數(shù)學(xué)雙專-常微分方程第四章

§

4.1General

Theory

of

Higher-Order

Linear

ODE§

4.1內(nèi)容回顧(4.2)21(t)x

a

(t)x

0nn1x(n)

a

(t)x(n1)

a解的性質(zhì)與結(jié)構(gòu)。?n階齊次線性方程的所有解構(gòu)成一個(gè)n

維線性空間方程(4.2)的一組n個(gè)線性無關(guān)解稱為它的一個(gè)基本解組。本節(jié)

要求/Requirements/熟練掌握常系數(shù)齊次線性方程的求解方法熟練掌握常系數(shù)非齊次線性方程的求解方法熟練掌握

方程的求解方法3§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE結(jié)構(gòu)齊線性方程的通解可由其基本解組線性表示。非齊線性方程的通解等于對應(yīng)齊次方程的通解與自身的一個(gè)特解之和。非齊線性方程齊線性方程非齊線性方程通解特解基解組表示關(guān)鍵常數(shù)變易法4§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE4.2.1復(fù)值函數(shù)與復(fù)值解/Complex

Function

and

Complex

Solution/—

定義z(t)

(t)

i

(t)t

[a,b]，(t)，

(t)是定義在[a,b]上的實(shí)函數(shù)。極限lim

z(t)

lim

(t)

i

lim

(t)tt0

tt0

tt0t0

[a,b]，連續(xù)lim

z(t)

z(t0

)

t0

[a,b]，tt0導(dǎo)數(shù)tt0lim

z(t)

z(t0

)

lim

(t)

(t0

)

i

lim

(t)

(t0

)tt50dt

dtdt

t

tt

t0t

t000

t

t0z(t)

z(t

)

dzlim

0

z(t

)

t

t0

t

t0

t

t0

d

iddtd212z

(t))

1(z

(t)

dt

dtdtdz

(t)

dz

(t)1d

[cz

(t)]

c

dz1

(t)dt

ddt如dtz2

(t)

z1(t)dtdz

(t)dz

(t)(z1

(t)

z2

(t))

21§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE易驗(yàn)證j

1,2

t

[a,b]，z

j

(t)

j

(t)

i

j

(t)1

2

1

1

2

2dt

dt

d

(z

(t)

z

(t))

d

(

(t)

i

(t)

(t)

i

(t))dt1

2

1

2

d

{[

(t)

(t)]

i[

(t)

(t)]}dt

dt1

2

1

2

d

[

(t)

(t)]

i

d

[

(t)

(t)]

i

d

2

)

dz1(t)

dz2

(t)dtdtdtdt

(

d1

i

d1

)

(

d2dt

dt6二

關(guān)于ekt表示k

i共軛復(fù)數(shù)，定義ekt

e(

i

)t

eteit

et

(cos

t

i

sin

t)e(

i

)t

et

(cos

t

i

sin

t)eit

cos

t

i

sin

teit

cos

t

i

sin

tk

ik

i

,為實(shí)數(shù)，t為實(shí)變量。ekt

e(

i

)t

e(

i

)t

et

(cos

t

i

sin

t)

et

(cos

t

i

sin

t)

ekt7§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODEekt的性質(zhì)1）

e(k1

k2

)t

ek1t

ek2t

kektdektdt2）

k

nekt8dt

nd

nekt3）結(jié)論實(shí)變量的復(fù)值函數(shù)的求導(dǎo)公式與實(shí)變量的實(shí)值函數(shù)的求導(dǎo)公式一致。實(shí)變量的復(fù)指數(shù)函數(shù)的求導(dǎo)公式與實(shí)變量的實(shí)指數(shù)函數(shù)的性質(zhì)一致?！?/p>

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODEdt

nnd

xd

n1xdt

n11

a

(t)n1dtn

a

(t)

dx

a

(t)x

f

(t)

(4.1)§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE三

線性方程的復(fù)值解/Complex

Solution

of

Linear

Higher-Order

ODE如果定義在

[a,

b]

