Methods of Mathematical Physics
Solution of Exercise Problems
Yan Zeng
March 9 , 2008
1
Contents
1
Complex N umbers and Complex Fu nctions
5
2
Analytic Fu nctions
2.1 Exercises in the t ext . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Exercises at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
5
5
7
3
Complex Integration
10
In且 nite Series
Exercises in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
Exercises at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
13
14
5
Local Expansion of Analytic Fu nctions
5.1 Exercises in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Exercises at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
18
19
6
Power Series Solution of Second Order Linear ODE
6.1 Exercises in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Exercises at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
24
24
4
4.1
4.2
7 Residue Theorem and It s Applications
7.1
7.2
Exercises in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Exercises at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
33
38
8
r Fu nction
52
9
Laplace Tr ansform
9.1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
55
55
56
10 ð Fu nction
62
11 Complex Fu nctions in Mathematica
68
12 Equations of Mathematical Physics
68
13 General Solutions of Linear PDE
13.1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
13.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
69
69
69
14 Separation of Variables
14.1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
14.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72
72
76
15 Orthogonal Curvilinear Coordinates
15.1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
83
83
84
16 Spherical Fu nctions
16.1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
16.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
85
85
87
2
17 Cylinder Functions
17.1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
17.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
87
87
88
18 Summary of Separation of Variables
18.1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..
18.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
88
88
88
19 Applications of Integral Tr ansforms
90
20 Method of Green's Function
94
21 Introduction to Calculus of Variation
94
21. 1 Exercise in the text . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 94
21. 2 The Rayleigh-Ritz method and its application to the Sturm-Liouville problem . . . . . . .. 96
21. 3 Solutions of the exercise problems from Gelfand and Fomin [4], Chapter 8 . . . . . . . . . .. 97
21. 4 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
22 Overview of Equations of Mathematical Physics
102
22 .1 Summary on the classification of second order linear PDE . . . . . . . . . . . . . . . . . . . . 102
22.2 Exercise at the end of chapter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
A The Black-Scholes partial differential equation
106
A.1 Derivation of the Bl ack-Scholes PDE and its boundary conditions . . . . . . . . . . . . . . . . 106
A.2 Simpli且cation of the Black-Scholes PDE via change-of-variable . . . . . . . . .. .. . . . . . 107
A.3 Solution of the simplified PDE and the Black-Scholes call option pricing formula . . . . . . . 108
3
This is a solution manual of selected exercise problems from Methods of Mathematical Physics , 2nd
Edition (in Chine叫 ， by 飞再Tu Chong-Shi (Peking University Press , Beiji吨 ， 2003).
4
1
Complex N umbers and Complex Fu nctions
Exercises are omitted since they are straightforward.
2
Analytic Functions
2.1
Exer cÌ ses in the text
2.1.
Proof.
.θf
t 一一=
. (θu
θu\θv
θuθfθu
θu
z 1 一一十 t 一一 1= 一 一一 +z 一一吨一一=一一 +z 一一
θx\θzθx)
θzθx
θνθυθU
So zgt= 去 if and 0叶 if Cau句-Riemann equations hold
口
2.2.
\飞tll/
门。大。
一。"
门。大。
z
\
/It-飞-
Therefore 茹= 0 if a nd only if Cau向-Riema
\飞tll/
U-u+
nd大U
飞'
、\
sa'
/，
，
qO言。
一一
+
nd玄。
\Iltl/ 1-2
1-2U-E 1-2U-u
U-x
门。大。
飞'
、\
sa'
-qtu
/，
，
+
J/
，』，，
tz
一。"
+
、、、
-i-9"
-t
仇-h
一-
\
/It-飞-
fJ一刊d
o-a
门向一句
P叫 8ince x = ~ (z + z*) and υ= 一言 (z - z*) , we have 先 =j and 茹=~ . 80
u-uu U-x
+
口
2.3.
Proof. Apply Cauchy-Riemann equations.
口
2.4.
Proof. Omitted since the proofs are based on the definition and are similar to those for functions of real
variables.
口
2.5.
丹阳阳
P
附
ro
叫
Oω
O
山 f
盯扣川
(μωz
斗)川臼川川
拍削
a
nl恼
叩叫
址
a
叫
叫呻
l忖
吻
yti
川川
川川川矶
C
时1]川
w
t吐川
h山川
1叫
f
mean value theorem does not apply to f(z). (This example is from [8].)
口
2.6.
Proof. 8uppose f(z) = u(x , y) 十 iv(凯 y) . Then by Ca旧hy- Riemann equations , f' (z) = 0 in G implies all
the partial derivatives of u and v with respect to x and y equal to 0 in G. Using the results for functions of
real variables , we conclude u and v are constants in G . 80 .f is a constant in G .
口
2.7.
Proof. This is a direct corollary of the Cauchy-Riemann equations.
2.8.
5
口
f. From the condition αu(x ， y) 十 bv(立 ， ν) = c , we 1
PTOO
111/
\、
b 一α
。unu
--一-
U
z
v
-u
门。玄。
一­
nO 玄。
u
-x
门。一
θ
一­
一一
冒。 一 α
、、飞z
''
''
』/
u - Uυ
一一
一α
tBE
U
X
飞'
飞'
\tI
''''
AO大。
一一
，。 一 α
u //ItI\
U
、、飞z
''
''
』/
『
AU 大。
1
一一
t/''
、、、、
ιv 一 α
飞'
飞'
\tI
''''
BB
v
-u
nO 大。
、、、『
、，，，，，
b 一α
一­
i.e.
v
-u
、\
/ti1
LV 一 α
-2
nO 言。
2
θ
θ -
b
(二十 1) 去=
、、、，，，，/
'''飞
l 飞\
U
ud
'''
门。 一θ
一一
uu
z
一一
飞t
、t
/zft
FOFO
++
θ θ
111//
\、
一α
-b一
b句
由
n
、
b
''1飞
' 飞\
U
z
'''
θ θ
一α
111//
\、
b
i.e.
、、\
，，，
B『E
，。 一 α
nL
，，
u
z
9"
一­
门。 一θ
80
h-h
一h
句
αα
/EEE
EE
，，，E、
.•
(三+ 1) 去=队时的 implies 去= o. Co毗I时 we conclude v is a constant in G. By Cauchy-
Riemann eql川阳1日， we can a1so co肌1ude u is a constant in G. 80 f(z) is a constant in G.
If α ， b, and c are non-zero comp1ex constants , the conclusion still ho1ds. Indeed , we have shown for the
rea1 case that when 二 十 1 兴。 ， f(z) is a constant in G. Now suppose 二十 1
0 , we have two cases:
b = ia and b =
肌 In the first case , we have α f(z) = αu(几 y) + iav(几 y) = αu(风 y) + 如何 ， y) = c. 80
f(z) = c/αis a constant. In the second case , we have α f* (z) = αu(x ， y) -iαυ (x ， y) = αu 怡， υ)+ 如何， υ) = c.
80 f(z) = (c/α)* is a constant.
口
2.9.
Proof. We write z in its po1ar form: z = r e' e = r( cos 8 + i sin 8). Then eZ = e'ω0νe'川
S归川
H:r
remains the same and the modu1us of e Z may tend to ∞， or 0 ,
or remain the same , depending on the sign of cos 8.
Denote by 8 the princip1e argument ofα ， then zη = 10g Iα1 + (8 + 2πη)i satis且es the requirement.
口
。∞
o with 丑X
对
ed 缸
a rg
伊
umen
时
1此t ， the argument of e Z
2.10.
Proof. All the equalities can be proved via the equa1ities for trigonometric functions and the re1ation between
hyperbo1ic functions and trigonometric fu日ctions. 认1e on1y prove the the two inequality.
First , we observe
~ le Y - e 叫 = 1 [e2ν+ε2ν2]1/2 :s ~[ε2ν+ε2y _ 2 + 4 日in 2 xf/2 = I si叫Z 十 ψ) I
I sinh yl
2
~ [ε2U + e2ν+2 - 4C0822lI/2<1[ε2U
+ e2ν+ 2f/2 = cosh y.
- 2
8imi1arly, we have
|ε νενI = ~[ε2ν+ε 2y _ 2f/2
sinh yl
[ε2y + ε2ν 十 2 - 4 日iII2 叫 1/2<
1
:s ~[e-2ν+ε
2y + 4cos 2 X - 2f/2 = I cos(x + iy)1
2
1
[ε2y + e 勾+ 2f/2 = cosh y
2
口
2.1 1.
Proof. We first assume If(z) 1 is a constant in G. If this constant is 0 , we have nothing to prove. 80 without
10ss of genera1ity, we assume If(z) 1 is a non-zero constant in G. 8叩pose f(z) = u(x , y) + iv(x ， ν) ， then
x If( zW =
y If( zW = 0 gives
!1
!1
(2u(ZU) 平 十 2v(x ， y) 旦1茸u!l = 0
μ (x ， y) 鸣叫 十 2巾， υ) 呜u!l = 0, 'v'(x ， y) εG
6
Using Cauchy-Riemann equations , we have
(卢 +v虹=;
θ /J
)一一
8y
e叫斗卡
|fh!1L M
川
θx
θu
-u 一-
~8x
川川)
川 1 = If川笋刮刊川
O们
o川
inn
削
θb
ψ叫(x，yω)
θ仇加削川川…
叫vy乓(μ
川川
巾川川
收川叫
Z川叫
Uω1 = 0
in G. That is , v(x , y) is a constant in G. Cauchy-Riemann equations imply u(x , y) is a constant in G as
well. 80 f(z) is a constant in G.
We then assume e = e(z) := argf(z) is a constant in G . Write f in polar coordinate: f(z) = r 忡， υ)ει
Then Cauchy Riemann equations become
( ￡乡趴扣
7叫(μZ
巾 ω
阶归川
叶
Cωω
O创吕
￡r(川)cωose =
If sin e = 0 , we have cos e 并 o and
才￡
t7(归
川
Z
， νyω) sine
!1c r = !1y = O. If cos e = 0 , we have sin e 并 O and ￡ T = ￡ r = O
T
e
If剖ne 并 o and co日 0 并 0 , 阴
we hav
刊e 才￡
tT = 才￡
}Tt阳
aMIM=
￡ T=O In either of 山 three cas旺 we always have
is a constant in G by the result of first half.
茬5tM
3
阳
创n 2 , which im
a
口
m
丑1甲
抖lies
p
臼吕 才￡
tr = O CO∞
ns日叫uent即
l忖
y，
!1 r = !1 r = 0 in G. 80 r is a constant in G and f (z)
y
r
口
2.12.
Proof. d ê, = 去 dx+ 弩句，向=告 dx+ 苟 dy . 80 by Cauchy-Riemann 叫1肌
θEθηθEθη/θη 飞 2/EV2
= ()山dy 十 ( ~" 1 dxdy = 1 f' ( zW 山dy.
θzθuθuθx
\θυ/\θy)
口
2.2
Exer cÌ ses at the end of chapter
1.
Proof. The basic method is to verify that Ref and Imf are differentiable as functions of real variables , and
that they satisfy Cauchy-Riemann equations .
口
2.
Proof Since
Jz=r …? , we have
I Y = rsm l1
国 = [冗:。 也] [f]:= 川自l
川sy to 叫 1(077) = fU;:; 一;yl miting 叫山…叫uat川
0
一
一-
uu
」
「IIIII-II
I-
「III-IIII-IL
θ-一
K句
θ
-EEEEEEEEEEJ
11
~] [f]v
。
，
一
一-
/
1
U
「lllIll-IIIL
自
-EBEEtEEE
」EE
、、
「lllIll-IIIL
哥。
-M
T
θ
A -i(
1inu
Aσ
一
1·4
-EEEEEEE
」EEE
0
、、，， /
T
「lllIll-IIIL
A
Aσ
U
一­
「lllIll-IIIL
[ilu=A(097)[ Z l14401)[1
-BEBEE-」
「lllIll-IIIJ
1IAU
θ句
啡
一
n。
Therefore , under the polar coordinate , the Cauchy-Riemann equations become
6] [! ]v
口
7
3.
f. Fix z = r- ei (J , then
PTOO
u( r- + !:l.r-, B) + iv( r- + !:l.r-， 的
[u(r-， B) + iv( r-, B)]
θ U .-i(J
θv-i (J
r- Iθuθvl
f f ( z ) = l i m ? = e z +i e = | + t |
• o
!:l.r- . ei (J
仇'θv
z 1 仇·仇.1
By the result of previous problem , we have
r- 18uθυ1 Iθυ
仇LI
zl 百- + z 否r-广二|苟 -z 页|
口
4. We use the followi吨 result from the theory of ordinary differential equations (吕ee ， for exam甲ple飞， Di吨 an
口l
Lμi [问
习]， Chapter 2, Theorem 1)
2
Theorem 1. S 'uppose function P(矶 y) αnd Q(立 ， ν)αr-e continuo ω on U = (风 ß) x (γ ， 0) ， αnd they hαve
continuous p州α 1 de r- ivatives t y P and tx Q. Then the l-fo r- m ω = P(x , y)dx + Q(x , y)dy is exα ct if and
or句 if tyP = !/c Q on U. Mo r-eove r-, the O-fo r- m wh附 d伽 r-ential is ωcαn be r-ep r-esented ω
1: P(叫
whe 陀 C is α constα n t.
(1)
P叫 P(川) = 生驴 =
吗且 = 2y , Q(x , y) = 生铲 = 吗且 = 2x + 1. 80
rx
I
U 忡， υ) =
rY
P(巳 y)df:，
JO
+ I
Q(O ， η )dη +C = 2xy+ υ +C，
JO
where C ε IR. is a constant , and
f(z) = u(x , y) + iv(x , y) = (x 2 _ y2 十 工) + i(2xy 十 U 十 C) = (x + iy)2 十 (x + iy) + iC = z2 + z + iC.
口
(2)
阶'oof. P(x , y) = 哈且 =一 吗型 = 在非扣 Q(x ， y) = 吗业 = 呜型 = 币制言
80 from some
constant C ε IR. ，
(川) =
lz J立丁dzJui土
J1 飞飞 + y2 )2 -." . JO (1 +η2 )2
uf dtrfb-d平l
y
[斗+斗+…
-2f
' 1 + y2 J
- JO
十
x2
y2
，
_~ VV _~~ "
t
IU
(1 十 七 an 2
B)2
ν[-J
+ JEl
+ mtmy - f1820 + 1)报刊
品 + y~
1+y
l
u[1
11
x 2 + υ2 +. 1 + y2 I
2
u[1
+
1: y2]
1
x 2 十 y2 ' 1 + υ2J
(2 叫十 C
叫叫
1 + tan 2(arctany)
十C
y τ + C.
x~ + y~
Therefore f(z) = u(x , y) + iv(川)=兰奇 +tC=t +C.
8
口
(3)
P叫 P(x ， y) = 业棋 =
吗型 =
业亏主主1 =
v(x , y) = J~c P( f;, y)df; + J~ Q(O , 7) )dη=
zeνsinx + iC = ey- t:c + iC = e- 1Z + iC.
内osx ， Q(x , y) = 生铲 = 呜，y) =
山川 80
♂ sinx + C where C εIR is a consta皿 80 f(z) = eY cosx
口
(4)
P叫'. P(x , y) = 哈且 =
吗且
-∞
c O创优…吕缸
i
v(μ
x飞， υyω) = - 兀
frOJI CO创吕 f; 剖
s inhνdf; 十 C =- 址
s i川 si阳
吐
n1让h叩
ν + C , where C εIR is a constant. 80 f (z) = cos x cosh y
isinxsinhy + iC = cosxcos(iy) - sinxsin(iy) + iC = cos(x + iy) + iC = cosz + iC
口
5. (1)
口
Proof-ff(z)=35+tit=gti37=1t
(2)
Pn叫 ff(z) = gtzgt = mzco削
i sin x sinh y = cos x co响) - sinsin(iy) = co巾 + iy) = cosz
口
6.
Proof. We note 告+窍 = 35
告 = (x 2 十 4xy 十的 + (x - y)(2x 十 4υ)and 3733 = 结结 =
一(泸 +4xy+ 的 + (x 一 υ )(4x + 勾). 801ving 山se two equations for ~~ and 苟 we get
生
= 6xy
旦旦 = 3(x2 _ y2)
θx
""'''''ò:J' θu
Then it's easy to 日ee 叫x ， υ) = 3x 2 υ - y3 + C for some constant C εIR and consequently, v = u 一 (x - y)(x 2 +
4xυ+ y2) = _ x 3 + 3xν2 + C. Therefore
f(z) = (3x 2y - y3 + C) + i( _x 3 + 3xυ2 + C) = (ωυ)3 + C( 1 + i) = -iz 3 + C( 1 + i).
口
7. (1)
Proof. We have 二亏工 = 芋， which is equivalent to (e iz )2 _ -1户 e 2Z - 1 = O. 801ving this quadratic
equation gives us
e 2Z
斗主 土 J 己俨三 +4
(-1+ 剖) 土 V8-6i
(-1+ 剖)土 (i - 3)
2 4 4
80 eiz = i - 1 = e~主 +(2r口τ+ 轩 )i or e" z = 平 =e lit主 +(2m+T)$9ηεZ. Therefore z = 轩+ 2mr
2 十 2mr +言 ln2 ， n εZ .
言 ln2 or
口
(2)
Proof. cos z
4 is equivalent to (俨? - 8e iz + 1
z=2mr ::l:: iln(4+ V15) (n εZ).
O.
80 eiz
4 土V15= e土 In(4+v'15).
Therefore
口
(3)
一 - ，z
主二-二三二一
Proof. tan z = i gives e i 王==
2， z
z 告司= i . 80 the equation becomes 产
1=
产
1 , which has no
2
口
solution.
(4)
9
PTOOf. The equation can be written as (2 cosh z - 1) (cosh z - 1) = O. 80 cosh z = 1 or ~ . Consider the
equation cosh z
二号二 = α. We reduce it to the quad川 ic equation of e Z : e2z - 2α e Z + 1
O. 80
e Z = a ::l: va亡1. If a = 1, we get e Z = 1, which implies z = 2mri (n E Z). If α = i?we get ez = i 土孚 Z ，
which implies z = (2η 土~)时 (n εZ).
口
15.
Proof. This is Exercise 2.11 .
3
口
Complex Integration
1. (1)
PTOOf. (i) J02 十 ZRezrlz = tzdz 十 J22+ i xdy = 2 + 2i . (ii) Jo2+ i Rezdz = Jo1 2t(2dt + idt) = 2 + i
口
(2)
Proof. (i) Jo 去 = 2 yÍz l11 = 2(å i - 1) = 2(i - 1). (ii) Jo 去 = 2 yÍz l11 = 2(e
号2 _
1) = - 2(i + 1)
口
2. (1)
Proof. By Cauchy integral formula ，允 1 =1 号 = 27r i . 1 = 2 7ri
口
(2)
Proof. 儿 1=1 号 =r fi号 =0
口
(3)
口
pqι
口
川可以
P叫·九 1=1 青 =f 4旦 = 0
(t
π
一­
飞J
今年
vd
Aσ
qA
f'J
汀
d)
1
一-
b-z
一­
f
吼叫
m( z
:
Proof. Denote 三百LTSi口号 by f(z).
阳m
口1
(i) f(z) is analytic on {z : Izl < D and continuous on {z : Izl = D. 80 by Cauchy integral theore
纠允zl= 吉 f(z)dz = 0
削S川
川
a
i i) f(υz) h削
one s
(叫
严叫
产E=
=1
轩子 dz严=2 汀刑i 节 IZ=l
tT
允 1归
同
1==1 兰击
孚争汀们1
(iii) f(z) has two singular points 土 1 in {z : Izl = 3}. 80 by Ca 山 hy integral theorem for ml山 iply
connected region ， 先 1=3 f(z)dz = 九 -11=0 f(z)dz 十先 +1 1=0 f(z)dz , where i5 > 0 is sufficiently small so that
{忡|怜z
叫
川
旧
ulC
cl可仙叭川
int
川叫
lt七
U
c 忡{ |问z斗|仅
<3玛}. ByρCωa
{Iz什川十刊叫
川
叫叫
11 三
υi5}川
巳 d叶刷川}川
M
川毗
巾z=2川汀们t 哇 Iz=
f 扣州
z斗)d
and ~z+ll=o 贝
bi口m削
led ， 阴
we conclude 儿
(μω
川
圳忡
严=-1 斗孚争汀刑i. Cωom
(iv) -/2时 The calculation is similar to that of (iii)
(2)
10
1=3 f(z)dz = -/27r i
口
f. Denote 三寺I e 口 by f(z)
PTOO
(i) f(z) has a singular point Zo = i in {Iz 一叫< 1} . 80 by Cauchy integral forml巾，扎 il = l f(z)dz =
2时 乓三
I~ =.;
互e
z •-1, 1""-"
(ii) f (z) has two singular points 土i in {Izl < 2}. 80 by Cauchy integral theorem for multiply connected
region and Cauchy integral formula , for ð > 0 sufficiently small ,
r
r
cþ
Jlz l= 2
( i叫
f(z)dz = cþ
r
f(z)dz + cþ
Jlz 十 i l =δJlz
2门i出 1.
e'Z .
e'Z
f(z)dz = 2刑| 一一气 Iz =-i+~ 1 内 1= 一如 sinh 1.
1
叫 =δ
Lz -
z 十
Z
The calcl由tion is similar to that of (iii)
must take all the values between 0 and 4 7r (see fig 盯 e). This closed curve
forms two contours , each of which contains the two singular points 土 i of f (z). 80 by a calculation similar
to that of (ii) and (iii) ， 九=川 T=3 一时 n f (z) dz = 2 . ( - 2汀 sinh 1) = - 4πsinh 1
(i\
忖川
叫
V
r) To have a closed curve ,
e
3
2
O
-2
-3
-3
-2
2
O
3
Fig盯e 1: r = 3 - sin 2 ~， e ε[0 ，如l
口
4. (1)
Proof. By Cau由川叩 al formula ， 九 1 = 2 守主 dz = 27r i cos zlz=o = 2付L
口
(2)
+
-h
川
9"
汀
11
dz
一­
+
川一川
fTh
户:
dz +
一
1 小
2z-z
二 十
f
一-
dz
￠州
二十
f
￠州
z-z
一川
P叫去t has two sing山 points 土i in {Izl < 2}. 80 by Cau由 integral theorem for multiply connected
region and Cauchy integral formula , for ð > 0 sufficiently small ,
--
OU
口
(3)
Proof. By Cauchy integral form
口
(4)
Proof. cosh z = 0 if and only if z = (~ + mr )i ， ηεZ. 80 {Izl < 2} contains two si吨ular points 土 ~i. By
Cauchy integral theorem for multiply connected region , for Ij > 0 sufficiently small , we have
无=2 斗zdz
Æ-~il=Iì丰dz + iZ+~il=1ì 本dz
ÆI=δωs;:于)dz+ 儿1 =δωs;二乞)
iZI=1ì 丰dz
4 iZI=1ì 五七dz
2
41| =d t z d z
U sing power series expansi
叫| = stIdz = 2π1: = 刑
</>.
J门 zl = 2
----'"二-dz
= lim，\~()
4 </>:寸ι--"dz = 41T i
coshz~~ HH~O → u ~ J门 zl = 1ì e"Z-l
口
5. (1)
Proof. By Cauchy integral formula ,
23三巾= 2 1T i. (sinz)'lz=o = 2 1T i
口
(2)
Proof. By Cauchy integral formula ,
Izle z
r
1
1
-7dz = 2·2mr ￠「lz = 4m(ezyh=o = kz
.1
L,1T Z Jlzl =2 r
口
(3)
Proof. By Cauchy integral formula ,
芒出 = 守扎扎=2 芝三lz = 1:(UIIZ)(3)|z=o = Et
z~
.3
.3
…
1协 note I 击叫二 | 月 二.1吨二:r- -1 1 三 zrf → o un
12
hen z
口
• 0
(4)
Proof. By Cauchy integra1 formu1a ,
正| = J(zf:16) = 2刑 (Z2 ~ 16)' Iz=o= 0
口
6. (1)
Proof. By Cauchy integra1 formu1a ,
1|J山 = 守去 t|J由 = 刑川 Iz=o = i
7r
口
(2)
Proof. F(z) is 旧
un
山
1
does not cωont
此ta
挝in 0, 吐
t h让is is true by Cauch可
y integra1 theor町
em
丑1. 80 we on1y need to consider the case where the
interior of C contains o. Without 10ss of generality, assume C = {I zl = 1} . Then by Ca uchy integra1 formu1a
i (~+去) = 生 [(Z2
eZ
dz
+ a)e Z j(2
80 whenα= -2 , F(z) is l山
4
In且 nite Series
4.1
Exer cÌ ses in the text
口
4. 1.
Proof. (1) 向 = 苟言 (2)α27z = ptT ， α2n+1
1 if n is odd; bn = 1αn.
击
(3)α2n
击 ， α2n+1
击. (4)αn
αn =
0 if n is even ,
口
4.2. (1)
Proof. If x = 0 , the series is clearly convergent. If x 并 0 ， then by noting the series is a geometric series , we
can calcu1ate it converges to 1. To see the convergence is not uniform , note 'Vn 主 1
1 1 豆叶 = 1 1 击 lJ± | = 市
No matter how big n is , we can a1ways 五 nd an x > 0 (dependent on n) , so that (1+拮古> ~. This shows 山
口
convergence is not uniform.
(2)
Proof. We note
+
| (1) 口(俨
--' (n+1)n .
11 = (n+1+x2)(n+x2)
(n+1+x~
n+x2
η+1+x 2
川i口m
川
刷
山削削
hi1让灿刷
icl
chωhi
h
w时咄M
呻1甲蚓p1i
乙 N 甘且剧;斗I~三 艺江:
乙二川T古古击击T百?俨俨川丁予川忖忡?川川川
叫| 艺江
∞.
口
13
4.3. (1)
Proof. min{R1' R 2 }. From the special case αη/ 三 o or b n 三 0 ， we can see this radius cannot be improved.
口
(2)
P叫 By Cau由 's criteri盹 R 1 = li且η →∞|去 11 / n and R 2
li皿n→∞|击 11 / n . 8ince for any positive
sequence (Xn) 吃 1 and (Yn) 吃 1 ，且旦n→∞ X n 且旦n →∞ Yn 三 li旦n→∞ (XnYn) ， the radius of convergence R for
2:=~二1αn
11 / n
R = 且旦 ←-一 |
n →∞ | α nOn I
11
1 1/口 1
三且旦| 一二 I
n →∞ |α n I
1
11/n
I~二
I On I
= R 1R 2 .
The special case where αn = bn == 1 shows the result R 二:: R 1R 2 cannot be improved.
口
(3)
1 11 / n
Proof. When lim n →∞ |去
exists ， the radius of convergence is 古 Otherwise ， the best result we can
,
obtain is R = limn--->oo 1α11 /n = _,. _ _1
口
li旦与→ ∞ | 古 11 / n •
n1
(4)
1
一∞
R
一-M
ME
>一
hE
α
z
>一
m 叫
一
1-Lh
--ι
Proof.
1 11/ n
When lim n →∞ | 古
exists ， the 鸣ht side of the above inequality becomes ~~
,
4.2
口
Exercises at the end of chapter
1. (1)
P叫 2二二21 舌 1 = 三二二 2 1忐>2:=二2 士 = ∞ 80 the series is not absolutely convergent . Meanwhil飞
主￡ = 主- [Cn(4:工2) 一 币+ 4)) 十 i (时工3) - ln(4: + 5))]
By Leibnitz's criterion for the convergence of alternating series ,
主 - [ln(4:再 一 可斗 = 主山
1日 convergent.
8imilarly,
去[材工3) 一 时工百] = 主 ztE
is convergent. Therefore 2:=二 2 1击 is ωnvergent
口
(2)
Proof. By argument similar to that of (1) , 2:=二 1 号 is converge时， but not absolutely convergent
2.
14
口
Pr时 8uppose Iz l < 1. Then (l-zn;ì':;~z叫) rv zn- 1. 80 2:二二1(1 flfJJZ川) is absolutely convergent on
1z 1 < 1. To find the sum function in 出is case , note for Iz l < 1,
立
z
主 zη1 击(二百二百)
1
二~(~二仨)
(l - z )2
Now suppose Izl > 1. Then n 俨;;二叫=川 (l-z n) (1_z- (n+1J)
去古 80 2:二二1(1
absolutely convergent on Iz l > 1. To find the sum function in this case , note for Iz l > 1,
立
二
2汇二
znfiJZ71+1)is
1
兰仨
击兰击
z ~ (刊
1 (τ
二古二仨云=可「
h
门
石γ
一11 ;二
ιιι(
仙川川
n叶叫+抖1
兰击干(信兰 fι
」主旦兰 1f二
1川川
冗
7
;:
!t1:ι0))
z(z-l )2'
口
3. (1)
Proof. 8ince for Iz l > 1 , Iz ln! > Iz l and L ~二 11 z 1 =∞， the series is divergent on Izl > 1. 8ince for Izl < 1 ,
Izln! < Iz ln and L ~二 11zl n <风 the series is convergent on 1z 1 < 1 .
口
(2)
…
叫 The series c
(3)
Proof. Iz2 + 2z + 21 < 1.
口
(4)
Proof. 8uppose z = x + yi . Then
e'Z - e-'z
1 ,_.,,-L ';'r
slnz = 一丁了一 =z(eHE
.,, _.;ry. ,
1
一
eu zz)=27 [EM(E U
EU)+eu(eu
e-zz)]=27EZZ(ε -y - eY ) + eY sinx.
Asz • 0 , eY sinx rv x and 去 e'X( εν ρ)rv 去( - 2ν)= 伊. 80 三二 二 1 2n [ 去 e' 示(已去 - e 头)] is convergent
if and 。由 if L汇 1 2n i7Jn i is convergent and L 汇 12 n e 头 sin 录 is co盯ergent if and only if L汇 12n j二
is convergent. 8ince 2:二二 12nS111 3云 = L二 1 2n [ 去 e' 头 (ε 头 e 头) 十 已去Esin jtL and since for arbitmy
x ， y εIR ， 2:二二1(iyy4and E二二 1 (~r x both conve耶， we conclude VzεC ， 2:二二12nsin j云 is conve耶nt .
口
4.
15
Proof. At z = 1 ，川川 equal 归之。(击一击) =丘。(击一击 ) =2(-1+~-i+ ~ +
"'+1) = 2(1 - 1n2). Forl问叫
z斗|仅< 1 , 2:艺:二0 7ι
旦 = 艺乙o Ir兀Jω
川n叮d仇咄仙
tvω = 兀;(应艺二
Ir
L0ω
旷川川
n叮
， )dω = Ir兀J 古击毛 S臼imil阳盯圳l忖y，
立0 吾丰号二 = 艺汇二0 兀
ρ2川
ω
2z - ln(l + z) for 材
Iz
刘I < 1. In 吕肌川
z
sur
umr
旧
ι
￡
工 /μz 叶
2νz
♂2趴n
叶川
+刊3 \
仙
lμ
2μz 一 ln(l
丑叫叫
1们(口l+z
补)，
阳I z
斗叫1<11
=oh 十 1
2η 十 3)
l2 - 21此
z = l
印
S
(μz)尸
= )， (一一 一 一一一 l←
= \
口
5.
Proof.
叶 Z) = fzt =- f 主川 =一 主 ι
So for a町 rε(-1 ， 1) ， by letti吨 z = T' e册， we get
M 十 re ie ) = 一 立叮些=立斗iv = 立 qE 川sn8 十 ir n sin n8)
Suppose 1叫 1 + T' eie ) = x + 归. Then e X cos y = 1 + T' cos 8 and e'E sinυ = T' sin 8. Solvi吨 for x andυgives
工 = ~ ln(l + 2r ωs8 十 r 2 ) ， y = 川aIlTft万 Matching the real part and irnaginary part in the equality
rsin8ι( _ l)n+l
一 ln(l 十 2T cos 8 十 泸) + i arctan 一一一一= ) . 一一一一 (rn cos n8 十 iT n sin n8)
1 + T' cos 8
亿
n
口
gives the desired equalities.
6. (1)
Proof. In the 且rst equality of Problem 5 , replace 8 with 8 +汀 and let T' • 1, we get
cos8 十
s 28
2
cos 38
+ ~~~ ~~
3
1
+ ... = - ~ ln(2 - 2 cos 8) = - ln(2 日 in ~).
2
To justify the process of taking lim扰， according to Abel's second theorem , it suffices to show the series
2:=~二 co铲旦 is co盯ergent. Since (士)吃 1 is monotone decreasi吨 to 0 , by Dirichlet's criterio凡 i t suffices to
show 2:二~= l cos k8 is bounded for any n. This can be seen by noting 与主拦 = 2:=~=o eike . After writing
both sides in terms of real and imaginary parts , we have by comparison
|去os 户
Similarly, in the second equality of Problem 5 , replace 8 with 8 + 1r and let r • 1 , we get
立自叫
一一一= arctan 一一一一= arctancot 一=一 (π
η1 - cos8
2
2
8).
口
(2)
16
PTOOf. Let T' • 1 in the 五rst equality of Problem 5 , we have 三二二 l( - l)n十1 旦旦?更 = ~ln(2 + 2cos8) . In (i) ,
we have shown ~二 1 旦铲旦 = - ln(2 日in ~). Add up these two equalities and divide both sides by 2 , we get
(note 1 十 cos 8 = 2 cos 2 ~)
s 38
cos 58
3
5
cos8 + ~ +
十
cos 78
7
1.
8
2
2
+ … = -=-lncot-=- .
Similar argument gives (note sin 8 = 2 sin ~ co曰~)
sin 38
sin 58
sin 78
由1& +一了+一了 +-7+ · ..
~ (arctan 击二百十 j(7T 一讨
~ [… (tan~) + ~川) ]
7r
4
口
(3)
Pmof. Integrating from ~ to 8 on both sides of
J 口 38
in 8 +-- ~ 3
+
si口 58
- -~
5
si口 78
-
+ --:..
