Calculus 4
CRAMSheet
Cramster.com - Learn More, Get Better Grades, Save Time, Help Others, On Your Schedule
Download Free at Cramster.com
CALCULUS 4
www.Cramster.com
FOURIER SERIES
Definition to know
The Fourier Convergence Theorem
A Fourier series is an expansion of a periodic function f(x)
in terms of an infinite sum of sines and cosines. The Fourier
series make use of the orthogonality relationships of the sine
and cosine functions.
The functions Sin ( mπ x L ) and Cos ( mπ x L ) ,
m = 1, 2, ---- form a mutually orthogonal set of functions on
the internal
− L ≤ x ≤ L, satisfying the following relations:
Suppose that f and f’ are piecewise continuous on the
interval − L ≤ x < L so that it is periodic with period 2L.
Then f has a Fourier series.
⎧0, m ≠ n
mπ x
nπ x
∫− L Cos L Cos L dx = ⎨⎩ L, m = n
L
L
∫ Cos
−L
L
∫ Sin
−L
mπ x
nπ x
Sin
dx = 0, all m, n
L
L
⎧0, m ≠ n
mπ x
nπ x
Sin
dx = ⎨
L
L
⎩ L, m = n
The Fourier series of a function f(x) is given by:
f ( x) =
a0 ∞ ⎛
mπ x
mπ x ⎞
+ ∑ ⎜ anCos
+ bn Sin
⎟
2 n =1 ⎝
L
L ⎠
L
1
where a0 =
f ( x ) dx
L −∫L
an =
1
nπ x
f ( x ) Cos
dx , n = 0, 1, 2, --∫
L −L
L
bn =
1
nπ x
f ( x ) Sin
dx , n = 1,2, 3, -----∫
L −L
L
L
L
and
Orthogonal Function
The standard inner product (u, v) of two real – valued
functions u and v on the interval α ≤ x ≤ β is defined
f ( x) =
a0 ∞ ⎛
mπ x
mπ x ⎞
+ ∑ ⎜ an Cos
+ bn Sin
⎟
2 m =1 ⎝
L
L ⎠
whose coefficients are:
L
1
a0 = ∫ f ( x ) dx
L −L
( nπ x ) dx
1
an = ∫ f ( x ) Cos
L −L
L
L
( nπ x ) dx
1
f ( x ) Sin
∫
L −L
L
L
bn =
The Fourier series converges to f(x) at all points where f is
continuous and to ⎡⎣ f ( x + ) + f ( x − ) ⎤⎦ 2 at all points
where f is discontinuous.
Even and Odd Functions
A function f is an even function if its domain contain the
point –x wherever it contains the point x and if f(-x) = f(x),
for each x in the domain of f.
A function f is an odd function if its domain contains
–x wherever it contains x and if f(-x) = -f(x) for each x in the
domain of f.
Fourier cosine series: The Fourier series of any even function consists only of the even trigonometric functions
Cos ( nπ x L ) and the constant term, such a series is called
Fourier cosine series. The series is given by:
( u, v ) = ∫ u ( x ) v ( x ) dx
α
The functions u and v are said to be orthogonal on
α ≤ x ≤ β if their inner product is zero that is
β
∫ u ( x ) v ( x ) dx = 0
α
2
nπ x
f ( x ) Cos
dx , n = 0, 1, 2,--∫
L0
L
L
where an =
β
by
a0 ∞
nπ x
+ ∑ an Cos
2 n =1
L
f ( x) =
Fourier series: The Fourier series of any odd function consists only of the trigonometric functions Sin ( nπ x L ),
such a series is called Fourier sine series and it is given by:
∞
f ( x ) = ∑ bn Sin
n =1
2
nπ x
f ( x ) Sin
dx , n = 1, 2, -----∫
L0
L
L
where bn =
nπ x
L
CALCULUS 4
www.Cramster.com
2
Heat Conduction Equation
The Wave Equation
Definition to know
Definition to know
Consider a straight bar of uniform cross section and homogeneous material. Let the x – axis be taken along the bar and
let x = 0 and x = L denote the ends of the bar. Suppose further that the sides of the cross sectional dimensions are so
small that the temperature u can be considered constant on
any given cross section. Then u is a function of x and time t.
