WTW 238 – Mathematics
Study Notes
Source: Class notes & Textbook
Theme 1: Linear systems of differential equations
Theme 2: Sequences and series of real numbers
Theme 3: Power series
Theme 4: Fourier series
Theme 5: Partial differential equations
Theme 1: Linear systems of differential equations
8 Lectures [week 1 & 2]
This theme will carry on from WTW 256 using the eigenvalue – eigenvector method to solve
systems of linear systems of differential equations. Special attention will be given to repeated
eigenvalues, non-homogenous systems and second order linear systems.
Theme 1.1: Eigenvalue – Eigenvector method
Theme 1.2: Repeated Eigenvalue solutions
Theme 1.3: Non-homogenous linear systems
Theme 1.4: Second-order linear systems
Pre-knowledge for this theme consists of linear systems (WTW 164), eigenvalue and
eigenvectors of a 𝑛 × 𝑛 matrix (WTW 256), linear systems of first order differential equations
(WTW 256) and finally, the method of undetermined coefficients (WTW 256)
Theme 1.1 – Eigenvalue-Eigenvalue method
𝑋̅ ′ = 𝐴 ∙ 𝑋̅
Where 𝑋̅ is the solution vector and 𝐴 is a 𝑛 × 𝑛 matrix.
det(𝐴 − 𝜆 ∙ 𝐼) = 0
Where 𝐼 is an 𝑛 × 𝑛 identity matrix and 𝜆 is the eigenvalues of the matrix 𝐴.
⃑⃑⃑⃑𝑛 is computed by solving the following
For an eigenvalue 𝜆𝑛 a corresponding eigenvector 𝑘
system such that the solution is a non-zero vector:
⃑⃑⃑⃑𝑛 = 0
⃑
(𝐴 − 𝜆𝑛 ∙ 𝐼) ∙ 𝑘
Let’s look at an example:
𝑥 ′ (𝑡)
𝑥 (𝑡)
10 5
(
) ∙ [ 1 ] = [ 1′ ]
𝑥2 (𝑡)
𝑥2 (𝑡)
3 8
1
10 5
det ((
)−𝜆∙(
0
3 8
10 − 𝜆
det ((
3
10 − 𝜆
det ((
3
0
10 − 𝜆
)) = det ((
1
3
5
))
8−𝜆
5
)) = (10 − 𝜆)(8 − 𝜆) − 15 = 𝜆2 − 18𝜆 + 65
8−𝜆
5
)) = (𝜆 − 13)(𝜆 − 5) = 0,
8−𝜆
𝜆1 = 13,
𝜆2 = 5
5
⃑⃑⃑⃑1 = (10 − 13
⃑⃑⃑⃑1 = ⃑0
)∙𝑘
(𝐴 − 𝜆1 ∙ 𝐼) ∙ 𝑘
3
8 − 13
−3 5
⃑,
(
) ∙ ⃑⃑⃑⃑
𝑘1 = 0
3 −5
⃑⃑⃑⃑1 = [5]
𝑘
3
5
⃑⃑⃑⃑2 = (10 − 5
⃑⃑⃑⃑2 = ⃑0
)∙𝑘
(𝐴 − 𝜆2 ∙ 𝐼) ∙ 𝑘
3
8−5
5
(
3
5 ⃑⃑⃑⃑
) ∙ 𝑘2 = ⃑0,
3
⃑⃑⃑⃑2 = [ 1 ]
𝑘
−1
⃑⃑⃑⃑1 = [5] and
Therefore, the eigenvalues 𝜆1 = 13 and 𝜆2 = 5 have corresponding eigenvectors 𝑘
3
1
⃑⃑⃑⃑
𝑘2 = [ ].
−1
Now solving for the functions
[
𝑥1 (𝑡)
1
5
𝑘1 + 𝑐2 𝑒 𝜆2 𝑡 ⃑⃑⃑⃑
𝑘2 = 𝑐1 𝑒13𝑡 [ ] + 𝑐2 𝑒 5𝑡 [ ]
] = 𝑐1 𝑒 𝜆1 𝑡 ⃑⃑⃑⃑
𝑥2 (𝑡)
−1
3
[
𝑥1 (𝑡)
5𝑐 𝑒13𝑡 + 𝑐2 𝑒 5𝑡
]
] = [ 1 13𝑡
𝑥2 (𝑡)
3𝑐1 𝑒 − 𝑐2 𝑒 5𝑡
Solving another problem with an IVP:
𝑥 ′ (𝑡)
𝑥 (𝑡)
7 −5
(
) ∙ [ 1 ] = [ 1′ ] ,
𝑥2 (𝑡)
𝑥2 (𝑡)
4 3
𝑥 (0)
1
[ 1 ]=[ ]
𝑥2 (0)
0
1
⃑⃑⃑⃑1 = [4 (2 − 4𝑖)]
The eigenvalues 𝜆1 = 5 − 4𝑖 and 𝜆2 = 5 + 4𝑖 have corresponding eigenvectors 𝑘
1
1
⃑⃑⃑⃑2 = [4 (2 + 4𝑖)].
and 𝑘
1
[
1
1
𝑥1 (𝑡)
] = 𝑐1 𝑒 𝜆1𝑡 ⃑⃑⃑⃑
𝑘1 + 𝑐2 𝑒 𝜆2𝑡 ⃑⃑⃑⃑
𝑘2 = 𝑐1 𝑒 (5−4𝑖)𝑡 [4 (2 − 4𝑖)] + 𝑐2 𝑒 (5+4𝑖)𝑡 [4 (2 + 4𝑖)]
𝑥2 (𝑡)
1
1
[
1
1
( (2 − 4𝑖)) 𝑐1 𝑒 (5−4𝑖)𝑡 + ( (2 + 4𝑖)) 𝑐2 𝑒 (5+4𝑖)𝑡
𝑥1 (𝑡)
4
]=[ 4
]
𝑥2 (𝑡)
(5−4𝑖)𝑡
(5+4𝑖)𝑡
𝑐1 𝑒
+ 𝑐2 𝑒
1
1
𝑒 5𝑡 (( (2 − 4𝑖)) 𝑐1 (cos(4𝑡) − 𝑖 ∙ sin(4𝑡)) + ( (2 + 4𝑖)) 𝑐2 (cos(4𝑡) + 𝑖 ∙ sin(4𝑡)))
𝑥1 (𝑡)
4
4
[
]=[
]
𝑥2 (𝑡)
5𝑡
𝑒 (𝑐1 (cos(4𝑡) − 𝑖 ∙ sin(4𝑡)) + 𝑐2 (cos(4𝑡) + 𝑖 ∙ sin(4𝑡)))
1
1
𝑒 5𝑡 ((( − 𝑖)) 𝑐1 (cos(4𝑡) − 𝑖 ∙ sin(4𝑡)) + (( + 𝑖)) 𝑐2 (cos(4𝑡) + 𝑖 ∙ sin(4𝑡)))
𝑥1 (𝑡)
2
2
[
]=[
]
𝑥2 (𝑡)
5𝑡
𝑒 (𝑐1 (cos(4𝑡) − 𝑖 ∙ sin(4𝑡)) + 𝑐2 (cos(4𝑡) + 𝑖 ∙ sin(4𝑡)))
1
1
1
1
𝑒 5𝑡 (𝑐1 (( − 𝑖) ∙ cos(4𝑡) − ( − 𝑖) ∙ 𝑖 ∙ sin(4𝑡)) + 𝑐2 (( + 𝑖) ∙ cos(4𝑡) + ( + 𝑖) ∙ 𝑖 ∙ sin(4𝑡)))
𝑥1 (𝑡)
2
2
2
2
[
]=[
]
𝑥2 (𝑡)
5𝑡
𝑒 (𝑐1 (cos(4𝑡) − 𝑖 ∙ sin(4𝑡)) + 𝑐2 (cos(4𝑡) + 𝑖 ∙ sin(4𝑡)))
[
𝑥1 (𝑡)
]
𝑥2 (𝑡)
1
𝑖
1
𝑖
𝑒 5𝑡 (𝑐1 (( cos(4𝑡) − 𝑖 cos(4𝑡)) − ( sin(4𝑡) + sin(4𝑡))) + 𝑐2 (( cos(4𝑡) + 𝑖 cos(4𝑡)) + ( sin(4𝑡) − sin(4𝑡))))
2
2
2
2
=[
]
𝑒 5𝑡 (𝑐1 (cos(4𝑡) − 𝑖 ∙ sin(4𝑡)) + 𝑐2 (cos(4𝑡) + 𝑖 ∙ sin(4𝑡)))
[
𝑥1 (𝑡)
]
𝑥2 (𝑡)
=𝑒
5𝑡
1
1
1
1
(𝑐1 (( cos(4𝑡) − sin(4𝑡)) − 𝑖 ( sin(4𝑡) + cos(4𝑡))) + 𝑐2 (( cos(4𝑡) − sin(4𝑡)) + 𝑖 ( sin(4𝑡) + cos(4𝑡))))
2
2
2
2
[
]
(𝑐1 (cos(4𝑡) − 𝑖 ∙ sin(4𝑡)) + 𝑐2 (cos(4𝑡) + 𝑖 ∙ sin(4𝑡)))
1
1
𝑥 (𝑡)
𝑐 ( cos(4𝑡) − sin(4𝑡)) + 𝑐2 ( sin(4𝑡) + cos(4𝑡))
[ 1 ] = 𝑒 5𝑡 [ 1 2
]
2
𝑥2 (𝑡)
𝑐1 cos(4𝑡) + 𝑐2 sin(4𝑡)
[
1
𝑥1 (0)
1
] = [𝑐1 (2) + 𝑐2 ] = [ ] ,
𝑥2 (0)
0
𝑐
1
𝑐2
1
[𝑐 ] = [ ]
1
0
1
𝑥1 (𝑡)
5𝑡 ( sin(4𝑡) + cos(4𝑡))
[
]=𝑒 [ 2
]
𝑥2 (𝑡)
sin(4𝑡)
Some properties of eigenvalues and eigenvectors:
➢
➢
In a lower- or upper triangular or diagonal matrix, the entries on the diagonal of the matrix are
the eigenvalues of the matrix.
An 𝑛 × 𝑛 matrix 𝐴 has eigenvalues 𝜆1 , 𝜆2 , … , 𝜆𝑛 . The inverted matrix, 𝐴−1 has the eigenvalues
1
,
1
𝜆1 𝜆2
➢
➢
,…,
1
𝜆𝑛
.
An 𝑛 × 𝑛 matrix 𝐴 has eigenvalues 𝜆1 , 𝜆2 , … , 𝜆𝑛 . The squared matrix, 𝐴2 has the eigenvalues
𝜆21 , 𝜆22 , … , 𝜆2𝑛 .
An eigenvalue or eigenvector can never be zero (only applies to theme 1.1).
Theme 1.2: Repeated Eigenvalue solutions
Multiplicity and completeness, what is it and what does it mean?
An eigenvalue with multiplicity of 𝑚 is said to be complete when the eigenvalue has 𝑚 unique associated
eigenvectors. The multiplicity, 𝑚, is the number of repetitions the eigenvalue has.
Example time:
𝑥 ′ (𝑡)
𝑥 (𝑡)
1 0
(
) ∙ [ 1 ] = [ 1′ ] ,
𝑥2 (𝑡)
0 1
𝑥2 (𝑡)
1
det ((
0
0
1
)−𝜆∙(
1
0
1−𝜆
det ((
0
[
𝑥1 (0)
1
]=[ ]
𝑥2 (0)
0
0
1−𝜆
)) = det ((
1
0
0
))
1−𝜆
0
)) = (1 − 𝜆)(1 − 𝜆) = 0
1−𝜆
𝜆1 = 𝜆2 = 1
⃑⃑⃑⃑1 = (1 − 1
(𝐴 − 𝜆1 ∙ 𝐼) ∙ 𝑘
0
0 0 ⃑⃑⃑⃑
⃑,
(
) ∙ 𝑘1 = 0
0 0
⃑⃑⃑⃑2 = (1 − 1
(𝐴 − 𝜆2 ∙ 𝐼) ∙ 𝑘
1
0
(
0
[
0 ⃑⃑⃑⃑
⃑,
) ∙ 𝑘2 = 0
0
0
) ∙ ⃑⃑⃑⃑
𝑘1 = ⃑0
1−1
⃑⃑⃑⃑1 = [1]
𝑘
0
0
⃑⃑⃑⃑2 = ⃑0
)∙𝑘
1−1
⃑⃑⃑⃑2 = [0]
𝑘
1
𝑥1 (𝑡)
1
0
𝑘1 + 𝑐2 𝑒 𝜆2 𝑡 ⃑⃑⃑⃑
𝑘2 = 𝑐1 𝑒 𝑡 [ ] + 𝑐2 𝑒 𝑡 [ ]
] = 𝑐1 𝑒 𝜆1 𝑡 ⃑⃑⃑⃑
𝑥2 (𝑡)
0
1
[
[
𝑥1 (𝑡)
𝑐 𝑒𝑡
] = [ 1 𝑡]
𝑥2 (𝑡)
𝑐2 𝑒
𝑐1
𝑥1 (0)
1
] = [𝑐 ] = [ ]
𝑥2 (0)
0
2
[
𝑡
𝑥1 (𝑡)
] = [𝑒 ]
𝑥2 (𝑡)
0
This matrix has complete eigenvalues.
What happens when the eigenvalues are incomplete or defective eigenvalues? Let’s have a look
at what to do when there is more than one defective eigenvalue. What is meant by a defective
eigenvalue?
When an eigenvector can not simply be calculated as in the previous example. By choosing a
different value for the eigenvector we could get two linearly independent eigenvectors. But this
is not the case in defective eigenvalues.
Let 𝐴 be the following matrix and solve the system of linear differential equations:
1
𝐴 = (0
4
𝑋̅ ′ = 𝐴 ∙ 𝑋̅,
1
det ((0
4
5
1
8
1
0
0) − 𝜆 ∙ ( 0
0
1
1−𝜆
det (( 0
4
5
1−𝜆
8
0
1
0
5
1
8
0
0)
1
0
1−𝜆
0)) = det (( 0
1
4
5
1−𝜆
8
0
0 ))
1−𝜆
0
0 )) = (1 − 𝜆)(1 − 𝜆)(1 − 𝜆) = 0
1−𝜆
𝜆1 = 𝜆2 = 𝜆3 = 1
1−1
(𝐴 − 𝜆1 ∙ 𝐼) ∙ ⃑⃑⃑⃑
𝑘1 = ( 0
4
0
(0
4
5
0
8
0
𝑘1 = ⃑0,
0) ∙ ⃑⃑⃑⃑
0
1−1
(𝐴 − 𝜆2 ∙ 𝐼) ∙ ⃑⃑⃑⃑
𝑘2 = ( 0
4
0
(0
4
5
0
8
0
𝑘2 = ⃑⃑⃑⃑
𝑘1 ,
0) ∙ ⃑⃑⃑⃑
0
0
(0
4
5
0
8
1−1
(𝐴 − 𝜆3 ∙ 𝐼) ∙ ⃑⃑⃑⃑
𝑘3 = ( 0
4
0
(0
4
5
0
8
0
⃑⃑⃑⃑3 = ⃑⃑⃑⃑
𝑘2 ,
0) ∙ 𝑘
0
5
1−1
8
0
(0
4
5
0
8
5
1−1
8
0
𝑘1 = ⃑0
0 ) ∙ ⃑⃑⃑⃑
1−1
0
⃑⃑⃑⃑
𝑘1 = [0]
1
0
⃑⃑⃑⃑2 = ⃑0
0 )∙𝑘
1−1
0
0
𝑘2 = [0] ,
0) ∙ ⃑⃑⃑⃑
1
0
5
1−1
8
1⁄
4
⃑⃑⃑⃑
𝑘2 = [ 0 ]
0
0
⃑⃑⃑⃑3 = 0
⃑
0 )∙𝑘
1−1
1⁄
0
4
⃑⃑⃑⃑
)
∙
𝑘
=
[
0
3
0 ],
0
0
− 1⁄10
⃑⃑⃑⃑
𝑘3 = [ 1⁄ ]
20
0
𝑥1 (𝑡)
𝑡2
⃑⃑⃑⃑2 + ⃑⃑⃑⃑
[𝑥2 (𝑡)] = [𝑒 𝑡 ⃑⃑⃑⃑
𝑘1 + 𝑒 𝑡 (𝑡 ∙ ⃑⃑⃑⃑
𝑘1 + ⃑⃑⃑⃑
𝑘2 ) + 𝑒 𝑡 ( ∙ ⃑⃑⃑⃑
𝑘 +𝑡∙𝑘
𝑘3 )]
2 1
𝑥3 (𝑡)
1⁄
1⁄
− 1⁄10
𝑥1 (𝑡)
2
0
0
0
𝑡
4
4
[𝑥2 (𝑡)] = 𝑒 𝑡 [0] + (𝑡 ∙ [0] + [ 0 ]) + ( ∙ [0] + 𝑡 ∙ [ 0 ] + [ 1⁄
])
2
20
1
1
1
𝑥3 (𝑡)
0
0
0
[
]
1⁄
1⁄
− 1⁄10
𝑥1 (𝑡)
0
0
𝑡2 0
4
4
𝑡
[𝑥2 (𝑡)] = 𝑒 [([0] + [ 0 ] + [ 1⁄ ]) + 𝑡 ([0] + [ 0 ]) + ( ∙ [0])]
2
20
1
1
1
𝑥3 (𝑡)
0
0
0
3⁄
0
1⁄
𝑥1 (𝑡)
20
4
𝑡
[𝑥2 (𝑡)] = 𝑒 [𝑐1 ([1⁄ ]) + 𝑐2 𝑡 ([ 0 ]) + 𝑐3 𝑡 2 ([ 0 ])]
1⁄
20
𝑥3 (𝑡)
1
2
1
Theme 1.3: Non-homogenous linear systems
𝑋̅ ′ = 𝐴 ∙ 𝑋̅ + 𝐹
Where 𝑋̅ is the solution vector and 𝐴 is a 𝑛 × 𝑛 matrix and 𝐹 is a forcing function.
We will do three examples:
2 −1 ̅
0
𝑋̅ ′ = [
]∙𝑋+[ ]
3 −2
3𝑡
1
1
𝑋𝑐 = 𝑐1 𝑒 𝑡 [ ] + 𝑐2 𝑒 −𝑡 [ ]
1
3
𝑎
𝑐
0
0
Guess the particular solution: 𝑋𝑃 = 𝑡 [ ] + [ ] , 𝐹 = 𝑡 [ ] + [ ]
𝑏
𝑑
3
0
𝑎
2
[ ]=[
𝑏
3
𝑎
𝑐
−1
0
] ∙ [𝑡 [ ] + [ ]] + [ ]
𝑏
𝑑
−2
3𝑡
𝑎
2 −1 𝑎𝑡 + 𝑐
0
[ ]=[
]∙[
]+[ ]
𝑏
3 −2 𝑏𝑡 + 𝑑
3𝑡
𝑎
2𝑎𝑡 + 2𝑐 − 𝑏𝑡 − 𝑑
0
[ ]=[
]+[ ]
𝑏
3𝑎𝑡 + 3𝑐 − 2𝑏𝑡 − 2𝑑
3𝑡
𝑎
2𝑎𝑡 − 𝑏𝑡 + 2𝑐 − 𝑑
[ ]=[
]
𝑏
3𝑎𝑡 − 2𝑏𝑡 + 3𝑡 + 3𝑐 − 2𝑑
𝑎
2𝑎 − 𝑏
2𝑐 − 𝑑
[ ] = 𝑡[
]+[
]
𝑏
3𝑎 − 2𝑏 + 3
3𝑐 − 2𝑑
0
2𝑎 − 𝑏
[ ] = 𝑡[
],
0
3𝑎 − 2𝑏 + 3
𝑎
3
[ ] = [ ],
𝑏
6
𝑎
2𝑐 − 𝑑
[ ]=[
]
𝑏
3𝑐 − 2𝑑
3
2𝑐 − 𝑑
[ ]=[
],
6
3𝑐 − 2𝑑
𝑐
0
[ ]=[ ]
𝑑
−3
3
0
1
1
𝑋̅ = 𝑋𝑐 + 𝑋𝑃 = 𝑡 [ ] + [ ] + 𝑐1 𝑒 𝑡 [ ] + 𝑐2 𝑒 −𝑡 [ ]
6
−3
1
3
Next example:
4
2 2 ̅
𝑋̅ ′ = [
] ∙ 𝑋 + [ 2𝑡 ]
1 3
𝑒
−2
1
𝑋𝑐 = 𝑐1 𝑒 𝑡 [ ] + 𝑐2 𝑒 4𝑡 [ ]
1
1
𝑎
𝑐
4
0
Guess the particular solution: 𝑋𝑃 = 𝑒 2𝑡 [ ] + [ ] , 𝐹 = 𝑒 2𝑡 [ ] + [ ]
𝑏
𝑑
1
0
𝑎
𝑐
4
2𝑎
2 2
0
𝑒 2𝑡 [ ] = [
] ∙ [𝑒 2𝑡 [ ] + [ ]] + 𝑒 2𝑡 [ ] + [ ]
𝑏
𝑑
2𝑏
1 3
1
0
𝑒 2𝑡 [
2𝑎
2
]=[
2𝑏
1
4
2 𝑎𝑒 2𝑡 + 𝑐
0
] ∙ [ 2𝑡
] + 𝑒 2𝑡 [ ] + [ ]
3 𝑏𝑒 + 𝑑
1
0
𝑒 2𝑡 [
2𝑡
2𝑡
4
2𝑎
0
] = [2𝑎𝑒 2𝑡 + 2𝑐 + 2𝑏𝑒2𝑡 + 2𝑑] + 𝑒 2𝑡 [ ] + [ ]
2𝑏
1
0
𝑎𝑒 + 𝑐 + 3𝑏𝑒 + 3𝑑
2𝑡
2𝑡
2𝑎
+ 2𝑑 + 4]
𝑒 2𝑡 [ ] = [2𝑎𝑒2𝑡 + 2𝑏𝑒2𝑡 + 2𝑐
2𝑏
𝑎𝑒 + 3𝑏𝑒 + 𝑒 2𝑡 + 𝑐 + 3𝑑
𝑒 2𝑡 [
2𝑎
2𝑎 + 2𝑏
2𝑐 + 2𝑑 + 4
] = 𝑒 2𝑡 [
]+[
]
2𝑏
𝑎 + 3𝑏 + 1
𝑐 + 3𝑑
2𝑎
2𝑎 + 2𝑏
𝑒 2𝑡 [ ] = 𝑒 2𝑡 [
],
2𝑏
𝑎 + 3𝑏 + 1
𝑎
−1
[ ] = [ ],
𝑏
0
0
2𝑐 + 2𝑑 + 4
[ ]=[
]
0
𝑐 + 3𝑑
𝑐
−3
[ ]=[ ]
𝑑
1
−1
−3
−2
1
𝑋̅ = 𝑋𝑐 + 𝑋𝑃 = 𝑒 2𝑡 [ ] + [ ] + 𝑐1 𝑒 𝑡 [ ] + 𝑐2 𝑒 4𝑡 [ ]
0
1
1
1
The final example:
2
𝑋̅ ′ = [
3
𝑡
−1 ̅
] ∙ 𝑋 + [ 𝑒 𝑡]
−2
−𝑒
1
1
𝑋𝑐 = 𝑐1 𝑒 𝑡 [ ] + 𝑐2 𝑒 −𝑡 [ ]
1
3
𝑎
𝑐
1
Guess the particular solution: 𝑋𝑃 = 𝑒 𝑡 [ ] + 𝑡𝑒 𝑡 [ ] , 𝐹 = 𝑒 𝑡 [ ] gives duplication.
