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7/23/2019 Sng n hi v ng dng trong a chn

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I HC THI NGUYNTRNG I HC KHOA HC

PHM THY LINH

SNG N HI V NG DNGTRONG A CHN

LUN VN THC S TON HC

THI NGUYN - 2014


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I HC THI NGUYNTRNG I HC KHOA HC

PHM THY LINH

SNG N HI V NG DNGTRONG A CHN

LUN VN THC S TON HC

Chuyn ngnh : TON NG DNGM s:60 46 01 12

Ngi hng dn khoa hc:

TS NGUYN VN NGC

THI NGUYN, 2014


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i

Mc lc

M u 1

1 Thit lp phng trnh sng n hi 41.1 Bin dng v ng sut n hi . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 Trng thi n hi ca vt . . . . . . . . . . . . . . . . . . . . . 41.1.2 Khi nim v ng sut (Stress) . . . . . . . . . . . . . . . . . . 41.1.3 Khi nim v s bin dng (Strain) . . . . . . . . . . . . . . . . 5

1.2 Cc hng s n hi v nh lut Hooke suy rng . . . . . . . . . . . . 61.2.1 Cc hng s n hi . . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 nh lut Hooke suy rng . . . . . . . . . . . . . . . . . . . . . 7

1.3 Mt nng lng bin dng . . . . . . . . . . . . . . . . . . . . . . . 81.4 Phng trnh cn bng sng n hi . . . . . . . . . . . . . . . . . . . 8

1.4.1 Lc to bi ng sut . . . . . . . . . . . . . . . . . . . . . . . . 81.4.2 nh lut hai Newton - H phng trnh cn bng Navier -

Cauchy - H phng trnh Lame . . . . . . . . . . . . . . . . . . 91.4.3 Ta tr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4.4 Ta cu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.5 iu kin u, iu kin bin v cc bi ton lin quan ca cc phngtrnh sng n hi. nh l v duy nht nghim . . . . . . . . . . . . . 111.5.1 iu kin u . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5.2 iu kin bin . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5.3 Bi ton Cauchy . . . . . . . . . . . . . . . . . . . . . . . . . . 121.5.4 Bi ton bin-gi tr ban u . . . . . . . . . . . . . . . . . . . 121.5.5 nh l v duy nht nghim . . . . . . . . . . . . . . . . . . . . 12

2 Sng iu ha-Cc sng n hi iu ha c bn 132.1 Mt s kin thc b tr . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.1 Khi nim v sng iu ha . . . . . . . . . . . . . . . . . . . . 132.1.2 Khi nim vhm Dirac v hm Heaviside H(x) . . . . . . . 14

2.2 Biu din nghim ca phng trnh sng n hi . . . . . . . . . . . . 152.2.1 H khng c ngun . . . . . . . . . . . . . . . . . . . . . . . . . 152.2.2 H c ngun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.3 Sng P, sng S, sng SV, sng SH v sng PSV . . . . . . . . . . . . . 192.3.1 Sng P v sng S (P-sng v S-sng) . . . . . . . . . . . . . . . 192.3.2 Sng SV, sng SH v sng PSV . . . . . . . . . . . . . . . . . . 20

2.4 Vn tc pha v vn tc nhm . . . . . . . . . . . . . . . . . . . . . . . 202.4.1 Vn tc pha . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20


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2.4.2 Vn tc nhm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.4.3 Vn tc tc pha v vn tc nhm ca mt s mi trng . . . . 22

2.5 Sng khi phng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.5.1 Pht biu bi ton . . . . . . . . . . . . . . . . . . . . . . . . 22

2.5.2 Bi ton gi tr ban u (Bi ton Cauchy)i vi sng khiphng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.5.3 Bi ton gi tr bin n gin ca sng phng . . . . . . . . . . 242.6 Sng cu i xng sinh bi h thun nht . . . . . . . . . . . . . . . . 252.7 Sng cu c sinh bi ngun im n . . . . . . . . . . . . . . . . . 28

2.7.1 Cc th v ca ngun . . . . . . . . . . . . . . . . . . . . . . . . 282.7.2 Phng trnh ca cc th v . . . . . . . . . . . . . . . . . . . . 292.7.3 Cng thc ca cc chuyn v . . . . . . . . . . . . . . . . . . . . 30

3 S phn x v khc x ca cc sng n hi phng 34

3.1 Cc phng trnh c bn . . . . . . . . . . . . . . . . . . . . . . . . . . 343.1.1 Cc phng trnh lin quan ti hai na khng gian . . . . . . . 343.1.2 Th v phng iu ha . . . . . . . . . . . . . . . . . . . . . . . 35

3.2 Phn x v khc x ca sngSH . . . . . . . . . . . . . . . . . . . . . 363.2.1 H s phn x v khc x . . . . . . . . . . . . . . . . . . . . . 363.2.2 Phn x ton phn . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.3 Phn x ca sng P ti mt b mt t do . . . . . . . . . . . . . . . . 40

Kt lun 45

Ti liu tham kho 46


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1

M u

Di tc dng ca ngoi lc mt vt rn no b bin dng. Nu vt c thkhi phc c hnh dng v kch thc ban u th trng thi bin dng nitrn c gi l trng thi n hi. Ngc li, trng thi bin dng c gi ldo, hay l d.

L thuyt n hi (LTH) l mn hc nghin cu v chuyn v, bin dngv ng sut xut hin trong vt th n hi di tc dng ca ti trng, hay vachm vi cc vt khc. Ni dung ca LTH bao gm cc vn chnh sau y:Thit lp quy lut vt l c bn ca LTH (nh lut Hooke) gia ng sutv bin dng.Thit lp cc phng trnh c bn ca LTH, cc iu kin bin v iu kinu.Nng lng n hi.nh l v tn ti v duy nht nghim, v.v..

L thuyt n hi l c s tnh ton bn bin dng v n nh trongk thut xy dng, trong ch to my, khai khong v cc lnh vc khc ca kthut, cng nh trong a chn v.v..

Mt trong nhng ng dng quan trng ca vic nghin cu sng n hi lng dng trong a chn hc. Sng a chn l nhng dng sng nng lng hnhthnh v lan truyn bi s va chm ca lp a tng khi xy ra ng t. Snga chn c nhiu dng vi nhiu cch lan truyn khc nhau, trong c thphn ra hai nhm ln l sng khi v sng b mt. Sng khi c th lan truyntrong cc tng t pha su, cn sng b mt ch lan truyn lp t pha trnca v tri t. C hai dng sng khi chnh l sng P (premier wave-sng scp) v sng S (secondary wave-sng th cp).

Thm d a chn l phng php a vt l nghin cu c im trngsng dao ng n hi trong mi trng t nhm gii quyt cc nhim v

a cht khc nhau, nh nghin cu cu trc v qu t, tm kim thm d dukh v ti nguyn khong sn, nghin cu nn mng cng trnh...Trong thm d


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a chn, bng cc kch ng nhn to nh n mn, rung, p...ngi ta kchthch vo mi trng a cht cc xung lc. S kch thch lc lm t rungng v lm xut hin sng n hi. Cc sng ny truyn qua hay phn x trn

cc lp t v cc my thu s ghi nhn thi gian cc sng phn x truynn dng cc bng a chn phn nh cc thng tin v lp t cn thm d.

Lun vn tp trung nghin cu v bn cht ca sng n hi, cc loi sngn hi v mt s ng dng trong cc vn ca ca a chn hc. B cc lunvn gm c 3 chng:

Chng 1: Thit lp phng trnh sng n hi tuyn tnh c in. Trongchng ny c trnh by mt s kin thc v bin dng v ng sut n hi,nh lut Hooke, mt nng lng v vn c bn nht ca chng ny,

l phng trnh cn bng ca sng n hi. trnh by cc iu kin bin viu kin u i vi h phng trnh sng n hi.

Chng 2: Sng iu ha - Cc sng n hi c bn. Ni dung chnh cachng ny gm c cc phn sau y.

Phn th nht trnh by cch tm nghim ca h phng trnh sng nhi bng phng php th v, a h phng trnh n hi v cc phng trnhsng tuyn tnh cp hai thuc lp phng trnh hyperbolic ca l thuyt ccphng trnh o hm ring.

Phn th hai trnh by v sng iu ha v cc loi sng n hi, nh sngP (sng s cp), sng S (sng th cp), cc sng th cp ng ( SV), th cpngang ( SH), sng s-th cp ng (PSV), v.v..

