Phuong Trinh Bessel

7/31/2019 Phuong Trinh Bessel

1/15

Bi ging:Phng trnh vi phn TS. Nguyn Hu Th

- 239 -

Bi s 4B(8)

PHNG PHP FROBENIUS.

PHNG TRNH BESSEL. NG DNG CA HM BESSEL

1. Phng php ca Frobenius: Cc trng hp ngoi l

Nhc li:

nh ngha:Mt chui v hn dang

1 20 1 2

0 0

( ) ...r n n r r r r n nn n

y x x c x c x c x c x c x

+ + +

= =

= = = + + +

c gi l chui Frobenius

Xt phng trnh h s hng sng chiu (c im k d chnh quy 0x = )2

( ) ( ) 0 + + =x y xp x y q x y (10)

Phng trnh xc nh: 0 0( 1) 0r r p r q + + = y 0 0(0), (0)= =p p q q (16

Nu chui Frobenius r nny x c x= l nghim ca phng trnh vi phn (10), ths mrphi l mt nghim ca phng trnh xc nh (16).

Xt phng trnh vi phn tuyn tnh bc hai c 0x = l mt im k d chnh quy

2

( ) ( )0

p x q xy y y

x x + + = (1)

tc l khi c hai hm ( )p x v ( )q x u gii tch ti 0x = .

Khi tm c hai s m 1r v 2r ( gi s l s thc v r1 r2) l nghim ca phng

trnh xc nh

0 0( ) ( 1) 0 = + + =r r r p r q (2)

+ Lun tn ti mt chui nghim Frobenius 1r nny x a x= tng ng vi 1; ( 0)>r x .+ Nu *1 2 + = r r k : tn ti thm chui nghim Frobenius th hai

2r n

ny x b x= .

Cu hi:Cc trng hp khc liu c tn ti nghim chui Frobenius hay khng?

a. Trng hp khng cha logarit vir1 = r2 + N

Th chui lu tha : ( ) = nnp x p x ; ( ) = nnq x q x v chui Frobenius:

0 0

( ) r n n r n n

n n

y x x c x c x

+

= =

= = , vi 0 0c (3)

vo phng trnh vi phn dng2 ( ) ( ) 0x y xp x y q x y + + = (4)

Qua phng php ng nht h s ta thy:chuiFrobenius l nghim ca phng trnh(4), nusm r v h s 0 1 1, ,..., ,n nc c c c tho mn phng trnh

0 1 1( ) ( ; , ,..., ) 0n n nr n c L r c c c + + = . (9)

y l cng thc truy hi can

c theo cc s hng 0 1 1, ,..., nc c c .


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Gi s rng 1 2r r N= + vi N l s nguyn dng, 1 2r r> :

+ Khi h s 1( ) 0, 1r n n +

+ Khi 0 1 1, ,..., nc c c c xc nh, ta c th gii c phng trnh (9) vi nc v tip tc

tnh ton cc h s trong nghim dng chui Frobenius tng ng vi s m 1r.

+ Tuy nhin khng phi lc no cng tn ti nghim dng chui Frobenius tng ng vi

s m nh hn 2r .

V d 1. Xt phng trnhx2y" + (6x + x2)y' + xy = 0. (11)

+ yp0 = 6v q0 = 0, nn phng trnh xc nh l2( ) ( 1) 6 5 0r r r r r r = + = + = (12)

vi nghim 1 0r = v 2 1 25 5r r r N = = = .

+ Ta th chui Frobenius: n rny c x+= vo (11) nhn c:

1 1

0 0 0 0

( )( 1) 6 ( ) ( ) 0

+ + + + + +

= = = =

+ + + + + + + = n r n r n r n r n n n nn n n n

n r n r c x n r c x n r c x c x .

21

0 1

( ) 5( ) ( ) 0

+ +

= =

+ + + + + = n r n r

n n

n n

n r n r c x n r c x .