𝑏
𝑑
−1
𝑎
𝑐
𝑐
2
𝑒 𝑡 [ ] + 𝑡𝑒 𝑡 [ ] + 𝑒 𝑡 [ ] = [
𝑏
𝑑
𝑑
3
𝑒𝑡 [
𝑒𝑡 [
𝑡
𝑎
𝑐
−1
] ∙ [𝑒 𝑡 [ ] + 𝑡𝑒 𝑡 [ ]] + [ 𝑒 𝑡 ]
𝑏
𝑑
−2
−𝑒
𝑡
𝑡
𝑡
𝑐
𝑎+𝑐
2 −1
] + 𝑡𝑒 𝑡 [ ] = [
] ∙ [[ 𝑎𝑒𝑡 + 𝑐𝑡𝑒 𝑡 ]] + [ 𝑒 𝑡 ]
𝑑
𝑏+𝑑
3 −2
𝑏𝑒 + 𝑑𝑡𝑒
−𝑒
𝑡
𝑡
𝑡
𝑡
𝑡
𝑐
𝑎+𝑐
] + 𝑡𝑒 𝑡 [ ] = [ 2𝑎𝑒𝑡 + 2𝑐𝑡𝑒𝑡 − 𝑏𝑒 𝑡 − 𝑑𝑡𝑒 𝑡 ] + [ 𝑒 𝑡 ]
𝑑
𝑏+𝑑
3𝑎𝑒 + 3𝑐𝑡𝑒 − 2𝑏𝑒 − 2𝑑𝑡𝑒
−𝑒
𝑒𝑡 [
𝑡
𝑡
𝑡
𝑡
𝑡
𝑐
𝑎+𝑐
] + 𝑡𝑒 𝑡 [ ] = [ 2𝑎𝑒𝑡 − 𝑏𝑒 𝑡 + 𝑒 𝑡 + 2𝑐𝑡𝑒 𝑡 − 𝑑𝑡𝑒 𝑡 ]
𝑑
𝑏+𝑑
3𝑎𝑒 − 2𝑏𝑒 − 𝑒 + 3𝑐𝑡𝑒 − 2𝑑𝑡𝑒
𝑐
𝑎+𝑐
2𝑎 − 𝑏 + 1
2𝑐 − 𝑑
𝑒𝑡 [
] + 𝑡𝑒 𝑡 [ ] = 𝑒 𝑡 [
] + 𝑡𝑒 𝑡 [
]
𝑑
𝑏+𝑑
3𝑎 − 2𝑏 − 1
3𝑐 − 2𝑑
𝑐
2𝑐 − 𝑑
𝑡𝑒 𝑡 [ ] = 𝑡𝑒 𝑡 [
],
𝑑
3𝑐 − 2𝑑
𝑐
2𝑐 − 𝑑
[ ]=[
],
𝑑
3𝑐 − 2𝑑
𝑐
𝑑
[ ] = [ ],
𝑑
𝑐
𝑒𝑡 [
𝑎+𝑐
2𝑎 − 𝑏 + 1
] = 𝑒𝑡 [
]
𝑏+𝑑
3𝑎 − 2𝑏 − 1
𝑎+𝑐
2𝑎 − 𝑏 + 1
[
]=[
]
𝑏+𝑑
3𝑎 − 2𝑏 − 1
𝑎
0
[ ] = [ ],
𝑏
−1
𝑐
2
[ ]=[ ]
𝑑
2
0
2
1
1
𝑋̅ = 𝑋𝑐 + 𝑋𝑃 = 𝑒 𝑡 [ ] + 𝑡𝑒 𝑡 [ ] + 𝑐1 𝑒 𝑡 [ ] + 𝑐2 𝑒 −𝑡 [ ]
−1
2
1
3
Theme 1.4: Second-order linear systems
𝑋̅ ′′ = 𝐴 ∙ 𝑋̅
Is a linear second-order homogeneous system of differential equations.
In this theme it will be wise to calculate eigenvalues as follows:
𝜆𝑛 = −𝜔𝑛2
As this will be easier to use in the standard solutions given the value of 𝜆.
𝑛
𝑋̅ = ∑ ((𝑎𝑖 cos(𝜔𝑖 𝑡) + 𝑏𝑖 sin(𝜔𝑖 𝑡)) ∙ 𝑘⃑𝑖 ) ,
𝜆𝑖 < 0
𝑖=1
𝑋̅ = (𝑎0 + 𝑡𝑏0 ) ∙ 𝑘⃑0 ,
𝜆0 = 0
Example:
0
𝑋̅ ′′ = [
2
0
] ∙ 𝑋̅
−3
0 0
1 0
−𝜆
det ((
)−𝜆∙(
)) = det ((
2 −3
0 1
2
−𝜆
det ((
2
0
)) = (−𝜆)(−3 − 𝜆) = 0,
−3 − 𝜆
𝜆0 = −𝜔02 = 0,
0
))
−3 − 𝜆
𝜆0 = 0,
𝜆1 = −3
𝜆1 = −𝜔12 = −3 ∴ 𝜔1 = √3
⃑⃑⃑⃑0 = (0 0 ) ∙ 𝑘
⃑⃑⃑⃑0 = ⃑0
(𝐴 − 𝜆0 ∙ 𝐼) ∙ 𝑘
2 −3
0
(
2
0
⃑⃑⃑⃑0 = 0
⃑,
)∙𝑘
−3
⃑⃑⃑⃑1 = (3
(𝐴 − 𝜆1 ∙ 𝐼) ∙ 𝑘
2
3 0 ⃑⃑⃑⃑
⃑,
(
) ∙ 𝑘1 = 0
2 0
3
⃑⃑⃑⃑
𝑘1 = [ ]
2
0 ⃑⃑⃑⃑
) ∙ 𝑘1 = ⃑0
0
⃑⃑⃑⃑1 = [0]
𝑘
1
0
3
𝑋̅ = (𝑎1 cos(√3𝑡) + 𝑏1 sin(√3𝑡)) ∙ [ ] + (𝑎0 + 𝑡𝑏0 ) ∙ [ ]
1
2
Theme 2: Sequences and series of real numbers
12 Lectures [week 3 & 4 & 5]
This theme will introduce series and sequences. The theme will address series of non-negative
numbers and convergence tests for various types of series.
Theme 2.1: Sequences of real numbers
Theme 2.2: Series of real numbers
Theme 2.3: Series with non-negative terms
Theme 2.4: Absolute- and conditional convergence
Pre-knowledge for this theme consists of limits to infinity (WTW 158) and L’Hospitals’ rule of
indeterminate forms of limits to infinity (WTW 158).
Theme 2.1 – Sequences of real numbers
The way a sequence can be written has many forms:
{𝑎𝑛 } = 2, 3, 8, 11, … ,
𝑎𝑛 = 3𝑛 − 1,
1 3 7 15
{𝑏𝑛 } = , , , , … ,
2 4 8 16
𝑏𝑛 = 1 −
𝑛∈ℤ
1
,
2𝑛
𝑛∈ℤ
Graphs of sequences are not continuous. Consider the second sequence:
𝑏𝑛 = 1 −
1
2𝑛
The limit can be easily seen from this sequence. The limit of a sequence is the term to which the
general term of a sequence tends to. In the previous demonstration it is easy to see that 𝑎𝑛 =
3𝑛 − 1 will tend to infinity or diverge. {𝑎𝑛 } diverges as when 𝑛 approaches ∞, {𝑎𝑛 } approaches
∞. {𝑏𝑛 } on the other hand, is convergent as when 𝑛 approaches ∞, {𝑎𝑛 } approaches 1, a real
number. Therefore, it is said that {𝑏𝑛 } converges to 1.
A sequence, {𝑎𝑛 } is said to converge when lim (𝑎𝑛 ) = 𝑎∞ = 𝐿 and 𝐿 ∈ ℝ, else the sequence
𝑛→∞
diverges. Either, 𝐿 = ∞ or 𝐿 does not exist.
Let’s refresh all the limit operations:
lim (𝑎𝑛 + 𝑏𝑛 ) = lim (𝑎𝑛 ) + lim (𝑏𝑛 ) ↔ lim (𝑎𝑛 ) & lim (𝑏𝑛 ) 𝑑𝑜𝑒𝑠 𝑒𝑥𝑖𝑠𝑡
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
lim (𝑎𝑛 − 𝑏𝑛 ) = lim (𝑎𝑛 ) − lim (𝑏𝑛 ) ↔ lim (𝑎𝑛 ) & lim (𝑏𝑛 ) 𝑑𝑜𝑒𝑠 𝑒𝑥𝑖𝑠𝑡
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
lim (𝐶) = 𝐶,
𝑛→∞
𝐶∈ℝ
lim (𝐶 ∙ 𝑎𝑛 ) = 𝐶 ∙ lim (𝑎𝑛 ) ,
𝑛→∞
𝑛→∞
𝑛→∞
𝐶∈ℝ
lim (𝑎𝑛 ∙ 𝑏𝑛 ) = lim (𝑎𝑛 ) ∙ lim (𝑏𝑛 ) ↔ lim (𝑎𝑛 ) & lim (𝑏𝑛 ) 𝑑𝑜𝑒𝑠 𝑒𝑥𝑖𝑠𝑡
𝑛→∞
lim (
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
lim (𝑎𝑛 )
𝑎𝑛
) = (𝑛→∞
) ↔ lim (𝑎𝑛 ) & lim (𝑏𝑛 ) ≠ 0 𝑑𝑜𝑒𝑠 𝑒𝑥𝑖𝑠𝑡
𝑛→∞
𝑛→∞
𝑏𝑛
lim (𝑏𝑛 )
𝑛→∞
𝑛
lim ((𝑎𝑛 )𝑛 ) = ( lim (𝑎𝑛 )) ,
𝑛→∞
𝑛→∞
lim (𝑎𝑛 )
lim ((𝐶)𝑎𝑛 ) = 𝐶 𝑛→∞
𝑛→∞
,
𝑙𝑒𝑡 𝑛 = 𝑥 𝑤𝑖𝑡ℎ 𝑥 ∈ ℝ
𝑙𝑒𝑡 𝑛 = 𝑥 𝑤𝑖𝑡ℎ 𝑥 ∈ ℝ
lim (ln(𝑎𝑛 )) = ln ( 𝑙𝑖𝑚 (𝑎𝑛 )) ,
𝑛→∞
𝑙𝑒𝑡 𝑛 = 𝑥 𝑤𝑖𝑡ℎ 𝑥 ∈ ℝ
𝑛→∞
The squeeze theorem:
Let 𝐿 = lim (𝑎𝑛 ) = lim (𝑏𝑛 ) and 𝑐𝑛 is such that 𝑎𝑛 ≤ 𝑐𝑛 ≤ 𝑏𝑛 then 𝑐𝑛 will converge to 𝐿.
𝑛→∞
𝑛→∞
The absolute value theorem:
Let {𝑎𝑛 } = |{𝑏𝑛 }| which implies {𝑎𝑛 } ≥ {𝑏𝑛 }. If {𝑎𝑛 } converges then {𝑏𝑛 } will also converge. If
{𝑏𝑛 } diverges then {𝑎𝑛 } will also diverge.
Geometric sequence:
{𝑎𝑛 } = 𝑎 ∙ 𝑟 𝑛
Will converge for the following values of 𝑟.
|𝑟| < 1: 𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
𝑟 = 1: 𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
𝑟 = 0: 𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
Any other value of 𝑟 will cause the sequence to diverge.
L’Hospital rule and the indeterminate form:
lim (
𝑛→∞
𝑎𝑛
𝑎′ (𝑥)
) = lim ( ′ ) ,
𝑛→∞ 𝑏 (𝑥)
𝑏𝑛
𝑎𝑛 , 𝑏𝑛 → 𝑎(𝑥), 𝑏(𝑥)
Remember to apply L’Hospitals rule it must be in the indeterminate form.
➢
➢
0
0
∞
∞
➢ 0 ∙ ∞, try 𝑎(𝑥) =
1
1
𝑎(𝑥)
➢ ∞−∞
➢ 00 , ∞0 , 1∞ try 𝑎(𝑥) → ln( 𝑎(𝑥))
Some proofs:
𝑛
𝑎
𝑎 𝑛
ln( (1+ ) )
𝑛
lim ((1 + ) ) = lim (𝑒
)
𝑛→∞
𝑛→∞
𝑛
𝑎
lim (𝑛∙ln(1− ))
𝑛
𝑒 𝑛→∞
𝑎
ln(1+ )
𝑛 )
1
𝑛→∞
𝑛
𝑒
lim (
= 𝑒 1(∞∙0) → 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙
𝑎
𝑎(𝑥) = ln (1 + ) ,
𝑥
0
= 𝑒 0 → 𝑟𝑒𝑎𝑙 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙,
𝑎
ln(1+ )
𝑛 )
1
𝑛→∞
𝑛
𝑒
lim (
=
1
𝑏(𝑥) = ,
𝑥
𝑎
𝑎
𝑥 2 (1+ )
𝑥
lim
1
𝑥→∞
𝑥2
(
)
𝑒
lim (
=𝑒
𝑥→∞
𝑎
𝑎)
1+
𝑥
𝑎′ (𝑥) =
−𝑎
𝑎
𝑥 2 (1 + 𝑥 )
𝑏 ′ (𝑥) = −
1
𝑥2
= 𝑒𝑎
𝑎 𝑛
lim ((1 + ) ) = 𝑒 𝑎
𝑛→∞
𝑛
Another proof:
1
𝑛
lim ( √𝑛) = lim (𝑛𝑛 )
𝑛→∞
𝑛→∞
1
lim (𝑒
ln(𝑛𝑛 )
𝑛→∞
𝑛→∞
1
lim ( ln(𝑛))
𝑛
𝑒 𝑛→∞
ln(𝑥)
lim (
)
𝑒 𝑥→∞ 𝑥
=
∞
𝑒0
1
) = lim (𝑒 𝑛 ln(𝑛) )
= 𝑒 0∙∞ → 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙
→ 𝑟𝑒𝑎𝑙 𝐿′ 𝐻𝑜𝑠𝑝𝑖𝑡𝑎𝑙,
ln(𝑥)
lim (
)
𝑥
𝑒 𝑥→∞
1
lim ( )
𝑥
= 𝑒 𝑥→∞
𝑎(𝑥) = 𝑥,
𝑏(𝑥) = ln(𝑥) ,
= 𝑒0 = 1
𝑛
lim ( √𝑛) = 1
𝑛→∞
These proofs may be used as standard forms in later paragraphs.
𝑎′ (𝑥) = 1
𝑏 ′ (𝑥) =
1
𝑥
Upper and lower boundaries:
An upperbound is a limit to which the sequence will not exceed. In other words, there exists a
constant such that:
{𝑎𝑛 } ≤ 𝐶1
In the case of the first proof if we let 𝑎 = 1, then the boundary is 𝑒.
A lowerbound is a limit to which the sequence will not exceed. In other words, there exists a
constant such that:
{𝑎𝑛 } ≥ 𝐶1
In the case of the second proof the limit is set at 1.
𝜋
Another case has an upper and lower bound: {𝑎𝑛 } = sin ( 4 𝑛)
The last case has no bounds (unbounded): {𝑎𝑛 } = −3𝑛 + 5
A convergent sequence is always bounded. In both proofs there were convergent sequences,
and both could be bounded either by an upper or lower bound. Do not get confused with the
inverse of this theorem: “A bounded sequence is convergent” as this is not always true as can be
𝜋
seen in the third figure. The sequence {𝑎𝑛 } = sin ( 4 𝑛) is not a convergent sequence as it does
not reach a real number at infinity but is bounded.
Monotonic sequences?
Sequences that are strictly increasing or decreasing is said to be monotonic. The following table
shows some tests to test whether sequences are monotonic.
Increasing
{𝑎𝑛 } = 𝑓(𝑥) → 𝑓 ′ (𝑥) > 0 for all 𝑥
{𝑎𝑛 } < {𝑎𝑛+1 } for all 𝑛
Decreasing
{𝑎𝑛 } = 𝑓(𝑥) → 𝑓 ′ (𝑥) < 0 for all 𝑥
{𝑎𝑛 } > {𝑎𝑛+1 } for all 𝑛
1<
1>
{𝑎𝑛+1 }
for
{𝑎𝑛 }
all 𝑛
{𝑎𝑛+1 }
for
{𝑎𝑛 }
all 𝑛
When a sequence is monotonic and bounded the sequence is said to be convergent.
In the first two cases the sequences were monotonic and bounded and therefore convergent.
The third was bounded but not monotonic and hence divergent. The last case was monotonic
but unbounded and hence divergent.
If from a sequence a finite number of terms is removed the limit will remain unchanged. Let’s
look at it graphically:
From each graph more and more terms were removed but the limit stayed unchanged.
Theme 2.2 – Series of real numbers
A series is a sum of a sequence of numbers.
𝑁
𝑆𝑛 = ∑{𝑎𝑛 }
𝑛=1
The partial sum only contains a finite number of terms of the sequence which is summed up. If
the partial sum of a sequence converges to a real value, 𝐿, then the sum will converge to the
same value.
Let 𝑆𝑛 = ∑𝑁
𝑛=1{𝑎𝑛 } be a partial sum.
If lim (𝑆𝑛 ) = 𝐿 then 𝑆𝑛 converges to 𝐿 and hence 𝑆𝑛 is convergent.
𝑛→∞
𝑁
𝑆𝑛 = ∑{ln(𝑛 + 1) − ln(𝑛)} = ln(2) − ln(1) + ln(3) − ln(2) + ⋯ + ln(𝑁 + 1) − ln(𝑁)
𝑛=1
= − ln(1) + ln(𝑁 + 1)
lim (𝑆𝑛 ) = − ln(1) + ln(∞) = ∞,
𝑛→∞
𝑁
1
1
1
1
1
𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
1
1
1
𝑆𝑛 = ∑ {𝑒 𝑡 − 𝑒 𝑡+1 } = 𝑒 1 − 𝑒 2 + 𝑒 2 − 𝑒 3 + ⋯ + 𝑒 𝑁 − 𝑒 𝑁+1 = 𝑒 1 − 𝑒 𝑁+1
𝑛=1
1
lim (𝑆𝑛 ) = 𝑒 1 − 𝑒 ∞ = 𝑒 − 1,
𝑛→∞
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
The above examples are known as telescoping series. One key feature of telescoping series lies
in the similar terms they produce that end up cancelling out when added. When testing for
convergence it is easier to calculate a partial sum and then taking an infinite limit.
𝑁
𝑁
𝑛=2
𝑛=2
1
1
1
𝑆𝑛 = ∑ { 2
− },
}= ∑{
𝑛 −𝑛
𝑛−1 𝑛
𝑁
∑{
𝑛=2
𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑠
1
1
1 1 1
1
1
1
− } = 1 − + − + ⋯+
− =1−
𝑛−1 𝑛
2 2 3
𝑁−1 𝑁
𝑁
lim (𝑆𝑛 ) = 1 −
𝑛→∞
1
= 1,
∞
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
Geometric series:
𝑁
𝑆𝑛 = ∑{𝑎 ∙ 𝑟 𝑛 } ,
𝑛=1
|𝑟| < 1,
|𝑟| ≥ 1,
𝑆∞ =
𝑎
1−𝑟
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
When linearly combining series care should be taken about the convergence and divergence of
the resulting series. When combining two convergent series the resulting series will be
convergent. When combine a convergent and divergent series the resulting series is always
divergent.
When combining a divergent and divergent series the resulting series has no conclusion.
𝑁
𝑁
𝑆𝑛 = ∑{ln(𝑛 + 1)} − ∑{ln(𝑛)} = ln(2) − ln(1) + ln(3) − ln(2) + ⋯ + ln(𝑁 + 1) − ln(𝑁)
𝑛=1
𝑛=1
= − ln(1) + ln(𝑁 + 1)
lim (𝑆𝑛 ) = − ln(1) + ln(∞) = ∞,
𝑛→∞
𝑁
𝑆𝑛 =
𝑁
1
1
1
1
𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
1
∑ {𝑒 𝑡 } −
∑ {𝑒 𝑡+1 } = 𝑒 1 + 𝑒 2 − 𝑒 2 − 𝑒 3 + ⋯ + 𝑒 𝑁 − 𝑒 𝑁+1 = 𝑒 1 − 𝑒 𝑁+1
𝑛=1
𝑛=1
1
lim (𝑆𝑛 ) = 𝑒 1 − 𝑒 ∞ = 𝑒 − 1,
𝑛→∞
1
1
1
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
Test for divergence:
𝑁
𝑆𝑛 = ∑{𝑎𝑛 }
𝑛=1
If lim {𝑎𝑛 } ≠ 0 then the series, 𝑆𝑛 , does diverge.
