10

Finite Fourier Sine and Cosine Transforms

“Mathematics compares the most diverse phenomena and discovers

the secret analogies that unite them.”
Joseph Fourier

“In the mathematical investigation I have usually employed such

methods as present themselves naturally to a physicist. The pure
mathematician will complain, and (it must be confessed) some-
times with justice, of deficient rigor. But to this question there
are two sides. For, however important it may be to maintain a
uniformly high standard in pure mathematics, the physicist may
occasionally do well to rest content with arguments which are fair-
ly satisfactory and conclusive from his point of view. To his mind,
exercised in a different order of ideas, the more severe procedure of
the pure mathematician may appear not more but less demonstra-
tive. And further, in many cases of difficulty to insist upon highest
standard would mean the exclusion of the subject altogether in
view of the space that would be required.”
Lord Rayleigh

10.1 Introduction
This chapter deals with the theory and applications of finite Fourier sine
and cosine transforms. The basic operational properties including convolution
theorem of these transforms are discussed in some detail. Special attention is
given to the use of these transforms to the solutions of boundary value and
initial-boundary value problems.
The finite Fourier sine transform was first introduced by Doetsch (1935).
Subsequently, the method has been developed and generalized by several au-
thors including Kneitz (1938), Koschmieder (1941), Roettinger (1947), and
Brown (1944).

407
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408 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

10.2 Definitions of the Finite Fourier Sine and Cosine

Transforms and Examples
Both finite Fourier sine and cosine transforms are defined from the corre-
sponding Fourier sine and Fourier cosine series.

DEFINITION 10.2.1 (The Finite Fourier Sine Transform). If f (x) is a

continuous or piecewise continuous function on a finite interval 0 < x < a, the
finite Fourier sine transform of f (x) is defined by

a  nπx 
Fs {f (x)} = f˜s (n) = f (x) sin dx, (10.2.1)
a
0

where n = 1, 2, 3, . . ..

It is a well-known result of the theory of Fourier series that the Fourier sine
series for f (x) in 0 < x < a
∞
2 ˜  nπx 
fs (n) sin (10.2.2)
a n=1 a

converges to the value f (x) at each point of continuity in the interval 0 <
1
x < a and to the value [f (x + 0) + f (x − 0)] at each point x of the finite
2
discontinuity in 0 < x < a. In view of the definition (10.2.1), the inverse Fourier
sine transform is given by
  ∞
2 ˜  nπx 
Fs−1 f˜s (n) = f (x) = fs (n) sin . (10.2.3)
a n=1 a

Clearly, both Fs and Fs−1 are linear transformations.

DEFINITION 10.2.2 (The Finite Fourier Cosine Transform). If f (x) is

a continuous or piecewise continuous function on a finite interval 0 < x < a,
the finite Fourier cosine transform of f (x) is defined by

a  nπx 
Fc {f (x)} = f˜c (n) = f (x) cos dx, (10.2.4)
a
0

where n = 0, 1, 2, . . ..

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Finite Fourier Sine and Cosine Transforms 409

It is also a well-known result of the theory of Fourier series that the Fourier
cosine series for f (x) in 0 < x < a

1 ˜
∞
2 ˜  nπx 
fc (0) + fc (n) cos (10.2.5)
a a n=1 a

1
converges to f (x) at each point of continuity in 0 < x < a, and to [f (x + 0) +
2
f (x − 0)] at each point x of finite discontinuity in 0 < x < a. By virtue of the
definition (10.2.4), the inverse Fourier cosine transform is given by
  1
∞
2 ˜  nπx 
Fc−1 f˜c (n) = f (x) = f˜c (0) + fc (n) cos . (10.2.6)
a a n=1 a

Clearly, both Fc and Fc−1 are linear transformations.

When a = π, the finite Fourier sine and cosine transforms are defined, re-
spectively, by (10.2.1) and (10.2.4) on the interval 0 < x < π. The correspond-
ing inverse transforms are given by the same results (10.2.3) and (10.2.6) with
a = π. The transform of a function defined over an interval 0 < x < a can be
written easily in terms of a transform on the standard interval 0 < x < π. We
πx
substitute ξ = to write (10.2.1) and (10.2.4) as follows:
a
a  nπx  π     ax 
a aξ a
f˜s (n) = sin f (x)dx = sin(nξ)f dξ = Fs f
a π π π π
0 0
a  nπx  π     ax 
a aξ a
f˜c (n) = cos f (x)dx = cos(nξ)f dξ = Fc f .
a π π π π
0 0

Example 10.2.1
Find the finite Fourier sine and cosine transforms of

(a) f (x) = 1 and (b) f (x) = x.