7
+…
4'
we get
s 38
cos 58
cos 78
5~
7~
7r ( 7 r \
s8 十 一一+一一了一 十 一一 … =~ I 一 - 81
3~
4\2
)
Replace 8 with 8 十 号 ， we get
in 38
sin 58
sin 78
sin 8 +- -~，，- - ---;,,-- + ---;::;;)… = ~( - 8).
3~
Therefore
si口 38
Slll tJ
5~
7~
sin 58
sin 78
72
十十 … =
32
52
7r _
σ
4υ
口
(4)
Pmof. From part (2) , we know
cos8 +
s 38
cos 58
cos 78
1.
8
十
+ --- + … = -=-lncot-=- .
3
5
7
2
2
Form part (3) , we obtain by differentiation
cos θ 一
cos 38
cos 58
.3
Q
cos 78
汀
一 + 一气?一 一 一气7一 +...= ~
也
.
口
7.
17
PTOOf. (1) R = 且旦n→∞ |ηnl1/n = +∞.
(2) R = ∞.
口m n -->
(阶
3剖) R = 恤
1 1m n →∞叫(归忡
H叶+叫1盯):V?(Cι;ι; 1)忖叶川r 斗1im
4
(性问
刽)R = ∞ .
(5) R = lim n→∞ (η lnn)l/n = lim n→∞ (ε lnn-lnn- l)l/n = lim n→∞ ehpE = 1
(6)R = 2.
(7) R = li叫→∞ lm+lut:ffJH+1)!=limn→∞(叶 1)(n+3113+1)= ∞
(8) R = lim n →∞ 1(1 - ~)-nI1/n = 1im n →∞ (1 - ~)-1 = 1.
5
Local Expansion of Analytic Fu nctions
5.1
Exer cÏ ses in the text
口
5. 1.
PTOOf. If g(z) 三 0 ， the prob1em is ill-posed; if f(z) 三 0 ， there is nothi吨 to prove. 80 without 10ss of
generality, we can assume Zo is a zero of f(z) of order n and a zero of g(z) of order m. Then f(z) can
be written as f(z) = (z - zo)n <þ (z) where <þ (z) is ana1ytic at Zo and is non-zero in a 时ighb。由ood of Zo.
8imi1arly, g(z) can be written as g(z)
(z - zo)mψ (z) whereψ (z) is ana1ytic at Zo and is ∞n-zero in a
neighborhood of Zo. Then we have
( (z - zo)n-m . 黑兰
f(z) 一 J ;P 对<r\~ )
g(z) 一 )1þ (z与
豆豆
ifn> m
ifn = m
if n < m ,
which imp1ies
o(时∞
一­
飞
/tttl
、 ttt
h叫
z-z
flu-od
z-z
if n > ηl
ifn = m
if n < m.
8imi1arly, we have
f' (z)
g'(z)
(n(z-zo )n-= <þ (z )+(z-zo)n-=+l φ '(z)
1
r巾 zo)n- <þ (z) + (z - zo)n <þ'(z ) _ J 时(咐 (fziVU)一句)川)
m(z - zo)m-1 ψ (z) 十 (z - zo)m ψ '(z)
1 nψ (z)+(z-zo) 1þ '(z)
n φ (z)+(z-zo) φ '(z)
1.. m(z-zo)= - n ψ (z)+(z-zo)= - n+l ψ '(z)
ifn > m
ifn = m
if n < m ,
which imp1ies
O 刷一圳∞
>=<
mmm
nnn
pfip+ipfi
z-Z
。 -o
一
­
飞
/tttj
飞 ttt
Q句
也→
iiz
z
Qd z
fJ
Therefore we must have
,,___
f(z)
,,___
f' (z)
• Zo g(z) z• Zo g'(z)
…一…-
z
口
5.2.
叫~ Theω1ution is sim阳 to that 川协m 5.1 once we write f(z) and g(z) as TZ吗?7aIIdt示，
respective1y.
口
18
5.3.
Proof.
z内
'J' ~n
Z'n
(2n+1) 7r i'η
1
' . z~
一
1一一 ， and z"
一
(2n+ 专)7r i ' ~U~ ~n (2n-1一一
~ )7r i .
2(n+1) 汀 i' ~n
口
5 .4.
Proo
旷
f. By cωonside
臼山
r'ir吨 t山
he 剖
s i吨u
吐la
创町川
川
r'i让七y of 0 for f(lν
/z
斗)， we can conclude the following results: if ∞ is a
removable singularity of f( 习， then f(z) has the form of L二。尹 in a neighborhood of ∞; if ∞ is a pole of
f( 斗， then f(z) has the form of L二o ~~: +汇丰 1 bk Zk for some positive integer m in a neighborhood of ∞
and b rn 并 0; finally, if ∞ is an essential si吨ularity of f( 斗， then f(z) has the form of L汇。尹 十 汇汇 1 bkz k
where infinitely many 饵 's are non-zero.
口
5.5.
\II/
z-2
z <
汀
/''‘
t飞、
ι-d
一­
z-9"
pupiu
∞汇叫
Proof. In formula (5 .48) , we alrea句 obtained
Replace ~ by iw with IωI < ~
5.2
口
Exer cÌ ses at the end of chapter
1. (1)
Proof. 1 - Z2 = 1 一 [ (z - 1) + 1]2 = -2(z - 1) 一 (z _1)2. Radius of convergence R = ∞ -
口
(2)
P叫 sin z = (- 1) n si呻
mr) = ( - l)n 丘。古击 (z
叫 2k十丘。结且 (z
叫 2k+ 1 . Radiu日 of
convergence R = ∞ .口
(3)
P叫 Solve the equation 1 十 z + z2 to get two roots: Zl = 归 = 二iTZi and z2 = 归 = 二l亏V3i. Then
1
1 + z + z2
古(士;士;)
兰卡古 t占)
Js[二三(二)n 一 ;二。 (:Jnj
方三叫e~(口十 1 )7ri - e 扫十 1 )7r i )
古兰 J-3 …J3(Tt叫
ιsin 3(η+ 1) 汀 n
z.
,
工o
Slll
37r
The radius of convergence R = 1.
口
(4)
19
f.
PTOO
芒 = 兰。品Z2k十1 兰ozm = 轧旦旦中 = 剖(nsl 品1
Zn
The radius of convergence R = 1.
口
(5)
Proof. It 's clear that 1 is a singularity. 80 the radius of convergence R = 1 and the function is analytic
」
inside the unit disc D(O , l). 8uppose e击 = 艺乙。 αkZk . Di丘erentiate both sides and we get 古亏古 =
α
η
+
∞艺时
η
+
α
∞艺问
捍α
句n
+
α
η
tα
灿
80 α
句0 刊
= tα
向
儿
11
1， α
向
a11 = 丛
2α
句2 - 2μ
向
11 削
an
时
d α
向n附
灿川
+1
叫
∞艺叫
α
+
α
∞艺问
∞艺问
ε;二 1 阳 kZk-1 . Therefore
号轩旨拙
廿t扫
T
市
αn-1卅
f扣O
旧
阳r η 三泣2. 岛
By …
reCl
川
m
C口
ursi
口r
α句2 = 3←巳' α句3 = 号乎轨巳飞， 削
a nd α向4 = 去巳 S如oe 击 = ε叫叶
(1 +忖z 十
咛i仨
γ泸
z2 十
+ 号非Z3 + 去 ε4飞
+0
叭(z♂5勺纠吩)月)
口
2. (1)
Proof. 8ince z
0 is a singularity for ln z , the radius of convergence R
ln(l - z) , we have
1. Using the power series of
￡工 [i(z 一仰汀
ι in
ln z = ln [i(l 一巾 - i))] = ln i - L 一 n 一一 =Et 一〉;7J(z-071.
口
(2)
Proof. The solution is similar to that of problme (1) , 01啡 that lM=-i 汀 ·
口
(3)
Proof. We note a时 an z is an od functio日， so its series expansion must have the form L二。 α2n十 1 Z2 η +1
Differentiate both sides and we get 1去言 =2二二。 α2n十 1(2η + 1)z2n. 80
1 = (1 + z2) 艺 α2n+1 (2n + 1 )z2n = 艺 α2n+1(2n + 1) 产+艺 α2川 (2n - 1)z2n
n=O
n=O
n=l
Thereforeα1 = 1 andα2n十
和tα2n-1 for n 主1. Thi日 implies a2n+1
which has 土 i as poles , the radius of convergence R = 1.
云兽 8ince we have used 市
口
(4)
f.
PTOO
+一 -
11
n -11
z
-z
ln( 川n(l+~) 一 叶 一 ~)
(2k + l) 7r i + ~二 [
( 俨 1 + 1] 三百二
协阳 + 三 tfn+1)
We used 山出sumptíon 击< 1 in 山 above derivation , so 山 domain of convergence is Izl > 1
20
口
3. (1)
Proof. Let f(z) = L:~二。去在 Then
1'
(z) = 去2n 占=打击+ 1 :z)
The时óre f(z) = ~ ln 芒~ with f(O) = O.
口
(2)
Proof. Inspired by definition of trigonometric functions via exponential function , we have
主品=;民币丁1 = ~川川shz
口
4. (1)
Proof. Assume π乞丁 =2二 二∞ α，，( z-l) ηThen by Theorem 5.4 ,
α
一一ft
If n 三
一一一一一一一一-
1
2时 J 1z - 11 =ρ 泸 (z - l)
一一一一一一一一一一=一一-1
(z - l) 叶
一一 - 一一一一一一一一一-
2πi J 1z - 1 1=ρ Z 2
(z - 1)n+2'
2 ， by Ca旧hy's theore盹 αn = O. If n = - 1 , by Cauchy's integral formula ， αn
去 Iz = l = 1. If
n 主 0 ， by Cauchy' 日 integral formula
1
dn • 1
|
a"
(Z-2) 1
α = 一一
i
(n+ 1)! d々 寸
|z=1
1
|
川) 1
( - 2)( - 3).. . [- (η
一一寸
\" 十 2)]z-(
飞川 1- 1)! \ -/ \ -/
,
-/j"
= (η 十 2)( - 1) 叫
80 π土耳 = 艺二。 (n 十 2)( - l)n 十 l(z - l)n + (z - l)-l = 艺汇 l( - l)n 十 l(n + 2)(z - l)n.
口
(2)
∞汇叫
∞汇叫
Proof.
口
(3)
Proof.
1
Z2 - 3z + 2
z- 2
2(1 - ~)
z- l
z(l - ~)
三 Gf 立击=三 zη 立三;二
口
(4)
21
f.
PTOO
Z2;z+2 = 击 一 击 = ~(占 一 击) = 兰r
口
(5)
Proof.
(z-1)(z-2)
(z - 3)(z - 4)
(z" - 3z + 2) ~ - 4(1 =- f) - z(l _ ~)
[-~ ∞ Z\n 二1 2二二
∞ (~) n]
(z2-3z+2) I-i~二 (D
3 ∞
n
∞
1-~~二 ι-2 ~二 3;十1
n=O
n = -l
口
(6)
Proof.
(z - l)(z - 2)
(z - 3)(z - 4)
(z" - 3z + 2)
- 一一
一一
z-4 一
z-3
(Z~4
Z~3)
(Z-3+~) [主(~)言。 )n]
cx)
1 + 艺 (3 沪 1 _ 2.3η1 )Z-n
n= l
口
6. (1)
o叫f
叫. 古 = 札古 一 E材
8 o ::l:α…附忡s川川
E百Z才斗]. 臼
f击古叫
singularity.
口
(2)
Proof.
cosaz
z2
艺汇。母在产
古工 ( - l)n ..2 n-2
但 (2n)! -
Z2
口
80 0 is a pole of order 2 and ∞ is an essential singularity.
(3)
Proof.
- 1 户口
cosω - cos bz =一
1 ~工(
( - l)n [α n _ b2n ] Z
2n = ι
[α 2n _ b2n ] z2n
)
、飞 一丁
~ 一丁
2 .江户/
岳飞…/
80 0 is a removable singularity and ∞ is an essential singularity.
(4)
22
口
f.
PTOO
sin z
1
sin z - z
2汇1 占弘T Z2n 十1
Z2
Z
Z2
Z2
毛 ( - 1) n
_.2 n-1
乞 (2η+ 1)!-
80 0 is a removable singularity and ∞ is an essential singularity.
口
(5)
Proof.
咔=主击;
80 ∞ is an essential singularity.
口
ρ
。
( )
Proof.
-.fi
sin -.fi
-.fi
ε二oti去主T zn . -.fi
1
2∞
iiiLzn.
n = O (2n +1)!
80 0 is a removable singularity. Meanwhile , (η汀 )2 (n ε 1':1) are poles of order 1 and ∞ is a non-isolated
singularity.
口
(7)
Proof. If we stipulate ln z lz = l 手。， then 1 is a pole of order 1. If we stipulate ln Z lz = l = 0 , then by l'Hospitale
rule for analytic functions (Exercise problem 5.1 and 5.2 in the text) , we have
lim 一三二上一 =lim z =l
z• i ln z - ln 1
z• 1
So IZzf了IE = IZJTFZMIasa pole oforder2 ∞ is a removable singularity
口
(8)
Proof.
1a z 哈马 = 1a z 兰￡为 d( = 三币LI71
80 the function is an entire function with ∞ an essential singularity.
口
7.
Proof. Z2 has ∞ as a pole of order of 2. ~ has ∞ as a removable singularity 守主 has ∞ as an essential
singularity (附 problem 6(2)) 孟zhM ∞ as a non-isolated singularity since mr +号 (n εZ) are poles.
乍1 has ∞ as an 臼sential singl削ty since lim z →∞宁 does not exist: lim z →叩 t兰主 = 0 并 ∞ =
lim z →∞，明争l 叫{ 一 去 } hω ∞ ωa removable 叫ularity since lim z →∞吨{ 一 去 } = e O = 1 忑拮
M 一 (mr 十 ~?削阶s(n EZ) ， ωit h削∞ ωa non-isol叫叫山ity. The function
V(~ - 1) (~ - 2) =
FTF写 has 0 川 pole of 叫r 1, so the funct
8.
P叫 The …s L~=o(az)n is conve明t in U1 = {z : Iz l <古}. The series 击丘。(俨ij苛J二 is
convergent in U2 = {z : 11 一 α11 z 1 < 11 - z I}. On U3 = U1 n U2 , both of the series represent the analytic
function 1_laz ' 80 the functions represented by these two series are analytic continuation of each other 口
9.
23
PTOOf. Denote by f(z) the analytic function represented by the series in Izl < 1 and by g(z) the analytic
function represented by the series in Izl > 1. 8ince the roots of zn = 1 (n εN) are singularities of the series
and consist of a dense s山set ofθD(O ， l) ， f(z) and g(z) cannot be analytic continuation of each other. Of
course , we need to verify that the series indeed co盯erges to analytic functions both in Izl < 1 and in Izl > 1.
When Izl 三 ρ< 1, we note
~n+
111 -- zn+11 - 11
~♂
zn 1I = 1(1
1 -州
1~ zn) 1 ~ρn
<
ρn
zn十 1) (1 - Zn) 一 (1
ρn)(l
ρ口十 1) 一 (1
ρ)(1
ρ)
By Weierstrass criterion , the series is uniformly convergent to an analytic function on Izl 三 ρ. 80 f(z) is
well-defined in Izl < 1. When Izl 三 R> 1 , we note
|己币 一 二百|
By Weierstrass criterion , the series is uniformly convergent to an analytic function on Izl 三 R. 80 g(z) is
Izl > 1.
口
well-曲直ned in
10.
Proof. Clearly the series is co日vergent in D(O , 1) and is divergent at z
1. 80 its radius of co日vergent
R = 1. Then f(z) is analytic in D(O , 1) and z = 1 is a singularity of f(z) . We note f(z) = z 十三二1Z2n=
z 十 2二二1Z2fl = z +艺汇 1(Z2)2← 1 = Z + f(Z2). Therefore , we have
f(z) = z + f(z2) = Z + z2 + f(z4) = Z + z2 + z4 十 f(Z8) = ….
80 the roots of equations z2
1 , z4
1 , z8
1，… 7 泸
1 ，…， etc. are all singularities of f . These
roots form a dense subset of θD(O ， 1) , so f(z) can not be analytically continued to the outside of its circle
of co盯ergence D(O , 1)
口
6
Power Series Solution of Second Order Linear ODE
6.1
Exer cÏ ses in the text
6.1.
PTOOf. It suffices to note we have the following linear equations
{:俨俨尸纠忡州川川
川川川
(υ
z才巾州)讪
Mω
叫;
p(υωz
斗)ω
叫~(υz)
+创
q (z
斗)ω2(υz) =- 1ωU;(υz
斗) .
Then we can apply results in linear algebra (i.e. Cram町、 rule).
6.2
口
Exer cÏ ses at the end of chapter
1. (1)
PTOOf. Let ω (z) = Co 叫 (z) + C1 叫 (z). Then 旷 (z) = Co + C1ez and w"(z) = C1ez. Then ω (z) - z旷 (z) =
C1ez - C1zez = (1 - z) ω" 80 the differential equation satisfied by ω (z) is
(z - 1)ω" - z 旷十 w = 0
口
24
(2)
ω
ω
句qMN
一
一
ω
+
泸
句一
Proof. Let ω (z) = Co 叫 (z) + C11归 (z). Then
9"
80 Z2 ω
一句 e+ + 2c1e-~ (*). Differentiati吨 both sides and multiply both sides by Z2 , we get 2z 3 w' +
z 4w" = coe+ + 4c1e-~ (**) . Combining (*) and (**) gives us 6c1e-~ = (Z2 + 2z 3 )旷 + z 4w" and 3coe+
(2z 3 - 2 泸)旷 + z4 ω" 80
w = i [(2♂ - 2z2)山川] + i [(z2 + 2州 十 川 = (z3 一 二)山二w"
This is eql巾alent to z 4w" + (2z 3 - Z2) 旷 - 2w = O.
口
(3)
Proof. Let w = Coω1 + C1 W2. Then we note 叫 =ftt归 and 叫 =itω1 . 80 w'
句 -f专 ω2C1 ·jt ω1 .
Multiplying both sides by z2 , we get z2 旷= Coαω2 - C1αω1 (*). Differentiati吨 both sides again , we have
2zω ， +z 2 ω" = Coα ( _ za2 ) ω1 - C1αC12)ω2 ， i.e. 2 z 3 旷 十 Z4ω"
一 句 α2ω1 - C1α2ω2 (忡). Combining (*) and
(**), we can get
(时1 Z2 ω 忖 0(2几血) = 仲 cô)a 2 w1
αcoz 2 旷 - C1 (2z 3 旷 + Z4ω") = (c6 + cî) α2 叫
Using the fact ω = Coω1 十 C1ω2 ， we can easily get z 4 w" 十 2 Z3 W ' + α2W = O.
口
(4)
Proof. Let ω= Co 叫+ C1 W2. Then (Z2 - 1)ω= 句 Z2 + C1 z. Differentiati吨 both sides of the equation gives
2zω 十 (Z2 - 1)ω ， = 2coz 十 C1. Di丘町entiate again , we get 2w + 2zw' 十 2zw' 十 (Z2 - 1)ω" = 2co . 801ving for
co , C1 , and plug the expressions into the equation ω = COW1 + C1W2 , we get
Z2(Z2 - 1) 旷， + 2Z(Z2 + 1) 旷 - 2w = O.
口
2. (1)
Proof. p(z) = ♂ and q(z) = 0 are both analytic. 80 the solution ω (z) is analytic and assumes the form
ω (z) = L~二。 αkZk Then
ω" - z 2w = 艺 αk . k(k 一 印k-2 _ ~二 αJ十2 = 2α2 十 6α3 Z 十 艺 [α k+2 . (k 十 2)(k + 1) 一 句 _2lzk = 0
k=2
k=O
k=2
Clearly， α。 =ω(0) andα
旷 (0). By the uniq时ness of power series representation of an analytic fu日ctio凡
we have α2=α3 = 0 , andα k+
一」主二L一
十2 (k+ 2)(k 十 1) .
αα 4(k-1)
α。
αJ(i )
42k.k!.(k - ~). ..(1 - ~)
24k . k! . r(k + ~)
α。
4k - 4k. (4k - 1) - 4k. (4k - 1) …
4.3
and
α
一
α 4(k-1)+1
α1α1
4k十 1 - (4k+1) .4k - (4k+1) .4k.. . . . 5 .4 - 42k.k!.(k+~). [ (k-1)+~l . ..(1+~)
α1 r(%)
24k . k! . r(k +
Let W1(Z) = 江od』(言 )4n and 叫 (z) = 丘。司主7(2) 47t+1TlmW)=MI(z)+ 川 (z)
口
25
%)"
(2)
Proof. p(z) = z and q(z) = 0 are both analytic. 80 the solutionω (z) is analytic and assumes the form
ω (z) = 艺乙。 α kZk Then
2二 αk . k(k
w 11 -zw
阳 k-2 _ ~二 α kZk+l
艺创+2(k + 2)(k + 叫 Zk _ ~二 αk_1 Zk
k=O
k=l
2α2+ 艺 [αk十2(k + 州+ 1)αkμ
k=l
O.
Clearly， α。 =ω(0) andα
旷 (0).
we have α L,--.J,
') = O. and α 丸十二
ι -ω
一」主仨
-1 一 .
(k+ 2)(k+l)
αι 一
By the u山
I
α 3(k-l)
α。
3k - 3k. (3k - 1) - . . . - 3k. (3k - 1)
α。
3.23 2k .k! . (k - 1) ... (1 - 1)
…
αor(~)
3 2k . k! . r(k + ~)
and
α
一
k+ l -
α 3(k- 1) 十 1α1α1
(3k 十 1).3k
(3k 十 1) . 3k …
32k .k! . (k+1).. . (1+1)
4.3
αlr(~)
32k.k!.r(k+~)
Let 叫z) = 艺汇oJ拼石在 削 ω2(Z) = 艺汇。司毕百字- Thenω (z) = aOwl(z) + 叭2(Z)
口
(3)
Proof. The equation can be written asω 11 + p(z) ω + q(z) = 0 with p(z) = z2~1 and q(z) =
万与 80 土 1
are two si吨ularities of the equation and the equation has an analytic solution in a 时ighborhood of 0 (e.g.
D(O , l) := {z: Izl < 1}). 8uppose ω= 艺;二。 α kZk . Then
(Z2 - 1)ω叫旷 一 ω = 艺 [ (k - l)(k 十 1)何 一 (k 十 l)(k + 2) 向十旷 一 (αo 十 2α2) - 6a3z = 0
k=2
Therefore ， α0 = ω (0) ， al = w'(O) , a2 =
呼1α3 = 0 , and for k 2': 2 ,
1ti--7k
lT
1κ 一
」川-2
(一
2k - 3
-
Moreover
2k - 3 2k - 5
α 2k = 丁γα 2k-2 = 丁「瓦习 α2k-4 =
曹扣儿一
Fromα3 = 0 , we concludeα2k十 1 = 0 for k 主1.
+
。牛
α
zk
... =
2k[(k - 1) - W(k - 2) - il... (0 - i
r(k 一 i)
2k . k!
α。=汇育气7αo
Let 叫z) = 丘。吕市户削 ω2(Z) = ιwe get w(z) = aOwl(z) + 川 (z)
Remar k 1. N ote 叫 (z) is the Taylor series of v'1丁芦，
口
(4)
26
P叫 The 叩ation can be wri阳 as w" + p(巾 + q(z) = 0 with p(z) = 誓旦 and q(z) = 百元言 80
4号!3i are two singularities of the equation and the equation has an analytic solution in D(O , 1) . 8uppose
w 二三乙。 αkZk . Then
(1+z+z2) ω" +2(1+ 却 )w'+zw = 艺 (k+1)(k+2)(αk+ 何十 1+αk+ 2)Zk+2(α0+α1+α2)+6(α1+α2+α3)Z = 0
k=2
Therefore , we have
to 十 α1 + a2 = 0
α1 十 α2 + α3=0
αk+ α k+ 1 + α k+2 = 0 (k 主 2) .
801vi吨 it gives usα3k
α0 ， α3k十
α1 ， andα 3k+2 = -(α。 +α1)' Plugging in the values of αk'S ， we get
after simplificationω (z) = αOW1(Z) + α1 叫(斗， whereω l(Z) = 叶无言 and 叫 (z) = 百合Z2' Eql盯ale叫y， we
can choose another basis with two linearly independent 时utions: 队=耳去z2 and W2(Z) = τ去?
Remark 2. We cα n solve the pTOblem more di r-ectly once we note
￡ [(1 十 Z+Z2)W(Z)] = 川 + z2川z) + 2川剧 (z) 十 如 (z)
So (1 + z + 泸 )ω (z) = α0+α 1Z ， i.e.
门
lZ
(z) = 一
τ+ 一二L一τ
1 + z + z :L
1 + z + z :L
口
3. (1)
P叫~ The 叩ation can be 叫
t 川叫
rar
圳
m
a创
ns
汹s山 m
丑l时
e仙
们七ωo w" 十 p( 圳F 十 q( 巾 = 0 , where p(z)
d
z(l~~) and q(z)
一 才芮 80 0 is the singularity of the equation. 8ince zp(z) = 节削 z2q(Z)
0< Izl < 1 , the equation has two regular solutions in 0 < Izl < 1:
- î~~ are analytic in
(俨叫叫川(υ…
z
叫2(ωz
ω
斗)=gω
叫1 (忡
z斗) ln z + zPρ2 艺
向0 笋 0
创)
2 二乙 o dμk z泸k ω( g 笋 00叮rd
for some constants g ， ρ1 and ρ2.ρ1 and ρ2 satisfy the index equation ρ(ρ - 1) 十 α。 ρ + bo
0 , where
α。= limz--+o zp(z)
1 and bo = limz--+o z2q(Z)
-1. 80 the equation for ρbecomes ρ2
1 andρis
therefore 1. 8uppose w(z) has the form of zP 2二二。 cnz飞 then
[(η 十 ρ)(n + ρ - 1) 十 αo(n + ρ) + b川 十 艺 [αn-z(l + ρ) 十 bn-zlcz = 0 (n 三 1 ) ，
where αn 's and bn 's come from the Laurent series of p( z) = 艺 汇。 αZZZ-l and q(z) = 艺 汇。 b z z Z - 2 ， respectively.
We then find the Laurent series expansion of p(z) and q(z) as follows
仲) = ; 一 2 忌 ， q(z) = 一 主 - 2 艺 ZZ-2
Therefore the recursive relation can be simplified to
[ (n + 阳 十 (η 十 1) - l]c n + ~二 卜 2(1 + 1) 十 ( - 2)]cz = 。
27
Define Çn = (η 十 2) 句.. (n 主 2) . Then this relation can be further simpli且ed to
tn=22:JZK
n
It 's easy to see by induction that Çn = (n + l)ço = (n + l)co . Therefore
∞∞ k
ω(Z) = zzc户
+ l _ k+ l
∞∞
Z k+l
CoZ
Co
where 叫 (z) = 击+旦子1 Fr om formula (6.27b) , we can conjecture 叫 (z) = ~ . Then it's easy to verify
this conjecture is indeed true.
口
(2)
Proof. The equation can be transformed toω叫 p(z)w' +q(z) ω= 0 , wh盯忡忡去 and q(z) = 击 +4Z2 .
80 0 is the si吨ularity of the equation. 8ince zp(z) =
and z2q(Z) = 古 + 4Z4 are analytic in 0 < z , the
equation has two regular solutions in 0 < 14
-%
1
1
Z k ， (co
{:俨
叫U川叫川川
ω
巾
l(川巾
(μz) = 心OC句k 泸
k ， (ωg 并 00
叫(怡
ωz
斗) = g叫 (μ
例
z斗) ln z 十 z户
ρ向2 艺汇 o dkz 飞
旧rd
白o '7兴f 川
0)
for some constants g ， ρ1 and ρ2.ρ1 and P2 satisfy the index equation
ρ(ρ - 1) + α。 ρ + bO = 0 ,
whereαo = lim z →∞ zp( z ) = -~ and bo = lim z • o z2q(z) = ~. 801ving the above equation gives ρ
向1=;a丑
向2
ρ
0; '
肌
wie note 七址山山
he
四e Lau
旧町r阳
!吧e川 白阳
seri
川j阳
e臼s 臼呵
归a
p
创
阳nsi
α0 = 干
才% and α n = 0 for n 主 1; bo = ~， b4 = 4 , and bn = 0 for n 主 1 and n 并 4. Therefore , the recurs 旧
equation
n-l
[(η 十 ρ)(η 十 ρ1) 十 αo(叶 ρ) + bo]c n + ~二 [αn-l (l + ρ) 十 bn-dCI = 0 (η 三 1)
1= 0
is simplified to
(n+p)2 一产+川l 句 +1川4Cn-4 = 0 川)
[ 8 7
and we can conclude
,10
Cn
4r~ _ ^
ι]-(… t J;177:4 们 t .，
Now let ρ = ρ1 =
if n = 1, 2, 3
if n 主 4.
f. Then for n 三 4 ，
Cη
4C n - 4
n(η+ 2)
Define Çk = C4k (k 三 0). Then for k 三 1 ，
EK =
4Çk1=-Zk-l
4k.(4k+2)
(2k).(2k+1)'
Byindu巾凡 it's 叫 to 附 Çk =苛吕TÇO =坛击 Co. Therefore
ι
(z) = 句Z 吉、
(_l)k
岳阳 + 1)!
4k
z."
_.1， ι (_l)k 4k+
句 Z " 、一一一?+2=co叫 (z) ，
岳飞山 t-- 1)!
28
whereω l(Z) = zt sin(z2). 8imila巾， by choosi吨 ρ = ρ2
~ ， the recursion equation can be written as
(n 2 - 2n)c n + 4ω= 川hich leads to tl随时ltion C4k = 甘 Co. Therefore ，山 other solution is
王三( - 1) 门λ
2(Z) = Z3 、 一-7z = zi cos(z2)·
t=OW川/
口
(3)
Proof. The equation can be transformed to ω"(z) ωf(z)+fω (z) = O. Let p(z) = -1 and q(z) = ~ . Then
zp(z) = - z and z2q(z) = Z are both analytic in Izl > O. By Theorer口 6.3 ， the equation 1削 two regular
solutions in Izl > 0:
(叫 (z) = ZPl 2:. ':=0Ck zk
阳
叫2(ωz
ω
斗) = gω
叫1 (μωz
斗)山
lnz + zρ向2 2:艺:乙 o d μk z泸k
(ωg 并 00
旧rd
白0 并 O川)
for some constants g ， ρ1 and ρ2.ρ1 andρ2 satisfy the index equation
ρ(ρ1) +α。 ρ + bO = 0 ,
whereα。= limz--+o zp(z) = 0 and bO= limz--+o z2q(Z) = O. 80 ρ1 = 1 and ρ2 = O.
8uppose 叫 z) has the form zP 2:.~二。 CnZηWe can get the following recursion equation
n-l
[(n+ ρ)(n + ρ1) +αo(n + ρ)+b队
l= O
8ince α。 = bO = 0 ， α
- 1 ， αn = 0 (η 主 2) ， b1 = 1 , bn = 0 (η 主 2) ， the above equation can be further
simplified: if n = 1, the equation becomes
(ρ+ 1)ρCl = 0;
if n 主 2 ， the equation becomes
(n+ ρ)(η+ρ
l)cn + (-n + 2ρ)C n -l = 0
We first let ρ=ρ1 = 1. Then Cl = 0 and (n + l)ncn = (n - l)C n-l for n 主 2. Therefore , Cn = 0 for
n 主 1 ， and one solution of the equation is
w(z) = ZPl 三二 ωn = CoZ
n=O
80 we can let 叫 (z) = z. We then let ρ=ρ2 = O. Then Cl can be any number and n(n - l)c n = (n - 2)C n-l
for n 主 2. This implies Cn = 0 for n 主 2. The corresponding solution is therefore
zρ2 2二句 zn = 句 + CI Z
n=O
80 we get back to the same solution ω (z) = z. This means we have to try the other form of the solution
叫 (z) = gWl(Z) lnz + zρ2 2二 dnz n = gz lnz 十 三二 dnz n ，
n=O
n=O
where g is a constant. Note
叫 (z) =glnz+g+ 汇仰zn-l 则叫 (z)=:+ 汇阳 (n
29
ψn-2
80
Ell-ll
vd z
n
」
f
冒
d
n
叮
zrT
口
十
n
·i
+
一
十
9"
‘.，‘们川才
川
二
叶叶
+
十
z Qdn n nl qd n z +
n+
ln
∞艺叫
1lu-」-
d1
∞-p=)
μ'η
\}/「 ll|lllL/
MW
n
-
47
z
-L4
一
g
一
十
一
‘
-
/1 、
ω1llll」
hv 叫∞
+斗斗
对
川dAn
42
470
ιuγ 才 叫 忡 忡
.、
飞
一
十 -Z
+d
二
机川也
PA
到
||llLg
+∞可 /-P
协「
H
的
Ug
QA7z2
Therefore the necessary and sufficient condition for 叫 ( z) 十 p(z) 叫 (z) 十 q(z) 叫 (z) = 0 to hold is
(g + 句 = 0
g = 2d 2
d n十 2(n
+ 2)(n + 1) = dn+1n (η 主 1).
We have g = - do and d2 = - ~do. By defining Çn = (η - l)d n (n 三 2) and working by induction , i扩 s easy
to ded uce d n = 一一土?寸
for n 一
> 2. 80 we have
飞川 i 尸 e
问 (z) = - dozlnz + do + d1z 一 、
do
. zn.
f=2(n1) - 71!
Note 叫 z) = z is already a solution , so we can 1吼叫 (z) = zlnz - 1 十 三二二 2 司仨η-
口
(4)
Proof. The equation can be transformed to w"(z) + p(z) ω '(z) + q(z) ω (z) = 0 , where p(υz吵)=1t a汩丑n1
z - 1 and z2q(z) = Z are both analytic in Izl > 0, the equation has two regular
q创(z
斗) = ~ . 8ince zp(z) =
solution in Izl > 0
(俨叫叫川(υ…
z
叫2(ωz
ω
补) = gω
叫1 (ωz
功)lr川 + zPρ2 汇
zk
二 o dk 泸
2二;二
(ωg 并 00
曰rd
山0 并 O叫)
for some constants g ， ρ1 and ρ2.ρ1 and ρ2 satisfy the index equation
ρ(ρ - 1) + α。 ρ + bo = 0 ,
o.
whereα。= limz--+o zp( z) = -1 and bo = lim z• o z2q(Z) =
80 ρ1 = 2 andρ2 = O.