上的實(shí)變量的復(fù)值函數(shù)

x

z(t)

滿足方程則稱

x

z(t)

為方程的一個(gè)復(fù)值解。d

n1xdtn19nd

x1

a

(t)dtn1

n

a

(t)

dx

a

(t)x

0

(4.2)dtn定理8

如果方程4.2中所有系數(shù)ai

(t)(i

1,2,,

n)都是實(shí)值函數(shù)，而

x

z(t)

(t)

i

(t)

是方程的復(fù)數(shù)解，則

z(t)

的實(shí)部

(t)，虛部

(t)

和共軛復(fù)數(shù)函數(shù)

z(t)也是方程4.2的解。d

n

x

dx定理9

若方程

dtn

a1(t)

dt

n1

an1

(t)

dt

an

(t)x

u(t)

iv(t)d

n1x有復(fù)數(shù)解x

U

(t)

iV

(t),這里ai

(t)(i

1,2,...,n)u(t)及v(t)都是實(shí)函數(shù)。那么這個(gè)解的實(shí)部U

(t)和虛部V

(t)

分別是方程dx

an1

(t)

dt

an

(t)x

u(t)d

n

x

a1

(t)

dt

n1d

n1

xdt

n和d

n

x

dx10

an1

(t)

dt

an

(t)x

v(t)a1

(t)

dt

n1d

n1

xdt

n的解。§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODEdx

an1

dt

an

x

0d

n1

xdt

n1L[x]

adt

n

1d

n

x…….(4.19)為常數(shù)。其中a1,a2

,...,an為了求方程(4.19)的通解，只需求出它的基本解組。n

階常系數(shù)齊次線性方程x

e

t1n1nn1

tL[et

]

net

a

e

aet

a

et

01nn1nF

()

a

n1

a

a

0

…….(4.21)etF

()

0

滿足etx

e

t結(jié)論：x

e

t

是方程(4.19)的解的充要條件

滿足F

()

0特征方程特征根§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE4.2.2

常系數(shù)齊線性方程和

方程/Coefficient

Linear

Homogenous

Higher-OrderODE

And

Euler

Equation/L[et

]

et

F()11§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE下面根據(jù)特征根的不同情況分別進(jìn)行

。1)特征根為單根的情況設(shè)1

,2

,,n

是特征方程（4.21）的n個(gè)互不相等的根，則相應(yīng)的方程（4.19）有如下n個(gè)解e1t

,

e2t

,

,

ent這n個(gè)解在區(qū)間

t

上線性無關(guān)，從而組成方程1

n1

nnn1

a

a

a

0F

()

的基本解組。事實(shí)上，1213nn

te1tn

t

e

eW

(t)

e121.............

ee.....21n1

t2e2t

ent

tn1

t2112

nn1n1n111.....................n1ent范行列1....12....n............

e1

2

n(

)te1t

,

e2t

,

,

ent是方程的基本解組。方程4.19的通解可表示為

x

c

e1t

c

e2t

c

ent2

n1(Vandermonde)式i

j

0(i

j)§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE14對共軛的出現(xiàn)，設(shè)1

i2

i則方程（4.19）有兩個(gè)復(fù)值解方程的一個(gè)特征根也是一個(gè)特征根e(

i

)

t

e

t

(cos

t

i

sin

t)e(

i

)

t

e

t

(cos

t

i

sin

t)§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE如果特征方程有復(fù)根，則因方程的系數(shù)是實(shí)常數(shù)。復(fù)根將成對應(yīng)兩個(gè)實(shí)值解e

t

cos

t,

e

t

sin

t例1求方程dt

415d

4

x

x

0的通解。解第一步：求特征根F()

4

1

0第二步：求出基本解組3,4

i1,2

1,et

,

et

,cost,

sin

t第三步：寫出通解x(t)

c

et

c

et

c

cost

c

sin

t1

2

3

4§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE例2求方程x

0dt3d

3

x

的通解。解第一步：求特征根F()