(a) We have
a  nπx  a
Fs (1) = f˜s (n) = sin dx = [1 − (−1)n ], (10.2.7)
a nπ
0
a  
 nπx  a, n=0
Fc {1} = f˜c (n) = cos dx = . (10.2.8)
a 0, n = 0
0

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410 INTEGRAL TRANSFORMS and THEIR APPLICATIONS
a  nπx  (−1)n+1 a2
(b) Fs {x} = x sin dx = .
a nπ
a
(10.2.9)
⎧ 2 ⎫
a  nπx  ⎪ a
⎨ , n = 0⎪
⎬
Fc {x} = x cos dx = 2 2 . (10.2.10)
a ⎪
⎩ a ⎪
0  0⎭
[(−1)n − 1], n =
nπ

10.3 Basic Properties of Finite Fourier Sine

and Cosine Transforms
As a preliminary to the solution of differential equations by the finite Fourier
sine and cosine transforms, we now establish the transforms of derivatives of
f (x).
 nπ 
Fs {f  (x)} = − f˜c (n), (10.3.1)
a
 nπ 2  nπ 
Fs {f  (x)} = − f˜s (n) + [f (0) + (−1)n+1 f (a)], (10.3.2)
a
 nπ  a
Fc {f  (x)} = f˜s (n) + (−1)n f (a) − f (0), (10.3.3)
a
 nπ 2
Fc {f  (x)} = − f˜c (n) + (−1)n f  (a) − f  (0). (10.3.4)
a

Similar results can be obtained for the finite Fourier sine and cosine transforms
of higher derivatives of f (x).
Results (10.3.1)–(10.3.4) can be proved by integrating by parts. For exam-
ple, we have
a  nπx 
Fs {f (x)} = f  (x) sin

dx,
a
0

which is, integrating by parts,

)  nπx *a nπ a  nπx 

= f (x) sin − f (x) cos dx
a 0 a a
0
 nπ 
=− ˜
fc (n).
a

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Finite Fourier Sine and Cosine Transforms 411

Similarly, we find that

a  nπx 

Fc {f (x)} = f  (x) cos dx
a
0
)  nπx *a a  nπx 
nπ
= f (x) cos + f (x) sin dx
a 0 a a
0
nπ ˜
= (−1) f (a) − f (0) +
n
fs (n).
a
This proves the result (10.3.3).
Results (10.3.2) and (10.3.4) and the results for higher derivatives can be
obtained by the repeated application of the fundamental results (10.3.1) and
(10.3.3).

DEFINITION 10.3.1 (Odd Periodic Extension). A function f1 (x) is said

to be the odd periodic extension of the function f (x), with period 2π if
 
f (x) for 0 < x < π
f1 (x) = . (10.3.5)
−f (−x) for −π < x < 0
Or, equivalently,
f1 (x) = f (x) when 0 < x < π,
f1 (−x) = −f1 (x), f1 (x + 2π) = f1 (x) for − ∞ < x < ∞.(10.3.6)
Similarly, the even periodic extension f2 (x) of f (x), with period 2π is de-
fined in −π < x < π by the equations
 
f (x) for 0 < x < π
f2 (x) = . (10.3.7)
f (−x) for −π < x < 0
Or, equivalently,
f2 (x) = f (x) when 0 < x < π,
f2 (−x) = f2 (x), f2 (x + 2π) = f2 (x) for − ∞ < x < ∞. (10.3.8)

THEOREM 10.3.1
If f1 (x) is the odd periodic extension of f (x) with period 2π, then, for any
constant α,
Fs {f1 (x − α) + f1 (x + α)} = 2 cos nα Fs {f (x)}. (10.3.9)
In particular, when α = π, and n = 1, 2, 3, . . . ,
Fs {f (x − π)} = (−1)n Fs {f (x)}. (10.3.10)

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412 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

Similarly, we obtain
Fc {f1 (x + α) − f1 (x − α)} = 2 sin(nα) Fs {f (x)}. (10.3.11)

PROOF To prove (10.3.9), we follow Churchill (1972) and write the right
hand side of (10.3.9) as
π
2 cos nαf˜s (n) = 2 cos(nα) sin(nx)f (x)dx
0
π
=2 cos nα sin nxf1 (x)dx
0
π
= [sin n(x + α) + sin n(x − α)]f1 (x)dx,
0

which is, since the integrand is even function of x,

π
1
= [sin n(x + α) + sin n(x − α)]f1 (x)dx,
2
−π

which is, by putting x + α = t and x − α = t,


π+α 
π−α
1 1
= sin nt f1 (t − α) dt + sin nt f1 (t + α) dt,
2 2
−π+α −(π+α)

which is, since the integrands are periodic function of t with period 2π, and
hence, the limits of integration can be replaced with limits −π to π,
π π
1 1
= sin nt f1 (t − α) dt + sin nt f1 (t + α) dt
2 2
−π −π
⎡ 0 ⎤
 π
1
= ⎣ + ⎦ {sin nt f1 (t − α)}dt
2
−π 0
⎡ 0 ⎤
 π
1⎣
+ + ⎦ {sin nt f1 (t + α)}dt. (10.3.12)
2
−π 0

Furthermore,
0 π
sin nt f1 (t − α) dt = sin nx f1 (x + α) dx
−π 0

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Finite Fourier Sine and Cosine Transforms 413

in which f1 (−x − α) = −f1 (x + α) is used.