+ S hng tng ng vi 0n = cho phng trnh xc nh trong (12), vi 1n ta c2

1( ) 5( ) ( ) 0n nn r n r c n r c + + + + + = (13)

Ch rng h s ca cn l (r + n).+ y: 1 2 2 5r r N r = + = + . Vi 2 5r = , phng trnh (13) trthnh

1( 5) ( 5) 0 + =n nn n c n c (14)

- Nu 5n , ta c th gii phng trnh ny thu c cng thc truy hi

1 , 5nnc

c nn

= (15)

Suy ra : c1 = -c0,01

2 ,2 2

ccc = =

023 ,3 6

ccc = = v 3 04 ,4 24

c cc = = (16)

- Vi 5n = , phng trnh (14) cho ta 50. 0 0c + = nn 5c l hng s th hai tu chn,vn s dng cng thc truy hi trong (15) ta nhn c:

56 ,6

cc = 6 57 ,7 6.7

c cc = = 7 58 ,8 6.7.8

c cc = = ...... (17)

+ Khi chng ta kt hp kt qu (16) v (17), ta c5

0

2 3 4 6 7 85 5 5

0 51 ...2 6 24 6 6.7 6.7.8

n

n

n

y x c x

x x x x x xc x x c x x

=

=

= + + + + +

vi hai hng s tu 0c v 5c .

+ Do chng ta tm c hai nghim chui Frobenius ca p/t (11)2 3 4

51 ( ) 1 2 6 24

x x xy x x x

= + +

v

.

21 1

( 1) ( 1)( ) 1 1 120 .

6.7....( 5) ( 5)!

n n n n

n n

x xy x

n n

= =

= + = +

+ +


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V d 2. Cho phng trnh2 2( 8) 0x y xy x y + = (18)

Kim tra xem phng trnh c hay khng hai nghim dng chui Frobenius LTT.

Gii. + Ta cp0 = -1 v q0 = -8, nn phng trnh xc nh l2( ) ( 1) 8 2 8 0r r r r r r = = =

c hai nghim l r1 = 4 v r2 = -2 c hiu lN = 6.+ Th n rny c x

+= vo phng trnh (18) ta c

2

0 0 0 0

( )( 1) ( ) 8 0.n r n r n r n r n n n nn n n n

n r n r c x n r c x c x c x

+ + + + +

= = = =

+ + + + =

22

0 0

( ) 2( ) 8 0.n r n r n nn n

n r n r c x c x

+ +

= =

+ + + =

+ H s 1rx + cho ta 2 1) 2( ) 8 0n r n r c + + = , suy ra 1 0c = .

+ Vi n 2 ta c phng trnh2

2( ) 2( ) 8 0.n nn r n r c c + + + = (19)

+ Xt nghim nhr = r2 = -2. Khi phng trnh (19) c th vit di dngn(n - 6)cn + cn-2 = 0 (20)

vi n 2. Vi n 6ta c th gii ra cng thc truy hi nh sau

2

( 6)n

n

cc

n n

=

vi n 2, n 6 (21)

+ Bi v c1 = 0, cng thc (vi 0 0c ) ny cho ta

02 ,8

cc = c3 = 0,

024 ,8 64

ccc = = v c5 = 0.

+ Khi phng trnh (20) vi n = 6trthnh

060. 064

cc + = .

+ V c0 0 do khng c cch chn c6 tho mn phng trnh. Nn khng c nghimdng chui Frobenius tng ng vi nghim nh r2 = -2.

+ tm nghim n dng chui Frobenius tng ng vi nghim ln r1 = 4, ta th r = 4vo phng trnh (19) thu c cng thc truy hi

2

( 6)n

n

c

c n n

= + vi n

2, (22)

t ta c: 0 02 2( 1) ( 1) 6

,2.4...(2 )8.10...(2 6) 2 !( 3)!

n n

n n

c cc

n n n n

= =

+ +

+ Vy nghim chui Frobenius ca phng trnh (18):2

41 2

1

( 1)( ) 1 6

2 !( 3)!

n n

nn

xy x x

n n

=

= +

+ .

Ch : Khi ch c mt nghim 1( )y x dng chui Frobenius ca PTVPTT thun nhtcp hai tn ti, ta s dng phng php h bc tm nghim th hai c lp tuyn tnh.

M t phng php h bc: Xt PTVPTT cp hai:( ) ( ) 0y P x y Q x y + + = (23)


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trn khong mIm trn khong P v Q l cc hm lin tc.+ Gi s ta bit mt nghimy1 ca phng trnh (23). Ta s tmy2 sao cho cng vi y1

to thnh h nghim LTT.

