Group #9
Yizhi Hong
Jiaqi Zhang
Nicholas Zentay
Sagar Lonkar
17.1 Introduction
Main Work:
Théorie analytique de la chaleur
(The Analytic Theory of Heat)
• Any function of a variable, whether continuous or
discontinuous, can be expanded in a series of
sines of multiples of the variable (Incorrect)
• The concept of dimensional homogeneity in
equations
• Proposal of his partial differential equation for
conductive diffusion of heat
Jean Baptiste Joseph Fourier
(Mar21st 1768 –May16th 1830)
Discovery of the "greenhouse effect"
French mathematician, physicist
http://en.wikipedia.org/wiki/Joseph_Fourier
even function: 𝑓 −𝑥 = 𝑓 𝑥
odd function: 𝑓 −𝑥 = −𝑓 𝑥
some characteristic:
even+even=even
even*even=even
odd+odd=odd
odd*odd=even
even*odd=odd
If f is even, then
If f is odd, then
𝐴
𝐴
𝑓 𝑥 𝑑𝑥 = 2 0 𝑓
−𝐴
𝐴
𝑓 𝑥 𝑑𝑥 = 0
−𝐴
𝑥 𝑑𝑥
Every function can be uniquely decomposed into the
sum of an even function, say fe, and an odd function, say
fo.
𝑓 𝑥 + 𝑓(−𝑥) 𝑓 𝑥 − 𝑓(−𝑥)
𝑓 𝑥 =
+
2
2
= 𝑓𝑒 𝑥 + 𝑓𝑜 (𝑥)
Periodic function
𝑓 𝑥 + 𝑇 = 𝑓 𝑥 for every x in the domain of f.
Then we say that f is a periodic function of x, with period
T. And f is T-periodic.
Examples:
Even function : Cosine function i.e. cos(θ)
Odd function: Sine function i.e. sin(θ)
Periodic Function: Both sine and cosine functions are
periodic with a period of 2 π
If f(x) is periodic, of period 2l, then we define the
Fourier series of f, say FS f, as
∞
𝐹𝑆𝑓 𝑥 = 𝑎0 +
𝑛=1
𝑎𝑛 cos
𝑛𝜋𝑥
𝐿
+ 𝑏𝑛 sin
𝑛𝜋𝑥
𝐿
Where the coefficients are given by the Euler formulas,
𝑎𝑜 =
𝑎𝑛 =
𝑏𝑛 =
1 𝐿
𝑓 𝑥 𝑑𝑥
−𝐿
2𝐿
1 𝐿
𝑛𝜋𝑥
𝑓 𝑥 cos
𝑑𝑥
𝐿 −𝐿
𝐿
1 𝐿
𝑛𝜋𝑥
𝑓 𝑥 sin
𝑑𝑥
𝐿 −𝐿
𝐿
For FS f to represent f we need the series to converge,
and we need its sum function to be the same as the
original function f(x).
Let f be 2L-periodic, and let f and f ‘ be piecewise
continuous on [-L,L]. Then the Fourier series converges
to f(x) at every point x at with f is continuous, and to the
mean value [f(x+)+f(x-)]/2 at every point x at which f is
discontinuous.
𝑓 𝑥 + ≡ lim 𝑓(𝑥 + ℎ)
ℎ→0
𝑓 𝑥 − ≡ lim 𝑓(𝑥 − ℎ)
ℎ→0
where h→0 through positive values. If f(x+)=f(x-)=f(x),
then f is continuous at x, otherwise it is discontinuous.
Even and odd
∞
𝑓 𝑥 = 𝑎0 +
𝑛𝜋𝑥
𝑎𝑛 cos
𝐿
𝑛=1
∞
+
𝑛=1
