dynamic green’s functions for a poroelastic half-space

7/30/2019 Dynamic Greens functions for a poroelastic half-space

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Acta Mech 224, 1739 (2013)DOI 10.1007/s00707-012-0720-2

Pei Zheng She-Xu Zhao Ding Ding

Dynamic Greens functions for a poroelastic half-space

Received: 19 October 2011 / Revised: 25 July 2012 / Published online: 10 October 2012 Springer-Verlag 2012

Abstract The dynamic responses of a poroelastic half-space to an internal point load and fluid source areinvestigated in the frequency domain in this paper. By virtue of a method of displacement potentials, the 3Dgeneral solutionsof homogeneous wave equations andfundamental singular solutions of inhomogeneous waveequations are derived, respectively, in the frequency domain. The mirror-image technique is then applied toconstruct the dynamic Greens functions for a poroelastic half-space. Explicit analytical solutions for displace-ment fields and pore pressure are obtained in terms of semi-infinite Hankel-type integrals with respect to thehorizontal wavenumber. In two limiting cases, the solutions presented in this study are shown to reduce toknown counterparts of elastodynamics and those of Lambs problem, thus ensuring the validity of our result.

1 Introduction

The problem of the dynamic responses of a fluid saturated porous medium under the action of internal sourcesis of interest to several disciplines such as geotechnical engineering, seismology, and geophysics. Severalpapers have appeared on the wave propagation in an unbounded poroelastic medium due to point and linesources. The full-space fundamental solution due to a point force was first published by Burridge and Vargas[1]. Later, Norris [2] derived the time-harmonic fundamental solutions for a point force in the solid as wellas a point force in the fluid. He also obtained a close-form solution in the time domain for an impulsivepoint load applied in a nondissipative medium by using Fourier transform. By drawing an analogy betweenporoelasticity and thermoelasticity in the frequency domain, Bonnet [3] and Cheng et al. [4] found the 2Dand 3D fundamental solutions for a time-harmonic concentrated load and fluid source, respectively. Chen[5,6] obtained the fundamental solutions for 2D and 3D problems in the Laplace transformed domain. Thecorresponding solutions in the time domain are achieved by using approximate methods, that is asymptoticexpansion techniques. Pilippacopoulos [7] also published fundamental solution in frequency domain but usinglike Burridge and Vargas [1] only a point force in the solid skeleton. Recently, Lu et al. [8] presented a so-called2.5D fundamental solution in the frequency domain. Such a 2.5D solution can be used if a 3D load is appliedto an infinitely long structure with a uniform cross-section, that is, the structure is essentially 2D.

By contrast, only limited work has been done for sources buried in poroelastic half-spaces or layeredporoelastic configurations (see also the review by Schanz [9]). Most geophysical applications involve an infi-nite boundary with zero traction, which is normally used to model the traction-free ground surface. Thus, thehalf-space fundamental solutions for fluid saturated porous solids seem more desirable in solving practical

P. Zheng (B) S.-X. ZhaoDepartment of Engineering Mechanics, Shanghai Jiaotong University, Shanghai 200240, Peoples Republic of ChinaE-mail: [email protected]

D. DingShanghai University, Shanghai 200444, Peoples Republic of China


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18 P. Zheng et al.

problems. Senjuntichai and Rajapakse [10] obtained the Greens function for a 2D poroelastic half-plane. Theapproach used in their study is similar to that employed by Pekeris [11]. The buried line source problem wastreated as a two-domain boundary value problem by requiring that stresses and pore pressure are zero at thesurface, while at a hypothetical interface passing through thesource, continuity anddiscontinuity conditions, in

terms of unit step functions, were imposed to ensure representation of a line source. The fundamental solutionsfor a vertical and horizontal point force buried in a poroelastic half-space were derived by Philippacopoulos[12] and Jin and Liu [13], respectively. Both solutions are found in the frequency domain by superpositionof the singular solutions corresponding to the inhomogeneous problem of the whole space and the solutionsof homogeneous wave equations representing relevant effects due to the presence of the free surface. Nev-ertheless, the fundamental solution for a fluid point source buried in a poroelastic half-space has never beenreported in previously published literature.

