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Lecture 1 Sujin Khomrutai – 1 / 22
Method of Applied MathLecture 1: Series Solutions of Second Order
Linear ODEs
Sujin Khomrutai, Ph.D.
Introduction
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 2 / 22
We solve second order linear ODE
y′′ + p(x)y′ + q(x)y = f(x) (1)
where p(x), q(x) are functions of a variable x by the power seriesmethod.
Recall
y′ =dy
dx= the first derivative of y,
y′′ =d2y
dx2= the second derivative of y.
Example 1 (Wedge heat sink)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 3 / 22
EX. The model equation for a wedge heat sink
is
x2 d2y
dx2+ x
dy
dx− µxy = 0
where y(x) is the temperature at x and µ > 0 is a constant.
Example 2 (Beam under loads)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 4 / 22
EX. The model equation for a beam of length 2L under loads
is
axd2y
dx2− Py =
1
2wx2 − wx
where y(x) is the deflection at x and a > 0 is a constant.
Outline
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 5 / 22
Plan.
1. Power series
2. Ordinary points v.s. Regular singular points
3. Power series solutions near an ordinary point.
Series & Functions
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 6 / 22
A power series is an expression of the form
∞∑
k=0
ck(x− a)k = c0 + c1(x− a) + c2(x− a)2 + · · · ,
ck are called coefficients and a is called the center.
The sum, where the series converges, defines a function of x, i.e.
f(x) =∞∑
k=0
ck(x− a)k.
Example 3
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 7 / 22
EX. The power series
3− 6x+ 12x2 − 24x3 + · · ·
has center: a = 0 and coefficients:
c0 = 3, c1 = −6, c2 = 12, . . . , ck = 3 · (−2)k, . . .
It sum is the function
f(x) = 3− 6x+ 12x2 − 24x3 + · · ·
= 3(1 + (−2x) + (−2x)2 + (−2x)3 + · · · )
= 3
∞∑
k=0
(−2x)k =3
1 + 2x
Taylor Series
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 8 / 22
An important group of power series is the Taylor series ofdifferentiable functions.
Definition. Let f be a differentiable function. The Taylor seriesof f(x) about a is
Ta(x) =∞∑
k=0
f (k)(a)
k!(x− a)k
or
Ta(x) = f(a)+f ′(a)(x−a)+f ′′(a)
2!(x−a)2+
f ′′′(a)
3!(x−a)3+· · ·
Example 4
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 9 / 22
EX. Let f(x) = ex. Since
(ex)′ = ex,
we have
f ′(0) = 1, f ′′(0) = 1, . . . f (k)(0) = 1 ∀ k.
So the Taylor series of f(x) = ex about 0 is
T0(x) =∞∑
k=0
1
k!xk = 1 + x+
x2
2!+
x3
3!+ · · ·
Analytic Functions
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 10 / 22
For a differentiable function f , if its Taylor series at a convergeson an interval I = (a−R, a+R), then
f(x) = Ta(x) =∞∑
k=0
f (k)(a)
k!(x− a)k ∀ x ∈ I.
We say that f is analytic at a.
Fact. Most functions are analytic on their domains.
• ex is analytic at any a ∈ (−∞,∞).
• A polynomial P (x) is analytic at any a ∈ (−∞,∞).
• A rational functionP (x)
Q(x)is analytic at any a where Q(a) 6= 0.
Ordinary and Singular Points
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 11 / 22
Definition. Consider the ODE
y′′ + p(x)y′ + q(x)y = f(x).
If p, q, f are analytic at a, then a is called an ordinary point ofthe ODE.
If p, q, or f is not analytic at a, then a is called a singular pointof the ODE.
Example 5
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 12 / 22
EX. For the ODE
3y′′ + 4xy′ − 5x2y = 0
find all ordinary points and singular points of this equation.
Example 6
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 13 / 22
EX. For the ODE
(x2 − 1)y′′ + xy′ − y = 0
find all ordinary points and singular points of this equation.
Series Solutions: An Ordinary Point
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 14 / 22
At an ordinary point, we have the following theorem.
Theorem. [Kreyszig, 5.1 Theorem 1] Let a be an ordinary pointof the ODE
y′′ + p(x)y′ + q(x)y = f(x).
Then any solution of the ODE is analytic at a, so
y = b0 + b1(x− a) + b2(x− a)2 + · · · =∞∑
k=0
bk(x− a)k,
for some constants b0, b1, b2, . . .
We will pay a special attention to the case a = 0 and f(x) = 0.