𝑛→∞
𝑁
1
𝑆𝑛 = ∑ {𝑒 𝑡 } ,
𝑛=1
1
lim {𝑒 𝑡 } = 𝑒 0 = 1 ≠ 0
𝑛→∞
1
Therefore, the series, ∑𝑁
𝑛=1 {𝑒 𝑡 }, diverges.
Be careful of the following statement: “ if the limit is zero the series converges”. This is not
always true. A convergent series will always have a limit equal to zero, but a zero limit does not
conclude convergence. Let’s look at an example of this.
Consider the following harmonic series:
𝑁
1
1 1 1 1 1 1 1
1
𝑆𝑛 = ∑ { } = 1 + + + + + + + + ⋯ +
𝑛
2 3 4 5 6 7 8
𝑁
𝑛=1
1 1
1
1
1 1
1 1 1 1
1 1 1
1 + + + ⋯ > 1 + + ( + ) + ( + + + ) += 1 + + + + ⋯
2 3
𝑁
2
4 4
8 8 8 8
2 2 2
𝑁
𝑁
𝑛=1
𝑛=1
1
1
∑{ }>1+∑{ }
𝑛
2
1
1
𝑁
By inspection it is easy to see that 1 + ∑𝑁
𝑛=1 {2} diverges, and since ∑𝑛=1 {𝑛} is larger than the
1
first then by the squeeze theorem ∑𝑁
𝑛=1 {𝑛} is divergent.
To use the test for divergence:
𝑁
1
𝑆𝑛 = ∑ { }
𝑛
𝑛=1
1
lim { } = 0
𝑛→∞ 𝑛
Theme 2.3 – Series with non-negative numbers
Let’s look at a few tests that may have a conclusion on the convergence of a series. The integral
test is one test one may use to determine convergence.
The integral test:
𝑏
∫ 𝑓(𝑥) ∙ 𝑑𝑥 ≈ ∑
𝑎
𝑏
𝑛=𝑎
∞
∫ 𝑓(𝑥) ∙ 𝑑𝑥 ∝ ∑
0
𝑓(𝑛) ∙ ∆𝑛
∞
𝑓(𝑛)
𝑛=0
This shows that the behaviour of an integral has the same behaviour as a series. Although this
test does not give the value of the series, it may be used to determine convergence.
The integral test may only be used if:
𝑓(𝑥) > 0,
𝑓 ′ (𝑥) < 0,
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 > 𝑥0 ,
𝑥0 ∈ ℝ
𝑓(𝑥) 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑎 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
In words, the function must be positive, decreasing and continuous.
∞
If the integral, ∫0 𝑓(𝑥) ∙ 𝑑𝑥, converges then the series will also converge. If the integral diverges,
the series will also diverge.
P-series:
∞
∑
𝑛=1
1
1
1
1
1
1
= 𝑝+ 𝑝+ 𝑝+ 𝑝+ 𝑝+⋯
𝑝
𝑛
1
2
3
4
5
𝑝
𝑝
𝑝
𝑝
Let 𝑝 < 0: ∑∞
𝑛=1 𝑛 = 1 + 2 + 3 + ⋯
Using the test for divergence: lim 𝑛𝑝 = ∞ ≠ 0, therefore this series is divergent.
𝑛→∞
Let 𝑝 = 0: ∑∞
𝑛=1
1
𝑛0
=
1
10
+
1
20
1
30
+
+⋯=1+1+1+1+⋯
Using the test for divergence: lim 1 = 1 ≠ 0, therefore this series is divergent.
𝑛→∞
Let 0 < 𝑝 < 1: ∑∞
𝑛=1
1
𝑛𝑝
=
1
1𝑝
+
∞ 1
𝑥𝑝
Using the integral test: ∫0
1
2𝑝
+
1
3𝑝
+⋯
𝑏 1
𝑥𝑝
∙ 𝑑𝑥 = lim (∫0
𝑏→∞
is divergent because the integral diverges.
Let 𝑝 = 1: Harmonic series. Divergent.
And finally let 1 < 𝑝: ∑∞
𝑛=1
1
𝑛𝑝
=
1
1𝑝
+
1
2𝑝
+
1
3𝑝
1
∙ 𝑑𝑥) = lim (1−𝑝 𝑏1−𝑝 ) = ∞ therefore this series
+⋯
𝑏→∞
∞ 1
𝑥𝑝
Using the integral test: ∫0
𝑏 1
𝑥𝑝
∙ 𝑑𝑥 = lim (∫0
𝑏→∞
−1
1 (𝑝−1)
∙ 𝑑𝑥) = lim ((𝑝−1) (𝑏)
𝑏→∞
) = 0 therefore this
series is convergent because the integral converges. So, after a long exercise, the p-series is only
convergent when 𝑝 > 1.
The next two tests are known as the direct comparison test and the limit comparison test. The
direct comparison test is like the squeeze theorem, but it is only focused on one side of the
inequality. The side will depend on what is being tested for.
Consider two series:
∞
∞
∑{𝑎𝑛 } ,
∑{𝑏𝑛 }
𝑛=1
𝑛=1
∞
∞
∞
If ∑∞
𝑛=1{𝑎𝑛 } converges and ∑𝑛=1{𝑎𝑛 } > ∑𝑛=1{𝑏𝑛 } then ∑𝑛=1{𝑏𝑛 } will converge.
∞
∞
∞
If ∑∞
𝑛=1{𝑎𝑛 } diverges and ∑𝑛=1{𝑎𝑛 } < ∑𝑛=1{𝑏𝑛 } then ∑𝑛=1{𝑏𝑛 } will diverge.
Makes sense, right? We may not however draw any conclusion from the following:
∞
∞
If ∑∞
𝑛=1{𝑎𝑛 } converges and ∑𝑛=1{𝑎𝑛 } < ∑𝑛=1{𝑏𝑛 }. (No conclusion)
∞
∞
If ∑∞
𝑛=1{𝑎𝑛 } diverges and ∑𝑛=1{𝑎𝑛 } > ∑𝑛=1{𝑏𝑛 }. (No conclusion)
The limit comparison test:
Suppose that the following series have only positive terms
∞
lim (
𝑛→∞
{
{𝑎𝑛 }
) = 0,
{𝑏𝑛 }
lim (
𝑛→∞
{𝑎𝑛 }
) = ∞,
{𝑏𝑛 }
∑{𝑎𝑛 } ,
∑{𝑏𝑛 }
𝑛=1
𝑛=1
∞
{𝑎𝑛 }
lim (
) = 𝑐,
𝑛→∞ {𝑏𝑛 }
𝐼𝑓:
∞
𝑐 ∈ ℝ>0 ,
∑{𝑎𝑛 } 𝑤𝑖𝑙𝑙 𝑏𝑒ℎ𝑎𝑣𝑒 𝑙𝑖𝑘𝑒 ∑{𝑏𝑛 }
𝑛=1
∞
∞
∞
𝑛=1
∑{𝑏𝑛 } 𝑖𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡,
∑{𝑎𝑛 } 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡
𝑛=1
∞
𝑛=1
∞
∑{𝑏𝑛 } 𝑖𝑠 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑡,
∑{𝑎𝑛 } 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑡
𝑛=1
𝑛=1
Let’s look at a few examples on how to apply these tests:
∞
∑
𝑛=1
∞
∫
1
∑∞
𝑛=1
1 1
𝑒𝑛
𝑛2
1 1
𝑒𝑛
𝑛2
𝑏
1
1
1 1
1 1
𝑥 ∙ 𝑑𝑥 = lim (∫
𝑒
𝑒 𝑥 ∙ 𝑑𝑥) = lim (−𝑒 𝑏 + 𝑒 1 ) = 𝑒 − 1
2
2
𝑏→∞
𝑏→∞
𝑥
1 𝑥
is therefore a convergent series by the integral test.
}
∞
∑
𝑛=1
1 > sin(𝑛) ,
5 − 2 ∙ sin(𝑛)
𝑛
−2 < −2 ∙ sin(𝑛) ,
3 5 − 2 ∙ sin(𝑛)
<
𝑛
𝑛
3
∞
∑∞
𝑛=1 𝑛 diverges by the p-series, and since it is less than ∑𝑛=1
5−2∙sin(𝑛)
,
𝑛
∑∞
𝑛=1
5−2∙sin(𝑛)
𝑛
will also
diverge by the direct comparison test.
∞
∑(
𝑛=1
∞
∑(
𝑛=1
2
√𝑛
2
√𝑛
+
(−1)𝑛
)
3𝑛+1
∞
)+ ∑(
𝑛=1
(−1)𝑛
)
3𝑛+1
2
∑∞
𝑛=1 ( 𝑛) is divergent by the p-series test.
√
∞
∞
∞
𝑛=1
𝑛=1
𝑛=1
(−1)𝑛
(−1)𝑛
1 1 𝑛
∑ ( 𝑛+1 ) < ∑ (| 𝑛+1 |) = ∑ ( ∙ ( ) )
3
3
3 3
(−1)𝑛
∑∞
𝑛=1 (
3𝑛+1
2
) converges by the direct comparison test and geometric series. Since ∑∞
𝑛=1 ( 𝑛) is
divergent and is added to the convergent series
(−1)𝑛
∑∞
𝑛=1 ( 3𝑛+1 ),
2
∑∞
𝑛=1 (√𝑛) +
√
(−1)𝑛
∞
∑𝑛=1 ( 𝑛+1 ) will
3
be a
divergent series.
The final example:
∞
1 𝑛
∑ (1 + )
𝑛
𝑛=1
1 𝑛
lim ((1 + ) ) = 𝑒 1 ≠ 0
𝑛→∞
𝑛
1 𝑛
∑∞
𝑛=1 (1 + 𝑛) diverges by the test for divergence.
Note through doing many problems these techniques will become more intuitive. There is no set
rule for when to use what.
Theme 2.4 – Absolute and conditional convergence
An alternating series is a series were terms are alternating between positive and negative
terms:
∞
∑(
𝑛=1
(−1)𝑛
(−1)1 (−1)2 (−1)3 (−1)5
1 1 1 1 1
+
+
+
+⋯=− + − + − +⋯
)=
𝑛
1
2
4
5
1 2 3 4 5
Note the terms are alternating hence the name alternating series. The big question is whether
the series converges or diverges. The following test may be used and is known as the alternating
series test. Consider the following alternating series:
∞
∑(−1)𝑛 {𝑎𝑛 } = −𝑎1 + 𝑎2 − 𝑎3 + ⋯
𝑛=1
The series will converge if {𝑎𝑛 } is strictly decreasing and have only positive terms and the limit
of the sequence is zero. In math terms
{𝑎𝑛 } → 𝑓(𝑥),
𝑓 ′ (𝑥) < 0
{𝑎𝑛 } > 0
lim ({𝑎𝑛 }) = 0
𝑛→∞
1 1
Looking at the demonstration, we can see that {𝑎𝑛 } = 𝑛. 𝑛 is decreasing an only has positive
1
𝑛→∞ 𝑛
terms. The limit lim ( ) does evaluate to zero. Therefore, the series converges by the
alternating series test.
The two graphs show the series and the sequence respectively. Note that the sequence does
converge to zero and the series does converge to a value as well, in this case − ln(2) ≈ −0.69.
What can we expect the error to be when we only consider the partial sums of these series? The
error will always be less than the absolute value of first term that is neglected of the sequence.
An example will demonstrate this well.
{𝑎𝑛 } =
(−1)𝑛
1 1 1 1 1 1
= −1, , − , , − , , − , …
𝑛
2 3 4 5 6 7
∞
(−1)𝑛
1 1 1 1
𝑆∞ = ∑
= −1 + − + − + ⋯ = − ln(2)
𝑛
2 3 4 5
𝑛=1
Consider the partial sum of the first 10 terms:
10
𝑆10 = ∑
𝑛=1
(−1)𝑛
1 1
1 1
= −1 + − + ⋯ − +
= −0.6456
𝑛
2 3
9 10
1
11
The next term to be added is 𝑎11 = − . This value will overshoot the series value.
1 1
1 1
1
𝑆11 = 𝑆10 + 𝑎11 = −1 + − + ⋯ − +
−
= −0.7365
2 3
9 10 11
We can then estimate the error as being less than the next term in the sequence:
|𝑆∞ − 𝑆𝑛 | = |𝑒𝑟𝑟𝑜𝑟| < |𝑎𝑛+1 |
|− ln(2) − (−0.6456)| = 0.0475 < 0.0909
This is always true for convergent alternating series. This is helpful to determine the required
terms to estimate the value of a sum.
Determine the number of terms required to ensure that the error is less than 0.01.
∞
𝑆∞ = ∑
𝑛=1
(−1)𝑛
1 1 1 1
= −1 + − + − + ⋯ = − ln(2)
𝑛
2 3 4 5
{𝑎𝑛 } =
(−1)𝑛
1 1 1 1 1 1
= −1, , − , , − , , − , …
𝑛
2 3 4 5 6 7
(−1)𝑛+1
1
|
|=
< 0.01,
𝑛+1
𝑛+1
𝑛 = 99
What is absolute convergence? When taking the absolute value of an alternating series, it is
obviously larger than the original series. If the absolute value of an alternating series is
convergent, then by the direct comparison test, the original series is also convergent. This
alternating series is then known as absolutely convergent. A series that is only convergent but
diverges when the absolute value of the series is taken is conditionally convergent. Let’s look at
an example:
∞
𝑆∞ = ∑
𝑛=1
(−1)𝑛
= − ln(2) ,
𝑛
∞
∞
𝑛=1
𝑛=1
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 (𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑠𝑒𝑟𝑖𝑒𝑠 𝑡𝑒𝑠𝑡)
(−1)𝑛
1
𝑆∞ = ∑ |
| = ∑ = ∞,
𝑛
𝑛
𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑠𝑒𝑟𝑖𝑒𝑠)
This series is referred to as conditionally convergent.
The next example:
∞
𝑆∞ = ∑ (
𝑛=1
(−1)𝑛
1
)=− ,
𝑛
3
4
∞
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 (𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑠𝑒𝑟𝑖𝑒𝑠 𝑡𝑒𝑠𝑡)
∞
(−1)𝑛
1 𝑛
1
𝑆∞ = ∑ |( 𝑛 )| = ∑ (( ) ) = ,
3
3
2
𝑛=1
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 (𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑒𝑟𝑖𝑒𝑠)
𝑛=1
This series is referred to as an absolutely convergent series.
Remember that all previous tests may be used to determine convergence.
Two new tests are introduced. Ratio and root test.
Ratio test:
∑{𝑎𝑛 }
𝑎𝑛+1
𝐿 < 1,
∑{𝑎𝑛 } 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
|) = 𝐿,
𝑛→∞
𝑎𝑛
𝑎𝑛+1
𝑖𝑓 lim (|
𝐿 > 1,
∑{𝑎𝑛 } 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
|) = 𝐿,
𝑛→∞
𝑎𝑛
𝑎𝑛+1
𝑖𝑓 lim (|
𝐿 = 1,
𝑛𝑜 𝑐𝑜𝑛𝑐𝑙𝑢𝑠𝑖𝑜𝑛
|) = 𝐿,
𝑛→∞
𝑎𝑛
𝑖𝑓 lim (|
Root test:
∑{𝑎𝑛 }
1
𝑖𝑓 lim (|𝑎𝑛 |𝑛 ) = 𝐿,
𝑛→∞
1
𝑖𝑓 lim (|𝑎𝑛 |𝑛 ) = 𝐿,
𝑛→∞
𝐿 < 1,
𝐿 > 1,
1
𝑖𝑓 lim (|𝑎𝑛 |𝑛 ) = 𝐿,
𝐿 = 1,
𝑛→∞
∑{𝑎𝑛 } 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠
∑{𝑎𝑛 } 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠
𝑛𝑜 𝑐𝑜𝑛𝑐𝑙𝑢𝑠𝑖𝑜𝑛
Example:
∞
∑
𝑛=1
𝑛!
2𝑛
2
(𝑛 + 1)!
2
(𝑛 + 1) ∙ (𝑛!) 2𝑛
(𝑛 + 1)
𝑎𝑛+1
(𝑛+1)2
2
lim (|
∙
|) = lim (| 2𝑛 |)
|) = lim (|
|) = lim (| 𝑛2 2𝑛
𝑛!
𝑛→∞
𝑛→∞
𝑛→∞
𝑛→∞
𝑎𝑛
2 ∙2
2 ∙ 2 ∙ 2 (𝑛!)
2
2𝑛
(𝑛 + 1)
(1)
lim (| 2𝑛
|) = lim (| 2𝑛
|) = 0 = 𝐿 < 1,
𝑛→∞
𝑛→∞ 2
2 ∙2
∙ 4 ∙ ln(2)
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡 (𝑟𝑎𝑡𝑖𝑜 𝑡𝑒𝑠𝑡)
Theme 3: Power series
10 Lectures [week 6 & 7 & 8]
This theme will introduce power series. The theme will address convergence and
representation of series as functions. Taylor series and polynomials will be addressed. The final
subtheme will solve differential equations using power series.
Theme 3.1: Sequences of real numbers
Theme 3.2: Series of real numbers
Theme 3.3: Series with non-negative terms
Theme 3.4: Absolute- and conditional convergence
Pre-knowledge for this theme consists of Taylor polynomials (WTW164).
Theme 3.1 – Convergence of a power series
What is a power series?
∞
∑ 𝑐𝑛 𝑥 𝑛 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 + ⋯
𝑛=0
𝑐𝑛 are constant coefficients of the polynomial terms. A power series about a point 𝑎 will have a
number 𝑅 which is known as the radius of convergence. Therefore, the series will converge for
values of 𝑥 on an interval 𝑥 ∈ (𝑎 − 𝑅, 𝑎 + 𝑅).
∞
∑ 𝑐𝑛 (𝑥 − 𝑎)𝑛 ,
𝑥 ∈ (𝑎 − 𝑅, 𝑎 + 𝑅)
𝑛=0
How can we determine the radius of convergence?
∞
2 𝑛
𝑆∞ = ∑ 𝑛 ( ) ∙ 𝑥 𝑛
3
𝑛=0
Always, always, always do a ratio test.
2 𝑛 𝑛
𝑐𝑛 = 𝑛 ( ) 𝑥 ,
3
𝑐𝑛+1
2 𝑛+1 𝑛+1
(𝑛
=
+ 1) ( )
𝑥
,
3
2
1
| ∙ 𝑥 ∙ lim (1 + )| < 1,
𝑛→∞
3
𝑛
3
3
− <𝑥< ,
2
2
−3
𝑓𝑜𝑟 𝑥 =
,
2
𝑓𝑜𝑟 𝑥 =
−3
,
2
2
−1 < 𝑥 < 1
3
𝑎 = 0,
∞
𝑐𝑛+1 2
1
= ∙ 𝑥 ∙ (1 + )
𝑐𝑛
3
𝑛
𝑅=
3
2
∞
2 𝑛
3 𝑛
𝑆∞ = ∑ 𝑛 ∙ ( ) ∙ (− ) = ∑ 𝑛 ∙ (−1)𝑛 ⟹ 𝐷𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (𝑎𝑙𝑡. 𝑠𝑒𝑟𝑖𝑒𝑠 𝑡𝑒𝑠𝑡)
3
2
𝑛=0
∞
𝑛
𝑛
𝑛=0
∞
2
3
𝑆∞ = ∑ 𝑛 ∙ ( ) ∙ ( ) = ∑ 𝑛 ⟹ 𝐷𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (𝑝 − 𝑠𝑒𝑟𝑖𝑒𝑠: 𝑝 = −1)
3
2
𝑛=0
𝑛=0
Therefore, the series will only converge on the interval:
3 3
𝑥 ∈ (− , ) ,
2 2
𝑅=
3
2
Theme 3.2 – Representation of functions as power series
Recall the geometric series:
∞
𝑆∞ = ∑ 𝑎𝑟 𝑛 =
𝑛=0
𝑎
1−𝑟
If we let 𝑎 = 1 and 𝑟 = 𝑥:
∞
𝑆∞ = ∑ 𝑥 𝑛 =
𝑛=0
1
1−𝑥
Some special (algebraic) manipulations:
1
1
=
|
1 − 𝛼𝑥 1 − 𝑋 𝑋→𝛼𝑥
1
1
=
|
𝛼
1−𝑥
1 − 𝑋 𝑋→𝑥 𝛼
1
1
1
1
=
= ∙
|
1
𝛼 − 𝑥 𝛼 (1 − 𝑥) 𝛼 (1 − 𝑋) 𝑋→ 1 𝑥
𝛼
𝛼
Through (calculus) manipulation different functions can be represented:
∞
𝑑
1
(𝑆∞ ) = ∑ 𝑛 ∙ 𝑥 𝑛−1 = −
(1 − 𝑥)2
𝑑𝑥
𝑛=0
∞
∫ 𝑆∞ ∙ 𝑑𝑥 = ∑
𝑛=1
1
1
𝑥 𝑛+1 = − ln(1 − 𝑥) + 𝐶 = ln (
)+𝐶
𝑛+1
1−𝑥
How does integrating and differentiating the power series influence the radius of convergence?