In the present work, first, based on Biots theory [1619], governing equations formulated with field vari-ables uw (solid displacement and relative solidfluid displacement) and up (solid displacement and porepressure), respectively, in the frequency domain are re-derived. Then, a method of displacement potentialsis developed for the uw formulation to reduce the homogeneous wave equations to four scalar Helmholtzequations, and thus, the general solutions are found. It should be mentioned that our method presented here isquite different from that used by Jin and Liu [13]. By Hankel transform and Fourier expansion with respectto radial coordinate r and azimuthal angle , respectively, Jin and Liu [13] found the general solutions ofhomogeneous wave equations formulated with up in a cylindrical coordinate system (r, ,z). On the otherhand, Philippacopoulos [14], employing the uw formulation, found the general solutions to axisymmetricproblems only, and thus, it fails to be able to be applied to asymmetric problems, for example, a horizontalpoint force buried in a half-space.

Subsequently, the fundamental solutions of the full space due to a point load and a fluid injection pointsource are derived, respectively, by following the methods of Norris [2] and Philippacopoulos [7]. By contrast,Jin and Liu [13] borrowed the Greens functions from Cheng et al. [4]. However, Cheng et al. [4] indirectlyobtained the spectral Greens functions by converting from the thermoelastic solutions of Nowacki [15].

Finally, instead of the early method used by Philippacopoulos [12] and Jin and Liu [13], the mirror-imagetechnique is applied to construct the dynamic Greens functions for a point load and fluid source buried ina poroelastic half-space. Compared to the prior published studies [12,13], our solutions consist of the threeparts: the full-space fundamental solutions for point sources in the half-space, the fundamental solutions of thefull space with corresponding image sources, and the solutions of homogeneous wave equations. Because theimage sources are located outside the half-space, the second part satisfies the homogeneous wave equationswithin the half-space. Therefore, our solutions actually have the same structure as those of the previous studies[12,13]. However, the solutions constructed in this way are very easily degenerated to the surface Greensfunctions due to a concentrated force at the half-space surface which provides a means of checking the validityof the half-space Greens functions by comparing with the known solution to this problem, that is Lambsproblem [14]. We will show this in Sect. 6.

2 Biots theory-government equations

Based on Biots theory [1619], the governing equations of dynamic poroelasticity consist of the following:the constitutive relations

= pI, (1a) = eI + , (1b)

p = Me + M, (1c)

the equations of motion

+ F =2

t2

u + fw

, (2a)

p + f =2

t2(u + mw) + b

w

t, (2b)

the strain-displacement relations

= 1/2 (u + u) , (3)


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Dynamic Greens functions 19

the continuity equation

t + q = , (4)

or the time-integrated continuity equation

+ w = , (5)

where

= total stress tensor; = effective stress tensor;p = pore pressure; = average strain tensor;I = second-order identity tensor; = variation of fluid content per unit volume;u = solid skeleton displacement;w = average fluid displacement relative to the solid skeleton;

e = u = dilation of the solid skeleton;q = w/t= fluid flux vector;F = body force per unit volume (including that on solid and fluid);f=body force acting on the fluid; = the rate of fluid supply per unit volume;

=t

0 = the total volume of supplied fluid;

= 1 K/Ks (K and Ks are the drained bulk modulus and bulk modulus of the solid grains, respectively);1/M = / Kf + ( )/Ks (Kf is the bulk modulus of the fluid);, = drained Lame constants of the bulk materials (c = +

2M = undrained Lame constants);b = / ( is the fluid viscosity and is the permeability coefficient); = density of the bulk materials;f = fluid density;m= f/ ( is the tortuosity coefficient and denotes porosity).

It is possible to use another equivalent set of material constants in the above formulation (see also Rice andCleary [20]). The expression for b is valid for the low-frequency range, that is


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20 P. Zheng et al.

where

m = m ib/. (10)

The superposed tilde that will be omitted from all quantities is used to denote the frequency-dependent part

after the removal of the factor eit.

3 General solutions for the homogeneous wave equations

The general solutions for the uw formulation will be derived in this section, which will be used in Sect. 5 tosatisfy the traction-free boundary condition.

