MAT3320 Notes - By Eric Hua
Contents
Review of methods in solving di↵erential equations 2
1.4 Orthogonal Functions 3
8.1 Analytic Functions and Power Series Solutions 3
8.2 Legendre’s Equation (see also XXXXXXXXXX
8.3 Orthogonality Relations for Pn(x) (see also XXXXXXXXXX
XXXXXXXXXXFrobenius Method 9
10.1 Bessel’s Equation: The Solution Jn(x) of the First Kind 13
10.2 The Solution Yn(x) of the Second Kind 17
10.3 The Functions J1/2(x) and J�1/2(x) 18
10.4. The Orthogonality Relation for Jn(x) 18
11.1. Divergence Form of a Di↵erential Equation 19
11.2. Sturm-Liouville Problems 19
11.3. Orthogonality Relations 23
11.4. Regular and Singular Sturm-Liouville Problems 24
3.1. Fourier Series 26
3.1.1. Fourier Series (FS), Fourier sine series (FSS), Fourier cosine series
(FCS) 26
3.1.2. Convergence of Fourier Series. 31
3.1.3. Particular Cases of Fourier Series Expansion. 32
3.1.5. Euler’s Form of Fourier Series. 32
3.3. Fourier Transform 33
1.2. The Mathematical Models 34
1.3. A First Idea: Separation of Variables 35
2.1. Wave Problems in One Dimension 35
2.2. Heat Problems in One Dimension 40
2.3. Wave Problems in Two Dimensions 45
2.4. Heat Problems in Two Dimensions 48
2.5. Potential Problems for a Rectangular Region 50
4.1. The Potential Problem for One Sphere 55
4.2. The Potential Problem for Two Concentric Spheres 56
4.3. The Temperature u(', t) of a Thin Sphere 57
6.1. Wave Problem 58
6.1.1. Solution by separation of variables. 59
6.1.2. Elementary computation with Bessel functions. 61
6.2. Heat and Di↵usion Problems 62
6.2.1. Solution by separation of variables. 62
6.2.2. The heat problem for a cylinder of finite length. 63
6.3. Potential Problems for one and two Circles 65
6.3.1. The mathematical model. 65
6.3.2. Solution by separation of variables. 65
7.2. The Two-Dimensional Case 68
7.3. The Three-Dimensional Case 68
7.4. In the n-Dimensional Case 68
1
2
Review of methods in solving differential equations
DE = ODE + PDE
(1) Separable DE: dydx = f(x)g(y),) dy/g(y) = f(x)dx.
(2) Homogeneous second order linear eqn: ay00+by0+cy = 0. Let r1 and r2 be
two solutions of the indicial eqn (or, characteristic eqn) ar2 +
+ c = 0.
• r1 6= r2: y = c1er1x + c2er2x.
• r1 = r2: y = (c1 + c2x)erx.
• r = ↵ + i�: y = e↵x(c1 cos �x+ c2 sin �x).
(3) Second-order Cauchy-Euler equation: x2y00 + Axy0 + By = 0, x 6= 0.
The indicial equation is r2 + (A� 1) r+B = 0. Let r1 and r2 be the two
solutions of the indicial equation.
• If r1 6= r2 are real, then y1 = |x|r1 and y2 = |x|r2 .
• If r1 = r2 (real), then y1 = |x|r1 and y2 = xr1 ln |x|.
• If r1, r2 = ↵ ± i� (complex), then y1 = |x|↵ cos(� ln |x|) and y2 =
|x|↵ sin(� ln |x|).
(4) Reduction of order: If y1 is a solution of y00 + p(x)y0 + q(x)y = 0. Then
y2 = u(x)y1, where u0 =
1
y21
e
�
R
p(x)dx.
(5) Trig Identities:
• cos2 x = 1 + cos 2x
2
, sin2 x =
1� cos 2x
2
, 2 sin x cos x = sin 2x.
• sin a sin b = cos(a� b)� cos(a+ b)
2
, cos a cos b =
cos(a+ b) + cos(a� b)
2
,
• sin a cos b = sin(a+ b) + sin(a� b)
2
.
(6) Integration:
• Fundamental Theorem of Calculus:
Z
a
f(x)dx = F (b)� F (a),
where F 0(x) = f(x).
• Integration by parts:
Z
u(x)v0(x)dx = u(x)v(x)�
Z
u
0(x)v(x)dx,
o
Z
udv = uv �
Z
vdu.
• Integration by substitution:
Z
a
f(g(x))g0(x)dx =
Z g(b)
g(a)
f(u)du, where
u = g(x).
• Trig integral:
Z
sinm x cosn xdx :
– If m is odd, then let u = cosx.
– If n is odd, then let u = sin x.
– If m and n are even, then use half-angle formula.
• Trigonometric substitutions:
–
p
a2 � x2: Let x = a sin ✓, �⇡2 ✓
⇡
2 ;
–
p
a2 + x2: Let x = a tan ✓, �⇡2 < ✓
⇡
2 ;
–
p
x2 � a2: Let x = a sec ✓, 0 ✓ < ⇡2 or ⇡ ✓
3⇡
2 .
Example 1. Solve the di↵erential equation: y00 � y = 0.
Solution: y1(x) = ex, y2(x) = e�x.
Example 2. Solve the di↵erential equation y00 � 2y0 + y = 0.
3
1.4 Orthogonal Functions
Definition 1. Functions y1, y2, · · · defined on an interval a x b are called
orthogonal on a x b with respect to a weight function p(x) > 0 if
(1)
Z
a
p(x)ym(x)yn(x) dx = 0 (m 6= n).
The norm kymk is defined by
(2) kymk =
sZ
a
p(x)y2m(x) dx.
The functions are called orthonormal on a x b with respect to a
weight function if they are orthogonal and have norm 1.