∞
𝑆∞ = ∑ 𝑥 𝑛 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯
𝑛=0
∞
𝑑
(𝑆 ) = ∑ 𝑛 ∙ 𝑥 𝑛−1 = 0 + 1 + 2𝑥 + 3𝑥 2 + ⋯
𝑑𝑥 ∞
𝑛=0
∞
∫ 𝑆∞ ∙ 𝑑𝑥 = ∑
𝑛=0
1
1
1
1
𝑥 𝑛+1 = 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 + ⋯
𝑛+1
2
3
4
Using the ratio test to find the radius of convergence of the following:
∞
𝑆∞ = ∑ 𝑥 𝑛 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯
𝑛=0
| lim (
𝑛→∞
𝑥 𝑛+1
)| < 1,
𝑥𝑛
−1 < 𝑥 < 1,
𝑅=1
The derivative:
∞
𝑑
(𝑆 ) = ∑ 𝑛 ∙ 𝑥 𝑛−1 = 0 + 1 + 2𝑥 + 3𝑥 2 + ⋯
𝑑𝑥 ∞
𝑛=0
| lim (
𝑛→∞
(𝑛 + 1)𝑥 𝑛
)| < 1,
𝑛𝑥 𝑛−1
−1 < 𝑥 < 1,
𝑅=1
∞
∑ 𝑛 ∙ (−1)𝑛−1 ,
𝑓𝑜𝑟 𝑥 = −1,
𝐷𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (𝑎𝑙𝑡. 𝑠𝑒𝑟𝑖𝑒𝑠 𝑡𝑒𝑠𝑡)
𝑛=0
∞
−1 < 𝑥 < 1,
𝑓𝑜𝑟 𝑥 = 1,
{
𝐷𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (𝑝 − 𝑠𝑒𝑟𝑖𝑒𝑠: 𝑝 = −1)
∑𝑛,
𝑛=0
And finally, the integral:
∞
∫ 𝑆∞ ∙ 𝑑𝑥 = ∑
𝑛=0
| lim (
𝑛→∞
1
1
1
1
𝑥 𝑛+1 = 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 + ⋯
𝑛+1
2
3
4
(𝑛 + 2)𝑥 𝑛+2
)| < 1,
(𝑛 + 1)𝑥 𝑛+1
∞
𝑓𝑜𝑟 𝑥 = −1,
∑
𝑛=0
∞
−1 ≤ 𝑥 < 1,
{
𝑓𝑜𝑟 𝑥 = 1,
−1 < 𝑥 < 1,
1
∙ (−1)𝑛+1 ,
𝑛+1
∑
𝑛=0
1
,
𝑛+1
𝑅=1
𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 (𝑎𝑙𝑡. 𝑠𝑒𝑟𝑖𝑒𝑠 𝑡𝑒𝑠𝑡)
𝐷𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (𝑙𝑖𝑚𝑖𝑡 𝑐𝑜𝑚𝑝𝑎𝑟𝑖𝑠𝑜𝑛 𝑡𝑒𝑠𝑡)
Note that the radius of convergence remained unchanged, but the interval did change.
The ol’reliable:
𝑓(𝑥) = tan−1 (𝑥)
𝑓 ′ (𝑥) =
1
1
1
=
=
|
2
2
1+𝑥
1 − (−𝑥 ) 1 − 𝛽 𝛽→−𝑥2
∞
∞
∞
𝑛=0
𝑛=0
𝑛=0
1
= ∑ 𝛽 𝑛 = ∑(−𝑥 2 )𝑛 = ∑(−1)𝑛 ∙ (𝑥 2𝑛 )
1−𝛽
∞
𝑓
′ (𝑥)
∞
𝑛
= ∑(−1) ∙
𝑛=0
(𝑥 2𝑛 )
,
−1 (𝑥)
𝑓(𝑥) = tan
= ∑(−1)𝑛 ∙
𝑛=0
1
∙ (𝑥 2𝑛+1 )
2𝑛 + 1
Theme 3.3 – Taylor-polynomials and Taylor-series
How can a polynomial be used to describe more complex functions such as exponentials and
trigonometric functions?
𝑓(𝑥) = 𝑒 𝑥
𝑓 ′ (𝑥) = 𝑒 𝑥 ,
𝑓 ′′ (𝑥) = 𝑒 𝑥 ,
𝑓 ′′′ (𝑥) = 𝑒 𝑥 ,
,… ,
𝑓 𝑛 (𝑥) = 𝑒 𝑥
𝑓 𝑛 (𝑎) = 𝑒 𝑎
∞
𝑇𝑛 (𝑥) = ∑
𝑛=0
𝑓 𝑛 (𝑎)
∙ (𝑥 − 𝑎)𝑛
𝑛!
∞
𝑥
𝑓(𝑥) = 𝑒 = ∑
𝑛=0
𝑒𝑎
∙ (𝑥 − 𝑎)𝑛
𝑛!
Let us approximate 𝑓(𝑥) = 𝑒 𝑥 about the point 𝑥 = 0 by using a few terms from the Taylor
polynomial (note that a Taylor polynomial centred about 𝑥 = 0 is known as a Maclaurin series:
1
𝑒𝑥 ≈ ∑
𝑛=0
1
∙ (𝑥)𝑛 = 1 + 𝑥,
𝑛!
1 1
𝑥 ∈ (− , )
2 2
Note that at the point 𝑥 = 0 the Taylor polynomial is equal to the function but further from the
point, the polynomial diverges. By adding more terms, the accuracy may be increased.
4
𝑥
𝑒 ≈∑
𝑛=0
1
1
1
1
∙ (𝑥)𝑛 = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4
𝑛!
2
6
24
The polynomial converges to the function the more terms is added.
Let’s look at some trigonometric functions:
𝑓(𝑥) = sin(𝑥)
𝑎 = 0,
𝑓 ′ (𝑥) = cos(𝑥) ,
𝑓 ′′ (𝑥) = − sin(𝑥) ,
𝑓 ′ (0) = 1,
∞
𝑇𝑛 (𝑥) = ∑
𝑛=0
𝑓(0) = sin(0) = 0
𝑓 ′′ (0) = 0,
𝑓 ′′′ (𝑥) = − cos(𝑥) ,
𝑓 ′′′ (0) = −1,
𝑓 𝑛 (𝑥) =?
,… ,
𝑓 𝑛 (𝑥) =?
,… ,
𝑓 𝑛 (0)
0
1
0
−1 3 0 4 1 5 0 6 −1 7
∙ (𝑥)𝑛 = 𝑥 0 + 𝑥 1 + 𝑥 2 +
𝑥 + 𝑥 + 𝑥 + 𝑥 +
𝑥 +⋯
𝑛!
0!
1!
2!
3!
4!
5!
6!
7!
∞
𝑇𝑛 (𝑥) = ∑
𝑛=0
𝑓 𝑛 (0)
1
1
1
1
∙ (𝑥)𝑛 = 𝑥 1 − 𝑥 3 + 𝑥 5 − 𝑥 7 + ⋯
𝑛!
1!
3!
5!
7!
∞
𝑓(𝑥) = sin(𝑥) = ∑
𝑛=0
(−1)𝑛
∙ (𝑥)2𝑛+1
(2𝑛 + 1)!
Similarly, for 𝑓(𝑥) = cos(𝑥):
𝑓(𝑥) = cos(𝑥)
𝑎 = 0,
𝑓 ′ (𝑥) = − sin(𝑥) ,
𝑓 ′′ (𝑥) = − cos(𝑥) ,
𝑓 ′ (0) = 0,
∞
𝑇𝑛 (𝑥) = ∑
𝑛=0
𝑓(0) = cos(0) = 1
𝑓 ′′ (0) = −1,
𝑓 ′′′ (𝑥) = sin(𝑥) ,
𝑓 ′′′ (0) = 0,
,… ,
𝑓 𝑛 (𝑥) =?
,… ,
𝑓 𝑛 (𝑥) =?
𝑓 𝑛 (0)
1
0
−1 2 0 3 1 4 0 5 −1 6 0 7
∙ (𝑥)𝑛 = 𝑥 0 + 𝑥 1 +
𝑥 + 𝑥 + 𝑥 + 𝑥 +
𝑥 + 𝑥 +⋯
𝑛!
0!
1!
2!
3!
4!
5!
6!
7!
∞
𝑇𝑛 (𝑥) = ∑
𝑛=0
𝑓 𝑛 (0)
1
1
1
1
∙ (𝑥)𝑛 = 𝑥 0 − 𝑥 2 + 𝑥 4 − 𝑥 6 + ⋯
𝑛!
0!
2!
4!
6!
∞
𝑓(𝑥) = cos(𝑥) = ∑
𝑛=0
(−1)𝑛
∙ (𝑥)2𝑛
(2𝑛)!
These approximations using only a few terms (four) has a high level of accuracy.
∞
𝑒𝑎
𝑥
𝑒 = ∑ ∙ (𝑥 − 𝑎)𝑛 ,
𝑛!
𝑛=0
∞
(−1)𝑛
sin(𝑥) = ∑
∙ (𝑥)2𝑛+1 ,
(2𝑛 + 1)!
𝑛=0
∞
(−1)𝑛
cos(𝑥) = ∑
∙ (𝑥)2𝑛
(2𝑛)!
𝑛=0
The last function if left for the reader as an exercise. Luckily the writer is also a reader.
𝑓(𝑥) = ln(𝑥)
𝑎 = 1,
𝑓 ′ (𝑥) =
∞
𝑇𝑛 (𝑥) = ∑
𝑛=0
1
,
𝑥
𝑓 ′′ (𝑥) = −
1
,
𝑥2
𝑓(𝑎) = 0
𝑓 ′′′ (𝑥) =
(1) ∙ (2)
,
𝑥3
,… ,
𝑓 𝑛 (𝑥) =?
𝑓 𝑛 (1)
∙ (𝑥 − 1)𝑛
𝑛!
=
(1) ∙ (2) 1
ln(1) 1 1
1 1
+ ∙ ∙ (𝑥 − 1) − 2 ∙ ∙ (𝑥 − 1)2 +
∙ ∙ (𝑥 − 1)3 + ⋯
0!
1 1!
1 2!
13
3!
1 1
1 1
1 1
= 0 + ∙ ∙ (𝑥 − 1) − ∙ 2 ∙ (𝑥 − 1)2 + ∙ 3 ∙ (𝑥 − 1)3 + ⋯
1 1
2 1
3 1
∞
=∑
𝑛=1
(−1)𝑛−1
1
1
∙ (𝑥 − 1)𝑛 = (𝑥 − 1) − ∙ (𝑥 − 1)2 + ∙ (𝑥 − 1)3 + ⋯
𝑛
2
3
ln(𝑥) =
𝑥 − 1 (𝑥 − 1)2 (𝑥 − 1)3
−
+
+⋯
1
2
3
Note that the partial Taylor polynomial can be used to approximate the function. If a remainder
term is incorporated into the equation such that:
𝑛
𝑓(𝑥) = 𝑇𝑛 (𝑥) + 𝑅𝑛 (𝑥) = ∑
𝑛=0
𝑓 𝑛 (𝑎)
𝑓 𝑛+1 (𝑐)
∙ (𝑥 − 𝑎)𝑛 +
∙ (𝑥 − 𝑎)𝑛+1
(𝑛 + 1)!
𝑛!
𝑐 ∈ (𝑥, 𝑎)
If the limit, lim (𝑅𝑛 (𝑥)) = 0 then that will result in 𝑓(𝑥) = 𝑇∞ (𝑥).
𝑛→∞
So, assume that 𝑓 𝑛+1 (𝑐) = 𝐾 does exist, then:
lim (
𝑛→∞
(𝑥 − 𝑎)𝑛
𝐾
∙ (𝑥 − 𝑎)𝑛+1 ) = 𝐾(𝑥 − 𝑎) ∙ lim (
)=0
𝑛→∞ (𝑛 + 1)!
(𝑛 + 1)!
From this it can be concluded that the Taylor polynomials will always converge to the function.
But how far will it converge? How can the radius of convergence be determined?
Remember the old friend, the ratio test:
∞
𝑥
𝑓(𝑥) = 𝑒 = ∑
𝑛=0
{𝑎𝑛 } =
𝑒𝑎
∙ (𝑥 − 𝑎)𝑛 ,
𝑛!
𝑒𝑎
∙ (𝑥 − 𝑎)𝑛
𝑛!
{𝑎𝑛+1 } =
𝑒𝑎
∙ (𝑥 − 𝑎)𝑛+1
(𝑛 + 1)!
{𝑎𝑛+1 } 𝑥 − 𝑎
=
{𝑎𝑛 }
𝑛+1
lim (|
𝑛→∞
𝑥−𝑎
|) = 0 ∙ (𝑥 − 𝑎)
𝑛+1
Therefore, for all real values of 𝑥 the limit will always be 0 and hence, less than 1. This means
that the radius of convergence is ∞. Irrespective about which point the polynomial is centred.
What is a binomial series?
𝑛
𝑘
(𝑘(𝑘 − 1)) 2 (𝑘(𝑘 − 1)(𝑘 − 2)) 3
∑ ⟨ ⟩ 𝑥 𝑛 = (1 + 𝑥)𝑘 = 1 + 𝑘𝑥 +
𝑥 +
𝑥 +⋯
𝑛
2!
3!
𝑛=0
(𝑘!)
𝑘
⟨ ⟩=
𝑛
𝑛! (𝑘 − 𝑛)!
Taylors theorem:
𝑛
𝑓(𝑥) = ∑
𝑛=0
𝑓 𝑛 (𝑎)
𝑓(𝑎) 𝑓 ′ (𝑎)
𝑓 ′′ (𝑎)
(𝑥 − 𝑎)1 +
(𝑥 − 𝑎)2 + ⋯ + 𝑅𝑛 (𝑥)
∙ (𝑥 − 𝑎)𝑛 + 𝑅𝑛 (𝑥) =
+
𝑛!
0!
1!
2!
𝑅𝑛 (𝑥) =
𝑓 𝑛+1 (𝑐)
∙ (𝑥 − 𝑎)𝑛+1
(𝑛 + 1)!
The error is estimated by using the remainder when the Taylor series is used to approximate
the function value. Let’s look at a Maclaurin series of ln(𝑥 + 1) if only the third order is used to
approximate the function at ln(1.1).
𝑓(𝑥) = ln(1 + 𝑥) ,
𝑓(0) = 0,
𝑓 ′ (𝑥) =
𝑓 ′ (0) = 1,
1
,
1+𝑥
𝑓 ′′ (𝑥) = −
𝑓 ′′ (𝑥) = −1,
1
,…
(1 + 𝑥)2
𝑓 ′′′ (0) = 2
𝑛
𝑛
𝑛=0
𝑛=1
𝑓 𝑛 (0)
𝑥2 𝑥3 𝑥4
𝑓 𝑛 (0)
∑
∙ (𝑥)𝑛 = 0 + 𝑥 − + − + ⋯ = ∑
∙ (𝑥)𝑛
𝑛!
2
3
4
𝑛!
𝑇3 (𝑥) = 𝑥 −
𝑥2 𝑥3
+
≈ ln(1 + 𝑥) ,
2
3
𝑇3 (0.1) = 0.1 −
𝑓(0.1) = ln(1.1)
0.12 0.13
+
= 0.09533 ≈ ln(1.1) = 0.09531
2
3
𝑛
𝑅𝑛 (𝑥) = 𝑓(𝑥) − ∑
𝑛=0
𝑓 𝑛 (𝑎)
∙ (𝑥 − 𝑎)𝑛 = 0.0953101 − 0.0953333 = −0.0000232
𝑛!
𝑅𝑛 (𝑥) = −0.0000232 ≈ −
0.14
= −0.000025
4
The error can be approximated using the first term that was neglected in the approximation.
Let’s do another example:
Approximate the value of:
0
2
∫ 𝑒 𝑥 ∙ 𝑑𝑥
−1
∞
∞
1
𝑒 𝑥 = ∑ ∙ (𝑥)𝑛 ,
𝑛!
𝑒
𝑥2
=∑
𝑛=0
0
∫ 𝑒
−1
𝑥2
𝑛=0
1
∙ (𝑥)2𝑛
𝑛!
∞
∞
𝑛=0
𝑛=0
0
1
1
1
−1
2𝑛+1
(𝑥)
∙ 𝑑𝑥 = ∑ ∙
∙
∙ (−1)2𝑛+1
| =∑ ∙
𝑛! 2𝑛 + 1
𝑛! 2𝑛 + 1
−1
∞
∑
𝑛=0
1
1
1 1
1
1 1
1
∙
=1+ +
+
+⋯≈1+ +
+
𝑛! 2𝑛 + 1
3 10 42
3 10 42
0
1 1
1
2
∫ 𝑒 𝑥 ∙ 𝑑𝑥 ≈ 1 + +
+
= 1.457
3 10 42
−1
Find a Taylor polynomial centred about 𝑥 = 2 for ln(3𝑥 + 2) and approximate the value of ln(7)
𝑓(𝑥) = ln(2 + 3𝑥) ,
𝑓 𝑛 (𝑥) =
𝑓 ′ (𝑥) =
(−1)𝑛−1
∙ 3𝑛 ∙ (𝑛 − 1)!,
(3𝑥 + 2)𝑛
3
,
2 + 3𝑥
𝑓 ′′ (𝑥) = −
1∙3∙3
,…
(2 + 3𝑥)2
3 𝑛
𝑓 𝑛 (2) = (−1)𝑛−1 (𝑛 − 1)! ∙ ( ) ,
8
𝑛≠0
𝑓(2) = ln(8)
∞
∞
𝑛=0
𝑛=1
𝑓 𝑛 (2)
𝑥2 𝑥3 𝑥4
𝑓 𝑛 (2)
∑
∙ (𝑥 − 2)𝑛 = ln(8) + 𝑥 − + − + ⋯ = ln(8) + ∑
∙ (𝑥 − 2)𝑛
𝑛!
2
3
4
𝑛!
∞
𝑓(𝑥) = ln(2 + 3𝑥) = ln(8) + ∑
𝑛=1
(−1)𝑛−1 3 𝑛
∙ ( ) ∙ (𝑥 − 2)𝑛
𝑛
8
3 1
−1 3 2
1
(𝑥
𝑓(𝑥) = ln(2 + 3𝑥) ≈ ln(8) + ( ) ∙ − 2) +
∙ ( ) ∙ (𝑥 − 2)2
8
2
8
1
2
5
3 1 5
−1 3 2 5
𝑓 ( ) = ln(7) ≈ ln(8) + ( ) ∙ ( − 2) +
∙ ( ) ∙ ( − 2) = 1.9466
3
8
3
2
8
3
Theme 3.4 – Power series solutions to differential equations
Let’s go back to WTW256:
𝑑𝑦
+ 2𝑦(𝑥) = 0
𝑑𝑥
Solving this separable DE, it yields the following function:
𝑦(𝑥) = 𝐶0 𝑒 −2𝑥
The power series method assumes that the solution is in the form:
∞
𝑦(𝑥) = ∑ 𝑐𝑛 𝑥 𝑛 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + ⋯
𝑛=0
Let’s attempt to solve it using the power series method:
∞
∑ 𝑐𝑛
∞
(𝑛)𝑥 𝑛−1
+ 2 ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=1
𝑛=0
Getting all indexes, the same:
∞
∞
𝑛
∑ 𝑐𝑛+1 (𝑛 + 1)𝑥 + 2 ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=0
𝑛=0
Make sure all the 𝑥 exponents are the same.
∞
∑[𝑐𝑛+1 (𝑛 + 1)𝑥 𝑛 + 2𝑐𝑛 𝑥 𝑛 ] = 0,
∴ 𝑐𝑛+1 (𝑛 + 1)𝑥 𝑛 + 2𝑐𝑛 𝑥 𝑛 = 0
𝑛=0
𝑐𝑛+1 (𝑛 + 1) + 2𝑐𝑛 = 0,
𝑐𝑛+1 = −
2𝑐0 (−1)1 1
=
∙ 2 𝑐0
1
1!
2𝑐1
2
2𝑐0 (−1)2 2
𝑐2 = −
=− ∙−
=
∙ 2 𝑐0
2
2
1
2!
2𝑐2 (−1)3 3
𝑐3 = −
=
∙ 2 𝑐0
3
3!
𝑛
(−1)
𝑐𝑛 =
∙ 2𝑛 𝑐0
𝑛!
2𝑐𝑛
← 𝑅𝑒𝑐𝑢𝑟𝑟𝑒𝑛𝑐𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛
𝑛+1
𝑛=0
𝑐1 = −
𝑛=1
𝑛=2
𝑛 ∈ [0, ∞),
∞
(−1)𝑛 𝑛
𝑦(𝑥) = ∑
∙ 2 𝑐0 ∙ 𝑥 𝑛
𝑛!
𝑛=0
∞
(−2𝑥)𝑛
𝑦(𝑥) = 𝑐0 ∑
= 𝑐0 𝑒 −2𝑥
𝑛!
𝑛=0
Does this solution converge to the actual solution? Use the ratio test:
𝑛∈ℤ
{𝑎𝑛 } =
(−2𝑥)𝑛
∙,
𝑛!
{𝑎𝑛+1 } =
(−2𝑥)𝑛+1
(𝑛 + 1)!
{𝑎𝑛+1 }
−2𝑥
=
{𝑎𝑛 }
𝑛+1
lim (|
𝑛→∞
𝑥−𝑎
|) = 0 ∙ (−2𝑥)
𝑛+1
Therefore, for all real values of 𝑥 the limit will always be 0 and hence, less than 1. This means
that the radius of convergence is ∞. The solution converges for all real values of 𝑥.
What are ordinary and singular points?
Consider the following DE:
𝐴(𝑥)𝑦 ′′ (𝑥) + 𝐵(𝑥)𝑦 ′ (𝑥) + 𝐶(𝑥)𝑦(𝑥) = 0,
𝐴(𝑥) = 1
On the interval where 𝐵(𝑥) and 𝐶(𝑥) can be evaluated and analysed, all these points are known
as ordinary points. When these functions cannot be evaluated at a point these points are called
singularities. The guaranteed radius of convergence can be determined by noting that all power
series solutions are centred around 𝑥 = 0. Let 𝑎 be a singularity point on a differential equation
series solution. The power series solution will converge on an interval 𝑥 < |𝑎|. The radius of
convergence is therefore 𝑅 = 𝑎.