In the absence of source terms, Eq. (8) becomes

(c + 2) u u + M w = 2

u + fw

, (11a)

M u + M w = 2

fu + mw

. (11b)

In cylindrical coordinates (r, , z), represent the displacement fields with six Helmholtz potentials as (see,

e.g., [21])

u = s + (sez) + (sez) , (12a)

w = f +

fez

+

fez

, (12b)

where

=

rer +

1

r

e +

zez, (13)

in which er, e and ez are the unit base vectors along coordinate curves. The components of the displacementfield and those of effective stress (zr, z, zz ) are given by

ur =s

r

+1

r

s

+2s

rz

,

u =1

r

s

s

r+

1

r

2s

z,

uz =s

z

1

r

r

r

s

r

1

r2

2s

2,

wr =f

r

+1

r

f

+2f

rz

,

w =1

r

f

f

r+

1

r

2f

z,

wz =f

z

1

r

r

r

f

r

1

r2

2f

2,

(14)

zr =

2

2s

rz+

1

r

2s

z+

3s

rz2

r

2s

r2+

1

r

s

r+

2s

2

, (15a)

z =

2

r

2s

z

2s

rz+

1

r

3s

z2

1

r

2s

r2+

1

r

s

r+

2s

2

, (15b)

zz = 2s + 2

2

sz2

z

2

sr2

+ 1r

sr

+ 2

s2

. (15c)

In addition, the pore pressure is given by

p = M2s M2f. (16)

After substitution of (12) into the equations of motion (11), the following sets of coupled Helmholtz equationswill be obtained:

Kp2 + 2M

= 0, (17a)

Ks2 + 2M

= 0, (17b)

Ks2 + 2M = 0, (17c)


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where the second-order symmetric dilatational and rotational stiffness matrix are defined as

Kp =

c + 2 MM M

; Ks =

00 0

, (18)

and the symmetric mass matrix is equal to

M =

ff m

. (19)

In addition,

T =

s, f

, T =

s, f

, T =

s, f

, (20)

and

2 =2

r2+

1

r

r+

1

r22

2+

2

z2. (21)

The characteristic equation of Eq. (17a) is k2pKp 2M = 0. (22)The solution of (22) yields two eigenvalues

kp1,2

2=

1

cp1,2

2=

1,2

+ 2

, (23)

where

1,2 =

b

b2 ac

,

a = + 2M

, b = 12

2f + c + 2

Mm

, c = m

f

2. (24)

kp1 and kp2 are the complex wavenumbers of two P-waves (P1 and P2) in the medium, while Re( Cp1) andRe(Cp2) are the propagation velocities of P1-waves and P2-waves, respectively. Two eigenvectors associatedwith two eigenvalues in (23) are taken as

aT1 = (1, 1) , aT2 = (1, 2) , (25)

where the amplitude ratios 1,2 are defined as

1,2 = M

kp1,2/

2 f

Mkp1,2/2 m. (26)

Eliminating 2f in (17a) and substituting

s = p1 + p2, f = 1p1 + 2p2 (27)

into (17a) we get

2p1 + k2p1p1 = 0, (28a)

2p2 + k2p2p2 = 0, (28b)

where use has been made of (22). The characteristic equation of (17b) and (17c) is

k2s Ks

2M = 0. (29)


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22 P. Zheng et al.

From (29), we have

ks

2=

1

cs

2=

3

, (30)

where

3 =

1

2f

m

. (31)

The eigenvector corresponding to (30) is taken as

aT3 = (1, 3) , (32)

where the amplitude ratios 3 are defined as

3 = f

m

. (33)

Substitution of

f = 3s, f = 3s (34)

into Eqs. (17b) and (17c), respectively, yields

2s + k2s s = 0, (35a)

2s + k2s s = 0. (35b)

By means of the technique of separation of variables of (28) and (35), we finally obtain the general solutionsof (17):

= a1

n=0

[A1n(r,z) cos n + B1n (r,z) sin n]Jn (kr) ep1z

+ a2

n=0

[A2n(r,z) cos n + B2n (r,z) sin n]Jn (kr) ep2z, (36a)

= a3

n=0

[Cn(r,z) cos n + Dn (r,z) sin n]Jn (kr) eqz , (36b)

= a3

n=0

[En(r,z) cos n + Fn (r,z) sin n]Jn (kr) eqz , (36c)

where

p1,2 =

k2 k2p1,2, q =

k2 k2s , Re p1,2 0, Re q 0. (37)

Jn () denotes the Bessel function of the first kind of order n [22] and A1n, A2n, B1n, B2n, Cn, Dn, En, Fn arearbitrary functions to be determined. The unbounded solutions as r = 0 or z have been ruled out.