Phn th ba trnh by ba bi ton c bn ca phng trnh sng. l biton Cauchy v bi ton bin gi tr ban u i vi sng khi phng; bi tonbin gi tr ban u i vi sng cu i xng thun nht khi bit cc iu kinu v cc iu kin bin trn mt mt cu nh cho trc; bi ton Cauchy ivi sng cu c ngun im gc ta .

Chng 3: S phn x v khc x ca sng n hi phng. Chng ny trnhby s phn x v khc x ca sng SH, sng P v sng SV trn b mt phncch z= 0ca hai na khng gian l nhng mi trng n hi khc nhau.

Cc vn c cp trong lun vn ny c th tm thy nhng ng dngca a chn hc v ch yu c hnh thnh t cc ti liu [1], [2], [3], [4], [5]v [6].

Lun vn ny c hon thnh ti trng i hc Khoa hc - i hc ThiNguyn. Tc gi xin by t lng bit n su sc vi TS Nguyn Vn Ngc, ngi

Thy hng dn tn tnh v ng vin tc gi trong sut thi gian nghincu va qua.


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Xin chn thnh cm n ti cc thy, c gio trong B mn Ton - Tin, Phngo to Khoa hc v Quan h quc t , cc bn hc vin lp Cao hc TonK6B trng i hc Khoa hc - i hc Thi Nguyn, v cc bn ng nghip

to iu kin thun li, ng vin tc gi trong qu trnh hc tp v nghincu ti trng.

Tc gi cng xin by t lng bit n su sc ti gia nh v ngi thn lunkhuyn khch, ng vin tc gi trong sut qu trnh hc tp v lm lun vn.

Mc d c nhiu c gng nhng lun vn kh trnh khi nhng thiu st vhn ch. Tc gi mong nhn c nhng kin ng gp qu bu ca cc thyc v bn c lun vn c hon thin hn.

Thi Nguyn, ngy 10 thng 9 nm 2014Hc vin

Phm Thy Linh


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Chng 1

Thit lp phng trnh sng n hi

Chng ny trnh by c s l thuyt ca l thuyt n hi tuyn tnh, bao

gm nh lut Hooke suy rng v mi lin h gia ng sut v bin dng, ccphng trnh cn bng, v.v.. Ni dung ca chng ny ch yu c hnh thnht cc ti liu [1], [2], [3] v [4].

1.1 Bin dng v ng sut n hi

1.1.1 Trng thi n hi ca vt

Di tc dng ca ngoi lc mt vt rn no b bin dng. Nu vt cth khi phc c hnh dng v kch thc ban u th trng thi bin dngni trn c gi l trng thi n hi. Ngc li, trng thi bin dng c gil do, hay l d. Xt v d: Gi s ta ko nh mt l xo khi v tr cn bngri bung th chic l xo vn c hnh dng v kch thc ban u, ta c trngthi n hi. Nu ta ko mnh hoc qu mnh th sau khi bung, chic l xokhng th ly li c hnh dng v kch thc ban u na, thm ch b phhu (hng) hon ton, ta c trng thi do (d), hoc ph hu.

1.1.2 Khi nim v ng sut (Stress)

Trong l thuyt n hi, ng sut l mt khi nim quan trng. Ta hnh dungqua im xca mi trng v mt mt nh S.Lc n hi gia cc phnca mi trng pha ny ca Stc ng ln cc phn t ca mi trng pha bn kia.

Hp lc ca cc lc ni trn bng nhau v ln nhng c chiu ngc nhau(lc trc i). ni v chiu ca cc lc , ta v php tuyn nn mt S.

Gi s rng, cc lc tc dng ln Stheo hng ca php tuyn nv tnh hc


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tng ng vi hp lc Tv moment ca cp lc M.Xt cc i lng

= T/(S), = M/(S).

Gi thit rng i vi cc i lng ny tn ti gii hn khi S0.Cc giihn ny ph thuc vo im xv php tuyn n.Nu chn hng ngc li canth cc gii hn s i du. Nu qua im xta chn mt khc S th ta lic hnh nh khc.

Ta gi thit rng cc gii hn trn y khng thay i i vi mi mt Si qua im xc cng php tuyn n.K hiu

(n) = lim

S0T/(S), (n) = lim

S0M/(S).

Cc i lng (n),(n) ph thuc vo im x v thi im t. i lng (n)

c gi l ng sut lc, cn i lng (n) c gi l ng sut moment tiim xtheo hng ca php tuyn n.

Trong khng gian ba chiu Oxyz, nu ti im x = (x,y,z)xt cc mt nh i qua xvung gc vi cc trc to vi php tuyn c hng l hngca cc trc to , th c cc vct ng sut:

x

= (xx, xy, xz),

y = (yx, yy , yz),

z= (zx , zy , zz),

trong xx, xy, xztng ng l to ca vect x trn cc trc to Ox,Oy,Oz. Tng t i vi cc vc t y v z.ng sut c tnh cht

xy =yx, xz=zx , . . . .

Trong nhiu ti liu dng k hiu

x = xx, y =yy , z=zz.

1.1.3 Khi nim v s bin dng (Strain)

Chuyn v.K hiu rMl bn knh vct ca im Mv P(rM, t)l ng sut,p sut (trong cht lu) ti imMca khng gian Euclid Rn thi imt.nv ca ng sut l N/m2,hay P a(Paxcan). Xt im M(x,y,z) R3 thuc i

tng nghin cu thi im t.Di s tc dng ca cc lc, trng thi cn


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bng im M s dch chuyn (rt nh) n im M(x, y, z).Vc t uM= MM

c gi l vc t chuyn v ca im M : uM = u(x,y,z; t).Ta c h thc:

rM = rM+ uM

.Bin dng. Trong gii hn n hi c ba loi bin dng c bn:1) Bin dng nn ( Compressional strain) = Th tch thay i/ Th tch banu.2) Bin dng trt n gin (Simple shear strain ) = S gia ca di/ diban u.3) Bin dng trt thun tu (Pure shear strain)= B nn + B ko (din tch

(area) khng thay i, gc thay i).

Gi s di tc dng ca lc vc t chuyn v ca im (x,y,z)lu= (u1, u2, u3).

bin dng theo cc phng Cauchy- Navier

ij =1

2

uixj

+ ujxi

, ij =ji. (1.1)

bin dng gin n

1) Th tch nguyn t ban u (Original Volume):V =xyz,

2) Th tch nguyn t bin dng (Deformed Volume):

V +V= (1 +xx)(1 +yy)(1 +zz)xyz,

3) Bin dng gin (n) ca th tch nguyn t:V

V = (1 +xx)(1 +yy)(1 +zz) 1

ii(ly tng theo i t 1 n 3) =iui= .u= divu.

Vy ta c cng thc

V =V.ii = V(xx+yy +zz) =V(u

x+

v

y+

w

z). (1.2)

1.2 Cc hng s n hi v nh lut Hooke suyrng

1.2.1 Cc hng s n hi

Modun Young E m t n hi dng ko, hoc nn dc theo trc(ca thanh,hay l xo) khi lc tc dng t dc theo trc , n c nh ngha bng t s


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gia ng sut ko (nn) cho bin dng. n v ca modun Young l N/m2. Modun n hi trt G miu t xu hng mt vt th ca mt vt th bct(trt, hnh dng bin dng vi th tch khng i) khi b tc ng bi cc

lc ngc hng, n c nh ngha bng ng sut ct (trt) chia cho bindng ko. n v ca modun trt G l N/m2.

Modun n hi (hng s Lame). Khi chu s tc ng ca mt ng sutko hoc nn, mt vt s phn ng bng cch bin dng theo tc dng ca lc.Trong mt gii hn bin dng nh, bin dng ny t l thun vi ng suttc ng. H s t l ny c gi l modun n hi v c xc nh theo cngthc

= ng sut (stress)bin dng (strain).

Cc hng s Lame: , , th nguyn N/m2,

= E

(1 +)(1 2), =

E

2(+).

Modun n hi ( Modun Young): E, th nguyn N/m2,

E=2(3+ 2)

+ .