+ Xt thng : 2

1

( )( )

( )

y xv x

y x= (24)

+ Nu ta bit v(x), thy2 s tm c bng cng thcy2(x) = v(x)y1(x) (25)

+ Ta bt u bng php th biu thc (25) vo phng trnh (23), vi

2 1 1y vy v y = + v 2 1 1 12y vy v y v y = + +

ta c

1 1 1 1 1 12 0vy v y v y P vy v y Qvy + + + + + =

+ Sp xp li ta c

1 1 1 1 1 12 0 + + + + + = v y Py Qy v y v y Pv y

+ Doy1 l nghim ca phng trnh (23) nn ta c phng trnh:

1 1 12 0 + + =v y vy Pv y (26)+ t v = u vi g/thity1(x) khng trit tiu trn I, khi phng trnh (26) trthnh

1

1

2 ( ) 0

+ + =

yu P x u

y(27)

+ Gii (27) ta nhn c:

( )2

21 1

exp - ( ).= = +

P x dxyv C dx K

y y

+ Chn C = 1 v K= 0, ta c

( )2 1 2

1

exp - P(x)dx.= y y dxy (28)

+ Hm s m khng trit tiu, y2(x) khng l bi s cay1(x), nny1 vy2 l hai nghimc lp tuyn tnh.

b. Trng hp logarithmic

By gita s nghin cu cng thc tng qut nghim th hai ca phng trnh

2

( ) ( )0,

p x q xy y y

x x

+ + = (1)

vi gi thit 1 2 + = r r N .

Gi s c nghim th nht dng chui Frobenius

11

0

( ) r nnn

y x x a x

=

= , (a0 0) (29)

vix > 0 tng ng vi s m ln r1.

t 2( ) ( )

( ) , ( )= =p x q x

P x Q xx x

, khi ta vit phng trnh (1) di dng

0y Py Qy + + = . (23) V phng trnh xc nh c hai nghim r1 v nghim r2 = r1 N, nn:

2 0 0 1 1

2 21 1 1

( 1) ( )( )( 2 ) ( ) 0

r p r q r r r r N

r N r r r r N

+ + = +

= + + =


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- 243 -

suy ra: p0 1 = N 2r1 , tc l:- p0 2r1 = -1 N (30)

Ta vit2

0 1 2 01 2

...( ) ...

p p x p x pP x p p x

x x

+ + += = + + +

th ta c

( ) 0 1 2exp - ( ) exp - ... = + + +

p

P x dx p p x dxx

0

20 1 2

21 2

1exp - ln ...

2

1exp ... ,

2

=

=

p

p x p x p x

x p x p x

nn

( ) ( )...1)(exp 2210 +++= xAxAxdxxPp (31)

Ta th (29) v (31) vo cng thc (28); vi la chn a0

= 1 trong (29), suy ra:

( )( )

....1

...122

212

221

121

0

+++

+++=

dxxaxax

xAxAxyy

r

p

Khai trin mu s v n gin, ta c

( )( )

( )

0 12 21 2

2 1 21 2

1 21 1 2

1 ....

1 ...

1 ... .

+ + +=

+ + +

= + + +

p r

N

x A x A xy y dx

B x B x

y x C x C x dx

(32)

Trng hp 1:S m bng nhau (r1 = r2 tc N=0),

ViN = 0 cng thc (32) trthnh

( )

( )

( )

1

1

22 1 1 2

21 1 1 2

21 1 1 2

2 31 0 1 2

1/ ...

1ln

2

1ln 1

2

ln

r

r

y y x C C x dx

y x y C x C x

y x x a x C x C x

y x x b x b x b x

= + + +

= + + +

= + + + + +

= + + + +

Nh vy trong trng hp s m bng nhau, dng tng qut ca y2 l

112 1

0

( ) ( ) ln

+

=

= + r nnn

y x y x x x b x . (33)

Trng hp 2: r1 = r2 +N (vi N > 0)

Lc ny (32) trthnh

( )1 22 1 1 21 .. ....N N

Ny y x C x C x C x dx = + + + + +


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2

11 1

11

1 1

1

1 0

1...

ln ( ....)1

1

ln ....1

N

N N

NN

N

r N n N

N nn

C Cy dx

x x x

C xxC y x y

N N

C x

C y x x a x x N N

+

+

+

=

= + + +

= + + + +

= + + + +

nn

2 10

( ) ( ) ln

=

= + Nr nN nn

y x C y x x x b x (34)

trong b0 = -a0/N.

NNH L 1.Trng hp ngoi l

Gi s rngx = 0 l mt im k d chnh quy ca phng trnh0)()(2 =++ yxqyxxpyx . (4)

Cho > 0 l gi tr nh nht ca bn knh hi t ca chui ly tha

=

=0

)(n

n

nxpxp v

=

=0

)(n

n

nxqxq .

Gi r1 v r2 l hai nghim, vi r1 r2 ca phng trnh xc nh0)1( 00 =++ qrprr .