Making a similar change of variables in the third integral of (10.3.12), we
obtain the formula
π π
2 cos nα f˜s (n) = sin nt f1 (t − α)dt + sin nx f1 (x + α)dx,
0 0

which gives the desired result (10.3.9).

Finally, f1 (x + π) = f1 (2π + x − π) = f1 (x − π) = −f1 (π − x), and when 0 <
x < π, f1 (π − x) = f (π − x). Thus, when α = π, result (10.3.9) becomes

π
f˜s (n) cos nπ = sin nx f (x − π)dx = Fs {f (x − π)},
0

which reduces to (10.3.10).

The proof of (10.3.11) is similar to that of (10.3.9), and hence, is left to the
reader.

THEOREM 10.3.2
If f2 (x) is the even periodic extension of f (x) with period 2π, then, for any
constant α,

Fc {f2 (x − α) + f2 (x + α)} = 2 cos nα Fc {f (x)}, (10.3.13)

Fc {f2 (x − α) − f2 (x + α)} = 2 sin nα Fc {f (x)}. (10.3.14)

This theorem is very much similar to that of Theorem 10.3.1, and hence, the
proof is left to the reader.
In the notation of Churchill (1972), we introduce the convolution of two
sectionally continuous periodic functions f (x) and g(x) defined in −π < x < π
by
π
f (x) ∗ g(x) = f (x − u) g(u) du. (10.3.15)
−π

Clearly, f (x) ∗ g(x) is continuous and periodic with period 2π. The convolution
is symmetric, that is, f ∗ g = g ∗ f . Furthermore, the convolution is an even
function if f (x) and g(x) are both even or both odd. It is odd if either f (x)
or g(x) is even or the other odd. We next prove the convolution theorem.

THEOREM 10.3.3
(Convolution). If f1 (x) and g1 (x) are the odd periodic extensions of f (x) and
g(x) respectively on 0 < x < π, and if f2 (x) and g2 (x) are the even periodic

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414 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

extensions of f (x) and g(x) respectively on 0 < x < π, then

Fc {f1 (x) ∗ g1 (x)} = −2 f˜s (n) g̃s (n), (10.3.16)

Fc {f2 (x) ∗ g2 (x)} = 2 f˜c(n) g̃c (n), (10.3.17)
Fs {f1 (x) ∗ g2 (x)} = 2 f˜s (n) g̃c (n), (10.3.18)
Fs {f2 (x) ∗ g1 (x)} = 2 f˜c(n) g̃s (n). (10.3.19)

Or, equivalently,
  1
Fc−1 f˜s (n) g̃s (n) = − {f1 (x) ∗ g1 (x)}, (10.3.20)
2
  1
Fc−1 f˜c (n) g̃c (n) = {f2 (x) ∗ g2 (x)}, (10.3.21)
2
  1
Fs−1 f˜s (n) g̃c (n) = {f1 (x) ∗ g2 (x)}, (10.3.22)
2
  1
Fs−1 f˜c (n) g̃s (n) = {f2 (x) ∗ g1 (x)}. (10.3.23)
2

PROOF To prove (10.3.16), we consider the product

π
2 f˜s (n) g̃s (n) = 2 f˜s (n) sin nu g(u)du,
0

which is, by using (10.3.11),

π
= g(u) [Fc {f1 (x + u) − f1 (x − u)}] du
0
⎡ π ⎤
π 
= g(u) ⎣ {f1 (x + u) − f1 (x − u)} cos nx⎦ du,
0 0

which is, by interchanging the order of integration,

⎡ π ⎤
π 
= cos(nx) ⎣ {f1 (x + u) − f1 (x − u)} g(u) du⎦ dx. (10.3.24)
0 0

Using the definition of convolution (10.3.15), introducing new variables of

integration, and invoking the odd extension properties of f1 (x) and g1 (x), we
obtain
π
f1 (x) ∗ g1 (x) = [f1 (x − u) − f1 (x + u)] g(u) du (10.3.25)
0
= I1 − I2 − I3 + I4 , (10.3.26)

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Finite Fourier Sine and Cosine Transforms 415

where
x π
I1 = f (u) g(x + u) du, I2 = f (u) g(u − x)du, (10.3.27ab)
0 x

π−x π
I3 = f (u) g(x + u) du, I4 = f (u) g(2π − x − u)du. (10.3.28ab)
0 x

In view of (10.3.25), we thus obtain the desired result (10.3.16) from (10.3.24).
This completes the proof.
The other results included in Theorem 10.3.3 can be proved by the above
method of proof.
As an example of convolution theorem, we evaluate the inverse cosine Fouri-
er transform of (n2 − a2 )−1 . We write, for n = 0,