Linearly independent solutions:
𝑦 ′′ (𝑥) + 𝑦(𝑥) = 0
∞
∞
𝑛=0
𝑛=0
𝑑2
(∑ 𝑐𝑛 𝑥 𝑛 ) + ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑑𝑥 2
∞
∑ 𝑐𝑛 (𝑛)(𝑛 − 1)𝑥
∞
𝑛−2
+ ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=2
𝑛=0
∞
∞
𝑛
∑ 𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1)𝑥 + ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=0
𝑛=0
∞
∑[𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1)𝑥 𝑛 + 𝑐𝑛 𝑥 𝑛 ] = 0
𝑛=0
𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1)𝑥 𝑛 + 𝑐𝑛 𝑥 𝑛 = 0,
−𝑐0
(1)(2)
−𝑐1
𝑐3 =
(2)(3)
−𝑐2
𝑐0
𝑐4 =
=
(4)(3) (1)(2)(3)(4)
𝑐2 =
𝑐𝑛+2 =
−𝑐𝑛
(𝑛 + 2)(𝑛 + 1)
𝑛=0
𝑛=1
𝑛=2
−𝑐3
𝑐1
𝑐5 =
=
(4)(5) (2)(3)(4)(5)
−𝑐4
−𝑐0
𝑐6 =
=
(6)(5) (1)(2)(3)(4)(5)(6)
−𝑐5
−𝑐1
𝑐7 =
=
(6)(7) (2)(3)(4)(5)(6)(7)
𝑛=3
𝑛=4
𝑛=5
∞
∑ 𝑐𝑛 𝑥 𝑛 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 + 𝑐4 𝑥 4 + 𝑐5 𝑥 5 + 𝑐6 𝑥 6 + 𝑐7 𝑥 7 + ⋯
𝑛=0
= 𝑐0 + 𝑐1 𝑥 −
= 𝑐0 (1 −
𝑐0 2 −𝑐1 3 𝑐0 4 𝑐1 5 −𝑐0 6 −𝑐1 7
𝑥 +
𝑥 + 𝑥 + 𝑥 +
𝑥 +
𝑥 +⋯
2!
3!
4!
5!
6!
7!
𝑥2 𝑥4 𝑥6
𝑥3 𝑥5 𝑥7
+ − + ⋯ ) + 𝑐1 (𝑥 − + + + ⋯ )
2! 4! 6!
3! 5! 7!
∞
∞
𝑛=0
𝑛=0
(−1)𝑛 ∙ 𝑥 2𝑛
(−1)𝑛 ∙ 𝑥 2𝑛+1
= 𝑐0 ∙ ∑
+ 𝑐1 ∙ ∑
(2𝑛 + 1)!
2𝑛!
𝑦(𝑥) = 𝑐0 ∙ cos(𝑥) + 𝑐1 ∙ sin(𝑥)
Let’s look at some examples to solve more complex differential equations.
(𝑥 − 3) (
𝑑𝑦
) + 2𝑦(𝑥) = 0
𝑑𝑥
∞
∞
(𝑥 − 3) ∑ 𝑐𝑛 𝑛𝑥
𝑛−1
+ 2 ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=1
∞
𝑛=0
∞
𝑛
∞
∑ 𝑐𝑛 𝑛𝑥 − 3 ∑ 𝑐𝑛 𝑛𝑥
𝑛=1
𝑛−1
+ 2 ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=1
∞
𝑛=0
∞
𝑛
∞
∑ 𝑐𝑛 𝑛𝑥 − 3 ∑ 𝑐𝑛+1 (𝑛 + 1)𝑥 + 2 ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=0
𝑛
𝑛=0
𝑛=0
∞
∑[𝑐𝑛 𝑛𝑥 𝑛 − 3𝑐𝑛+1 (𝑛 + 1)𝑥 𝑛 + 2𝑐𝑛 𝑥 𝑛 ] = 0
𝑛=0
𝑐𝑛 𝑛 − 3𝑐𝑛+1 (𝑛 + 1) + 2𝑐𝑛 = 0,
2
𝑐1 = 𝑐0 ( 1 )
3
3
2 3
3
𝑐2 = 𝑐1 ( ) = 𝑐0 ( ) ( ) = 𝑐0 ( 2 )
6
3 6
3
4
3 4
4
𝑐3 = 𝑐2 ( ) = 𝑐0 ( 2 ) ( ) = 𝑐0 ( 3 )
9
3
9
3
𝑛+1
𝑐𝑛 = 𝑐0 ( 𝑛 )
3
𝑐𝑛+1 =
−2𝑐𝑛 − 𝑛𝑐𝑛
𝑛+2
= 𝑐𝑛 (
)
−3(𝑛 + 1)
3𝑛 + 3
𝑛=0
𝑛=1
𝑛=2
𝑛 = 0,1,2,3, …
∞
∞
𝑛
𝑦(𝑥) = ∑ 𝑐𝑛 𝑥 = 𝑐0 ∙ ∑ (
𝑛+1 𝑛
)𝑥
3𝑛
𝑛=0
𝑛=0
𝑛+1 𝑛
)𝑥 ,
3𝑛
{𝑎𝑛+1 } = (
Ratio test:
{𝑎𝑛 } = (
𝑛 + 2 𝑛+1
)𝑥
3𝑛+1
{𝑎𝑛+1 } 𝑛 + 2 𝑥
=
∙
{𝑎𝑛 }
𝑛+1 3
𝑛+2 𝑥
𝑥
lim (|
∙ |) = | | < 1
𝑛→∞ 𝑛 + 1 3
3
−3 < 𝑥 < 3,
𝑅=3
∞
𝑥 = −3,
∑(−1)𝑛 (𝑛 + 1) ,
𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (𝑎𝑙𝑡. 𝑠𝑒𝑟𝑖𝑒𝑠 𝑡𝑒𝑠𝑡)
𝑛=0
∞
−3 < 𝑥 < 3 ∶
{
𝑥 = 3,
∑(𝑛 + 1) ,
𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑠 (𝑡𝑒𝑠𝑡 𝑓𝑜𝑟 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒)
𝑛=0
Testing (separable):
𝑑𝑦
2
+
𝑦(𝑥) = 0
𝑑𝑥 𝑥 − 3
ln(𝑦(𝑥)) = −2 ln(𝑥 − 3) + 𝑘
𝑦(𝑥) =
𝑘0
(𝑥 − 3)2
Finding a power series to represent the solution:
𝑌(𝑥) = ∫ 𝑦(𝑥) ∙ 𝑑𝑥 = −
𝑘0
+𝐶
𝑥−3
∞
𝑘0
𝑘0
𝑘0
𝑥 𝑛
−
+𝐶 =
+ 𝐶 = 𝐶 + ∙ ∑ ( ) = 𝑌(𝑥)
1
𝑥−3
3
3
3 (1 − 3 𝑥)
𝑛=0
∞
∞
∞
𝑛=0
𝑛=1
𝑛=0
𝑑𝑦
𝑘0
𝑥 𝑛
𝑘0
1
𝑘0
1
𝑦(𝑥) =
∙ ∑ 𝑛 𝑛𝑥 𝑛−1 =
∙ ∑ 𝑛+1 (𝑛 + 1)𝑥 𝑛
[𝐶 + ∙ ∑ ( ) ] =
𝑑𝑥
3
3
3
3
3
3
∞
(𝑛 + 1) 𝑛
𝑘0
𝑦(𝑥) =
∙∑
𝑥
9
3𝑛
𝑛=0
They have similar results.
Let’s look at another problem.
𝑦 ′′ (𝑥) + 𝑦(𝑥) = 𝑥
∞
∞
∑ 𝑐𝑛 (𝑛)(𝑛 − 1)𝑥
𝑛−2
+ (∑ 𝑐𝑛 𝑥 𝑛 ) = 𝑥
𝑛=2
𝑛=0
∞
∞
𝑛
(∑ 𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1)𝑥 ) + (𝑐0 + 𝑐1 𝑥 + ∑ 𝑐𝑛 𝑥 𝑛 ) = 𝑥
𝑛=0
𝑛=2
∞
∞
𝑛
(2𝑐2 + 6𝑐3 𝑥 + ∑ 𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1)𝑥 ) + (𝑐0 + 𝑐1 𝑥 + ∑ 𝑐𝑛 𝑥 𝑛 ) = 𝑥
𝑛=2
𝑛=2
∞
∞
𝑛
(∑ 𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1)𝑥 ) + (∑ 𝑐𝑛 𝑥 𝑛 ) = −𝑐0 − 𝑐1 𝑥 − 2𝑐2 − 6𝑐3 𝑥 + 𝑥
𝑛=2
𝑛=2
𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1) + 𝑐𝑛 = 0,
𝑐0 = −2𝑐2 ,
−𝑐0 − 2𝑐2 = 0,
𝑐1 = 1 − 6𝑐3 ,
𝑐𝑛+2 = −
−𝑐1 − 6𝑐3 + 1 = 0
𝑐𝑛
(𝑛 + 1)(𝑛 + 2)
𝑐0 (−1)1 𝑐0
=
2
2!
(−1)1 (𝑐1 )
𝑐1
𝑐3 = −
=
(2)(3)
3!
(−1)2 (𝑐0 )
2𝑐2
𝑐4 =
=
(2)(3)(4)
4!
(−1)2 (𝑐1 )
6𝑐3
𝑐5 =
=
(2)(3)(4)(5)
5!
(−1)3 (𝑐0 )
𝑐4
2𝑐2
𝑐6 = −
=−
=
(5)(6)
(2)(3)(4)(5)(6)
6!
(−1)3 (𝑐1 )
𝑐5
6𝑐3
𝑐7 = −
=−
=
(6)(7)
(2)(3)(4)(5)(6)(7)
7!
𝑐2 = (−1)1
𝑛=0
𝑛=1
𝑛=2
𝑛=3
𝑛=4
𝑛=5
∞
∑ 𝑐𝑛 𝑥 𝑛 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 + 𝑐4 𝑥 4 + 𝑐5 𝑥 5 + 𝑐6 𝑥 6 + 𝑐7 𝑥 7 + ⋯
𝑛=0
(−1)0 (𝑐0 ) (−1)1 (𝑐0 ) 2 (−1)1 (𝑐1 ) 3 (−1)2 (𝑐0 ) 4 (−1)2 (𝑐1 ) 5
= 𝑐1 𝑥 +
+
𝑥 +
𝑥 +
𝑥 +
𝑥
0!
2!
3!
4!
5!
(−1)3 (𝑐0 ) 6 (−1)3 (𝑐1 ) 7
+
𝑥 +
𝑥 +⋯
6!
7!
1
= 1𝑥 1 +
(−1)0 (𝑐0 ) (−1)1 (𝑐0 ) 2 (−1)2 (𝑐0 ) 4 (−1)3 (𝑐0 ) 6
(−1)0 (𝑐1 ) 1
+
𝑥 +
𝑥 +
𝑥 + ⋯+
𝑥
0!
2!
4!
6!
1!
(−1)1 (𝑐1 ) 3 (−1)2 (𝑐1 ) 5 (−1)3 (𝑐1 ) 7
+
𝑥 +
𝑥 +
𝑥 +⋯
3!
5!
7!
∞
∞
𝑛=0
𝑛=0
(−1)𝑛 2𝑛
(−1)𝑛
𝑦(𝑥) = 𝑥 + 𝑐0 ∙ ∑ (
) 𝑥 + 𝑐1 ∙ ∑ (
) 𝑥 2𝑛+1
(2𝑛)!
(2𝑛 + 1)!
1
𝑦(𝑥) = 𝑥 1 + 𝑐0 ∙ cos(𝑥) + 𝑐1 ∙ sin(𝑥)
The final example:
𝑦 ′′ (𝑥) − 2𝑥𝑦 ′ (𝑥) − 4𝑦(𝑥) = 0
∞
∞
∑ 𝑐𝑛 (𝑛)(𝑛 − 1)𝑥
𝑛−2
∞
− 2𝑥 ∑ 𝑐𝑛
𝑛=2
(𝑛)𝑥 𝑛−1
− 4 ∑ 𝑐𝑛 𝑥 𝑛 = 0
𝑛=1
∞
𝑛=0
∞
𝑛
∞
(2𝑐2 + ∑ 𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1)𝑥 ) − 2 (∑ 𝑐𝑛
𝑛=1
(𝑛)𝑥 𝑛
) − 4 (𝑐0 + ∑ 𝑐𝑛 𝑥 𝑛 ) = 0
𝑛=1
𝑛=1
∞
∑ 𝑥 𝑛 [𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1) − 2𝑐𝑛 (𝑛) − 4𝑐𝑛 ] = −2(𝑐2 ) + 4(𝑐0 )
𝑛=1
−2𝑐2 + 4𝑐0 = 0,
𝑐𝑛+2 (𝑛 + 2)(𝑛 + 1) − 2𝑐𝑛 (𝑛) − 4𝑐𝑛 = 0
𝑐2 = 2𝑐0 ,
𝑐𝑛+2 =
2𝑐𝑛
(𝑛 + 1)
2𝑐0
(1)
2𝑐1
𝑐3 =
(2)
2𝑐2
2 2𝑐0 (2)2 (𝑐0 )
𝑐4 =
=
∙
=
(3) (3) (1)
(1)(3)
2𝑐3
2 2𝑐1 (2)2 (𝑐1 )
𝑐5 =
=
∙
=
(4) (4) (2)
(2)(4)
(2)3 (𝑐0 )
2𝑐4
2
2 2𝑐0
𝑐6 =
=
∙
∙
=
(5) (5) (3) (1) (1)(3)(5)
(2)3 (𝑐1 )
2𝑐5
2
2 2𝑐1
𝑐7 =
=
∙
∙
=
(6) (6) (4) (2) (2)(4)(6)
𝑐2 =
∞
𝑛=0
𝑛=1
𝑛=2
𝑛=3
𝑛=4
𝑛=5
∑ 𝑐𝑛 𝑥 𝑛 = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + 𝑐3 𝑥 3 + 𝑐4 𝑥 4 + 𝑐5 𝑥 5 + 𝑐6 𝑥 6 + 𝑐7 𝑥 7 + ⋯
𝑛=0
= 𝑐0 + 𝑐1 𝑥 +
= 𝑐0 (
(2)3 𝑐0 6
(2)3 𝑐1 7
2𝑐0 2 2𝑐1 3 (2)2 𝑐0 4 (2)2 𝑐1 5
𝑥 +
𝑥 +
𝑥 +
𝑥 +
𝑥 +
𝑥 +⋯
(1)
(2)
(1)(3)
(2)(4)
(1)(3)(5)
(2)(4)(6)
(2)1 𝑥 2 (2)2 𝑥 4
(2)3 𝑥 6
(2)1 𝑥 3 (2)2 𝑥 5
(2)3 𝑥 7
+
+
+ ⋯ ) + 𝑐1 𝑥 + 𝑐1 (
+
+
+ ⋯)
(1)
(1)(3) (1)(3)(5)
(2)
(2)(4) (2)(4)(6)
1 ∙ 3 ∙ 5 ∙ 7 ⋯ (2𝑛 − 1) = 1 ∙ 3 ∙ 5 ∙ 7 ⋯ (2𝑛 − 1) ∙
(2𝑛)!
2 ∙ 4 ∙ 6 ∙ 8 ⋯ (2𝑛)
= 𝑛
2 ∙ 4 ∙ 6 ∙ 8 ⋯ (2𝑛) 2 (𝑛)!
2 ∙ 4 ∙ 6 ∙ 8 ⋯ (2𝑛) = (2𝑛 )(1 ∙ 2 ∙ 3 ∙ 4 ⋯ (𝑛)) = 2𝑛 (𝑛!)
∞
∞
𝑛=0
𝑛=0
(𝑛!)(2𝑥)2𝑛
(𝑥 2𝑛+1 )
= 𝑐0 ∙ ∑ (
) + 𝑐1 ∑ (
)
(2𝑛)!
(𝑛!)
Theme 4 – Fourier series
6 Lectures [week 8 & 9]
This theme will introduce Fourier series and the Fourier transforms. The convergence of the
Fourier series and the special cases like Fourier-sine and -cosine series.
Theme 4.1: General Fourier series and convergence
Theme 4.2: Fourier-sine and -cosine series.
Pre-knowledge for this theme consists of factor integration (WTW 164) and even and odd
functions (WTW 158).
Theme 4.1: General Fourier series and convergence.
Recap: Even and odd functions
When 𝑓(𝑥) = 𝑓(−𝑥), 𝑓(𝑥) is known as an even function. An example is 𝑓(𝑥) = cos(𝑥)
When −𝑓(𝑥) = 𝑓(−𝑥), 𝑓(𝑥) is known as an even function. An example is 𝑓(𝑥) = sin(𝑥)
Orthogonal functions:
Functions are orthogonal when their inner product is zero. The inner product is the integral of
the product of the functions on that interval.
𝑏
∫ (𝑓1 (𝑥) ∙ 𝑓2 (𝑥)) ∙ 𝑑𝑥 = 0
𝑎
Let’s look at sin(𝑥) and cos(𝑥).
𝑝
𝑛𝜋𝑥
𝑛𝜋𝑥
) cos (
) 𝑑𝑥
𝑝
𝑝
−𝑝
𝑝
1
2𝑛𝜋𝑥
∫
sin (
) 𝑑𝑥
𝑝
−𝑝 2
1 𝑝
2𝑛𝜋𝑥 𝑝
(− ∙
∙ cos (
))|
2 2𝑛𝜋
𝑝
−𝑝
𝑝
(−
) ∙ (cos(2𝑛𝜋) − cos(2𝑛𝜋))
4𝑛𝜋
=0
𝑛𝜋𝑥
𝑛𝜋𝑥
sin ( 𝑝 ) and cos ( 𝑝 ) are
∫ sin (
𝑝
𝑛𝜋𝑥
𝑚𝜋𝑥
) sin (
) 𝑑𝑥
𝑝
𝑝
−𝑝
𝑝,
𝑚=𝑛
={
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
∫ sin (
𝑝
𝑛𝜋𝑥
𝑚𝜋𝑥
) cos (
) 𝑑𝑥
𝑝
𝑝
−𝑝
𝑝,
𝑚=𝑛
={
0,
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
∫ cos (
orthogonal on the interval 𝑥 ∈
[−𝑝, 𝑝].
A Fourier series transform takes a given function and approximates it by summating a finite or
infinite number of sine and cosine terms. The Fourier transform has the following standard
form:
∞
∞
𝑛=1
𝑛=1
𝑎0
𝑛𝜋𝑥
𝑛𝜋𝑥
ℱ(𝑥) =
+ ∑ 𝑎𝑛 cos (
) + ∑ 𝑏𝑛 sin (
)
2
𝑝
𝑝
Let us derive formulas for the three coefficients.
Let
∞
∞
𝑓(𝑥) = 𝑎0 + ∑ 𝑎𝑛 cos(𝑛𝑥) + ∑ 𝑏𝑛 sin(𝑛𝑥) ,
𝑛=1
−𝜋 < 𝑥 < 𝜋
𝑛=1
Integrate the equation:
𝜋
𝜋
𝜋 ∞
𝜋 ∞
∫ 𝑓(𝑥) ∙ 𝑑𝑥 = ∫ 𝑎0 ∙ 𝑑𝑥 + ∫ ∑ 𝑎𝑛 cos(𝑛𝑥) ∙ 𝑑𝑥 + ∫ ∑ 𝑏𝑛 sin(𝑛𝑥) ∙ 𝑑𝑥
−𝜋
−𝜋
𝜋
𝜋
−𝜋 𝑛=1
−𝜋 𝑛=1
∞
∞
𝜋
𝜋
∫ 𝑓(𝑥) ∙ 𝑑𝑥 = ∫ 𝑎0 ∙ 𝑑𝑥 + ∑ 𝑎𝑛 ∫ cos(𝑛𝑥) ∙ 𝑑𝑥 + ∑ 𝑏𝑛 ∫ sin(𝑛𝑥) ∙ 𝑑𝑥
−𝜋
−𝜋
−𝜋
𝑛=1
−𝜋
𝑛=1
𝜋
∞
𝜋
∫ 𝑓(𝑥) ∙ 𝑑𝑥 = 𝑎0 (𝜋 + 𝜋) + 2 ∑ 𝑎𝑛 ∫ cos(𝑛𝑥) ∙ 𝑑𝑥
0
𝑛=1
−𝜋
𝜋
∞
1
∫ 𝑓(𝑥) ∙ 𝑑𝑥 = 2𝜋𝑎0 + 2 ∑ 𝑎𝑛 ( (sin(𝜋𝑛) − sin(0)) )
𝑛
𝑛=1
−𝜋
𝜋
𝜋
1
𝑎0 =
∙ ∫ 𝑓(𝑥) ∙ 𝑑𝑥
2𝜋
∫ 𝑓(𝑥) ∙ 𝑑𝑥 = 2𝜋𝑎0 ,
−𝜋
−𝜋
For the term 𝑎𝑛 , multiply all terms with cos(𝑚𝑥):
∞
∞
𝑓(𝑥) cos(𝑚𝑥) = 𝑎0 cos(𝑚𝑥) + ∑ 𝑎𝑛 cos(𝑛𝑥) cos(𝑚𝑥) + ∑ 𝑏𝑛 sin(𝑛𝑥) cos(𝑚𝑥)
𝑛=1
𝑛=1
Integrate the equation:
𝜋
𝜋
∞
∞
𝜋
𝜋
∫ 𝑓(𝑥) cos(𝑚𝑥) ∙ 𝑑𝑥 = ∫ 𝑎0 cos(𝑚𝑥) ∙ 𝑑𝑥 + ∑ 𝑎𝑛 ∫ cos(𝑚𝑥) cos(𝑛𝑥) ∙ 𝑑𝑥 + ∑ 𝑏𝑛 ∫ sin(𝑚𝑥) cos(𝑛𝑥) ∙ 𝑑𝑥
−𝜋
−𝜋
−𝜋
𝑛=1
𝜋
∞
𝑛=1
−𝜋
𝜋
∫ 𝑓(𝑥) cos(𝑚𝑥) ∙ 𝑑𝑥 = 0 + ∑ 𝑎𝑛 ∫ cos(𝑚𝑥) cos(𝑛𝑥) ∙ 𝑑𝑥 + 0
𝑛=1
−𝜋
𝜋
−𝜋
∞
∫ 𝑓(𝑥) cos(𝑚𝑥) ∙ 𝑑𝑥 = ∑ 𝑎𝑛 ({0,
−𝜋
𝜋,
𝑚=𝑛
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒)
𝑛=1
The summation falls away as the only non-zero term will be the term where 𝑚 = 𝑛.