8uppose 叫 z) has the form zP L:~二。 Cnz n . We can get the following recursion equation
[(n+ ρ)(n + ρ1) +αo(n + ρ)+b队
8ince α。 = - 1 ， α1 = 1 , and αn = 0 (n 三 2); bo = 0 ， 的 = 1 , and bn = 0 (n 三 2) ， the above recursion equation
can be further simpli且ed to (η + ρ - 2)c n + Cn-l = O.
We first letρ = ρ1 = 2. Then i山 easy to see Cn = 与L. co . 80 we can let Wl(Z) = z2 L:汇。与Lη =
z 2e- z . We then let ρ=ρ2 = 0 , and the 四
r eCl盯吕ion equation becomes (n-2)c n +Cn-l = 0 (n 主 1). It 's easy to
see Co = Cl = 0 and Cn = 扩是 C2 fo川主 2. Ph
阳
吨
u1吃g臼
咄
聪gin
叩
1
we get ω
叫(z
斗) = 句
C 2 Z2马ε z This is the same as the first solution , so we have to try the other form of the solution
叫 (z) = gWl (z) ln z + 户 汇 dnz n = gz 2e- z ln z + 汇 dnz n ，
n= O
30
n= O
where g is a constant. Note
叫 (z) = ge- Z (2z1nz 一 z 2 1nz + z) 十 汇 d r川
and
W~(z) = ge- Z [3 - 2z + (2 - 4z + ♂) ln z] 十三二 (η 一 1)ηd n z n - 2
80
dü - d 1
ι
W"(z) + p(Z) 旷 (z) + q(Z) ω (z) = - ge- Z ( - 2 十 z) 十 一一一
+γ (n + 2)(叫+2 十 dn+1)zn.
z
ι~
n=ü
8et 旷'(z) +p(z)w' (z) +q(z) ω (z) = O. Then it's easy to see g = 0 and by induction we must have do = d1 = 0 ,
4+1 = 》主ld2 for n 三 2. Thi
怡
1让i由s 阳
1e侃ad
削由吕 us
川
sb阳ack to 出
t川
h1e 且ff irst
Ther时
e吐t扣
or飞 we apply Lμio盯i让11e'、s formula , formula (6.30) , to get the second solution: (note p(z) = 1 - ~ )
叫 (z)
1- r
ωl(Z) }I ~一」一I?
exp
p(ç)dçl ~向
l[叫 (7])]2 ~._y l )
n"/-"J
J
l(Z) }I
z 2 e- z I
J
~l[一土丁
Wl( 7])]2 叫[一η., 十' ln 7]]>η
J
~--Yl
---IJ
芋i;工dη
ητ e
~"
z 2 e- z JZ 二dη
Der
we have
e Z (l+z)+ ♂ Ei(i)
1
2
+ z) + z 2e- zEi(z)]
2Z2
Remark 3. Ver伽 the W2(Z) 陀presented in thω form is 仇 e sαmeαs the 叫 (z) gωen by the textbook's
solution.
ω2(Z) =
z 2e- z .
~ \~
,
~j 0 '
'"
~.\.j = ~[-(1
口
4.
Proof. In the given equation , we have p(z) = ~ and q(z) = m 2. 8ince zp(z) = 2 and z2q(Z) = m 2 泸 are both
analytic in Izl > 0 , the equation has two regular solution in Izl > 0:
{:俨俨叫叫川川
1叫(υ
忡
巾z吵←户川)尸)
气扩
=吨
zPρ
叫2(忡
ω
z斗) = gω
叫1 (υz)灿
lnz + zρ向2 2:艺:乙 o dkz泸k
(ωg 并 00ωrd
山0 并 O川)
for some constants g ， ρ1 and ρ2.ρ1 and ρ2 satisfy the index equation
ρ(ρ - 1) + α。 ρ + bü = 0 ,
where α。 = limz-> o zp(z) = 2 and bo = limz-> o z2q(z) = O. 80 ρ1 = 0 and ρ2 =- 1.
Let 叫 (z) = ZPl 2:二二o cnz n = 2::二。 Cnz n . Then
叫 (z) =
L Cn+l (n + l)zn and 叫 (z) = 汇 Cn +2(叶 2)(η+ 印H
n=ü
n=ü
Therefore
川 (z) + 忡忡 )+ψ
刷川)灿川
ω
n=ü
31
This implies
(C1=o
Cn+2(n + 2)(n + 3) + m 2 句 = 0 , n 三 O
If m = 0 , the equation has only a constant solutionω (z) = Co. If m 并 0 ， we must have
crz+2 =
T~
m
(η +2)(n+3)
cn ， η> o.
一
Define Çk = C2k (k 主 0). Then the above relation can be written as
Çk+ 1
- 7η2
- 7η2
C2 k+ 2 = (2k + 3)(2k + 2) 句 k = (2k+3)(2k+2) 七
一~2
…、 2
一句 2
(2k + 百否E士37.(2k + 1)-2K
( 1) k+ 1 m 2 ( k+ 1 )
(2(k + 1) + 1)! 可·
EZ 句
Therefore
川) = 岳阳 = co + COZ 哥哥z2kt[mz + 27217JK+1l = co非
Let W2(Z) = Zρ2 2二二 od川口 = EF 十 三二二 o d n 十 1Zn. 80 叫 (z) =
争 +2二二 2 dn+1n( n - 1 )zn-2 . Therefore
吉 + 艺汇 1 d n + 1 7旷， -1 andω ~(z) =
2d 2 + m 2ι
叫 (z) + p(z)w~(z) + q(z) ω2(Z) = 一一一一 + 予 丁 (η + 2)(n + 3)d n+3 + m 2dn+1V \
z
Then we have the equations for d n 's:
{~川=。
(n + 2)(n + 3)dn+ + m 2dn+1 = 0 , n 主。
3
Following a procedure similar to that of the 且rst solution , we can easily find dn's and prove the second
solution is 叫 (z) = -'-'节去三
口
5.
Proof. In the given equation , we have p(z) = ~ and q(z) = _m 2. 8ince zp(z) = 1 and z2q(Z) = _m 2 泸 are
both analytic in Izl > 0 , the equation has two regular solution in Izl > 0:
(俨叫叫川(υ…
z
叫2(忡
ω
z斗) = gω
叫1 (μωz
斗)山
lnz + zρ向2 艺
2 二乙 o dkz泸k
ω( g 兴 00
旧rd
白0 兴 O川)
for some constants g ， ρ1 and ρ2 . ρ1 and ρ2 satisfy the index equation
ρ(ρ1) +α。 ρ + bo = 0 ,
whereα0= limz->o zp(z) = 1 and bo = limz->o z2q(Z) = o. 80 ρρ2 = 0
Let ω (z) = zP L 汇o cnz n = 2二三 o cnz n . Then
w"(z) + p(z)ω'(z忡忡)ω (z) = 旦 + γ [C n 十 2(η+ 2? - m 2 叫 zn.
z
32
80 we have the equations for Cηk
(C1 =0
2
Cη+2 = 帚2)2Cn ， n 2': 0
Working by induction , it's easy to see (k 三 0)
-
、1/-
/ t 飞、
-2一伙
， m
AU
一一
l'
、
飞
/ttJ
飞 tt
W
ρ
n = 2k + 1
'k
qA-qA
-
punu
n = 2k .
Therefore , one solution of the ordinary differential equation is
叫 (z) = 主古(子r k = 川) ,
where 10: (z) is the modi且ed Bessel fu配tion
I o: (z)
= 乞!mJα + 1) G)2问
N ote the series representation for Bessel function J n is
已( _l)k (~)2川
J ,, (z) = ) .
叫 (z) can also be written as Jo(imz) .
To get the other solution , we let ω ( z ) = gIo(mz)ln z 十 二二o dn z飞 where 9 is a constant. Plug this
representation into the equation ω气功 + p(z) ω '(z) + q(z) ω (z) = 0 , we can get equations for 9 and dn 's.
Now the computation becomes really messy, so we omit the details for this version. Mathematica command DSolve[w"[z]+w'[z]/z - m^ 2w[z]
0 , w[z] , z] gives the two solutions as Jo( - imz) = Jo(imz)
and YO ( -imz). Here 飞 (z) is Bessel function of the second kind and is defined as
==
一
α-
(-)Z
一 了/
α 一川
QO-co
TM一
叫一川
一
pu 一
α-
(-)-o-([
Z
Remark 4.
Td一
ya ( z)
.
Verify ω2(Z) repr臼 ented in tl旧 form i8 the 8αme α8 the ω2 (z)
given by the textbook '8 8011而on .
口
7
Residue Theorem and It s Applications
7.1
Exer cÏ ses in the text
7.1.
Proof. 8uppose f(z) has Laurent series 三二三 ∞ αnzn in a neighborhood of O. Then
f(Z) = ju(Z) + f 叫 = j 艺川l + (- l)n] = 立句k Z2
80 ，
fO
臼
缸
r ρ 吕s咀
u1 值舱
cien
臼
e时
n1此t
句
叫ly 吕I口
ma
integral ，fo扣r口m
丑mla
口
33
7.2. f(z) can be written as zng(z) in a neighborhood of 0 , where g(z) is ana1ytic near 0 and g(O) 兴 。­
Without 10ss of genera1ity, we assume n 三1. Otherwise , all the residues be10w equa1 to O.
(1)
Proof.
川口 19(Z) + zng'(zn
f' (z)
f(z)
zng(z)
g'(z)
g(z)
,
Z
I
80 Res (f' / f , 0) = n.
口
(2)
Proof. If n = 1,
[g(z) + zg'(z)]'
f(z)
f" (z)
f(z)
2g'(z) + zg"(z
zg(z)
g' (z ) , g" (z )
g(z) . g(z)
2
一.一一-
z
80 Res (f "/f , O) = 2g'(0)/g(0). If n 三 2 ，
n(n - 1)zn-2g(z) + 2nzn-1g'(z) + zn g 气 z
f(z)
f气 z)
f(z)
n(n -
1)
2n g'(z)
' z g(z)
,
,
g气 z)
一一一一一一
z2
,
g(z)
80 Res (f "/f , O) = 2ng'(0)/g(0) . Combined , we conclude Res (f"汀， 0) = 2ng'(0)/g(0).
口
(3)
Proof. If n = 1,
f" (z)
f' (z)
2 g' (z) + z g" (z )
g(z) + zg'(z)
is ana1ytic near O. 80 its residue is equa1 to O. If n ~三 2 ，
f气 z)
n (n - 1)zn-2g(z) 十 2nzn-1g'(Z) + zng"(z)
1
f' (z)
nzn-1g(Z) + zng'(z)
z
n(n - l)g(z) + 2ηzg'(z) + z2g气 z)
叼 (z) + zg'(z)
Note 山 second term in the prodl叫s ana1抖ic near 0 , so Res (fγ ， 0) =巾 l )g 叮叮，rvu)L=。=
n - 1. Combined , we conclude the residue is equa1 to n - 1
口
(4)
Proof. If η = 1,
(n - 1) f' (z) - z f" (z
f(z)
2 g' (z) + z g" (z)
f" (z)
一一一(
~
f(z)
~
g(z) 十 zg'(z)
is ana1ytic near 0 , so its residue at 0 is equa1 to O. If n 主 2 ，
(η
1) f' (z) -
f(z)
z f" (z
(
,\
f' (z)
f" (z)
=\ . (
n
1 ) z~ f(z)
0
~ J f(z)
=n(η1)η (n-1)=0 ，
where the 日econd to 1ast equa1ity is due to part (1) and (2)
口
7.3. f(z) can be written as z -n g (斗， where g is ana1ytic near 0 and g(O) 并 O. Without 10ss of generality, we
assume n 主1. Otherwise , all the residues be10w are equa1 to O.
(1)
Proof.
f' (z)nz-n-1g(Z) + z- ng ,(z)
-ng( z) + zg'(z)η ， g' (z)
zg(z)
Z ' g(z)'
二
f(z)
80 Res (f' / f , 0) =
z-ng(z)
η.
一
口
(2)
34
PTOOf.
f气 z)
f(z)
n(n + 1)z-n-2g(z) - 2nz-n-1g'(Z) + Z-n g 气 z)
z- ng(z)
叫 n + 1)
2n g' (z)
Z g(z)
,
9气 z)
一一一一一-
z2
,
g( z )
川…idl盯quals to -2n辅
口
(3)
Proof.
f" (z)
f' (z)
1η(η + l)g(z) - 2ηzg'(z) + Z2 g"(Z)
η(η + 1)z-n-2g(z) - 2η z-n-1g'(Z) + z-ng"(z)
nz-n-1g(Z) + z-ng'(z)
- ng(z) + zg'(z)
z
Note the second term in the above product is analytic near 0 , so the residue is equal to
2nzg'(z) + Z2 g"(Z)
Iz=ü =一 (n + 1).
ng(z) + zg'(z)
η(η + l)g(z) -
口
(4)
Proof. (n + 1) f' (z) + z f" (z ) = 一 (n + 1)nz-n- 1g(z ) + (n + l)z-ng'(z) + n(n + l)z-n 一 19(Z) - 2nz-n g'(z) +
l)z-n g'(Z) + z-n+1g"(z). 80
z-n十 1 g" (z) = 一 (n -
(n + 1) f' (z) 十 z f" (z)
- (n - l)z-ng'(z) 十 z-n+1g叮z
=
， \ g' (z)
- (n - 1)
g(z)
,
十
,
Zg" (z)
g(z) '
口
80 the residue is equal to O.
7 .4.
PTOOf.
function
given conditions
7151
Zü are zeros of g( 功 ， f (z) and has the same order
type of singularity
removable
gf((zz))
g(zü) 并 0 ， f(zü) = 0 , f' (zü) 手 O
pole of order 1
旦
f(旦zi)
Zo is zero of g(z) of order m and zero of f(z) of order m + 1
pole of order 1
4
f(三zi)
g( zo) 弄 。 ， f( zo) = f' (ZO) = 0 , f" ( ZO) 弄 。
pole of order 2
记和
g(zo) 并 O
pole of order 2
gf((zz))
Zo is zero of f(z) of order m and g(zo) 兴 。
Zo is zero of g(z) of order m and zero of f( z ) of order m + n
pole of order m
且
f(三zi)
residue
。
fgr((zzoo))
(m十 l )g\m) (zo)
f (-"'+ 1) (ZO)
g'(zo)
pole of order n
口
7.5.
Proof. P(z) = c6 +与年十二~P(z) ， where P( z ) is a power 阳ies of~. By formula (7.12) , Res (f，∞)=
- 2C OC1.
口
7.6.
Proof. 8uppose the singular让 ies of f areα1 ， … ， α n' Let R be large enough so that all the 五ni te singular让 ies
of f fall in the disc Izl < R. Then Cauchy's theorem implies
zlTl
川 J lzl= R
f( z)dz
>
i = l ，川 <∞
Rm(fA)=O ?
i.e. 2二二 1 Res (f，向) + Res (f，∞) = O. 80 the sum of residues of f on t is O.
35
口
7.7.
Proof. Let Bk
a鸣 ßk (k = 1, 2 ,… , m) andρbe a positive number that is su伍cie川l忖
y 吕ma
址11. De丑
fin
Cρ = {卡z : 问I z斗叫1 = 1 , 问
I argz 一 0
此kl ;三三 ρ , 1 三 k 三 m}. De且ne γk as the arc that starts from ei(e k ρ ) ， ends
at ei(e k +ρ) ， has ßk = e向 as the center , and dents toward origin. Then by Residue Theorem , for ρ> 0
sufficiently sma11 , we have
m 生ldz = 2汀 |E 什年)
Then for each k ε {1 ， 2 γ . . , m} , we have
r f(z) ~'-
J- Îk
where 归 (ρ)
白 (ρ) →汀 asρ → O.
r 归 (ρ)f(ßk + ρe"" )
~~~
Z
股上
此 +ρ e 2 0!
)φ tC ρ)
Since βk is a pole of order 1 , in a neighborhood of ßk , we can write f( z )
ω 去 where g( z) is analytic near ßk. Then (note T什 LFA ) = 非 )
r
rφ2(p)
f(Z). ,
人 Îk τ-a z = 九 (ρ)
g(此 +ρε叫
α
_.g(βk)
(战 十 ρεi O! )ρεzα ，-~…一→ m 丁k
~ ~ (f(z)
0
ì
… e3 1.. τ- ， Pk J ' 一 r
υ
Therefore
Jhifρ 号由 =2汀 lEm( 年} + }~~[Îk 生ldz
fRMmM=
2汀|呈阳(平}+名 Tes { 王子l ， ßk }
Remark 5.
The α bove Tesult and the trick of indenting the contour cα n be found in Whittaker α nd Watson
μ1}， 36.23, pαge 117.
口
7.8.
P叫 We compute a more general integral Jooo 击右 where p ε(1 ， (0). Choose two positive numbers r and
R such that 0 < r < 1 < R. Let γl={z:r 三 Izl 三 R ， argz = O} ， γ2={z:r 三 Izl 三 R ， argz = 2叶，
ÎR
= { z : Izl = R , O < arg z < 2叶 and γr = {z : Izl 习， 0 < arg z < 21f}. Define f (z) = 甘示 where
Zl/p = ε 守主 is defined on C \[0 ，∞)-Note by substituting ut for z , we get
f JL
1 + xP
= flfL
Jo
p(ν+ 1)υ
By Residue Theorem ,
( 1 )t
If f(z)dz = 2汀Res(f?1)=2m·=E
J Î1 十 ~- n - ~
P=2tαε削'
-p
where α = 互.
We have the estimates
P
11R f(z)dzl 1a P(~:i:12川。牛 tt → 0
2 7r
=
as R →∞，
1
|ly(Z)dz| =|fp::;;/ 牛 tk → 0
36
as r • 0 , and
r
rf
rR (\~;
xe 2 ~ dzfRztdz2αz
，/ ，\~~
f
, ,\ .e
\.1
7r i )
f f(z)dz=-f
J ^l 2 J \-r-
Therefore by letting r •
Jr
p(X 十 l)x
o and R →∞， we have
fJL=f
」工=主主
2 i
P
l +x
1
p(X 十 l)x
Jr
Jo
1 - e o:
p(x + 1)x
-2iaeO:
一 2 sin ae( 号 +α)i
土=生 csc (~)
sinαp\ p)
口
7.9.
Proof. This is a special case of exercise 7.10. Answer:
;0 (verified via Mathematica: Sum[l/n^ 4 , n , 1 ,
4
Infinity]).
口
7.10.
Proof. Let C N be the contour used in Lemma 7.2: C N = [N + ~， (N + ~ )i , -(N + ~)， -(N + ~)飞 N+H
Then by Residue Theorem ,
iN 守三dz = 2刑兰NRES 冉升
z = 0 is a pole of order 1 for cot z. 80 we can aSSllme the Lallrent series of cot z near 0 is 2:汇 o b 2n _ 1 z 2 η -1
Then
?=艺叮z户扛ι
Therefore , forρ> 0 sufficiently small ,
如(乎， 0) =赳1=ρ 于dz = b 2 k-1产
For n 并 0 ， take ρ> 0 sufficiently large , we have
Res I/汀 cot 7r.. Z. • n \ 1
\
Z2k
,.)
r
1
(-l)nc创盯·汀
\ -/ ...
. (z
\ . - n)
. . / . - - - dz =
27r i J lz -n l=ρ
Z2k sin 汀 (z-n)
z-n
n 2k
Combined , we can conclude
iN 守三巾= 2 i (b 2k - 1 产 +2 主去)
7r
By Lemma 7.2 , limN→∞ :fCN ~鄂主 dz =
= =
o. 80ε二 1 :;卢=一些?土产
b_ 1
1, b 1
- ~， b 3 = 岩 ， b 5 = 击，etc. In particular , for k
which gives the answer to exercise 7.9 .
From form由 (5.25) ， we have
= =
2， 协 1
b3 =
古 80 2:二1 才可=菇，
口
7.1 1.
Proof. We 且rst deduce the Laurent series of 芷z near o. 0 is a pole of order 1 for 芷z Note sin z is an odd
function , we can assume its Laurent series near 0 is 2: ~ o b21_1Z21-1. Then
1 = sin
z 寄2n_1 Z2n - 1 - 主 tkk+1 去21 产=兰[辛21-1 川
37
80 L 1
=1 则立obl 1#尘元 = 0 川三1. Usir
灿吕臼阻
叫
m
1口吨
n皿l电g 吐山l让is ， we cωa
创削叩
n1 创削叫
S创
sil让句
l忖y ge叫忧叫Lau
创削urer
(忖
ve
曰n诅丑e
叫
d by Ma
川thema
川tica: Series[zjSin 间， 杠， 0 , 10}])
1
1
Z
7z 3
360
31z 5
15120
127z 7
604800
73z 9
3421440
?一=一+一+一 十 一一 十 一一一 十 一一一 十 o( 产)
6
Let CN be 山
t he ∞
c O川O盯 used in Ler
创mr
by Residue Theorem
丑1 ，
tN 三年二dz = 2πih二EEEZA)
7τ Z '
2
一一
AU
何一6
EtIJ'
\飞
QU
、
/，，
，
飞飞
EE
Re
汀一川
Fr om the Laurent series of _._1一‘
we can deduce
Sln
For n 手。， we can findρ> 0 su伍 ciently small , so that
如(斗y) = 材-nl=ρ12: 斗士由 = 于
Th
叫
l
叫
ωm
眈
n1比
ωclude
叫毗
1巾
d由eL
立
二 1 咛i乒:二二
ι二
7.2
舌非(阳刚i丑b
刨
e叫db忡yM
胁at山
he
1em川
览
a川tica: Su叫
叫毗[
m
(叫
阳(←仙川
-1
μ1盯) ^ (归n - 叫
1 )j川n^ 2 , n , 1 , Infi
lfin
咀血
缸
灿
fini
时n础lÏ山峙蚓
i让t句均均叫
刮
句时圳
yT1们])
讪
口
Exer cÌ ses at the end of chapter
1. (1)
Pn时
口
(2)
Pn叫
口
(3)
叫 1 - cosz =立1 喘户2n 80 0 is a pole of order 2 for (市 f . g…1伍cle由 smallρ> 队
we have
r
z4
dz
I . 0 I = - -- I
.
飞 \1 - cos z) '~J
27r i J lzl =ρ(1- cosz )2 Z2
((
Res I I
Z
\2\1
--;0-
d
I
1.
z4
1I
.~ I I
dz l(l- cω Z)2 J Iz=O
By repeatedly applying I'Hospitale'日 rule for analytic functions (see exercise 5.1, 5.2 in the text , page 62)
and the fact lim户口 号主 = 1, we can conclude the residue is equal to O.
口
(4)
Proof. By repeatedly usi吨 I'Hospitale's rule for analytic funct
Res
(本~， O) = 叫去 z779 0) = ￡(丰)|J i
口
(5)
38
f.
PTOO
R臼(云1， 1) =叫 (z 一 ι 十 lJ=ttFL=1=0
口
户。
( )
P叫 Let zn = 一(号与) 2. If we take the convention that ，j习=飞 we have ,;z;; = (mr +号 )i. It 's easy
to 附 cosh ,;z;; = cosh [( n 7r +刊= co巾汀+号) = 0 , and
lim (cosh
h ,jZ
si口 (n7r + 号
( - l)n
vz) = lim 一一一一二
=一一一一
2 ,jZ
(2n + 1) 汀
(2n + 1)π.
I
Z--->Z"
80 z~ is a Dole of order 1 for 一气L丰 and
…
cosn 、
z
s(羊言 'Zn) = Res (孟号士n 牛
= (- 1)n(2n
+ 1)π
口
2. (1)
Proof. Let f(z) =
z3( 1-;)(Hz) . Then
1 d2
1 I
Res (fρ) = ;， τ 一一τ
2! dz~ 1 - z~ Iz=ü
Res (f, 1) =
各 and Res (f, - 1) = - ~.∞ is a removable si吨ularity.
口
(2)
Proof. ∞ is a removable singularity. Let f(z) = (Z2南市T
1 dm
Restf, i)=ZIZE [(z +
百步古古萨古 Then
~)m+l ] IZ=i =击 [-(m + 1川 (m+2)] . . . (-2m) 川 2m- 1 1
臼
川
叫
川
山川
叫呻
ml
1让
叫
h
i让
世恤
la
h
S
i阳伽
M
(2m)!
2 2m 十 1
(m!)2
口
(3)
Proof. 1 - cosz = 2sin2~. 80 zn = 2n 7r (n εZ) is a pole of order 2 for the function f(z) = τtz -Note
for each n ,
( ζ严 ) 2
1
2 z ( ζ严 )2
f(z) = ~τ一， 0
一一一一
2 sin 言
2 (ζ严 )2 旷亏mf
(z - 2n 7r )2
sin 2ζ严.
Define h( w) = 一旦一 . 8ince the Laurent series of 一上一 near 0 is
飞
~W
~W
1
1
Z
7z 3
360
31z 5
15120
127z 7
604800
73z 9
3421440
?一 = 一十一十一一 十 一一一十一一一 十 一一一一 十 o(z 叫 ，
6
we have the Luarent series of h(ω) near 0:
w2
7w 4
31ω 6
127w 8
6
360
15120
604800
73w10
3421440
h(ω) = ~一 =1+ 一+一一+一一+一一一+一一一+ o(ω11 ).
39
Therefore , h(O) = 1 and h' (0) = O. Moreover ,
如(~衬"2' 2zh (斗旦) , 2nK)
Res (f， 2η汀)
2
JEm ￡[缸h 2 G- nK)]
λ弘[叫~ - nK) 十 2z 叫:叫 h'G 叫~]
口
(4)
P叫 Let f(z) = 岳母 Then 由
灿
m
1址川
h由l yÍz =O川
i证f 川 on
呻
1
(μ
zn
川)nE Z are si吨1山rities of f(z). For n = 0 , Zo = 0 is a removable si吨ularity since
lim 一丘一
Z
1
• ô sinh YÍZ
cosh 0
Therefore Res (f, 0) = O. For n 并 0 ， we have (suppose ρ> 0 is sufficiently small)
r
yÍz(yÍz - ~)(YÍZ + ~)，
~(~ +~)
dz =
= 2zn =- 2(m)2.
sinh YÍZ - sinh ~
cosh V瓦
1
Res (f二句) = ,,=-. I
一一一·
i Jlz 川 =ρ z - zn
Remark 6. This solution hα8α diffe陀nt 陀sult 卢om that of the textbook's solution. Check.
口
(5)
Proof. Both 0 and ∞ are essential singularities of f (z) = exp [ ~ (z - ~)]， as can be seen by letting z → ∞
and z • o along positive and negative real axis.
To find the residue of the function at 0 , we note
exp
D(z - ~ )] =立古 (z_~)n
For each n ,
(z
-~)二立(~) zk(_z-l)n-k
k= O
(z - ~) n co阳ins Z-l term if and only if n is an odd 川由 er ， and in this case , the
叫
ωω
e田险C阳川
O巾
f Z-l气斗川怀
钊计
isS叫仆(卜
川
H
一→1)厅平叫(nf) 川 川p归阳阳川川
川
a创m
ns臼S
80 the expansion of
ι1
(m\
忐 ( _ 1)m+1 (~)2m+1
、飞 (1)m+() 二、飞、，二
中。
2 2m十 1(2m + 1)!\2m + 1)
巾。
J1 (1) ，
where Ja(z) is Bessel function of the 五rst kind:
占 (z) = 立 m!ι1: + 川~rr问
80 Res (f, 0) = -J1 (1). 8ince the above expansion is valid in C \{O} , by remark on page 87 (forml由 (7.12)) ，
we conclude Res (f， ∞) = -Res (f, O) = J 1 (1).
40
Remark 7. The results in the α bo 时 solution h α ve szgns οpposite to those of the textbook 's solution. Check.
口
口υ
( )
Proof.
have
00 is a
removab1e singu1arity and 0 is a po1e. Using the Tay10r series of cosine function near 0 , we
081 = 亏 (_l)n 1
Fz~ (2n)! zn'
一.一-
80 the resid ue is equa1 to the coefficient of ~， w hich is - ~.
口
(7)
Proof. Let f(z) = (z_ ll了TnZ' Then 1 is a po1e of f(z) and ∞ is a removab1e singu1arity. When 1n 1 = 2mri
(n εz \{O}) , for ρ> 0 su血cient1y small ,
Res (f, 1)
=去 lZ-ll=ρ f(柏=卢=古
When 1n 1 = 0 ，音~ is ana1ytic near O. 80 forρ> 0 s咀cient1y small , by app1ying l'Hospita1e's ru1e , we have
Res (f, 1)
= 去 lZ-ll=ρtF t山 = 二 [itll=1 = i
口
(8)
[1
P叫'. Let f (z) = ~ +击 十 甘市 十 十 古τ]. The山nd - 1 叫01叫 f， whi1e ∞ is a remova协
singu1arity of f. For ρ> 0 sufficiently small ,
半 lZI=p ~ [1 十 击 +dF++[击石] dz
[1 +击 十 dF 十十 (z: l)n] Iz=o
Res (f, O)
n+ 1.
8ince f(z) can be written as
11 F于+T主 |(z + 1) - J一|
1--;;丰T
川
(z+l)nJ
we have
r
- 1
.1.
1
dn - 1
Res (f， 一 1) = - l
一一一-，- dz = 一 一一一 τ=TZ
i Jlz+11 =ρ Z2(Z + l)n ~(n - 1)! dz n - 1 -
_2
1
Iz =-l
|
=一
- 2)... ( - n)z-C叶 1)
(n - 1)!
1
Iz =-l
|
口
3. (1)
Proof. Note ~ is ana1ytic on C \{吗， so Res(~ ，∞ ) = - Res( 己的 = - 1
口
(2)
Proof. Note 守主 is ana1ytic on C\ { 叶， so Res( 守主?∞ ) = -Res( 守主， 0) =-1.
41
口
(3)
Proof. 8ince cos(2η7r +~ ) = 0(η εZ ) ，∞ is not an isolated singularity.
口
(4)
Proof. 8ince (Z2 + l)e Z is analytic on C , Res((z2 + l)e Z ， ∞) = 0
口
(5)
Proof. e 去 = ~二。崇苦 ， which has no Z-l term. 80 Res(ε 之∞) = 0
( )
口
ρhu
Proof. Recall Res( 飞/( z-1)(z-2) ， ∞) is equal to the coefficient of the term z in the power series expansion
of -V(~-l叶) (~ - 功2) nea
飞山/八
/(1口1 一 z才)(←
1卜一 2却z功纠) = 1 一 2主2三一三8 一 2艺16
三兰: 一 .
D e叩
问
pen
创叫
di问
吨
n
g川
川e br叫创
叫l比叫
m
ch
叫咄川川…
附e盯町
W
叫叫
chωhoω
…
∞
O
oos
创附毗吕肥盹
se飞，， 附
weh
阳
盯
a
刊
veeVZ
θ
豆歹2 = 土扫z 臼川山问
e 川…
near 0 i吕
平(~ - ~ - ~ 一 芸 一
)
Therefore , Res(\/(z - l)(z 一 句 ，∞ ) = :!:去
口
4. (1)
Proof. The equation Z4 + 1
0 has four roots: Zl
ε 卡 ， z2
ε 一芋，均 = ε 子乱， and z4
e 号飞 The
intersection points of Iz - 11 = 1 and Izl = 1 are 已土 引 800日ly Zl and z2 fall within the disc 1z - 11 < 1. By
Residue Theorem , we have
~一
九 1 1= 1 1 十 泸
2时[叫击 ， e2，f i) + Res (击4 ， e- 2，f i) ]
2{-z-Z1+l-z-Z2}
7r i I lim 一一一一 十
丑 一一一一 |
飞 z → :;1 z4 + 1 ' Z • :;2 z4 + 1 )
2刑(走 +去)
27r i. Zl + Z2
们一ττ
7r~
y'2 '
口
(2)
Proof. Usi吨 the same notation as in part (1) , we first show all the roots of Z4 + 1 = 0 fall within the circle
Iz - 11 = 2. lndeed , for any e ε [0 ， 2π) ， 1e'!:Ì - 11 = 21 sin 1 三 2 ， where the equality holds if and only if e = 7r .
80 z二日 (i = 1 , 2 , 3 , 4 ,) all fall within the circle Iz - 11 = 2. Then by Residue Theorem and an argument
similar to part (1) , we have
!
[
dz
CJ)
一一一一一--，-
Jlz-1 1=2 1 + Z4
= L 7r Z •
zI 十 Z2 十 Z3 十 Z4
--------
υ
-4
口
42
(3)
Proof. By Residue Theorem ,
r
1
sin 号
7r Z
2i
九归王Imzu=2mET= 否
口
(4)
Proof. By Residue Theorem ,
1
1
川川
~，._ 7r Z
一-
几 1= 3 z2 -
,,_: (sin .;f ， 时n (-.;f)\仄 A
J.
-... \ 1 + l '
../…
1ι4 ~-
二
一
-1 - 1 }… u
一
.…，
口
(5)
Proof. The singularities that fall wi扰川拙
由
tlhin
Re
臼sidue Theorem
丑1 ， we have
lbn 阳
zd巾z = 扣2汀刑i 豆
tm
…汀盯讪
Iz斗| =n
lim
k= nz→ k+ 吉
1卡叫一斗(川恻
z卡川
cos(盯)
k= n
nin(kπ 十 打)
一
口
ρ
。
( )
Proof. Let zn (1 三 η 三 10) be the n~th root of the equation z lO = 2. For example , we can let zn = 2fo e 斗「.