Here, we care about two special cases when n takes the value 0 and 1 in (36). For n = 0 the displacementfields possess axial symmetry with respect to the z axis. Thus, the component u and w vanish and (36)reduce to

(r,z) =

a1Aep1z + a2Be

p2z

J0 (kr) ,

(r,z) = a3CeqzJ0 (kr) . (38)


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After substitution of (38) into (14) and (16), integrate the displacement fields and pore pressure with respectto kfrom 0 to , which yields

ur =

0

Aep1z Bep2z + Cqeqz J1

(kr)kdk,

uz =

0

Ap1e

p1z Bp2ep2z + Ck2eqz

J0(kr)dk,

p = M

0

Ag3e

p1z + Bg4ep2z

J0(kr)dk,

(39a)

where

g3 = ( + 1) k2p1, g4 = ( + 2) k

2p2. (39b)

The stresses and pore pressure at the surface z = 0 are given by

zr =

0

2 (Ap1 + Bp2) C

k2 + q2

J1 (kr) kdk,

zz =

0

Ag1 + Bg2 2Ck

2q

J0 (kr) dk,

p = M

0

(Ag3 + Bg4) J0 (kr) dk,

(40a)

where

g1 =

+ 2

k2p1 + 2k

2, g2 =

+ 2

k2p2 + 2k

2, (41)

Performing over (40a) the Hankel transform defined by the relations

(n) (k, ,z) =

0

(r, ,z) Jn (kr) rdr,

(r, ,z) =

0

(n) (k, ,z)Jn (kr) kdk,

(42)

where the superscripts in parentheses denote the order of the Hankel transform, we have

(1)zr = 2 (Ap1 + Bp2) C

k2 + q2

,

(0)zz = Ag1 + Bg2 2Ck

2q

k,

p(0) = M(Ag3 + Bg4)/k. (43)

For n = 1, without loss of generality, (36) becomes

(r, ,z) =

a1Aep1z + a2Be

p2z

J1 (kr) cos ,

(r, ,z) = a3DeqzJ1 (kr) sin ,

(r, ,z) = a3CeqzJ1 (kr) cos .

(44)


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24 P. Zheng et al.

Similarly, the displacement components and pore pressure are given by urcos

+u

sin

(2)= Aep1z + Bep2z Cqeqz Deqz , (45a)

urcos

u

sin (0)

= Aep

1z

+ Bep

2z

Cqeqz

+ Deqz

, (45b) uzcos

(1)=

Ap1ep1z Bp2e

p2z + Ck2eqz

k, (45c) pcos

(1)= M

Ag3e

p1z + Bg4ep2z

k, (45d)

where we have made use of the following identities:

2d

dJv () = Jv1 () Jv+1 () ,

2v

Jv () = Jv1 () + Jv+1 () . (45e)

The stresses and pore pressure at the surface z = 0 are given by

zrcos

z

sin (0) = 2Ap

1+ 2Bp

2 Ck2 + q2 + Dq , zr

cos +

z

sin

(2)=

2Ap1 + 2Bp2 C

k2 + q2

Dq

,

(1)

zz

cos =

Ag1 + Bg2 2Ck

2q

k,

p(1)

cos = M (Ag3 + Bg4)

k.

(46)

4 Fundamental singular solutions

In this section, the 3D fundamental solutions of an infinite medium due to a time-harmonic concentrated point

force and fluid source will be derived, respectively, in the frequency domain.

4.1 Point force

We seek a solution for an infinite region subject to a time-harmonic concentrated force. Thus, we have to solvethe equations of

(c + 2) u u + M w + 2

u + fw

= F (r r0) e, (47a)

M u + M w + 2

fu + mw

= 0, (47b)

where (r) is the three-dimensional delta distribution, e is a constant unit vector, and Fdenotes the magnitudeof the concentrated force. Noting that the source term can be decomposed as

F (r r0) e = F2

14R

e = F

e

4R e

4R

, R = |r r0| , (48)

we represent the solution in terms of (see, e.g., [7])

u = F[ (se) + (se)] , (49a)

w = F

fe

+

fe

. (49b)

Substitution of (49) into (47) yields

Kp

2 + 2M =

1

4R

10

, (50a)

Ks

2 + 2M

=

1

4R 10

. (50b)


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The problem is therefore reduced to the solution of a pair of scalar inhomogeneous Helmholtz equations in aninfinite region. The solution can be written as the sum of the particular solution of the inhomogeneous equa-tion, plus the general solution of the counterpart homogeneous equation. The latter must satisfy the radiationcondition at infinity.