The variation of temperature in the bar is governed by a partial differential equation,
α 2u xx, = ut 0 <, x < L
Suppose that an elastic string of length L is tightly stretched
between two supports at the same horizontal level, so that
the x-axis lies along the string.
Suppose that the string is set in motion so that it vibrates in vertical plane. Let u(x, t) denotes at the point x at
time t. If the damping effects are neglected and if amplitude
of motion is not so large then u(x, t) satisfies the partial differential equation.
a 2u xx = in
uttdomain 0 < x < t, t < 0
(where α2 is a constant known as thermal diffusivity)
This equation is known as the heat conduction equation.
Solution of Heat Conduction Equation
The solution of heat conduction equation
0< x<L ,t<0
α 2u xx = ut ,
This equation is known as one – dimensional wave equation.
The constant coefficient a2 is given by
T where T
a2 =
is the tension in the string and ρ is mass per unit length.
Solution of Wave Equation
with boundary conditions
u ( 0, t ) = 0 , u ( L, t ) = 0 , t > 0
The wave equation is: a 2u xx = utt
and the initial condition
u ( x, 0 ) = f ( x ) , 0 ≤
and initial conditions
Initial position: u ( x, 0 ) =
is given by
∞
with boundary conditions
x≤L
u ( x, t ) = ∑ C n e − n π α
2 2
2
t L2
n =1
Sin
nπ x
L
nπ x
dx
L
0
∞
nπ x
f ( x ) = ∑ cn Sin
L
n =1
ρ
u ( 0, t ) = 0, u ( L, t ) = 0 , t ≥ 0
f ( x) , 0 ≤ x ≤ L
Initial velocity: ut ( x, 0 ) = g ( x ) , 0 ≤ x ≤ L
Where f and g are given functions with
g ( 0) = g ( L ) = 0
f ( 0) = f ( L ) = 0 ,
L
where cn =
and
∫ f ( x ) Sin
Then the solution of the wave equation is given by:
∞
Special cases:
1. Non – homogeneous boundary conditions:
If the boundary conditions are: u ( 0, t ) = T1 ,
u ( L, t ) = T2 , t > 0 then the solution is given by:
u ( x, t ) = (T2 − T1 )
1. If the initial displacement is non – zero but initial velocity
is zero then the initial conditions are:
u ( x, 0 ) = f ( x ) , ut ( x, 0 ) = 0 , 0 ≤ x ≤ L
∞
2 2 2
2
x
nπ x
+ T1 + ∑ cn e − n π α t L Sin
L
L
n =1
2 ⎡
x
nπ x
⎤
where cn = ∫ f ( x ) − (T2 − T1 ) − T1 Sin
dx
⎢
⎥
L0⎣
L
L
⎦
L
2. Bar with insulated ends:
If the bar has insulated ends, then in this case the boundary
conditions are: u x ( 0, t ) = 0 , u x ( L, t ) = 0 , t > 0
u ( x, t ) = ∑ cn Sin
n=0
2
nπ x
f ( x ) Cos
dx , n = 0, 1, 2, -----∫
L0
L
L
where cn =
2
( nπ x ) dx , n = 1, 2, -----2
f ( x ) Sin
∫
L0
L
L
where cn =
2. If the initial conditions are:
u ( x, 0 ) = 0 , ut ( x, 0 ) =
g ( x) , 0 ≤ x ≤ L
then the solution of wave equation is given by:
∞
u ( x, t ) = ∑ kn Sin
n =1
Then the solution is given by:
∞
2 2 2
2
c
nπ x
u ( x, t ) = 0 + ∑ cn e − n π α t L Cos
2 n =1
L
nπ x
nπ at
Cos
L
L
where k n =
2
L
nπ x
nπ at
Sin
L
L
g ( x ) Sin
nπ a ∫
0
nπ x
dx , n = 1, 2, -----L
CALCULUS 4
www.Cramster.com
3
LAPLACE’S EQUATION
