In civil engineering, analyzing and solving boundary value problems—especially those involving heat transfer, wave motion, or vibrations—often requires converting a function from the spatial domain to a frequency domain. Fourier transforms are indispensable in this context. However, when the problem is defined on a semi-infinite domain (e.g., x ≥ 0), Fourier Cosine and Sine Transforms are preferred over the general Fourier transform. These transforms allow decomposition of functions into orthogonal sine and cosine basis functions, effectively handling problems with specific boundary conditions.

For a function f(x) defined on [0,∞), the Fourier Cosine Transform is defined as:

r2 Z ∞
F (s)= f(x)cos(sx)dx
c π
0
Where:
• f(x) is piecewise continuous on every finite interval in [0,∞),
• f(x) is absolutely integrable on [0,∞),
• s is the transform variable (frequency domain variable).

This restores the original function from its cosine transform.

r2 Z ∞
f(x)= F (s)cos(sx)ds
π c
0

1. Linearity:
   F {af(x)+bg(x)}=aF {f(x)}+bF {g(x)}
   c c c
2. Scaling:
   F {f(ax)}= F , a>0
   c a c a
3. Differentiation: If f(x) is differentiable and vanishes as x→∞,
   r2 Z ∞
   F {f′(x)}=−s f(x)sin(sx)dx
   c π
   0
4. Parseval’s Identity:
   Z ∞ Z ∞
   f(x)2dx= F (s)2ds
   c
   0 0

Example 1: Fourier Cosine Transform of f(x)=e−ax, where a>0

r2 Z ∞ r2 a
F (s)= e−axcos(sx)dx= ·
c π π a2+s2
0

For a function f(x) defined on [0,∞), the Fourier Sine Transform is defined as:

r2 Z ∞
F (s)= f(x)sin(sx)dx
s π
0
Inverse Fourier Sine Transform
r2 Z ∞
f(x)= F (s)sin(sx)ds
π s
0

1. Linearity:
   F {af(x)+bg(x)}=aF {f(x)}+bF {g(x)}
   s s s
2. Scaling:
   2
   1 (cid:16)s(cid:17)
   F {f(ax)}= F , a>0
   s a s a
3. Differentiation: If f(x) is differentiable and vanishes as x→∞,
   r2 Z ∞
   F {f′(x)}=s f(x)cos(sx)dx
   s π
   0
4. Parseval’s Identity:
   Z ∞ Z ∞
   f(x)2dx= F (s)2ds
   s
   0 0

Example 2: Fourier Sine Transform of f(x)=e−ax, where a>0

r2 Z ∞ r2 s
F (s)= e−axsin(sx)dx= ·
s π π a2+s2
0

Fourier sine and cosine transforms are applied in the following civil engineering contexts:
• Heat Conduction in Semi-Infinite Slabs: Boundary conditions at one end (e.g., insulated or fixed temperature) often lead to cosine or sine transform solutions.
• Deflection of Beams with One Fixed End: Solving beam bending equations using Fourier cosine transforms when the displacement or slope is known at one end.
• Wave Propagation in Strings or Rods: For rods fixed at one end and free at the other, sine transforms help solve the partial differential equations.

The full Fourier transform is defined over (−∞,∞), but for even and odd extensions of functions defined only on [0,∞), the sine and cosine transforms correspond respectively:
• If f(x) is even:
F{f(x)}↔Cosine Transform
• If f(x) is odd:
F{f(x)}↔Sine Transform
Thus, these transforms are not just computational tricks—they carry deep connections to symmetry properties and domain constraints of physical systems.

Fourier Cosine Fourier Sine Transform
Function f(x) Transform F (s) F (s)
c s
q q
1 2 · 1 2 · 1
π s π s
q q
e−ax 2 · a 2 · s
π a2+s2 π a2+s2
q q
x 2 · s 2 · 2a
π s2 π (a2+s2)2
q q
sin(ax) 2 · a 2 · s
π s2+a2 π s2+a2

Many engineering problems reduce to solving partial differential equations (PDEs) with boundary conditions. Fourier sine and cosine transforms are highly effective in solving such problems, particularly when domains are semi-infinite.