In view of the point symmetry of the displacement fields, the general solutions corresponding to thehomogeneous equations of (50a) and (50b) can be found as

= 1

4 k2s

AIa1

eikp1R

R+ AIIa2

eikp2R

R

, (R = 0), (51a)

= 1

4 k2sAIIIa3

eiksR

R, (R = 0) , (51b)

where kp1,2, ks, a1, a2, a3 are defined in (23, 30, 25), and (32). AI, AII, AIII are constant to be determined bythe regularity conditions at R = 0. The particular solutions of (50a) and (50b) can be taken as

= =1

4 k2s

1

Ra3. (52)

Thus, the solutions of (50a) and (50b) can be written as

=1

4 k2s

1

Ra3 AIa1

eikp1R

R AIIa2

eikp2R

R

, (53a)

=1

4 k2sa3

1

R AIII

eiksR

R

, (53b)

where

AI =2 3

2 1, AII =

1 3

1 2, AIII = 1. (54)

1,2 and 3 are defined in (26) and (33). Substituting (53) into (49) and then introducing the results into (1c)and making use of

2

1

R

= 4 (r r0) ,

2 + k2

eik RR

= 4 (r r0) . (55)

we arrive at

u = FGfu e, (56a)

w = FGfw e, (56b)

p = FGfp e, (56c)

where

Gfu =

2sI + (s s)

=

1

4 k2s

k2s

eiksR

R I + eiksR

R AIeikp1R

R AIIeikp1R

R

, (57a)

Gfw =

2fI +

f f

=1

4 k2s

3k

2s

eiksR

RI +

3

eiksR

R 1AI

eikp1R

R 2AII

eikp1R

R

, (57b)

Gfp =

M

4 k2s

k2p1 ( + 1) AI

eikp1R

R+ k2p2 ( + 2) AII

eikp2R

R

. (57c)

Noting that

k2p1 ( + 1) AI = k2p2 ( + 2) AII =

M

k2s

k2p1 k2p2

f m

+ 22, (57d)


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26 P. Zheng et al.

(57c) can be further expressed as

Gfp =

1

4

f m

( + 2) k2p1 k2p22

eikp1R

R

eikp2R

R

. (57e)

Gfu , G

fw and G

fp are respectively tensor-valued and vector-valued spectral Greens functions that give the solid

matrix displacement, relative solidfluid displacement, and excess pore pressure due to a point force withmagnitude F, applied at r0, and in the direction e by means of(56).

4.2 Fluid injection point source

We seek a solution for an infinite region due to a fluid injection point source. Therefore, we have to find thesolution of

( + 2) u u + 2

2f

m

u

f

mp = 0, (58a)

2p + 2m

Mp 2

f m

u = im (r r0) , (58b)

where .denotes the source strength. Consider the following representation:

u =

s +

s

, (59a)

p =

M2s M2f

. (59b)

Substitution of (59) into (58), yieldsKp

2 + 2M =

1

4R

0

im

, (60a)

2s + k2ss = 0, (60b)

where

Kp =

c + 2

fm

M

fm

M

M M

, (61a)

M =

2f

m0

f m

. (61b)

Obviously, the second-order dilatation stiffness matrix and mass matrix are no longer symmetric under upformulation. However, the characteristic equation corresponding to homogeneous equation of (60a) is also(22). With the notation defined in (23) and (25), the general solution corresponding to homogeneous equationof (60a) can be found as

=1

4

i

BIa1

eikp1R

R+ BIIa2

eikp2R

R

, (R = 0) . (62)

The particular solution of(60a) can be taken as

= i

4R

01

. (63)

Hence, the solutions of(60a) and (60b) can be written as

= 1

4

i

1

R

01

Ba1

eikp1R

R+ Ba2

eikp2R

R

, (64a)

s = 0, (64b)


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where

B = BI = BII =1

1 2. (65)

Substituting (64) into (59) and then introducing the results into (2b) and making use of (55) and (26), we have

u = Gsu, (66a)

w = Gsw, (66b)

p = Gsp, (66c)

where

Gsu =B

4

i

eikp1R

R

eikp2R

R

, (67a)