Then similar to our solution of part (1) , Residue Theorem gives
正|刊
N侃2: ~~1 e一半 =e一平 Uζ工;乒:旦斗O. 如
S ot山
he int
盹
口M
t归e叩
g
口
υε
(7)
Proof. By Residue Theorem ,
1|j由 = 2们去去|恒。 = π1
口
(8)
Proof. 叭f屿e note 乒
e 2盯Z泸3- 1 =Oi证f and only if for some k ε Z ， z3 = k. Sin
丑 ceη < R 3 < n + 1, a 丑
r1urr
口
m
丑1be盯r 儿z* is
T刊'I，旧泸
e 2纣7刑
Z -1 = 0 wi让t址h出in
a root of 户
roots are Zj. Then by Residue Theore盹 we have (assumeρ> 0 is sufficiently small)
r
φ
Z2
凡 I=R e叫一 1
~
z7
r
Z2
dz = 2刑、，，_ ，:_3 + cþ
JfzoM3号产~Zj
. ',
1
f~
几 1 =ρ27riz 3 +击 (2πiZ 3 )2 十
•
Remark 8. Nο te 0:盯 Tesult is difJeTent from the textbook 's solution.
43
'"''
dz
1
~
、
3 NAd三口
2η
1+1 = --:::-+1.
口
5. (1)
f.
PTOO
1)2n
Ih1)2ndz
2汀川
z2n+1
2 i l
iz =
)
J 1z1 =1 巳
=1 \ 2
1 2K 川创
2 2η (2η)!
d二(♂+ 1) 凶 |
27r
2n(2 η) !山
汀
I z=o
7r
=JL 旦 (η ) -z 2n
22n (2 η)! dz 2n \ 2n)
l
Iz=o
(2n)!
22n - 1 (n! )2'
口
(2)
Proof.
J 1 到; =1 (α + bz+~- l
九 (α + b cos x )2
f iz 九 1=1 (的 2+2ω +
The equation bz 2 + 2α z + b = 0 has two solutions: Zl = 平与俨王 and Z2
b)2
-a- ，俨主
C1ear1y I 句 1> 1
and IZ11 < 1. 80 bγResid时 Theorem ， we have (f (z) := (bz 2 + 豆叶 b )2 i and ρ> 0 is sufficiently sm址 1)
filo
。
一α
一+
7
0
v
句 J - pi
JL
27r iRes (f, Zl)
r
a
JIZ- Zl l = ρ b 2 (z -
4z
dz
Zl)2(Z - Z2)2
一
21
勾
b的
d
z
川
盯
LF
π1 Ez
)2E斗 口白
二石 - Z2)
牙刑
列
=习
87汀r
z l+ 句
Z2
乳
4
Iz=zZ
b2 (Z2 - zd 3
2α 7r
(α2 _ b2 )3 /2 '
口
(3)
Proof. We note sin 2 B = ~[1- cos(2B)] . Using the substitution ru1e B = 护， we have
fEtftr24|=1Z21 十 1
The eql川 ion Z2 - 6z + 1 = 0 has two roots: Zl = 3 + 2 vÍ2 and z2 = 3 - 2 vÍ2. It 's clear that 扫
I zll > 1 an
|问Z2址I < 1. By Residue Theore盹
l
K
1+韦 = 斗31| =1(z - JL - Z2) = 一 '凹
τ
Z21Z1 一 朵
r. 一一一
一一一 一 -
口
(4)
44
f. 8imi1ar to prob1em (3) , we 1
PTOO
do
dα
(1 十 日in 2 8)2
(3 - cosα)2
r
zdz
i Jl zl=l (z2 - 6z + 1)2
8
d |z
| | -一 M
Z1 十 Z2
1 0 1f -c.----,--;c
dz l (z - zd 2J Iz=z2 ~~" (Zl - Z2)3
167「
r
zdz
2时 J lzl = l (z - Zl)2(z - z2 )2
1
~---，-~ = 16π ，---
I
3汀
4 V2
口
6. (1)
Proof. This is a specia1 case of (3) , with n = 2 and m = 1. 8ee the solution there .
口
(2)
Proof. Let f(z) = 百奇古 Then for C R = {z : 0 三 arg z :::; 1f, Izl = R} (R > 0). When R > 1 , we have by
Residue Theorem and Prob1em 2 (2) of this chapter
i ( 2n ) !汀 (2n)!
IrR f(z)dz 十 IrCR f(z)dz = 2πiRes (f， i) = 2刑.-----22n+1(η !
22n (n! .
,. f
\
,
,. f
1
\
,
f
1
('
"\
-
)2
j-R
)2
Furthermore , we note
ILR
as R →∞
r
R ei (J id8
1 + R2 川口 +1 I 三 人
K
f(z)dz l =
(1
R d8
(R2 - 1)n+1
1f R
(R2 - 1)n+1
• 11
80 J工目击tTdz=Etj部
口
(3)
一二-~-dx
1 十 x '2 n
l
j一 ∞
r OO
=2 I
Jo
mp
Ju
Z
∞
币 2盯
一­
r OO
22
一十
Proof. We note
f't'-OU 旦旦土L
I-n
11
2饥
(1 + y) υ
伽川
忡
b
y 山 S
巾
山川日"由州灿
u
阳阳川
山1
bsti
tí1此t饥
we a1ready 日 howed
而 dy = 汀创(~)
~2 :~
-α
80 JIr ∞∞ 一一一
d x = -"-霄η/ CSC 飞I(2~+1
1f"!)I
i 十二c
2n 1f
~.~.
:l 7n 山
,
口
~
(4)
Proof. We note cosh( 号 z) = 0 if and on1y if z = (2η + l)i (n εZ). Define f(z) = 1 纠 + 1)ιh( 卡) ' Then for
n 兴 。? 一 1 ， zn is a po1e of order 1 for f (z) and we have (assume ρ> 0 is sufficiently sm 址 1)
r
I
i
R 臼(f， zn) = ,,=-.
dz
~z 号 z n
Jlz 川 =ρ z zn cosh(~z) cosh(~zn) ~(Z2 十 1)
1
一一一-
1
1
sinh(~zn) ~(z~ + 1)
( - 1) 叶 1
2 汀 η(η 十 1)"
To find the residue of f(z) at i , define h(z) = 号叫号汇 Ls 时 i)' Then h( i) = sin市可 = - i. App1ying
l'Hospita1e's ru1e , we have
/，t飞、
h )
-qtu
l
号 cosh(~z) - ~(z - i)~ sinh(~z) _
[cosh( ~ z ) ]2
z-, i
122(z
z)
= 1
11 日I
户
汀2
4
"…÷十
二lπsinh( 号 z)
45
手 sinh(~z) -
1r42
[sinh(~z) + (z 一何 cosh (~z)l
2 cosh( 号 z H sinh( 号 z)
Z
l
); (-)-)
1
一
'n 一
φL-
Z
一 -2
+一十
一
Z 一/飞
Z
立
V八 一π
80
一
9" 一
Res(叶
([-fh
Therefore
z z
27r i
21
∞ t 川叫
1
1
21n
岛es(巾
R
fλ
川
f，叫
， z句
znη川川)恒
Zn
=一十 2
艺二
一一一一 = 一 + 一 (μ1+
叶21n2
川
1吐2功叫)= 一
n= 12汀叫 n + 1)
27r i ' 27r i
Now we consider the path C N = [- N， N， N 十 4Ni ， - N 十 4Ni ， - N]. Then
iNh)flz = I 十 H 十 111 + 11 ,
where
1= 汇只 z 忱
We note
dυ2
(IN +iy l2 -1). ♂ N/21 甘 /
4N
<---一一一一一一一一
e~t' - e-~t'
N2 -1
-•
川
as N → ∞. 8imilarly, we can show IWI 一→ o as N → ∞. Meanwhile , we have
ρN
1II1 < /ω
工 N ( I x 十 4Nil 2 - 1) 兰号二三
2N
-1
<一一一一一一一 → 。
一 16N2
as N →∞ . Therefore , J二 f(x)dx = limN →∞ :fCN f(z)dz and by Residue Theorer丑 ， we have
1: 川
2N-1
∞
A 唱
A
J(Z) 由 = JVm E二 Res (f， zn) = 2 汀 i~二 Res (f， zn) = 2刑手~ = 21n2
J叫
川b
ν o
Remark 9. 1n the p 'rO of， ωe used the following facts from cα lculus (see , for ex αmple ， Shen [9), page 221):
立吁二 = ln2
、1 l/ /
- 4li
9d
十
且
\ell，
l，
/
+
、、、‘J
，'，
'『，，
-i
一
十
一
1111
引
2
一­
1-3
1i -A哇
/'''』"'飞飞、
- 2+
吐
、111/1i-A
十
-
飞
-2
tft
-ITA--3J 一
nj
1
- -1
-n
n/il\1-2
1
11
11/il\
L
-f\
口/\-11/''12
飞
←
{
乞 η (n + 1)
x
-p。「」
ι ( _l )n
J\
αnd
优
1
C
阳叫叩 O旷f αanα吻
恍E 旧阳e 加
The ωculωu
旧S p 'rO of 旷 2汇二二 (- 1乒- = ln2 附E协
阳
ds α l耐
1付t材耐州价
t削lle 仰
b 1付t 例t 俨问ick. Ho川
t挝ω
ue1肌 吁 ω
th肚EωO
l均y肘tμ优
叭仰
τ
仇
加
「叫
比
时
ecωomη
时
，
附
阳
比
ep
oof
旧
臼
tf
d
.
1ndeed
un
l
ction
s
then
t
h
'rO
b
t
s
stm
α
t
μO
α
we
note
in
the
unit
disc
gh~
扣
陆，
严
叨
,
,
f
ι (_1)n-1 z n
ln(l + z) = )
，一一百一一·
The series L 二 1 己号:三二 clea 吻 com旷'[J es at z = 1. So by Abel's Second Theorem (see , for example , Fang
[3), p αge 121) , we must have
ln2
= 且 1ln(1 + z) = 点 1 艺出严 = 艺吁:
46
口
7. (1)
Proof. Let f(z) = 舌头 and C R = {Z : Iz l = R ， O 三 argz 三汀}. The equation 1 +泸= 0 has four roots:
Zl = e~i ， Z2
ε 号 Z3
ε 号飞 and Z4 = ε 子飞 where Zl and Z2 fall in the upper ha1f p1ane. For R 1arge
enough , we have by Residue Theorem
汇 f(z)dz + 儿川z= 加(Res(f， z r) + Res (f, Z2))
It 's easy to see Res (f, z r) = 1im z → u 市 e'z
80
寺 =-jthZ1=-j去 ε 二步 and Res (f, Z2) = 一古ie 二f
27f i(Res (f, z r) + Res (f，非
Meanwh川r z E CR , I 古| 三古→ o as R → ∞ By by Jord川
J马儿Rf(Z)dz = 。
Combined , we conclude
I ~~
cos X
Jlow 1 +
1
dz = 二2 l咀异乞dx
+ x 4 = ~2 . 2_..7fi(Res (f, Zl) + Res (f, Z2)) = 工卢卡os 土 + sin 土)
xr~
J 一∞ 1
~~
'\~~~~\J' - i J
•
~~~~\J' -~JJ
2V豆
\J豆ý'2)
口
(2)
Proof. Let 贝
f 陀削
(υωz
均)恒= 舌兰元元丁节古 Th
阳
l臼
1览
eenn sim阳 tω00删
盯r sol
u
仙
1uti
f
J
I 山
o
cosx
ff7
dz
(1 + X 2 )3 ~~
=
1
r飞叩
一~(叩
且
cos巳ιι?τ
三二
才才讪
ττ才
τ~伽
dZ户←
= 汀刑巾
4诅阳
Re
伽e臼叫吕
2 J一∞ (1 + x 2 )3
口
(3)
叫 Let f(z) = 迂石 Define C R = {z : Iz l = R ,O :S;叩 ο}. Then for z 山，同才三
R2-各才 → o as R • O. 80 by Jordan's 1emma (Lemma 7.1) and Residue Theorem
汇 f(z)dz = 27f iRes (f, 1 + i) = 2刑
r~
ZEZZ|
1 I
("1
_' \
k 一 (1 - i)]lz=H i
作
一 [ (cos 1 - sin 1) + 巾os 1 + sin 1)].
e
Compare the rea1 and imaginary parts of both sides of the equality, we have J~(X) x:};E;去VZ =5(c051+
sin 1)
口
8. (1)
47
f.
PTOO
v.p.l: x(x- 乱 2)
Jh[IJzJ)川 十 1 1 一\(工 :22)+rx(Z 222)+L( 工 2225]
一 1一
We note 一一一」一一一
(x-1)(x-2) - 2(x-2)
~+土.
80
x -1
2x
I
JUJ[ 仕写占+去] dx
工二 FZ(2 2)
I (N + 1)2 1 , 1 , 5(5 十 2)
--矿工马 2
l N(N + 2) J ' 2 一 (5 + 1)2
，， ~~ 1
L
1 , 5(5+2)
2 --- (5 + 1)2 '
ZZ2) 才11tr 占 十 去|叶 ln [~1 ;二:!JS 2] ,
1
1\(Z
and
j:~ 二(工 fLrr[tr 占+去 lω = jlnbJF 剖，
J叫 :0 [2(X ~ 2) -占+去] dx
1:FZ(iu2)
1 主二2 旦土旦旦二 l
limN → oo~
~-2 l
(N-1)22+5J
5
(1 +的 2
1L
2 --- 5(2 + 5)
一+
AU 一
俨
τλυ
-zυ
+
-2
1/'-zυ
、
2-wj
/t 飞 -zυ
的一+
/f 飞
俨一-
-zυ
1
mλυ一
27/
/f 飞
d 一一
十一
-zυ
一lT
/f 飞
l-2
n
-z
俨
AU 一
d,
-1
i
俨
AU 一/i 飞、
-z -z
·斗
m
fL
z-V
八
V P
十一俨
Therefore
ov
口
(2)
Proof. Note 日 in(x+α) sin(x 一 α) = -~[cω (2x) 一 ωs(2α)] . 80
a
f 叫 十 α) si巾 α) dx = _.::.1 J()Ir ∞ cos(2x)ωs(2α) dx = _.::.1 Ir∞ ωs(2x)ω(2α)
.,
x2 一 α 2
---\ -~~/
- = -\-~./
x 2 一 α2
2
.,
4 J- ∞
x2 -
2
Define f(z) = 号手工 Let C R = {z : 1z 1= R , 0 :::; arg z :::; R} (R > 0) ， ι(α)={ z :l zα 1= 叭。三 arg z 三
叶 ， Cr ( α)={z:lz+α1 = r ， O 三 argz 三 R}. Then by Residue Theorem ,
I
J(-R ，
f(z)dz = 0
α
俨 )Uc ，， (α)U(α 十 r ， a -r )Uc r ( α)U(α +r ， R)UC R
8ince αis a pole of order 1 for f (z) , f (z) can be written 制 33mrαwhere 作) is anal芦 ic nearαand
g( α) 并 O. 80
川z=l) …~~~dZ
z z = I(g( α +Eα附) 白协 =tfg( 川口 )dα → i1rg( ← i7fRes (f， a)
1， (α)l(α
i
48
as r • O. 8imilarly, Jc r ( α) f(z)dz = - i7rRes (f，
r • o and R → ∞ ， we have
α) . It 's easy to see limR→ ∞ JCR f(z)dz = O. 80 by letting
L:
二〉〉贝扣川
(υz斗圳忡←)川灿户
d巾z产←归
=可
…i叫
巾咖刷啊叫
蚓叫
Res
e臼S叫叫叫
(οfμ
咐川
， α叫a) 巾叫叫叫(f队川
fι'
z+ α1-卸
z
α| 卸
80 Jo= ~MZZTFz -rz Ldx = 才~ sin(2α)
口
(3)
丹'Oof. Define f(z) = 天前 Let R> r > 0 , C R = {z : Izl = 托巴叫 z ::; 7r} , and Cr = {z : Izl = 叭。 三
argz 三 pi}. Then by Residue Theorem
兀 f(z)dz + 儿川 z + JR f( 忡
Note Res (f, i)
=先gt LZ = 一与二 and Res (f, -i) =先写了 |z=-f l f s o Re毗 i) + R叫， -i) =
丐二 = - cosh 1. Also , we note
儿川 z = 才方吃
By repeatedly using l'Hospitale's rule , we have
二…
Z re 'l， α
ια
J•
二
e 'l T e
II 日、
Ô r 2ε2ω (1 十 r 2ε2ω)
11 日1
二→ Ô
ie U>: -
ie' α e 'I， Te
也α
2re 2 ω
1
2
80 li口
1日冉
11I
let忱ting R →∞ and r
• 0, we have
汇 f(z) 巾 = 2πi. (- cosh 1) 十 iz
By comparing the real and imaginary parts of both sides of the equality, we obtain
1= 元427dz=ijt 玉tf亨丁 dx=~(~-e-~)
Remark 10. The α bove result is difJerent from the textbook's solution. 1 think 1 mα de αcαlculational mistake
somewhere. Check.
口
(4)
Proof. We shall use the following result: ifα 并 o and (β/α) 并 土 1 ， ::1: 2 ,…, then
L04= 主 {na~ ß - ~α 十 L βy}
创
For a proof, see Conway [1], Chapter V , Exercise 2.8 (page 122) , or my solution mar
anu
3 , Exercise 11 (iii) (page 119)
49
We have
1:EFPI
~~
100 ~一 (1 了二川省山 十 100 ~一;二三qudu
严 αx
1 - eX
qLLCf)zdz
ftt'10
dz
∞汇叫
nit-o
fE
∞汇叫
100~ 丁二77)xdz
dz
主 1 00 [ε 川 - e-(口十1- p )Xjdx 一 主 1 00 [e 川 _ e-(n十1- q )Xjdx
亘古 一 计可) 一 号。(击 一 计可)
汀 cot(p汀)汀 cot(q汀)
口
9. A class of integration problems can be solved by the followi吨 general reSl山 (Whittal町 and Watson [11 ],
36.24 , Evaluation of integrals of the fO 'rm Jooo xα 1Q(x)dx)
Theorem 2. Let Q(x) be αTα tionα1 function of x such that it hαs no poles on the pos州时 pa'rt of the 'real αxzs
αnd xaQ(x) • o both when x → 0α阳i when x → ∞ . 1f L: T denote the sum οIf the Tes时ues of ( - z) α 1Q(Z)
αtα II its poles , then
00 α 1
Z 哈咄
1 ∞飞
1Q(X
αω(μ
川
Z
Corollary 1. 1.扩fQ(μZ
叫) hα8α7ηìU川
7ηmηî加b}屹e'r 旷
0If sim
ηîpl扣e poles on 仇
t h比e positive pαT
忖t 旷
0'f 伪
t he 陀
'r，E侃
α
al αω
xz吨
8鸟， 仿
z tηmα
叼
υb加e 执
s hωozωurηî by
tη de η tiη9 the contou 'r th αt
叫∞ x a - 1 Q(x)dx
ωhe're
L: 'r' is the sum of the 附idues of za-1Q(z) αt these poles.
(1)
Pmof. By the above theorem , we have
1 T二jdz =-Mot叫叫f二 1) = 川ot
00
口
(2)
Pmof. If 8 = 1, then
f∞
xdx
1 r∞
dy
1
∞
l
Jo
(1 + x 2 )2
2 Jo
(1 +υ)2
2(1 +υ) 10
2
To calculate the case where 8 并 1 ， we choose T and R such that 0 < T < R.
Let Î1
{z
T 三
Izl 三 R ， arg z
O} ， η
{z
T 三 Izl 三 R ， argzπ} ， γR
{z
Izl
R , O < argz <叶 ， and
协= {z : Izl = 'r, 0 < arg z < 汀} . Define f(z) = π舌η . Suppose 'r is s吐ficiently small and R is s吐ficiently
large so that all the poles of f (z) are contained in the conto盯 formed by γhγ2 ， γr-， andγR. Then
Ir
d
Z8
8 - 1
f(z)dz= 2汀iRes (J， i) = 21f i. ~ I '" . \') I
= 21f i . ð - ,.L e 号 ( 8 印=
人1 +ìR+ ì2-ìγ
dz (z + i)2 Iz=i
-4
50
主 (8 - l)e 号 圳
We have (note 一 1 < s < 3)
|IRf(Z)dz| =|f(1320)2REZO 牛茹→ 0
as R → ∞，
|Irf(Z)dz|= |f J
as T • 0, and
f
川s
f(z)dzz = f
I I~~~
~\0 dx = e
2
γ(1
Since s 并 1 ， e S 1I:' 并
1.
+ X )2 ~~
zf
J
+
zs
X 2 )2
o
(1
汀
l-s
Therefore
f∞
九
28J27riRes (f， i)
(1 + x2 )2 …
1 + eS 1I:'
4 cω (~s)'
Combining all the cases and regar也ng the cases where s εZ as limit case of the forml山 2 」二L
(号 S) ,
conclude the int
叫
t
Remark 1 1. We could hαve used the geneml theoTem , but ωe still go to the spec忻c solz巾。n so that some
insight cαn be ShE过i 0η how the gener α 1 theo 陀 m is proved.
口
(3)
Pmof. We choose T and R such that 0 < T < R. Let γ1 = {z: T S; Izl S; R ， a咆 z = O} ， γ2
{z: T 三
Izl 三 R , argz = 2叶 , γR = {z : Izl = R , O < argz < 2叶 , and γγ = {z : Izl = T, O < argz < 2汀叶} . Defi
且n
只扣川
f
(μωz
斗) = 4于俨三 Su
叩1ψ
朋
p凹
啊l阿 T 比吕 S咄
u咀曲毗田险归
臼阳叫叫
cii扣
削
m
刨州
en
川t址
n
叫ly 叩叫
sma址
all 削
annd 们
R i比吕 吕峭
u咀曲刷白胎归
臼阳叫叫
cii扣
削
m
刨州
en
川叫七
n
由
d
ly 阳1 町叫即
g悖e 吕so that all the poles of f (z) are
contained in the contour formed by 节， γ2 ，机， andγR- Then by Residue Theorem
I
f(z)dz = 2刑Res (f， - 1) = 2刑( - 1)α 11n( - 1) = - 2何与 (α 加Z
J ìl + ìR-ì2- ìγ
We note
11R f(Z)dZI S;
as R → ∞，
11 2 11:叫:27EZO)Rezoz牛
1 1 川z l S; 11 211: (r- e
as T • 0, and
l 川z
lR 川
l+x
l Rxa -le 2 (a 一叫nx+ 如
圳
M
i)
l+x
)ZIL一
11tlv-1
主-JEL币
+ e 2 ( a-1 ) 1I: i 2 7r i
一一_rI
l+x
L l+x
51
It's not hard to show Jooo ~:~1' dx = 汀 csc(cm). 80 by letting r •
1
h-2
2-
-14
一十
Z 一
仕
∞
fllo
o and R →风啊 have
-2 汀2ε( o: -l) 7r i + e2( 0: -1 ) 7r i2汀 z . 汀 csc(α汀)
1 - e 2 ( α1 )7r i
2汀 2e 0 7r i +
2汀2 csc(α汀 )e 2 0: 7r i i
1ε2α们
2汀 2 1 十 四 c(απ)i [COs(α汀) 十 1 日in(α汀) ]
e 一 α7τ '，.
-
eαπz
-
2 COS(α7r)
" sin 2 (α汀
,
口
r Function
8
1. (1)
Proof. (2n)!! = (2n) . (2η2) … 2 = 2n . n! = 2n r(n + 1) .
口
(2)
pmofln1)!! JH(2zzr♂2321 = 冉 = 旦时7
口
(3)
Proof. r(n 十 U 十 1) = (n 十 v) r( n 十 v) = … = (n 十 v)(η1 十 v) … (1 +v)r(l + 1小 80 (1 十 v)(2 十 v) … (n 十 v) =
马草草ii.
r(v+ 川
口
(4)
Proof.
II[l(l + 1) - v(v + 1)]
II[(l - v)(l + v 十 1) ]
1= 0
II (l - v). II (l 十 U 十 1)
1=0
1=0
F (n-v+1)
r( - v)
r(n+υ+ 2)
r(v 十 1)
r(η+υ+ 2)r(n -
TF
v + 1)
旦旦旦 r(η+υ +2)r(n-v+1).
7r
口
2. (1)
Proof. We 且rst assumeα ε (0 ， 1) . Let CR = {z: Izl = R ， O 三 argz 三~ } ， Cr = {z : Izl = 飞 O 三 argz 三~ } ，
and assume R > T. Then by Residue Theorem
JR 二dz+ ι 二dz+ 汇二dz 十二二由 =0
52
Then it's easy to see
|儿二dzl ~ Iho 币川牛 fr1α e- T S 业 271α → 0
as r • 0, and
半 R 俨 ide l ~ 1~ R 1 - a e 川 e ~
1 J1 旦出 1 ~ 1I 1~
Jo
Jo
一
CR Z α
(Reilì ) α
as R → ∞ . 80 by letting T •
I ----=
1
2
R 1- a R-R子 de = 士 (1 - e- R ) • o
o and R → ∞， we have
f 二 dz = ft;d忡
This implies
f/α
x - a sin xdx = (sin 叶 i COS ~1f ) f(l 一 α)
Z αcos xdx + i I
\2
f 。门
Compare and equal the real and imaginary parts of the two sides , we get OO x 一 asinxdx = f(l 一 α) cos 乍
and JoOO x - a cos xdx = f(l 一 α) sin %1f. For αε(1 ， 2) ， we note
Jo
1
00
;r + (α
叫飞 αsi 叫= (α1)
x-( α 灿 xdx = x-( α1) 剖 nx l
Jo
80 forαε (1 ， 2) ， oo x- a sinxdx = 兰I f (l 创me
a
Jooo x 一白日 inxdx is the 日阳阳
(α
1)) sin 号与= f(l
1
00
x
α 川Z
α) cos 号1f. That 盹 the fo口nula for
巾
巾
创1口
阳e
ch cannot 悦
a
m
w hi抽
b eo
bt阳
x= 汀汇， 咄
Joooο 吧主 d
(2)
Proof. Let CR = {z : Iz l = R ， O 三 argz 三。 }， Cγ = {z : Iz l = r ， O 三 argz 三。} and assume R > r. Then by
Residue Theorem
fzfIezdHLRZα1ε Zdz + /;川 α 1e- xeie d 川 十 i. z α 1 e - Z dz = 0
Note cos fun
| LRZO1e 切 1 =
11 (Re ) e
iç α 1 -Re jÇ d 叫三
8
1 8 Rα 1e- R 叫 Rdç =
1 Rae8
川
h
耻
R C∞
叫
m
O田川
S叫叫E
On the interval [- "Í, "Í], cos ç = sin (ç + 号)三 3(5+ 号). 80
|止一 1 ε 叫三 1 Rae 一手叫ç= 车 [fR-fR(1+ 号。) ]
8
8ince e E (- "Í , "Í) , 1 十 号。>。如 limR →∞丐二 [e- R - e- R 叶
l i， z 寸咋 1 8 r'-'e- R 吨三 er 臼→ 0
as r • O. 80 by letting r •
o and R →∞， we have
1 00xα 1e- x 由
rCX二
I
x a - 1 e i8 ( α l)e - xe ' 飞 i8 dx =
JO
x a - 1 e i8αε
xe ' θ dx
JO
roo
E仇
rCX二
I
I
X ,-, - l
e- x cos 8 [cos(x sin e) -
J(]
53
i sin(x sin e) ]dx .
Let 1 =
Jooo X"-l ε X cos {} co巾 sin e)dx and 11 = Jooo x
r(α) = (cos eα+ i sin eα)(1 -
c>- l
e- x
COS{}
sin(x sin e)dx . Then we have
i1I) = (1 cos eα+ 11 sineα) + i(1 sineα11 coseα) .
Equating the real and imaginary parts of the terms on both sides of the equation , we can get two equations
of 1 and 11. 801ving these two equations gives us
1 = r(α) cosαe ， 11 = r(α) sinα0
口
3. (1)
Proof. 8ince r(z + 1) = zr(z) , we have r'(z + 1) = r(z) 十 zr'(z). 80
中 (z
+ 1)
r'(z + 1)
r (z) + zr'(z
r( z + 1)
zr(z)
~f\~
;~:
~ \~J
nf
\
\~J
1
+ 审 (z)
口
(2)
Proof. 业 (z + η) = Z+仨T 十 业 (z + n - 1) = Z+仨T 十 古仨2+ 业 (z + η
+ 审 (z).
f
2) = ... = Z+仨T 十 z仨2+ 00 .+
口
(3)
Proof. By r(z)r(l- z) = 亘古言 ， we have lnr(z) + lnr(l- z) = ln 7r -ln(sin 叫. Differentiating both sides ,
we have 审 (z) 审 (1 - z) = 击ff 汀 80 审 (1 - z) 审 (z) = 汀 cot 盯口
(4)
Proof. By the formula r(2z) = 2 2z - 1 汀 ~r(z)r(z +冬 we have
lnr(2z) = (2z - 1)ln2 -
1
2
;:;-ln 7r + lnr(z) 十 ln r( z + ;:;-)
Differentiati吨 both sides , we get 2 审 (2z) = 21n 十 审 (z) 十 审 (z 十 ~ ).
口
4. (1)
Proof. Use the substitution x = 2υ1 ， we get
五川
口
(2)
Proof. Let p 二 lfand q= 气Q: Then p , q > 0 , p + q = 1, and
f 川de =
1 s川 cos 。刷。= 1 si户 1 肚os2q-1 贮
2
2
口
5. (1)
54
f. We note
PTOO
三 i(tI 一元1) =兰 Cn~l 一丰-丰+舌1)
·一一
口1η(4η2 一 1)
言~(击;+击)
By formula (8.27) , we conclude 艺汇 1 丽击17 = ;问( - ~) - 2 \[1 (0) 十 业(~) ] . By the formula 审 (2z) =
抖 (z) + 抖 (z + ~) + ln2 , we get 业 (0) = 审(~) + 21n2 =一γ ， where γ is the Euler constant. This implies
、一 1 一一
= 1[(γ
21n2 十 2) - 2(γ) + ( γ
1)
2
2 ln 2) 1 = 2 ln 2 - 1.
乞叫时
口
(2)
Proof. Using Mathematica command Apart[lj(z^ 2 - a^ 2)^ 2 , we have
、
2 、
n←∞ (n 2 十 1 )2三~ (n 2 + 1)2
一 一
1=2 、
1-
- ~ l 4(z - i )2
4(z - i)
一一一一二一 -
2 亿 lz - i '
(z - i)2
z + i' (z + i)2J
一
~
+
4(z 十 i)2 '
.1
4(z + i)J
-
By forml由 (8.29c) ，
V 一 1 一 =- 1.(- 1)-W(-t) - V(-t) 一 烟。 - \[I '(i)] - 1.
nfT∞ (n 2 十 1 )2
By the forml由审 (z)
审( - z) = - ~ - Jr cot 7叫 we have 中( -i)
审 (i) =士们ot( 刑) = i(-l
And by \[I '(z) + \[1'( - z) = 去Jr2 csc 2(Jr z) , we have \[1' (i) + \[1' ( - i)
Therefore
汀 coth 汀) .
- 1 - Jr 2 cs巾) = - 1
忐言
导
-L=lh+
汀 cot 盯 +1+ 」U1 = ZcothH
」L
L-∞ (η2 + 1)2
2\sinh~π)
2
' 2 sinh~ 汀
口
9
Laplace Tr ansform
9.1
Exer cÌ se in the text
9. 1.
f.
PTOO
叫f川} = 1= e 呐 - T川 一 材 = ε 才∞ ε p(t-T) f(t - T川 一 材 = ε pT F(p)
叩(αt)} = feptfM)dt = :fe 叫
ι川叶 1= e-(p 叩)dt = F(p - po)
口
55
9.2.
Proof.
1
1
ι{1 00 州， 叫 = 1 00 ε pt 00 f(t， T川 = 00 dT 1∞ εm 中 1 00F 川7
Define g(t) = JtOO 早 dT ， then g' (t) = 一 半 By Property 4， ι {g'(t)} = pι {g(t)}-g(O). 80 - Jooo e- pt f~t) dt =
μ{g(t)} - Jooo 阜 dT ， which implies
均收)} = 1l∞ ζ卢 f(t)dt = ! /∞ (e 叫f(t)dt = 1lp fce moatlq = 1·pF归)dq
Jo
P Jo
T
Jo
P Jo Jo
P Jo
口
9.2
Exer cÏ se at the end of chapter
1. (1)
、、自'''-
η -P
He1'e we have used Rep > 0 to conclude e-pttn+llü = O. By induction , we proved our claim
叶
十一
几
一­
一p
川
J/
BB 『，，
-t4 i'LU
一一
ιι
η
ρEU
i'LU
、、、
∞
∞0
ρU
a''
4
i'ι
i'LU
一
一=
一一
、LtJ
FJ飞 I、
十
ι
、 、BF/
一一
i'LU
/，
，， 飞\
∞
Pn叫 Fn(P) = ι{叫 = #T for p εC with Rep > O. To prove this , we work by induction. When n = 0 ,
this is just Example 9. 1. Àssume the formula is true for k = 0 , 1 , . .. , n. Then
filo
filo
1i
1
+14
n
P n+ ,
P
P
F +--4 PA
F p
+ vd
E
,
-9"
n
P
口
(2)
P叫~ F(p) = 平担 for p εC with Rep > O. Ind时，
ι{t CX } = (00 e 川t = 土"1 (00 e-Pt(pt)CXd(抖) = 丰 I e-tt(cx+l)-l 巾，
JO
JO
jJ
jJ
JL
where L is the radial straight line that goes from 0 to ∞， with angle arg p. By the extended definition of r
function (8.3) , we have F(p) = 平#(Rep >0)
口
(3)
Proof. The problem and its solution in the textbook do not match . 80 we calculate the Laplace t 1' ansform
both for eλt sinwt and eλt sinwt.