1 n(−1)n+1 (−1)n+1
= 2 · = f˜s (n) g̃s (n),
(n2 − a ) (n − a2 )
2 n

(−1)n+1
where f˜s (n) = n(−1)n+1 (n2 − a2 )−1 and g̃s (n) = so that
  n
sin ax x
f (x) = and g(x)= .
sin aπ π
Evidently,
  x
1 sin ax
= f˜s (n) g̃s (n) = Fs Fs .
(n2 − a2 ) sin aπ π

According to result (10.3.20),

   
1 ˜s (n) g̃s (n) = − 1 f1 (x) ∗ g1 (x),
Fc−1 = F −1
f (10.3.29)
(n2 − a2 ) c
2

where f1 (x) is the periodic extension of the odd function f (x) with period 2π
x
and g1 (x) = . Thus, it turns out that
π
  π π
1 1 1
Fc−1 =− f1 (x − u) g1 (u) du = − f1 (x − u) u du.
(n2 − a2 ) 2 2π
−π −π

This integral can easily be evaluated by splitting up the interval of integration

or, by using (10.3.26), and hence,
 
−1 1 cos{a(π − x)}
Fc =− . (10.3.30)
(n2 − a2 ) a sin aπ

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416 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

10.4 Applications of Finite Fourier Sine and

Cosine Transforms
In this section we illustrate the use of finite Fourier sine and cosine transforms
to the solutions of boundary value and initial-boundary value problems.

Example 10.4.1
(Heat Conduction Problem in a Finite Domain with the Dirichlet Data at the
Boundary). We began by considering the solution of the temperature distri-
bution u(x, t) of the diffusion equation
ut = κ uxx, 0 ≤ x ≤ a, t > 0, (10.4.1)
with the boundary and initial conditions
u(0, t) = 0 = u(a, t), (10.4.2ab)

u(x, 0) = f (x) for 0 ≤ x ≤ a. (10.4.3)

Application of the finite Fourier sine transform (10.2.1) to this diffusion
problem gives the initial value problem
d ũs  nπ 2
+κ ũs = 0, (10.4.4)
dt a
ũs (n, 0) = f˜s (n). (10.4.5)
The solution of (10.4.4)–(10.4.5) is
  nπ 2 
ũs (n, t) = fs (n) exp −κ
˜ t . (10.4.6)
a
The inverse finite Fourier sine transform (10.2.3) leads to the solution
∞   nπ 2   nπx 
2  ˜
u(x, t) = fs (n) exp −κ t sin
a n=1 a a
∞   nπ 2   nπx 
2 
u(x, t) = exp −κ t sin
a n=1 a a
a  
nπξ
× f (ξ) sin dξ. (10.4.7)
a
0

If, in particular, f (x) = T0 = constant, then (10.4.7) becomes

  ∞   nπ 2   nπx 
2 T0  1
u(x, t) = [1 − (−1)n ] exp −κ t sin . (10.4.8)
π n=1
n a a

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Finite Fourier Sine and Cosine Transforms 417

This series solution can be evaluated numerically using the Fast Fourier
transform which is an algorithm for the efficient calculation of the finite
Fourier transform.

Example 10.4.2
(Heat Conduction Problem in a Finite Domain with the Neumann Data at the
Boundary). We consider the solution of the diffusion equation (10.4.1) with
the prescribed heat flux at x = 0 and x = a, and the associated boundary and
initial data are

ux (0, t) = 0 = ux (a, t) for t > 0, (10.4.9)

u(x, 0) = f (x) for 0 ≤ x ≤ a. (10.4.10)

In this case, it is appropriate to use the finite Fourier cosine transform

(10.2.4). So, the application of this transform gives the initial value problem

dũc  nπ 2
+κ ũc = 0, (10.4.11)
dt a
ũc (n, 0) = f˜c (n). (10.4.12)

The solution of this problem is

  nπ 2 
ũc (n, t) = f˜c (n) exp −κ t . (10.4.13)
a

The inverse finite cosine transform (10.2.5) gives the formal solution
∞   nπ 2   nπx 
1 ˜ 2 ˜
u(x, t) = fc (0) + fc (n) exp −κ t cos
a a n=1 a a
⎡ a ⎤
 a ∞   
1 2 ⎣ nπξ
= f (ξ)dξ + f (ξ) cos dξ ⎦
a a n=1 a
0 0
  nπ 2   nπx 
× exp −κ t cos . (10.4.14)
a a

Example 10.4.3
(The Static Deflection of a Uniform Elastic Beam). We consider the static
deflection y(x) of a uniform elastic beam of finite length  which satisfies the
equilibrium equation

d4 y W (x)
= = w(x), 0 ≤ x ≤ , (10.4.15)
dx4 EI

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418 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

where W (x) is the applied load per unit length of the beam, E is the Young’s
modulus of the beam, and I is the moment of inertia of the cross section of
the beam. If the beam is freely hinged at its ends, then
y(x) = y  (x) = 0 at x = 0 and x = . (10.4.16)
Application of the finite Fourier sine transform of y(x) to (10.4.15) and
(10.4.16) gives
 4