𝑏𝑛 sin
= 𝑓𝑒 𝑥 + 𝑓𝑜 (𝑥)
∞
And
𝑓𝑒 𝑥 = 𝑎0 +
∞
𝑓𝑜 (𝑥) =
𝑛=1
𝑛𝜋𝑥
𝑎𝑛 cos
𝐿
𝑛=1
𝑛𝜋𝑥
𝑏𝑛 sin
𝐿
𝑛𝜋𝑥
𝐿
The results:
𝐿
𝑚𝜋𝑥
cos
−𝐿
𝐿
cos
𝐿
𝑚𝜋𝑥
sin
−𝐿
𝐿
sin
𝐿
𝑚𝜋𝑥
cos
−𝐿
𝐿
sin
𝑛𝜋𝑥
𝐿
𝑛𝜋𝑥
𝐿
𝑛𝜋𝑥
𝐿
𝑑𝑥 =
𝑑𝑥 =
0
L
2L
0
L
m≠n
m=n≠0
M=n=0
m≠n
m=n≠0
𝑑𝑥 = 0 for all m, n, where m and n are integers.
For the reason that:
𝑐𝑜𝑠 𝑐1 𝑥 𝑐𝑜𝑠 𝑐2 𝑥𝑑𝑥 =
|c1|≠|c2|.
sin
2
𝑐1 −𝑐2 𝑥
𝑐1 −𝑐2
+
sin 𝑐1 +𝑐2 𝑥
2 𝑐1 +𝑐2
1
𝑐𝑜𝑠 𝑐𝑥 𝑑𝑥 = + sin 𝑐𝑥 cos 𝑐𝑥
2𝑐
sin 𝑐1 −𝑐2 𝑥
sin 𝑐1 +𝑐2 𝑥
𝑠𝑖𝑛 𝑐1 𝑥 𝑠𝑖𝑛 𝑐2 𝑥𝑑𝑥 =
−
2 𝑐1 −𝑐2
2 𝑐1 +𝑐2
2
|c1|≠|c2|.
2
𝑠𝑖𝑛 𝑐𝑥 𝑑𝑥 =
for
𝑥
2
𝑥
2
−
𝑠𝑖𝑛 𝑐1 𝑥 𝑐𝑜𝑠 𝑐2 𝑥𝑑𝑥
|c1|≠|c2|.
1
sin 𝑐𝑥 cos 𝑐𝑥
2𝑐
cos 𝑐1 +𝑐2 𝑥
=−
2 𝑐1 +𝑐2
𝑠𝑖𝑛 𝑐𝑥 𝑐𝑜𝑠 𝑐𝑥 𝑑𝑥 =
1
𝑠𝑖𝑛2
2𝑐
𝑐𝑥
−
for
cos 𝑐1 −𝑐2 𝑥
2 𝑐1 −𝑐2
for
For definiteness, let us solve FS f for a2.
𝐿
𝑓 𝑥 cos
−𝐿
2𝜋𝑥
𝑑𝑥 =
𝐿
∞
𝐿
𝑎0 +
−𝐿
𝐿
𝑛=1
𝑛𝜋𝑥
𝑛𝜋𝑥
+ 𝑏𝑛 sin
𝐿
𝐿
cos
2𝜋𝑥
𝑑𝑥
𝐿
𝐿
2𝜋𝑥
𝜋𝑥
2𝜋𝑥
𝜋𝑥
2𝜋𝑥
= 𝑎0
cos
𝑑𝑥 + 𝑎1
cos cos
𝑑𝑥 + 𝑏1
sin cos
𝑑𝑥
𝐿
𝐿
𝐿
𝐿
𝐿
−𝐿
−𝐿
−𝐿
𝐿
𝐿
2𝜋𝑥
2𝜋𝑥
2𝜋𝑥
2𝜋𝑥
+ 𝑎2
cos
cos
𝑑𝑥 + 𝑏2
sin
cos
𝑑𝑥
𝐿
𝐿
𝐿
𝐿
−𝐿
−𝐿
𝐿
3𝜋𝑥
2𝜋𝑥
+ 𝑎3
cos
cos
𝑑𝑥 + ⋯
𝐿
𝐿
−𝐿
= 0 + 0 + 0 + 𝑎2 𝐿 + 0 + ⋯
= 𝑎2 𝐿
So
𝑎2 =
𝐿
𝑎𝑛 cos
1 𝐿
𝑓
𝐿 −𝐿
𝑥 cos
2𝜋𝑥
𝑑𝑥
𝐿
Example:
The function given by:
y=-x
- π≤x≤0
y=x
0≤x≤ π
The period of the above function is 2π.
Thus 2l = 2π
Therefore l= π
1
𝑎𝑜 = 2𝐿
1
𝑎𝑛 = 𝐿
1
𝑏𝑛 = 𝐿
𝐿
1 0
1 𝜋
𝜋
𝑓
𝑥
𝑑𝑥
=
−𝑥𝑑𝑥
+
𝑥𝑑𝑥
=
−𝐿
2𝜋 −𝜋
2𝜋 0
2
𝐿
0
𝑛𝜋𝑥
1
𝑛𝜋𝑥
1 π
𝑛𝜋𝑥
2 (−1)𝑛 −1
𝑓 𝑥 cos 𝐿 𝑑𝑥 = π − π −𝑥 cos 𝐿 𝑑𝑥 + π 0 𝑥 cos 𝐿 𝑑𝑥 =π 𝑛2
−𝐿
𝐿
𝑛𝜋𝑥
1 0
𝑛𝜋𝑥
1 π
𝑛𝜋𝑥
𝑓
𝑥
sin
𝑑𝑥
=
−𝑥
sin
𝑑𝑥
+
𝑥
sin
𝑑𝑥 = 0
−𝐿
𝐿
π −π
𝐿
π 0
𝐿
∞
𝐹𝑆𝑓 𝑥 = 𝑎0 +
n=0
n=1
n=2
n=3
n=4
n=5
𝑛𝜋𝑥
𝑎𝑛 cos 𝐿
𝑛=1
=
𝜋
2
∞
+
2 (−1)𝑛 −1
cos( nx)
𝑛2
𝑛=1 π
𝜋
2
𝜋
4
FS f = − cos(𝑥)
2
𝜋
𝜋
4
FS f = 2 − 𝜋 cos(𝑥)
𝜋
4
4
FS f = 2 − 𝜋 cos 𝑥 − 9𝜋 cos(3𝑥)
𝜋
4
4
FS f = 2 − 𝜋 cos 𝑥 − 9𝜋 cos(3𝑥)
𝜋
4
4
4
FS f = 2 − 𝜋 cos 𝑥 − 9𝜋 cos 3𝑥 − 25𝜋 cos(5𝑥)
FS f =
s0
s1
s3
s5
k
m
F(t)
m = mass
c = damping factor
k = spring constant
F(t) = 2L- periodic forcing function
mx’’(t) + cx’(t) + k x(t) = F(t)
http://www.jirka.org/diffyqs/
Differential Equations for Engineers
The particular solution xp of the above equation is periodic
with the same period as F(t) .
The coefficients are k=2, and m=1 and c=0 (for simplicity).
The units are the mks units (meters-kilograms- seconds).
There is a jetpack strapped to the mass, which fires with a