ι {e- Át sinωt} = /
e -pt e λt sinwtdt = /
e - (P+λ)t sinwtdt =
e-pteλt siωtdt = /
e一 (p一 λ)t siωtdt =
九九
ω
(p + λ )2 + ω2
whe1'e we 1'equi 1'e Rep> 一 λ ， and
ι{eλt sinωt} = /
九九
ω
(p 一 λ )2 + ω2
口
whe1'e we 1'equi 1'e Rep> λ.
(4)
Proof. By the formulaι{ 平} = Jpoo F(例， we have
ι{ 生叫 = JI ∞ ι{sinωt}句 = JI ∞ τ寸dq = 川创1X 1主 = 石 a1'ctan '=-- = a ctan
iω
汀
pω
1'
lτ
)
p
p
q~ 十 ω"ω4ωp
where we require Rep > O.
56
Remark 12.
matica.
The αbove result differs fTOm the textbook 's solution , but matches with the result οIfMathe­
口
(5)
P叫 Byappl间让le formulaι{ 平} = Jp= F(q)dq twice , we have
ι( 斗斗
fι( 斗斗 dq = 川才
1= f
ρ
俨
Yι
∞飞
町叩{札川
1卜川一寸叩
Cωosω
叫叫叫
ω叫叫州咐
t叶咐协}问忡
d价r
(~ι;L一舌百才)d价rd句q
1 = /=
l=J民 [lnN 一 l吁 :ln户巾:叫 十 ω2)] dq
J叫N [~叫
J万E工τ2
r∞
lim Nln 立￥工 +l
川
Jp
ωPτ
p2 + ω2
→∞
A
ω2
fτ dq + plnp - ~pln(p2 +ω2)
q"' 十旷
L
w arctan - - - m 一一一一τ一一一
p
2
p '2
口
ρ
。
( )
f 川 (1= 写~dT) dt
fi--t
JUT
ltl/
\飞
By integration-by-parts formula , we have
p
/'''』"'飞飞、
俨
r'''ο'
、，，
-斗
m
EEt IEEJ
、，
/'t、E
J
BEa
E、
S
ι
dT
町
一7
> O. Then
ftft
町
-T
Proof. 认1e require p satisfy Rep
JU
~ [e- 1= ~守主dT[ + !;= e- pt 平 dt]
Pt
~ !;= 川 e- PT ) 平iT
80 by the forml山 Jo= F(p)dp = Jo= 午战， we have
，
α
7
、l飞rlJ
T 一
C 一
‘、
ω-7
∞
〈 lt
ι
fl
ft''1't
:f(1εμ) 竿dt = jfι{(1 - eμ)ωst}dq
~1a=[古 ι川 ωst}] dq
:f[tI 一 (pfJLl 句
JU 1a [~dln(q2 + 叶仙(川川)
N
57
]
Remark 13. 扩 ωeα:pply the result 旷 Exercise 9.2, JtOO f~l dT 迦叫 F(q)dq， 加
th加e cal
αal叫αω
仰伽
仇
tμ
1ωOη1ωs 0吻 Oη
pOO
step . The function - J; 哼工 dT is called cosine integml function.
二
口
2.
filo
rJ
α
p
十
d
ρEu
E P ,tT
十
f'''ln+
一
-
，
α
ρEu
P rJ
∞艺叫
flIgo
-
一
一
FP
∞艺叫
Proof.
主 100: e- pt f(t)dt . e- apn = 10 e 叮(t)dt 主 (e 中己石 100: e- pt f(t)dt
0:
口
3. (1)
Proof. Isinwtl has period 5. Using reSl山 of Problem 2, we have
叫 I sinwtl} = 己=叶飞 μ 川
By applying integration-by-parts formula twice , we can easily verify
PA
-ω
2
一lT
ρ七 一
1
P
十一
ω 一
，
α
n
一一
QU
ω
ι{I sinwtl} =
ρU
fi--nυ
80
P
-
(1+ε 击 p)
τ
τ一一τ = τ一一τcoth ~.. .
1 - e- "'::; P p~ 十 ω~
p~ 十 ω~
2ω
口
(2)
α [i 1 has periodα80 by Problem 2, we have
Proof. f(t) = t
ι {t α[~]} = 己叶飞呼叫~]) dt = 己刊α e-pttdt = ιι石
口
4. (1)
Proof. By the formulaι{( - t)n f(t)} = [ι {f (t)}](n l ， we have
古J
一α
飞lLIJ
ι
rtz4
】 tt、
d
- - -l
pPT
W)-3ι1 { (卢y2 l } +αιl{(卢)'} - eαt7)(t)
[1 - e- at
(1 十 时 十 字)] 7) (t)
口
58
(2)
Pn时
口
(3)
ProoJ'f. We note
句- ~
1P + ，2.3 _
1 _ 十 13 ~
U~Uv (p2+p)( 4 p2_1) p+1
p_~
"v
I
1:- 1
~
. Therefore
p+ 言
I
{仙:AJ 一 寸 (1 十 :e飞 id川e一中
口
(4)
Proof. We note
p2 + ω2
一一一一一=一一一一
十 一一一一=一一2l~L
[ι{ e~ wt }]' 一一2 [ι{ε wt}]' = (ι{ - coshωt} )' .
(p2
ω2 )2
2(p
ω)2 ' 2(p + ω )2
Jj
80 by the formula ι {(-t)nf(t)} = [ι {j (t)W叫， we have
ι( 二 与专}=同一 c川
口
(5)
P叫 We note ι{1山}=千 80 by 山 formulaι{ 们 (s)ds} = F~) ， we have
ι1:-一-11{ιE
二弓守二二芸
;F: }←=JOF
户
t\盯与
l{s川
{υ8
P
口
户。
( )
Proof. We have shown in Problem 3(2) that
斗 α[~]} =卢 -7ιz
By the forml山 ι{( - t)n f(t)} = [ι {j (t)}] 叫
均}一 ι {t 一 α[~]} = ~ιF
t
η
'TL
，
一α
」
「III-IIII
-l
「III-IIII-IJ
ttj
59
〉
5. (3)
一一
一 巳
一一
p
、 tt
e
1·EA
(
p
叮
一
mv
ι
'i
、J
rEE、E
EE
EE
，
and
口
f. Denote by F(p) the Laplace transform of y(t) . By the Convolution Theorem (Theorem 9.1) , F(p) =
p乍 一 2F(p) p1乍 80 F(p) = lP音)2 and
PTOO
y(t)
= ι-1 { (p: 1)2 } =-aJ:- (卢)'}=αte- t
1
{
口
(4)
P叫 Denote by F(p) the Laplace transform of f(t) . Then F(p) + 2F(叫声=击 . Therefore
Fω) = /
Hence
9 (p2 +1 6 5 4
=一 一一一一 十 一一 十 一一 .
(p 十 月 2'
p - 2'p 十 1
f(t) = 6ι-1 { (卢)'}+仇
口
6. (1)
oo F(p)dp = Jooo 午战 ， we have
P叫 By the formula Jo
100~γMCOB川
1 ∞00 ι 作
~
l Iα + p+ c
加叫阳
)cCω
叫
O
b 十 P|
P
l( α+ p) 2
1 唱
b2
2
(b + p )2 + c2 J
+ c2
2 山歹士ZE·
Here the Laplace transform inside the integral is obtained by applying integration-by-parts formula twice to
integrals of the form oo e 叫 cos(cx)dx
口
Jo
μ
P叫 B
酌y 山 f阳
O旧r口
r川川
ml
I口n孔旧ula
叫
JoOO 盯
F邱附
(ωωp
叫)冲
d句p = Jooo 王平乎乎弘阳
与忧
lc
h
d耐
川
ita
川
t盯
arn吐归川
们川
d
t仕he
时
eforα
川川
r口
r口ml
I
f 气;冉
E兰E生坐出
L
主与勾
与d白Z户
←= 1 00ι刊(斗坐斗}仆
φ=
Note
f(j-tF)dq=JUln7 一 jln 生二) = ~ln 号旦
and
口口门
f
i←←ln 平牛 =→~ (p
f t百沪d:φp 叫 i |; 。 j
=b
n
=
b
口
8. (1)
60
f.
PTOO
兰 ti = 主( 叫∞ e-te 叫 =ft t 主(-e 挝)ndt=fIf二二百dt
-U υ
lT
2 一
fl110
日
1
fl110
一
句
叫
叫 一+
一-n
∞艺叫
Substituting e- t for y , we have
vdNud
ifhJJJL-~1)2 + ~] dy
~ {ln川 i 咐 ~r 十 ;l + dmm(7))| =o
~ (1中 2V3 arctan 去)
~(山先)
口
(2)
Proof.
1
言时; = 呈阳 1 00e 一(韧 +l)tdt = 00 e古(JYdt=f ￡与dt
Substituting e- t for y , we have
ι ( - l)n
(dy
1
t=04n+1jo1+υ4
r Iν - V2
y 十 v2
I.J
y2 + V2y + 1 I 叮
1
V2y + 1
2 V2 λI y2 -
一丰{ ~ 1n[(也 一 1川 - arctan(巾 一 叶川y + 1川 - arctan(巾 + 1)}[
一土
1n 立卫生 一 --arctan(V2
2V2 ~I ~
2 ------\ . - - 1)
-/ - arctan(
-- ----\V2
. - + 1) I~
2 十
v2
~
,
~
- /
252[jln(3 川) - ~]
白21n(v2 + 1) +汀l
口
(3)
d
PA--
1-P
h''''''''HU
Mu--T
，
-Lr ,
一ν
，
61
h'''''''"U
d,
α
∞艺叫
filo
α
十
内
-P
'''''''"U
n-q
I--T
∞艺叫
/γ-n
∞艺叫
Proof. Suppose p , q ε 1'1 and q 三 p， we have (substituteet tor y)
Ud
川剧y 川rify (we have pr叫 Jo1 命 = i (ln2 +去)川川(川d note 命=一拮甘甜牛节)
r ln 2
IlL(l叮
1 + yP υl
一 ln2 十元
p = q = 1;
p = 3 , q = 1;
p = 3, q = 2.
Therefore
立
川口
去一1)ntrt页 一 古l
气?
1
i 兰 tjf 十 ;主 tji 一 ;唁兰百占
i七卜←ln山
1
νln2 一 bC=
;〉〉
口
6 Fu nction
10
1. Let ψ (x) be any test function tl川 satisfies certain regularity conditions
(1)
f. f二内 )c5 ( - x)dx = f二 ψ ( - x) 的 )dx = ψ(0) = J二内)的 )dx. 80 c5( - x) = c5 (x).
PTOO
口
(2)
P叫 f二 ψ (x) . x c5 (x)dx = f工 (Xip(x)) 的 )dx = 0 ψ(0) = O. 80 x c5 (x) = O.
口
(3)
Proof. f二仰) . f(x) 州市 =ψ (O)f(O) = J工 ψ (x) . f(O) 的 )dx. 80 f(x) c5 (x) = f(O) c5 (x).
口
(4)
Proof.
1:
ip(X) . x c5 '(x)dx
汇川 )c5'(x)dx = Xip(x川
一ψ(0) = -
1:
ip川)dx
80 x c5'(x) = - c5 (x)
口
(5)
P叫'. J工 ψ (x) c5(ω)dx = J工 ψ(ν/α) c5(列宁 = 扫 (0). 80 c5( ax) = ~c5 (x)
户。
( )
62
口
PTOOf.
λ 忡)i5(X2 - a2)dx +
汇 f(X)i5(X2
1 f(圳工2
00
_
a2)dx
J~a f(- Jy + a 2 川 - Jy + a 2 ) 十 汇 f(日)的)dJy + a2
∞ f( 口) y)dy
口丁 的)叫山歹们
汇气日)
\/y 十 α4
川、 y + α2
打(α忡忡) ]
where the last equality is due to the fact 0 ε( α2 ∞). Therefore , i5 (x2 - a 2)
=
2~ [i5 (x
α) + i5 (x + α) ].口
Remark 14. More genemlly， 阴 h αve the following 'useful r 臼 ult
Proposition 1. S叩pose ψ (x) is α continuously differentiαble function αηid the equαtion ψ (x) = 0 has βη巾 ly
many TOOtS (Xk) f:= l ωith ψ (Xk) 并 O. Then
￡L d(z
i5 [<p(X)] = 、
zU
已|旷 (Xk) I
Proof. For each k ε {1γ.. , N} , we prove i5 (ψ (X)) is C k i5 (x - Xk) for some constant C k in a neighborhood of
Xk. lndeed , we can find ε> 0 such that Xlγ.. , Xk-l , Xk+l , … , XN éj [盯了 一 ε ， Xk+ ε]. Clearly i5 (ψ (Xk)) = ∞­
Furthermore , we have by the cha吨e-of-variable fo口nula (y : = ψ (X) )
1~~: ε i5(ψ(X忡 = f['P ("k 一吨 ψ ("k+εJ I 旷 刮到dy
==
(ψ l(υ)) I
I 旷 (ψ l(ψ (xk)))1
I 旷 (Xk) I
80 in a su刷e臼
侃
m
I丑1此时tly small neigl阳山od of Xk , i5 (ip(x)) = 苦而
口
2. (1)
丹刚
P
ro
阳
叫
Oω
O
By 七d
山
he continuity property of g(x; t) at X = t (formula (10.36a) and formula (10.36b)) , we have
r h d t )[
kCl(t)ekt - kC2(t) ε kt = 1.
801ving this 叩ation ， we get Cl (t) = 孚 and C2(t) =
have
去 Therefor飞 combining with formula (10.39) , we
川)恒=→[卡
[Cl
川
1川((t)
Remark 15. The α bove r时ult difJers fTOm the αnsω er of the textbook. But α ccording to the textbook's α ns ωo
to Exercise Problem 3(2) of this cf叩 ter， we see the correct αnsωer is indeed sinh k(x - t) η (X - t)
î
口
(2)
Proof. 酌
B yExe
叫
m
对
凹
e
rcis日 ePro
时州
ble
臼
err
川
I
i5 (X 一 t功) (x > 功t) is Cl (t) 叫 (X) + 句 (t) ω 2(X) where
￡工
r(3/4
王三
(X\ 4n
r(5/4( 叭创刊
(z)= 〉()，叫 (X) = 、(~ )
=odr(η 十 3/4)
\2}
' ~"'\~J
63
~ n!f(η 十 5/4)\2}
By the continuity property of g(x; t) at x = t (formula (10.36a) and formula (10.36b)) , we have
f10)ω1 (t) + C2 (t )W2 (t) = 0
Cl (t) 叫 (t) + C2 (t) 叫 (t) = 1
| 叫 (x)
叫 (x) 1
Usi吨 the hint that I ~.~)::( ~.:)::( I
|ω
叫~ (x
叫)叫 (x
叫) 1
~ ， we can solve the above equations to get Cl(t)
-
-2ω2 例
(tt功) 汩
a口
n1
2'
C2(t) = 2叫 (t). Therefore , combining with formula (10.39) , we have
g(x; t) = 2[ω2(X) ω1 (t)
ω1 (x) ω2 (t)] η (x -
Remark 16. To see ωhy the hint is true , we note [Wl(X)W;(x)
叫 (x) . x 2 ω2(X) - x 2 ω 1(X)W2(X)
叫 (0)ω2(0) = 1/2.
O.
t) .
叫忡忡2(X)]' = ω l(X)W~(X)
Tl川、efore， 叫 (x) 叫 (x) 一叫 (x )W2 (x)
叫， (x) 叫 (x) =
const
叫 (0) 叫 (0) 一
口
(3)
Proof. By Exercise Problem 2(4) of Chapter 6 , the homogenous equation
[川+泸)it 十 2(1 + 2 工)二 十 2]g(x; t) = 。
has solution Cl (t) 叫 (x) + C2(t) ω 2(X) ， where
1 寸 'ω 2(X) = 一 x τ
ωl(X) =一
l 十 x 十 x
1 十 x 十 x
"，J'
"，J
To use the continuity conditions of the Green's function at x = t , we note
l
(1 +x 十 22); 十 2(1 + 2x) 一十
21 g(x;t) = 工
d
d 2 [(1 + x + x 2)g(x; t)] .
[f
80 by the condition 吗~;t) I
= 0 and integrating both sides of the 叫 uatio 民 we have
I x<t
山
-1
dx
[(1 + x + 泸 )g(x; t)]li:+:g = 1,
乒
ax
i.e.
(1 十 2t) [ C l(t) 叫 (t) +
川七
阳
山
U
ir川
nl
C2 (t) ω2 (t)] 十 (1 十 t + t 2)[Cl (t ) 叫 (t) 十 C2(t) 叫 (t)] = 1. And by the cωon
t叶圳)川Ix < t = 0 , we have Cl(t) 叫 (t) 十 C2(t) 叫 (t) = O. Combined , we have the system of
g(x
叫;汁t叶叫) at七 x = t and g(x
叫;汁叫
equations
臼(川) = 0
f10)W10)+
Cl (t) 叫 (t) + C2(t) 叫 (t) = τ t+t2
|叫 (t)
Let DO)=tu;(t)
叫 (t) 1
叫住 ) 1 . Then it 's 叫 to 附 D(t) = 四拮 Then
-EEEEEEE
」EEE
一一
「lllIll-lll」
-9-
「IIIll-III-J
一
+
--A
01 一忏
-i
「lllIll-lll」
' ι
?
1'/
、、
9"/i 飞
?ι
'
/tt \
、
「lllIll-lllJ
一
ω1
ω
，
?ι
宁iuJ' 'h 飞
，
(rI
、、，，
/
'i
叫 ω
J
D
ιι
一-
-q
1EEEEEEE
」EEE
一
+
，
+ι
-1i
一
抖耳
O1i
币i
ιι
「lllIll-lllJ
­
i'七
ZE
斗一材
凹川wlh/
、/\/
，
+ι
ωωη
))T-r1(
tMVZ
2r2
=
ιι
u
1||」
ιι
、、 BF /
3. (1)
/川/叭一­
「lllIll-」
lll
口口
O QU Z
pup
.
-1
'Tb
口
64
f. By Example 10.5, the solution to the homogeneous equation
PTOO
( 豆气结毕毕
3步萨轩铲告岳驴
μi
兰
f且i忏川叩叫+叫耐咐
』川川
且川
k川2亏内印川
9剑讪切协
仰如
仙
(μ
Z川;
州州;汁t功)=恒= 0，吗且 | z=。 =0
is g(x ; t) =
í sin k(x - t) η (x - t). By forml由 (10.60) ，
fz
Ig(x;t)l
州=儿 g(x; t)f(刷一 IA 丁厂 -Bg(川)J 问
fiMIlkht)f(M [ Amkht)?shIkht)]时
i f dIIkh 一 t)f(t)dt + A 叫x+ 手inkx
口
(2)
Proof. By Exercise Problem 2(1) of this chapter , the homogeneous equation
(旦在i!l 川) = o(x - t) , x , t > 0
州;t) = 0， 吗ili| J O
has solution g(x ;t) =
í sinh k(x - t) η (x - t) . By formula (10 .60) , we have
(X~/~~ . ~\ J: /~\J~
以帅州￡叫)恒= 儿 g
仰如(怡阳
叫;圳
Z
t功旷呐)川
f
I Adg(X;t)
TL(~~. ~J
A__ _ LL~~
B _ :~~ LL~~
1 (X
口
(3)
Proof. By Exercise Problem 2(2) of this chapter , the Green's function is
g(x ;t) = 2[ω2(X) 叫 (t) 一 叫 (x) ω2 (t)] η (x - t) ,
whereωl(X) = L二od冉写 (~)4n andω2(X) = 艺汇ot陆( ~) 4n+ 1. Define
| 叫 (t)
ω 2(t) 1 n (~ ~\ _ 1ω1 (t)
ω2 (t) 1
1(tJ) = |
|?
D2(tJ)=
||
| 叫 (x)
ω2(x) l ' ~~\"' ~ !-I 叫 (x)
叫 (x) 1
Then g(x; t) = 2Dl(t, x) η (x - t) and by formula (10.60)
y(x)
l
X
g(x; t)f(t)dt
- :1\;"
Bg(x ; t) to
t)f(t)dt -- l A 句:;t)l
21
l X 叫1
Dl 川 (帧
ρ
" j
1
A
-
阳制川
(Z叫μ巾圳)川川州
川
ω叫咱;川(创0) 叫叫川川(何例
Z叫叫叫)川川
ω叫i红
州 (仲 B 阳制(川O叫) 叫叫川川
例
(μ
Z叫川川)川川
ω叫2扎叫(叫
X
〉
D1 川
21ρz
Remark 17. This result difJers from the αηsω er of the textbook. Check.
口
4. (1)
65
PTOOf. By Exercise Problem 2(1) of this chapter , the general solution to the equation
出_ k 2 ] 川) = b(x - t)
is
t sinh k(x - t) η (x - t) + C(t)e kx 十 D(t) ε kx. Pl吨gi吨 this formula into the boundary conditions , we get
(俨叫
ω
C(t)均川)
t S剖inhk(口1 - t) 十 C(t) 巳μk + D(t)诈ε k = O.
801ving it gives us D(t) = 去坐拮pi and C(t)=
去出击FL Thmfore
g(x ;t) = 7 sinh k(x - t) η (x - t) -
t)
\
sinh kx
~
k sinh k
n hk(l -
~---J--~:
- /
L
口
(2)
Proof. By Exercise Problem 2(2) of this chapter , the general solution to the equation
[￡ - z2l 川 ) = 州b(归Z
臼 2[忡
ω2(归
叫) ω
x
叫1川(位仰
t叶)
叫1川(x
ω
叫) ω 2纣( 叶t训)川]77η (归
x - 叶t) + C(t
例
t均) ω
叫1川(归例
叫) + D(t)1ωj)归
Z
)2 例
(tt均)， w here
叫 (x) = ♀!;14)(UHJ2(Z) = ♀414)(iY十1
Plugging this formula into the boundary conditions , we get (note W1(0) = 1 and W2(0) = 0)
?(t)=0
2[w2(1) 叫 (t) 一叫 (1) 叫 (t)] + C(t) ω1(1) + D(t) ω2(1) = O.
801ving it gives us C(t) = 0 and D(t) =
毛j;于1 . 80
(x)
g(x ;t) = 一 一一-D1071) 十 2D 1 (t , x) η (x - t) ,
ω2(1)
|叫( t)
|叫 (x)
where D 1 (t , x) = [:lì:~
ω2 (t) I
"~Zì:~1
ω2(X) I
Remark 18. The α bove result differs fTOm the α nsω er in the textbook. But α ccording to the textbook's α ns ωo
to Exercise Problem 盯3) ， the α bove result is the correct one.
口
(3)
Proof. By Exercise Problem 2(3) of this chapter , the general solution to the quation
[(1+zd) ￡ +2(1+22) 二十 2] g(x; t) = b(x - t)
1臼吕 τι刊卅(x - t) + I击2 盯
+ τ且2 X . 阳
P lu吨咄
1皂g阻g酬
and D(t
例
t叶) =- (l - t)/
μ1. 80
- t (1 - t)x
g(x; t) = 一一一一τη
(x - t) 一
。
l 十 工 十 X2' 1\~
VJ
1(1 + x + x2)'
66
Remark 19. The textbook's α nsω er zsωmng， αs seen eαsily by checking the boundα ry condition at x = 1.
口
5. (1)
PTOOf. By Example 10.7 , the solution to the equation
([』lg(叫) = Ó(x - t) (0 < x , t < 1)
g(O; t) = 0 , g(l ; t) = 0
is g(x ; t) = Î; sin k(x - t)η (x - t) - Î; 且是1k-t) sin kx. By formula (10.67) , we have
1
1
y(x)
g(x; t)f(t)dt + B cos
--- '".k(l
\ ~1
sin k
~ / sin kxl t1[
- A 1- cos k(x - t) + 叫10l kO
0|
--- 1.
I
,
- -
_,__ L( __
~\ r/~\ J~ sin kx [1 _,__ L
~\ r(~\ _1~ , n sin kx ,
-;- hI[ "; sin
k(x - t)f(t)dt 一一一~ I sin k (1 - t) f (t )dt + B ~~~一一
、 sin k ) 0
sin k
f1
A
Sink(l-x)
sin k
口
(2)
Pmof. By Exercise Problem 4(1) of this chapter , the solution to the equation
{l示
k 2 ] g(x ;t) = Ó(x - t)
(0 < 川< 1 , k > 0)
g(O; t) = 0 , g(l , t) = o.
is g(x ; t) =
Î; sinh k(x - t)η (x - t)
y(x)
咛去~~t) sinhkx. Using formula (10.6) , we have
K2 f1 .
cosh k(l - t) _' 1. 1
If z sinhk(x - t)f(t)dt - SInh
;~ ~~~~;" ~ I sinhk(l - t)f(t)dt 十 B
SInh K2|
1
k sinh k )0
)0
sinh k
It= l
11 _..
~\，+ cosh
k(1-t ) _'__L1J
A I___L
cosh k(
x - t)
-~~ ~~ . E ~sinhk x l
1
smnκJt = O
_.__ LL (_
~\j'/~\-，~ sinhkx [1 _'_ _LL/1 ~\j' li\_1i ， nsinhkx
ASinh k(x + 1)
I[X sinh
k(x - t)f(t)dt
- k; ~sinh
~~~~ ;" ~ I sinh k(l - t)f(t)dt + B -~~~~~'"，~ - A
) u - ----- \ --/
r
k )u
-/
sinh k
sinh k
J
Remark 20.
\ -
J
The αbo肥 result dijJers fTOm the textbook's αnsωer.
\ - /
•
-
Check.
口
(3)
Pmof. The solution to the equation
{j养
牛 (x; t) = Ó( x - t) (0 < x , t < 1)
g(O ; t) = 0 , g(l; t) = 0
67
1S 仰 ; t) =
y(x)
号高LD 1 (t ， 1) +皿1 (t , x) η (x - t). By formula (10.67)
1g川 (t)dt B 呼气=1 - A 呼气。
1叫加)灿ω
(υ问阳川
21、
拮i b 位
拮2 [w~(1)w2(1) 叫川
1
+
, 1)f(t)dt - B
1(t , x )f(t)dt
A刊( 击扭i 川
2叫
D 1川J
kkD
叭川时川川川
队队仰川
(tt仁巾
冽
，川川
Z叫x) 川 一 叮斟i 仆
f飞』川
川
ρμ
D乌圳圳圳川川
1川忱阳川
(t山队川
t仁川刊，叫)川削州川
1) 附…
)dd巾t
俨
1 x、〉、
Z叫斗)]
(怡川叫
叫5卅:二去拮扑
jt击萨协
;3非←
j←ω叫扣忖忖川
叫川川
2纣μ怡(Z叫川) 一 叫叫呻川叫
l
Remar k 2 1. The α bove result differs from the textbook's αnsω er. Check.
口
11
Complex Fu nctions in Mathematica
There are no exercise problems for this chapter.
12
Equations of Mathematical Physics
1.
Proof. For the points inside the bar , we can apply the partial differential equation (12 .10) . For the boundary
conditions , the end where x = 0 is 且xed ， so u(O , t) = 0; the end where x = 1 has no external stress , so by
Hooke's law (forn叫a (12.9)) 吗尹
…
Ix = l
= 0 (see formula (12 .36)) . For the initial conditions , the initial
velocity of every point on the bar is 0 , so 些铲ll .
= 0; at time 0 , Hooke's law implies E 旦LZL旦L = P = 号，
It = O
so ult=o = 忐 x . Combined , we have
(θ
'U
万t2
~2θ
瓦言 一
一
u
叫 /: = 0 = 0,
θtι
u It=
l+- n
!: ,
o = Êsx
θ 'U I
_ (ì
θ:C Ix = l 一 叫
一 。
Dt It= o -
v
口
2.
Proof. Let D be the rate of diffusion. Then from formula (12.20) , we conclude
旦旦 =D 寸2 u+ au
θt
口
3.
Proof. By Fourier's law (formula (12.15)) , we have
θu I
ql
θu I
θ创汇 = 0
k' θX Ix=l
q2
k'
口
68
4.
Proof. We choose polar coordinate and place the OI地in of the coordinate at the center of the ball , with axis
pointing to the sun. Then by formula (12 .4 1) , the boundary conditions are
ru
EulTO H (苦叫 j;<0<
三。巧，汀 ，
否;:+k "一 =Euo=io，
一
where H is the proportional constant in Newton's law of cooling.
13
General Solutions of Linear PDE
13.1
Exer cÏ se in the text
口
13.1.
PTOOf. 8叩pose the solution has the form u(x , y) = g(υ )qy(y 十 αx). Then
(Dx 一 αDν - ß)u = g(υ )(Dx 一 αDy)qy(υ + αx) 十 份 (y + αx)( 一 α Dy - ß)g(υ) .
8ince (D";
α Dy)qy(y + αx) = 0, we obtain the ODE for g(υ):
g(υ)=e- 3 1J So 叫x ， y) = e一 星 。(υ+αx) .
13.2
(αDν+ 应 )g(υ)
= 0, which has a solution
口
Exercise at the end of chapter
1. (1)
Proof. The auxiliary equation isα2_2α- 3 = 0 , which has roots 3 and -1. 80 the general solution has the
form of f(3x + y) + g(x - y) , where f and g are two independent C 2 (twice differentiable) functions
口
(2)
Proof. The auxiliary equation is α2 - 2α 十 2 = 0, which has roots 1 土 ι80 the general solution has the form
of f(x + υ + ix) 十 g(x 十 ν - ix) , where f and g are two independent C2 functions.
口
(3)
Proof. The auxiliary equation isα2 一 α= O. 80 the general solution has the form of f(y) + g(ν+ 叫， where
f and g are two independent C 2 functions.
口
(4)
Proof. We consider the PDE for ru(t , r} The original PDE gives us D;u = 兰 (2rD，u 十 r 2 D~u) ， which is
equivalent to
[D; - c2 D;](ru) = rD;u - 2c 2 Dru - c2 rD;u = 0
80 the auxiliary equation for ru( t , r) isα2 _ c 2 = 0 , which has roots 土 c. 80 ru( t , r) has the general form of
f(r + ct) + g(r - ct). Ther吐'ore u(t , r) has the general form of ~[f (T + ct) + g(r - ct)]
口
(5)
P 叫 The 创a川川…
山
u应∞
u
xi
对i山
X
l且
山
ia
缸
a
阳ry
川
叫
e
qu
川
a
叫灿灿
ω
lu七
Ui归
阳
川
m
ω
创I
O
f and g are two independent C 2 func时tior丑1日.
口
户。
( )
69
PTOOf. The auxiliary equation isα4 _ 1 = 0 , which has roots 土飞土1. 80 the general solution has the form
口
of 白 (υ 十 工) + 归 (υ - x) + 向 (υ + ix) 十 归 (υ - ix) .
2. (1)
Proof. The general solution to the homogeneous equation
θ 2U
θ 2U
十一一言 = 0
. 8y~
θx
has the form of f(x + 叩) + g(x - iy) where f and 9 are linearly independent (formula (13.11)). To find a
special solution , we note
儿立
/η2 飞 nl
T3们
T4
百可 (x 2 + xy) = 主 |χ(-1)" t 副 I (x 2 + xy) = 古 (x 2 + xy) = 元+亏
X
In=ü \
Xj
80 the general solution to the original equation has the form of
~4
~3 …
f(x + iy) + g(x - iυ)+ 二 十 二三.
12
6
口
(2)
Proof. The general solution to the homogeneous equation
θ2U
θ 2U
θzθυ2
has the form of f(x + υ) + g(x - y) where f and 9 are linearly independent (formula (13.11)). To find a
special solution , we note
t可xy 叫=击[主 (2)](ZU 叫=古xy ←宁三
80 the general solution to the original equation has the form of
加 y) + g(x - y) 十 ;13(U1)
Remark 22. The textbook's αnswer is f(x+ υ )+g(x-y)+i X3 (υ+1) ， ωhich cαn be easily verified as ωTOηg.
口
(3)
Proof. The auxiliary equation of the homogeneous equation
。 2U
θ 2U
θ 2U
一----;;--2~+ 一一才 =0
8x~
θzθy
8y~
isα2 - 2α 十 1 = O. 80 the general solution to the homogeneous equation has the form of 呻 (x + ν) + ψ (x + y) ，
where ø and ψare linearly independent functions. To find a special solution , note
12(z2+υ)
(D" - Dy)
击址[兰 (去剖
蚓
g
川
rη1仆
l
1 (, . 2ηU\2 .
;)'î~l + 可) (x" + y) =
2 ,\
1 I 2 .
80 the general solution to the original equation has the form of
~4
m3
m2川
呻(工 十 y) + ψ (x 十 y) 十二一 十 二一 十 二三 .
12
70
x4
.
x 2y . x 3
iJî (x" + Y + 万;1) = E + 7 十 丁F
3
2
Remark 23.
The αbove Tesult is diJJeTent fmm the textbook's answer二 Check.
口
3. (1)
PTOOf. Using the transformation x = e t and y = ♂， we have
x
28 2 u
θ2U
x2
θzθ y
_2θ2U
θU
θu
+ y 一一
θzθυ
τ - 2xυ 一一一一 +υ 一一可 + x -;:;,
v
8y2
Dt(Dt - 1) - 2DtDs + Ds(Ds - 1) + Dt + Ds
(Dt - Ds)2.
80 the general solution has the form
U(x ， υ )=t<Þ (t+s)+ ψ (t + s) = lnx <þ (lnx + lny) +ψ (ln x + ln y) =
ln x . f ( xy) + g ( xy ) ,
where <þ and ψ(or equivalently, f and g) are linearly independent functions.
口
(2)
Proof. It 's easy to see sin(xy) is a special solution to the inhomogeneous equation and f(x 十 y) + g(x - y) is
the generals solution to the correspondi吨 homogeneou日 equatio凡 where f and g are linearly independent
functions . 80 the general solution to the inhomogeneous eql川ion is f(x + υ) + g(x - y) + sin(xy)
口
4.
