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Module 8: The Fourier Transform Methdos for PDEs

In the previous modules (Modules 5-7), the method of separation of variables was used

to obtain solutions of initial and boundary value problems for partial differential equa-

tions given over bounded spatial regions. The present module deals with partial differen-

tial equations defined over unbounded spatial regions. The mathematical tools used for

solving initial and boundary value problems over unbounded spatial regions are integral

transforms: The Fourier transform (FT), the Fourier sine transform (FST) and the Fourier

cosine transform (FCT).

The outline of this module is as follows. The first lecture introduces the necessary back-

ground for the definition of FT. The FST and FCT are introduced in the second lecture.

Applications of FT, FST and FCT for the three types of PDEs viz., heat equation, wave

equation and Laplace equation are described in third, fourth and fifth lectures, respec-

tively.

1

MODULE 8: THE FOURIER TRANSFORM METHDOS FOR PDES 2

Lecture 1 Fourier Transform

Recall that if f is a periodic function with period 2L on R then f has a Fourier Series(FS)

representation of the form

f(x) =1

2a0 +

∞∑n=1

(an cos(

nπx

L) + bn sin(

nπx

L)), (1)

where

an =1

L

∫ L

−Lf(x) cos(

nπx

L)dx, n = 0, 1, 2, . . .

and

bn =1

L

∫ L

−Lf(x) sin(

nπx

L)dx, n = 1, 2, . . .

Fourier series are powerful tools in treating various problems involving periodic functions.

Many practical problems do not involve periodic functions. Therefore, it is desirable to

generalize the method of Fourier series to include non-periodic functions. If f is not

periodic then we may regard it as periodic with an infinite period, i.e., we would like to

see what happens if we let L → ∞. We shall do this for reasons of motivation as well

as for making it plausible that for a non-periodic function, one should expect an integral

representation (Fourier integral) instead of Fourier series.

Set ωn = nπL . Then (1) can be rewritten as

f(x) =1

2a0 +

∞∑n=1

[an cos(ωnx) + bn sin(ωnx)].

=1

L

∫ L

−Lf(x)dx+

1

L

∞∑n=1

[cos(ωnx)

∫ L

−Lf(t) cos(ωnt)dt

+sin(ωnx)

∫ L

−Lf(t) sin(ωnt)dt

].

Note that

ωn+1 − ωn =(n+ 1)π

L− (n)π

L=π

L.

Setting ∆ω = ωn+1 − ωn = πL , we write the Fourier series in the form

f(x) =1

L

∫ L

−Lf(x)dx +

1

π

∞∑n=1

[cos(ωnx)∆ω

∫ L

−Lf(t) cos(ωnt)dt

+sin(ωnx)∆ω

∫ L

−Lf(t) sin(ωnt)dt

].


This representation is valid for any fixed L, arbitrary large, but finite. Letting L → ∞and assuming that the resulting nonperiodic function

f(x) = limL→∞

f(x)

is absolutely integrable over (−∞,∞), i.e.,∫ ∞

−∞|f(x)| <∞

it seems plausible that the infinite series (1) becomes an integral from 0 to ∞, i.e.,

f(x) =1

π

∫ ∞

0

[cos(ωx)

∫ ∞

−∞f(t) cos(ωt)dt+ sin(ωx)

∫ ∞

−∞f(t) sin(ωt)dt

]dω.

The above equation can be put in the form

f(x) =1

π

∫ ∞

0[A(ω) cos(ωx) +B(ω) sin(ωx)]dω, (2)

which is called the Fourier integral of f , where

A(ω) =

∫ ∞

−∞f(t) cos(ωt)dt, B(ω) =

∫ ∞

−∞f(t) sin(ωt)dt,

This motivates the following result.

THEOREM 1. If f is piecewise continuous in every bounded interval of R and∫ ∞

−∞|f(x)| <∞,

then f can be represented by a Fourier integral

f(x) =1

π

∫ ∞

0[A(ω) cos(ωx) +B(ω) sin(ωx)]dω, (3)

where

A(ω) =

∫ ∞

−∞f(t) cos(ωt)dt, B(ω) =

∫ ∞

−∞f(t) sin(ωt)dt, (4)

The Fourier integral of f converges to f if f is continuous. If f is discontinuous at a

point x the integral converges to

f(x+) + f(x−)

2

i.e., the average value of the left and right hand limits of f(x) at that point.


In order to motivate the definition of Fourier transform we first express complex form

of the Fourier integral as follows. Using the identity cos(a − b) = cos a cos b + sin a sin b,

we write the integral (3) as

f(x) =1

π

∫ ∞

0

[∫ ∞

−∞f(t) cos(ωx− ωt)dt

]dω. (5)

Since cos θ = eiθ+e−iθ

2 , we obtain

f(x) =1

2π

∫ ∞

0

∫ ∞

−∞f(t){ei(ωx−ωt) + e−i(ωx−ωt)}dtdω

=1

2π

∫ ∞

0

∫ ∞

−∞f(t)ei(ωx−ωt)dtdω +

1

2π

∫ ∞

0

∫ ∞

−∞f(t)e−i(ωx−ωt)dtdω

Replacing w by −w in the second term on the right hand side and adjusting limits from

−∞ to 0, we obtain

f(x) =1

2π

∫ ∞

−∞

∫ ∞

−∞f(t)ei(ωx−ωt)dtdω

=1√2π

∫ ∞

−∞eiωx

[1√2π

∫ ∞

−∞f(t)e−iωtdt

]dω,

which is the complex form of the Fourier integral of f . This leads to following pair of

transforms.