H s Poison (Poissons ratio): ,khng th nguyn:

=

2(+), 0<


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Trong h trc ta Oxyzv vi vct chuyn v u = (u,v,w), cng thc(1.3) c dng

xx = u

x +

v

y +

w

z

+

u

x , (1.4)xy =

u

y+

v

x

, (1.5)

xz=

u

z+

w

x

, (1.6)

yy =

u

x+

v

y+

w

z

+

v

y, (1.7)

yx =

v

x+

u

y

, (1.8)

yz=v

z+w

y

, (1.9)

zz=

u

x+

v

y+

w

z

+

w

z, (1.10)

zx = w

x+

u

z

, (1.11)

zy =w

y +

v

z

. (1.12)

(1.13)

1.3 Mt nng lng bin dngMt nng lng bin dng ca vt n hi c xc nh bi cng thc

E=1

2

i,j=x,y,z

ijij (1.14)

=1

2

i,j=x,y,z

[ijkk + 2ij ]ij (1.15)

Nhn xt rng, mt nng lng bin dng l mt dng ton phng xcnh dng i vi su i lng c lp ij ; i, j = 1, 2, 3.

1.4 Phng trnh cn bng sng n hi

1.4.1 Lc to bi ng sut

Xt th tch nguyn t V nh. Hp lc tc dng ln V l vc t F =(Fx, Fy, Fz) = (F1, F2, F3),trong

Fx =

(xx+xx

x x) xx

yz+

(xy+ xy

y x) xy

xz


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+

(xz+

xzz

z) xz

xz+fxxyz=

xxx

+ xy

y +

xzz

+fx

xyz.

Vy, vi i= x, y,zth

Fi=ix

x +

iyy

+ iz

z +fi

V, (1.16)

trong fil to ca vc t mt lc khi f= (f1, f2, f3).

1.4.2 nh lut hai Newton - H phng trnh cn bng Navier- Cauchy - H phng trnh Lame

V2ui

t2 =Fi, i= x, y,z= 1, 2, 3. (1.17)

Thay (1.16) vo (1.17) ta c h phng trnh cn bng ng sut - chuyn vca l thuyt n hi ng

2u1t2

=xx

x +

xyy

+xz

z +f1,

2u2

t2

=yx

x

+yy

y

+yz

z

+f2,

2u3t2

=zx

x +

zyy

+zz

z +f3,

(1.18)

trong xy =yx, yz=zy , zx =xz. (1.19)

Ch (1.19), t (1.3) v (1.18) ta c h 9 phng trnh i vi 9 n hm lu1, u2, u3, xx, xy, xz, yy , yz, zz.H (1.18) c gi l h phng trnh cn bngNavier - Cauchy.

a (1.16) vo (1.17), ch (1.1) v a cc h thc (1.4)-(1.12) vo (1.18),ta c

ui= (+)(.u)

xi+.(ui) +fi, i= 1(x), 2(y), 3(z), (1.20)

hay l ta c phng trnh sng n hi i vi chuyn v dng vc t

u= (+)(.u) +.u + f. (1.21)

H phng trnh cn bng dng chuyn v (1.21) c gi l h phngtrnh Lame.


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1.4.3 Ta tr

Trong ta tr r, , zta c cc phng trnh sau y.H thc gia bin dng v chuyn v

rr =ur

r, =

1

r

u

+ur

, zz=

uzz

, zr =1

r

urz

+ uz

r

, (1.22)

r =1

r

1

r

ur

+u

r

ur

, z=

1

r

uz

+1

r

uz

. (1.23)

Phng trnh cn bng ng sut-chuyn v

rrr

+1

r

r

+rz

z +

1

r(rr ) +fr =

2urt2

, (1.24)

r

r +1

r

+z

z +2

r r +f =

2u

t2 , (1.25)zrr

+1

r

z

+ zz

z +

1

rzr +fz=

2uzt2

. (1.26)

1.4.4 Ta cu

Trong ta cu r, , ta c cc phng trnh sau y

H thc gia bin dng v chuyn v

rr =ur

r, =

1

r

u

+ur

, =

1

r sin

u

+ursin +ucos

, (1.27)

r =1

r

1r

ur

+ u

r

ur

, =

1

2r

1sin

u

+ u

ucot

, (1.28)

r =1

r

1

r sin

ur

+ u

r

ur

. (1.29)

Phng trnh cn bng ng sut-chuyn v

rrr +1r r

+ 1r sin r

+1r (2rr +rcot ) +fr =

2

urt2 , (1.30)rr

+1

r

+ 1

r sin

+1

r

( )cot + 3r

+f =

2ut2

, (1.31)

rr

+1

r

+ 1

r sin

+1

r

2cot + 3r

+f =

2ut2

. (1.32)


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1.5 iu kin u, iu kin bin v cc bi tonlin quan ca cc phng trnh sng n hi.

nh l v duy nht nghim1.5.1 iu kin uVi phng trnh n hi dng chuyn v th cc iu kin u s l

ui(x,y,z; 0) =i(x,y,z), ui

t(x,y,z; 0) =i(x,y,z), i= 1, 2, 3. (1.33)

iu kin th nht trong (1.33) biu th chuyn v ban u ca cc im, cniu kin th hai biu th vn tc ban u cc cht im ca mi trng.

1.5.2 iu kin bin

Gi s phng trnh cn bng n hi c cho trong min R3 vi binl .iu kin bin loi mt.Trn cho bit gi tr ca cc chuyn v

ui(x,y,z; t)

=g(x , y, z, t), (x,y,z) ; t= 1, 2, 3. (1.34)

ngha vt l ca cc iu kin trong (1.34) l cho bit chuyn v trn binca min m ta quan tm.iu kin bin loi hai.Trn cho bit gi tr ca cc ng sut. minh ha xt cc v d n gin sau y.

V d 1.1.Mt khi bao gm 2 na khng gian, ngn cch bi mt mt phngti z= 0.

Cc chuyn v l ux, uy, uz, v cc ng sut l pxx, pxy, pxz, pyy , pyz, pzz

iu kin bin z= 0cho cc kt hp khc nhau ca na khng gian c:rn - rn: ux, uy, uz, pxz, pyz, pzzlin tc,rn - lng: uz, pzzlin tc; pxz=pyz= 0,lng - lng: uz, pzzlin tc,rn - cng: ux = uy =uz= 0,lng - cng: uz= 0,chn khng - rn: pxz=pyz=pzz= 0,lng - chn khng: pzz= 0.

Nu ti mt b mt vi vector php tuyn ni, ng sut khng phi l khng(vector ng sut Pi), vector ng sut trong khi c c gi tr bin ny.


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12

p(r)ij nj =Pi, (1.35)

p(r)ij l nhng gi tr bin ca cc thnh phn ca ng sut tensor b mt,

v n c th c tnh t (1.35).

V d 1.2. P(t)trn mt mt phng. i vi trng hp p lc

Hnh 1.1: p lc ln mt b mt phng

n = (1, 0, 0) = (n1, n2, n3),P = (P(t), 0, 0) = (P1, P2, P3).

Tng t nh vy, s dch chuyn c th c quy nh trn b mt cakhi.

iu kin bin loi ba (hn hp).Gi s = 1 2, 1 1 = v trn1cho bit gi tr ca cc chuyn v, cn trn 2th cho iu kin v ng sut.

1.5.3 Bi ton CauchyCc phng trnh c cho trong ton b khng gian Rn.Ch c cc iu kin ban u.

1.5.4 Bi ton bin-gi tr ban u

Cc phng trnh c cho trong min Rn.Cc iu kin ban u.

Cc iu kin bin tng ng trn bin ca min .

1.5.5 nh l v duy nht nghim

Mt vn t ra l nghim ca cc bi ton ca l thuyt n hi gii theochuyn v hay ng sut c duy nht khng? Ngi ta chng minh c nhl v duy nht nghim bng nhiu phng php khc nhau.

nh l 1.1.Nu tha nhn v trng thi t nhin ca vt v nh lut v c

lp tc dng ca cc lc th nghim ca cc bi ton bin c bn th nht v thhai ca l thuyt n hi l duy nht.


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13

Chng 2

Sng iu ha-Cc sng n hiiu ha c bn

Chng ny trnh by cch gii ca h phng trnh cn bng ca sng nhi thit lp c trong Chng 1 bng cch a ra cc hm th v bini h phng trnh sng n hi v cc phng trnh sng cp hai c lp.Trn c s trnh by khi nim v sng iu ha v cc loi sng P, sng Sv mt s loi sng khc. Xt mt s bi ton v sng khi phng v sng khicu. Ni dung ch yu ca chng ny c hnh thnh t cc ti liu [2],[3],[4], [5] v [6].