force of 1 newton for 1 second and then is off for 1 second, and
so on. We want to find the steady periodic solution.
The equation is:
x’’ + 2x = F(t)
Where F(t) =>
0 if
1 if
-1<t<0
0<t<1
We expand F(t) in the Fourier series
𝑎𝑜 =
𝑎𝑛 =
𝑏𝑛 =
1
2
1
1
1
1
1
1
1𝑑𝑥 =
0
2
1
1 ∗ cos(n𝜋t) 𝑑𝑡 = 0
0
1
1−cos(𝑛𝜋)
1 ∗ sin(n𝜋t) 𝑑𝑡 =
0
𝑛
1
2
𝐹𝑆𝑓 𝑥 = +
∞
1−cos(𝑛𝜋)
sin(𝑛𝜋𝑡)
𝑛
𝑛=1
=
1
2
∞
+
2
sin(𝑛𝜋𝑡)
𝜋𝑛
𝑛=1,3,5..
To find the particular solution we take
x = A cos(n𝜋t) + Bsin(n𝜋t)
x’ = - n𝜋Asin(n𝜋t)+ n𝜋Bcos(n𝜋t)
x’’ = - n2𝜋2x
x’’ + 2x = sin(n𝜋t)
(2- n2𝜋2)x = sin(n𝜋t)
x=
sin(n𝜋t)
(2− n2𝜋2)
Particular solution corresponding to ½ is given by
(2- n2𝜋2)x = ½ at n=0
Thus x = ¼
Thus the desired steady state response is, by linearity and
superposition,
1
4
xp = +
∞
2
2 2 sin(𝑛𝜋𝑡)
𝜋𝑛(2−
n
𝜋)
𝑛=1,3,5..
Plot of steady periodic solution thus obtained is:
Using the definitions
𝑒 𝑖𝜃 +𝑒 −𝑖𝜃
2
𝑒 𝑖𝜃 −𝑒 −𝑖𝜃
2𝑖
𝑐𝑜𝑠𝜃 =
and 𝑠𝑖𝑛𝜃 =
it is possible to re-express the Fourier series
formula in terms of complex exponentials, as
follows:
𝑖𝑛𝜋𝑥 𝑙
FS 𝑓 = ∞
𝑐
𝑒
𝑛=−∞ 𝑛
where
𝑐𝑛 =
1 𝑙
−𝑖𝑛𝜋𝑥 𝑙
𝑓(𝑥)𝑒
𝑑𝑥
2𝑙 −𝑙
Although the cn's and exponentials in FS f are
complex, the series does have a real-valued sum.
The usual definition
∞
𝑁
𝐴𝑛 ≡ lim
𝑁→∞
𝑛=1
𝐴𝑛
𝑛=1
does not apply because the lower limit is infinite as
well. It can be shown that the appropriate meaning
of the above series is
∞
𝑁
𝑐𝑛 𝑒 𝑖𝑛𝜋𝑥
𝑛=−∞
𝑙
𝑐𝑛 𝑒 𝑖𝑛𝜋𝑥
≡ lim
𝑁→∞
𝑛=−𝑁
𝑙
With this we can proceed: FS 𝑓 = 𝑎0 +
𝑛𝜋𝑥
𝑛𝜋𝑥
∞
+ 𝑏𝑛 sin
𝑛=1 𝑎𝑛 cos
𝑙
= lim
𝑁→∞
= lim
𝑁→∞
=
lim
𝑁→∞
𝑙
𝑛𝜋𝑥
𝑁
𝑛=0 𝑎𝑛 cos 𝑙
𝑁
𝑛=0
𝑎𝑛 −𝑖𝑏𝑛
𝑁
𝑛=0
2
𝑎𝑛
+
−
𝑒 𝑖𝑛𝜋𝑥 𝑙 +𝑒
𝑒 𝑖𝑛𝜋𝑥 𝑙
2
+
𝑛𝜋𝑥
𝑏𝑛 sin
𝑙
𝑖𝑛𝜋𝑥
𝑙
+ 𝑏𝑛
𝑎𝑛 +𝑖𝑏𝑛
𝑁
𝑛=0
2
−
𝑒 𝑖𝑛𝜋𝑥 𝑙 −𝑒
2𝑖
𝑒 −𝑖𝑛𝜋𝑥
Changing n to -n for the second sum gives:
FS 𝑓 =
𝑎𝑛 −𝑖𝑏𝑛
𝑁
−𝑁 𝑎−𝑛 +𝑖𝑏−𝑛
𝑖𝑛𝜋𝑥
𝑙
lim 𝑛=0
𝑒
+ 𝑛=0
𝑒 𝑖𝑛𝜋𝑥
2
2
𝑁→∞ 𝑁
𝑖𝑛𝜋𝑥
𝑙
lim 𝑛=−𝑁 𝑐𝑛 𝑒
𝑁→∞
=
∞
𝑖𝑛𝜋𝑥 𝑙
𝑛=−∞ 𝑐𝑛 𝑒
𝑖𝑛𝜋𝑥
𝑙
𝑙
𝑙
=
For n = 0,
𝑐0 = 𝑎0 =
1 𝑙
𝑓
2𝑙 −𝑙
𝑥 𝑑𝑥,
For n > 0,
𝑎𝑛 −𝑖𝑏𝑛
𝑐𝑛 =
=
=
1 𝑙
𝑓
2𝑙 −𝑙
2
𝑥 𝑒
1 𝑙
𝑓
2𝑙 −𝑙
−𝑖𝑛𝜋𝑥 𝑙
𝑥
𝑛𝜋𝑥
cos
𝑙
𝑛𝜋𝑥
− 𝑖 sin
𝑙
𝑑𝑥,
For n < 0,
𝑎−𝑛 +𝑖𝑏−𝑛
=
2
1 𝑙
𝑛𝜋𝑥
𝑓 𝑥 cos −
2𝑙 −𝑙
𝑙
𝑐𝑛 =
=
1 𝑙
𝑓
2𝑙 −𝑙
𝑥 𝑒 −𝑖𝑛𝜋𝑥 𝑙 𝑑𝑥
+ 𝑖 sin
𝑛𝜋𝑥
−
𝑙
𝑑𝑥
𝑑𝑥
Example: Find the Fourier series for the function defined by
f ( x)  e x (  x   )
Solution:
Where
cn ( f ) 
an ( f ) 
bn ( f ) 
1
2
1




