ct  1
for x = - ct results that u = 0
 ct  x  1  ct
for x = 1 - ct results that u = 1
Let’s represent “u” versus “x” at three different time instances: t = 0, τ, and 2τ
u
for t = 2τ → u = x + 2c τ, with -2cτ < x <1- 2cτ;
1
for x = -2cτ → u = 0;
t=2τ
-2cτ
1 - 2cτ
0
BACKWARD propagating wave
(Incoming wave)
x
u
for t = τ → u = x + c τ, with -cτ < x <1 - cτ;
1
for x = -cτ → u = 0;
t=τ
-cτ
1 - cτ
x
u
for x = 1- cτ → u = 1
for t = 0 → u = x, with 0 < x <1;
1
for x = 0 → u = 0;
t=0
0
for x = 1- 2cτ → u = 1
1
for x = 1 → u = 1
x
The wave is steadily translated BACKWARD, unchanged in shape, along
the x-axis with the speed “c”
Propagating waves
For different time instances a wave is translated unchanged with a speed “c” along
“x” axis (in space). One can have:

FORWARD propagating (Outgoing) waves for u = x – ct

BACKWARD propagating (Incoming) waves for u = x + ct
For an outgoing wave, u = x – ct defines a phase front of the wave:

x – ct = 1, signifies the head of the wave

x – ct = 0, signifies the tail of the wave

x – ct = 1/2, signifies the mid-point of the wave
For an incoming wave, u = x + ct defines a phase front of the wave:

x + ct = 1, signifies the tail of the wave

x + ct = 0, signifies the head of the wave

x + ct = 1/2, signifies the mid-point of the wave
Notice that the limitation 0 < x ± ct < 1 is a very important part of the description
of the wave. Without it, the disturbance would have no beginning or end; it would
stretch from - to +.
Characteristics Diagram
A simple way to keep track of waves is to use a Characteristics Diagram, which is a
graphical display in which the axes are the independent variables time (t) and
space (x):

using the forward and backward propagating waves, by drawing a line connecting the
tails in the in the sketches, we obtained the path followed by the tail in the (x, t) plane

the paths along which various phase fronts travel are called characteristics
u = x – ct
u = x + ct
t
BACKWARD propagating wave
(Incoming wave)
FORWARD propagating wave
(Outgoing wave)
Left running characteristic
Right running characteristic
0
½
1
x
Characteristics for the wave equation
Properties of the characteristics:

their slope is the inverse of the propagation speed (1/c)

each characteristic is associated with a specific value of the dependent variable (u)
In a sense the characteristics diagram contains the entire solution of the problem in graphical form. Let’s
consider the characteristics x - ct and x + ct and the solution of the Wave Equation for the point where the
characteristics intersect, of coordinates (x0 , t0):
u ( x0 , t0 ) 
t
x0  ct 0
f ( x0  ct0 )  f ( x0  ct0 )
1


g ( s )ds, its careful inspection shows that:
2
2  c x0 ct0

the solution at the intersection point (x0 , t0) depends only on
the initial conditions in the interval (x0 – ct0 , x0 + ct0) and does
Zone of Influence
not depend on the initial conditions outside of this interval;
t0
LEFT running
characteristic. It
has the slope -(1/c)
(x0 , t0)
RIGHT running
characteristic. It
has the slope +(1/c)
Zone of Dependence

the region enclosed between the x-axis and the two
characteristics is termed domain of dependence; physically this
is caused by the limited propagation speed “c” of the mutual
influences through the solution domain;

tan(α)=t0/ct0=1/c
α
(x0 – ct0 , 0)
x
x0
(x0 + ct0 , 0)
changes at the point (x0, t0) influence events at latter time in
the domain of influence, also bounded by the characteristics.
Characteristics for the wave equation
The solution of the wave equation: u ( x, t ) 
x  ct
f ( x  ct )  f ( x  ct )
1

  g ( s )ds
2
2  c x ct
the 1st term of the solution represents propagation of the initial data along the
characteristics

the 2nd term represents the effect of the data within the closed interval (x – ct , x + ct) at
t=0


The limited domain of dependence bounded by the characteristics that pass through the
point (x0 , t0) is a fundamental property of hyperbolic PDEs
t
Zone of Influence
t0
LEFT running
characteristic. It
has the slope -(1/c)
(x0 , t0)
RIGHT running
characteristic. It
has the slope +(1/c)
Zone of Dependence
tan(α)=t0/ct0=1/c
α
(x0 – ct0 , 0)
x
x0
(x0 + ct0 , 0)
Finding D’Alembert’s solution:
 2u
 0  u  0
  

integrating with respect to  gives:
 2u
      0, where f(ξ) is an arbitrary function of ξ and
u
 f()


integrating with respect to ξ gives:
 2u
      0
u
 g ( ) , where g(η) is an arbitrary function of η