Proof. The key is to note that
￡ [(1;)22l 主 (1;)22
(1 - x / lI
卢 1- 2 (1 庐x 十' (l\ - ir)2;
-- / 二主
a \ - -- 信
|θ
__
ι [2二
(
8x~
;:, 2
8t~
~
;:， 2
、
18x 2 [(1 - x) 叫主声 [ (1 - x)u] r
Define v(x , t) = (1 - x)u(x , t) , we can get a new system of equations
(和阳)圭和叶。
vlt=o (1 - x) <þ( 叫，去 I t=o (1 - x) ψ (x).
=
=
By Example 13.9 , we conclude
rX+ αt
巾 ?hj [(l-z+ 削
80
rX+ αt
(l - Ç)功(Ç) dç.
一一 [(1 - x 十 时)制r - at) 十 (1 - x - at) 功 (x 十 α t) ] 十一~
叫 29 t) =-J
I
2α (1 - x) 儿时
2(1 - x)
也
Remark 24. The α bove Tesult is diffeTent fTOm the textbook's αnsweT. Check.
口
71
14
Separation of Variables
14.1
Exercise in the text
14. 1.
Proof. If we solve for T(t) directly, we can conclude the general solution to the equation T"(t) + λα2T(t) = 0
is T(t) = A 吼叫 V>，at) 十 B cos( V>，αt). Boundary conditions give A = B = O. 80 T(t) 三 O.
If we apply theory of ordinary differential equations , we note by Theorem 6.1 ,
(Tf(川凹) = 。
T(O) = 0, T'(O) = 0
has a unique solution in (0 ，∞)， which has to be O.
口
14.2.
Proof. Assume u is C 2 , then
(句:。
飞
θ U(1"t) I
θx
。 < x < l ,t > 0,
x
θ ui""t) I
。
~，
θx
= 0
叫时 = cþ(功， 37| 问 =ψ (x ),
t;::: 0 ,
0< x < l
implies
åu(x , t) I
å I I
å叫工 ， t) I
= 0 ，旷 (l) = lim .; (叫 = o) = lirn， ~~~~， V!I
8x I;c = O - , T \. I
X → lθx
→ oθx
I
J / ( 7\
4(0) = lim(u|t=o) = lim|
z → oθx
→Õ
and
叭。
We should extend cþ(x) and 圳工) in such a way that the extended functions are at least C 1 . 80
I cþ( - x) , - l 三 x 一'
~ 0 , ，T， I ...\
ψ ( - x) ，
由 (x) = <
I cþ(x) , 0 三 z 三 l ，
ψ (x) ，
φ (x) = ~
一l 三 Z 三 0，
0 三 z 三 l，
and then extendφ (x) and 审 (x) to (∞?∞) as periodic functions with period 2l.
口
14.3.
Proof. Assume u is C 2 , then similar to previous exercise problem ,
(给咱 = 0 ,
0 < x < l, t > 0
川) I叫=队 a.，去ifiL=l=O? 川
ul 同 = cþ(x) ， 告|时 =ψ (x) ，
0三z三l
implies
功 (0) = 0 , cþ'(l) = 0 ， ψ(0) = 0 ， ψ'(l) = O.
We should extend cþ(x) and ψ (x) in such a way that the extended functions are at least C 1 . Therefore , we
sho由1 且rst extend cþ(x) as
Icþ(2l - x) , l 三 x 三 2l ，
φ (x) = ~
I cþ(叫
o 三 Z 三 l;
then we extendφ (x) to [- 2l , 2l] so that it becomes an odd function; finally, we extend φ (x) to (∞?∞) as
a periodic function with period 4l. ψ (x) should be extended similarly.
口
72
14 .4.
Proof. If we take choice (1) , then the general solution to equation (14.21) will have the form
川=主 [Anexp { 斗~7rY} + 川P(ZLu)lmZ1时
Plug this ，fo
扣 rm
口
m
丑mla i时o (14.21c) , we get
FL …)叩门 = f(x)
2二二。斗士l 汀 [An exp {弓俨忖} - Bnexp { 一驾护付} 1sin 弓flm=O
Using the ortl时onality of the system (sin 弓Fm) 工。 over [0 , a] , we have
(AJB = 刊 f川守门dx
An exp {弓士与b} - Bn exp {一号fl 忖 }=o
Solving the equations gives us
n
J
~ oa f(x) sin 与土1 7rxdx
a J U
0 ，...，L.~ ,
, An
1+ 叫{一咛M} ?
exp { 一 坦土与 b} ~
α
Ja f(x) sin 弓flmdz
O
1+ 叫{一咛1 7rb}
H
The reSl山 is the same as choosing the following form for Y n (υ) :
2n + 1
,
2n + 1
川的 = Cn sinh 丁7 叼 + Dn cosh 丁7 叼But clearly choice (1) makes the result look messie r.
If we take choice (2) , the general solution to equation (14.21) will have the form
λ1. . , 2n + 1
, 2n + 1 "
, 1.
2n + 1
u(x , y) = ) : IAnsinh
叼 + Bn cosh
汀 (b - y) I sin ~7rX
~L
汩汩
2α
Plug this forml由 into (14.21c) , we get
(Unc咋付日中x= 州)
艺了。 An 与土l
汀 cosh 马土与
bsin 穹土1
7rX = 0
La
> W
> W
7ι = u
川= 0 and Bn = 叫:;平「伽 Th肌 choice (2) 叫es 山时o s川
14.5.
Proof. Note in equation (14.21) , the roles of x and y are symmetric , so the problem can be reduced to solving
the following two problems:
o<x< α ， 0 < y < b,
(H=o
ulx=o = 0 ， 告 1:(;=α = 0 ,
ulν=0 =仰) ，告 l厂
and
。三 υ 三 b ，
ψ (x ),
(在 十 步 = 0
o<x< α.
。 <x< α ， 0 < y < b,
u|z=o=f(u) ， 33|z=α = g(ν) ,
ulν=0 = 0，去 |υ=b = 0 ,
O 三 υ 三 b，
o<x< α.
These two problems can be solved via separation of variables. Then the solution to the original problem is
the sum of the respective solutions to the two new problems.
口
73
14.6.
Proof. Suppose v 忡 ， t) is a special solution to the problem
( 在 _ a 2 宗 = f(x , t)
。 < x < l ,t > 0
t > O.
ulx=o = 0 , Ulx=l = 0 ,
Let w(x , t) = X(x)T(t) be a solution to the homogeneous problem
(仇
万t 2
~2 否
仇。
x2 v,
0< x < l , t > 0
ulx=o = 0 ， 叫 =1 = 0 , t > O.
u
Then ω (x ， t) must have the form of 艺汇 1 (Cn sin 平时 + Dn cos 平时) sin 平 x. Let u(时) = v(x , t)+ ω (x ， t) ,
then the initial value condition becomes
(立1 川平r =
巾。) 十 仲)
江1 叫叫x =
旦去LELL。 +ψ (x)
Using the orthogonality of the eigenfunctions , we have
Cn
= 丰 1 1 (一鸣叫=。+帅))54矶 Dn = 可IM
口
14.7.
Proof. Yes , we can. See 314.6 for detailed discussion. Once we obtain a solution Ul (x , t) to equation (14.81)
and a solution U2(X , t) to equation (14.49) , u(x , t)
叫 (x ， t) + U2(X , t) will be a solution to the problem
u日der consideration.
口
14.8.
Proof. No , because we want {Xn(x)} to be complete. The boundary condition (14 .49b) or (14.74) guarantees
the self-adjointness of the differential operator associated with the eigenvalue problem satisfied by X川叫，
which implies the completeness of {X川 x)} . See Chapter 18 , the discussion after Example 18.6 as well as
Proposition 4.
口
14.9.
P叫~ v(立 ， t) = 一 乌?iiμ (t) + 圣 ν (t)
口
14.10.
P叫叫x ， t) = 毕 μ (t) + 和 (t)
口
14.11.
Proof. Step 1. Find a function v(工， t) satisfyi吨 the boundary value condition
v(x , t)1 户。 = μ(吟 ， v(x , t) Ix=l = ν (t).
Step 2. Find a solutionω (x ， t) to the problem
l 在一←)一[在一在l
ωIx= o = 0 ， ω Ix = l = 0 ,
ωIt= o =
cþ(x) - v(队。) , 智 It=o = ψ (x) 一 生铲1 1 问
74
。 < x < l ,t > 0
t>O
。 < x
< l.
This step can be further divided into the following sub-steps.
Step 2.1. Solve the following eigenvalue problem
{~俨俨
X川
川(归川
Z
X(仰0) = 0, X(l) = 0,
and obtain a system of orthogonal eigenfunctions {Xn(x)} 早 1 = {sin 平 X}~= l'
Step 2.2. Ex阳d w(x , t) and f(x , t) - [~:~ - a 2 在 l 配 cording to eigenf川tions {Xn(x)}:
(叫巾川
川叫
Z
t功叮)恒=2:汇二
川
υ
1♂
川(阳
Tn ( t
f州(怡工叽
川'工均t) - [综穿: αa 2 告剖卜
υ
19川川
gnn川(付t)X
几n(例
Z叫) ,
l 二 艺汇 州训
sin 平尹xd缸x. 凹
P lu叩
1
w咖
he盯
r盯
e叼gn叫
n(t付例t功)=斗?叫耳叫{卡扣
f贝州(怡2叽川川
， 功t) 一斗[丢综窑iz: 一 αd2 3￡窑纠:=刮号斗μ)
l川←
扫除吕剖
h
址1 equation for ω( 立 ， t) :
tia
艺 [可 (t) 十 λ n a2Tn(t)]Xn(x) = 三二 gn(t)Xn(x)
n=l
n= l
Usi 吨。 rthogonality of {Xr，( X)} 注 1 ，
T~' (t) + 入 n a2T，， (t)
如何). Note ω (x ， t) automatically
X ,, (l) = O. To make it further satisfy the initial value condition ,
we conclude
satis且 es the boundary conditions as Xn(O) =
响 need …川)一川) andψ (x) 一呼叫 t=O 配 cord…伊 nfu 日 ctions {几位)}
x ， O)
(仰)叫
1anXn(x)
ψ (x) 一 鸣叫 =02: ::=
汇
1 bnX (x) ,
=
,,
… t J~[Ø(x) v(川)]
wh
n=
have an equation for Tr,( t) :
sin nr xdx and bn
= 叫 [ψ (x) 一鸣叫 =J sin 平 xdx. Combined , we can
山 (t) gn(t)
(mt)
Tn(O) + 川 T~(O) bn
=
=
= α
Once we find Tn(t) , the solution to the original PDE can be written as 叫 x ， t) = v(x , t) 十 三二三 1 Tn(t) sin 平 x.
Step 2.3. There are many methods to solve the equation
山 (t) gn(t)
(mt)
Tn(O) + 川 T~(O) bn
=
=
= α
We apply method of Green's function to refresh our memory of Chapter 10. The Green's function corre
sponding to the above initial value problem is
(和 ( t; 川俨) =
o(t
g(t; s)1 ω =09 24ifi |=0.
Wψ
剧川叫日川叫
has
tl
a
川 ) 川川
It <s
…
0
t 三 O
rm 问问川(怡
m
S功州)汩日川
忡
ω
…m HfoO臼
川
叫
时n1此t创叫时im叫
时
ω
1且1扰肚耐
t甘町y 川川 =s川川叫
a旧m
n1叫d 吗平主丘斗i斗ι|lCL::
二+ 川e havee
叫川川川川
凶仙阳川
阳山
川r口川
口1
';<0
刊
V
B(s)cos 0Ças = 0
矶时
(45)S111
0Ça[A(s)
cos 0Ças - B(s) sin 0Ças] 1 ,
=
75
. By the
which implies
A(s) = 坐土巴 ， B(s) =
坐土坐
飞 λnα
飞 λnα
80
g( t; s)
7L[cosAZMIld;αt - sin VD山 cos Fn at] T/ (t - s)
L
λ nα
7Lmd山(t - s) T/ (t - s)
Lλ n α
Therefore by formula (10.60) (recall λn = (平 )2)
时
r
t ~(.L ~L U J~
I ~ dg(t; s)
~(.L ~J I
Jo g(t; s)gn(s)ds 一 |αn 丁7-M(们) J18=0
L
zizfsinht - s)gn(S)ds + 向 cos 平时 十 元 m 宁αt
Step 3. u(尘 ， t) = v(x , t) 十 三;二 1 Tn(t) sin 平 x is the solution to the original problem , where
Tn(t) = - "-
n 7r
b…I
Ir sin n 7r α (t - s)gn(s)ds + αncos -7
时 + ~ sin ~ at
t
n 7rα J()
n 7rαl
withαn = f J~ 忡忡) - v(x , 0)] sin nr x巾， b n = 叫 [ψ (x) 一呼叫=olsiII 平 xdx， and
岛， (片 f(f(XJ)[24lhn?xdz
口
14.2
Exer cÏ se at the end of chapter
Proof. The partial differential equation under consideration i日
α
(θ2δu
θ
ulx=o = 0 ,
ult=o =毒工?
θt 2
;[; 2
v
,
θ ，， 1
θx
1
。
,,, =1
-
v
θ ，， 1
n
θt
v
It=o -
,
The corresponding eigenvalue problem is
(川)+λX(x) = 。
X(O) = 0 , X'(l) = O.
The solution is Xn(x)
sin 旦斗苛铲1 汀门
7rX
rx (归
η 三 O
创) with 哺咿
ei 伊
gee创盯
n1
{X
兀η7μ)门}~二
~ 1 ， we have the coefficient
t
α2- . 一一~T"in~η
+ l_~~L
8Fl
(_l)n
.、… 一一一一一- 7r xax = 一一一一一一一一一一一一n -
l Jo ES ~
UH"
2l
…山
ES7r 2 (2η 十 1)2
80 Tn(t) satis且es the ODE
(俨俨勾
T~'(t川
t
Tn 仰
( 0) = αn ， T~(O) = O.
76
Therefore T" (t) = αn cos ,JY;,at and
(
王三
8Fl ι
-1) η
2η + 1
2η 十 1
咐 ， t) = 乞
、 Tn(t)Xn(x) = ESπ2 但
、 (21几 十 1)2 ~sin
--.-at
~~~
2l ..~X cos
~~~
2l
--.-1T
口
2.
Proof. The partial di丘'erential equation to be solved is
( 万t2
θ~2
ä 一
- u 否Z言，
0< x < l , t > 0
t ?: 0 ,
Ulx= D = ux = l = 0 ,
Ult= D =
%x 1{皑皑 c} + I~Jl - X)l{c <õ x <Õ I}' ~~ I 问 = 0 ,
=
This is a special case of Exercise Problem 14.11 , with f(x , t)
。 < x
< t.
0 ， μ (t) = ν (t)
=
0 ， ψ (x)
=0 , and cþ(x) =
( ~。
<iE<C Applying the formula we obtained for Exercise Problem 14.11 , we have u(x , t) =
\ (,
击 (l-x)
c 三 z 三 l
E二二 1αn COS 平 αt sin 平风 where
n n
汀
c 一 '''b
,
EG
一汀
M
-
归
2 一
n
一
nr
x
一
Therefore 巾工)=ft;FZ二1 和i吁 ccos 平时 sin
c
2
f''''IC
叶
h 一
+
立
m
n -l z
二
- x
e
h
f--'0
'HM
。" 一'···ι
,
EG
，
M
-
尘
α
n
巾曲
一一
α
A Z
V
9 年 一'···ι
ft''10
27···b
口
X
3.
一伊
OE
nυ
1
。
'
、
J
瓜。==
/EEEF BEt〈
归一创叫叫
Proof. The partial differential equation under consideration is
Plug T(t)X(x) into the di丘'erential equation , we get T'(t)X(t) - T(t)X"(t) = O. 80 the eigenvalue problem
associated with this PDE is
(rr(2) 十 圳工) = 0
X(O) = X( l) = 0
If λ = 0 , X 三 O队.If λ 并 0 , 山
t he ω
叫lu
叫七
Uωi
归
∞
O
nm
川
u吕炕t ha
la\飞V
恼
刊
re 山
t he 阳
fO
旧rm
缸
S
8uppos回eu
叫(μ
x ， 功t) = 汇
2二:
二
"= 1 7'，丑口(例
t功)X
兀叫n，(μZ
叫). Then1
( 江口(阳[盯阳阳川瓦叩旧(
1 [T;, t
王且i
艺二 1 7'，丑~(仰0)川
Xn 例
( Z叫) = b呜云严严
U sing orthogonality of {X" (x)} 二 l' we conclude
(10) 十 川) = 0
Tn(O) = αm
where
古 J~ x(l/ - ~;--"\(...
x)X,, (x)dx = 4b
十
J
; ~~[1 + ( - 1) 叶 Il = {OL
J~ X~(x)dx
n" 俨
l 示告
77
if n is even
if n is odd.
0v-n
if n is even
if n is odd.
κ
、
ρlv
8-3
λ
、
..
b-π
一-
EE
，，，E、
Z
rEEE
Therefore ,
Finally, we can conclude
(州) = 咛 -L e( 节)2κt sin 旦时
3 亿 (2n
+ 1)3
口
5.
P叫 Plug u(川) = X(叫 Y (y) into the differential 叫川叽 we get 毛咛 =
引f. 80 the associated
eigenvalue problem for Y(υ) is
Y气υ)+λY(υ)=0
Y'(O) = Y'(b) = O.
Ther
创陀r冒它
eforαr
陀冒
enon此巾
t
we have
(江 1 [X;: (工) 一 川川y) = 0
江1 Xn(O)凡 (y) = 句 ， 江1 Xn (α)Yn(y) = 叫3 (t)2 - 2 (t)3]
Usi吨 the orthogonality of {瓦 (y)} 二 l' we have
(xfh)
川) = 0
X n (α) =仇，
Xn(O) = α口'
where
if n is even
αn=fho凡(ωdu=0 b J d Uo [3 (扩- 2(t)刊(州= [0
J; Yn (y)dy
J; Y;(υ )dy
l 一 走告 if n is odd
,15>"
2
801ving the ODE for X叫 x) ， we get
Xn(x) = ~ 0
l
48u
~:~ r川
h
(rm) 4 sinh 工手 α
U'UU
b 自
if n is even
if n is odd.
Combined , we conclude
48Un ~
1
(川) =一 τ王三、一一一
sinh ~平i 7rX
2n + 1
υ
台 (2n + 1)4 sinh 咛1 7ra
口
6.
Proof. We apply the formula developed in Exercise Problem 14.11. In this problem's context , we have
cþ(x) = ψ (x) = 0 , v(x ,t) = 0 , and f(x ,t) = bx(l - x). 80
9n (t)
2flbz(l-min
+ 1] = ~ 08bl
Jo --- \- -; .--- 巳dx
l ---- = 旦与(_1)n+1
(n 7r )3
l\
78
-;
'-j
if n is even
2
百布
if n is odd.
This implies
r t n汀
4bZ 2 ,
4bZ 4 "
, ( \
T" (t) = .. ~ α.. )/0 sin
一
一一[
叶 1 +， 1]
= 一一一[
( -1) 叶 1 + 1] 飞/
(1 - cos ~ at 1
~ ...
Zα
- \(t-s)ds
~ -~
mf )3 l (\ -1)
~/
~J
( mf ) 5α2
v
/
(
Therefore
￡工
巾， t) = )
,
乞
mf
8bZ 4 ι
l
向2 恒。 (2n
T n (t) sin ~ x =
1
、
. 2n + 1
(_
'"
\
2n + 1 \
+ 1)5 sin ~7rX ( 1 - cos ~7rat )
V
/
Remark 25. The textbo仙 mω er has a 7r r川
口
7. (1)
f. Let u 怡， υ) = X(x)Y(y) . Then the original equation becomes
PTOO
川
UU
{~仁仁
川川川…川
X
例川川
(μ
Z叫圳明川…)厅川川
Y盯叫(ωυ
X(仰
O)Y( υω) = X(归例
叫)Y(ωUω)=0
α
X(x)Y(%) = X(x)Y( - %) = O.
80 we can consider the eigenvalue problem
(川) 十 川) = 0
X(O) = X(α) = 0 ,
which has the solution Xn(x) = sin 0Çx with 儿= (子 )2 (n ε 1':1). 8uppose u(x , y) = L二 1 Xn(X)Yn(υ) .
Then by the orthogonality of {Xn(X)} ~= l' we have
(可 (y) 一 川 = 击刮剖[(μι
←归(←
一-1
E瓦7年
凡， (←一 3妇) =}飞
7♀， (传%) = O.
To find the expression for }年 (υ) ， we note the Green 's function associated with }年 (υ) satis五四
ιι
、 、BF/
-
俨
AU
i'LU
) - ( NU
白。
、
、1/
飞
门3
-2
NU
(
Hb
斗川 =
-
、l/
--
门3·· J
If
i'LU
λ/
·?ιι
/t 飞
川ud- 7
、
-dQu
J川叫 b -2
引
一
，
户LUM
fll'
l、
l 飞
，，
..,
Therefore
jA(t)sinh 0Ç (y+%) ,
-% < y<t
lB(t)sinh 0Ç (y- %),
t < ν<j
g(υ ; t) = <
where the coefficient function A(t) and B(t) are determined by the continuity property of g(υ ; t) at y = t :
{~剧(川 ♂叩叫(收叫
t
♂~B(收例
t叶) cωO创
sh ♂二
川
(0t
i幻) - Y'X
兀;
江
A(仰
t叶) cωO州
sh y'X汇
二 (收t + %封) 斗
= 1..
801ving this equation gives
r(t) = 叫 巾灿而(t - %)
B(t) = 而 84JXEb mhdk+3)Therefore
g(υ ; t) =
Sinh 0Ç (t - %) sinh 0Ç (y + %)1{ _% < 川} 十 sinh 0Ç(t 十 j)mh ♂二 (y - %)1{ 叫什}
d二 sinh 0Çb
79
and
r
汇 ω; t)dt
L%
si~h v'X二(υ - 3)mhdk+E)dt+f mh v'X二(U + i)sMdk
v'X二 sinh
J:Ç"b ~..... v .." \ ~ , 2
人J:Ç"
sinh J:Ç" b
E)
sinhn 而(υ-i~
Iωsh 而(UA
- ILsinh 矶时
~h
- ~)l
sinh
'2
sinh
l- 一 ∞sh J:\:(Y
v
2/ J
v'x二b
l~~~..
V
- Jλ n
f
""'\û
v'x二b
~ ~~..
. . " , \û
Mnh v'X二b - sinh J:Ç" (y - ~) + sinh J:Ç"(y 叶)
λn sinh J:Ç"b
-sinh 矶山 sinh 平 cosh v'x二ν
λn si出 d二b
1
cosh J:Ç" y
λn
入n c创号
Hence
一叫
r∞
二 [(-1)n-1]I
凡 (y)
/'0
n7r l \
-
/
2
n
2 ""ch 旦主旦
|
+ 1]1 一工τ 一山…一 α 丰|
,
- J 1 (叫 2
(n 7r)2 ωsh 守|
、
、
d3
nn
dmn
皿
、
Combined , we have
一川。
EE
，，，E、
/EEE
..
λ∞
A
g(y;t)dt= 二[( _1)n+1
n
￡工
8α2 ￡工
乞
汀3 巾。 (2n + 1)3
1
u(X , t) = 、 Xn( X)}年 (ν) = 一一、一一一一 1 1 一
l-
ω日h 咛均 l
{ 二一
1
cosh 驾护汀 bJ
. 2η 十 1
sin 一一-7rX.
口
(2)
Proof. 8imilar to part (1) , we find the solution to the eigenvalue problem
as Xn(X) = sin J:Ç"x with 入口= (子) 2 (n εN). Expanding u (x , y) according to {X叫X)}~= l' we have
u(X ， y)
汇瓦 (y)Xn(叫，
where
3飞 (u) =
fJz2UXAz)d X
rαv
r
[(
2
..2
(a 2 ( - 1)口十
2α2 r f 1 \n 11[
一
I x2yX n(x)dx
+ 一:\3
_ l)n - 1] j~
α 儿
n\~J~~ =
- 2υ{
~ iJ l 一一一一一
n 7r
(n 川
'v
(\1.
..
…(
80 the eige盯alue problem associated with Yn(x) becomes
(「俨俨俨尸
附川叶(归例忡…
ηm
Z叫卜川叫
叶川)
一→λ川叮
瓦
4啃唱孚苛
『
ζι
4i
##::L:
E瓦?♀1( -~幻) =}凡
7♀1(ω3
妇) = O.
The Greer内 function associated with }年 (υ) is the same as that of part (1):
g(ν; t) =
mh J:Ç" (t - ~) sinh ♂二(y+~)l{_%< 川} + sinh J:Ç" (t + ~) sinh J:Ç" (y - ~ )l{t< 叫}
J:Ç" sinh J:Ç"b
80
Therefore
Yn(Y)
巳(咛主+苟言 [ι 中 t)dt
在(← 1)'叶1+ 击[问
Note
sinh y(Ç(y+~) r~L.: .. L
汇 tg(y;t)dt
/ - t sinh
~sinh y(Çb 九
I\
A,(t
( L
b
,
一 )dt+
UL
sinh y(Ç(y-~) [y
l
2
♂二 sinh y(Ç b ) _ ~
hinhd二(t 十一 )dt
sinh y(Ç(y+~) I b
y ..~L 1\(.. b\ , 1 . :..L 1\(.. bJ
|」-coshJ元
(U)+mtIJ元
(y
- ~)
仅二 sinh y(Ç b l2 J:\二
d二
2
λ n ~- ---\v
2 }JI +
V
sinh y(Ç (y - ~) I
♂二 sinh y(Çb
b
|
12 仅二
， ycosh y(Ç(y 叶
十
ρ;
b [sinh ♂二(υ+ ~) 十 sinh y(Ç (y - ~)1
,
• '11,
1 .:..L 1\(.. , b\ I
sinh A，(y 十 ) I
|
入口
y sinh( - b) 仅二
且2
2λn sinh y(Çbλ n sinh y(Çb
b sinh y(Çy
y
u
、
λ
、
Therefore
Yn (ν)
24( 俨1 十二τ[( _ l)n 川(土豆旦豆豆斗
mr l \ ~J
'(mr )2
~J
~J J \ 2 入 n sinh 平凡/
l\
出一
+LiJ 甲)
v2SHIll 42 /(m)2m九ZM2SInh
42 /
tz(υ-i 击最)
if n is odd
川 even
Hence
u(x ， υ)
2二 Y，，(ν )Xn(X)
n= l
2α4ι( _l)n (
, M V
HK1
b sinh 乎叭
一一一一一-1 1)一一一一一一一一一
mr
8α4 ￡1
α
7r
1
/
, (2n + 1)5γ
5~
1 sm 一一-X 十一一一
2sinh 去时
一一一一一一一一一
1 1)一
b sin 主TLTU\
2η 十 1
1 sm 一一一一一-7r X.
2 sinh 弓俨 πb)
α
Remark 26. The α bove solution differs from the textbook's α nswer by α sign. Check.
口
9.
Proof. We choose υ叫(归
叽时，叶)
Z
t) = cωO吕巧干 Z
川C∞
c创
oss 于 αωt and su
叩
lp
即
po
创吕eu
叫(归
凯，叫t) = 叫
x
v(归
叽州，叶t) + ω
Z
叫(μx
υ
凯
， 叶t). Then
equation1
(θ万t2
四
a 2é万;:(;2
t '!/ 一
= 0
u
|θ~
I l
wlx=o
= 0 ，万:r lx=
ω (X ， O) = 0 ，誓 |时 = sin 击 Z
Solving the associated eigenvalue problem
(川) 十 圳工) = 。
X(O) = O, X'(l) = 0
81
gives the 哺nfunctions {Xn(x)} = {sin~x} ， with 机= (呼叫 2 . SupposeWJ)=Z汇。 Tn(t)Xn(x) ，
then we must have
(丘。(可 (t) + 入的Xn(x) = 0
L~二。 Tn(O)Xn(x) = 0, 2二二ü T;,(O)Xn(x) = Xü(x)
Therefore Tn(t) 三 o for η 三 1and To(t) = ft sin 击时. Combined , we conclude u(x , t) = cos 于 xcos 干时十
三tm 击 xsin 击时.
口
10.
Proof. We choose 巾 ， t) = L于'Ae α2 时 + 'f Be β2κt and suppose 叫川) = v(x , t) + 叫x ， t) . Note v(O , t) =
Ae- a2 h: t and v(l , t) = Be β2 h: t , we have the followi吨 equation for ω (x ， t):
(Ult
铲卜一→
+κ
叫Ix = ü =ω I x = l = 0
ω
ω(川) =巾， 0) =毕A
严=仰)
Solving the eigenvalue problem
(川) 十 川) = 0
X(O) = X(l) = 0
gives the 鸣enfunctions {Xn (x)}早 l' where Xn(X) = sin v'Ã山 withλn = (平) 2. Expand f(川) according
to {X川 x)} ， we have f(x , t) = 艺 汇 1 gn(t)Xn(X) with
2
rl Il - x
ι
x _ ~ ra
02
._~ I
ÿ Jü l T A α 怡。川 7BPMPκ 艺 I Xn(x)dx
川
Note 卫 Xn(x) 伽=才tcosdh|L=o= 马午=止号与1 and
JIJJlz c叫:l 丁;;
r = 一 -一 = -" (-1) 附 1
xXn(x)dx = 一?产
Therefore
仇 (t)
=
fl|202MBPκε 一 β2 时 -A α2κε_a 2κt
7儿
IAακ e-a- h:txn(x) + ~ r-' ." ~
2"" Aα2ε
山 +2EBPe 内 ( _1)n+1
n 7r
n 7r
Expand cþ( x) according to {Xn (x)} 二 1 ， we have cþ(x) = 汇 汇 1αnXr，( X) where
rl II - A+~xl
A- B 1
I
Xn(x)dx =
L J()
向=;
一一 ( _ 1)n+1.
Iηπηπ
If we letω (x ， t) = 二二 1 Tn(t)Xn(X) , then
( 汇口阳
1 [盯阳阳巧阳阳(收川
t
2汇二二 1 'T，丑~(例
O川)X叫
叫)=2
艺二:
二
= 1α
向nXn叫(x
叫)
n (x
By orthogonality of {Xn(x)} , we have
(阳川卫 (t) = gn(t)
丑 (0) = αm
82
l
xXn(x) I dx
which implies
巳 川μ
1t 川
gn(
n川S(川
扑
丑时例
(tt叶
2κ|εα2κtεκλn t
|α29
饥汀|
一 α2κ+κλn
11
e β2κt
巳 κληn t I
+ BF2(1)n+19|
一 β2κ+κλ n
I
Combined , we can wr山 the solution 叫 x ， t) to the original problem as
[1
山 e一 κ 附
, 十
~ x Ae
Ae- 臼 κt
+ T
Be- ß2 "t]
+~
le 内产(平)2叮}
主 Be~'~ IAα 2 e
~ + Bß 2(_ 1)n+lL
, ___,?
1 + αJ sin
γ ~{剖
叫+
a2 ä
ß2
J
乞~
Lmr l
-0: 2κ+κ( 平 r
_ß2 κ+κ( 平 r
J
-J
Remark 27. If we choose v(川) = A乓击乌山 +B 吉普 ε 内少 then it's easy to verify ~~κgiE=O
So this choice of v(x , t) simultαneoωly homogenizes boundαry cond州 on αnd the differential equation， ωhich
mαkes the solution much eαszer.
口
15
Orthogonal Curvilinear Coordinates
15.1
Exercise in the text
15.1
Proof. Suppose 入m and λn are two distinct eigenvalues of the eige盯alue problem (15 .41). Let φm be an
eigenfunction associated withλm andφηan eigenfu日ction associated with λ n. Then
φ ~n(CÞ) + λmφm( 功) = 0 ， φ~( 功)+入nφ n(CÞ) = O.
Therefore
o =
2Wρ
如飞冗霄、
φ町吼阳;二山川川川
;μμ
[1
~ 忡例忡￠创咐)陆
(cþ) φn(CÞ
耐忡，(柏刷忡(忡俐肿￠的仲仰)川【
[阳队阳川叫
φ也轧幻
川
M
州川礼，(
n(CÞ
刷(忡俐忡￠的州)陆
cþ φ眨mμ
~n阳川
刷(忡例￠的)
川J
(CÞ 一 川)町的ll~W + (λm 一 儿) 1 2霄川)驯的"
(λmMJKM) 川 )c岭
This implies φ rn and φ n are orthogonal to each other
口
15.2.
Proof.
f …。…
口
15.3.
Proof.
1 2we叫
口
83
Z
15.2
Exer cÏ se at the end of chapter
3.
Proof. By forml山 (15 .48) ， the general solution to the equation is
r，f川
叫忡(价叭叫
u
， 仲)
cþ = α
问0+ 局
川
ω
怕
lr川
n1川盯r+ 汇
2二 (C
川
ωm
m1俨
价， = 1
价， = 1
8uppose the expansion of f( 的 and g( 的 for the given eigenfunctions {O , sin m功 ， cosm￠}23= 1 缸'飞 respectively，
f(cþ) = Ao + 艺 (Am cos mcþ + Bm sin m呐 ， g(功)=α+ 艺 )(Cm cosmcþ + Dm sin 时)
m=l
m=l
The boundary conditions 叫α ， cþ) = f(cþ) and 叫 b ， cþ) = g( cþ) gives the equations
(αo 十 ßolnα = Ao
α。 十 ßolnb =
co,
(CItf + D 川盯 = Bm
C m1 bm + D m1 b- m = Dm '
(川 + D m2 a- m = Am
Cm2 bm + D m2 b- m = Cm .
801ving these equations gives us the expression of u(r飞功) as
u(r飞的
=
Ao
ln b - ln T
ln b 一 lna
忐 (t)m(?)m
十 、
…~ (%)m _ (%)
m(Am cosmcþ 十 Bm sin mcþ)
lna - ln T
ι(~)m 一 (%)m
ln b - ln a
r~l ( 扩(*)
、~ ~ m ( C m cos mcþ
+ Dm sin m的.
口
4.
Proof. All the subproblems of this exercise share the same feature. 80 we 且 rst deal with a general problem:
(在 + 告 = f(x , y) , x 2 + y2 < a 2
Ul X 2 十y2 =α2 = 0
8imilar to the calculations carried out in 315.4 , we can use polar coordinate to transform the above problem
into the following one:
O<r< α ， 0<cþ<2何?