f ( x)e  inx dx ( n  Z )
f ( x) cos nxdx




1




(n  N )
f ( x) sin nxdx (n  Z  )
Reference: Fourier Analysis (Author: Eric State, Pure and Applied Mathematics: a
Wiley-Interscience Series of Texts, Monographs, and Tracts ) P 11
We’ll compute the cn(f) first, we get
So
We also have
and
so
It often happens in applications, especially when we solve
partial differential equations by the method of separation of
variables, that we need to expand a given function f in a
Fourier series, where f is defined only on a finite interval.
We define an “extended function”, say fext, so that fext is
periodic in the domain of -∞< x < ∞, and fext=f(x) on the
original interval 0<x<L. There can be infinite number of such
extensions.
Four extensions: half- and quarter- range cosine and sine
extensions, which are based on symmetry or antisymmetry
about the endpoints x=0 and x=L.
fext is symmetric about x=0 and
also about x=L. Because of its
symmetry about x=0, fext is an
even function, and its Fourier
series will contain only cosines, no sines. Further, its period is
2L, so L is half the period.
∞
𝐹𝑆 𝑓 𝑥 = 𝑎0 +
1
𝑎𝑜 =
𝐿
𝐿
0
𝑛𝜋𝑥
𝑎𝑛 cos
𝐿
𝑛=1
, ( 0< x < L)
2
𝑓 𝑥 𝑑𝑥 𝑎𝑛 =
𝐿
𝐿
0
𝑛𝜋𝑥
𝑓 𝑥 cos
𝑑𝑥
𝐿
Proof:
For the half-range cosine case the period is 2L, and
𝑎𝑜 =
𝑎𝑛 =
𝑏𝑛 =
1 𝐿
1 𝐿
𝑓𝑒𝑥𝑡 𝑥 𝑑𝑥 = 0 𝑓 𝑥 𝑑𝑥,
−𝐿
2𝐿
𝐿
1 𝐿
𝑛𝜋𝑥
2 𝐿
𝑓 𝑥 cos
𝑑𝑥 = 0 𝑓
𝐿 −𝐿 𝑒𝑥𝑡
𝐿
𝐿
1 𝐿
𝑛𝜋𝑥
𝑓𝑒𝑥𝑡 𝑥 sin
𝑑𝑥 = 0.
−𝐿
𝐿
𝐿
𝑥
𝑛𝜋𝑥
cos
𝑑𝑥,
𝐿
fext is antisymmetric about
x=0 and x=L, the period is
2L, and we have the
half-range sine extension.
∞
𝐹𝑆 𝑓 𝑥 =
𝑛𝜋𝑥
𝑏𝑛 sin
𝐿
𝑛=1
𝐿
2
𝑏𝑛 =
𝐿
0
, ( 0< x < L)
𝑛𝜋𝑥
𝑓 𝑥 sin
𝑑𝑥
𝐿
fext is symmetric about
x=0 and anti-symmetric
about x=L, the period is
4L (so L is only a quarter
of the period), and we have the quarter-range cosine
extension.
∞
𝑛𝜋𝑥
𝑓 𝑥 =
𝑎𝑛 cos
2𝐿
𝑛=1,3,…
2 𝐿
𝑛𝜋𝑥
𝑎𝑛 = 0 𝑓 𝑥 cos
𝑑𝑥.
𝐿
2𝐿
, ( 0< x < L)
fext is anti-symmetric about
x=0 and symmetric about
x=L, the period is, and we
have the quarter-range sines extension.
∞
𝐹𝑆 𝑓 𝑥 =
𝑏𝑛 =
2 𝐿
𝑓
𝐿 0
𝑥
𝑛𝜋𝑥
𝑏𝑛 sin
2𝐿
𝑛=1,3,…
𝑛𝜋𝑥
sin
𝑑𝑥.
2𝐿
, ( 0< x < L)
Example: F(x) = sin(x)
L= π
HRC:
π
2
sin(𝑥)𝑑𝑥
=
=
0
π
2 π
sin(𝑥) cos(nx) 𝑑𝑥
π 0
0<x<π
1
𝑎𝑜 = π
𝑎𝑛 =
𝐹𝑆 𝑓 𝑥 =
𝜋)
2
+
π
2
=π
π 1
( (sin
0 2
2 1+cos(𝑛π)
n2−1
𝑥 + 𝑛𝑥 + sin 𝑥 − 𝑛𝑥 )𝑑𝑥 = − π
∞
2 1+cos(𝑛π)
− π n2−1 cos(nx)
𝑛=1
=
2
4
−
π
π
∞
cos 𝑛x
𝑛=2,4,6.. n2−1
( 0< x <
HRS:
∞
𝐹𝑆 𝑓 𝑥 =
𝑛=1
2
𝑏𝑛 = π
𝑏1 = 1
𝑏𝑛 sin
𝑛𝜋𝑥
𝐿
π
sin(𝑥) sin(nx) 𝑑𝑥
0
𝐹𝑆 𝑓 𝑥 = sin(x)
, ( 0< x < L)
= 0 for n>1
( 0< x < 𝜋)
QRC:
∞
𝑓 𝑥 =
𝑛=1,3,…
𝑎𝑛 =
𝑎𝑛 cos
2 𝜋
𝑛𝑥
sin(𝑥)
cos
𝑑𝑥
𝜋 0
2
∞
𝑓 𝑥 =
𝑛=1,3,…
< 𝜋)
𝑛𝜋𝑥
2𝐿
=
, ( 0< x < L)
2 π 1
( (sin
π 0 2
𝑥
𝑛𝑥
+ 2
𝑛𝜋
8 cos 2 +1
𝑛𝑥
− 𝜋 n2−4 cos 2
=
8
𝜋
𝑛𝜋
)𝑑𝑥
8 cos 2 +1
=− 𝜋 n2−4
1
𝑛𝑥
cos
2
𝑛=1,3,… 4−n2
( 0< x
+ sin 𝑥
𝑛𝑥
− 2
∞
QRS:
∞
𝐹𝑆 𝑓 𝑥 =
𝑛=1,3,…
𝑏𝑛 =
𝑏𝑛 sin
2 𝜋
𝑛𝑥
sin(𝑥)
sin
𝑑𝑥
𝜋 0
2
∞
𝑓 𝑥 =
8
𝜋
𝑛=1,3,…
𝑛𝜋𝑥
2𝐿
=
𝑛𝜋
sin 2
4−n2
, ( 0< x < L)
2 π 1
( (cos
π 0 2
𝑛𝑥
sin( 2 )
𝑥−
𝑛𝑥
2
− cos 𝑥 +
( 0< x < 𝜋)
𝑛𝑥
2
𝑛𝜋
)𝑑𝑥 =
8 sin 2
𝜋 4−n2
Uniform convergence
THEOREM 17.5.1Weierstrass M-Test
If ∞
𝑛=1 𝑀𝑛 is a convergent series of positive constants
and |𝑎𝑛 |≤𝑀𝑛 on an x interval I, then ∞
𝑛=1 𝑎𝑛 is
uniformly (and absolutely) convergent on I.
THEOREM 17.5.2Termwise Differentiation of Series
Let
𝑑
𝑑𝑥
∞
𝑛=1 𝑎𝑛 converge on an x interval I. Then
∞
∞
𝑑
𝑎
(𝑥)
=
𝑎𝑛 (𝑥), if the series on
𝑛=1 𝑛
𝑛=0 𝑑𝑥
right converges uniformly on I.
the
THEOREM 17.5.3 Uniqueness of Trigonometric Series
∞
If 𝑎0 +
𝐴0 +
𝑛𝜋𝑥
𝑛𝜋𝑥
+ 𝑏𝑛 sin
𝐿
𝐿
𝑛=1
∞
𝑛𝜋𝑥
𝑛𝜋𝑥
𝐴𝑛 cos
+ 𝐵𝑛 sin
,
𝐿
𝐿
𝑛=1
𝑎𝑛 cos
=
Where the trigonometric series on the left- and right-hand
sides converge to the same sum for all x, then a0=A0, an=An,
and bn=Bn for each n.
THEOREM 17.5.4Termwise Integration of Fourier Series
If a Fourier series is integrated termwise between any finite
limits, the resulting series converges to the integral of the
periodic function corresponding to the original series.
Example:
Verify the convergence of the series on the given interval:
∞
𝑛=1
exp −𝑛𝑥 sin(𝑛𝑥) on 2<x<5
By Weierstrass M-test:
an(x) = exp −𝑛𝑥 sin(𝑛𝑥) ≤ exp −𝑛𝑥 ≤ exp −2𝑛 on 2<x<5
∞
𝑛=1 exp −2𝑛 is a convergent geometric series.
Therefore the given series
on the given interval.
∞
𝑛=1
exp −𝑛𝑥 sin(𝑛𝑥) is a convergent
Some definitions:
 Function spaceCp[a,b] of all real-valued piecewise-continuous
functions defined on [a,b].
f=f(x) and g=g(x) be any two functions in Cp[a,b], and let α be
any (real) scalar.
f + g ≡f(x) + g(x), αf≡α f(x).
Observe that if f and g are piecewise continuous on [a,b] then
f+gand αfare also piecewise continuous, so Cp[a,b] is closed under
vector addition and scalar multiplication.