Gsw = B4i

1 eik

p1R

R 2 e

ikp2

R

R

, (67b)

Gsp = MB

4

i

( + 1) k

2p1

eikp1R

R ( + 2) k

2p2

eikp2R

R

. (67c)

Gsu, Gsw and G

sp are respectively vector-valued and scalar-valued spectral Greens functions due to a fluid

injection point source.As a final remark, we also find the correspondence between the pressure solution due to a point force and

the displacement solution due to a fluid injection point source, that is

Gfp = iG

su, (68)

where use has been made of (22) and (26). This relation was noticed by Cheng et al. [4]. Due to the timevariation factor assumed to be eit instead ofeit, Chengs result has a negative sign in front ofiGsu .

5 Half-space Greens functions

The Greens functions for a poroelastic half-space subject to a buried point force and fluid source are presentedin this section. For simplicity, expressions for the solid displacement and excess pore pressure are given. Thesurface of half-space is assumed to be traction free and permeable, that is

zr = 0, z = 0, zz = 0, p = 0. (69)

The solutions can be constructed as follows:

Ui (r, ,z) = ui (r, ,z) + ui (r, ,z) + ui (r, ,z) , (70a)

P (r, ,z) = p (r, ,z) + p (r, ,z) + p (r, ,z) , (70b)

where i = r, , z. ui and p, appearing on the right side of (70), are the displacement component of solidskeleton and excess pore pressure due to a single source placed at (0, 0, h) in an infinite medium. The sec-ond terms, ui and p

, correspond to contributions from an image source placed at (0, 0, h) in the infinitemedium. The last terms on the right side of (70) are constructed from general solutions obtained in Sect. 3 andcorrespond to a distribution of forces and excess pore pressure on the half-space surface so that the boundarycondition (69) will be satisfied.


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The second terms in (70) are given by

ur = Fz

4 k2s

2

rz Gs AIG

p1 AIIG

p2 , (78a)

uz = Fz

4 k2s

2

z2

Gs AIG

p1 AIIG

p2

+ k2s G

s

, (78b)

p =Fz

4k2s

M

z

k2p1 ( + 1) AIG

p1 + k

2p2 ( + 2) AIIG

p2

, (78c)

where

Gp1 =eikp1R

R, Gp2 =

eikp2R

R, Gs =

eiksR

R, (79)

R =

r r0

=

r2 + (z + h)2. (80)

It is noted that the expression (78) has the general form as the solution corresponding to a vertical point forceof magnitude Fz applied at (0, 0, h), along the negative z direction. It is easily shown that

zr (r, 0) +

zr (r, 0)

(1)= 0, (81a)

zz (r, 0) +

zz (r, 0)(0)

= Fz

2k2sMz, (81b)

p (r, 0) + p (r, 0)

(0)=

Fz

2k2s

M

Nz, (81c)

where

Mz = g1AIep1h

+ g2AIIep2h

2k2

eqh

, (82a)Nz = g3AIe

p1h + g4AIIep2h. (82b)

In view of (81), we must add further solutions to remove the boundary stress and excess pore pressure, andthese solutions must introduce no new singularities in the region 0 z. This requirement can be fulfilled byadding a solution of homogeneous wave equations obtained in Sect. 3. Let

(1)zr (r, 0) = 0, (0)

zz (r, 0) =

kMz, p

(0) (r, 0) = M

kNz . (83)

Substituting (83) into (43), we can determine the constants to be

A =1

(k)

(k2 + q2) (g4Mz g2Nz) + 4Nzk2p2q

,

B = 1

(k)

(k2 + q2) (g3Mz g1Nz) + 4Nzk

2p1q

,

C=2

(k)

Mz (p1g4 p2g3) Nz (p1g2 p2g1)

,

(84)

where

(k) = (g1g4 g2g3)k2 + q2

4k2 (p1g4 p2g3) q, (85)

is the corresponding Rayleigh function for the case of a poroelastic half-space.