些年且|
一旦出丑11
。φ|φ=。一
θφ| 中 =2 71"
。 <r< α ，
ul r =α = 0 ，
0< 功< 27r,
叫r ， cþ) I 俨= 0 is bounded ,
0<cþ<2 汀 .
O<r< α ，
J
=2P
i))
(f =
AV
=
。气/汀
2
7
AY町
飞
/tttj
、 ttt
φ-30 、。
dZTφφ
For the eigenvalue pr由 lem
协 (rg?)+ 古拉 = f(r , cþ) ,
u(川的 |φ= 0 = u(r , CÞ)Iφ= 2们
we have eigenfunctions {1 , sin mcþ , cos mcþ} 泣= 1 with eigenvalues {m 2 } 泣'= 0' Expand u(r， 的 accordi吨 to this
system of eigenfunctions , we obtain
u(时) = A(r) + 汇 Bm(叶归时+汇 Cm (巾0日时
门， = 1
m=l
84
Expand also f (r飞 的 according the above system of eigenfunctions , we have
f(时) =α(，) +汇 b rn (,) sin 川+汇 C rn (叶 cosmcþ
m=l
m=l
Then the 叩ation 记(啡)+去告 = f(时) becomes
;三(， d~~，)) + 主[~二 (4斗-弘(r)lsimm￠ +tt(7 守主)一 μ(，)] cos 川
α (T) + ~二比(，) sin mcþ + 汇 Cm ( T) cos mcþ
竹， = 1
m=l
Therefore , we have the following equations
fAF(7) 叫) = 仰)
J [r-A'
r eauivalentlv <
A(α) = 0 ,
V
{~叩灿
UB
玩叽叩附
以m(ρ
川(卡例
川川，
7忖)忖
叫叫叫叫
+
吼附蜘;二
B;μ
川
川川(们例
;μ
川
7叶)
(T)]' = rα (T) ，
I A(α) = 0,
队
μ
畏
州
ι
J川忏…
γ川
←
7叶
叶(
)尸
=
竹1> 1
#队乌(
m> 1
Bm(归
叫) = 0 ,
α
{t灿川
仰(价例
qω
T叶)忡+叫
附
ω
C;;，(，)
价例
T叶)
Cm 归
( α) =
O.
(1) f(T ， 的 = - 4. In this case ， α (T) = - 4 and b r叫 T) = Cm(T) = 0 (m εN) . It 's easy to see A(T) = - T 2 + α2
By Theorem 6.3 and the boundedness of u(" cþ) at , = 0 , we conclude Bm(') and Cm (,) are analytic in
{X2 + 泸 <α2} (mεN) . 8ince bm ( = Cm ( = 0 , we can deduce Bm(') = Cm ( = O. Combined , we
conclude u( T, cþ) = α2 _ T2
,)
,)
,)
(2) f(飞 cþ) = - 4Tsin 功 . 80 α( ，) = 0 ，的 (T) = 缸" bn(T) = 0 (η 主 2) ， and Cm(T) = 0 (m εN) . Then it's
easy to see A(T) = Bn(T) = Cm(T) = 0 (m εN ， n 三 2) by an argument similar to that of part (1) . To find
B 1 (叶， we consider a general power seriesψ(，) = ~二。 αn ， n . Then
α 1(1 - m 2 )
m 2α 川
ι
旷的 + 扩 (T) 一寸 仲) =一 -7+
+ 〉:创刊[ (η 十 2)2 - m2]T n
，~
，~
，
ι~
By letti吨 m = 1 and the above formula equal to - 4T , we concludeα0 = 0 ， α1 is arbitrary， α2 = 0 ， α3 = i
and αη= 0 for n 主 4 . 80 B 1 ( ,) = αl' - ~， 3 By the boundary condition B 1 (α) = 0 , we have α1=jd.So
B 1 ( = ~(α2 _ 2) and 叫尘 ， t) = ~(α2 _ 2) sin cþ.
,)
, ,
, ,
(3) 8imilar to the argument in part (1) and (2) , we have U(T， 的 = 去 (α2 _ T2)T 2 sin 2cþ
(4) 8imilar to the argument in part (1) and (2) , we have 叫 7飞。) = ~(α2 _
16
Spherical Functions
16.1
Exercise in the text
, 2) , (sincþ + coscþ)
口
16.1
f. M由 iply both sides of (16.12a) by y*(x) and integrate from - 1 to 1, we have by integration-by-parts
PTOO
formula
。 = ν(ν + l)[ll IY 川
85
When y(样 0，儿 (1 一 泸) 1 去 |2dx>OandfJU(工)1 2 dx > O. 80
Ill-----VL
，、
11/-vlu
、
-f
qA-
口
一-
rL一
ν
ν
+
1·i
9"JU--JU-q
dz-Mr U'z-vd
z> AU
,"
口
16.2.
Proof. We note
可(x) = 主 ljF且ln(干)
n-1
P['(x)
= 主古巴忏- 1) (干)
80 可 (1) =叫 = (1 + 1)1 and P{'(l) = 1才74骂 2=~(1+2)(1+ 川一 1)
n-2
口
16.3.
Proof. By formula (16.19) ，月 ( - x)( - l) = (- l)lp{(x) and P{'( - x) = (- l)lp{'(x). 80
P{( -1) = (-1) 旧 P{(l) = (-1)1+11(1 + 1)
and
(- 1)1
P{'( - l) = (- l)lp{'(l) = 一一 ( 1 十 2) (l + 1)1(1 - 1).
2
口
16.4.
Pr时~ P21 (x) = ε;.=o( - ly 221 + 1 习22JS; 匀 )122l2飞 so P~I(O) = 0
俨
(x) = ￡「
γ( - 1)
(
含522l 十 1r!(21
+罚
qA-1-
-q4
一十
1·i
+
一
ou
一­
十
0
四叫"
00
41 + 2 一川
21十1-2俨
+ 1 - r)!(21 + 1 - 2r)!
口
16.5.
Proof. This is straightforward from (16.24)
口
16.6.
Proof. By separation of variable民 we have the following three eigenvalue problems:
(￡ lr2 中l 一入R(r) = 0
lim r →∞ R(r) = 0 ,
(俨古材乖刮叶叶[←←←日剖
sin
叫
n
8(0) , 8( 汀叫) are bounded ,
(…=。
For the third eigenvalue problem , by the calculations in 315.4 , page 216 , the eigenvalues are m 2 (mεN)
with corresponding eigenfunctions sin mcþ and cos mφ. For the second eigenvalue problem , by 316.8 , the
eigenvalues areλ 1 = 1(1 十日 ， 1 = 肌 m + 1, m + 2, '" with corresponding eigenfunctions P1m(cos8) , where
p;m are the associated Legendre's polynomials. Finally, for each given λ 1 ， the calculation on page 236 shows
the 且rst eigenvalue problem has eigenfunction T- 1- 1 . Combined , we conclude the general solution has the
form of
∞
I
u(r , 8 , cþ) = 艺艺 T- 1 - 1 P1m(cos8)[A 1m cosmcþ + B 1m sin mcþ]
1= 0 m = O
To determine A 1m and Blm' we expand f(8 ， 的 according to {Pím(cos 8)e 'm φ } ， then use the fact u( α ， 8 ， cþ) =
f(8 ， 的 and the orthogonality of {月叮
86
16.7.
Proof. Similar to the calculations in 316.7 and Exercise Problem 16.6 , we can conclude the general solution
has the form of
u(忱的 = 三二三二 P川
1=0 口， = 0
Then we expand f(B , cþ) and g(B ， 的 accordi吨 to {P1m(cosB)e， m中} to determi
16.2
Exercise at the end of chapter
17
Cylinder Fu nctions
17.1
Exercise in the text
17. 1.
Proof.
c os2(Z/1r )Jv(z) - cos( 川)J-v (z) + sin 2 (V'7r)Jv (z)
cos(ν汀 )Nv(z) + si口 (v汀 )Jv(z)
sin(ν汀)
J v(z) - cos(ν汀 )J-v(z)
sin( 川)
=
N-v(z)
如otingι(川叫=丘。时母η( 言 )2k土νe :f::ν… =J土 (z) 户…， we have
N v(ze m7r ' )
cos(ν 的 J，Aze m 叫
J_ v (ze m 7r ')
sin(v7r )
cos( 川 )Jv(z)e m 7r ' -
co日 (ν
川汀 )J，ν (z)e
νm
川霄 '+cωOω
叫吕叫(ν汀叫)J，ν (z)e
川7阳
ν
川7
吕剖in( ν 汀叫)
2i sin(mν汀) cot(ν汀 )Jv(z) + e- mv 7rl N v (z).
The equality N_v(ze rrm ) = ε -n~ν7r 'N_ v (z) + 2isin(mv汀) csc(ν汀 )Jv(z) can be proved similarly.
口
17.2.
……,
Proof. 1川叫巾川
ar川Z
叫川叫)川忡队山
阳
ha
1恼a
肌
阳V
刊
y吧
…飞
刊
e
川
a c
tl叩
zero , which contradicts with the fact that J，νand N v are linearly independent (see 36 .4, page 79).
口
17.3.
Proof. We note
￡川ν(工) ]
d
I cos( 川 )x'/ Jv(x) - XVLv(x )l
dx l
cos( 川 )Xv 1.ν -l(X) - d~ [x-( ν)J( ν ) (x)]
sin(ν汀
sin( V 7r)
cos(νπ )xv 1.ν1 (x) + 旷 工 ν+l(X)
sin(νπ)
JCos(ν - 1) 汀 JV-1(X) - L( ν1) (x)
中山 (ν- 1) 汀
xV Nv _ 1 (x) ,
and
二扛呐v(x)]
d
I cos( 町 )x- V 1.ν (x) - x- v Lv(x )l
dx l
cos( 町)( - x-v )J，叶 l(X) -
sin(ν汀
[v LV-1(X)
sin( V 7r)
- cos(ν + 1)πx- v Jν +l(X) + X-vJ一 ν-1 (x)
sin(ν+ 1) 汀
= -x-VN.ν +1 (x)
口
87
17.2
Exer cÏ se at the end of chapter
18
Summary of Separation of Variables
18.1
Exer cÏ se in the text
18. 1.
Proof. For αεC ， we have
。 三 Ilf 一 αgl12 = IIfl1 2- 2Re (f， αg) + 1α1211g112.
问判 we pickα=itp?fromwhIUIe 仕hwarz inequality is们川山
川
III
I口I
口
18.2.
P叫 胁肋川口
Diivi
飞vi怆
也
d
内b
e
协
山
otl
b忡y e J *捕
1告t封拮+d气飞 w
啊eg酬
时
e
t
扑J 捕dzzl+(入 [-;jije喘吁一 [-ijijeJ 如如
Therefore , we can set
p(x) = e J 结jdz ， ρ (x) = 一豆豆 J 岳阳= 一豆豆 p( 叫 ， q(x) = 一旦王l e I 提十 dx
α (x)
α (x)
α (x)
一豆豆 p(x)
α (x)
口
18.2
Exer cÏ se at the end of chapter
1. We apply the result obtained in Exercise Problem 18.2 .
(1)
Proof. Multi 阳
(2)
Proof. We have p(x) = e I 去世音 dx
俨 (1 - x)b α. Soq(x) = Oand ρ(x) = - xa-l(l - x)b α
口
(3)
Proof. M 帅 lying both sides 川 e equat 川ye 一川
口
(4)
血刷
吐
I址蚓
Proof. Mu
lti
ti
口
2.
88
PTOOf. The key idea is to make a change of variable 工 = iu(r)SO that rtf = 峦. This implies dx = ~r- ， so
x(r) = ln r . Plugging r = ë into the or昭nal equation , we have
1 d ( dR \λ1 d dR
I r ~.~ ~ 1 + 丁 R
~.~ - + λ e- L， XR = O.
r dr 飞 dr J r~
e"; e";dx dx
-=-_...-
which is simply the equation 驾驭 十 周 (x) = O. This reminds us of the 吨enval叫roblem (14.3) (314.1 ,
page 186) , only thatαis not zero. 80 we make a further change of variable: 工 = z + lnα ， then we have the
following eigenvalue problem
(号;三1+λR(z) = 0
R(O) = 0 , R (l n b 一 lnα) = 0
The 由ove new 咿叫ue 川m is 川 in 314.1 to have 山回lutionλn = (击曰叶)\ 瓦凡叫灿川，(巾川
川
(μ
z斗)尸=
Cαωhan
问叫
叫叫
lan
叫吨
m
a创
n1鸣呻
gi叭ck
昭昭
ω
山
王川巾
tωO 仕出山
1览e 吨inal …
vari a仙
1讪ble
训蛐 r ,
叫击曰讨叫
z→)，
今户
n= 川
口
3.
Pmof. 8uppose we have two dωis创tin
叫眈
lC
时t 创
e 1ge盯ah时 S λ 1 旧
a I时 λ 2 , with their cωorrespo 时II吨 e
创1ge时ur町t创ion吕 υ
仇1(μ
x)
an川2纠加州(归例
叫)川
Z
， 叫
I
时叫巾七
e
削州叫
ctU
仙
ivel
刊f咫吻
飞V
el
叫呻
叫l忖
y. 咀
Tl
T
her
l创
忧
erI此
l叫t
l
o
b
{Y2(X)
扑 (x ) 中 ] + [入 ρ (x) 仰州州川)川加川
]Yν
1
归州(归例川
叫)
Z
By symmetry, we have
的 (x川;(x)l~ + lb [À 2 P(X) 一 q(X)]Y1(X川dx = 0
Taking the differe眈e of these two equations and using the condition p(α) = p(b) , we have (Po := p(α) = p(b))
o
叫的川叫 一 的 (x)Y1(x)]I~ 十 lb (入1 À 2 川 1 (x川由
-
咐:: ;二 1 - 1) y~ 川α) 一 呻
"巾
口
4.
Pmof. 8uppose u = 1二kαk 岛 ， then
v 2 u = 汇 αk V2 岛=汇 α k( 儿)岛 = -f = - LAk <Þ k
Cor叩旧 ing the coefficien
口
5.
89
f. We first solve the eigenvalue problem
PTOO
(刊 (x 川 (x 的 =0
φ (0 ， y) = φ(α ， y) = φ (x ， O) = φ(凯的= O.
8upposeφ(川) = X(x)Y(ν) ， then we have 毛咛 十 号? 十 λ = O. 801ving two se川te eigenval叫roblems
(川) 十 αX(x) = 0
(川) 十 βYω = 0
Y(O) = Y(b) = 0,
X(O) = X(α) = 0 ,
we have αm = (平 )2 with Xm(X) = sin 咛主 and ßn = (平 )2 with Yn(Y) = sin 平 (m ， n εN). 80 the
eigen叫
""'
.fi mn
m 1f X
nπu
u(x , y) = ) , ~
叹曰n 一-SIn-7
751( 芋 r + ( 平 rα
。
where Amn = 去 Joa sin 咛主 dx J~ f(x , y) sin 平 dy
口
Applications of Integral Tr ansforms
19
f. Let U(x , p) = Jooo u(x , t)e-ptdt. Then
PTOO
O
θu(x ， t) _-vt
_JL
r∞ 一一一~e
θ2 叫x ， t) -ptdt
。
lf∞ 一一?一~e
-ptdt 一 κl
Jo
ät
äx 二
Jo
叫 x ， t)e-ptlgo
f∞
+p I
e-ptu(x , t)dt
θ2U(X ， p)
κ 一亏了一
J(]
U(吼叫一 κ
υ x~
θ2U(风 p)
8x 2
80 the general solution for U(x , p) is C 1 (p)e .jIx + C 2(p)e- .jIx. By the boundedness of ul x 叫 ， we have
the boundedness of U(x ， p)lx →∞ . 80 we must have C 1 (p) 三 O. 8ince U(O , p) = uo Jooo e 一μdt=?we have
U(x , p) = ~e-~:r: . By Example 9.8 , u(川) = uoerfc "2言t' which satisfies Ult=o = o .
口
2.
Proof. The partial di丘"erential equation for the problem is
θuθ2U
θt
> O.
-, t > 0
κ-一一~
8x '2 = O-,龟 x-- -
Ult=o = 0 , x < 0
u I t=O = uo , x > 0
Let U(x , p) = Jooo 叫 x ， t) ε -ptdt. Then (intuitively, u(x , t) should approach to T as t →∞ and is hence
bounded)
。
f∞ θu(x ， t) n-ptri手，~ r∞ θ2U(X ， t) n-ptri手
儿
θt
川 )e-ptlgo + p
…… j 口
θx 2
10f e-ptu(川 )dt
u(x , 0) + pU(风 p)
∞
κ
QdOV
θ2U(X ， p)
8x 2
…
θ 2U(X ， p)
κ 丁7
80 we have the ODE for 【U
叭
I叫(x叽叩
， 叫p) :勾乒L fFU(μZ叽叩
， p)恒=一旦铲且 Using 山
t he method of Green'日 fur川
I
we can obtain (see Examp1e 10.8)
叭队 p) = 土 (00 e- y!Il x-tl 旦出 dt=lfε y!Ilx-tl dt = ~如{而}
κ J- ∞
κ2vrm Jo ~
l ~ - ~吨 {-v!x}
x < 0,
x > O.
By Examp1e 9.8 , we can obtain the formu1a for u(x , t) as
u(川) = 户erfc (一古)
x < 0,
卜。 一 ?他(市)
x > O.
口
3.
P叫 Let U(x , p) = Jooo 帅， 如 -ptdt. Then the equation 去 一 κ 告 = 0 gives us 呜PL - fU(XJ) = 09
whose genera1 solution is C 1 (p)e y!Ix + C 2 (p)e- y!I x. For convenience of app1ying the boundary conditions ,
we can write U(x , p) as C1 (p)sinh V!x 十 C2 (p) sinh V! (l - x). Then
/v.
U(O , p) = C2 (p) sinh 1 与
r∞
2ιιA
I
Ae-1w-te 一叩=一一τ
hκ
Jo
p 十 κα"
and
川=川inh 刊
(00 Be-K ß2te-Ptdt = 兰百
JO
Vfù
Therefore?U(ZJ)
A
inh 飞 1κ (l-x )
P
U
川 2κsinh ，;pr;;，l
p T fùμ
B
sinh Vp j K
+， 叫2κ
一」卫工
sinh ，;pr;;，l .
口
4.
阳
h
丹
P
附
ro
叫
Oω
O
( 守 + α k~U川…(伙阳川……
k川川叶…
t吟巾卢)恒问
=F
叭(川川
U
川
M
k丸
劝
， O叫0) = 古
7tξ
言 fι
工 ￠仲州(怡例
叫)吵e 忱 "dx := φ (k)
Z
吗?且 |问 = 左 f工 ψ (x) ε 'ikxdx : = 川)
The solution of the above prob1em can be obtained by superposition of solutions to the following two
prob1ems:
2叮
向
川川
U
( 主气号j严~
十忖叫叫叫
αa 2马kμ
k干呐
O创呻
才φ町<Þ蚓州(忱
=
(kk川) , 吗FEL
川
m
H
尸
耻川……)
and
LY ( L
K
{~I毕业 + a2 k 2 U(k , t) = F(k , t)
U(k , O) = 队吗严 L=。 =0
To solve the 五rst prob1e阻 we note the genera1 solution to the homogeneous differentia1 equation is U(k , t) =
C1 ( 的 sin(耐) 十 C2 (k) cos(akt). The initia1 conditions dictate C2 (k)
邮) and C 1 (k) = 非 80 the
sol u tion to the 且rst prob1em is
、l/
,hh
ιι
,K) CO (
Q口
，
，
91
+
α
i'LU
J，。飞、
α
n( ,K
、l/
QU
φ
)-
'K 一'k
w「一α
J，。飞
ιι
U 'K ) =
To solve the second problem , we apply the method of Greer山 function and obtain (see formula (10.34))
U(KJ) = UtF(ki) 叫(t 一 刊
Combined , we conclude the solution to the original non-homogenous second order ODE with non-homogeneous
initial conditions has the form of
r
t
审 (k) _ =_. I . u \ , 1
U(k , t) = φ (k) cos(α kt) + 一i-!- sin(α kt)+~ I F(k ,f:, )sin(ak(t- f:, ))df:,
ak
ak Jo
「 Mω
21币 1
叫
M
一
J一 一
由
=
川 I1
叮
1
MU
川 Ih
旦
m
W
M
+-A
忡如"
Aψ
ωKK
+
寸寸叫
AV
z-za
归
1
+
α-d2
』「飞/
+
||LMmu
「
汇汇叮
「W
M
币 1一
一
r
忱
L尸，
AV
||L
uwm
xr
归叫句
「
1212
吐
ρ
且
=
F
1-2
vkt
αV
l
FR
α
U叫 -m
U
φM
口D
O
∞
巳 -
川 一
φ (k) cos(αkt)
j二毗仰
而
1
一 川加州μ
主 汇∞LL
To find the inverse Fo旧ier transform of φ (k) cos(α kt) ， we note
X
Th n
蚓) cos(akt) = 二[守旧川
Using the result for φ (k) cos(α kt) ， we have
θ h(x ， t)
ψ (x 十 αt) 十 ψ (x 一 时)
θt
2
80 there exists some function 1(x) such that
r+? ψ (f:，)df:，
h(x , t) = J E \
十 l(x).
Lα
I
ψ(ç- )代|
也削
0日ce we "guessed" out the form of h(x , t) , we can verify easily that F ←丁τ一 1= 号Zl sin(αkt) ， 日O
l 创刊削 F- 1 (非州州)) = 去 1:JJJψ (Ç) df:，
To find the inverse Fourier transform of a1k J~ F(k , f:,) sin(ak(t - Ç) )df:" we suppose
F-1[UtF
Then
F(斗斗
去F附， t)) 二 l tF 川叫 k(t - T))dT
jtFMEiak (t-T) + e- iak(t 讨
By the convolution theorem of Fourier transform , we have
F 伏 ， T)e i ak(t-T)
F (f (x , T))F(b(x + 州一叶)) = F ( [:
F (f (x + α (t - T) , T)) ,
92
f(x 一 f:" T)b( f:, + α( t 一叫
只f(x， T))F(的州叶)) = F (汇 f(x - ç , T)ð(ç - a(t 讨 )d ç )
F 伏 ， T)e-iak(t-T)
F (f (x 一 α (t -
80 we have
F( 斗斗 =
l
which implies
T) , T)) .
t F(f (x 十 α (t - T) , T)) + F(f 川t - T) , T))
rt f(x + α (t-T) ， T)+f(x 一 α (t-T) ， T ) ~1
θH(x ， t)
Jo
θt
2
…
Therefore , there exists 吕ome functio日 l(x) 吕uch that
rt r X十α ( t-T )
H(x , t) = l(x) + 云 I I
f(己 ， T)dçdT.
L, U
Jo
Jx
α (t-T)
Now let's wave hand and assume l(x) 三 0 ， we then have the solution to the original problem
r t f'c + α (t-T)
f'r+ α1
1
u(尘 ， t) = ~[cþ(x+ αt) + cþ (x - at)] + 了 l
L- a 扎r
ψ (ç)dç +了 I
L- a Jo
α
I
Jx
α (t-T)
f(已 T)dçdT
Remark 28. It'll be ni臼 if we can find an eαsy ωαY to show l(x) 三 o in the α bove cαlculation. We leave
this to the next υ ersion of the solution mαη ual .
口
5.
f. We define
PTOO
U川工) =去 12 u(x , y , t)e- ikx -川叫?
忱 m)=UyhJ川叫xd弘
and
帕， m)= ιlψ(x， y)ε -ik:r-imy伽dy
J JR 2
L,/I
Then the original problem is converted after Fourier transform into the following problem
(牛+… ]U(k， m 功 =0
U(k ， 叽 0) =φ (x ， y) ， 等 |问 =审 (k ， m)
Then it's easy to deduce that
U(k , m , t) = 帕， υ)cos( Vk'2士百2at) +
审 (x， y) ~ sin( Vk'2τ百时)
α vk2τ百五
] =h
泊即
盯
lave
a刊
veeb忡ycωO臼onvo
(x 品圳圳
S忡
u叩pp
啊阳
Oω日 eFγm飞;骂汇tE
三时
叫L) 问
∞口飞
=忡
均
， U弘y 川川
， t功). Then we have
U(k , m , t)
F
到仰刷
[忡￠圳(川(
阳
t功们)忡+ ψ川川*叶叫h(ι， .,冽斗
t功)什l斗
h((ι， .,均们
屹
F[卡￠们什*亏+知二卦扣轩
Hence , u(x , y , t) can be written as
(x , y , t) =
{
I
θ
{
, t)dx'dy 十 一 I cþ (x' , y')h(x - x' , y - y' , t)dx'dy'.
I ψ (x' ， y')h(x - x' 9, y - y' ?θ
t
II
I
1\ 1 I
I
.\
1
I
1
jR 2
11
I
J JR 2
To find h(x , y , t) γ. (to be continued)
口
93
20
Method of Green's Function
1.
PTOOf.
口
21
Introduction to Calculus of Variation
2 1. 1
Exercise in the text
21.1.
f. We follow the line of reasoning in Gelfand and Fomin 间 ， 335. Assume the integration region R stays
PTOO
且xed while the function U(Xl γ. , Xn ) goes into
旷 (Xl""
, Xn ) = U(Xl γ··?Zn) +ε cþ(Xl γ··?Zn)+ …
where the dots denote terms of order higher than 1 relative to c . By the variation 5J of the function
J [U]
J. . .J F(Xl , . . . , Xn , U, U'Cl' . . . , U,c JdXl . . . dx n , correspondi吨 to the transformation U →旷， we
mean the principle linear part (inε) of the differe配es J[叫 一 J[叫. For simplicity, we write u(x) ， 功 (X)
instead of U(Xl γ.. ， X n ) ， 圳工 1γ.. , x n ) , dx instead of dXl … dx川 etc. Then , using Taylor's theorem , we find
that
J[U
旷叫*吁l 一 J[U
儿扣护『问[扛札忡
川
M
Z凯呻
?川叩川叫
， 叫Uu(L(巾(归
叩Z叫川)忡
十忖ε￠刷忡(
:归川
X
ε儿
忡(←←忏凡
μ
昕
ω
￠忻唁+古言凡川
x. 肌Xi)
CÞx 护忏
d巾叶+...
x+ ,
where the dots again denote terms of order higher than 1 relative toε. It follows that
J= 记(阳+生凡如 ) dx
is the variation of the functional J [u].
Next , we try to represent the variation of J[叫 as an integral of an expression of the form
G(X) 功 (X) 十 div(...) ，
i.e. , we try to transform 5J in such a way that the derivatives CÞXi only appear in a combination of terms
which can be written as divergence. To a仙ve this , we replace FU Xi 仇 i (x) by 去[凡X;cþ(X)]
and obtain
主子。(X)
户儿(凡一主￡凡 ) cþ(x)dx + ε 居￡川叫]
This expression for 5J has the important feature that its second term is the integral of a divergence , and
hence can be reduced to an integral over the boundary r of the integration region. In fact , let dσbe the
area of a variable element of r , regarded as an (n - 1) 吐 imensional Sl时ace. Then the n-dimensional version
of Green' 日 theorem states that
LZ ￡川州 = ρ(X)川
94
where G = (Fh1 9… , F,
(伊队
叫1 ， '" , l/n) is the unit outward norma1 to f , and (G , ν ) denotes the sca1ar prod uct of G and ν . Therefore
ν
5J = E
l (凡 一 生￡凡Xi) ø川 十 ε[ø川 ν)dσ
In order for 山
t he functiona1 J[u
川
u
叫
l] tω01
in pa1'川时
ticu
吐
1让1a
巾
盯r ， that 5J = 0 for all admissib1e Ø(x) which vanishes on the bounda1'Y f. For such functions , 5J
1'educes to
J=
l (凡 -S凡 ) Ø(x
and then , because of the a 1'bitrariness of Ø(x) inside R , 5J = 0 implies that
凡 - z ￡ι = 0
for all x ε R.
口
2 1. 2.
Aυ
f'''1α
Proof. We follow the 1ine of reasoning in Ge1fand and Fo1min [4] , 36. The prob1em can be formu1ated as
follows: Among αII cur时 s whose end points lie on two given vertical lines x = ααnd x = b, find the curve
for 也!hich the functional
Td
F Z UU dz
uυ
Uυ
一一
hαsαη extremum.
We begin by ca1cu1ating the variation 5J of J [y]. As before , 6J means the p 1'incip1e linea1' pa1't of the
inc 1'ement
ð. J = J[y + h]- J [y] =
l 川 + 阶 hν的f丁) 一 川f丁')]d川阳丁]d
b
[F
F
u 吕剖ing Tay1or' 日 theoαr冒em丑1 to expand the in时1此tegr冒刀田
创nd ， we obtain
a
ð. J =
l b 叩 十 凡 Ih')dx 十
?
where the dots denote te 1' ms of orde 1' higher than 1 re1ative to h and h' , and hence
6J =
l b 叩+凡 Ih')dx
Here , unlike the fixed end point prob1em , h(x) need no 10吨e1' vanish at the points αand b, so that integ1' ation
by parts now gives
6J
=
l (凡扣) h(x)dx + 凡Ih川二:
b
~b (马 一 手凡 ) h(x)dx + 乌 IX=bh(b
一­
95
U
nυ
M月
υ
一由
d
We 且1'st conside 1' function h(x) such that h(α) = h( b) = O. The rationa1e is that if 旷 is an ext 1'ema1
among all admissib1e function , then y* must be an extrema1 among the smaller class of functions whose
va1ues at end points ag 1'ee with those of 旷. Then as in the simp1est variationa1 prob1em , the condition
6J = 0 imp1ies that
F
F
Therefore , in order for the curve y = 圳工) to be a solution of the variable end point problem , y must be an
extremal , i.e. a solution of Euler's equation. But if y is an extremal , the integral in the above expression for
5J vanishe吗， and then the condition 5 J = 0 takes the form
Fy'I"=bh(b) - Fy'I'E=α h( α) = 0,
from which it follows that
凡 'I x=α = 0 , Fy'I"'=b = 0 ,
since h(x) is arbitrary. Thus , to solve the variable end point proble盹 we must first find a general integral
of Euler 's equation , and then use the natur αl boundα叩 cond州 ons F宫 'Ix=α = Fy' Ix=b = 0 to determine the
values of the arbitrary constants.
口
2 1. 2
The Rayleigh- Ritz method and its application to the Sturm- Liouville
problem
The textbook's explanation of the Rayleigh-Ritz method is a bit ambiguous. We therefore give a summary
of this method , as presented in Gelfand and Folmin 间， Chapter 8.
The idea of the Rayleigh-Ritz method consists of two parts. First , convert a di丘érential equation to a
variational problem , in that the solution of the differential equation is the extremal of a variational problem.
Second , in a certain function space , use a set of complete functions to approximate the extremal , so that
each approximation is reduced to a finite-dimensional optir丑ization problem.
As an example , consider the Sturr口 Liouville problem: Let P = P(x) > 0 and Q = Q(x) be two given
functions , where Q is continuous and P is continuously differentiable , and consider the Sturm-Liou飞rille
equation
(Py')' + Qy = λy ，
Sl创 ect to the boun出ry conditions y(α) = ν (b) = O. It 's required to find the eige时unctions and eigenvalues
of the above boundary value problem.
The following result converts the above problem of solving a differential equation into a problem of 五nding
variational extremal (Gelfand and Fomin [4 ], 312 , Theorem 1; also see 341.1):
Theorem 3. Given t!叫unctior
以← A， y(b) = B , K[y] =
1G川')dx
b
= l,
ωhe1'e K [y] is αnothe1' functional， αnd let J[y] hαMαn ext1'emum fo 1' y = υ (x).
Then , if υ=υ (x) is not αn
ext1'emal of K[y ], the 1'e exi础 α constαnt 入 such that y = y(x) is αn ext1'emal of the functional (F + 入G)dx ，
z.e. y = υ (x) satisfies the difJe1'entω equαtion
J:
Fy 一￡Fur+λ(Gy 一￡叫 =0
With the above result , the Sturm-Liou飞rille problem is reduced to finding an extremum of the quadratic
functional
I (py'2 + Qy2)dx ,
J[y] =
d
α
吕ubject to the boundary conditions y(α) 二 y(b) 二 o and the s由sidiary condition
J: y dx 二 1.
2
2
Then we can apply the Rayleigh-Ritz method as follows. Suppose we are looking for the minimum of a
functional J [y] defined on sOl丑e space M of admissible functions , which for simplicity we take to be a normed
linear space. Let Y? 1 , Y? 2, . . . be an infinite sequence of functions in 儿1 ， and let 儿1" be the n-dimensional
linear subspace of M spanned by the 且rst n of the funct
fu
un
旧
丑lctional
眈
J凶
M
[ν
叫]lead吕 to a function J[O: l Y? 1 + . . . + α n ψ n] of the n variables α 1 ,
α
2Use Theorern 3 , changingλto 一 λ
96
N ext , we choose α1 ，
αηin s山h a way as to minimize J [α1ψ1 + ... 十 αn ψn] ， denoting the minimum
byμn and the element of M n which yields the minimum by Yn ' We say the 叫uence (叫:二 1 is complete (in
M) if given any Y ε M and any ε> 0 , there is a finite linear combination 'rJ n of the formηn
such that 11ηn - yll <ε(where n depends on ε). Then we have the fo11owing theorem
L ~ lαzψz
Theorem 4. lf the functional J[y] is continωω in the norm of M , and if the sequence (叭) 三 1 is complete ,
then
limμn
μ?
n----+oo
t仇 ere μ= infνJ[υ]
In the particular case of the Sturm-Liouville problem , we can choose 内 (x) = sin(kx). Then (仙
lμ州川
L n)~
'f 二
L1
Cωon
盯
ve
臼rg
阱创 tωo the sma
叫
a11
丁让训
11e
倒
s创t 吨
e i厄
睽e
g
创
阴I盯a
址lu
归
时
1忧
e of the 缸
S t饥
t灿
urr
S叩
ponding 臼
e 19
萨
en
吐
1让f目UI
山
mc七tion .