2.1 Mt s kin thc b tr

2.1.1 Khi nim v sng iu ha

n gin, xt phng trnh sng mt chiu khng gian

1

c22u

t2 =

2u

x2. (2.1)

y xl bin khng gian, cn tl bin thi gian, cl hng s dng, c thnguyn m/s, nu u = u(x, t) c n v l mt. Nghim tng qut ca phngtrnh sng (2.1) c cho bi cng thc dAlambert

u(x, t) =F(x ct) +G(x+ct), (2.2)

trong F(x)vG(x)l nhng hm ty hai ln kh vi lin tc.

nh ngha 2.1.Nu nghim ca phng trnh sng c dng

u(x, t) =A(x, t)cos(t kx+)hoc= A(x, t)sin(t kx+), (2.3)


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th n c gi l sng iu ha, trong Al bin , = kcl tn s gc,

k = /cl s sng, = constc gi l pha ban u gc ta ,i lng t kx+c gi l pha ca sng thi im tti im x.

Nu A(x, t) =A(x),ngha l nu bin ca sng khng ph thuc vo thigian, th sng c gi l sng dng.

Mt s i lng khc c trng cho sng iu ha. f=/(2)l tn s: s dao ng ton phn trong mt giy, T = 1/f l chu k: thi gian thc hin mt dao ng ton phn,

= 2/k l bc sng: khong cch ngn nht gia hai dao ng ng pha.

Ni chung, sng iu ha c th c biu din dng

u(x, t) =Aexp

i(t x

c)

=Aexp[i(t kx)],

trong bin Ac th thc hoc phc. Biu din sng iu ha dng mi khi rt thun tin cho tnh ton.

2.1.2 Khi nim v hm Dirac v hm Heaviside H(x)hm Dirac l hm suy rng quan trng do nh vt l ngi Anh a ra

vo khong 1926. C th hiu hm Dirac mt cch hnh thc nh sau: hml mt " hm " c cc tnh cht1. (0) = ,2. (x) = 0, x = 0,

3.Rn

(x)(x)dx= (0)i vi mi hm lin tc (x).

Tnh cht th 3 thng thng c ly lm nh ngha hm suy rng Diracnh l phim hm tuyn tnh lin tc trn khng gian cc hm c bn, cxc nh theo cng thc [6]

< , >:=(0). (2.4)

Cc tnh cht 1 v 2 ni ln rng, gi ca hm Dirac tp trung ti gc tox = 0,tc l - hm c gi compact. Ch rng, tnh cht 3 tng ng vi

Rn (x a)(x)dx= Rn (x)(x a)dx= (a). (2.5)


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Cng vi delta hm, hm Heaviside (x)l nhng hm suy rng c bn. HmHeaviside c nh ngha nh sau

(x) =1, x >0,

0, x


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Nhn bn tri ca (2.12) vi .(tch v hng) v ch ti (2.13) ta c

2 = (+)2[2] +22. (2.15)

Do

2[ (+)2] = 0.

suy ra phng trnh sng

(+)2 = 0, (2.16)

hay l

c2p

2 = 0,

(2.17)trong

cp=

+ 2

, (2.18)

l vn tc ca sng s cp (premary wave). tm vc t th v ,ta nhn hai v ca (2.14)vi (nhn vc t) v ch

tnh cht = 0, (2.19)

ta c phng trnh sng 2= 0, (2.20)

hay l c2s

2= 0, (2.21)

trong

cs =

, (2.22)

l vn tc ca sng th cp (second wave) Phng php th hai[3]. Trong h ta Cartesian, ta c cc h thc

(2u1, 2u2,

2u3) = 2u = .u u . (2.23)

S dng cng thc (2.23), chng ta vit li h phng trnh (2.9) dng

2u

t2 (+ 2).u + u = 0. (2.24)

t (2.10) vo (2.24) ta c h phng trnh

2

t2 +

2

t2 (+ 2)2 +

= 0. (2.25)


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Vit li h (2.25) dng

2

t2 (+ 2)2

+

2

t2 +

= 0. (2.26)

Cc hm , s c tm, sao cho cc phng trnh sau y c tha mn

2

t2 (+ 2)2 = 0, (2.27)

2

t2 +

= 0. (2.28)

H trn y c th c vit li dng

1

c2p

2

t2 = 0, (2.29)

1

c2s

21t2

1+21

x2 +

22xy

+23xz

= 0, (2.30)

1

c2s

22t2

2+21yx

+22

y2 +

23yz

= 0, (2.31)

1

c2s

23t2

3+21zx

+22zy

+23

z2 = 0. (2.32)

Ch 2.1.Cc hm th v ,c biu din qua vc t chuyn v utheocc cng thc ([3], tr.35):

= 14

R3

u.rr3

dxdydz, (2.33)

= 1

4

R3

u r

r3 dxdydz, (2.34)

trong r= (x,y,z), r=

x2 +y2 +z2.

2.2.2 H c ngun Xt h c ngun h khng thun nht . Biu din vc t f dng

f=f = +

. (2.35)

Khi , khi iu kin (2.11) c tha mn, th cc th v , tha mn ccphng trnh sng khng thun nht sau y

1

c2p 2 =

+ 2, (2.36)

1

c2s 2=

, (2.37)


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Vc t f=fbiu din dng

f=f = +

, (2.38)

trong cc hm , c xc nh qua vc t f theo cng thc

= 1

4

R3

f.r

r3dxdydz, (2.39)

=

1

4

R3

f r

r3 dxdydz, (2.40)

trong r= (x,y,z), r=

x2 +y2 +z2.

Khi cc hm , s tha mn cc phng trnh sng sau y

1

c2p

2

t2 =

+ 2, (2.41)

1

c2s

21t2

1+21

x2 +

22xy

+23xz

=1

, (2.42)

1c2s

22t2

2+ 21

yx+ 22

y2 +23

yz = 2

, (2.43)

1

c2s

23t2

3+21zx

+22zy

+23

z2 =

3

. (2.44)

Nhn xt 2.1.Phng trnh(1.21)c th vit thnh h sau y

2u1t2

u1 (+)

x

u1x

+ u2

y +

u3z

=f1,

2u2

t2 u

2 (+)

yu1

x +

u2

y +

u3

z =f

2,

2u3t2

u3 (+)

z

u1x

+ u2

y +

u3z

=f3.

(2.45)

Nhn xt 2.2.Dng v hng ca biu din nghim(2.10)s l

u1=

x+

3y

2

z , (2.46)

u2=

y+

1z

3

x, (2.47)

u3= z +2

x 1

y . (2.48)


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2.3 Sng P, sng S, sng SV, sng SH v sng PSV

2.3.1 Sng P v sng S (P-sng v S-sng)

Thayu= sei(kxt), (2.49)

tc lu1= s1e

i(kxt), u2= s2ei(kxt), u3= s3e

i(kxt),

trong k, sl nhng vc t hng cha bit vo phng trnh sng (2.9)ta c

2s (+)(s.k)k k2s= 0, (k2 =k21+ k22+ k

23). (2.50)

Cc vc t k, stng ng c gi l vc t s sng v vc t bin . Mi quanh hnh hc gia hai vc t ny gn vi cc loi sng. Nhn v hng hai v caphng trnh trn y vi sta c

2|s|2 (+)(s.k)2 k2|s|2 = 0. (2.51)

Xt trng hp s.k= 0.Khi t (2.51) suy ra

/k = /= cs. (2.52)Trong trng hp ny cng thc (2.49) cho ta sng S c lan truyn vi vntc cs.Xt trng hp s||k.Khi t (2.51) suy ra

/k =

(+ 2)/= cp. (2.53)

Trong trng hp ny cng thc (2.49) cho ta sng P c lan truyn vi vntc cp.R rng l cp> cs.

Sng P (primary or compressional wave- sng s cp) l sng dc (tng tsng m) c lan truyn vi vn tc gia 1 n 14 km/s. Sng dc l sng nhi do bin dng ca th tch, trong cc ht vt cht dao ng theo phngtrng vi phng truyn sng. Sng S (Shear or secondary-Sng th cp) l sng ngang (tng t sng nc)c lan truyn vi vn tc gia 1 n 8 km/s. Sng ngang l sng trong ccht vt cht dao ng theo phng vung gc vi phng truyn sng.Trong a chn thm d, sng P c ngha hn c. V nguyn tc th sng

S cng c ch, nhng to sng S c nng lng cao nh sng P l rt kh. Ccsng khc, nh sng Love, sng Rayleigh v.v.. v khng mang thng tin ca cc


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20

lp t di su nn t c ngha i vi cng tc thm d a chn. S cc mt ca cc sng ny gy tr ngi cho qu trnh thu sng c ch, nn chngc coi l nhiu.