DEFINITION 2. (The Fourier Transform)

Let f : (−∞,∞) → R or C. The Fourier transform (FT) of f(x) is defined by

F(f)(ω) = f(ω) =1√2π

∫ ∞

−∞f(x)e−iωxdx (−∞ < ω <∞) (6)

provided this integral exists.

REMARK 3. Note that all functions may not have Fourier transform. For example, the

constant function C (C = 0), sinx, ex, x2 do not have FT. Only functions that tend to

zero sufficiently fast as |x| → ∞ (rapidly decreasing to zero functions) have FT.

DEFINITION 4. (The Inverse Fourier Transform)

The inverse Fourier transform of a function f(ω) (−∞ < ω <∞) is defined as

F−1(f) = f(x) =1√2π

∫ ∞

−∞f(ω)eiωxdω (−∞ < x <∞) (7)

provided this integral exists.

EXAMPLE 5. Find the FT of the function

f(x) =

{1 for |x| ≤ L,

0 for |x| > L.


Solution. Using the definition of FT, we have

f(ξ) =1√2π

∫ ∞

−∞f(x)e−iξxdx =

1√2π

∫ L

−Le−iξxdx =

1√2π

e−iξx

−iξ

∣∣∣∣L−L

=1√2π

e−iξL − eiξL

−iξ=

1√2π

2 sin(ξL)

ξ.

Note that even though f(x) vanishes for x outside the interval [−L,L], the same is not

true of f(ξ). In general, it can be shown that if f and f vanish outside [−L,L], thenf ≡ 0.

Some Basic Properties of Fourier Transforms:

• Linearity: F is a linear transformation. For any two functions f1 and f2 with FTs

F [f1] and F [f2], respectively and any constants c1 and c2, we have

F [c1f1 + c2f2] = c1F [f1] + c2F [f2].

• Conjugation: Let f(x) be a function with FT F [f ]. Then the FT of the function

f(x) (complex conjugate) is given by

F [f(x)] = F [f ](−ω).

• Continuity: Let f(x) be an absolutely integrable function with FT f(ω). Then

f(ω) is a continuous function.

• Convolution: We know the convolution of the functions f and g, denoted be f ∗ g,is defined by

(f ∗ g)(x) =∫ ∞

−∞f(x− t)g(t)dt,

provided the integral exists for each x. (i.e., if f is bounded and g is absolutely

integrable). Let f(ω) and g(ω) be the FTs of f and g, respectively. Then

F(f ∗ g) = F(f)F(g) = f(ω) g(ω).

Note: In general, F [f(x)g(x)] = F [f ]F [g].

• Parseval’s identity: For any two functions f(x) and g(x) with FTs f(ω) and g(ω),

respectively. Then ∫ ∞

−∞f(x)g(x)dx =

∫ ∞

−∞f(ω)g(ω)dω.

In particular, ∫ ∞

−∞|f(x)|2dx =

∫ ∞

−∞|f(ω)|2dω.


• Transformation of partial derivatives:

(i) Let u = u(x, t) be a function defined for −∞ < x <∞ and t ≥ 0. If u(x, t) → 0

as x→ ±∞, and F [u](ω, t) = U(ω, t), then

F [ux](ω, t) = iωF [u] = iωU(ω, t).

If, in addition, ux(x, t) → 0 as x→ ±∞, then

F [uxx](ω, t) = −ω2F [u] = −ω2U(ω, t).

(ii) If we transform the partial derivative ut(x, t) (and if the variable of integration

in the transformation is x), then the transformation is given by

F [ut](ω, t) =d

dt{F [u]}(ω, t) = d

dtU(ω, t).

The Fourier transform of a time derivative equals the time derivative of the Fourier

transform. This shows that time differentiation and the FT with respect to x com-

mute.

Practice Problems

1. Compute the complex FS for each of the function f(x) = eax cos(bx).

2. Show that if f(x) is absolutely integrable on (−∞,∞), then

F [eibxf(ax)] =1

aF [f((ξ − b)/a)], a, b ∈ R, a = 0.

3. Find the FT of

(A) f(x) = e−cx2, where c is a constant.

(B) f(x) = e−|x|

(C) f(x) = sinx2

4. Verify the following properties of FT:

(A) F [ux] = iξF [u]

(B) F [uxx] = −ξ2F [u]

(C) F [u(x+ c)] = eicξF [u], for any c ∈ R

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