Laplace’s Equation
Dirichlet’s Problem for a Circle
The Laplace’s equation is given by:
u + u = 0 (in two dimensions)
If u is function satisfying Laplace’s equation in a circular
region r < a subject to boundary condition u ( a,θ ) = f (θ )
xx
yy
and u + u + u = 0 (in three dimensions)
xx
yy
zz
where f is a given function on 0 ≤ θ ≤ 2π , then
Laplace’s equation in polar co-ordinates is
Dirichlet Problem for a Rectangle
1
1
urr + ur + 2 uθθ = 0
r
r
If the function u satisfies the Laplace’s equation
u xx + u yy = 0 in the rectangle 0 < x < a, 0 < y < b and also
satisfying boundary conditions
u ( x, 0 ) = 0 , u ( x, b ) = 0 , 0 < x < a
and its solution is given by:
u ( 0, y ) = 0 , u ( a, y ) = f ( y ) , 0 ≤ y ≤ b
where a n c =
n
where f is a given function on 0 ≤ y ≤ b then its solution
is given by
nπ a
nπ y
u ( x, y ) = ∑ cn Sinh
Sin
b
b
n =1
∞
nπ a 2
nπ y s
= ∫ f ( y ) Sin
dy
b
b0
b
b
where cn Sinh
Normalized Eigenfunctions
The Eigenfunction φn ( x ) associated with Eigen values
λn of Strum Liouville problem are said to be normalized if
they satisfy the condition
'
∫ r ( x )φ
2
n
0
Sturm Liouville’s Boundary Value Problem
It consists of a differential equation of the form
⎡⎣ p ( x ) y '⎤⎦ '− q ( x ) y + λ r ( x ) y = 0
α1 y ( 0 ) + α 2 y ' ( 0 ) = 0
Note: If we introduce the linear homogeneous differential
operator defined by L y = − ⎡ p ( x ) y '⎤ '+ q ( x ) y
⎣
⎡⎣ p ( x ) y '⎤⎦ '+ q ( x ) y + λ r ( x ) y = 0
∫ f (θ ) Cosnθ dθ
π
1
, n = 0, 1, 2, ----
0
2π
∫ f (θ ) Sinnθ dθ
π
, n = 1, 2, -----
0
Self– Adjoint Boundary Value Problem
A boundary value problem is said to be self – ad joint if
every pair of functions say u and v satisfying boundary conditions satisfy the condition
( L [u ] , v ) − (u, L[v]) = 0
(where ( u , v ) = ∫ u ( x ) v ( x ) dx )
b
a
If u and v be functions having continuous second derivatives
on the interval 0 ≤ x ≤ 1 and satisfying boundary conditions α1u ( 0 ) + α 2u ' ( 0) = 0 , β1u (1) + β 2u ' (1) = 0
and α1v ( 0 ) + α 2 v ' ( 0 ) = 0 , β v (1) + β v ' (1) = 0
1
2
then the Lagrange’s identity is
∫ {L [u ] v − uL [v ]} dx = 0
⎦
Results to know:
• All the Eigen values of Sturm – Liouville problem are
real.
• If φ1 and φ2 are two Eigen functions of the Strum –
Liouville problem corresponding to Eigen values λ
1
and λ2 respectively, and if λ1 ≠ λ2 then
'
∫ r ( x ) φ ( x ) φ ( x ) dx = 0
1
L [ y] = λr ( x ) y
2
0
•
3
n
2π
0
β1 y (1) + β 2 y ' (1) = 0
can be written as
a kn =
and
1
'
on the interval 0 < x, 1, together with boundary conditions
then the differential equation
c0 ∞ n
+ ∑ r ( cnCosθ + kn Sinnθ )
2 n =1
Lagrange’s Identity
dx = 1 , n = 1, 2, -----
[ ]
u ( r ,θ ) =
The Eigen values of the Strum – Liouville problem are
all simple, i.e. to each Eigen value there corresponds
only one linearly independent Eigen function.