2.3.2 Sng SV, sng SH v sng PSV

Trong mi trng a tng, c th c hai loi sng S, l sng SV (V-vertical)chuyn ng song song vi nhau trong nhng mt phng thng ng v sngSH(H-horizontal) chuyn ng song song vi nhau trong nhng mt phng nmngang. Trong mi trng a tng sng SH c th xut hin gia cc mt t dophn cch cc a tng. Ta tng tng rng, nu mt sng phng c lantruyn trong mt phng song song vi mt phng trang sch trong mi trnga tng vi cc mt phn cch nm ngang, th SH c th c th hin biphng trnh sng v hng

y =c2s

2y, (2.54)

trong mi tng vi vn tc cs.Ngc li, cc sng phng PSV c th c cho bi cc phng trnh

x = c2s

2x, z=c2s

2z, (2.55)

trong yl to ca trc vung gc vi trang sch ny.

2.4 Vn tc pha v vn tc nhm

2.4.1 Vn tc pha

Khi nim. Vn tc pha l vn tc dch chuyn ca im c pha dao ngkhng i trong khng gian theo hng cho trc, thng c xem trng vi

hng ca vc t sng.Khi nim vn tc pha ch c s dng khi m t sng iu ha, hay cn gil sng n sc (sng c mt tn s duy nht, tc l sng c dng cos , sin , ei)v nhng sng c hnh dng tng t.

Cng thc. Chng ta bit rng, i vi cc sng iu ha vi vn tc truynsng c

P(x, t) =Aoei(kxt), P(x , y, z, t) =Aoe

i(k.xt),

th c h thcc=

k, k=

k2x+k2y +k

2z,


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21

trong kc gi l s sng. Cng thc trn y chnh l cng thc ca vntc pha sau y

vp =

k, = vpk, (2.56)

c trc tip suy ra t cng thc pha ca sng phng trong khng gian mtchiu = kx t, hoc = k.x t trong khng gian ln hn mt chiu.Vn tc pha ca sng De Broglie.Sng De Broglie l sng c sinh rado cc ht chuyn ng vi vn tc cao. Khi vn tc pha c tnh theo cngthc

vp=dx

dt =

E

p =

c2

u,

trong x- ta , t- thi gian, E- nng lng, u- vn tc ht, c- vn tc nh

sng trong chn khng.Vn tc pha trong mt s cht.- Vn tc pha ca sng trong cht rn n hi ng hng

v=

G

,

trong Gl modun trt, cn l mt ca mi trng.- Vn tc ca sng dc trong mt thanh mng

v=

E

,

trong El sut Young, lin quan n ng sut php =F

Stheo cng thc

= El

l .

- Vn tc sng ngang trn si dy cng mnh c mt ,sc cng Fv titdinS

v=

FS

.

Vn tc pha c th vt qu vn tc nh sng trong chn khng

2.4.2 Vn tc nhm

m t sng khng iu ha, c bit l cc chm sng, ngoi khi nim vntc pha, ngi ta cn s dng khi nim vn tc nhm m t chuyn ng

khng phi ca nhng ngn sng ring bit, m l ca mt nhm sng, hay bsng (cc sng iu ha c tn s gn bng nhau).


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Vn tc nhm. Vn tc nhm l s m rng ca vn tc pha v c xc nhtheo cng thc

vg =d

dk , d=vpdk. (2.57)Nu tn s gc l tuyn tnh i vi s sng, th vn tc pha v vn tc nhmtrng nhau. Hm (k)biu th s bin i ca tn s gc theo s sng kcgi l mi quan h ca s tn sc nh sng. S tn sc nh sng c biu thbi s ph thuc ca vn tc nhm vo bc sng.

2.4.3 Vn tc tc pha v vn tc nhm ca mt s mi trng

Mi trng Vp(km/s) Vs/VpKhng kh 0.31 - 0.36t trng phong ha 0.1 - 0.5 0.5 - 0.6Ct kh 1.5 - 1.6 0.1 - 0.3Ct t 1.5 - 1.6 0.1 - 0.3St m 1.5 - 2.5 0.1 - 0.3Nc 1.4 - 1.6 0.1 - 0.3Ct kt 1.5 - 4.0 0.4 - 0.6Mui m 4.5 - 6.0 0.5 - 0.6 vi 2.6 - 6.5 0.5 - 0.6

Gnanit 4.0 - 6.0 0.46 - 0.62Bazan 5.0 - 6.5 0.56 - 0.62

2.5 Sng khi phng

Sng khi l sng c pht sinh bn trong vt th (bn trong qu t).Trong mc trn cp ti mt s dng sng phng nh sng PV, sng SHv sng PSV c pht sinh trong cc mi trng n hi. Trong mc ny trnhby cch tm nghim ca cc phng trnh sng khi phng.

2.5.1 Pht biu bi ton

Xt mt s trng hp ring ca sng a chn khi vc t chuyn v u cdng c bit, v d nhu= (ux(x, t), 0, 0),hoc u= (0, uy(x, t), 0),trong ccthnh phn th nht hoc th hai ch ph thuc vo bin khng gian xv binthi gian t.

Trong trng hp th nht ta c phng trnh

c2p2uxx2

=2ux

t2 , c2p =

+ 2

. (2.58)


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Trong trng hp th hai ta c phng trnh

c2s2uyx2

=2uy

t2 , c2s =

. (2.59)

Cc phng trnh trn y l phng trnh sng mt chiu, l nhng trng hpring ca phng trnh

c22u

x2 =

2u

t2. (2.60)

Nghim tng qut ca phng trnh (2.60) c cho bi cng thc

u(x, t) =F(x ct) +G(x+ct), (2.61)

trong F, Gl nhng hm ty hai ln kh vi. Mt dng khc ca nghimtng qut l

u(x, t) =F

t x

c

+G

t+x

c

. (2.62)

Cc s hng th nht v th hai trong (2.61) v (2.62) tng ng biu ths truyn sng theo chiu dng v chiu m ca trc Ox. minh ha, xt shng th nht trong (2.62) vi x= x1

u(x1, t) =Ft

x1c

=F1(t).

Tng t, vi x2> x1,ta c

u(x2, t) =F

t x2 x1

c

x1c

=F1

t x2 x1

c

.

iu ny c ngha l, cng mt thi im t, sng ta x2 ging htsng ta x1 thi im t (x2 x1)/c.Vn tc truyn sng l c.Cc mtsng (wavefronts) ca sng, ngha l cc mt nm gia vng ri (perturb) vvng m (unperturb), chnh l cc mt x = const. Do , y l cc mt

phng sng. Nu G(x)trong (2.61) hoc G(t)trong (2.62) bng khng, th haimt phng truyn sng l i din nhau (opposite direction).

Trong trng hp u = ux, chng ta c sng dc (longitudial wave), trongtrng hp u= uy,chng ta c sng ngang (transverse wave).

2.5.2 Bi ton gi tr ban u (Bi ton Cauchy)i vi sngkhi phng

Xt phng trnh sng (2.60) vi cc iu kin ban u sau y

u(x, 0) =f(x), i vi chuyn v,


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u

t(x, 0) =g(x), i vi vn tc ht.

Trong trng hp ny, nghim ca phng trnh (2.60) c cho bi cng thc

dAlambertu(x, t) =

1

2[f(x ct) +f(x+ct)] +

1

2c

x+ctxct

g(s)ds. (2.63)

Trng hp g(x) 0Trong trng hp ny ta c cng thc

u(x, t) =1

2[f(x ct) +f(x+ct)],

l tng hp ca sng thun (theo chiu dng caOx) v sng ngc (theo chium ca Ox).Trng hp f(x) 0Trong trng hp ny, gi s hm g(x)c dng g(x) = Qo(x), trong Qo =const,cn (x)l hm Dirac. Ta c

u(x, t) =Qo

2c

x+ctxct

(s)ds=

Qo2c

H(t xc ), x >0,

Qo2c

H(t+ xc ), x


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Ly tch phn ng thc trn ta c

F(t) = 1

cp t

P()d= 1

cp t

0

P()d.

y ta gi thit rng P(t) = 0 khi t < 0. Do , chuyn v ux c cho bicng thc

ux(x, t) = 1

cp

tx/cp0

P()d. (2.64)

Xt trng hp c bit khi P(t) =Po(t),t cng thc trn suy ra

ux(x, t) = Pocp

H

t x

cp

. (2.65)

Trong cng thc trn, Poc th nguyn l ng sut nhn vi thi gian. Ti thiimt= x/cp,tt cc cc im trong na khng gian u b di ch tc th biPo/cptheo chiu +x.Xt trng hp ring sau y.