ỹs (n) = w̃s (n). (10.4.17)
nπ
Inverting this result, we find
∞
23  1  nπx 
y(x) = 4 4
sin w̃s (n)
π n=1 n 
∞  nπx     
23  1 nπξ
= 4 sin w(ξ) sin dξ. (10.4.18)
π n=1 n4  
0

In particular, if the applied load of magnitude W0 is confined to the point

x = α, where 0 < α < , then w(x) = W0 δ(x − α) where W0 is a constant.
Consequently, the static deflection is
∞
23 W0  1  nπx   nπα 
y(x) = sin sin . (10.4.19)
π 4 n=1 n4  

Example 10.4.4
(Transverse Displacement of an Elastic Beam of Finite Length). We consider
the transverse displacement of an elastic beam at a point x in the down-
ward direction where the equilibrium position of the beam is along the x-axis.
With the applied load W (x, t) per unit length of the beam, the displacement
function y(x, t) satisfies the equation of motion
∂4y 1 ∂ 2 y W (x, t)
+ = , 0 ≤ x ≤ , t > 0, (10.4.20)
∂x4 a2 ∂t2 EI
where a2 = EI/(ρα), α is the cross-sectional area and ρ is the line density of
the beam.
If the beam is freely hinged at its ends, then
∂2y
y(x, t) = =0 at x = 0 and x = . (10.4.21)
∂x2
The initial conditions are
∂y
y(x, t) = f (x), = g(x) at t = 0 for 0 < x < . (10.4.22)
∂t

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Finite Fourier Sine and Cosine Transforms 419

We use the joint Laplace transform with respect to t and the finite Fourier
sine transform with respect to x defined by
∞   nπx 
−st
ũ¯s (n, s) = e dt u(x, t) sin dx. (10.4.23)

0 0

Application of the double transform to (10.4.20)–(10.4.22) gives the solution

for ỹ¯s (n, s) as
 2
¯ sf˜s (n) + g̃s (n) a W̃ s (n, s)
ỹs (n, s) = + , (10.4.24)
(s2 + c2 ) EI (s2 + c2 )
 nπ 2
where c = a .

The inverse Laplace transform gives
g̃s (n)
ỹs (n, t) = f˜s (n) cos(ct) + sin(ct)
c
 2  t
a 1
+ sin c(t − τ )W̃s (n, τ )dτ. (10.4.25)
EI c
0

Thus, the inverse finite Fourier sine transform yields the formal solution as

2
∞  πnx 
y(x, t) = ys (n, t) sin ,
 n=1 
2
∞  nπx   g̃s (n)

= sin f˜s (n) cos(ct) + sin(ct)
 n=1  c
⎤
 2  t
a 1
+ sin c(t − τ ) W̃s (n, τ )dτ ⎦ , (10.4.26)
EI c
0

where
     
nπξ nπξ
f˜s (n) = f (ξ) sin dξ, g̃s (n) = g(ξ) sin dξ. (10.4.27ab)
 
0 0

The case of free vibrations is of interest. In this case, W (x, t) ≡ 0 and hence,
W̃s (n, t) ≡ 0. Consequently, solution (10.4.26) reduces to a simple form
∞    nπx 
2 ˜ g̃s (n)
y(x, t) = fs (n) cos ct + sin ct sin , (10.4.28)
 n=1 c 

where f˜s (n) and g̃s (n) are given by (10.4.27ab).

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420 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

Example 10.4.5
(Free Transverse Vibrations of an Elastic String of Finite Length). We con-
sider the free vibration of a string of length  stretched to a constant tension
T between two points (0, 0) and (0, ) lying on the x-axis. The free transverse
displacement function u(x, t) satisfies the wave equation

∂2u 2
2∂ u
= c , 0 ≤ x ≤ , t > 0, (10.4.29)
∂t2 ∂x2
T
where c2 = and ρ is the line density of the string.
ρ
The initial and boundary conditions are

∂u
u(x, t) = f (x), = g(x) at t = 0 for 0 ≤ x ≤ , (10.4.30ab)
∂t
u(x, t) = 0 at x = 0 and x =  for t > 0. (10.4.31ab)

Application of the joint Laplace transform with respect to t and the finite
Fourier sine transform with respect to x defined by a similar result (10.4.23)
to (10.4.29)–(10.4.31ab) gives

˜
¯s (n, s) = sfs (n) + g̃s (n) ,
ũ (10.4.32)
(s2 + a2 ) (s2 + a2 )
 nπc 2
where a2 = .

The inverse Laplace transform gives

g̃s (n)
ũs (n, t) = f˜s (n) cos at + sin at. (10.4.33)
a
The inverse finite Fourier sine transform leads to the solution for u(x, t) as
∞    nπx 
2 ˜ g̃s (n)
u(x, t) = fs (n) cos at + sin at sin , (10.4.34)
 n=1 a 

where f˜s (n) and g̃s (n) are given by (10.4.27ab).