𝒖 ,  = 𝒇  + 𝒈 
D’Alembert’s solution:
𝒖 ,  = 𝒇  + 𝒈 
  ( x  ct )
  ( x  ct )
𝒖 𝒙, 𝒕 = 𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕
This is known as D’Alembert’s Solution
Now want to find D’Alembert’s Solution that Satisfies Initial Conditions
D’Alembert’s Solution that Satisfies Initial Conditions :
𝒖 𝒙, 𝒕 = 𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕
𝝏𝒖
𝒙, 𝒕 = 𝒄𝒇′ 𝒙 + 𝒄𝒕 − 𝒄𝒈′ 𝒙 − 𝒄𝒕
𝝏𝒕
Where primes denote derivatives with respect to the
entire argument (x-ct) & (x+ct)
𝒖 𝒙, 𝟎 =  𝒙
𝝏𝒖
𝒙, 𝟎 = 𝝋 𝒙
𝝏𝒕
𝒖 𝒙, 𝟎 = 𝒇 𝒙 + 𝒈 𝒙 =  𝒙
𝝏𝒖
𝒙, 𝟎 = 𝒄𝒇′ 𝒙 − 𝒄𝒈′ 𝒙 = 𝝋 𝒙
𝝏𝒕
D’Alembert’s Solution that Satisfies Initial Conditions :
𝒖 𝒙, 𝟎 = 𝒇 𝒙 + 𝒈 𝒙 =  𝒙
𝝏𝒖
𝒙, 𝟎 = 𝒄𝒇′ 𝒙 − 𝒄𝒈′ 𝒙 = 𝝋 𝒙
𝝏𝒕
Divide second equation by c and integrate wrt (x)
𝒇 𝒙 − 𝒈 𝒙 = 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟏
𝒇 𝒙 =  𝒙 +𝝋 𝒙
𝟐
𝟏
+
𝒄
𝒙
𝝋 𝒔 𝒅𝒔
𝒙𝒐
𝟏
+ 𝒇 𝒙 𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
+
𝟐𝒄
𝒙
𝝋 𝒔 𝒅𝒔
𝒙𝒐
D’Alembert’s Solution that Satisfies Initial Conditions :
𝒖 𝒙, 𝟎 = 𝒇 𝒙 + 𝒈 𝒙 =  𝒙
𝝏𝒖
𝒙, 𝟎 = 𝒄𝒇′ 𝒙 − 𝒄𝒈′ 𝒙 = 𝝋 𝒙
𝝏𝒕
Divide second equation by c and integrate wrt (x)
𝒇 𝒙 − 𝒈 𝒙 = 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟏
𝒈 𝒙 =  𝒙 −𝝋 𝒙
𝟐
𝟏
+
𝒄
𝒙
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝟏
− 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
−
𝟐𝒄
𝒙
𝒉 𝒔 𝒅𝒔
𝒙𝒐
D’Alembert’s Solution that Satisfies Initial Conditions :
𝟏
𝒇 𝒙 =  𝒙
𝟐
𝟏
+ 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
+
𝟐𝒄
𝟏
𝒈 𝒙 =  𝒙
𝟐
𝟏
− 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
−
𝟐𝒄
𝟏
𝒇 𝒙 + 𝒄𝒕 =  𝒙 + 𝒄𝒕
𝟐
𝟏
+ 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
+
𝟐𝒄
𝟏
𝒈 𝒙 − 𝒄𝒕 =  𝒙 − 𝒄𝒕
𝟐
𝟏
− 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
−
𝟐𝒄
𝟏
𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕 =  𝒙 + 𝒄𝒕 +  𝒙 − 𝒄𝒕
𝟐
𝒖 𝒙, 𝒕 = 𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕
𝟏
+
𝟐𝒄
𝒙
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙+𝒄𝒕
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙−𝒄𝒕
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙+𝒄𝒕
𝒉 𝒔 𝒅𝒔
𝒙−𝒄𝒕
D’Alembert’s Solution that Satisfies Initial Conditions :
𝟏
𝒇 𝒙 =  𝒙
𝟐
𝟏
+ 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
+
𝟐𝒄
𝟏
𝒈 𝒙 =  𝒙
𝟐
𝟏
− 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
−
𝟐𝒄
𝟏
𝒇 𝒙 + 𝒄𝒕 =  𝒙 + 𝒄𝒕
𝟐
𝟏
+ 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
+
𝟐𝒄
𝟏
𝒈 𝒙 − 𝒄𝒕 =  𝒙 − 𝒄𝒕
𝟐
𝟏
− 𝒇 𝒙𝒐 − 𝒈 𝒙𝒐
𝟐
𝟏
−
𝟐𝒄
𝟏
𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕 =  𝒙 + 𝒄𝒕 +  𝒙 − 𝒄𝒕
𝟐
𝟏
+
𝟐𝒄
𝒙
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙+𝒄𝒕
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙−𝒄𝒕
𝒉 𝒔 𝒅𝒔
𝒙𝒐
𝒙+𝒄𝒕
𝒉 𝒔 𝒅𝒔
𝒙−𝒄𝒕
D’Alembert’s Solution that Satisfies Initial Conditions :
𝟏
𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕 =  𝒙 + 𝒄𝒕 +  𝒙 − 𝒄𝒕
𝟐
𝟏
+
𝟐𝒄
𝒙+𝒄𝒕
𝒉 𝒔 𝒅𝒔
𝒙−𝒄𝒕
By definition:
𝒖 𝒙, 𝒕 = 𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕
𝒖 ,  = 𝒇  + 𝒈 
If initial velocity is zero:
𝟏
 𝒙 + 𝒄𝒕 +  𝒙 − 𝒄𝒕
𝟐
𝟏
𝒇  +𝒈  =   + 
𝟐
This also shows that f & g are similar functions
𝒇 𝒙 + 𝒄𝒕 + 𝒈 𝒙 − 𝒄𝒕 =