𝜋
∫ 𝑓(𝑥) cos(𝑚𝑥) ∙ 𝑑𝑥 = 𝑎𝑛 𝜋,
−𝜋
For the term 𝑛𝑛 , multiply all terms with sin(𝑚𝑥):
𝜋
1
𝑎𝑛 = ∫ 𝑓(𝑥) cos(𝑚𝑥) ∙ 𝑑𝑥
𝜋
−𝜋
∞
∞
𝑓(𝑥) sin(𝑚𝑥) = 𝑎0 sin(𝑚𝑥) + ∑ 𝑎𝑛 cos(𝑛𝑥) sin(𝑚𝑥) + ∑ 𝑏𝑛 sin(𝑛𝑥) sin(𝑚𝑥)
𝑛=1
𝑛=1
Integrate the equation:
𝜋
𝜋
∞
∞
𝜋
𝜋
∫ 𝑓(𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥 = ∫ 𝑎0 sin(𝑚𝑥) ∙ 𝑑𝑥 + ∑ 𝑎𝑛 ∫ cos(𝑚𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥 + ∑ 𝑏𝑛 ∫ sin(𝑚𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥
−𝜋
−𝜋
𝑛=1
−𝜋
−𝜋
𝑛=1
𝜋
∞
𝜋
∫ 𝑓(𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥 = 0 + 0 + ∑ 𝑏𝑛 ∫ sin(𝑚𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥
𝑛=1
−𝜋
𝜋
−𝜋
∞
∫ 𝑓(𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥 = ∑ 𝑏𝑛 ({0,
𝜋,
𝑚=𝑛
𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒)
𝑛=1
−𝜋
The summation falls away as the only non-zero term will be the term where 𝑚 = 𝑛.
𝜋
𝜋
∫ 𝑓(𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥 = 𝑎𝑛 𝜋,
−𝜋
1
𝑏𝑛 = ∫ 𝑓(𝑥) sin(𝑚𝑥) ∙ 𝑑𝑥
𝜋
−𝜋
Note that these formulas only apply to a function that has a period of 2𝜋. When the most general
case is considered, the period must be adjusted accordingly:
𝑏
1
𝑎0 =
∙ ∫ 𝑓(𝑥) ∙ 𝑑𝑥,
2𝑝
𝑎
𝑏
1
𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos (
) ∙ 𝑑𝑥,
𝑝
𝑝
𝑎
𝑝=
𝑏
1
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin (
) ∙ 𝑑𝑥
𝑝
𝑝
𝑎
(𝑏 − 𝑎)
2
Let’s look at a simple example:
𝑓(𝑥) = 𝑥,
𝑝=
−1 < 𝑥 < 1
1 − (−1)
=1
2
∞
∞
𝑛=1
𝑛=1
𝑎0
𝑛𝜋𝑥
𝑛𝜋𝑥
ℱ(𝑥) =
+ ∑ 𝑎𝑛 cos (
) + ∑ 𝑏𝑛 sin (
)
2
𝑝
𝑝
1
1
1
1
𝑎0 =
∙ ∫ 𝑥 ∙ 𝑑𝑥 = 𝑥 2 | = 0
2(1)
4
−1
−1
1
1
𝑎𝑛 = ∫ 𝑥 cos(𝑛𝜋𝑥) ∙ 𝑑𝑥 = 𝑜𝑑𝑑(𝑓𝑛. )|𝑎−𝑎 = 0
1
−1
1
1
(−1)𝑛+1 2
2
𝑛
(−1) =
𝑏𝑛 = ∫ 𝑥 sin(𝑛𝜋𝑥) ∙ 𝑑𝑥 = 2 ∫ 𝑥 sin(𝑛𝜋𝑥) ∙ 𝑑𝑥 = −
𝑛𝜋
𝑛𝜋
−1
∞
ℱ(𝑥) = ∑
𝑛=1
0
(−1)𝑛+1 2
2
1
2
1
sin(𝑛𝜋𝑥) = sin(𝜋𝑥) − sin(2𝜋𝑥) +
sin(3𝜋𝑥) −
sin(4𝜋𝑥) + ⋯
𝑛𝜋
𝜋
𝜋
3𝜋
2𝜋
𝑁
𝑓(𝑥) ≈ ∑
𝑛=1
(−1)𝑛+1 2
sin(𝑛𝜋𝑥)
𝑛𝜋
𝑁=1
𝑁=4
𝑁 = 50
It is expected that there are only sine terms as sine is an odd function like 𝑓(𝑥) = 𝑥. If 𝑓(𝑥) is an
even function only cosine and the constant term will play a role in the Fourier transform. Note
that the transform converges to the function (if it didn’t, that would be embarrassing). The
transform at a jump discontinuity will converge to the average value between the two points. As
in the above case, there is a jump at 𝑥 = 1 from 𝑓(1− ) = 1 to 𝑓(1+ ) = −1, the Fourier
transform, ℱ(𝑥), converges to 0 as it is the average between the two values.
Let’s look at a function that is neither odd nor even.
𝑓(𝑥) = 𝑥 2 + 1,
𝑝=
0<𝑥<2
2−0
=1
2
∞
∞
𝑛=1
𝑛=1
𝑎0
𝑛𝜋𝑥
𝑛𝜋𝑥
ℱ(𝑥) =
+ ∑ 𝑎𝑛 cos (
) + ∑ 𝑏𝑛 sin (
)
2
𝑝
𝑝
2
2
1 3
8
14
𝑎0 = ∫(𝑥 + 1) ∙ 𝑑𝑥 = ( 𝑥 + 𝑥)| = ( + 2) =
3
3
3
0
2
0
𝑎0 =
2
14
3
2
2
𝑎𝑛 = ∫(𝑥 + 1) cos(𝑛𝜋𝑥) ∙ 𝑑𝑥 = ∫(𝑥
0
0
2
2
2
2)
cos(𝑛𝜋𝑥) ∙ 𝑑𝑥 + ∫ cos(𝑛𝜋𝑥) ∙ 𝑑𝑥
0
2
2
𝑥2
2𝑥
1
𝑥
sin(𝑛𝜋𝑥)| − ∫
sin(𝑛𝑥) ∙ 𝑑𝑥 + ( sin(𝑛𝜋𝑥))| = 2 ∫
sin(𝑛𝑥) ∙ 𝑑𝑥
𝑛𝜋
𝑛𝜋
𝑛𝜋
𝑛𝜋
0
0
0
0
2
2
2𝑥
2
−4
4
(−
cos(𝑛𝜋𝑥))|
−
∫
cos(𝑛𝜋𝑥)
∙
𝑑𝑥
=
−
(
cos(2𝜋𝑛))
=
(𝑛𝜋)2
(𝑛𝜋)2
(𝑛𝜋)2
(𝑛𝜋)2
0
0
𝑎𝑛 =
2
4
(𝑛𝜋)2
2
2
2
𝑏𝑛 = ∫(𝑥 + 1) sin(𝑛𝜋𝑥) ∙ 𝑑𝑥 = ∫(𝑥
0
2)
sin(𝑛𝜋𝑥) ∙ 𝑑𝑥 + ∫ sin(𝑛𝜋𝑥) ∙ 𝑑𝑥
0
2
0
2
2
2
𝑥2
𝑥
−1
−4
𝑥
cos(𝑛𝜋𝑥))| + 2 ∫
cos(𝑛𝜋𝑥) ∙ 𝑑𝑥 + ( cos(𝑛𝜋𝑥))| =
+ 2∫
cos(𝑛𝜋𝑥) ∙ 𝑑𝑥
(−
𝑛𝜋
𝑛𝜋
𝑛𝜋
𝑛𝜋
𝑛𝜋
0
0
0
0
2
2
−4
2
2
−4
4
−4
(sin(2𝑛𝜋) − 0)) =
+(
sin(𝑛𝜋𝑥))|
+∫
sin(𝑛𝜋𝑥) ∙ 𝑑𝑥 =
+(
2
2
2
(𝑛𝜋)
(𝑛𝜋)
(𝑛𝜋)
𝑛𝜋
𝑛𝜋
𝑛𝜋
0
0
𝑏𝑛 =
−4
𝑛𝜋
∞
∞
𝑛=1
𝑛=1
7
4
−4
ℱ(𝑥) = + ∑
cos(𝑛𝜋𝑥) + ∑
sin(𝑛𝜋𝑥)
2
(𝑛𝜋)
3
𝑛𝜋
ℱ(𝑥) =
7 4
1
−4
−2
+ 2 cos(𝜋𝑥) + 2 cos(2𝜋𝑥) + ⋯ +
sin(𝜋𝑥) +
sin(2𝜋𝑥) + ⋯
3 𝜋
𝜋
𝜋
𝜋
𝑁=1
𝑁=4
𝑁 = 50
Find the Fourier series of the following piece wise function (square wave):
0,
𝑓(𝑥) = {
𝑥,
𝑝=
−1 < 𝑥 < 0
0<𝑥<1
1 − (−1)
=1
2
∞
∞
𝑛=1
𝑛=1
𝑎0
ℱ(𝑥) =
+ ∑ 𝑎𝑛 cos(𝑛𝜋𝑥) + ∑ 𝑏𝑛 sin(𝑛𝜋𝑥)
2
0
1
1
1
1 2 1 1
𝑎0 = ∫ 0𝑑𝑥 + ∫ 𝑥𝑑𝑥 = ( 𝑥 )| =
2
2
4
4
0
−1
0
𝑎0 =
1
4
1
1
1
𝑥
1
𝑎𝑛 = ∫ 𝑥 cos(𝑛𝜋𝑥) ∙ 𝑑𝑥 = ( sin(𝑛𝜋𝑥))| − ∫
sin(𝑛𝜋𝑥) 𝑑𝑥
𝑛𝜋
𝑛𝜋
0
0
0
0,
𝑛 = 𝑒𝑣𝑒𝑛
1
(−1)𝑛
1
1
(
cos(𝑛𝜋𝑥))| =
−
= { −2
,
𝑛 = 𝑜𝑑𝑑
(𝑛𝜋)2
(𝑛𝜋)2 (𝑛𝜋)2
0
(𝑛𝜋)2
𝑎𝑛 =
−2
2
((2𝑛 − 1)𝜋)
1
1
1
𝑥
1
𝑎𝑛 = ∫ 𝑥 sin(𝑛𝜋𝑥) ∙ 𝑑𝑥 = (−
cos(𝑛𝜋𝑥))| + ∫
cos(𝑛𝜋𝑥) 𝑑𝑥
𝑛𝜋
𝑛𝜋
0
0
0
1
(−1)𝑛+1
(−1)𝑛+1
1
+(
cos(𝑛𝜋𝑥))|
=
(𝑛𝜋)2
𝑛𝜋
𝑛𝜋
0
𝑏𝑛 =
(−1)𝑛+1
𝑛𝜋
∞
∞
𝑛=1
𝑛=1
(−1)𝑛+1
1
−2
ℱ(𝑥) = + ∑
sin(𝑛𝜋𝑥)
2 cos((2𝑛 − 1)𝜋𝑥) + ∑
4
𝑛𝜋
((2𝑛 − 1)𝜋)
ℱ(𝑥) =
1 −2
−2
1
−1
+ 2 cos(𝜋𝑥) + 2 cos(3𝜋𝑥) + ⋯ + sin(𝜋𝑥) +
sin(2𝜋𝑥) + ⋯
4 𝜋
9𝜋
𝜋
2𝜋
𝑁=1
𝑁=4
𝑁 = 50
Remember that absolute valued functions may be re-written as piecewise functions:
𝑓(𝑥) = |sin(𝑥)|,
𝑓(𝑥) = {
− sin(𝑥) ,
sin(𝑥) ,
𝑥 ∈ (−𝜋, 𝜋)
−𝜋 < 𝑥 < 0
0<𝑥<𝜋
sin(𝑛𝜋) = 0,
cos(𝑛𝜋) = (−1)𝑛 ,
𝑛∈ℤ
We can use the Fourier series to calculate the value of a series. Use the previous example to
determine the value of the series given below.
0,
𝑓(𝑥) = {
𝑥,
∞
∞
𝑛=1
𝑛=1
(−1)𝑛+1
−2
−1 < 𝑥 < 0 1
= +∑
cos((2𝑛
−
1)𝜋𝑥)
+
∑
sin(𝑛𝜋𝑥)
2
0<𝑥<1
4
𝑛𝜋
((2𝑛 − 1)𝜋)
Determine the value of:
∞
∑
𝑛=1
1
(2𝑛 − 1)2
cos((2𝑛 − 1)𝜋𝑥) = cos(2𝜋𝑛) = 1,
sin(𝑛𝜋𝑥) → sin(0) = 0,
∞
𝑥=0
𝑂𝐾
1
−2
0= +∑
2
4
((2𝑛 − 1)𝜋)
𝑛=1
∞
𝜋2
1
=∑
2
8
𝑛=1 ((2𝑛 − 1))
Theme 4.2 – Fourier sine series and Fourier cosine series
The product of even and odd functions:
Let 𝑓(𝑥) be an even function and 𝑔(𝑥) an odd function:
ℎ(𝑥) = 𝑓(𝑥) ∙ 𝑓(𝑥)
ℎ(−𝑥) = 𝑓(−𝑥) ∙ 𝑓(−𝑥)
ℎ(−𝑥) = 𝑓(𝑥) ∙ 𝑓(𝑥)
ℎ(−𝑥) = ℎ(𝑥)
ℎ(𝑥) = 𝑥 2 cos(2𝜋𝑥)
ℎ(𝑥) = 𝑔(𝑥) ∙ 𝑔(𝑥)
ℎ(−𝑥) = 𝑔(−𝑥) ∙ 𝑔(−𝑥)
ℎ(−𝑥) = (−1)2 ∙ 𝑔(𝑥) ∙ 𝑔(𝑥)
ℎ(−𝑥) = ℎ(𝑥)
ℎ(𝑥) = 𝑥 sin(𝜋𝑥)
ℎ(𝑥) = 𝑓(𝑥) ∙ 𝑔(𝑥)
ℎ(−𝑥) = 𝑓(−𝑥) ∙ 𝑔(−𝑥)
ℎ(−𝑥) = −𝑓(𝑥) ∙ 𝑔(𝑥)
ℎ(−𝑥) = −ℎ(𝑥)
𝜋
ℎ(𝑥) = 𝑥 cos ( 𝑥)
2
As stated previously, odd functions only have sine terms. The even functions have a constant
term as well as cosine terms.
𝑝
𝑝
1
1
1
𝑎0 =
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 = (𝐹(𝑝) − 𝐹(0))
2𝑝
𝑝
𝑝
−𝑝
0
𝑝
𝑝
1
2
𝑎𝑛 = ∫ 𝑓(𝑥) cos(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) cos(𝑥) 𝑑𝑥
𝑝
𝑝
−𝑝
0
𝑝
1
𝑏𝑛 = ∫ 𝑓(𝑥) sin(𝑥) 𝑑𝑥 = 0
𝑝
−𝑝
And for an odd function 𝑔(𝑥):
𝑝
1
𝑎0 =
∫ 𝑔(𝑥)𝑑𝑥 = 0
2𝑝
−𝑝
𝑝
1
𝑎𝑛 = ∫ 𝑔(𝑥) cos(𝑥) 𝑑𝑥 = 0
𝑝
−𝑝
𝑝
𝑝
1
2
𝑏𝑛 = ∫ 𝑔(𝑥) sin(𝑥) 𝑑𝑥 = ∫ 𝑔(𝑥) sin(𝑥) 𝑑𝑥
𝑝
𝑝
−𝑝
0
For even expansions of functions, the Fourier cosine series is used. The odd expansion uses only
sine terms. What is meant by even and odd expansions?
Good question. Consider the following function:
𝑓(𝑥) = 𝑥 2 + 1,
𝑓(𝑥) = 𝑥 2 + 1,
0<𝑥<2
𝑓(𝑥) = 𝑥 2 + 1,
0<𝑥<2
−2 < 𝑥 < 2,
𝑒𝑣𝑒𝑛
𝑓(𝑥) = 𝑥 2 + 1,
−2 < 𝑥 < 2,
𝑜𝑑𝑑
−2 < 𝑥 < 2,
𝑜𝑑𝑑
Let’s look at the Fourier expansions:
𝑓(𝑥) = 𝑥 2 + 1,
𝑓(𝑥) = 𝑥 2 + 1,
−2 < 𝑥 < 2,
𝑒𝑣𝑒𝑛
0<𝑥<2
𝑓(𝑥) = 𝑥 2 + 1,
Solving DE’s using Fourier series. There are two different types of problems, Dirichlet
conditions and Neumann conditions:
Dirichlet
𝑎𝑦 ′′ + 𝑏𝑦 = 𝑓(𝑥),
𝑥(0) = 𝑥(𝐿) = 0
Use the Fourier sine series to solve
Neumann
𝑎𝑦 ′′ + 𝑏𝑦 = 𝑓(𝑥),
𝑥 ′ (0) = 𝑥 ′ (𝐿) = 0
Use the Fourier cosine series to solve
Consider the following DE:
Dirichlet:
𝑦 ′′ (𝑥) + 3𝑦(𝑥) = 3 − 𝑥,
𝑦(0) = 𝑦(3) = 0
𝑝 =3−0=3
∞
∞
𝑛=1
𝑛=1
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑦(𝑥) = ∑ 𝑏𝑛 sin (
) = ∑ 𝑏𝑛 sin (
)
𝑝
3
∞
𝑦
′ (𝑥)
∞
𝑛𝜋
𝑛𝜋𝑥
= ∑ 𝑏𝑛
cos (
),
3
3
𝑦
′′ (𝑥)
𝑛=1
𝑛𝜋 2
𝑛𝜋𝑥
= − ∑ 𝑏𝑛 ( ) sin (
)
3
3
𝑛=1
𝑓(𝑥) = 3 − 𝑥
∞
∞
𝑛=1
𝑛=1
𝑛𝜋𝑥
𝑛𝜋𝑥
ℱ𝑠𝑖𝑛𝑒 {3 − 𝑥} = ∑ 𝐵𝑛 sin (
) = ∑ 𝐵𝑛 sin (
)
𝑝
3
𝑝
3
2
𝑥𝑛𝜋
2
𝑥𝑛𝜋
6
𝐵𝑛 = ∫(3 − 𝑥) sin (
) 𝑑𝑥 = ∫(3 − 𝑥) sin (
) 𝑑𝑥 =
𝑝
𝑝
3
3
𝑛𝜋
0
0
∞
ℱ𝑠𝑖𝑛𝑒 {3 − 𝑥} = ∑
𝑛=1
6
𝑛𝜋𝑥
sin (
)
𝑛𝜋
3
𝑦 ′′ (𝑥) + 3𝑦(𝑥) = 3 − 𝑥
∞
∞
∞
𝑛=1
𝑛=1
𝑛=1
𝑛𝜋 2
𝑛𝜋𝑥
𝑛𝜋𝑥
6
𝑛𝜋𝑥
− ∑ 𝑏𝑛 ( ) sin (
) + 3 ∑ 𝑏𝑛 sin (
)=∑
sin (
)
3
3
3
𝑛𝜋
3
∞
∞
∞
𝑛=1
𝑛=1
𝑛=1
𝑛𝜋 2
6
− ∑ 𝑏𝑛 ( ) + 3 ∑ 𝑏𝑛 = ∑
3
𝑛𝜋
𝑛𝜋 2
6
−𝑏𝑛 ( ) + 3𝑏𝑛 =
,
3
𝑛𝜋
𝑏𝑛 =
∞
𝑦(𝑥) = ∑
𝑛=1
6
2
𝑛𝜋
𝑛𝜋 (3 − ( 3 ) )
,
𝑏𝑛 =
54
𝑛𝜋𝑥
sin (
)
2
𝑛𝜋(27 − (𝑛𝜋) )
3
54
𝑛𝜋(27 − (𝑛𝜋)2 )
Neumann:
2𝑦 ′′ (𝑥) − 𝑦(𝑥) = 3 − 𝑥,
𝑦 ′ (0) = 𝑦 ′ (3) = 0
∞
∞
𝑛=1
𝑛=1
𝑎0
𝑛𝜋𝑥
𝑎0
𝑛𝜋𝑥
𝑦(𝑥) =
+ ∑ 𝑎𝑛 cos (
)=
+ ∑ 𝑎𝑛 cos (
)
2
𝑝
2
3
∞
𝑦
′ (𝑥)
𝑛𝜋
𝑛𝜋𝑥
= − ∑ 𝑎𝑛
sin (
),
3
3
∞
𝑦
′′ (𝑥)
𝑛=1
𝑛𝜋 2
𝑛𝜋𝑥
= − ∑ 𝑎𝑛 ( ) cos (
)
3
3
𝑛=1
𝑓(𝑥) = 3 − 𝑥
∞
∞
𝑛=1
𝑛=1
𝑛𝜋𝑥
𝑛𝜋𝑥
ℱ𝑐𝑜𝑠𝑖𝑛𝑒 {3 − 𝑥} = 𝐴0 + ∑ 𝐴𝑛 cos (
) = 𝐴0 + ∑ 𝐴𝑛 cos (
)
𝑝
3
𝑝
3
1
1
3
𝐴0 = ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 3 − 𝑥𝑑𝑥 =
𝑝
3
2
0
0
𝑝
3
0,
𝑛 = 𝑒𝑣𝑒𝑛
2
𝑛𝜋𝑥
2
𝑛𝜋𝑥
𝐴𝑛 = ∫ 𝑓(𝑥) cos (
) 𝑑𝑥 = ∫(3 − 𝑥) cos (
) 𝑑𝑥 = { 12
,
𝑛 = 𝑜𝑑𝑑
𝑝
𝑝
3
3
(𝑛𝜋)2
0
0
∞
3
12
𝑛𝜋𝑥
ℱ𝑐𝑜𝑠𝑖𝑛𝑒 {3 − 𝑥} = + ∑
cos (
)
2
(𝑛𝜋)
2
3
𝑛=1
2𝑦 ′′ (𝑥) − 𝑦(𝑥) = 3 − 𝑥
∞
∞
∞
𝑛=1
𝑛=1
𝑛=1
(2𝑛 − 1)𝜋𝑥
𝑛𝜋 2
𝑛𝜋𝑥
𝑎0
𝑛𝜋𝑥
3
12
−2 ∑ 𝑎𝑛 ( ) cos (
) − − ∑ 𝑎𝑛 cos (
)= +∑
cos (
)
2
3
3
2
3
2
3
((2𝑛 − 1)𝜋)
−
𝑎0 3
= ,
2
2
𝑎0 = −3
(2𝑛 − 1)𝜋𝑥
𝑛𝜋 2
𝑛𝜋𝑥
𝑛𝜋𝑥
12
−2𝑎𝑛 ( ) cos (
) − 𝑎𝑛 cos (
)=
cos (
)
2
3
3
3
3
((2𝑛 − 1)𝜋)
12
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑛𝜋 2
cos (
),
𝑛 = 𝑜𝑑𝑑
2
−𝑎𝑛 cos (
) (2 ( ) + 1) = {(𝑛𝜋)
3
3
3
0,
𝑛 = 𝑒𝑣𝑒𝑛
𝑛𝜋𝑥
𝑛𝜋 2
12
𝑛𝜋𝑥
−𝑎𝑛 cos (
) (2 ( ) + 1) =
cos (
),
2
(𝑛𝜋)
3
3
3
𝑎𝑛 =
∞
−108
(𝑛𝜋)2 (2𝑛2 𝜋 2
+ 9)
,
𝑛𝜋 2
12
−𝑎𝑛 (2 ( ) + 1) =
(𝑛𝜋)2
3
𝑛 → 2𝑛 − 1
(2𝑛 − 1)𝜋𝑥
−3
108
𝑦(𝑥) =
−∑
cos (
)
2
2
3
((2𝑛 − 1)𝜋) (2(2𝑛 − 1)2 𝜋 2 + 9)
𝑛=1
Theme 5 – Partial differential equations
13 Lectures [week 10 & 11 & 12]
This theme will introduce eigenvalue problems and partial differential equations. The wave and
heat equation Fourier solutions.
