Download Wave Equation - Wave Phenomena - Lecture Slides and more Slides Microwave Engineering and Acoustics in PDF only on Docsity!
Phys 3750
Fourier Transforms and the Wave Equation
Overview and Motivation : We first discuss a few features of the Fourier transform
(FT), and then we solve the initial-value problem for the wave equation using the
Fourier transform.
Key Mathematics : More Fourier transform theory, especially as applied to solving
the wave equation.
I. FT Change of Notation
In the last lecture we introduced the FT of a function f (^ x )through the two equations
( ) ( ) ∫
∞
−∞
f x = f k e dk ˆ ikx 2
, (1a)
( ) (^) ∫ ( )
∞
−∞
− f k = f xe dx
ikx
. (1b)
Note that we have changed notation compared to the last lecture. Hereafter we
designate the FT of any function by the same symbol, but with an overhead caret
included. That is, the FT of f ( ) x we now write as f ˆ^ ( k ). As we shall see, this is useful
when dealing with equations that include FTs of several functions.
1
II. Some Properties of the Fourier Transform
We now discuss several useful properties of the Fourier transform.
A. Translation
The first property has to do with translation of the function f ( x ). Let's say we are
interested in f ( x − x 0 ), which corresponds to translation of f ( x )by x 0. Then, using
Eq. (1a) we can write
( ) ( )
( )
[ ( ) ] ∫
∫
∞
−∞
−
∞
−∞
−
f k e e dk
f x x f k e dk
ikx ikx
ikxx
0
0
0
(^1) This notation is fairly common practice. At some point you may even see the FT of f ( ) x written as f ( k ).
At least we won't be doing that here!
Phys 3750
Thus, we see that the FT of f ( x − x 0 )is ˆ^ ( ) 0
ikx f k e
−
. In other words, translation of f ( x )
by x 0 corresponds to multiplying the FT f ˆ^ ( k )by 0
ikx e
−
B. Differentiation
The second property has to do with the FT of f ′( x ), the derivative of f ( ) x. Again,
using Eq. (1a) we have
( ) (^) ∫[ ( )]
∞
−∞
f ′^ x = ikf k e dk ˆ ikx
2
So we see that FT of f ′( )^ x is ik f ˆ^ ( k ). That is, differentiation of f ( ) x corresponds to
multiplying f ˆ^ ( ) k by ik.
C. Integration
Let's consider the definite integral of f ( x ),
∫ ( )^ ∫ ∫ ( )
∞
−∞
2
1
2
1
x
x
ikx
x
x
dxf x dx dkf k e
Switching the order of integration on the rhs produces
( ) ( )
( ) ∫ (^ )
∫ ∫ ∫
∞
−∞
∞
−∞
2 1
2
1
2
1
ˆ
ikx ikx
x
x
ikx
x
x
e e ik
f k dk
dx f x dkf k dxe
So if we define If ( ) x to be the indefinite integral of f ( x ), we can rewrite Eq. (5) as
( ) ( )
( ) ( ) ∫
∞
−∞
− =^2 −^1
2 1
ikx ikx e e ik
f k If x If x dk
Phys 3750
( ) ( ) (^) ∫ ( )
∞
−∞
− ′
( f *ˆ g ) k = f ˆ k dx ′ gx ′ eikx. (10)
Recognizing the integral as 2 π g ˆ( k )we finally have
( f *ˆ g ) ( ) k = 2 π f ˆ( ) k g ˆ( ) k. (11)
So we see that the FT of the convolution is the product of the FT's of the individual
functions (along with a factor of 2 π ). One way you may hear this result expressed
is that convolution in real space ( x ) corresponds to multiplication in k space.
Equation (11) is known as the convolution theorem.
III. Solution to the Wave Equation Initial Value Problem
Way back in Lecture 8 we discussed the initial value problem for the wave equation
( ) ( ) 2
2 2 2
2 , ,
x
qxt c t
qxt
∂
on the interval − ∞< x <∞. For the initial conditions
q ( x , 0 ) = a ( ) x , (13a)
( (^) x ) (^) b ( ) x t
q
∂
, 0 , (13b)
we found that the solution to Eq. (12) can be written as
( ) ( ) ( ) ( )
∫
−
x ct
xct
bx dx c
q xt ax ct ax ct
With the help of the Fourier transform we are now going to rederive this solution,
and along the way we will learn something very interesting about the FT of q ( x , t ).
We start by defining the (spatial) FT of q ( x , t )as
( ) ( ) ∫
∞
−∞
− q kt = qxt e dx
ikx , 2
, (15a)
Phys 3750
so that we also have
( ) ( ) ∫
∞
−∞
q xt = qkt e dk
ikx ˆ , 2
π
. (15b)
We also define the FT of Eq. (13), the initial conditions,
q ˆ (^ k , 0 ) = a ˆ( ) k , (16a)
( (^) k ) (^) b ( ) k t
q (^) ˆ , 0
. (16b)
Now each side of the Eq. (12) is a function of x and t , so we can calculate the FT of
both sides of Eq (12),
( ) ( )
∫ ∫
∞
−∞
−
∞
−∞
−
∂
e dx x
qxt e dx c t
q xt ikx ikx 2
2 2 2
2 , ,
On the lhs of this equation we can pull the time derivative outside the integral. The
lhs is then just the second time derivative of q ˆ(^ k , t ). The rhs can be simplified by
remembering that the FT of the ( x ) derivative of a function is ik times the FT of the
original function. Thus the FT of ( )
2 2
∂ q x , t ∂ x is just
2 − k times q ˆ(^ k , t ), the FT of
q ( x , t ). Thus we can rewrite Eq. (17) as
( ) kcq ( kt ) t
q kt ˆ ,
2
2 = − ∂
This equation should look very familiar to you. What equation is it? None other than
the harmonic oscillator equation! What does this tell us about q ˆ ( k , t )? It tells us that
q ˆ ( k , t )(for a fixed value of k ) oscillates harmonically at the frequency ω = kc. Thus
we can interpret the function q ˆ( k , t )as the set of normal modes coordinates for this
problem. This further means that the FT has decoupled the equations of motion for
this system [as represented by Eq. (12), the wave equation.] Notice also that the
dispersion relation ω = kc has also fallen into our lap by considering the FT of Eq.