Example 10.4.6
(Two-Dimensional Unsteady Couette Flow). We consider two-dimensional un-
steady viscous flow between the plate at z = 0 at rest and the plate z = h in
motion parallel to itself with a variable velocity U (t) in the x direction. The
fluid velocity u(z, t) satisfies the equation of motion

∂u P (t) ∂2u
=− +ν 2, 0 ≤ z ≤ h, t > 0, (10.4.35)
∂t ρ ∂z

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Finite Fourier Sine and Cosine Transforms 421

and the boundary and initial conditions

u(z, t) = 0 on z = 0, t > 0; (10.4.36)
u(z, t) = U (t) on z = h, t > 0; (10.4.37)
u(z, t) = 0 at t ≤ 0, for 0 ≤ z ≤ h; (10.4.38)
where the pressure gradient px = P (t) and ν is the kinematic viscosity of the
fluid.
Application of the double transform defined by (10.4.23) to this initial
boundary value problem gives the solution for ũ¯s (n, s) as
 
ν n2 π 2 ¯ hP̄ (s)
s+ ũs (n, t) = − [1 + (−1)n+1 ]
h2 nπρ
ν nπ
+ (−1)n+1 Ū (s). (10.4.39)
h
The inverse Laplace transform yields
t  
h ν n2 π 2 τ
ũs (n, t) = − [1 + (−1)n+1 ] P (t − τ ) exp − dτ
nπρ h2
0
t  
ν nπ ν n2 π 2 τ
+ (−1)n+1 U (t − τ ) exp − dτ. (10.4.40)
h h2
0

Finally, the inverse finite Fourier sine transform gives the formal solution

2
∞  nπz 
u(z, t) = ũs (n, t) sin . (10.4.41)
h n=1 h

If, in particular, P (t) = constant P and U (t) = constant = U , then (10.4.41)

reduces to
∞  3  nπz    
2P  h ν n2 π 2 t
u(z, t) = − [1 + (−1) n+1
] sin 1 − exp −
μh n=1 nπ h h2
∞    nπz    
2U  h ν n2 π 2 t
+ (−1)n+1 sin 1 − exp − . (10.4.42)
h n=1 nπ h h2

This solution for the velocity field consists of both steady-state and transient
components. In the limit as t → ∞, the transient component decays to zero,
and the steady state is attained in the form
∞  3  nπz 
2P  h
u(z, t) = − [1 + (−1)n+1 ] sin
μh n=1 nπ 
   nπz 
∞
2U  h2
+ 2 (−1)n+1 sin . (10.4.43)
h n=1 nπ h

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422 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

In view of the inverse finite Fourier sine transforms

   
3
−1 h
Fs 2 [1 + (−1)n+1
] = z(h − z), (10.4.44)
nπ
  2 
−1 h
Fs (−1) n+1
= z, (10.4.45)
nπ

solution (10.4.43) can be rewritten in the closed form

 
Uz h ∂p  z
u(z, t) = − 1− z. (10.4.46)
h 2μ ∂x h

This is known as the generalized Couette flow. In the absence of the pressure
gradient term, solution (10.4.46) reduces to the linear profile of simple Couette
flow. On the other hand, if U (t) ≡ 0 and P (t) = 0, the solution (10.4.46) rep-
resents the parabolic profile of Poiseuille flow between two parallel stationary
plates due to an imposed pressure gradient.

10.5 Multiple Finite Fourier Transforms

and Their Applications
The above analysis for the finite Fourier sine and cosine transforms of a func-
tion of one independent variable can readily be extended to a function of
several independent variables. In particular, if f (x, y) is a function of two
independent variables x and y, defined in a region 0 ≤ x ≤ a, 0 ≤ y ≤ b, its
double finite Fourier sine transform is defined by

a b  mπx   nπy 
Fs {f (x, y)} = f˜s (m, n) = sin sin dx dy. (10.5.1)
a b
0 0

The inverse transform is given by the double series

    ∞  ∞  mπx 
4
Fs−1 ˜
fs (m, n) = f (x, y) = f˜s (m, n) sin
ab m=1 n=1 a
 nπy 
× sin . (10.5.2)
b
Similarly, we can define the double finite Fourier cosine transform and its
inverse.
The double Fourier sine transforms of the partial derivatives of f (x, y) can
easily be obtained. If f (x, y) vanishes on the boundary of the rectangular

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Finite Fourier Sine and Cosine Transforms 423

region D{0 ≤ x ≤ a, 0 ≤ y ≤ b}, then

 2   2 
∂ f ∂ 2f 2 m n2 ˜
Fs + = −π + fs (m, n). (10.5.3)
∂x2 ∂y 2 a2 b2

Example 10.5.1
(Free Vibrations of a Rectangular Elastic Membrane). The initial value prob-
lem for the transverse displacement field u(x, y, t) satisfies the following equa-
tion and the boundary and initial data
 2 
2 ∂ u ∂2u ∂2u
c + = , for all (x, y) in D, t > 0, (10.5.4)
∂x2 ∂y 2 ∂t2

u(x, y, t) = 0 on the boundary ∂D for all t > 0, (10.5.5)

u(x, y, t) = f (x, y), ut (x, y, t) = g(x, y) at t = 0, for (x, y) ∈ D . (10.5.6ab)