3

1

02

23

1

i2,31

1,

第二步：求出基本解組e

t

,sin

ttte

cos

t,

e

32

321212第三步：寫出通解t16t

32sin

t3t

c

e

32121x(t)

c

et

c

e2

cos1

2§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE17§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE2)特征根有重根的情況設(shè)1

,2

,,m

是特征方程（4.21）的m個(gè)互不相等的根。dt

nL[x]

d

n

xdx

a1

an1

dt

an

x

0dt

n1d

n1

x…….(4.19)1

n1

nn1

a

0

a

aF

()

n…….(4.21)重?cái)?shù)k1

,

k

2

,

,

km

k1

k

2

km

n,

ki

1I.

設(shè)1

0

是

k1

重特征根1nk1n1

a

0n

k1

a

1nk

1n

n1a

a

a

0

0111n1d

n1x

a1

ankkd

k

xnd

n

xdtdtdt

01an

k顯然

1,t,t

2

,,tk1

1是方程的k1

個(gè)線性無關(guān)的解，方程(4.19)有

k1

重零特征根方程恰有k1

個(gè)線性無關(guān)的解1,t,t

2

,,tk1

1II.

設(shè)1

0

是

k1

重特征根令x

ye1

tdt

ndtn1

nn

n1L[x]

d x

a

d x

a

dx

a x

01

dt

n1…….(4.19)118d

k1

x

ank

01d

n1xdtn1a1d

n

x

dtkdtn§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODEy

b

y)

0

b2

n1e1t

(y(n)

b

y(n1)

b

y(n2)n特征方程

G()

n

b1n1

bn1

bn

0(4.24)19L[ye1t

]

y

b y

0n…….(4.23)L

[

y]

y(n)11

b

y(n1)1

b

y(n2)

b2n11L[

ye1t

]

e1t

L

[

y]§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODEx(m)

(

ye1t

)(m)

111111m

t(m)

t

ye

yem(m

1)1

2!y

e

my

e

2

(m2)

t(m1)

tF(

1)

G()(4.19)的k1重特征根1(4.23)的k1

重特征根零,

j

1,2,,

k1120dG

j()d

jd

jjdF

(

)11(

)tF

(

)e

L[e(1

)t

]

L[e

te1t

]11

t

t

e L

[e

]1(

)t

e

G()1L[

ye1t

]

e1t

L

[

y]1j

1,2,,

k

11(j

)F

(

)

0,11(k

)F

(

)

0,§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE方程(4.23)恰有k1

個(gè)線性無關(guān)的解1,t,t

2

,,tk1

112mk1k

2kme1t

,te1t

,t

2e1t

,,tk1

1e1te2t

,te2t

,t

2e2t

,,tk2

1e2temt

,temt

,t

2emt

,,tkm

1emtk1

k

2

km

n,

ki

1由

x

ye1

t方程(4.19)恰有

k1

個(gè)線性無關(guān)的解

e1t

,te1t

,t

2e1t

,,tk1

1e1t類似地基21本解組(4.26)§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE22j證明

假若這些函數(shù)線性相關(guān)，則存在不全為零的數(shù)

A

(

r

)使得0

1

k1

1(A(1)

A(1)t

A(1)

t

k1

1

)e1t(A(2)

A(2)

t

A(2)

t

k2

1)e2t0

1

k2

1(A(m)

A(m)

t

A(m)

t

km

1

)emt

00

1

km

11P

(t)e

1t

P

(t)e

2t

P

(t)e

mt

02

m(4.27)假定多項(xiàng)式

Pm(t)

至少有一個(gè)系數(shù)不為零，則

Pm(t)不恒為零，2

1m

P

(t)e

(m

1

)t

0(

)tP1

(t)

P2(t)e微分k1

次§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE[P

(t)e

(r

1

)t

](k1

)rrkr

1

r

1

r(k

)r

11(

)t(k

1))