 We define the zero vector 0 as the function which is identically
zero, so that f + 0 = f(x) + 0 = f
 We define the negative inverse of f = f(x) as –f≡ -f(x), in
which case we have f + (-f) = f(x) + [-f(x)]=0 = 0
 Inner product for Cp[a,b], the inner product of f and g as
< 𝑓, 𝑔 >≡
𝑏
𝑓
𝑎
𝑥 𝑔 𝑥 𝑑𝑥.
f and g are orthogonal if <f,g>=0, that the norm ||f|| is
defined as ||𝐟|| = < f, f >, and f is said to be normalized if
||f||=1.
 { e1 , e2 ,…, en } is ON (orthonormal ) set in S, then the best
approximation of f within span { e , e ,…, e } is given by the
orthogonal projection of f onto span { e1 , e2 ,…, en } , namely, by
1
2
n
To apply these results to Fourier series, let S be Cp[a,b], with
the inner product and norm defined above, let a=-L and b=L,
and consider the vectors
e1=1, e2=cos
𝜋𝑥
𝜋𝑥
, e3=sin ,
𝐿
𝐿
… , e2k=cos
𝑘𝜋𝑥
𝑘𝜋𝑥
, e2k+1=sin
𝐿
𝐿
Cp[-L,L].
{ e1 , e2 ,…, en } is orthogonal
by virtue of the inner product
definition and the integrals:
in
So that the normalized en’s are:
Thus we can approximate a givenf=f(x), in Cp[a,b], in the
form:
Equivalent, we can write
If f(x) is piecewise continuous on [-L,L] and a0, a1, b1, a2, b2, …
are the Fourier coefficients, then
∞
𝐹𝑆𝑓 𝑥 = 𝑎0 +
𝑛=1
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑎𝑛 cos
+ 𝑏𝑛 sin
𝐿
𝐿
Holds in the sense of vector (least-square) convergence,
namely,
2
k
L 
n x
n x  

lim   f  x   a0    an cos
 bn sin
  dx  0
k   L
L
L 
n 1 

Example:
Find the error for the following function:
F(x) = |𝑥| defined on the interval –π<x<π
||E||2=
𝑙
2
𝑓(𝑥)
𝑑𝑥
−𝑙
− 𝑙 2𝑎0 2 +
𝑘
2
1 (𝑎0
+ 𝑏0 2
l=π
𝑎𝑜 =
𝑎𝑛 =
𝑏𝑛 =
1 𝜋
𝜋
|𝑥|𝑑𝑥
=
2𝜋 −𝜋
2
1 𝜋
2 cos 𝑛𝜋 −2
|𝑥|
∗
cos(n𝑥)
𝑑𝑥
=
𝜋 −𝜋
𝜋𝑛2
1 𝜋
|𝑥| ∗ sin(n𝑥) 𝑑𝑥 = 0
𝜋 −𝜋
for n= odd 𝑎𝑛 =
−4
𝑛2 𝜋
𝜋
2
|𝑥|
𝑑𝑥
−𝜋
||E||2
K
1
2
3
4
5
6
7
=
2𝜋3
3
=
2𝜋3
3
− 𝜋
||E||2
0.0747
0.0747
0.0118
0.0118
0.0037
0.0037
0.0015
𝜋2
2
4
+
16
𝑘
1,3,5.. 𝑛4 𝜋2
=
𝜋3
6
− 𝜋
||E||
0.27
0.27
0.11
0.11
0.06
0.06
0.04
The error goes on decreasing as k increases.
16
𝑘
1,3,5.. 𝑛4 𝜋2
Let 𝜆𝑛 and 𝛷𝑛 (𝑥) denote any eigenvalue and corresponding eigenfunction
of the Sturm-Liouville eigenvalue problem (1), respectively.
(a). The eigenvalues are real
(b). The eigenvalues are simple. That is, to each eigenvalue there
corresponds only one linearly independent eigenfunction. Further, there
are an infinite number of eigenvalues, and they can be ordered so that
𝜆1 < 𝜆2 < 𝜆3 < 𝜆4 < 𝜆5 < 𝜆6 < ⋯where 𝜆𝑛 → ∞as n → ∞ .
(c). Eigenfunctions corresponding to distinct eigenvalues are orthogonal.
That is, if 𝜆𝑗 ≠ 𝜆𝑘 , then < 𝛷𝑗 , 𝛷𝑘 >= 0 .
<𝑓,𝛷 >
𝑛
(d). Let f and f' be piecewise continuous on a≤x≤b. If 𝑎𝑛 =
, then
<𝛷𝑛 ,𝛷𝑛 >
the series ∞
𝑛=1 𝑎𝑛 𝛷𝑛 (𝑥)converges to f(x) if f is continuous at x, and to the
mean value [f(x+)+f(x-)]/2 if f is discontinuous at x, for each point x in the
open interval a < x < b.
Example:
Solve the Sturn-Liouville problem
(  R '(  )) '  1R(  )  0 (a    1) , R(a)  R(1)  0 ,
There a is a constant in the interval (0,1). Also write down the
expansion of an arbitrary element of the appropriate “ ” space in
terms of the eigenfunctions of the problem.
Solution:
We put
R(  )  Y ( y )  Y (ln  ) .
and
So
Whose general solution is
if not
Reference: Fourier Analysis (Author: Eric State, Pure and Applied Mathematics: a
Wiley-Interscience Series of Texts, Monographs, and Tracts ) P 224
Consequently, our original differential equation has general solution
Now for the boundary conditions. We note first that zero is not an
eigenvalue, for the following reason: Putting the boundary condition
R(1)=0 into the equation
gives
;
putting R(a)=0 into the resulting equation
then gives
so that
is identically zero.
,
So we assume λ ≠ 0; the first then gives sin λ ln 𝑎 = 0 , meaning λ must
𝜋
be a nonzero integer multiple of
.
ln 𝑎
We may as well take this integer to be positive; negative n’s give nothing
extra. In sum, our problem has the following eigenvalues and
corresponding eigenfunctions:
So finally we have the following expansion of an
Where
:
We assume elementary knowledge of the
Cartesian representation of complex number 𝑧 =
𝑥 + 𝑖𝑦 and the complex conjugate
𝑧 = 𝑥 − 𝑖𝑦.
𝑏
Inner product: < 𝑓, 𝑔 >= 𝑎 𝑓 𝑥 𝑔 𝑥 ѡ 𝑥 𝑑𝑥
Some properties:
< 𝑓, 𝑔 >=< 𝑔, 𝑓 >
< µ𝑓, 𝑔 >= µ < 𝑓, 𝑔 >
< 𝑓, µ𝑔 >= µ < 𝑓, 𝑔 >
< 𝑓, 𝑔 >≠< 𝑔, 𝑓 >
Then we can express 𝑝 𝑥 𝑦 ′ ′ + 𝑞 𝑥 𝑦 + 𝜆ω x y = 0, (a<x<b) in
operator form as 𝐿 𝑦 = 𝜆𝑦,
Where L is the differential operator
1 𝑑
𝑑
𝐿=−
𝑝
+𝑞
ѡ 𝑑𝑥
𝑑𝑥
And let u and v be any functions having continuous second
derivatives on [a,b] and satisfying the homogeneous boundary
conditions: 𝛼𝑦 𝑎 + 𝛽𝑦′ 𝑎 = 0 𝛾𝑦 𝑏 + 𝛿𝑦′ 𝑏 = 0
Then, < 𝐿 𝑢 , 𝑣 >=
𝑏
1
−
𝑎
ѡ
𝑝𝑢ʹ ʹ + 𝑞𝑢 𝑣𝑤𝑑𝑥 = −
𝑏
𝑎
𝑝𝑢ʹ ʹ +
And the boundary term is zero because u and v satisfy
the homogeneous boundary conditions. For any 𝛼 and β
(not both zero) and any γ and δ ( not both zero), we can
conclude:
< 𝐿 𝑢 , 𝑣 >=< 𝑢, 𝐿 𝑢 >, which is known as the
Lagrange identity.
Example:
Prove Green’s formla:
Since
L