V d 2.1.ViPo = 1barsec = 9.81Nsec/cm2 106dynsec/cm2, = 3g/cm3, cp =6km/sec,th chuyn v xux 0.5cm.

2.6 Sng cu i xng sinh bi h thun nht

Trong cc phn trc, chng ta xem xt s m rng v hn ca sng. Chngl mt s l tng ho, bi v nhng hin tng nh th khng th c hnhthnh trong thc t khi yu cu m rng ngun v hn. Loi sng n gin nhtt cc ngun vi hu hn m rng l sng hnh cu, tc l sng c xut x timt im (im ngun) v lan truyn ra ton khng gian. Mt sng ca chngl hnh cu. Trong trng hp n gin nht, cc vector dch chuyn c totrn nh hng v cng i xng xuyn tm ty theo cc ngun im, nghal s dch chuyn xuyn tm trn mt hnh cu xung quanh ngun im l nh

nhau khp mi ni. Nu mt v n hnh cu trong mi trng ng nht xamt phn cch c kch hot, kt qu dch chuyn cui cng c hai tnh cht.Do , chng ta gi l nhng v n im ngun. Cc kt qu thu c vi lthuyt n hi tuyn tnh ch c th c p dng vi sng hnh cu t cc vn trong phm vi khong cch, trong iu kin trc ht ca l thuyt (bindng v cng, mi quan h tuyn tnh ng sut v bin dng) l tha mn.Trong min do, min b ph hy v min phi tuyn (y l phn loi th vi sgia tng khong cch t trung tm ca v n) cc yu cu ny khng c p

ng. i vi mt v n ht nhn ca 1 Megaton TNT tng ng (khong mb= 6,5-7,0), khu vc tan v l khong rng t 1 ti 2 km.


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Chng ta d nh gii quyt vn bin nh sau: cho cc di chuyn xuyntm khong cch r = r1 t ngun im U(r1, t) = U1(t), chng ta mun tmU(r, t)cho r > r1.

Chng ta bt u t phng trnh chuyn ng (1.21) vi

f = 0.K hiu

(r, t) =

rU(r, t)dr.

Trong to cu (r,,)ta c

=

r,

1

r

,

1

rsin

= (U(r, t), 0, 0) .

Trong ta cu i xng phng trnh sng (2.17) c dng n gin

2 =2

r2 +

2

r

r =

1

r

2(r)

r2 =

1

c2p

2

t2,

hay l cn n gin hn2(r)

r2 =

1

c2p

2(r)

t2 . (2.66)

Trong trng hp i xng xuyn tm, phng trnh sng c th c gimxung cn hnh thc ca mt phng trnh sng mt chiu cho ta cho cchm r

2ux2

= 1c2p

2ut2

.

Nghim ca phng trnh sng (2.66) c cho bi cng thc dAlambert

(r, t) =1

r

F

t

r

cp

+G

t+

r

cp

.

M t s chng cht ca hai sng p, mt lan truyn ra ngoi t ngun imv lan truyn khc vo bn trong v pha ch ngun. Trong cc vn thc t,

s hng th hai lun lun l s khng v

(r, t) =1

rF

t

r

cp

. (2.67)

HmF(t)thng c gi l hm kch thch hoc gim chuyn th nng. Ccmt sng l nhng mt cu r= const. Th nng nh mt hm ca thi gian chnh thc tng t khp mi ni, v bin gim vi khong cch nh1/r.S dch chuyn xuyn tm ca ln sng hnh cu bao gm hai ng gp vi sph thuc khc nhau theo r

U(r, t) =

r =

1

r2F

t

r

cp

1

rcpF

t r

cp

. (2.68)


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Do hai s hng thay i vi khong cch tng. Ni chung,iu ny ngvi s dch chuyn ca sng t im ngun. S hng u tin trong (2.68) cgi l s hng trng gn v n th hin r rt i vi r nh(gn ngun). S

hng th hai c gi l s hng trng xa v n m t sng kh chnh xc tinhng im cch ngun ln hn vi bc sng. iu c ngha l di chuyngim t l thun vi 1/r

T iu kin bin r= r1, do

1

r1cpF

t r1cp

1

r21F

t

r1cp

=U1(t).

Chng ta chn thi im gc U1(t)ch bt u c khc khng cho t= r1cp .Do , xut hin nh ln sng bt u ti thi im t = 0ti ngun im r = 0.Nu U1(t r1cp ) = U1(t), n giU1(t) cho khc 0 vi t > 0. Vi = t

r1cp

ktqu cho

1

r1cpF() +

1

r21F() = U1(). (2.69)

Nghim ca phng trnh (2.69) c cho bi cng thc [3]:

F() = r1cp

0

U1()e

cp

r1()

d.

T , ta c

F() = r1cpU1() +c2p

0

U1()e

cp

r1()

d.

S dch chuyn xuyn tm cho r > r1sau c th c vit bng cch sdng (2.68) nh

U(r, t) =r1

rU1() +cp1r 1r1

0

U1()e

cp

r1

()

d , (2.70)

vi thi gian tr =t rcp .Bi ton bin t ra c gii quyt.

ng dng1)U1(t) =U0

t r1cp

, hay l,U1(t) =U0(t).n v caU0l thi gian nhn

di.Thay biu thc trn vo cng thc vo cng thc (2.70), ta c

U(r, t) =r1

rU0

() +cp

1

r

1

r1

ecp

r1

H(), =t r

cp.


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2) U1(t) =U0H

t r1cp

, hay l U1(t) =U0H(t). Th nguyn ca U0l chiu

di. T (2.70), ta c:

U(r, t) = r1rU0

H() +cp

1r 1r1

ecp

r1

r1cpecp

r1==0

H()

= r1

rU0H()

1 +r1

r 1

1 e

cp

r1

= r1

rU0H()r1

r +

1 r1

r

e

cp

r1

.

Hnh 2.1:U(r, t)l mt hm thi gian

2.7 Sng cu c sinh bi ngun im n

Mc ny xt bi ton Cauchy i vi h phng trnh chuyn v c nguntrong h ta cu vi iu kin ban u bng khng:

u (x , y, z, t)t=o

=u

t

t=o

= 0.

2.7.1 Cc th v ca ngun

Xt phng trnh n hi vi ngun

f= f = (0, 0, (x)(y)(z)K(t)),

t tm ca h ta Oxyznh hnh v.Biu din lc ny qua cc hm th v v theo cng thc (2.10) i vi

cc chuyn v:f = +

,

trong

(x , y, z, t) = 1

4

1

r3(z )()()()K(t)ddd=

K(t)z

4r3 ,

r2 =x2 +y2 +z2, r2 = (x )2 + (y )2 + (z )2. (2.71)


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Hnh 2.2: H ta Decartes vi lc khi ph thi gianK(t)

(x , y, z, t) = 14

1r3

{(y ), x , o}()()()K(t)ddd

= K(t)

4r3(y,x, 0). (2.72)

2.7.2 Phng trnh ca cc th v

Nu (2.71) v (2.72) c s dng trong cc phng trnh vi phn (2.31) cacc th v, ta c th v trt

= (x, y, z), z= 0.Vi gi thit cc hm

x, y tha mn iu kin xx

+y

y = 0. (2.73)

(xem iu kin (2.11)). Khi ta c cc phng trnh khng thun nht c lpsau y

2 1

c2p

2

t2 =

K(t)z

4c2pr3

, (2.74)

2x 1

c2s

2xt2

= K(t)y

4c2sr3

, (2.75)

2y 1

c2s

2yt2

= K(t)x

4c2sr3

. (2.76)

Nghim ca phng trnh sng khng thun nht

2a 1

c22a

t2 =f(x , y, z, t),

trong ton b khng gian R3 vi cc iu kin ban u bng khng, c chobi cng thc [4]:

a(x , y, z, t) = 1

4

1

rf

, , , t

r

c

ddd, (2.77)


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vir2 = (x )2 + (y )2 + (z )2.

S dng cng thc (2.77), c th tm c nghim ca cc phng trnh sng

(2.74)-(2.76) theo cc cng thc sau y:

(x , y, z, t) = z

4r3

r

cp0

K(t )d,

x(x , y, z, t) = y

4r3

r

cs0

K(t )d,

y(x , y, z, t) = x

4r3

r

cs0

K(t )d,

z(x , y, z, t) = 0,

r2 = x2 +y2 +z2.