P

(t)]e(t)

(r

1(t)

k

(

)P1

[Pr

Q

(t)e

(r

1

)tQm

(t)

不恒為零，

02mQ

(t)e

(2

1

)t

Q

(t)e

(m

1

)t(Qm

(t))

(Pm

(t))(Qr

(t))

(Pr

(t))

0mR

(t)e

(m

m1

)t(Rm

(t))

(Pm

(t))Rm

(t)

不恒為零，e(m

m1

)t

0！(4.26)

中函數(shù)線性無關(guān)，其構(gòu)成的解本解組。23241

i2

i§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE方程的一個(gè)k

重特征根也是一個(gè)k

重特征根它們對應(yīng)2

k個(gè)線性無關(guān)的實(shí)解是e

t

cos

t,

te

t

cos

t,,

t

k

1e

t

cos

t,e

t

sin

t,

te

t

sin

t,,

t

k

1e

t

sin

t,例3

求方程

d

3

x

3

d

2

x

3

dx

x

0dt3

dt

2

dt25的通解。解第一步：求特征根F()

3

32

3

1

0第二步：求出基本解組et

,

tet

,

t

2et

,第三步：寫出通解1,2,3

1,x(t)

c

et

c

tet

c t

2et1

2

3§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE例4求方程d

4

x

d

2

x

dt

4

2

dt226x

0的通解。解第一步：求特征根F

()

4

22

1

0第二步：求出基本解組1,2

i

二重根cos

t,

t

cost,

sin

t,

t

sin

t,第三步：寫出通解x(t)

c1

cost

c2

t

cost

c3

sin

t

c4

t

sin

t§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE27作業(yè)：P.113，第4，6，7，8，9題P.145，第2，3，4，5，6題§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE可化為常系數(shù)線性方程的方程------- (Euler)

方程(4.29)

a1xn1

d

n1

ydxn1dy

an1x

dx

an

y

f

(x)dxnxn

d

n

y其中a1,a2

,...,an引入自變量代換為常數(shù)。x

etln

x

tdx

et

dtdy

dy

dt

et

dy

1

dy(edx

dt

dxtdt

x

dtd

2

y

d

dy

)

dx2

dx

dtdt

2

dx(et

edy

t

d

2

y

dtdt)d

2y

dy2t1

d

2

y

dy

)dt

e(

dt

2

dt

)

x2

(

dt

2§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE28假設(shè)dxm

xm

(

dtmdtm1m1d

ymdm

y

1

d

y

1

m1dtdy

)自然數(shù)m

有以下關(guān)系式成立，1,2

,,m1為常數(shù))]dxm1dm1

y

ddtdydtdx

x

dtmm1

d

y[

m

(m1m1d

y

1

m1

m1

dtdy

)]

dtdxdtmmd

ydmtdtm1m1d

y[e

(

1dtdt

dt

mdtmmd

ym1d

ydtm1d

m1

ydt

m1mtdtdy)]et

m1d d

m

yemt

m1dtdy

)

[me

(

(

11)291

xm1

(

dtm1

1dm1

ydtdydtmdm

y

m§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE對一切自然數(shù)m

均有以下關(guān)系是成立，dm

y

1

dm

ydxm

xm

(

dtmdm1

y

dtm1

1dy

m1dt

)原方程(4.30)d

n1

yb1

dt

n1tn

b

y

f

(e

)dt

nd

n

y

dtdy

bn1可化為常系數(shù)線性方程(4.29)30

a1xn1

d

n1

ydxn1dy

an1x

dx

an

y

f

(x)dxnxn

d

n

yx

et§

4.2

Solving

Method

of

Constant

Coefficients

Linear

ODE方程x

et常系數(shù)線性方程y

ek

ty

xk

F

(k)

0F

(

k

)

0確定求解方程的過程

a1xn1

d

n1

ydxn1dy

an1x

dx

an

y