(a) Nu r1 = r2, th phng trnh (4) c hai nghimy1 vy2 dng

=

=0

11)(

n

n

n

rxaxxy vi a0 0 (35a)

v

=

++=0

112

1ln)()(n

n

n

rxbxxxyxy . (35b)

(b) Nu 1 2 + = r r N , th phng trnh (4) c hai nghimy1 vy2 c dng

=

=0

11)(

n

n

n

rxaxxy vi a0 0 (36a)

v

22 1 0

0

( ) ( ) ln ; 0

=

= + r nnn

y x Cy x x x b x b . (36b)

V d 3. Chng ta s minh ha cho trng hp r1= r2 bng cch suy ra nghim th hai caphng trnh Bessel bc khng,2 2 0x y xy x y + + = (37)

+ Vi r1= r2 = 0. ta tm c nghim th nht ca phng trnh l

=

==

022

2

01 )!(2

)1()()(

nn

nn

n

xxJxy (38)

+ Theo phng trnh (35b) nghim th hai ca phng trnh c dng

2 11

( ) ( )ln nn

n

y x y x x b x

=

= + (39)

+ Tm cc h sn

b : o hm cay2 ta c


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112 1

1

( ) ( ) ln nn

n

yy x y x x nb x

x

=

= + +

v

21 12 1 2

2

2( ) ( ) ln ( 1) n

n

n

y yy x y x x n n b x

x x

=

= + +

Ta th vo phng trnh (37) v s dng tnh chtJ0(x) cng tha mn phng trnh ny nnta thu c:2 2

2 2 2

2 21 1 1 1

2

2 1 1

0

ln 2

( 1) ,n n nn n n

n n n

x y xy x y

x y xy x y x xy

n n b x nb x b x

+

= = =

= + +

= + + + +

+ + +

t ta c:

.)(2)!(2

2)1(20

32

222

21

122

2

=

=

++++

=n

n

nn

nn

nn

xbbnxbxbn

nx(40)

+ Do: n2bn + bn-2 = 0 nu n l s l, nn tt c cc h s c ch s l trongy2u trit tiu.+ H s vi ch s chn trong phng trnh (40): vi n 2, tha mn cng thc truy hi

222222

)!(2

2)1.(2)2(

n

nbbn

n

n

nn

=+ (42)

+ Xt php th:

222

1

2 )!(2

)1(

n

cb

n

n

n

n

+= (43)

v hy vng tim c cng thc truy hi cho c2nn gin hn b2n.+ Ta chn (-1)n+1 thay v (-1)n , v b2 = 1/4 > 0 nn vi n = 1 trong (43) ta c c2 = 1.+ Th (43) vo (42) ta c

12 2 2 2

2 2 22 2 2 2 2 2 2

( 1) ( 1) ( 2)( 2) 2 1(2 )

2 ( !) 2 (( 1)!) 2 ( !)

+

+ = = +

n n n

n nn nn n n

c c nn c c

n n n n,

+ Do : nn Hn

c =+++++=1

....4

1

3

1

2

112 (44)

trong ta nh nghaHn l tng ring thn ca chui iu ha )/1( n .+ Ch rng h s ca cc ch s l l bng 0, ta th (43) v (44) vo (39) thu c

nghim th hai1 2

2 0 2 21

2 4 6

0

( 1)( ) ( ) ln

2 ( !)3 11

( ) ln ...4 128 13824

+

=

= +

= + +

n n

n

nn

H xy x J x x

n

x x xJ x x

(45)

ca phng trnh Bessel bc khng.+ Chui ly tha trong (45) hi t vi mi x. Ta thng cng thc:

;2

)2ln(2

)( 210 yyxY

+=

tc l,

;)!(2

)1()(

2ln

2)( 22

21

00

+

+=

+

n

xHxJ

xxY

n

n

n

n

(46)

trong c k hiu l hng sEuler: lim( ln ) 0.57722n

nH n

= .

Hm Y0(x) c gi l Hms Bessel bc khng loi hai.