Theme 5.1: Eigenvalue problems.
Theme 5.2: The heat equation.
Theme 5.3: The wave equation.
Pre-knowledge for this theme consists of linear second order differential equations (WTW256).
Theme 5.1: Eigenvalue problems.
What is a boundary value problem? A boundary value problem is a type of differential equation
whose conditions are at the boundary.
Examples hereof:
➢ 2D-flow over a stationary body: Due to the no-slip condition, the flow at the boundary
(the body) will be zero and at some velocity at the free surface.
𝑑𝑢
𝜏=𝜇∙
,
𝑢(0) = 0,
𝑢(𝛿) = 𝑉
𝑑𝑦
➢ Fixed-fixed vibrating string: Due to the fixed-fixed condition the endpoints will not
displace, hence the displacements at the boundaries are zero.
𝜕2𝑦
𝜕2𝑦
2
=
𝑎
∙
,
𝑦(0, 𝑡) = 𝑦(𝐿, 𝑡) = 0
𝜕𝑡 2
𝜕𝑥 2
➢ Cantilever beam with loading: The cantilevered side will have a zero slope and zero
displacement.
𝑑2 𝑣
𝑀(𝑥) = 𝐸𝐼 ∙ 2 ,
𝑣(0) = 𝑣 ′ (0) = 0
𝑑𝑥
Consider the following eigenvalue problem:
𝑦 ′′ (𝑥) + (𝑓(𝜆))𝑦 ′ (𝑥) + 𝑔(𝜆)𝑦(𝑥) = 0
The characteristic equation:
𝑚2 + 𝑓(𝜆)𝑚 + 𝑔(𝜆) = 0
[𝑓(𝜆)]2 − 4𝑔(𝜆) < 0,
𝜆 ∈ ℂ,
𝑐𝑜𝑚𝑝𝑙𝑒𝑥
2
{[𝑓(𝜆)] − 4𝑔(𝜆) = 0,
𝜆 ∈ ℝ,
𝑟𝑒𝑎𝑙 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑
[𝑓(𝜆)]2 − 4𝑔(𝜆) > 0,
𝜆 ∈ ℝ,
𝑟𝑒𝑎𝑙 𝑢𝑛𝑖𝑞𝑢𝑒
Find the eigenvalue as well as the associated eigenfunction:
𝑦 ′′ (𝑥) + 𝜆𝑦(𝑥) = 0,
𝑦(0) = 𝑦(𝐿) = 0
𝜆 = 0,
2
𝑚 + 𝜆 = 0,
{𝜆 > 0,
𝑦(𝑥) = 𝐴𝑡 + 𝐵
𝑦(𝑥) = 𝐶1 cos(√𝜆𝑥) + 𝐶2 sin(√𝜆𝑥)
𝜆 < 0,
𝑦(𝑥) = 𝐶1 𝑒 √𝜆𝑥 + 𝐶2 𝑥𝑒 √𝜆𝑥
For 𝜆 = 0 (the boring one, not an eigenvalue):
𝑦(0) = 𝐵 = 0,
𝑦(𝐿) = 𝐴𝐿 = 0,
𝑦(𝑥) = 0
For 𝜆 > 0:
𝑦(0) = 𝐶1 = 0,
𝑦(𝐿) = 𝐶2 sin(√𝜆𝐿) = 0,
sin(√𝜆𝐿) = 0 = sin(𝑛𝜋) ,
𝑛𝜋 2
𝜆=( )
𝐿
𝐶2 ≠ 0
𝑛𝜋𝑥
𝑦(𝑥) = 𝐶2 sin (
)
𝐿
For 𝜆 < 0 (not an eigenvalue):
𝑦(𝐿) = 𝐶2 𝐿𝑒 √𝜆𝐿 = 0,
𝑦(0) = 𝐶1 = 0,
𝐶2 = 0
𝑦(𝑥) = 0
Another example:
𝑦 ′′ (𝑥) + 𝜆𝑦(𝑥) = 0,
𝑦 ′ (0) = 𝑦 ′ (𝐿) = 0
𝜆 = 0,
𝑚2 + 𝜆 = 0,
{𝜆 > 0,
𝑦(𝑥) = 𝐴𝑡 + 𝐵
𝑦(𝑥) = 𝐶1 cos(√𝜆𝑥) + 𝐶2 sin(√𝜆𝑥)
𝜆 < 0,
𝑦(𝑥) = 𝐶1 𝑒 √𝜆𝑥 + 𝐶2 𝑥𝑒 √𝜆𝑥
For 𝜆 = 0:
𝑦 ′ (0) = 𝐴 = 0,
𝑦 ′ (𝐿) = 𝐴 = 0,
𝑦(𝑥) = 𝐵,
𝐵∈ℝ
𝑦(𝑥) = 1
For 𝜆 > 0:
𝑦 ′ (𝑥) = 𝐶2 √𝜆 cos(√𝜆𝑥) − 𝐶1 √𝜆 sin(√𝜆𝑥)
𝑦(0) = 𝐶2 √𝜆 = 0,
𝑦(𝐿) = −𝐶1 √𝜆 sin(√𝜆𝐿) = 0,
sin(√𝜆𝐿) = 0 = sin(𝑛𝜋) ,
𝐶1 ≠ 0
𝑛𝜋 2
𝜆=( )
𝐿
𝑛𝜋𝑥
𝑦(𝑥) = cos (
)
𝐿
For 𝜆 < 0 (Not an eigen value):
𝑦 ′ (𝑥) = 𝐶1 √𝜆𝑒 √𝜆𝑥 + 𝐶2 √𝜆𝑥𝑒 √𝜆𝑥 + 𝐶2 𝑒 √𝜆𝑥
𝑦 ′ (0) = 𝐶1 √𝜆 + 𝐶2 = 0,
𝑦 ′ (𝐿) = 𝐶2 √𝜆𝐿𝑒 √𝜆𝐿 = 0,
𝐶2 = 0
𝑦(𝑥) = 0
The last example:
𝑦 ′′ (𝑥) + 2𝑦 ′ (𝑥) + (1 + 𝜆)𝑦(𝑥) = 0,
𝑚2 + 2𝑚 + (𝜆 + 1) = 0,
𝑚=
𝑦(0) = 𝑦(1) = 0
−2 ± √4 − 4(1 + 𝜆)
= −1 ± 𝑖√𝜆
2
For 𝜆 = 0 (Not an eigen value):
𝑦(𝑥) = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑥𝑒 −𝑥
𝑦(1) = 𝐶2 𝑒 −1 = 0,
𝑦(0) = 𝐶1 = 0,
𝑦(𝑥) = 0
For 𝜆 > 0:
𝑦(𝑥) = 𝐶1 𝑒 −𝑥 cos(√𝜆𝑥) + 𝐶2 𝑒 −𝑥 sin(√𝜆𝑥)
𝑦(1) = 𝐶2 𝑒 −1 sin(√𝜆) = 0,
𝑦(0) = 𝐶1 = 0,
sin(√𝜆) = 0 = sin(𝑛𝜋) ,
𝐶2 ≠ 0
𝜆 = (𝑛𝜋)2
𝑦(𝑥) = 𝑒 −𝑥 sin(𝑛𝜋𝑥)
For 𝜆 < 0 (Not an eigenvalue):
𝑦(𝑥) = 𝐶1 𝑒 (−1+√𝜆)𝑥 + 𝐶2 𝑒 (−1−√𝜆)𝑥
𝑦(1) = 𝐶1 𝑒 (−1+√𝜆) + 𝐶2 𝑒 (−1−√𝜆) = 𝐶1 (𝑒 √𝜆 − 𝑒 −√𝜆 ) = 0
𝑦(0) = 𝐶1 + 𝐶2 = 0,
𝑒 √𝜆 − 𝑒 −√𝜆 ≠ 0
𝐶1 = 0,
𝑦(𝑥) = 0
Standard eigenvalue problems’ solution:
𝑦 ′′ (𝑥) + 𝜆𝑦(𝑥) = 0,
𝑦
′′ (𝑥)
+ 𝜆𝑦(𝑥) = 0,
𝑦(0) = 𝑦(𝐿) = 0,
𝑦
′ (0)
=𝑦
′ (𝐿)
= 0,
𝑛𝜋 2
𝜆=( ) ,
𝐿
𝑛𝜋𝑥
𝑦(𝑥) = sin (
)
𝐿
𝑛𝜋 2
(
𝜆={ 𝐿) ,
0
𝑦(𝑥) = {
𝑛𝜋𝑥
cos (
)
𝐿
1
2
𝑦 ′′ (𝑥) + 𝜆𝑦(𝑥) = 0,
𝑦 ′ (0) = 𝑦(𝐿) = 0,
𝜆=(
(2𝑛 − 1)𝜋
) ,
2𝐿
𝑦(𝑥) = cos (
2
𝑦
′′ (𝑥)
+ 𝜆𝑦(𝑥) = 0,
𝑦
′ (0)
= 𝑦(𝐿) = 0,
(2𝑛 − 1)𝜋
𝜆=(
) ,
2𝐿
𝑦(𝑥) = sin (
(2𝑛 − 1)𝜋𝑥
)
2𝐿
(2𝑛 − 1)𝜋𝑥
)
2𝐿
Theme 5.2 – The heat equation
The heat equation is an equation used to describe heat flow in an laterally insulated, solid, thin,
homogenous material (although in this case that is the assumption). The heat function, 𝑢(𝑥, 𝑡),
describes the temperature in the solid in terms of the position on the solid, 𝑥, as well as at a
time, 𝑡. The heat equation problem will consist of the actual heat equation, two boundary
conditions and an initial condition. There are two standard heat equations. The first, Problem A,
has a boundary condition that keeps the ends at a constant temperature.
𝑢𝑡 (𝑥, 𝑡) = 𝑘𝑢𝑥𝑥 (𝑥, 𝑡),
0<𝑥<𝐿
0<𝑡
{ 𝑢(0, 𝑡) = 𝑢(𝐿, 𝑡) = 0,
𝑢(𝑥, 0) = 𝑓(𝑥),
0<𝑥<𝐿
The second, Problem B, has a boundary condition that keeps the ends insulated (no heat
transfer).
𝑢𝑡 (𝑥, 𝑡) = 𝑘𝑢𝑥𝑥 (𝑥, 𝑡),
0<𝑥<𝐿
0<𝑡
{ 𝑢𝑥 (0, 𝑡) = 𝑢𝑥 (𝐿, 𝑡) = 0,
𝑢(𝑥, 0) = 𝑓(𝑥),
0<𝑥<𝐿
How does one go about solving this? Consider problem A:
𝑢𝑡 (𝑥, 𝑡) = 𝑘𝑢𝑥𝑥 (𝑥, 𝑡),
0<𝑥<𝐿
0<𝑡
{ 𝑢(0, 𝑡) = 𝑢(𝐿, 𝑡) = 0,
𝑢(𝑥, 0) = 𝑓(𝑥),
0<𝑥<𝐿
Assume the equation is separable such that the temperature is a product: 𝑈(𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇(𝑡)
𝑈(𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇(𝑡)
𝑈(0, 𝑡) = 𝑋(0) ∙ 𝑇(𝑡) = 0,
𝑋(0) = 0,
𝑈(𝐿, 𝑡) = 𝑋(𝐿) ∙ 𝑇(𝑡) = 0
𝑋(𝐿) = 0,
𝑈𝑡 (𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇 ′ (𝑡),
𝑇(𝑡) ≠ 0
𝑈𝑥𝑥 (𝑥, 𝑡) = 𝑋 ′′ (𝑥) ∙ 𝑇(𝑡)
𝑋(𝑥) ∙ 𝑇 ′ (𝑡) = 𝑘 ∙ 𝑋 ′′ (𝑥) ∙ 𝑇(𝑡),
𝑋 ′′ (𝑥) 𝑇 ′ (𝑡)
=
𝑋(𝑥)
𝑘𝑇(𝑡)
Looking at this equation when we change 𝑥, the left-hand side will change as it depends on 𝑥.
The right-hand side will remain unchanged if 𝑥 is changed. The same will happen if 𝑡 changes.
The only way this equation can be satisfied, is when both are equal to a constant.
𝑋 ′′ (𝑥) 𝑇 ′ (𝑡)
=
= −𝜆
𝑋(𝑥)
𝑘𝑇(𝑡)
𝑋 ′′ (𝑥)
= −𝜆,
𝑋 ′′ (𝑥) + 𝜆𝑋(𝑥) = 0
𝑋(𝑥)
This is an eigenvalue problem from theme
5.1.
𝑋 ′′ (𝑥) + 𝜆𝑋(𝑥) = 0,
𝑋(0) = 𝑋(𝐿) = 0
𝑇 ′ (𝑡)
= −𝜆,
𝑇 ′ (𝑡) = −𝜆𝑘𝑇(𝑡)
𝑘𝑇(𝑡)
This is a separable differential equation from
WTW256.
𝑑𝑇
= −𝜆𝑘𝑇(𝑡)
𝑑𝑡
Solving the eigenvalue problem:
𝑋 ′′ (𝑥) + 𝜆𝑋(𝑥) = 0,
𝑛𝜋 2
𝜆=( ) ,
𝐿
𝑋(0) = 𝑋(𝐿) = 0
𝑛𝜋𝑥
𝑦(𝑥) = sin (
)
𝐿
Solving the separable differential equation:
𝑑𝑇
= −𝜆𝑘𝑇(𝑡)
𝑑𝑡
𝑇(𝑡) = 𝑒
−𝑘𝑡(
𝑛𝜋 2
)
𝐿
Putting all the pieces together:
𝑈(𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇(𝑡)
𝑈(𝑥, 𝑡) = 𝑒
−𝑘𝑡(
𝑛𝜋 2
)
𝐿
𝑛𝜋𝑥
∙ sin (
)
𝐿
For the Fourier solution:
𝐿
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 𝑓(𝑥) sin (
) 𝑑𝑥
𝐿
𝐿
0
∞
𝑈(𝑥, 𝑡) = ∑ (𝑐𝑛 𝑒
−𝑘𝑡(
𝑛𝜋 2
)
𝐿
𝑛=0
𝑛𝜋𝑥
∙ sin (
))
𝐿
Problem B is solved using the same steps. The eigenvalue problem will have a different solution
𝑛𝜋𝑥
) and a constant term 𝑐0 .
𝐿
and 𝑐𝑛 is computed using cos (
Problem A Fourier series solution:
∞
𝑈(𝑥, 𝑡) = ∑ (𝑐𝑛 𝑒
−𝑘𝑡(
𝑛=0
𝑛𝜋 2
)
𝐿
𝐿
𝑛𝜋𝑥
sin (
)) ,
𝐿
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 𝑓(𝑥) sin (
) 𝑑𝑥
𝐿
𝐿
0
Problem B Fourier series solution:
∞
𝑈(𝑥, 𝑡) = 𝑐0 ∑ (𝑐𝑛 𝑒
𝑛=0
−𝑘𝑡(
𝑛𝜋 2
)
𝐿
𝑛𝜋𝑥
cos (
)) ,
𝐿
𝐿
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 𝑓(𝑥) cos (
) 𝑑𝑥 ,
𝐿
𝐿
The standard heat equation
The end points are held at a constant temperature
The end points are insulated
The thermal diffusivity of the material
The initial condition or initial temperature
distribution
The length of the rod
0
𝐿
1
𝑐0 = ∫ 𝑓(𝑥)𝑑𝑥
𝐿
0
𝑢𝑡 (𝑥, 𝑡) = 𝑘𝑢𝑥𝑥 (𝑥, 𝑡),
0<𝑥<𝐿
𝑢(0, 𝑡) = 𝜃1 ,
𝑢(𝐿, 𝑡) = 𝜃2 ,
0<𝑡
𝑢𝑥 (0, 𝑡) = 𝑢𝑥 (𝐿, 𝑡) = 0,
0<𝑡
𝑘
𝑢(𝑥, 0) = 𝑓(𝑥),
0<𝑥<𝐿
𝐿
Two steel rods 𝑘 = 0.15
𝑐𝑚2
𝑠
are both 25 𝑐𝑚 in length. Rod A is having an initial uniform
temperature of 100℃ while rod B is 0℃. At time 𝑡 = 0 the ends of the rods are put into contact.
Determine:
a) The Fourier series solution when the outer ends are kept at 0℃
b) The Fourier series solution when the outer ends are insulated.
𝑢𝑡 (𝑥, 𝑡) = 0.15𝑢𝑥𝑥 (𝑥, 𝑡),
0 < 𝑥 < 50
𝑢(0, 𝑡) = 𝑢(50, 𝑡) = 0,
0<𝑡
100,
0 < 𝑥 < 25
𝑢(𝑥, 0) = {
0,
25 < 𝑥 < 50
𝑈(𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇(𝑡)
𝑈(0, 𝑡) = 𝑋(0) ∙ 𝑇(𝑡) = 0,
𝑋(0) = 0,
𝑈(50, 𝑡) = 𝑋(50) ∙ 𝑇(𝑡) = 0
𝑋(50) = 0,
𝑈𝑡 (𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇 ′ (𝑡),
𝑈𝑥𝑥 (𝑥, 𝑡) = 𝑋 ′′ (𝑥) ∙ 𝑇(𝑡)
𝑋 ′′ (𝑥)
𝑇 ′ (𝑡)
=
= −𝜆
𝑋(𝑥)
0.15 ∙ 𝑇(𝑡)
𝑋(𝑥) ∙ 𝑇 ′ (𝑡) = 0.15 ∙ 𝑋 ′′ (𝑥) ∙ 𝑇(𝑡),
𝑋 ′′ (𝑥)
= −𝜆,
𝑋 ′′ (𝑥) + 𝜆𝑋(𝑥) = 0
𝑋(𝑥)
𝑋 ′′ (𝑥) + 𝜆𝑋(𝑥) = 0,
𝑋(0) = 𝑋(𝐿) = 0
𝑛𝜋𝑥
𝑋𝑛 = sin (
),
50
𝑛𝜋 2
𝜆=( )
50
𝑈(𝑥, 𝑡) = ∑ (𝑐𝑛 𝑒
−0.15𝑡(
𝑛=0
25
𝑛𝜋 2
)
50
𝑇 ′ (𝑡)
= −𝜆,
𝑇 ′ (𝑡) = −𝜆𝑘𝑇(𝑡)
𝑘𝑇(𝑡)
𝑑𝑇
= −𝜆𝑘𝑇(𝑡)
𝑑𝑡
𝑇𝑛 = 𝑒 −
𝑈𝑛 (𝑥, 𝑡) = 𝑒 −
∞
𝑇(𝑡) ≠ 0
0.15𝑛2 𝜋2
𝑡
2500
𝑛𝜋𝑥
sin (
)) ,
50
0.15𝑛2 𝜋2
𝑡
2500
𝑛𝜋𝑥
∙ sin (
)
50
𝐿
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 𝑓(𝑥) sin (
) 𝑑𝑥
𝐿
𝐿
0
50
1
𝑛𝜋𝑥
1
200
𝑛𝜋
𝑐𝑛 =
∫ 100 sin (
) 𝑑𝑥 +
∫ 0𝑑𝑥 =
(1 − cos ( ))
25
𝐿
25
𝑛𝜋
2
0
25
∞
𝑈(𝑥, 𝑡) = ∑ (
𝑛=1
𝑡 = 0𝑠
𝑡 = 10𝑠
𝑛𝜋 2
200
𝑛𝜋
𝑛𝜋𝑥
−0.15𝑡( )
50 sin (
(1 − cos ( )) 𝑒
))
𝑛𝜋
2
50
𝑡 = 200𝑠
𝑡 = 1800
𝑡 = 7200𝑠
𝑢𝑡 (𝑥, 𝑡) = 0.15𝑢𝑥𝑥 (𝑥, 𝑡),
0 < 𝑥 < 50
(0,
(50,
𝑢𝑥 𝑡) = 𝑢𝑥
𝑡) = 0,
0<𝑡
100,
0 < 𝑥 < 25
𝑢(𝑥, 0) = {
0,
25 < 𝑥 < 50
𝑈(𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇(𝑡)
𝑈𝑥 (0, 𝑡) = 𝑋 ′ (0) ∙ 𝑇(𝑡) = 0,
𝑋 ′ (0) = 0,
𝑈𝑥 (50, 𝑡) = 𝑋 ′ (50) ∙ 𝑇(𝑡) = 0
𝑋 ′ (50) = 0,
𝑈𝑡 (𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇 ′ (𝑡),
𝑈𝑥𝑥 (𝑥, 𝑡) = 𝑋 ′′ (𝑥) ∙ 𝑇(𝑡)
𝑋 ′′ (𝑥)
𝑇 ′ (𝑡)
=
= −𝜆
𝑋(𝑥)
0.15 ∙ 𝑇(𝑡)
𝑋(𝑥) ∙ 𝑇 ′ (𝑡) = 0.15 ∙ 𝑋 ′′ (𝑥) ∙ 𝑇(𝑡),
𝑋 ′′ (𝑥)
= −𝜆,
𝑋 ′′ (𝑥) + 𝜆𝑋(𝑥) = 0
𝑋(𝑥)
𝑋 ′′ (𝑥) + 𝜆𝑋(𝑥) = 0,
𝑋 ′ (0) = 𝑋 ′ (𝐿) = 0
𝑛𝜋𝑥
𝑋𝑛 = cos (
),
50
𝑛𝜋 2
𝜆=( )
50
𝑈𝑛 (𝑥, 𝑡) = 𝑒
∞
𝑈(𝑥, 𝑡) = 𝑐0 + ∑ (𝑐𝑛 𝑒
−0.15𝑡(
𝑛𝜋 2
)
50
𝑛=0
𝑇(𝑡) ≠ 0
𝑇 ′ (𝑡)
= −𝜆,
𝑇 ′ (𝑡) = −𝜆𝑘𝑇(𝑡)
𝑘𝑇(𝑡)
𝑑𝑇
= −𝜆𝑘𝑇(𝑡)
𝑑𝑡
𝑇𝑛 = 𝑒 −
−
𝑛𝜋𝑥
cos (
)) ,
50
0.15𝑛2 𝜋2
𝑡
2500
0.15𝑛2 𝜋2
𝑡
2500
𝑛𝜋𝑥
∙ cos (
)
50
𝐿
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 𝑓(𝑥) cos (
) 𝑑𝑥 ,
𝐿
𝐿
0
25
𝐿
1
𝑐0 = ∫ 𝑓(𝑥)𝑑𝑥
𝐿
0
50
1
𝑛𝜋𝑥
1
200
𝑛𝜋
𝑐𝑛 =
∫ 100 sin (
) 𝑑𝑥 +
∫ 0𝑑𝑥 =
(sin ( ))
25
𝐿
25
𝑛𝜋
2
0
25
50
50
0
0
1
1
𝑐0 =
∫ 100𝑑𝑥 +
∫ 0𝑑𝑥 = 50
50
50
∞
𝑈(𝑥, 𝑡) = 50 + ∑ (
𝑛=0
𝑡 = 0𝑠
𝑡 = 10𝑠
𝑛𝜋 2
200
𝑛𝜋
𝑛𝜋𝑥
−0.15𝑡( )
50 cos (
(sin ( )) 𝑒
))
𝑛𝜋
2
50
𝑡 = 200𝑠
𝑡 = 1800
𝑡 = 7200𝑠
Notice the plots of each case. In case a) the points where held at a constant temperature even
though heat exchange is happening meaning that heat was constantly rejected at the endpoints
and in doing so maintained constant temperature (heat transfer at the ends where allowed).