Remark 1. For “orthogonal with respect to a weight function p(x) = 1” we
simply say “orthogonal”. Thus functions y1, y2, · · · are orthogonal on a x
if Z
a
ym(x)yn(x) dx = 0 (m 6= n).
The norm kymk is then simply defined by
kymk =
sZ
a
y2m(x) dx.
The functions are called orthonomal if they are orthogonal and have norm 1.
Example 3. (orthonomal set) The functions ym(x) =
sinmxp
⇡ , m = 1, 2, ... form
an orthonomal set on the interval �⇡ x ⇡.
In fact, for m 6= n, we have
Z ⇡
�⇡
ym(x)yn(x)dx =
1
⇡
Z ⇡
�⇡
sinmx sinnx dx =
1
2⇡
Z ⇡
�⇡
[cos(m�n)x�cos(m+n)x] dx = 0,
kymk2 =
1
⇡
Z ⇡
�⇡
sin2 mxdx = 1.
8.1 Analytic Functions and Power Series Solutions
A power series is an infinite series of the form
(3)
1X
m=0
am(x� x0)m = a0 + a1(x� x0) + a2(x� x0)2 + · · · ,
where a0, a1, · · · are coe�cients, x0 is the centre. The radius of convergence
of (3) can be obtained from the following formula:
(4) r =
1
limm!1 m
p
|am|
= lim
m!1
����
am
am+1
���� .
Remark 2. If r = 0, then the series (3) only converges at x0; if r > 0, then
the series (3) converges in the interval |x � x0| < r (convergence interval) and
diverges in |x� x0| > r. We write
s(x) =
1X
m=0
am(x� x0)m (|x� x0| < r).
4
Example 4.
1
1� x =
1X
m=0
x
m = 1 + x+ x2 + · · · , (|x| < 1).
e
x =
1X
m=0
x
m
m!
= 1 +
x
1!
+
x
2
2!
+ · · · , (|x| < 1).
Remark 3. Some useful equalities:
1X
m=0
am(x� x0)m =
1X
j=0
aj(x� x0)j
1X
m=1
mam(x� x0)m�1 =
1X
m=0
(m+ 1)am+1(x� x0)m,
1X
m=2
m(m� 1)am(x� x0)m�2 =
1X
m=0
(m+ 2)(m+ 1)am+2(x� x0)m.
Power series has the following properties:
(1) Termwise di↵erentiation: If
s(x) =
1X
m=0
am(x� x0)m (|x� x0| < r),
then it may be di↵erentiated term by term:
s
0(x) =
1X
m=1
mam(x� x0)m�1 (|x� x0| < r),
s
00(x) =
1X
m=2
m(m� 1)am(x� x0)m�2 (|x� x0| < r).
(2) Termwise addition and Termwise multiplication: If
f(x) =
1X
m=0
am(x� x0)m (|x� x0| < r1),
g(x) =
1X
m=0
m(x� x0)m (|x� x0| < r2),
then for |x� x0| < r = min(r1, r2),
f(x) + g(x) =
1X
m=0
(am + bm)(x� x0)m
f(x)g(x) =
1X
m=0
(a0bm + a1bm�1 + · · ·+ amb0)(x� x0)m
(3) Vanishing of all coe�cients (VAC): If
0 =
1X
m=0
am(x� x0)m (|x� x0| < r),
then a0 = a1 = a2 = · · · = 0.
5
Definition 2. A real function f(x) is said to be analytic at x = x0 if it can
e represented by a power series in powers of x � x0 with positive radius of
convergence R.
Now we consider the following di↵erential equation:
(5) h(x)y00 + p(x)y0 + q(x)y = r(x).
Theorem 1. (Existence of power series solutions) If h, p, q and r in (5) are
analytic at x = x0 and h(x0) 6= 0, then each solution of (5) is analytic at x = x0
with radius of convergence R > 0.
Definition 3. If h(x0) 6= 0, then x0 is called ordinary point; if h(x0) = 0, then
x0 is called singular point.
Example 5. Consider the DE: y00 � 2xy0 + y = 0. Note that x0 = 0 is ordinary
point, we shall attempt to find a series solution in the form:
y =
1X
n=0
anx
n
.
1) Find the recursive relation of the coe�cients in the series solution about
x0 = 0.
2) Solve the recursive relation.
3) Find the particular solution with y(0) = 1, y0(0) = 2.
Solution: 1) From y =
P1
n=0 anx
n we have
y
0 =
1X
n=1
nanx
n�1
,) xy0 =
1X
n=1
nanx
n
,
and
y
00 =
1X
n=2
n(n� 1)anxn�2 =
1X
n=0
(n+ 2)(n+ 1)an+2x
n
.
Substitute all of them into the DE, we imply that
an+2 =
2n� 1
(n+ 1)(n+ 2)
an.
2) From the recursive relation above,
a2n =
3 · · · 7 · 11 · · · (4n� 5)
(2n)!
a0,
a2n+1 =
1 · · · 5 · 9 · · · (4n� 3)
(2n+ 1)!
a1.
We have
y = a0
1� 1
2!
x
2 �
1X
n=2
3 · · · 7 · 11 · · · (4n� 5)
(2n)!
x
2n
!
+a1
x+
1X
n=1
1 · · · 5 · 9 · · · (4n� 3)
(2n+ 1)!
x
2n+1
!
3) From y(0) = 1 we have a0 = 1; by y0(0) = 2, we get a1 = 2. Thus
y =
1� 1
2!
x
2 �
1X
n=2
3 · · · 7 · 11 · · · (4n� 5)
(2n)!
x
2n
!
6
+2
x+
1X
n=1
1 · · · 5 · 9 · · · (4n� 3)
(2n+ 1)!
x
2n+1
!
Example 6