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V d 4. Nh mt phng php khc ca php th, ta minh ha trng hp r1 r2 = Nbngcch dng k thut rt gn bc thu c nghim th hai ca phng trnh Bessel bc mt,

0)1( 22 =++ yxyxyx (48)+ Phng trnh xc nh tng ng c nghim l r1 = 1 v r2 = -1.+ Ta bit: mt nghim ca phng trnh (48) l

...18432384162)!1(!2

)1(

2)()(

753

02

2

11 ++=+

==

=

xxxx

nn

xx

xJxyn

n

nn

(49)+ Vi P(x) = 1/x t (48), s h bc cng thc truy hi ca (28) cho ta

2 1 21

1 3 52

1

1

( ...)2 16 384

y y dxxy

y dxx x x

x

=

=

+

1 2 43 2

1 2 4 63

2 4 6

1 3

3

1 3

2 4

1 1 2

4

(1 ...)8 192

14

5(1 ...)

4 192 4608

1 74 1 ...

4 192 4608

1 1 7 194 ...

4 192 4609

1 7 19ln 4 ... .2 384 18432

=

+

=

+ +

= + + + +

= + + + +

= + + + +

y dxx x

x

y dxx x x

x

x x xy dx

x

x xy dx

x x

x xy x yx

Do 3 5

2 1

1 11( ) ( ) ln ...

8 32 4608

x x xy x y x x

x= + + + (50)

Ch rng k thut h bc d thy mt vi s hng u ca chui, nhng khng d tmra mt cng thc truy hi c th xc nh c s hng tng qut ca chui.

2. Phng trnh Bessel

Xt mt s trng hp ca phng trnh Bessel bc p 0,0)( 222 =++ ypxyxyx (1)

Nghim ca n c gi lHm Bessel bc p.Phng trnh xc nh ca (1) l r2 p2 = 0, p/t xc nh c hai nghim l r1 = p, v r2 = -p.Nu chng ta th += rmmxcy vo phng trnh (1), ta tm s hng sao cho c1 = 0 v

2 22( ) 0; 2 + + = m mm r p c c m . (2)

a. Trng hpr = p > 0

+ Nu ta r = p v th phng trnh (2) cho ta cng thc truy hi

2 ; : .(2 )

= =+

m

m m m

aa a c

m p m(3)

+ Bi v a1 = 0, suy ra am = 0 cho mi m nhn gi tr l.+ Vi cc s hng chn: cng thc tng qut l


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))...(2)(1(!2

)1(2

02

mpppm

aa

m

m

m+++

=

nn vi nghim ln r = p ta c nghim

=

+

+++

=

02

2

01 ))...(2)(1(!2

)1()(

mm

pmm

mpppm

xaxy (4)

+ Nup = 0y l l mt nghim dng chui Frobenius;+ Vi a0 = 1 ta c hmJ0(x) m ta bit.

b. Trng hpr = -p < 0 :

Xt: r = -p , phng trnh (2) c dng

2( 2 ) 0; : + = =m m m mm m p b b b c (5)

vi m 2 cn b1 = 0.+ Nup l s nguyn dng hoc l mt s nguyn dng l l bi ca s c cc kh nng

sau:- Khi m = 2p, phng trnh (5) l 0.bm + bm-2 = 0; do nu bm-2 0, th khng c gi tr

no ca bm c th tha mn phng trnh.- Nup l s nguyn dng bi l ca : Gi s rngp = k/2 trong kl s nguyndng l, khi ch cn chn bm = 0 cho mi gi tr l ca m v ta c:

2( ) 0; + =k kk k k b b

v phng trnh ny tho mn v bk= bk-2 = 0.+ T nup khng l s nguyn dng, c th ly bm = 0 vi m l s l v xc nh h s ca

ca cc ch s chn theo b0 bng cng thc truy hi

)2(2

pmm

bb mm

= vi m 2 (6)

+ So snh (6) v (3), ta thy (6) s cho kt qu tng t nh (4):

=

+++=

02

2

02 ))...(2)(1(!2)1()(

mm

pmm

mpppmxbxy . (7)

Cc chui trong (4) v (7) hi t vi mix > 0 vx = 0 l mt im k d chnh quy caphng trnh Bessel.

c. Hm Gamma

+ Xt Hm gamma(x), c nh ngha vix > 0 bi cng thc

=0

1)( dttex xt (8)

+ Nhn thy tch phn suy rng hi t vi mix > 0. rng

00

(1) lim 1b

t t

be dt e

= = = (9)

+ p dng tch phn tng phn hai v vi u = tx v dv = e-tdt:

1 1

00 0 0

( 1) lim lim limb b b

bt x t x t x t x

b b bx e t dt e t xe t dt x e t dt

+ = = + =

tc l,(x + 1) = x(x). (10)

+ Kt hp (9) v phng trnh (10) th ta c(2) = 1.(1)=1!, (3) = 1.2.(2)=2!, (4) = 3.(3)=3!,