14/23

30 P. Zheng et al.

We therefore arrive at the solution for a vertical point force in a poroelastic half-space:

U

z

r =

Fz

4k2s

0

k

e

q|zh|

e

q(z+h)

AI

e

p1|zh|

e

p1(z+h) AII

e

p2|zh|

e

p2(z+h)+ 2

Aep1z Bep2z + Cqeqz

J1(kr)kdk, (86a)

Uzz =Fz

4k2s

0

k2

q

eq|zh| eq(z+h)

p1AI

ep1|zh| ep1(z+h)

p2AII

ep2|zh| ep2(z+h)

+ 2

Ap1e


J0(kr)kdk, (86b)

Pz = Fz

4k2s

M

0

g3

AI

ep1|zh| ep1(z+h)

+ 2Aep1z

+ g4AII

ep2|zh| ep2(z+h)

+ 2Bep2z

J0(kr)kdk, (86c)

where

= sgn (z h) =

+1, z > h,1, z < h.

(86d)

5.2 Horizontal point force

Let a point force of magnitude Fx applied at z = h (h > 0), along the positive xdirection. After setting e = exin (56a) and (56c), the component of the displacement field ui and pore pressure p becomes

ur =Fxcos

4 k2s

k2s Gs +

2

r2

Gs AIGp1 AIIGp2

, (87a)

u = Fxsin

4 k2s

k2s Gs +

1

r

r

Gs AIGp1 AIIGp2

, (87b)

uz =Fxcos

4 k2s

2

rz

Gs AIGp1 AIIGp2

, (87c)

p = Fxcos

4k2s

M

rk2p1 ( + 1) AIGp1 + k2p2 ( + 2) AIIGp2 . (87d)

Introducing (87) into (1b) we have

zr =Fxcos

4k2s

z

k2s Gs + 2

2

r2

Gs AIGp1 AIIGp2

, (88a)

z = Fxsin

4k2s

z

k2s Gs +

2

r

r

Gs AIGp1 AIIGp2

, (88b)

zz =Fxcos

4k2s

r

k2s Gs +

2 + 2

2

z2

Gs AIGp1 AIIGp2

. (88c)


15/23


Therefore

zrcos

z

sin

(0)=

Fx

2k2sq2eq(zh) + k2 AIep1(zh) + AIIep2(zh) , (89a) zr

cos +

z

sin

(2)=

Fx

2k2sk2eq(zh) AIe

p1(zh) AIIep2(zh)

, (89b)

(1)

zz

cos =

Fx

4k2sk

g1

AI

p1ep1(zh) + g2

AII

p2ep2(zh) 2qeq(zh)

, (89c)

p(1)

cos =

Fx

4k2s

M

k

g3

AI

p1ep1(zh) + g4

AII

p2ep2(zh)

. (89d)

Similarly, we have

zr (r, , 0)

cos

z (r, , 0)

sin

(0)+

zr (r, , 0)

cos

z (r, , 0)

sin

(0)=

Fx

k2sk2Mx, (90a)

zr (r, , 0)

cos +

z (r, , 0)

sin

(2)+

zr (r, , 0)

cos +

z (r, , 0)

sin

(2)=

Fx

k2sk2Nx, (90b)

zz (r, , 0) +

zz (r, , 0)(1)

= 0, (90c)p (r, , 0) + p (r, , 0)

(1)= 0, (90d)

where

Mx =q

k

2eqh AIe

p1h AIIep2h, (91a)

Nx = eqh AIe

p1h AIIep2h. (91b)

Let

zr (r, , 0)

cos

z (r, , 0)

sin

(0)= Mx,

zr (r, , 0)

cos + z (r, , 0)

sin

(2)= Nx,

(1)zz (r, , 0) = p(1) (r, , 0) = 0.

(92)

Substituting (92) into (46), we can determine the constants to be

A =1

(k)(Mx + Nx) k

2qg4, B = 1

(k)(Mx + Nx) k

2qg3,

C=1

2(k)(Mx + Nx) (g1g4 g2g3) , D =

1

2q(Mx Nx) .

(93)


16/23

32 P. Zheng et al.

Thus, the solution for a horizontal point force in a poroelastic half-space has been found as

Uxr =Fxcos

8 k2s

0

u(0)J0(kr)kdk+

0

u(2)J2(kr)kdk

, (94a)

Ux = Fxsin

8 k2s

0

u(0)J0(kr)kdk

0

u(2)J2(kr)kdk

, (94b)

Uxz =Fxcos

4 k2s

0

k

eq|zh| eq(z+h) (94c)

AI

ep1|zh| ep1(z+h)

AII

ep2|zh| ep2(z+h)

+ 4

Ap1e


J1(kr)kdk, (94d)

Px

=

Fxcos

4k2s

M

0

k

g3AI

p1

ep1|zh|

ep1(z+h)