Consider a semi-infinite rod x ∈ [0,∞) initially at zero temperature, and for t > 0, the end x = 0 is held at a constant temperature T. The governing equation is the heat conduction equation:
∂u ∂2u =α2 , x>0, t>0
∂t ∂x2
Subject to:
u(0,t)=T , u(x,0)=0, lim u(x,t)=0
0 x→∞
Solution Method:
We take the Fourier Cosine Transform with respect to x:
Let U(s,t)=F {u(x,t)}= 2 R∞ u(x,t)cos(sx)dx
c π 0
Using properties of derivatives under transforms:
(cid:26) ∂2u(cid:27)
F =−s2U(s,t)
c ∂x2
∂U
⇒ =−α2s2U(s,t)
∂t
This is a first-order linear ODE in t with solution:
U(s,t)=A(s)e−α2s2t
From initial condition u(x,0)=0⇒U(s,0)=0⇒A(s)=0, but this contradicts the boundary condition. To fix this, we transform the nonhomogeneous boundary via suitable substitution (e.g., Duhamel’s principle or separation of variables), which leads to:
(cid:18) (cid:19)
x
u(x,t)=T erfc √
0 2α t
Where erfc(z) is the complementary error function.

A cantilever beam of length L, fixed at x = 0, subjected to a load q(x). The Euler-Bernoulli beam equation is:
d4y q(x)
=
dx4 EI
With boundary conditions (fixed end at x=0):
y(0)=0, y′(0)=0
We apply the Fourier Cosine Transform to both sides:
Let Y(s)=F {y(x)}, then:
c
5
F =s4Y(s)
dx4
Thus,
1 1
s4Y(s)= F {q(x)} ⇒ Y(s)= F {q(x)}
EI c EIs4 c
Taking the inverse transform yields the deflection y(x).

Now consider problems where the solution vanishes at the boundary x = 0, which is ideal for Fourier Sine Transform.

Let us consider the one-dimensional wave equation:
∂2u ∂2u
=c2
∂t2 ∂x2
With boundary conditions:
∂u
u(0,t)=0, lim u(x,t)=0, u(x,0)=f(x), (x,0)=0
x→∞ ∂t
Apply the Fourier Sine Transform:
∂2U
U(s,t)=F {u(x,t)}⇒ =−c2s2U
s ∂t2
This is a second-order ODE in t:
⇒U(s,t)=A(s)cos(cst)+B(s)sin(cst)
Initial conditions give:
(cid:12)
U(s,0)=F s{f(x)}=A(s), ∂ ∂U
t
(cid:12)
(cid:12)
(cid:12)
=0⇒B(s)=0
t=0
Thus,
U(s,t)=F {f(x)}·cos(cst)⇒u(x,t)= F {f(x)}cos(cst)sin(sx)ds
s π s
0
This integral solution gives the displacement of a vibrating rod fixed at one end.

Another practical use of Fourier sine and cosine transforms is evaluating improper integrals.
Example 3: Evaluate R∞ xsin(ax)dx
0 x2+b2
Let:
f(x)=⇒FST of f(x)= dx
x2+b2 π x2+b2
0
Using integral tables or residue calculus:
Z ∞ xsin(ax) π
dx= e−ab
x2+b2 2
0
Hence, Fourier transforms provide a powerful way to evaluate such definite integrals analytically.

Knowing how to handle derivatives within the transforms is crucial for PDEs.
1. Cosine Transform of First Derivative
F {f′(x)}=−sF {f(x)}
c s
2. Sine Transform of First Derivative
r2
F {f′(x)}=sF {f(x)}− f(0)
s c π
These relations help simplify many boundary-value problems in civil engineering (e.g., thermal gradient, slope in beam deflection).