(2.78)

Ch rng, hm K(t)trong cng thc (2.78) khng th c cho trc ty ,mphi c chn, sao cho cc hm x, ytha mn iu kin (2.73).

2.7.3 Cng thc ca cc chuyn v

tnh ton cc cng thc cho chuyn v, chng ta chuyn sang ta cu(r,,)c ch ra trn hnh 2.3 :

Hnh 2.3: Ta cu (r,,)


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x = rsincos,

y = rsinsin,z = rcos.

Trong ta cu , th trt khng c r-thnh phn v -thnh phn, cn-thnh phn c xc nh theo cng thc

= xsin+ ycos

.

Hnh 2.4: Mt phng x, yca hnh 2.3

Ta c

(r,,t) = cos

4r2

r

cp0

K(t )d,

(r,,t) = sin

4r2

r

cs

0

K(t )d.

(2.79)


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Phng trnh ny khng ph thuc vo . Vc t chuyn v u = + .

c th c vit trong ta cu nh sau:

ur =

r +

1

rsin

(sin),

u = 1

r

1

r

r(r),

u = 0.

(2.80)

iu ny cho thy sng Pc suy ra t khng phi l hon ton theochiu dc, n cha mt thnh phn ngang (trong u). Tng t, sng Scsuy ra t khng phi l hon ton ngang v ur cha mt thnh phn sng

trt . Thnh phn u tin trong uv th hai trong url nhng thnh phnca trng gn (near field terms [3]). y chng ta tnh ton ch i vi ccthnh phn xa ca ur v ca u :

ur cos

4c2prK

t

r

cp

(sng P - theo chiu dc),

u sin

4c2srK

t

r

cs

(sng S - theo chiu ngang).

(2.81)

Nh vy, cc trng chuyn v xa t lc K(t)s gim theo 1r

.Ngun lc im

n gy ra cc c tnh bc x c ch ra trn hnh 2.5

Hnh 2.5: c tnh bc x lnh vc xa ca ngun lc im

Cc c tnh bc x ( sng PvS) u l hai vng trn, sng Sc bn knh

l

c2p

c2s ln hn sng P. Nu gc bc x thay i cho rc nh, th chuyn v url t l thun vi khong cch OP1, cn chuyn v ul t l thun vi khong


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cchOP2. Cc du hiu ca s chuynurthay i trong qu trnh chuyn i tvng trn u tin n vng trn th hai. c im bc x 3-D y c thhin trong hnh 2.5 bi xoay quanh s ch o ca lc lng ny. Trong khun

kh ca cc phng trnh trng xa ( 2.81).Trong thc t cc ngun lc im n thng c t vung gc vi b

mt t do ca mi trng cn kho st. V vy, m hnh xt trn y kh phhp vi vn kch thch rung - a chn bi cc v n gn b mt t do.


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Chng 3

S phn x v khc x ca cc sngn hi phng

Chng ny trnh by s phn x v khc x ca cc sng n hi phngtrn mt phn cch ca hai na khng gian. Cc sng ny l nhng sng nhi iu ha. thun tin, trong chng ny, cc sng iu ha s c cho dng hm m vi i o. Ni dung ch yu ca chng ny c hnh thnh tti liu [3].

3.1 Cc phng trnh c bn

3.1.1 Cc phng trnh lin quan ti hai na khng gian

Chng ta s xt hai na khng gian ngn cch nhau bi mt phng z= 0.Gi thit rng sng khng ph thuc vo bin y, nn chng ta s xt bi tonti mt mt phng bt k vung gc vi trc Oy.

Ta k hiu cc thng s ca cc na khng gian l (j , j , j , j , j),vij= 1ta c na khng gian trn, j = 2ng vi na khng gian di, trong j , jtng ng l cc vn tc s cp v vn tc th cp, j l mt khi, j , j l

cc hng s Lame:j =

j+ 2

j, j =

jj

.

Vc t chuyn v

u = (u,v,w); u= u(x, z), v= v(x, z), w= w(x, z).

Biu din chuyn v theo cc th v

u = + .


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Ta c

u=

x

2z

,

v= 1z 3

x ,

w=

z +

2x

.

R rng l thnh phn vch ph thuc vo 1v3,cn cc thnh phn u, wch ph thuc vo v2.

Vi gi thit1x

+3

z = 0,

v vit thay cho 2,ta c

2 = 1

22

t2,

2 = 1

22

t2 ,

2v= 1

22v

t2, (3.1)

2

=

2

x2 +

2

z2 ,

u=

x

z, (3.2)

w=

z +

x. (3.3)

Cc cng thc cn thit ca ng sut

zz=z=

22

t2 + 2

2

z2 +

2

xz

, (3.4)

zx =

22

xz+2

x2 2

z2

, (3.5)zy =

v

z. (3.6)

3.1.2 Th v phng iu ha

Th v iu ha dng phc

=A exp it x .

k

, (3.7) =B exp

i

t x .k

n . (3.8)


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y gi thit A, B l cc hng s (thc hoc phc),

k , n l cc vc t hngn v, x l bn bnh vc t a phng, l tn s gc, il n v o, , tng ng l cc vn tc n hi s cp v th cp, tl bin thi gian.

Mt s biu thc vi phn ca cc th v iu ha

= i

A exp

i

t x .

k

k , (3.9)

=

i

B exp

i

t x .

k

k n. (3.10)

3.2 Phn x v khc x ca sng SH

3.2.1 H s phn x v khc x

Chuyn v v0trong s c ca sng SHti mi mt phng vung gc vi trcOy l

v0 = C0exp

i

t

sin

1x

cos

1z

. (3.11)

y k hiu C0l bin ca sng, 1(2)l vn tc sng th cp, l tns gc ca sng, l gc ti ca sng (xem hnh 3.1).

Hnh 3.1: S c nhiu x v phn x sng SH

Gi s hai mi trng n hi khc nhau chim l hai na khng gian phn

cch nhau bi mt phng z= 0.Trong mi trng ng nht ng hng mitia sng c truyn thng, khi gp b mt phn cch vi mi trng khc th


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mt phn ca sng ti b phn x tr li mi trng c, mt phn truyn sangmi trng khc vi hng truyn thay i gi l tia khc x. Trn hnh 3.1, v1v1tng ng l tia v gc phn x, cn v2v2tng ng l tia v gc khc

x.

Phn x: v1 = C1exp

i

t

sin11

x+cos1

1z

, 1=

11

. (3.12)

Khc x : v2 = C2exp

i

t

sin22

x cos2

2z

, 2=

22

. (3.13)

Cc n s l nhng gc 1v 2, h s phn x rss = C1/C0v h s khcx bss =C2/C0.

Cc iu kin bin yu cu ti z= 0tnh lin tc ca chuyn v (l mt yucu hp l ) v lin tc ca ng sut php tuyn v ng sut tip tuyn. iuny dn n:

v0+v1 = v2,

1

z(v0+v1) = 2

v2z

.

ti z= 0. (3.14)

Cc thnh phn ng sut zzv zx bng khng khp mi ni, v khng c

sng P hoc sng SV xy ra. a (3.11), (3.12) v (3.13) vo (3.14) . T iukin bin u tin dn n:

C0exp

i

t

sin

1x

+C1exp

i

t

sin11

x

=C2exp

i

t

sin22

x

. (3.15)

nh lut Snell i vi sng

1=, sin

1 =sin1

1 =sin2

2 . (3.16)

Suy rasin2sin

=21

.

T (3.15) v (3.16) suy ra

C2 C1=C0. (3.17)

iu kin bin th hai trong (3.14) cho:

1i

cos

1C0+

cos11

C1

= 2i

cos22

C2.


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Vi 1 = v 1,2/1,2 = 11, 21,2, suy ra 11cos(C1 C0) = 22cos2C2,hoc

22coss2

11cos C2+C1 = C0. (3.18)

T (3.17) v (3.18), suy ra cc h s phn x v khc x:

rss = C1

C0=

11cos 22cos211cos+22cos2

, (3.19)

bss = C2

C0=

211cos

11cos+22cos2. (3.20)

Ch n (3.16) ta c

cos2= (1 sin22)

1

2 =

1

2221

sin2

12

. (3.21)

i vi trng hp vung gc (= 0):

rss =11 2211+22

, bss = 211

11+22.