Application of the double finite Fourier sine transform defined by
a b  mπx   nπy 
ũs (m, n) = u(x, y) sin sin dxdy, (10.5.7)
a b
0 0

to the system (10.5.4)–(10.5.6ab) gives

 2 
d2 ũs 2 2 m n2
+ c π + ũs = 0, t>0 (10.5.8)
dt2 a2 b2
 
dũs
ũs (m, n, 0) = f˜s (m, n), = g̃s (m, n). (10.5.9)
dt t=0
The solution of this transformed problem is
ũs (m, n, t) = f˜s (m, n) cos(cπωmn t)
−1
+ (cπωmn ) g̃s (m, n) sin(cπωmn t), (10.5.10)
where
  12
m2 n 2
ωmn = + 2 . (10.5.11)
a2 b
The inverse transform gives the formal solution for u(x, y, t) in the form
  ∞  ∞  mπx   nπy  +
4
u(x, y, t) = sin sin f˜s (m, n) cos(cπωmn t)
ab m=1 n=1 a b
−1 ,
+ (cπωmn ) g̃s (m, n) sin(cπωmn t) , (10.5.12)

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424 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

where
a b    nπη 
mπξ
f˜s (m, n) = f (ξ, η) sin sin dξdη, (10.5.13)
a b
0 0
a b    nπη 
mπξ
g̃s (m, n) = g(ξ, η) sin sin dξdη. (10.5.14)
a b
0 0

Example 10.5.2
(Deflection of a Simply Supported Rectangular Elastic Plate). The deflection
u(x, y) of the plate satisfies the biharmonic equation

∂4u ∂4u ∂ 4 u w(x, y)

∇4 u ≡ + 2 + = ,
∂x4 ∂x2 ∂y 2 ∂y 4 D
in D{0 ≤ x ≤ a, 0 ≤ y ≤ b}, (10.5.15)

2Eh3
where w(x, y) represents the applied load at a point (x, y) and D =
3(1 − σ 2 )
is the constant flexural rigidity of the plate.
On the edge of the simply supported plate the deflection and bending mo-
ments are zero; hence, equation (10.5.15) has to be solved subject to the
boundary conditions
⎫
u(x, y) = 0 on x = 0 and x = a⎪
⎪
⎪
⎪
u(x, y) = 0 on y = 0 and y = b ⎪
⎪
⎪
⎪
⎬
∂2u
=0 on x = 0 and x = a⎪ . (10.5.16)
∂x2 ⎪
⎪
⎪
⎪
∂2u ⎪
⎪
⎪
=0 on y = 0 and y = b ⎭
∂y 2

We first solve the problem due to a concentrated load W0 at the point (ξ, η)
inside D so that w(x, y) = P δ(x − ξ) δ(y − η), where P is a constant.
Application of the double finite Fourier sine transform (10.5.7) to (10.5.15)–
(10.5.16) gives
 2      nπη 
4 m2 n 2 P mπξ
π + 2 ũs (m, n) = sin sin ,
a2 b D a b
or,    
P mπξ  nπη 
ũs (m, n) = sin sin , (10.5.17)
D π 4 ωmn
4 a b

© 2007 by Taylor & Francis Group, LLC

Finite Fourier Sine and Cosine Transforms 425

where ωmn is defined by (10.5.11).

The inverse transform gives the formal solution
 ∞  ∞     nπη 
4P −4 mπξ
u(x, y) = ω sin sin
π 4 ab D m=1 n=1 mn a b
 mπx   nπy 
× sin sin .(10.5.18)
a b
For an arbitrary load w(x, y) over the region α ≤ x ≤ β, γ ≤ y ≤ δ inside the
region D, we can replace P by w(ξ, η) dξdη and integrate over the rectangle
α ≤ ξ ≤ β, γ ≤ η ≤ δ.
Consequently, the formal solution is obtained from (10.5.18) and has the
form
⎧ β ⎫
  ∞  ∞ ⎨ δ    nπη  ⎬
4 mπξ
u(x, y) = w(ξ, η) sin sin dξdη
π 4 abD m=1 n=1 ⎩ a b ⎭
α γ
 mπx   nπy 
−4
× ωmn sin sin . (10.5.19)
a b

10.6 Exercises
1. Find the finite Fourier cosine transform of f (x) = x2 .
2. Use the result (10.3.2) to prove
a3  a 3
(a) Fs {x2 } = (−1)n+1 − 2 [1 + (−1)n+1 ],
nπ  nπ 
a4 6 1
(b) Fs {x3 } = (−1)n 2 − .
π n3 π 3 nπ
3. Solve the initial-boundary value problem in a finite domain
ut = κ uxx , 0 ≤ x ≤ a, t > 0,
u(x, 0) = 0 for 0 ≤ x ≤ a,
u(0, t) = f (t) for t > 0,
u(a, t) = 0 for t > 0.