+ Tng qut ha ta c : (n + 1) = n!, vi n 0 v l s nguyn. (11)+ Mt gi trc bit quan trng ca hm gamma l


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- 248 -

===

00

2/1 22)2/1( duedtte ut (12)

+ Nu -1 < x < 0, th

x

xx

)1()(

+= ;

v phi ca ca phng trnh c xc nh v 0 < x + 1 < 1. Cng thc tng t c thc sdng mrng cho nh ngha ca hm (x) trn khong m( -2, -1), th i vi khong m(-3, -2) ta cng lm tng t, v c nh vy. th ca hm gamma c ch ra trong Hnh3.5.1.

Hnh 3.5.1. th ca hm Gamma mrng.

d. Hm Bessel loi mt

+ Xt:

=

+

+++

=

02

2

01 ))...(2)(1(!2

)1()(

mm

pmm

mpppm

xaxy (4)

+ Nu ta chn a0 = 1/[2p(p +1)] vip > 0, v ch rng

(p + m + 1) = (p + m)(p + m 1)(p + 2)(p + 1) (p + 1)

khi Hm Bessel loi mt bcp rt ngn gn vi trgip ca hm gamma nh sau:

=

+

++=

0

2

2)1(! )1()( m

pmm

p xmpm

xJ (13)

+ Nup > 0 khng l s nguyn: xt

=

+++

=

02

2

02 ))...(2)(1(!2

)1()(

mm

pmm

mpppm

xbxy

v chn b0 = 1/[2-p(- p +1)] ta sc:

=

++

=

0

2

2)1(!

)1()(

m

pmm

p

x

mpmxJ (14)

ca phng trnh Bessel bc p. Nup khng l s nguyn, ta c nghim tng qut)()()( 21 xJcxJcxy pp += (15)

vi x > 0;xp

phi c thay th bng |x|p

trong phng trnh (13) n phng trnh (15) cc nghim ng vix < 0.+ Nup = n l s nguyn khng m, th phng trnh (13) cho ta

2

0

( 1)( )

!( )! 2

+

=

=

+

m nm

n

m

xJ x

m m n(16)

i vi phng trnh Bessel loi 1 vi bc nguyn. Do 2 2 4 6

0 2 2 2 2 2 2 2 20

( 1)( ) 1 ....

2 ( !) 2 2 4 2 4 6

=

= = + +

m m

mm

x x x xJ x

m(17)

v3 52 1

1 2 10

( 1) 1 1

( ) ....2 ( !)( 1)! 2 2! 2 2!3! 2

+

+=

= = + +

m m

mm

x x x x

J x m m (18)


11/15


- 249 -

th caJ0(x) vJ1(x)c ch ra trong Hnh 3.5.2. , khng im ca hmJ0(x) vJ1(x)l an xen nhau.

+ Vi n ln:- Khng im thn caJ0(x) xp x l: (n 1/4) ,- Khng im th n caJ1(x) xp x l (n + 1/4) .- Do khong gia khng im lin tip caJ0(x) hocJ1(x) l xp x tng t

cosx v sinx.

Hnh 3.5.2. th ca Hm Bessel J0(x) v J1(x)

Lu rngJp(x) l hm s cbn nu bc ca n l mt na ca mt s nguyn, V d

xx

xJ sin2)(2/1

= v xx

xJ cos2)(2/1

=

e. Hm Bessel loi hai

Hm s:221

20 0

( 1) ( )2 1 2 ( 1)! 1( ) ln ( )

2 ! !( )! 2

n mmn mnn m n

n n n mm m

H Hx n m xY x J x

m x m m n

+ +

= =

+ = +

+ (20)

c gi l hm Bessel loi hai vi bc nguyn n 0.

Nghim tng qut ca phng trnh Bessel c bc nguyn n l: )()()( 21 xYcxJcxy nn += f. Hm Bessel ng nht

Hm Bessel l tng t nh hm lng gic, chng u tho mn phn ln nhng tin ch phbin, c bit trong php tnh tch phn lin quan n Hm Bessel. Vi phn ca

=

+

++

=

0

2

2)1(!

)1()(

m

pmm

p

x

mpmxJ , (13)

vi p l s nguyn khng m cho ta: [ ] )()( 1 xJxxJxdx

dp

p

p

p

= . (22)

+ Tng t ta c

[ ] )()( 1 xJxxJxdx

dp

p

p

p

+

= . (23)

+ T ta c: )()()( 1 xJx

pxJxJ ppp = , (24)

v : )()()( 1 xJxJx

pxJ ppp += .