As we should know by this point, the solution to Eq. (18) can be written as
( ) ( ) ( )
ikct ikct q kt Ak e Bk e
−
Phys 3750
( ) [ ( ) (^ )^ ( ) (^ )]
( ) (^) ( ) ( )
( e e ) dk
ik
bk
c
q xt ak eikxct ak eikxct ikxct ikxct
∞
−∞
The first two terms we recognize as ( 1 2 )[ a ( x + ct ) + a ( x − ct )], while we can use Eq. (5)
to recognize the second half of the rhs of Eq. (24) as ( )∫ ( )
−
x ct
x ct
1 2 c bx dx. Thus Eq. (24)
can be re-expressed as
( ) ( ) ( ) ( )
−
x ct
xct
bx dx c
q xt ax ct ax ct
which is identical to Eq. (14).
Exercises
- 17.1 FT Properties. If the FT of f ( x )is h ( k ),
( a ) show that the FT of [ e f ( ) x ]
ik 0 x is h ( k − k 0 );
( b ) show that the FT of [ x f ( ) x ]is ih ′(^ k );
( c ) show that the FT of [ x f ( ) x ]
2 is − h ′′( k ).
- 17.2 Show that ( f * g )( ) x = ( g * f )( ) x by
( a ) directly by manipulating Eq. (7), the definition of convolution;
( b ) by using Eq. (11), the result for the FT of ( f * g )( ) x.
* 17.3 Convolution and the Gaussian. The function that has the same form as its
Fourier transform is the Gaussian. Specifically if ( )
x^2 σ^2 f x e
−
= , its FT is given by
( )
σ k σ
h k e
− =. Using this fact, show that the convolution ( f (^) 1 * f 2 )( ) x of two
Gaussian functions ( )
2 1
2 1
x σ f x e
− = and ( )
2 2
2 2
x σ f x e
−
= is proportional to the Gaussian
function
− x^2 (σ 12 + σ 22 )
e. [Hint: you need not calculate any integrals to do this problem.]
Phys 3750
** 17.4 The Rectangular Pulse.
Here you will explore the convolution theorem as it applies to a rectangular pulse and
its convolution, a triangular pulse. We start with the function f (^ x ), a rectangular
pulse of height H and width 2 L centered at x = 0. Elsewhere the function is zero.
( a ) Graph f ( ) x
( b ) Calculate Find f ˆ^ ( ) k , the FT of f ( x ). (This is a real function; express it as such.)
( c ) Graph f ˆ^ ( ) k.
( d ) The function ( f * f )( ) x , the convolution of f ( x )with itself, is a triangle function
of height
2 LH^2 and base 4 L centered at zero. It is zero elsewhere. Graph this
function.
( e ) Write down the mathematical expression for ( f * f )( ) x [that you graphed in (d)].
Then directly calculate ( f *ˆ^ f )( ) k using your functional form for ( f * f )( ) x. (Do not set
H and L to specific values. Again, this is a real function; express it as such.)
( f ) Graph your calculated transform ( f *ˆ^ f )( k ).
( g ) Lastly, use f ˆ^ ( ) k and the convolution theorem to find ( f *ˆ^ f )( k ). Show that this is
equal to the result in part (e).
** 17.5 FT Solution to the 1D Wave Equation. Eq. (23),
( ) ( )
( ) (^) ( ) ( )
( ) (^) ( ) e dk ikc
bk e ak ikc
bk q xt ak
ikxct ikxct ∫
∞
−∞
π
is the formal solution to the initial-value problem.
( a ) What kind of waves are described by the functions exp[ ik ( x + ct )]and
exp[ ik ( x − ct )]? Be as specific as possible!
( b ) From a vector-space point of view, the functions exp[ ik ( x + ct )]and exp[ ik ( x − ct )]
can be considered basis functions for the vector space that consists of solutions to the
wave equation. Given this, what do the terms ( )
( )
kc
ibk a k
and ( )
( )
kc
ibk a k
represent?
( c ) Given your answer in (b), how would you describe the solution q ( x , t )as written
above? [Hint: the term linear combination should appear in your answer.]
( d ) How are a ˆ( )^ k and b ˆ( )^ k related to the initial conditions q ( x , 0 )and ( x , 0 ) t
q
∂
? That
is, write down expressions for a ˆ^ ( ) k and b ˆ(^ k )in terms of q ( x , 0 )and ( x , 0 ) t
q
∂