We can continue this procedure to find the rest of the eigenvalues and eigenfunctions
(see Gelfand and Fomin 间， 34 1. 4) . This is summarized in the fo11owi吨 result
Theorem 5. The Sturm-Liouville problem hαsαn infinite sequence of eigenvalues 入 (1) ，入 (2)7 …， αnd to
eα ch eigenvα lue λ (n) there corresponds αη ezger飞function y(叫 which is unique to within α constαnt f，α ctor.
2 1. 3
Solutions of the exercise problems from Gelfand and Fomin 间， Chapter
8
2.
Proof. We first calcl出te the exactωlution. Let F(立 ， υ ， y') = 俨 - y2 _ 2xy. Then
乌兰凡f = - 2y - 2x - !f- (2y') = - 2(y + x 十 y")
dx
To solve the second order linear ODE
we can employ at least four methods: the Green's function for boundary value problems , Fourier expansion
over the interval (0 , 1) , operator calc山lS (including Laplace transform , see Ding and Li [2]) , and reduction
to a system of 五rst order ODEs (also see Ding and Li [2]) . However ， 旷日 very easy to see directly the
general solution of y" + υ = -x is -x + C 1 cosx + C 2 sinx. Using boundary condition , we conclude y(x) =
-x + csc 1 sinx. In this case , it's easy to calculate J[y] = ~.
We then use the Ritz method to find an approximate solution. For this purpose , we need to show the
completeness of {lPn}~= l in the space M = U ε V 1 (0 ， 1) : f(O) = f(l) = O} , whereψn = xn(l- x) and the
norm is that of V 1 (0 , 1) (i.e. Ilfll = max吧 t9lf(叫 1 + maxo:<:;x 三 1 1f' (x)I).3 We have the fo11owi吨 lemmas .
Lemma 1. The set of polynomiα ls is dense in V 1 (0 , 1) .
Pmof. By the Weierstrass approximation theorem , for any element f ε V 1 (0 , 1) , we can find a sequence
(Pn)~= l of poly∞mials ， such that
♂ap|ff(z) 凡 (x) 1 三 1
<x<l
ηJ
Let Qn(x) = fc~ Pn( Ç) d f, + f(O). Then Qn is still a polynomial and
O~:~l 1 1 x 川f， + f(O) - [l f'(f, )df, + f(O)] I+四川)-f' (x)1
IIQηfll
X
<战112|PAU f(￡)|dZ+:
<
一
n
口
This proves the lemma.
3For de缸山 ion of the norrned space Dn ( α ， b) ， see Gelfand and Fornin [4 ], page 7
97
Lemma 2. The linear spα ce spanned by { ψn}~=l is dense iη M under the norm ofV 1(O , 1).
Proof. It 's clear eachψηελ.1/. Hence any 且nite linear combination of 叭1 's belongs to 儿1. For any f ε 儿1 ，
by Lemma 1, we can find (ε n)~= l with εn ↓ o and a 肥quence (Pn) 二 1 of poly∞mials s旧h that
IlPn - fll 三 ε n'
Each Pn can be written in the form of Qn(X) + αn+ 点 (1 - X) , where Qn(X) is a finite linear combination
of 叭 's. 8ince IIPn - fll • o as n • ∞， we have αn=1凡气气丑n(
削
a
Sη → ∞. Therefore , we have by triangle inequality
IIQn - fll 三 11αn+ 风 (1 - x)11 + IlPn - fll 三句 +1αnl + 21ßnl • 0
as n → ∞. This shows the linear space spanned by {'Pn} 二 1 is dense in M.
口
Combining Lemma 1 and Lemma 2 , we can by Theorem 4 find the approximate solution of the original
variational problem. lndeed , consider the n-th degree approximate solution Yn
L~= l α时向 (X). For
剖I叩licity of notation , we writeαk for each αnk. Then finding the extremum of J[Ynl = Jo1(y;; - y~ - 2叩门 )dx
becomes the minimization of a function of n variables α1 ， α2 ，…， α n'
argmin(α 1 ，... ,an)EIR ,j [Ynl.
Note for i = 1 , . . . ，凡 we have
1 1 主 ω-d 一 2xYn) 伽
21 (咕υLh咕￡υ口￡hψ)
主
J[Ynl
uai
θαz
1
dx
Z
21 ω (x叫) 仇川叭叫(付Z叫) 叫川(归川
1
2
[主 αkt才 一哇!产+且主 +dE 一 z占)一川 ;t 十 3) 1
80 the extremal Yn = L~=l 阳山 (x) are given by solving the linear equation
α1
3.4
I
4.5
I _ I
.. … . .
α2
A[)
αnl
J
I
,
1Un+2
7一一气川7η+3)
一一气
2 丝土且主+业土且主 +-i
L一
where Aik = 一主
一一一一」一一
k十 i+1
k+i-1
i+k
'k+i+2
k+i+ 3 '
Therefore , to find the η:-th degree approximation of the extremal Y as well as the corresponding extreme
J [ 叫， we first solve the above system of n linear equations to get Yn , and then apply either the quadrature
method or an explicit forml由 to evaluate J[Ynl
/c
-
Remark 29. Note the mα 衍 'ix A is ill-conditioned as n gTOWS. 80 we postpone αη umerical illustmtion of
the αbo 时 αppTOxzmα tion to next versio η of the solution mαη ual.
口
8.
98
f. 8ince on a 缸lite interva1 , L 2 -co盯ergence imp1ies L 1 -convergence (by the Ca山hy-8chwarz inequality) ,
PTOO
we can without 10ss of genera1ity assume the meaning of "in the mean" is in L 2 -sense. By extending h"(x)
to an odd function on [一 汀，叫 (not necessari1y continuous at 0) , we can write h"(x) on [0 ,7r] as
h"(x) = 汇 A， sin(叫
γ= 1
where Ar = ~ Jo1r h"(x) sin(rx)dx. Defin
4万
汇
n 叫
Then h~(x) → h" (x) in L 2 (0 , 7r) as 71 →∞­
Meanwhi1e , we have (note h'(O) = 0)
忡忡 ) - 们印州川(怡川川
h' Z叫叫)川1
1a 叫h: 帆 +h叭帆
川川(仰例
;μ
O叫)_[1ρ
艺〉〉hν"气飞(任ω
帆
ç)
X
=
I
s;
1a Ih~(ç) - h"(ç)1句+ Ih~(O)1
s;
Ilh~: - h"IIL2(0 ,1r)V7f + Ih~(O)I.
X
80 Ilh~ - h'IIL2(阳)三 Ilh~ - h"IIL2(阳)汀+ Ih~(O)I v'王 We further note
~ k
~ 2
~ 2
(1r _" _
h~(O) = 一 至 于 =三 百 10 h" 川
(1r
By extendi吨 h'(x) to a continuous even function on [汀，叫， we can cωO眈II时 e that the Fou
盯ner cωO日i丑
expa
ans剖ior丑1 of 叭
创
h'
叫/气(叫)
Zν
x (note h(O) = h(汀) = 0)
主 ;fhfh)cosh)dz 叫
rr
、
-
户户、
卢、
句
t、
nυ
句
t、
converges to h' (x) pointwise as n →∞ In particu1ar , h~(O) = ~~= 1 ~ Jo1r h'(x) cos(rx)dx . co日 (0. x) •
h'(O) = 0 as n →∞. Combini吨 with the inequa1ity Ilh~ - h'II L2 (o ，1r)三 1 叫 h" 1 L2 (O ，1r) 汀+ Ih~(O)I y'7T， we
conclude h~， (x) • h'(x) in L 2 (0 ,7r) as n → ∞
Finally, we have
ftt'to
filo
n
it-o
J
'h
'd < h
h z
h z -h
h
h ou <
h L
+h
-
80 Ilhn - hllu(o ，1r) 三 Ilh~ - h'IIL2(O ， 叫作 → o asη → ∞.
In summary, we conclude as n →∞ ， h n → h ， 叫z → h' ， and h~ → h" all in L 2 (仰0矶?川，7r汀7r) .
一 F土
2ξ
王f
兀
J fhWr
川
川
f
1扩
f川f气勺(例
叫) sin(rx
Z
川
叫)dx 恒
Z
is 阳巾 independent 叶
0 fη.
口
Q山
P----r
υ
hvnj
PTfuuu
JU
户
Zz-(
-rJ-rJ
lj'α
、
工一
川
Z
,
dL
到=
b
jqr"
2L
α
q4
，。
川川
门日
-M
α
宁。
'hυ
气
川川川
十
fLd 川
+ιJ
Qd
八JH
J'tt 飞
、，
州
丑/t
-
ιvfJ
、
1
703
hw
)(
一叫
η
iJα
<
hM 叫
位仇
i
<一
,-Z
'.E，们
Z Qd
币曲￡
z
JU
/''飞、
Z
ρ+''
、飞
/''1
、ll'
rh z Qd
M
》M
i!α
fjlh
ιu
ort'1α
I卜
Ti-ob
8ince ι/ → f in mea口， (11fn11L2 怡 ， b))~= l is bounded. As n →∞， the RH8 of the above inequa1ity goes to 0 ,
so we can conclude
i:
fn(x)gn(x)dx → J~ f(巾 (x)dx
QdQU
口
2 1. 4
Exer cÏ se at the end of chapter
1. (1)
P叫巾 ， y ， y') = 0τ y2y'2 . 80 the Eu1er-Lagrange equation becom
θF
d θF
uy' 2
θy
dx θy'
J士 y2y'2
d
y'y2
dx 0士 y2y'2
uy'2
(时 +2时 )0可严内22:但7;ty"
J士 y2y'2
1 + y2y'2
After simplification , we have 0
y" y + y'2
(~y2)". Therefore , the genera1 solution is y2
(y' y)'
αx + b .
口
(2)
Proof. F(轧机 y') = 乒 十 沪 ， so the Eu1er-Lagrange equation becomes
θF
d θF
θy
dx 8y'
一丁 = 2ν
dx
(2y') = 2y - 2y" = 0
Therefore y = α eX+be- x .
口
(3)
Proof. F(x , y, y') = 详;Sothe E1巾-Lagrange equation becomes
θF
d θF
d
θy
dx θy'
dx (x + y')2
x
80 there exists some constant C s川 that 百命节 = C , which gives y' = - x ::l:
Æ. Use a change-of二 variab1飞
we have y' = 主l: X 1 / 2 - x. Hence y = α23/24 十 b
口
(4)
Proof. F(x , y , y') = y'fτx0τ y'2. 80 thc Eu1cr-Lagrangc cquatio日 bccomcs
。 一 ~I
ý'f石「主导
1 =0
ax I
2 飞 1 + y'~ I
There must exist a constant αs叫 that 、' l +
x Ji号古 = α
飞
、
801ving this equation gives y'
万苦?
Therefore y = 2a ý'fτ x - a 2 + b
口
2.
Proof. The point on the cone x 2 十 泸 = z2 can be 出scribed by the parametric coordinate (z cos () , z sin () , z)
(0 三。 <2汀).
For any given curve γon the cone , assuming γis parametrized by () , the 1e吨th of γis
川)
(去r+(~~r+(去)LL:;:)F而平d() = 川)肌
100
where F(e , z , z') = Yi，Z2+去巧2 80 the Euler-Lagrange equation becomes
θF
d θF
θz
de θz'
d可写开
de vzπ五开
γ J Z2 + 2(z')言一巧苦学
z
Z2 + 2(Z')2
J可写开
z
2 z"[z2+2(矿产] - 2( 矿产 (z + 2z勺
d可写开
[Z2 + 2(Z')2] Vz可写苟言
o.
8implifying the above equation , we have
Z[Z2 + 2( 矿产] - 2z" z2 - 4(矿产 z" + 2( 矿产 z + 4(z')2 Z" = z[z2 + 4(z')2 - 2zz"] = 0
80 a non-trivial 创刊on must satisfy the 叫uation z" - ~ (斤 -~=Oor 叫uivalently 子一巧立=~. N侃
(矿 = 与Jiii = 于 一 (~f. Us川1e substitution 兰~ = (~)' 十 (~f ， we 叫órm the original
equation to
(二)'-(二r =~
Define h( e) = 组 =ιwe get a R叫叫uation of h:
。牛
一
­
十
1-2
vhM
'hM
It is well-known that a general Riccati equation can always be reduced to a second order linear ordinary
differential equation (附 Remark 30 below). For our particular case , we use the substitution h = - ~， then
1
1_'
1_2
w"
旷?
(ωf\2ω
80 ωsatisfies 山 equation 川+抄 = 0 , which has a general 时utionω = C1 删去 +C2 日 in 去 By usi吨
trigonometric identitimmcm write tu mtcos 节
αsec -e-+;!!
ý'2.
Note 一号 =h= 乞 we conclude z(e)
市=
Remark 30. In mathem α tics ， α Riccα ti equα tion is α ny ordin α ry differential equ α tion th α t has the form
y' = qo(x) + q1(X)Y + 的 (X)y2 时旷e qo(x) 并 O α时的 (x) 并 o. It cαn be reduced to α second order lineαT
equation by the followi叼 procedure (see , for exαmple， wikipediα). First , use the substitt而on v
的 (x)y.
Then the original equ α tion becomes
v' = v 2 + R(x)v + S(x) ,
ωf旷'e S(x) = qo(X)q2(X) αnd R(x) = q1(X)
+ q~(x)j的 (x). Then s伪stitute v = -u' ju αnd it follows that u
Sα tisfies the lineα r second order ODE
u" - R(x)u' 十 S(x)u = 0
A solution of this equαtion will leα d to α solution y = - u' j(uq2(X)) of the original Riccαti equαtion.
口
3.
101
PTOOf. Points on the cylindrical surface are described by the parametric coordinates (7' cos B, 7' sin B, z) where
r > 0 is a constant and B ε [0 ， 2叫 For any curveγon the cylindrical surface , assuming γ is parametrized
by B, then the length of Î is
U FHO=UU
where F(矶 z ， z') = vr言τ石')言 and the E由-Lagrange 叫uation becomes
z" 卢布平+节市
zr
θF
d θF
d
θz
dBåz'
dB P王 ( Z ')2
r 2 + (z' )2
After simplification , we get z叮r 2 + 2( 矿产 1 = 0 and hence z" = O. This implies z (B)
constants αand b .
22
Overview of Equations of Mathematical Physics
22.1
Summary on the classi自 cation of second order linear PDE
α + bB for some
口
Suppose we have a general second order linear partial differential equation
生生 A ij 在; 十 会去 + Ou = 卫
If we choose a non-singular change of variable
ll/ll/
一一一
一一一
-J
、
t
F 占
Z
怕川口'叮
"7
Uυudtb
1止。
UυUU
194η
ttttt
/tttty
飞
((1
ZZ 22
1
1
11
1
then the original equation becomese
主主 Akl忐-+主亘古 +Cu= 卫
B
句h
+
nγ
与国
b
2ν
A
大U
B
冯υ -x
汇
n H
H 问
汇
A
伽h
h
伽-
A
汇
n M
汇
n
国
with
In m atrix form , we have
A=(JZKI)=JAJT?
陀旦L
lθX ，
且显饥
where
主ι
θX o
J=θ (Yl ，... , Yn ) = |23125j
θ(Xl '.. . , X n )
|主主L
旦旦丘
1θ;]; 1θ二C2
•• •
旦旦立
θ;E n
Since A is a re址 symmetric matrix , we can find an orthogonal matrix U(= U(Xl , X2 , . . . , X n )) such that
fJAIlT is adiagonal matrix Then we obtaiI1712equations 337 = 川 (Xl ，. .. , X n ) (i , j = 1 ,.. . , n)
In the special case of n = 2, we suppose the second order di 丘町 ential operator in the original equation is
2U
θ 2U
θ 2U
Lu 三 A 一一一 十 2B 一一一 + C 一一
θxθuθzθuθυ2.
102
Then with the change of variable ç = ç(立， ν) ， η = η(凯 y) ， the operator Lu takes the form
2U
θ 2U
θ2U
Lu = A 1 一τ +2B1; ~n +C1 一τ +
θuθu
--,, =Lç +
句，
。与 "θZθη0η"θZθη
where
/θE\2θZθZ
+ 2B
A , = A I ~ ." 1
i\θx)
，
~ ." ~ ."
θzθη
+ C门 I/θZ\21 .,
，
-
~ ."
\θυ)
B , =A豆豆豆豆 +B( 坐 22+222EV+c豆豆豆旦
iθzθx
\θzθυ1θzθy)
,
-θυθυ'
/η\2θ77θη/θη\2
C , = A I ~ 1 + 2B~~ + C I ~ 1
A\θx)
，
θzθη!\θν/
川
A
S川
…
a
Cωcor
臼∞
叫
nl
(22.13) in the textbook , we also have the following more explicit results
1) B 2 = AC. In this ca盹 put k = 主=丢 and we have
A1 = A
(主+旬 ， B = A (主+嗲) (去+旬，← A(三十 KZ)
1
In order that both B 1 and C1 vanish it is su自 cient to put
'此
η-u
ndu玄。
，
η -z
nd大
u。
+
-
OU
For the soll时on of 直rst order linear PDE , see Di吨 and Li [2]. ç can be cho日en arbi让tra
缸
叫
.r
r
far as the change of variable (x ， υ) ←→ (Ç ， 77) is no坠 singular.
2) B 2 > AC. First assume A does not vanish (the case C 并 0 ， A = 0 can be treated in a similar way;
the case A = C = 0 we shall deal with separately). We put ç = x ， η=ψ (x ， y) . Then the condition that B 1
should vanish becomes
θηθη
A 一一十 B 一一 = 0.
θzθy
801ving for η ， we can obtained the canonical form of a hyperbolic second order linear PD E. If A = C = 0 ,
Use the substitution ç = x + 机
then 山 original equation will have a principle term of the form 就
η = x - y , the equation takes the canonical form
θ2U
θ 2U
一一一一
θxi 2
22.2
θη 2
'
.
.
.
Exercise at the end of chapter
1. (1)
,
, ,
,
,
PTOOf. Using the notation of Theorem 22.1 we have α = 1 b = 0 c = y. If b 2 一 αc > 0 i.e. y < 0 the ODE
for characteristics is
dy
"一
丁一=土、 -y.
ax
80 the two integral curves are FY - ~ = C 1 a nd V=百十号 = C 2 , or equivalently, x 二 C1 and 2FY 二 C2 .
U nder the change 川able ç =川=归， the equation is simplified t。在存= O. If b 2
1. e.ν> 0 , the ODE for characteristics is
dν
α c< 队
「
丁一=土凡 υ.
αx
By similar argument , we should choose the change of variable ç = x ， η = 2Vfj. Then the original equation
is reduced to 窍+窍 =0
口
103
(2)
Proof. Using the notation of Theorem 22.1 , we have α = 1 + x 2, b = 0 , and c = 1 + y2. 80 the ODE for
characteristics is
dy
,. I
?一 = 土 i ， /{l + 泸 )(1 + 112) .
αx
Writing it in the form of separated variables , we have
dy
dx
J可 一 :í:Z \/1 + x2
We can solve these equations to obtain the change of variable
(?「￡←= 缸肌川……
创
ω
C臼
csin
剖白口1
η = 缸
a rc
臼sin让由
hy ，
则er which 山 original 叩ation is redu创 togi￥+存 =0
口
(3)
Proof. Using the notation of Theorem 22.1 , we have αtan 2 x , b
b 2 一 α c = O. 80 the ODE for characteristics is
dy
dx
- ytanx , and c
y2. Then D.
y
tanx'
η
，
U 一η
门。-o
十
，
J
户户、
U-2
uu-t
」
θ-o
9"
η
卢、
十
qt"
-
AU
，
门。-qo
which has the general solution y = C sin x. Let ç = y sin x andη = y cos x. The original equation is simplified
to
9"-qf
口
(4)
Proof. Using the notation of Theorem 22.1 , we have α
1， b
characteristic becomes
dy
?一
- Slnx ,
- sinx , c = - cos 2 x. 80 the ODE for
αx
which has a general solution υ= cos x + C. Calculation shows the following change of variable simplifies the
original equation most
{~←= 叶
x +忖
川ν ∞
η =
x - y + cosx ,
川
u
I
口
2. (1)
Proof.
「
υ1
22=e一(创刊ν) I 一巾， ν)+ 一|
门2U
(__~
(θvlθuθ2V 1
」 = ε一 (α叫) ~一 α| 一 αυ+ 一|一 α 一+一↓
θx l'θx 2
θx
I
L. .Î
1 - Iθxlθzθx 2 I
By symmetry, we have
OU
「
θvl
门2U
(~_'L.\(_Iθvlθuθ2V 1
|θυl 'θυ2
L - 1θυ|θυθυ2 J
ε (ax十by) I- bv(x , y) + 一 |
」 = ε(ω均) ~ - b I - bv 十一 I - b一 + 一 }
104
Therefore
qθuθ1
寸 ~u
+ 2α
+ 2b
θzθν
「
θuθuθuθvl
e-(ax +b Y) I (α2 + b2)v - 2α 一 - 2b 一 + '\iv - 2(α2 + b2)v 十 2α 一 + 2b 一 |
|θzθυθzθυ|
ε(α r+bν)
[V2V - (α2 十 b 2 )v1.
80 the original equation is transformed into 气7 2V 一 (α2 十 b 2 )v = o.
口
(2)
Proof. Follow the hint and use the transformation u(x , y) = εαx+bν v(x ， y).
(3)
口
,,
Proof. Let 叫iE?UJ) = 叫尘 ， υ ， t)h(α ，风 γ ， x ， y t) where h(风 β?γ ， x ， y ， t) = εα x+β凹十， t Then it's easy to see
主 =h(αυ+ 去)，去=巾)+2) ，主=小 +2)?
and
+
内的
一
U
uu
θ
o
z
θ
BE
，，，
、、、、，，，，
+ U
- +
α
θ o
ν
α
飞'
飞t
\I
/'''
aμ
一-
b 句
一
所 h
-
h
句
与一
尘 = h [α(们十二)+击+剖，在=巾 (ßV+ 去)+吭+刮
Therefore
h-
, Iθ2U
θ2U
θ2U
θuθu
i
Iα 一τ+ 2b: : + c 一τ + d ~ + e ~ + fu -
I
åx~
θxθy
åy~
θx
åul
-", I
θυθt I
θu
(αα2 + Cß2 + 2baß + dα + eß)v + (2αα + 2bβ + d) 一 + (2cß + 2bα + e) τ+ (f一 γ)υ
θx
θ2V
θ2V
+α 一~ + 2b ~- :
åx~
θxθy
θ2V
+ c一τ
,
-
åy~
θt
一-
「lllIll-IIIL
1EEEEEEEEEEJ
α aμ
「lllIll-IIIL
「lllIll-lllJ
，。
9年
α
vhvpu
-EEEEEEEEEEJ
vdE
"2桂芳丁 ， ß = "2桂芳丁. Note
-EEEEEEE
」EEE
「lllIll-IIIL
」
-EEEEEEE
」EEE
lllIll-lll
αaμ
「lllIll-IIIL
，。
Lυc
「
α
「lllIll-IIIL
8μ
一-
α
、J
rEE、E
EE
EE
，
de
1EEEEEEE
」EEE
「lllIll-IIIL
α
+
aμ
αaμ
-EEEEEEE
」EEE
「lllIll-IIIL
，。
日川J叮
α
4μ
「lllIll-IIIL
一-
α
+
ρU
JU
aμ
+q"EO +
α
qt"
α
FL
aμ
+
4μ
α
α
qt"
「lllIll-IIIL
801ving the equation
we haveα
ov
θυ
15IJ
+ de -- ou
,
80 by choosing the above αand ß, and by settingγ = f , we can transform the original equation into the
form
δ2V
δ2V
a 一一τ 十 2b ~- :
åx '2
θzθy
δ2υδu
+ c 一τ
=
åy '2
θt
Remark 3 1. Something wrong in the calculation? Check!
口
3.
105
f. By Example 13.1 (p172) , the general solution of
PTOO
θ2U
一
θt 2
-
~ 2θ2U
--
..
θx 2
can be written as f(工 十 时) 十 g(x 一 时). Using the boundary condition , it's easy to see f(x) = ø (~) - g(O)
and g(x) = ψ(~) - f(O). 80 the solution must be
AV
nυ
2
\、飞，，，，，/
二
ι'ι-
Z 一
山V
+
，，，
『、
B
E 、\
1
α 一
,
+iu-
，，
+-2
\、 ，』，/'
Z 一
飞\
/II1
A
V
口
4. (1)
Proof.
θU
一一 =
θx
- e
叫
θ2U
Sln
v 一一一 = ε
~... '"θ x 2
叫
一 门
θu
smv
一一 = e
~..." ，
θy
内
cosv
θ2U
一一一 =
~~~'"θν2
- e
叫
一中
Slnv
~... "
80 V2u = O. ulν = 0 = x is obvious.
口
(2)
Proof. Let (凯 y) approach (0 , 1) along the y-axis , we have
川) =川)=~τ
→∞
1 y~
80 as along as u(x , y) has a definite value at (0 , 1) , it cannot be continuous at (0 , 1). The case of (0 ， 一 1) can
be proved similarly.
口
A
The Black-Scholes partial differential equation
In mathematical finance , the following PDE appears in the deri飞ration of the Black-8choles call option pricing
formula (K > 0):
lw) +(t) 寸
。) = TC(t , x) 中)主 O
c(T , x) = (x - K)+ ,
c(t ， O) = O ， t ε [0 ， 町 ，
limx →∞ [c(t ， x) - (x
ε r(T-t) K)] = 0 , t ε [O ， T]
Many more PDEs important in mathematical 且m配e can be found in Kohn [6]. For now , we focus on the
explicit solution of the Black-8choles PDE using the techniques from the current textbook. " Ho ω do you
deTi时 αη d solve the Black-Scholes PDE' is a frequently asked question in Wall 8treet job interviews. The
presentation below is essentially that of 飞再Tilmott et a l. [12 ], 35 .4.
A.l
Derivation of the Black-Scholes PDE and its boundary conditions
For the derivation of the main PDE , we refer to 8hreve [10 ], 34.5.3 (see also Remark 32); for the derivation
of the boundary conditions , we refer to 8hreve [10] , 34.5.4.
106
A.2
Simpli且cation of the Black-Scholes PDE via change-of-variable
The first 叫 of simplification is to convert 哈 and x 2 茹 to %T) and 券 ， respectively, via some change of
variable x = f(7)) . This is the trick presented in 313 .4 of the textbook. With the change-of-variable x = ♂
and ω (t ， 7)) = c(t ， 叫 ， we have
ωη (t ， 7)) = x cx(t , x) , ω叮叮 (t ，7)) = xcx(t , x) + x 2 cxx (t ， 工) .
80 the original PDE becomes
叫 十 (T - ~()2
= r 队∞ < 7)<∞， t ε [O ， T).
2 ìω'1
J '/ + 与2Wηη
2
飞
叮叮
Or equivalently,
ω
一-
η1
+ lF
w
'比
1·i
ω
n川，
ω
"- + k
where k = T / ~ σ2
The second simpli且cation is to convert the equation into a form as close as possible to the standard heat
equation
学
α6.v = [...],
OT
which means we need to normalize the coefficient ofωt to - 1. Define V(T , 7)) = V(T(t) ， η) = ω (t ， 7)). Then
ω叮叮 (t， 7)) + (k - l)wT)(川)+占叫 (t， η) 川， η)
豆 σ"
V r山， η ) + (k
1 dT
阳η (T， η)+FEEJUT(7dku(Td
80 we want to set T = ~σ 2(T - t). Then V(T， η) satisfies the following PDE:
VT = VT) 'I + (k - 1川
kv ， ∞ <η< ∞， 0 < T 三 jdT
The third 阳p of simplification is to remove 山 first order differential operator %n so that we have 出
To do so , we use the trick introduced in Exercise Problem 2 of Chapter 22
of the textbook. More precisely, rewriting the PDE for V in the following form
吕创tanda
汩rd form of heat equa tion.
VT + kv = V叮叮 +(k-1)vη?
and motivated by the exponential integrating factor employed in solving first order ODE , we try the function
U(T ,7)) = e叫十βTV(T， η). Then
U'I = εαT) +βT(αυ + v ,J , uT)η = e白 T) +βT(α 2V + αUη + αV'I + 叫η) ， UT = εαT) +βT(ßV + vT).
80 we have
UT - u T)T) = ε川1+βt [νT 十 (ß
Comparing with the PDE for V: VT + kv - (k - 1)νη
~ (k + 1?
In summary, U satisfies the PDE
UT(T， η) = UT)T) (T , 7)) ，
α 2)V - 2αUη
Vηη].
VT)T) = 0 , we want to set a =平 and ß = α2+k=
∞ < η< ∞ ， 0 < T 三 iσ2T，
咐 η) = (ε平η - Ke导干 ，
li叫→∞ U(T ， η) = 0, T ε [0 ， ~σ 2T]，
limη→∞ [厅η 年1川) - (♂
107
Ke- rT /! o- 2) ] = 队内 [0， ~()2T]
and u is related to c( t , x) in the followi吨 way
t主
坠土豆
(~
T
U(T， η) = 户十';[-'- T V (叩7) = e " ;"'1+" .-"4~T 叫 T
T
~\
/\2~
/
\
t主
坠土豆
平 ， 1] ) = ε2771
\2~
(~
飞T
平 ， e'1 )
or equivalently
巾) = u (~o-2(T - t) , lnx) ε 导 l川 i主主u.:.扩 (T-t)
Remark 32. We have used change-of-vari α ble throughout to simplify the Black-Scholes PD E. There is αn
observα tion that cαη simplify the PDE to begin with. Recα II during the deriv α tion of the Black-Scholes PDE,
we used the fact that 'under the risk-neutral meα8旷已 the discounted cαII optioη price e-rtc( 品 ， t) should
be αmαrtingα le， ωhere the underlying αsset price process St satisfies the SDE dSt = r-Btdt 十 σ StdWt (Wt
zs α Broωmαη motion under risk-neutral measure). Note X t
ε rt St is αmαrtingale sαtisfying the SDE
dX t = σXtdWt ， so instead of writing cα8α function of St αnd t , we suppose c is α function of X t α nd t.
Then
2 C XX II dt + martingαle pαrt
牛一时c(Xt ， t)] = ε -Tt I-rc + Ct + ~σ2X
2
.~ C
So the PDE has α simpler for即 Ct 十 iσ2X 2 C XX = rc. To remo时 x 2 ， 悦的II need to set x = ♂ Mz2L
becomes 茹
￡ Thtmul问叩-ession is ηwt
r， 陀
叫
m
αl闯
l旬Uα旧s s叫l扣Eα旧Sω
恍εlωLωO阳
p网d
叫
臼i， 幻
s ince
叩
O
pe
旷
Tα
时to
旷
r stμ
耐
1刽IIp
训 严
er，
附
'sz衍st臼s.
But 伪
t he cω
oe.庐庐 cient臼s become simpler. The rest of the simplificα tion should proceed αs
before.
A.3
Solution of the simpli且ed PDE and the Blad• Scholes call option pricing
formula
There are many methods to solve the initial value problem of heat equation over an in五 nite line. For example ,
we could use Fourier's transform. However , due to the messy form of the boundary value function , which
is probably not easy for inverse Fourier transform , we employ the method of Green 's function. Recall the
fundamental solution of the equation Ut = uη叮 is
一一一τ~ e
4T
2 、7rt
80 叫风 T) can be obtained thro吨h the convolution forml山:
U(T， η)
古汇 u仰， x)e 叫z
古汇 (m-KF小王旦旦旦 dx
After some tedious calculation , we can get the Black-8choles call option pricing formula
c(St , t) = StN(d I) - K e-r(T-t) N(d 2 ) ,
where
d1
log~ 主+
(r + ~o2)l '(T - t)
n.
L,
d.,2
log 去 + (r - ~o-2) (T - t)
__
~---，，-n.-，----_-，-----:=
Remark 33. An αlternαtive， αnd much eαszer， method, is viα the Feνmαnn-Kac formula , see 0ksendal l吃l
for detω ls. It CO Tr臼ponds directly to the so-called 内政-neutral pricing methodology.
108
References
[1] John B. Conway. Functions of one complex vα叫α ble， 2nd Edition. Spri吨er ， 1978.
[2] Ding Tong-Ren and Li Che吨 Zhi. A course in ordinαry difJer 巳ntial 呵uations (in Chinese). Higher
Education Press , Beijing , 2004.
[3] Fang Qi-Qin. A course on complex αnalysis (in Chinese) , Peking University Press , 1996.
[4] 1. M. Gelfand and S. V. Fomin (translated and edited by Richard A. Silverma叫 . Calculus of var创wns，
Dover Publications , 2000.
[5] Gong She吨 and Gong You-Hong. C川C脱 complex αnalys叫 2nd Edition. World Scientific , 2007.
[6] Robert V. Kohn. PDE for Finαnce ， NYU Master of Science Program: Mathematics in Fina配e. Spring
2003. http://www.math.nyu.edu/faculty /kohn/pde且nance . html
[7] B. 0ksendal. Stochαstic differential equαtions: An introduciton with α.pplicαtions. Sixth edition. SpringerVerlag , Berlin , 2003.
[8] M.A. Qazi. The mean value theorem and analytic functions of a complex variable. J. Mαth. Anal. Appl.
324 (2006) 30-38.
[9] Shen Xie-Chang. Mathematical analys叫 Volume 2 (in Chinese). Higher Education Press , Beijing , 1985.
[10] S. Shreve. Stochαstic cα lculus for finance II. Continuous-time models. Springer- Ve血g ， New York , 2004.
[11] E. T. Whittaker and G. N. Watson. A course of modern analys以 4th Edition. Cambridge University
Press , 1927.
[12] P. Wilmott , S. Howison , and J. Dewynne , The mαthematics of finαncω 1 derivα tives: A student introduction , Cambridge University Press , Cambridge , UK , 1995.
109