+Ta c th biu din hm Bessel bc cao hn thng qua hm Bessel bc thp hn:

)()(2

)( 11 xJxJx

pxJ ppp + = , (26)

+ V c th biu din hm Bessel bc l s m ln thng qua hm Bessel bc l s m nhhn


12/15


- 250 -

)()(2

)( 11 xJxJx

pxJ ppp + = (27)

Ch : khng khi no c hm Bessel c bc l s nguyn m xut hin. Ni chung, chngtn ti vi mi gi trp khng l nguyn.

V d 1. + Vi p = 0 phng trnh [ ] )()( 1 xJxxJxdxd

pppp = cho ta

+= CxxJdxxxJ )()( 10

+ Tng t, vi p = 0, phng trnh [ ] )()( 1 xJxxJxdx

dp

p

p

p

+ = cho ta

+= CxJdxxJ )()( 01 .

V d 2. Trc tin s dng p = 2 v sau s dng p = 1 trong phng trnh

)()(2

)( 11 xJxJx

pxJ ppp + = , ta c

),()()(24

)()(4

)( 101123 xJxJxJxx

xJxJx

xJ

==

sao cho

)(18

)(4

)( 1203 xJx

xJx

xJ

+= .

+Tng t, mi hm Bessel c bc l s nguyn dng u biu din thng qua J0(x) v J1(x).

V d 3. tnh c vi phnxJ2(x),u tin ta ch rng

CxJxdxxJx += )()( 11

21

theo phng trnh [ ] )()( 1 xJxxJxdx

dp

p

p

p

+ = vip = 1.

+ Do ta vit

= dxxJxxdxxxJ ))(()( 2

122

V tch phn tng phn hai v viu = x2 , dv = x-1J2(x)dx

du = 2xdx v v = -x-1J1(x)

ta c

( ) ( ) ( )2 1 1 1 0( ) 2 ( ) 2= + = + xJ x dx xJ x J x dx xJ x J x C.

3. Phng trnh tham s Bessel

Phng trnh tham s Bessel c bc n l0)( 2222 =++ ynxyxyx , (28)

trong l tham s dng.+ Php tht = x vo phng trnh (28) bin phng trnh (28) thnh phng trnh Bessel

22 2 2

2( ) 0

d y dyt t t n y

dt dt + + = (29)

vi nghim tng quty(t) = c1Jn(t) + c2Yn(t).

+ Do nghim tng qut ca phng trnh (28) ly(x) = c1Jn(x) + c2Yn(x) (30) By gita xt bi ton gi tr ring


13/15


- 251 -

( )

2 2 2( ) 0, [0, ]

0

+ + =

=

x y xy x n x L

y L(31)

Ta tm > 0 sao cho tn ti mt nghim khng tm thng ca (31) v n lin tc trn [0, L].

+ Nu chng ta vit = 2, th phng trnh vi phn (31) l phng trnh vi phn (28), nnnghim tng qut ca phng trnh c cho bi :

+ S lin tc cay(x) i hi c2 = 0. Do y(x) = c1Jn(x).+ iu kin biny(L) = 0 suy raz = L phi l mt nghim dng ca phng trnh

Jn(z) = 0. (32)+ Gi tr ring thkca bi ton trong (31) l

2

22 )()(

L

nkkk

== (33)

+ Hm ring tng ng l

= x

LJxy nknk

)( . (34)

4. Cc ng dng ca hm Bessel

Ta bit rng: nghim ca rt nhiu phng trnh vi phn tuyn tnh bc hai khc c thbiu din thnh hm Bessel.

Chng hn, xt phng trnh Bessel bcp c dng

0)( 222

22 =++ wpz

dz

dwz

dz

wdz , (1)

v php th, = =w x y z kx (2)

Sau bin i thng thng v dng0)()21( 2222222 =+++ yxkpyxyx ,

tc l,0)(2 =+++ yCxByAxyx q , (3)

trong A = 1 - 2, B = 2 2p2, C =2k2 v q = 2. (4)

+ Tc l:

q

BAp

q

Ck

qA 4)1(,

2,

2,

2

1 2 ===

= (5)

+ Gi s rng cn bc hai trong (5) l s thc, t nghim tng qut ca (3) l)()()( kxwxzwxxy == ,

trong )()()( 21 zYczJczw pp +=

l nghim tng qut ca phng trnh Bessel trong (1).