The rod eventually cooled down to zero. In case b) the endpoints were insulated meaning no
heat transfer could happen to the surroundings from the ends. Slowly heat is distributed
uniformly within the rod. The ends aren’t kept at a constant temperature.
These PDE’s are considered as the standard heat equation. Non-standard equations may need
transformations to help solve the equation.
2𝑢𝑡 (𝑥, 𝑡) = 𝑢𝑥𝑥 (𝑥, 𝑡) + 8,
0<𝑥<1
𝑢(1, 𝑡) = 30,
0<𝑡
{𝑢(0, 𝑡) = 10,
𝑢(𝑥, 0) = 10 − 4𝑥 2 ,
0<𝑥<1
Use the transformation: 𝑈(𝑥, 𝑡) = 𝑊(𝑥, 𝑡) + ℎ(𝑥)
𝑈𝑥𝑥 = 𝑊𝑥𝑥 + ℎ′′ (𝑥)
𝑈𝑡 = 𝑊𝑡 ,
2𝑊𝑡 = 𝑊𝑥𝑥 + ℎ′′ (𝑥) + 8
The first objective is to find ℎ(𝑥) such that it will result in a standard heat equation.
ℎ′′ (𝑥) + 8 = 0
ℎ(𝑥) = −4𝑥 2 + 𝐶𝑥 + 𝐷
ℎ(0) = 𝐷,
ℎ(1) = −4 + 𝐶 + 𝐷
𝑈(0, 𝑡) = 10 = 𝑊(0, 𝑡) + ℎ(0),
𝑊(0, 𝑡) = 0,
ℎ(0) = 10
𝑈(0,1) = 30 = 𝑊(0,1) + ℎ(1),
𝑊(0,1) = 0,
ℎ(1) = 30
ℎ(0) = 10,
ℎ(1) = 24,
ℎ(𝑥) = −4𝑥 2 + 24𝑥 + 10
Find the initial temperature distribution 𝑊(𝑥, 0):
𝑈(𝑥, 0) = 𝑊(𝑥, 0) + ℎ(𝑥),
10 − 4𝑥 2 = 𝑊(𝑥, 0) − 4𝑥 2 + 24𝑥 + 10
𝑊(𝑥, 0) = −24𝑥
The new transformed problem is:
2𝑤𝑡 = 𝑤𝑥𝑥 ,
0<𝑥<1
0<𝑡
{𝑤(0, 𝑡) = 𝑤(1, 𝑡) = 0,
𝑤(𝑥, 0) = −24𝑥,
0<𝑥<1
Some non-standard heat equations can still be solved using the separation of variables.
𝑢𝑡 (𝑥, 𝑡) = 2𝑢𝑥𝑥 (𝑥, 𝑡) − 3𝑢(𝑥, 𝑡),
0<𝑥<2
{
𝑢(0, 𝑡) = 𝑢(2, 𝑡) = 0,
0<𝑡
𝑢(𝑥, 0) = 4𝑥,
0<𝑥<2
𝑈(𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇(𝑡)
𝑈(0, 𝑡) = 𝑋(0) ∙ 𝑇(𝑡) = 0,
𝑋(0) = 0,
𝑈(2, 𝑡) = 𝑋(2) ∙ 𝑇(𝑡) = 0
𝑋(2) = 0,
𝑈𝑡 (𝑥, 𝑡) = 𝑋(𝑥) ∙ 𝑇 ′ (𝑡),
𝑇(𝑡) ≠ 0
𝑈𝑥𝑥 (𝑥, 𝑡) = 𝑋 ′′ (𝑥) ∙ 𝑇(𝑡)
𝑋(𝑥) ∙ 𝑇 ′ (𝑡) = 2𝑋 ′′ (𝑥) ∙ 𝑇(𝑡) − 3𝑋(𝑥) ∙ 𝑇(𝑡)
3
𝑋 ′′ (𝑥) 𝑇 ′ (𝑡) 3
𝑋(𝑥) ∙ 𝑇 ′ (𝑡) = 2𝑇(𝑡) ∙ (𝑋 ′′ (𝑥) − 𝑋(𝑥)) →
=
+ = −𝜆
2
𝑋(𝑥)
2𝑇(𝑡) 2
𝑋 ′′ (𝑥)
= −𝜆
𝑋(𝑥)
𝑋 ′′ (𝑥) + 𝑋(𝑥) = 0,
𝑋(0) = 𝑋(2) = 0
𝑛𝜋𝑥
𝑋𝑛 = sin (
),
2
𝑛𝜋 2
𝜆=( )
2
𝑇𝑛 =
𝑈𝑛 (𝑥, 𝑡) =
∞
𝑈(𝑥, 𝑡) = ∑ (𝑐𝑛 𝑒
𝑛2 𝜋2
−𝑡(
+3)
2
𝑛=0
𝑇 ′ (𝑡) 3
3
+ = −𝜆,
𝑇 ′ (𝑡) = −2 (𝜆 + ) 𝑇(𝑡)
2𝑇(𝑡) 2
2
𝑑𝑇
3
= −2 (𝜆 + ) 𝑇(𝑡)
𝑑𝑡
2
𝑛 2 𝜋2
−𝑡(
+3)
2
𝑒
𝑛 2 𝜋2
−𝑡(
+3)
2
𝑒
𝑛𝜋𝑥
∙ sin (
)
2
𝐿
𝑛𝜋𝑥
∙ sin (
)) ,
2
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 𝑓(𝑥) sin (
) 𝑑𝑥
𝐿
𝐿
0
2
(−1)𝑛+1 ∙ 16
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 4𝑥 sin (
) 𝑑𝑥 =
2
2
𝑛𝜋
0
∞
𝑈(𝑥, 𝑡) = ∑ (
𝑛=0
2 2
(−1)𝑛+1 ∙ 16 −𝑡(𝑛 𝜋
2
𝑒
𝑛𝜋
Always keep the eigenvalue problem standard.
+3)
𝑛𝜋𝑥
∙ sin (
))
2
Theme 5.3: The wave equation.
The next PDE we’ll be discussing is the wave equation.
𝑦𝑡𝑡 (𝑥, 𝑡) = 𝛼 2 𝑦𝑥𝑥 (𝑥, 𝑡),
(0 < 𝑥 < 𝐿,
0 < 𝑡)
The heat equation has boundary conditions, an initial velocity and an initial displacement.
(0 < 𝑥 < 𝐿,
𝑦𝑡𝑡 (𝑥, 𝑡) = 𝛼 2 𝑦𝑥𝑥 (𝑥, 𝑡),
0 < 𝑡)
(0 < 𝑡)
𝑦(0, 𝑡) = 𝑦(𝐿, 𝑡) = 0,
(0 < 𝑥 < 𝐿)
𝑦(𝑥, 0) = 𝑓(𝑥),
(0 < 𝑥 < 𝐿)
𝑦𝑡 (𝑥, 0) = 𝑔(𝑥),
{
The first case is when the initial displacement is a non-zero function and the initial velocity is
zero (Problem A).
(0 < 𝑥 < 10,
𝑦𝑡𝑡 (𝑥, 𝑡) = 5𝑦𝑥𝑥 (𝑥, 𝑡),
0 < 𝑡)
(0 < 𝑡)
𝑦(0, 𝑡) = 𝑦(10, 𝑡) = 0,
2
(0
𝑦(𝑥, 0) = 𝑥 − 10𝑥,
< 𝑥 < 10)
(0 < 𝑥 < 10)
𝑦𝑡 (𝑥, 0) = 0,
Assume that 𝑦(𝑥, 𝑡) is separable:
𝑌 =𝑋∙𝑇
𝑌𝑡𝑡 = 𝑋 ∙ 𝑇 ′′ ,
𝑌𝑥𝑥 = 𝑋 ′′ ∙ 𝑇
𝑋 ∙ 𝑇 ′′ = 5 ∙ 𝑋 ′′ ∙ 𝑇 →
𝑦(0, 𝑡) = 0 = 𝑋(0) ∙ 𝑇(𝑡),
𝑋 ′′
𝑇 ′′
=
= −𝜆
𝑋
5∙𝑇
𝑇(𝑡) ≠ 0 ∴ 𝑋(0) = 0
𝑦(0,10) = 0 = 𝑋(10) ∙ 𝑇(𝑡),
𝑇(𝑡) ≠ 0 ∴ 𝑋(10) = 0
𝑦𝑡 (𝑥, 0) = 0 = 𝑋(𝑥) ∙ 𝑇 ′ (0),
𝑋(𝑥) ≠ 0 ∴ 𝑇 ′ (0) = 0
Solve the two differential equations:
𝑋 ′′
= −𝜆 → 𝑋 ′′ + 𝜆𝑋 = 0,
𝑋
𝑛𝜋𝑥
𝑋𝑛 = sin (
),
10
𝑋(0) = 𝑋(10) = 0
𝑛𝜋 2
𝜆=( )
10
𝑇 ′′
= −𝜆 → 𝑇 ′′ + 5𝜆𝑇 = 0,
5∙𝑇
𝑚 2 + 5𝜆 = 0
𝑇 ′ (0) = 0
𝑚 = ±𝑖√5𝜆
𝑇(𝑡) = 𝐶1 cos(𝑡√5𝜆) + 𝐶2 sin(𝑡√5𝜆)
′ (𝑡)
𝑇
= −𝐶1 √5𝜆 sin (𝑡√5𝜆) + 𝐶2 √5𝜆 cos (𝑡√5𝜆)
𝑇 ′ (0) = 𝐶2 √5𝜆 = 0 ∴ 𝐶2 = 0
𝑇(𝑡) = cos(𝑡√5𝜆)
𝑛𝜋𝑡
𝑇(𝑡) = cos (√5
)
10
𝑛𝜋𝑥
𝑛𝜋𝑡
𝑌𝑛 (𝑥, 𝑡) = sin (
) cos (√5
)
10
10
𝐿
∞
𝑛𝜋𝑥
𝑛𝜋𝑡
𝑌(𝑥, 𝑡) = ∑ (𝑐𝑛 sin (
) cos (√5
)) ,
10
10
2
𝑛𝜋𝑥
𝑐𝑛 = ∫ 𝑓(𝑥) sin (
) 𝑑𝑥
𝐿
𝐿
𝑛=0
0
10
2
𝑛𝜋𝑥
800
𝑐𝑛 =
∫ (𝑥 2 − 10𝑥) sin (
) 𝑑𝑥 = −
3
10
10
((2𝑛 + 1)𝜋)
0
∞
𝑌(𝑥, 𝑡) = ∑ (−
𝑛=0
800
𝑛𝜋𝑥
𝑛𝜋𝑡
) cos (√5
))
10
10
3 sin (
((2𝑛 + 1)𝜋)
The second case is when the initial velocity is a non-zero function (Problem B).
(0 < 𝑥 < 10,
𝑦𝑡𝑡 (𝑥, 𝑡) = 5𝑦𝑥𝑥 (𝑥, 𝑡),
0 < 𝑡)
(0 < 𝑡)
𝑦(0, 𝑡) = 𝑦(10, 𝑡) = 0,
2
(0
𝑦(𝑥, 0) = 𝑥 − 10𝑥,
< 𝑥 < 10)
(0 < 𝑥 < 10)
𝑦𝑡 (𝑥, 0) = 0,
𝑋 ∙ 𝑇 ′′ = 5 ∙ 𝑋 ′′ ∙ 𝑇 →
𝑋 ′′
𝑇 ′′
=
= −𝜆
𝑋
5∙𝑇
𝑦(0, 𝑡) = 0 = 𝑋(0) ∙ 𝑇(𝑡),
𝑦(0,10) = 0 = 𝑋(10) ∙ 𝑇(𝑡),
𝑦(𝑥, 0) = 0 = 𝑋(𝑥) ∙ 𝑇(0),
𝑇(𝑡) ≠ 0 ∴ 𝑋(0) = 0
𝑇(𝑡) ≠ 0 ∴ 𝑋(10) = 0
𝑋(𝑥) ≠ 0 ∴ 𝑇(0) = 0
Solve the two differential equations:
𝑋 ′′
= −𝜆 → 𝑋 ′′ + 𝜆𝑋 = 0,
𝑋
𝑛𝜋𝑥
𝑋𝑛 = sin (
),
10
∞
𝑋(0) = 𝑋(10) = 0
𝑛𝜋 2
𝜆=( )
10
𝑇 ′ (0) = 0
𝑚 = ±𝑖√5𝜆
𝑇(𝑡) = 𝐶1 cos(𝑡√5𝜆) + 𝐶2 sin(𝑡√5𝜆)
𝑇(0) = 𝐶1 = 0 ∴ 𝐶1 = 0
𝑇(𝑡) = sin(𝑡√5𝜆)
𝑛𝜋𝑡
𝑇(𝑡) = sin (√5
)
10
𝑛𝜋𝑥
𝑛𝜋𝑡
𝑌𝑛 (𝑥, 𝑡) = sin (
) sin (√5
)
10
10
𝑛𝜋𝑥
𝑛𝜋𝑡
𝑌(𝑥, 𝑡) = ∑ (𝑐𝑛 sin (
) sin (√5
)) ,
10
10
𝑛=0
𝑇 ′′
= −𝜆 → 𝑇 ′′ + 5𝜆𝑇 = 0,
5∙𝑇
𝑚 2 + 5𝜆 = 0
𝐿
2
𝑛𝜋𝑥
𝑐𝑛 =
∫ 𝑓(𝑥) sin (
) 𝑑𝑥
𝑛𝜋𝑎
𝐿
0
10
𝑛𝜋𝑥
200(−1)𝑛+1
𝑐𝑛 =
∫ 𝑥 sin (
) 𝑑𝑥 =
10
𝑛2 𝜋 2 √5
√5𝑛𝜋
2
0
∞
𝑌(𝑥, 𝑡) = ∑ (
200(−1)𝑛+1
𝑛=1
𝑛2 𝜋 2 √5
𝑛𝜋𝑥
𝑛𝜋𝑡
sin (
) sin (√5
))
10
10
0
1
2
3
5
6
7
8/1
4
The wave equation may be asked in a non-standard form which will need a transformation to
solve:
(0 < 𝑥 < 2,
𝑦𝑡𝑡 (𝑥, 𝑡) = 3𝑦𝑥𝑥 (𝑥, 𝑡) + 2𝑥,
(0 < 𝑡)
𝑦(0, 𝑡) = 𝑦(2, 𝑡) = 0,
(0 < 𝑥 < 2)
𝑦(𝑥, 0) = 0,
(0 < 𝑥 < 2)
𝑦𝑡 (𝑥, 0) = 0,
0 < 𝑡)
Use the following transformation: 𝑦(𝑥, 𝑡) = 𝑤(𝑥, 𝑡) + ℎ(𝑥)
𝑦𝑡𝑡 = 𝑤𝑡𝑡 ,
𝑤𝑡𝑡 = 3𝑤𝑥𝑥 + 3ℎ′′ (𝑥) + 2𝑥,
𝑦𝑥𝑥 = 𝑤𝑥𝑥 + ℎ′′ (𝑥)
3ℎ′′ (𝑥) + 2𝑥 = 0,
ℎ(𝑥) = −
𝑥3
+ 𝐶𝑥 + 𝐷
9
𝑦(0, 𝑡) = 𝑤(0, 𝑡) + ℎ(0) = 0,
𝑤(0, 𝑡) = 0 ∴ ℎ(0) = 0
𝑦(2, 𝑡) = 𝑤(2, 𝑡) + ℎ(2) = 0,
𝑤(2, 𝑡) = 0 ∴ ℎ(2) = 0
ℎ(0) = 𝐷 = 0,
8
ℎ(2) = − + 2𝐶 = 0,
9
4
𝐶= ,
9
→ ℎ(𝑥) = −
𝑦(𝑥, 𝑡) = 𝑤(𝑥, 𝑡) + ℎ(𝑥) = 0
𝑦(𝑥, 𝑡) = 𝑤(𝑥, 𝑡) −
𝑥3 4
𝑥3 4
+ 𝑥 = 0 → 𝑤(𝑥, 0) =
− 𝑥
9 9
9 9
𝑦𝑡 (𝑥, 0) = 𝑤𝑡 (𝑥, 𝑡) = 0 → 𝑦𝑡 (𝑥, 0) = 𝑤𝑡 (𝑥, 0) = 0
(0 < 𝑥 < 2,
𝑤𝑡𝑡 = 3𝑤𝑥𝑥 ,
0 < 𝑡)
(0 < 𝑡)
𝑤(0, 𝑡) = 𝑤(2, 𝑡) = 0,
𝑥3 4
(0 < 𝑥 < 2)
𝑤(𝑥, 0) =
− 𝑥,
9 9
(0 < 𝑥 < 2)
𝑤𝑡 (𝑥, 0) = 0,
𝑥3 4
+ 𝑥
9 9
Another method used to solve the wave equation is by using d’Alembert’s solution.
𝑦𝑡𝑡 = 𝑎2 𝑦𝑥𝑥 ,
0 < 𝑥 < 𝐿,
𝑦(0, 𝑡) = 𝑦(𝐿, 𝑡) = 0
𝑦(𝑥, 0) = 𝑓(𝑥),
0 < 𝑥 < 𝐿,
0 < 𝑥 < 𝐿,
{𝑦𝑡 (𝑥, 0) = 𝑔(𝑥),
𝑡>0
𝑡>0
𝑡>0
Problem A:
1
𝑦𝐴 (𝑥, 𝑡) = [𝐹(𝑥 + 𝑎𝑡) + 𝐹(𝑥 − 𝑎𝑡)]
2
The 𝐹 function refers to the odd extension of 𝑓, the initial displacement function. This method is
effective when the displacement function is requested at a specific timestamp. This is also
helpful when evaluating exact points on a string at any time.
Problem B:
𝑥+𝑎𝑡
1
1
[𝐻(𝑥 + 𝑎𝑡) − 𝐻(𝑥 − 𝑎𝑡)],
𝑦𝐵 (𝑥, 𝑡) =
∫ 𝐺(𝑠)𝑑𝑠 =
2𝑎
2𝑎
𝑥
𝐻(𝑥) = ∫ 𝐺(𝑠)𝑑𝑠
𝑥−𝑎𝑡
0
The 𝐺 function refers to the odd extension of 𝑔, the initial velocity function.
In the case where both the initial velocity and displacement is non-zero, the solutions to
problem A and B may be added together.
1
1
[𝐻(𝑥 + 𝑎𝑡) − 𝐻(𝑥 − 𝑎𝑡)]
𝑦(𝑥, 𝑡) = [𝐹(𝑥 + 𝑎𝑡) + 𝐹(𝑥 − 𝑎𝑡)] +
2
2𝑎