Series Solutions: An Ordinary Point
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 15 / 22
Thus, solutions to y′′ + p(x)y′ + q(x)y = 0 have the form ofpower series, about each ordinary point a.
Method. Step 1. Set the solution y as a series about a
y = b0 + b1(x− a) + b2(x− a)2 + b3(x− a)3 + · · · ,
that is
y =∞∑
k=0
bk(x− a)k
Step 2. Termwise differentiation:
Series Solutions: An Ordinary Point
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 16 / 22
y′ = b1 + b2(2)(x− a) + b3(3)(x− a)2 + · · ·
y′′ = b2(2)(1) + b3(3)(2)(x− a) + b4(4)(3)(x− a)2 + · · ·
that is
y′ =∞∑
k=0
bk+1(k + 1)(x− a)k
y′′ =∞∑
k=0
bk+2(k + 2)(k + 1)(x− a)k.
Step 3. Solve for b0, b1, b2, . . .
Example 7 (Airy’s equation)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 17 / 22
EX. Show that 0 is an ordinary point of the ODE
y′′ − xy = 0.
Then find the general solution of the ODE as power series about 0.
Sol. It is easy to verify that a = 0 is an ordinary point.
According to the series solution method, we set
y =∞∑
k=0
bkxk
y′′ =∞∑
k=0
bk+2(k + 2)(k + 1)xk
Example 7 (Airy’s equation)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 18 / 22
∞∑
k=0
bk+2(k + 2)(k + 1)xk − x
∞∑
k=0
bkxk = 0
∞∑
k=0
bk+2(k + 2)(k + 1)xk −
∞∑
k=0
bkxk+1 = 0
Shift the index the second series to align the powers of x(j = k + 1):
∞∑
k=0
bkxk+1 =
∞∑
j=1
bj−1xj dummy
=∞∑
k=1
bk−1xk
(we throw away j finally, and re-use k).
Example 7 (Airy’s equation)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 19 / 22
Thus
∞∑
k=0
bk+2(k + 2)(k + 1)xk −
∞∑
k=1
bk−1xk = 0
Split terms in the first series (to align the starting sum):
b2 · 2 · 1 +∞∑
k=1
bk+2(k + 2)(k + 1)xk
So
b2 · 2 · 1 +∞∑
k=1
bk+2(k + 2)(k + 1)xk −
∞∑
k=1
bk−1xk = 0
Example 7 (Airy’s equation)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 20 / 22
Combine series
b2 · 2 · 1 +∞∑
k=1
[bk+2(k + 2)(k + 1)− bk−1]xk = 0
Note.∑
∞
k=0 ck(x− a)k = 0 then ck = 0 for all k. So
2b2 = 0, bk+2(k + 2)(k + 1)− bk−1 = 0 for all k ≥ 1.
Thus
b2 = 0, bk+2 =bk−1
(k + 1)(k + 2)for all k ≥ 1.
These equations are recurrence equations.
Example 7 (Airy’s equation)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 21 / 22
Solve recurrence equations
• b2 = 0, b5 = 0, b8 = 0, b11 = 0, . . .
b3n+2 = 0 (n ≥ 4)
• b3 =b0
2 · 3, b6 =
b0
2 · 3 · 5 · 6, b9 =
b0
2 · 3 · 5 · 6 · 8 · 9
b3n =b0
2 · 3 · 5 · 6 · · · (3n− 1)(3n)(n ≥ 4)
• b4 =b1
3 · 4, b7 =
b1
3 · 4 · 6 · 7, b10 =
b1
3 · 4 · 6 · 7 · 9 · 10
b3n+1 =b1
3 · 4 · 6 · 7 · · · (3n)(3n+ 1)(n ≥ 4)
Example 7 (Airy’s equation)
Introduction
EX 1.
EX 2.
EX 3.
EX 4.
Ord & Sing Points
EX 5.
EX 6.
Series Sol Method
EX 7
Lecture 1 Sujin Khomrutai – 22 / 22
Set back to the solution:
y =∞∑
k=0
bkxk = b0
[
1 +x3
2 · 3+
x6
2 · 3 · 5 · 6+ · · ·
]
+ b1
[
x+x4
3 · 4+
x7
3 · 4 · 6 · 7+ · · ·
]
This is the general solution where the fundamental solutions are
y1 = 1 +x3
2 · 3+
x6
2 · 3 · 5 · 6+ · · · ,
y2 = x+x4
3 · 4+
x7
3 · 4 · 6 · 7+ · · ·