NNH L 1.Nghim theo hm BesselNu C > 0, q 0, v (1 - A)24B th nghim tng qut ( vix > 0) ca phng trnh (3) l:

1 2( ) ( ) ( )p py x x c J kx c J kx

= + (6)

Trong ,, kvpc cho bi phng trnh (5). Nup l s nguyn, thJ-pc thay thcho Yp.

V d 1. Gii phng trnh0)3(84 42 =++ yxyxyx (7)


14/15


- 252 -

Gii. + Ta vit li (7) di dng

2 41 32 ( ) 04 4

x y xy x y + + =

+ V ta thyA = 2, B = -3/4, C = 1/4, v q = 4.+ Do phng trnh (5) cho ta = -1/2,= 2, k = 1/4 v p = 1/2.+ Vy nghim tng qut trong (6) ca (7) l

1/ 2 2 21 1/ 2 2 1/ 2

1 1( ) ( ) ( )4 4

y x x c J x c J x

= +

+ Ta nhli rng:

( ) zz

zJ sin2

2/1

= v ( ) zz

zJ cos2

2/1

= ,

+ Do nghim tng qut ca phng trnh (7) c thc vit di dng

.4

sin4

cos)(22

2/3

+=

xB

xAxxy

V d 2. Gii phng trnh Airy09 =+ xyy (8)

Gii. + u tin ta vit li phng trnh di dng09 32 =+ yxyx .

+ y l mt trng hp c bit ca phng trnh (3) viA = B = 0, C = 9, v q = 3. Ttheo phng trnh (5) ta c = 1/2 , = 3/2 , k = 2 vp = 1/3 . Do nghim tng qut caphng trnh (8) l

1/ 2 3/ 2 3/ 21 1/3 2 1/3( ) (2 ) (2 )y x x c J x c J x = + .

Scong ca ct thng ng

+ Xt mt ct thng ng khng thay i s b on di sc nng ca chnh n.+ Chng ta lyx = 0 ti im t do trn nh ca ct vx = L > 0 ti im y ca n.+ Gi sim y c gn cnh trn mt phng; xem Hnh 3.6.1

Hnh 3.6.1. Ct cong

+ C ngha l gc lch ca ct ti imx l (x).

+ T l thuyt v sn hi ta c: 02

2

=+

xgdx

dEI , (9)

trong El sut Young ca cht lm nn ct,Il momen qun tnh ca mt ct ngang, l mt

tuyn tnh ca ct v g l gia tc trng trng.+ iu kin bin l:

(0) = 0, (L) = 0 (10)


15/15


+ Vi

EI

g == 2 , (11)

ta c bi ton gi tr ring02 =+ x ; (0) = 0, (L) = 0 (12)

+ Ct ch b on ch khi phng trnh (12) c mt nghim khng tm thng, nu khng ct

s trv trng thi c ca n khng b lch khi mt thng ng.+ Phng trnh vi phn (12) l phng trnh Ariy viA = B = 0, C = 2v q = 3.

+ Theo (5) ta c = 1/2 ,= 3/2,2

3k = vp = 1/3, nn nghim tng qut l:

1/ 2 3/ 2 3/ 21 1/3 2 1/3

2 2( )

3 3x x c J x c J x

= +

(13)

+ Ta thp = 1/3 vo

=

+

++

=

0

2

2)1(!

)1()(

m

pmm

p

x

mpmxJ ,

v sau n gin ha ta c

+

+

+

= ...

18061

)3/2(

3...

50412)3/4(3)(

6432

3/1

3/12

7442

3/1

3/11 xxcxxx

cx

+ Tiu kin im cui (x) = 0 suy ra c1 = 0, nn

1/ 2 3/ 22 1/3

2( )

3x x c J x

=

. (14)

+ / kin (L) = 0 cho ta

3/ 21/ 3

20

3J L

=

. (15)

+ Do ct s on ch khi 223

z L= l nghim ca phng trnhJ-1/3 = 0. th ca

+

=

=

1

2

23/1

3/1)13.(5.2!2

3)1(1

)3/2(

)2/()(

mm

mmm

mm

zzzJ (16)

c ch ra trong Hnh 3.6.2, ta thy rng khng im nh nht dngz1 l nh hn 2.

Bi tp v nh: Cc bi tp lTr332n 335; 344n 347; 352, 353c trc cc Mc: 4.1, 4.2 v 4.3 chuNn b choBi s5A(9)

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Phuong Trinh Bessel