b
a
(uLv  vLu)dx  [ p(uv ' vu ')] a
1d  d  
  p   q
w  dx  dx  
d  d 
First simplify it as L   p   r
dx  dx 
Then
b
where r   qw


d
d
uLv  vLu  u   pv '  rv   v   pu '  ru   u  p ' v ' pv '')  v( p ' u ' pu '')
 dx
 dx


 p  uv ' vu '   p '(uv ' vu ')  p(uv '' u ' v ' vu '' v ' u ')  u  p ' v ' pv '')  v( p ' u ' pu '')

So
uLv  vLu   p (uv ' vu ')  '
b
Integrate and get a (uLv  vLu)dx  [ p(uv ' vu ')] a
b
By a Sturm-Liouville problem we mean a linear homogeneous
second-order differential equation:
𝑝 𝑥 𝑦′ ′ + 𝑞 𝑥 𝑦 + 𝜆ω x y = 0, (a<x<b) (1)
With homogenous boundary conditions of the form
𝛼𝑦 𝑎 + 𝛽𝑦′ 𝑎 = 0
𝛾𝑦 𝑏 + 𝛿𝑦′ 𝑏 = 0
Where a, b are finite, where p, p', q, ω are continuous on [a, b], and
where p(x)>0 and ω x > 0 on [a, b]. 𝛼 𝑎𝑛𝑑 𝛽are not both zero,
𝛾 𝑎𝑛𝑑 𝛿 are not both zero. a, b, p(x), ω(x), α, β, γ, δ are all real.
We say that the boundary conditions are separated since on
condition applies at x=a and the other at x=b.
Periodic boundary condition.
In this case we have, in place of the separated boundary conditions, the
nonseparated conditions
y(a)=y(b) and y'(a)=y'(b)
Singular case.
In this case p(x) [and possibly w(x)] vanishes at one or both endpoints, so
that p(x)>0 and w(x)>0 holds on the open interval (a,b) rather than on
the closed interval [a,b]. Further, the boundary conditions are modified
as follows.
1. p(a)=0[and p(b)≠0]: Then the boundary conditions are
y bounded at a, and 𝛾𝑦 𝑏 + 𝛿𝑦′ 𝑏 = 0
2. p(b)=0[and p(a)≠0]: Then the boundary conditions are
𝛼𝑦 𝑎 + 𝛽𝑦′ 𝑎 = 0
and y bounded at b.
3. p(a)=p(b)=0: then the boundary conditions are
y bounded at a, and y bounded at b.
by y being bounded at a, for example, we mean that lim 𝑦(𝑥) exists (and
𝑥→∞
is therefore finite).
For these cases we have the following results:
Let 𝜆𝑛 and 𝛷𝑛 (𝑥) denote any eigenvalue and corresponding eigenfunction
of a Sturm-Liousville problem with periodic boundary conditions, given
by (2), or a singular Sturm-Liouville problem (as defined above)
(a)The eigenvalues are real.
(b)If q(x)≤0 on [a,b] and 𝑝 𝑥 𝛷𝑛 𝑥 𝛷′ 𝑛 𝑥 |𝑏𝑎 ≤ 0 for the
eigenfunction𝛷𝑛 (𝑥), then not only is 𝜆𝑛 real, it is also nonnegative: 𝜆𝑛 ≥
0.
(c)Eigenfunctions corresponding to distinct eigenvalues are orthogonal.
That is, if 𝜆𝑗 ≠ 𝜆𝑘 , then < 𝛷𝑗 , 𝛷𝑘 >= 0 .
As for the regular case, positive statements can be made about the
completeness of the sets of orthogonal eigenfunctions generated by these
problems, in the sense of their being based for the eigenfunction
expansion representation of sufficiently well behaved functions on the
interval a<x<b.
Example:
Expand f ( x)  H ( x) , on -1<x<1, in terms of the eigenfunctions of the
Sturm-Liouville problem
(1  x 2 ) y '' 2 xy '  y  0
Where y(-1) and y(1) are bounded.
According to Section 4.4, the Legendre equation are bounded on 1  x  1
are possible only if   n(n  1) for n=0,1,2…., and those nontrivial
solutions are the Legendre polynomials
Thus, the eigenvalues and eigenfunctions are
Pn ( x)
n ( x)  Pn ( x)
n  n(n  1)
For n=0,1,2…
The eigenfunction expansion of a given function f is given by