Trong trng hp ny, rss v bss ch ph thuc vo tr khng 11 v 22ca hai na khng gian . i vi gc ti l l mt phn cch (= /2), rss = 1vbss = 0. Cc gi tr tuyt i ca bin ca sng phn x khng bao gi lnhn ca sng ti ; ca sng khc x c th ln hn nu 22 < 11(v d cho= 0).

Nu rss m, iu ny c ngha l ti mt im ca mt phn cch s dchchuyn vector ca sng phn x theo hng y, nu s dch chuyn vector casng ti theo hng+y. Vi kch thch xung (xem thm sau ), iu ny c nghal hng chuyn ng u tin ca sng ti v cc sng phn x l ngc nhau.

Cc con s sau y cho thy |rss

|

nh mt hm ca

cho t l vn tc khcnhau1/2 > 1 v1=2.


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Hnh 3.2:|rss|nh mt hm cacho t l vn tc khc nhau

3.2.2 Phn x ton phn

Hin tng, trong sng khc x i l l mt phn cch hai mi trng(2= /2) c gi l phn x ton phn.

Nu2< 1, nh trong hnh 3.2 ,cos2l thc cho tt c cc gc ti , cngng cho rssv bss. Phn x ton phn, tc l gi tr ca gc ti l |rss| = 1.Nu 2 > 1, cos2ch c gi tr thc khi

=arcsin12

.


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Hnh 3.3:|rss|nh mt hm cacho t l vn tc khc nhau

l gc ti hn (gc gii hn phn x ton phn). Theo (3.21), = c

gn vi cc trng hp lan truyn l l ca sng trong na khng gian th hai(2=/2).

3.3 Phn x ca sng P ti mt b mt t do

Vic nghin cu cc phn x ca sng s cp P t mt b mt t do c tmquan trng thit thc cho a chn hc. Sng P t trn ng t v cc v n

lan truyn xuyn qua Tri t v tc ng n ti trm a chn t bn di.Di di ng nm ngang v ng thng ng c thay i bi cc b mtt do. Hn na, phn x sng P v sng S c phn x xung v ghi nhn nhng khong cch ln hn, i khi vi bin ln. Do , hu ch v cn thit bit h s phn x ca b mt tri t. Cho thi im ny, chng ta b quabn cht phn tng ca v tri t trong m hnh ca chng ta, do , ch ara nhng xp x u tin vi thc t.


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Hnh 3.4: Sng ti P v sng phn x P,SV

Tng t nh i vi sng iu ha SH, ta cSng ti P

0=A0exp

i

t sin

x

cos

z

. (3.22)

Sng phn x P

1=A1exp i t sin1

x+cos1

z . (3.23)

Sng phn x SV

1=B1exp

i

t

sin1

x+cos1

z

. (3.24)

y, thun tin k hiu , tng ng l cc vn tc s cp v sngth cp.

Cc iu kin bin ti z= 0yu cu trit tiu cc ng sut php v ng sut

tip zz=zx = 0.Khng i hi iu kin bin trn mt z= 0ca cc chuynv. Vi (3.4), (3.5) v = 0+ 1, = 1suy ra cc iu kin bin sau y:

1

22

t2(0+ 1) +

2

2

z2(0+ 1) +

21xz

= 0, z= 0, (3.25)

2 2

xz(0+ 1) +

21x2

21

z2 = 0, z= 0. (3.26)

nh lut Snellsin

=

sin1

=sin1

. (3.27)


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T , c kt qu 1= v 1=arcsin(sin)< . Vi (3.22)-(3.24) v

=

+ 2 2=

2

2 22 =

2

2 22,

t (3.25) dn n1

2(A0+A1)(i)

2

+2 2

2 22

(A0+A1)(

i

cos)2 +B1(

i

sin1)(

i

cos1)

= 0.

Suy ra

A0+A1+ 2 22

2 22 (A0+A1)cos2

2 B1

sin1cos1

2 = 0.Vi

1 + 22

2 22cos2 =

22

2 22

2

22 1 +cos2

= 22

2 22

2

22 sin2

= 2

2 22

2

2 2sin2

= 2sin2 2

,

v

=2

2 >2

, ta c

2sin2

2 (A0+A1)

2sin1cos1

2 B1= 0.

T :

( 2sin2)A1

A0 2sin( sin2)

1

2B1A0

= 2sin2 . (3.28)

Phng trnh (3.26) cho:

2A0

i

sin

i

cos

+ 2A1

i

sini

cos

+B1

i

sin1

2 B1

i

cos1

2= 0,hoc

2sincos

2 (A0 A1) +

sin21 cos21

2 B1= 0.


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Phng trnh (3.27) cho:

2sincosA1A0

+ ( 2sin2)B1A0

= 2sincos. (3.29)

T (3.28) v (3.29) ta c t s cc bin l

A1A0

= 4sin2 cos ( sin2 )

1

2 ( 2sin2 )2

4sin2 cos ( sin2 )1

2 + ( 2sin2 )2, (3.30)

B1A0

= 4sin cos ( 2sin2 )

4sin2 cos ( sin2 )1

2 + ( 2sin2 )2. (3.31)

Bin dch chuyn ca sng ti P l iA0,ca sng phn x P l iA1.

iu ny sau cung cp cho ta cc h s phn x PP:

Rpp=A1A0

. (3.32)

Phng trnh (3.10) cung cp cho cc bin dch chuyn ca sng phn xSV nh iB1. Nh vy, h s phn x PS l

Rps =

B1A0

. (3.33)

RppvRpsl thc v l tn s c lp cho tt c cc gc ti .Rpslun lunl dng. Cho = 0v = 2 , Rpp = 1v Rps = 0, tng ng, v ch c mtsng P c phn x.

Hnh 3.5: H s phn x v khc x ca sng P cho gc ti khc nhau

ngha ca du m trong h s phn x tr nn r rng, nu vector chuyn


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v ca sng ti v sng phn x c biu din thng qua (3.9) v (3.10)

u 0 = 0= i

A0exp it

sin

x

cos

z

k 0 , (3.34)

u 1 = 1= i

A0Rppexp

i

t

sin

x+

cos

z

k 1 , (3.35)

u 1 = = i

A0Rpsexp

i

t

sin 1

x cos1

z

k 1

n ,(3.36)

1 = arcsin

sin

.

Hnh 3.6: Chiu phn cc ca sng phn x P v SV

Rpp < 0, c ngha l nu s dch chuyn ca sng phn x Pti mt imtrn mt phn cch(z= 0)c ch trong hng ca

k1, sng ti c ch trong

hng cak0.i vi trng hp tia sng ti, Rps


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KT LUN

Lun vn ny trnh by cc vn sau y:1. Thit lp phng trnh sng n hi tuyn tnh c in dng ng sut v dng chuyn v trong ta Decarthes, ta tr v ta cu. Trnh bycc iu kin u, iu kin bin v nh l duy nht nghim ca cc bi tonbin c bn.

2. Trnh by cch tm nghim ca h phng trnh sng n hi bng phngphp th v, a h phng trnh n hi v cc phng trnh sng tuyn tnhcp hai thuc lp phng trnh hyperbolic ca l thuyt cc phng trnh ohm ring.

3. Trnh by v sng iu ha v cc loi sng n hi c bn, nh sng P(sngs cp), sng S (sng th cp), cc sng th cp ng (SV), th cp ngang (SH),sng s - th cp ng (PSV), v.v..

4. Xt ba bi ton c bn ca phng trnh sng. l bi ton Cauchy vbi ton bin gi tr ban u i vi sng khi phng; bi ton bin gi tr banu i vi sng cu i xng thun nht khi bit cc iu kin u v cc iukin bin trn mt mt cu nh cho trc; bi ton Cauchy i vi sng cu cngun im gc ta .

5. Xt s phn x v khc x ca sng n hi phng: SH v sng P trn bmt phn cch z= 0ca hai na khng gian l nhng mi trng n hi khcnhau.


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Ti liu tham kho

[1] o Huy Bch (1979) ,L thuyt n hi, NXB H v THCN.

[2] Aki, K. and P.G. Richards (2000) , Quantitative Seismology: Theory andMethods,Freeman and Co.,San Francisco.

[3] Gerhard Muller (2007),Theory of Elastic Waves, Germany.

[4] Gerard T. Schuster (2003),Basic Seismic Wave Theory,University of Utah.

[5] Internet: vi. wikipedia.org; bachkhoatrithuc.vn, vatlyphothong.net, vat-lyphothong.vn.

[6] V. S. Vladimirov (1971), Equations of Mathematical Physics, MarcelDekker, INC, New York.

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