4. Solve Exercise 3 above by replacing the only condition at x = a with the

radiation condition
ux + hu = 0 at x = a,

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426 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

where h is a constant.
5. Solve the heat conduction problem

ut = κ uxx , 0 ≤ x ≤ a, t > 0,

ux (0, t) = f (t)
for t > 0,
ux (a, t) + h u = 0
u(x, 0) = 0 for 0 ≤ x ≤ a.

6. Solve the diffusion equation (10.4.1) with the following boundary and
initial data

ux (0, t) = f (t), ux (a, t) = 0 for t > 0

u(x, 0) = 0 for 0 ≤ x ≤ a.

7. Solve the problem of free vibrations described in Example 10.4.4, when

the beam is at rest in its equilibrium position at time t = 0, and an
impulse I is applied at x = η, that is,
 
I
f (x) ≡ 0 and g(x) = δ(x − η).
ρα

8. Find the solution of the problem in Example 10.4.4 when

(i) W (x, t) = W0 φ(t) δ(x − η), 0 < η < ;
(ii) a concentrated applied load is moving along the beam with a con-
stant speed U , that is, W (x, t) = W0 φ(t)δ(x − U t) H(U t − ), where
W0 is a constant.
9. Find the solution of the forced vibration of an elastic string of finite
length  which satisfies the forced wave equation

1 ∂2u ∂2u
= 2 + F (x, t), 0 ≤ x ≤ , t > 0,
c2 ∂t2 ∂x
with the initial and boundary data

u(x, t) = f (x), ut = g(x) at t = 0 for 0 ≤ x ≤ ,

u(0, t) = 0 = u(, t) for t > 0.

Derive the solution for special cases when f (x) = 0 = g(x) with
(i) an arbitrary non-zero F (x, t), and
P (t)
(ii) F (x, t) = δ(x − a), 0 ≤ a ≤ , where T is a constant.
T

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Finite Fourier Sine and Cosine Transforms 427

10. For the finite Fourier sine transform defined over (0, π), show that
x  1
(a) Fs (π − x) = 3 [1 + (−1)n+1 ]
2 n
 
sinh a(π − x) n
(b) Fs = 2 , a = 0.
sinh aπ (n + a2 )
11. For the finite Fourier cosine transform defined over (0, π), show that
2π 2 π3
(a) Fc {(π − x)2 } = for n = 1, 2, . . . ; F s {(π − x) } = for n = 0.
n2 3
a sinh(aπ)
(b) Fc {cosh a(π − x)} = 2 for a = 0.
(n + a2 )
12. Use the finite Fourier sine transform to solve the problem of diffusion of
electricity along a cable of length a. The potential V (x, t) at any point
x of the cable of resistance R and capacitance C per unit length satisfies
the diffusion equation

Vt = κ Vxx , 0 ≤ x ≤ a, t > 0,

where κ = (RC)−1 and the boundary conditions (the ends of the cable
are earthed)
V (0, t) = 0 = V (a, t) for t > 0,
and the initial conditions
⎧   ⎫
⎪
⎪ 2V0 a⎪
⎪
⎨ x, 0≤x≤ ⎪ ⎪
a 2⎬
V (x, 0) =   ,
⎪
⎪ 2V0 a ⎪
⎪
⎪
⎩ (a − x), ⎪
≤ x ≤ a⎭
a 2
where V0 is a constant.
13. Establish the following results
 
d
(a) Fs {f1 (x) ∗ g1 (x)} = 2nf˜s (n)g̃s (n),
dx
⎡ x ⎤

2
(b) Fs ⎣ {f1 (u) ∗ g1 (u)} du⎦ = f˜s (n)g̃s (n).
n
0

14. If p is not necessarily an integer, we write

π π
f˜c (p) = f (x) cos px dx and f˜s (p) = f (x) sin px dx.
0 0

Show that, for any constant α,

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428 INTEGRAL TRANSFORMS and THEIR APPLICATIONS

(a) Fc {2 f (x) cos αx} = f˜c (n − α) + f˜c (n + α),

(b) Fc {2 f (x) sin αx} = f˜s (n + α) − f˜s (n − α),
(c) Fs {2 f (x) cos αx} = f˜s (n + α) + f˜s (n − α),
(d) Fs {2 f (x) sin αx} = f˜c (n − α) − f˜c (n + α).
15. Solve the problem in Example 10.5.2 for uniform load W0 over the region
α ≤ x ≤ β and γ ≤ y ≤ δ.
16. Solve the problem of free oscillations of a rectangular elastic plate of
density ρ bounded by the two parallel planes z = ±h. The deflection
u(x, y, t) satisfies the equation

D ∇4 u + ρh utt = 0, 0 ≤ x ≤ a, 0 ≤ y ≤ b, t > 0,

where the deflection and the bending moments are all zero at the edges.

© 2007 by Taylor & Francis Group, LLC