f ( x)   an Pn ( x)
n0
an


1
1
H ( x) Pn ( x)dx

1
1
H ( x) 
Pn 2 ( x)dx

2n  1 1
Pn ( x)dx
2 0
1
3
7
P0 ( x)  P1 ( x)  P3 ( x)  ....
2
4
16
If a function f defined on -∞< x <∞ is periodic (and
sufficiently well-behaved), then it can be represented by a
Fourier series. Sometimes we work with functions, defined on
-∞< x <∞, that are not periodic, we cannot expand such
functions in Fourier series if they are not periodic. Yet, we can
think of f as periodic but with an infinite period.
Frequency spectrum: 0, π/L, 2 π/L, 3 π/L, …
L= π: nπ/L=0, 1, 2, 3, 4, …
L=2π: nπ/L=0, 0.5, 1.0, 1.5, 2.0, …
L=3π: nπ/L=0, 0.1, 0.2, 0.3, 0.4, …
Observe that as L increases the discrete spectrum becomes more
and more dense, and approaches a continuous spectrum (from 0 to
∞) as L→∞. Therefore, we can expect that as L→∞ the
summation of trigonometric series, on the discrete variable n, will
give way to an integration on a continuous variable, say ω.
𝑓 𝑥 =
∞
0
𝑎( ω)cosωx + b ω sinωx]dω,
( Fourier integral of f, FI f).
1 ∞
𝑓(𝑥) cosωx 𝑑𝑥,
𝜋 −∞
1 ∞
= −∞ 𝑓(𝑥) sinωx 𝑑𝑥.
𝜋
Where 𝑎 ω =
𝑏 ω
Let f be defined on -∞< x <∞, let f and f´be piecewise
continuous on every finite interval [-L, L] (i.e., for L
∞
arbitrarily large), and let −∞ |𝑓 𝑥 | 𝑑𝑥 be convergent. Then
the Fourier integral of f converges to f(x) at every point x at
which f is continuous, and to the mean value [f(x+)+f(x-)]/2
at every point x at which f is discontinuous.
𝑆𝑖 𝑥 =
𝑥 𝑠𝑖𝑛𝑡
𝑑𝑡,
0 𝑡
Using the sine integral function we can evaluate the partial
Ω 𝑠𝑖𝑛ω
integral 𝑓Ω 𝑥 = 0
𝑐𝑜𝑠ωx dω.
ω
And 𝑓Ω 𝑥 =
1
{𝑆𝑖
𝜋
Ω(𝑥 + 1) 𝑆𝑖 Ω(𝑥 − 1) }
Example:
Find the Fourier sine coefficients bk of the square wave SW(X)
For k=1,2,….between 0 and 
a( ) 
1


b( ) 


1



f ( x) cos  xdx 
f ( x) sin  xdx 
1

1

0


0
(   cos  xdx   cos  xdx)  0
0


0
(   sin  xdx   sin  xdx) 

f ( x)   [a( ) cos  x  b()sin  x]d
0
For w=1,2,…
f ( x) 
4 sin x sin 3x sin 5 x
[


 ....]
 1
3
5
2 1 cos 
( 
)
 

One term
Four terms
Two terms
Five terms
Three terms
Ten terms
As the increasing of the number of terms, the function is approaching
SW(x)
Transition from Fourier integral to Fourier transform
𝑓 𝑥 =
∞
0
𝑎( ω)cosωx + b ω sinωx]dω. (1a)
Where 𝑎 ω
𝑏 ω
1 ∞
= −∞ 𝑓(𝑥) cosωx 𝑑𝑥,
𝜋
1 ∞
= −∞ 𝑓(𝑥) sinωx 𝑑𝑥.
𝜋
(1b)
Put (1b) into (1a):
1 ∞ ∞
𝑓 𝑥 =
{
𝑓 𝜉 𝑐𝑜𝑠ω𝜉𝑐𝑜𝑠 ωx + 𝑠𝑖𝑛ω𝜉𝑠𝑖𝑛ωx]d𝜉}𝑑ω
𝜋 0 −∞
1 ∞ ∞
=
𝑓 𝜉 𝑐𝑜𝑠ω(𝜉 − 𝑥)d𝜉𝑑
𝜋 0 −∞
Since cos(A-B)=cosAcosB+sinAsinB and introduce complex exponentials:
1
𝑓 𝑥 =
𝜋
1 ∞
=
2𝜋 0
∞
∞
𝑒 𝑖ω 𝜉−𝑥 + 𝑒 −𝑖ω 𝜉−𝑥
𝑓 𝜉
d𝜉𝑑ω
2
0
−∞
∞
∞ ∞
1
𝑓 𝜉 𝑒 𝑖ω 𝜉−𝑥 d𝜉𝑑ω +
𝑓 𝜉 𝑒 −𝑖ω
2𝜋 0 −∞
−∞
𝜉−𝑥
d𝜉𝑑ω
To combine the two terms on the right-hand side, let us change the
dummy integration variable from ωto – ω:
𝑓 𝑥
1
=
2𝜋
1
=
2𝜋
−∞
∞
𝑓 𝜉 𝑒 −𝑖ω
0
∞
−∞
∞
𝜉−𝑥
1
d𝜉(−𝑑ω) +
2𝜋
𝑓 𝜉 𝑒 −𝑖ω𝜉 d𝜉]𝑒 𝑖ω𝑥 𝑑ω
−∞
−∞
∞
∞
𝑓 𝜉 𝑒 −𝑖ω
0
−∞
𝜉−𝑥
d𝜉𝑑ω
Thus,
𝑓 𝑥 =𝑎
∞
𝑐
−∞
ω 𝑒 𝑖ωx 𝑑ω 𝑐 ω = 𝑏
∞
𝑓
−∞
𝜉 𝑒 −𝑖ω𝜉 d𝜉ab=1/2π
There is no longer a need to distinguish x and 𝜉, so:
𝑓 𝑥
𝑐 ω
1 ∞
=
𝑐
2𝜋 −∞
∞
= −∞ 𝑓 𝑥
ω 𝑒 𝑖ωx 𝑑ω
𝑒 −𝑖ω𝑥 d𝑥.
𝑓 𝑥
𝑐 ω
1 ∞
=
𝑐
2𝜋 −∞
∞
= −∞ 𝑓 𝑥
ω 𝑒 𝑖ωx 𝑑ω, (1)
𝑒 −𝑖ω𝑥 d𝑥. (2)
They can be a transform pair: (2) defines the Fourier
transform, c(w), of the given function f(x), and (1) is called
the inversion formula.
𝐹 𝑓 𝑥
=𝑓 ω =
𝐹 −1 𝑓 ω
=𝑓 𝑥
∞
−𝑖ωx 𝑑x,
𝑐
ω
𝑒
−∞
1 ∞
𝑖ωx 𝑑ω.
=
𝑓
ω
𝑒
2𝜋 −∞
Example:
 
Derive the result F e
a x

2a
 2  a2
(a>0)
Solution:

According to the definition F  f ( x)  f ( )   f ( x)e  i x dx

Then
 
F e
a x

 e

a x
e
 i x
0
dx   e e

ax  i x

dx   e  ax e i x dx 
0
1
1
2a


a  i a  i a   2
1.Linearity of the transform and its inverse.
F{af+bg}=aF{f}+bF{g}
F-1{a𝑓+b𝑔}=aF-1{𝑓}+bF-1{𝑔}
2.Transform of nth derivative.
𝐹 𝑓
𝑛
𝑥
= (𝑖ω)𝑛 𝑓 ω .
3.Fourier convolution.
(𝑓 ∗ 𝑔)(x) =
∞
𝑓(𝑥
−∞
− 𝜉)𝑔(𝜉)𝑑𝜉.
Then the Fourier convolution theorem:
F{𝑓 ∗ 𝑔} = 𝑓 ω 𝑔(ω) and F-1{𝑓 𝑔}=f*g
4. Translation formulas, x-shift and ω-shift
𝐹 𝑓 𝑥 − 𝑎 = 𝑒 −𝑖aω 𝑓 ω
𝐹 −1 𝑓 ω − a = 𝑒 𝑖aω 𝑓 𝑥
Example:
Solve the wave equation c 2u xx  utt ; u ( x, 0)  f ( x) and ut ( x, 0)  g ( x)
Take the Fourier Transform of both equations. The initial condition gives


u ( , 0)  f ( )
 
u t ( , 0)  u ( x, t )
t


t 0
 g ( )
And the PDE gives
2 
c ( u ( , t ))  2 u ( , t )
t
2
2

Which is basically an ODE in t, we can write it as

2 
2 2
u
(

,
t
)

c

u
( , t )  0
2
t
Which has the solution, and derivative

u ( , t )  A( ) cos ct  B( )sin ct
 
u ( , t )  c A( )sin ct  c B( ) cos ct
t
Reference: Steven Bellenot; Fourier Transform Examples
http://www.math.fsu.edu/~bellenot/class/f09/fun/ft.pdf

So the first initial condition gives A( )  f ( ) and the second gives

c B( )  g ( )

and make the solution


u(, t )  f ( ) cos ct 
g ( ) sin ct

c
Let’s first look at

eict  eit
1 
ict
f ( ) cos ct  f ( )(
)  ( f ( )e  f ( )eict )
2
2


Then

F [ f ( ) cos ct ] 
1
1
( f ( x  ct )  f ( x  ct ))
2
The second piece


g ( ) sin ct g ( ) sin ct


c
i
ic
And now the first factor looks like an integral, as a derivative with respect
to x would cancel the iw in bottom. Define:
h( x)  
x
s 0
g ( s)ds
By fundamental theorem of calculus

h '( x)  g ( x)

g ( )  i h( )


g ( ) sin ct 
eict  eict 1
1 
ict
 h( )(
)
 (h( )e  h( )eict )

c
2i
ic 2c
So
F 1[
x  ct
1 
1
1 x  ct
1 x  ct
g ( ) sin ct ]  ( h( x  ct )  h( x  ct ))  (  g ( s)ds   g ( s)ds) 
g ( s)ds
0
c
2c
2c 0
2c x ct
Putting both piece together we get the solution
1
1 x  ct
u ( x, t )  ( f ( x  ct )  f ( x  ct ))   g ( s )ds
2
2c x ct
17.11.1 Cosine and sine transforms
1.Fourier cosine transform
𝐹𝐶 𝑓 𝑥
= 𝑓𝐶 ω =
∞
𝑓
0
𝑥 𝑐𝑜𝑠ωx𝑑x,
And its inverse:
𝐹𝐶−1
𝑓𝐶 ω
=𝑓 𝑥 =
2 ∞
𝑓𝐶
0
𝜋
ω 𝑐𝑜𝑠ωx 𝑑ω.
2.Fourier sine transform
𝐹𝑆 𝑓 𝑥 = 𝑓𝑆 ω =
And its inverse:
𝐹𝑆−1 𝑓𝑆 ω
∞
𝑓
0
2
=𝑓 𝑥 =
𝜋
𝑥 𝑠𝑖𝑛ωx𝑑x,
∞
𝑓𝑆 ω 𝑠𝑖𝑛 ωx 𝑑ω
0
Properties:
𝐹𝐶 𝑓´ 𝑥 = ω𝑓𝑆 ω − 𝑓(0)
𝐹𝑆 𝑓´ 𝑥 = −ω𝑓𝐶 ω .
𝐹𝐶 𝑓´´ 𝑥 = −ω2 𝑓𝐶 ω − 𝑓´(0).
𝐹𝑆 𝑓´´ 𝑥 = −ω2 𝑓𝑆 ω + ω𝑓(0)
Example:
Solve heat transfer equation
u  2u

t x 2
B.C: (1) u(0,t)=0
(2)u(x,0)=P(x)
1 x  2
or
P(x)=1,
Solution with Fourier Sine Transform:
Fs {u '' u '}  Fs {0}
Fs {u ''}  Fs {u '}

fs
 2 f s   f 0   f c ( ) 
t


According to the B.C, we can get
f0  0
Then


 f s (, t )
 f s (, t ) 
t
2



0
0


f s ( , t )  f s ( , 0) exp( 2t )
f s ( , 0)  (2 /  )  u ( x, 0) sin( x) dx  (2 /  )  P( x) sin( x) dx  (2 /  )[cos   cos 2 ]


f s ( , t )  f s ( , 0) exp( 2t )  (2 /  )[cos   cos 2 ]exp( 2t )
Inverse

f s ( , t )
Gives the complete solution
 

0
0
u ( x, t )   f s ( , t ) sin(t )d    (2 /  )  cos( )  cos  2   exp(  2t ) sin( x) d
Laplace transform:
𝐿 𝑓 𝑡
=𝑓 𝑠 =
∞
−𝑠𝑡 𝑑𝑡.
𝑓(𝑡)𝑒
0
Where we changed the dummy integration variable from Ƭ to
t, then gives the inversion formula as
1
−1
𝐿 {𝑓 𝑠 } = 𝐻 𝑡 𝑓 𝑡 =
2𝜋𝑖
ɤ+𝑖∞
𝑓 𝑠 𝑒 𝑠𝑡 𝑑𝑠
ɤ−𝑖∞
Greenberg, Advanced Engineering Mathematics 2nd
Edition
Jain and Iyengar, Advanced Engineering Mathematics
(2007).
Jiri Lebl, Differential Equations for Engineers (October
2011). Available at http://www.jirka.org/diffyqs/.