+ Aep1z

+ g4

AII

p2

ep2|zh| ep2(z+h)

+ Bep2z

J1(kr)kdk, (94e)

where

u(0) =

k2

q 2q

eq|zh| eq(z+h)

+ k2

AI

p1

ep1|zh| ep1(z+h)

+AII

p2

ep2|zh| ep2(z+h)

+ 4k2

Aep1z + Bep2z Cqeqz + Deqz

, (95a)

u(2) = k2 1qeq|zh| eq(z+h) AI

p1ep1|zh| ep1(z+h) AII

p2ep2|zh| ep2(z+h)

4Aep1z + Bep2z Cq eqz Deqz

. (95b)

5.3 Fluid injection point source

Suppose that there acts a concentrated fluid source at (0, 0, h) in the half-space. From (66) we have

ur = B

4

i

r

Gp1 Gp2

, (96a)

uz = B

4

i

z Gp1 Gp2

, (96b)

p = M B 4

i

k2p1 ( + 1) Gp1 k2p2 ( + 2) Gp2

. (96c)

Introducing (96) into (1b) we have

zr = B

2

i

2

rz

Gp1 Gp2

, (97a)

zz = B

4

i

2 + 2

2

z2

Gp1 Gp2

, (97b)

Therefore

(1)zr = B

2

i

ke

p1(zh) ep2(zh) , (98a)


17/23


(0)zz = B

4

i

g1

1

p1ep1(zh) g2

1

p2ep2(zh)

, (98b)

p(0) = M B

4

i

g31

p1ep1(zh) g4

1

p2ep2(zh) . (98c)

Similarly, we have zr (r, 0) +

zr (r, 0)

(1)= B

i

kMs, (99a)

zz (r, 0) +

zz (r, 0)(0)

= 0, (99b)p (r, 0) + p (r, 0)

(0)= 0, (99c)

where

Ms = ep1h ep2h. (100)

Let

(1)

zr

(r, 0) = , (0)

zz

(r, 0) = 0, p(0) (r, 0) = 0. (101)

Substitution of (101) into (43) leads to

A = 1

(k)2k2qg4, B =

1

(k)2k2qg3, C=

1

(k)(g1g4 g2g3) . (102)

We therefore arrive at the solution for a fluid injection point source in the half-space:

Usr = B

4

i

0

k

1

p1

ep1|zh| ep1(z+h)

1

p2

ep2|zh| ep2(z+h)

4Ms

Aep1z Bep2z + Cq eqz

J1(kr)kdk, (103a)

U

s

z = B

4

i

0

e

p1|zh|

e

p1(z+h)

e

p2|zh|

e

p2(z+h) 4Ms

Ap1e


J0(kr)kdk, (103b)

Ps = M B

4

i

0

g3

1

p1

ep1|zh| ep1(z+h)

+ 4AMs

g4

1

p2

ep2|zh| ep2(z+h)

4BMs

J0 (kr) kdk. (103c)

6 Verification of the solutions

In order to provide a means of checking the validityof the solutions presented in Sect. 5, two limiting cases willbe considered. First, investigate the solutions form as h approaches zero, to see whether they would exactlytake the form as the solutions of Lambs problem. Next, examine the solutions as , f, m, b, M approachzero, to see whether they would reduce to the known counterparts of elastodynamic solutions. For simplicity,the solution for a vertical point force is presented. Clearly, the solution for a horizontal point force can also bederived in the same way for the above-mentioned two limiting cases.

6.1 Limiting case 1: Lambs problems

The displacement field due to a vertical concentrated force in a poroelastic half-space, that is the expression(86), should be reduced to the known solution of Lambs problem by letting h 0. When h 0, (82)becomes


18/23

34 P. Zheng et al.

Mz = k2s , (104a)

Nz = 0, (104b)

where use has been made of

AI + AII = 1, (104c)kp1kp2

2

2=

3m

( + 2) M, (104d)

g3AI = g4AII =

f m

kp1kp2

2k2p1 k

2p2

m

, (104e)

g1AI = 2AIk2

k2s

Mk2p1 m

M

k2p1 k

2p2

, (104f)

g1AI = 2AIIk

2

+

k2s Mk2p2 m

M

k2p1 k2p2

. (104g)

Substitution of (104a) and (104b) into (84), one has

A = 1

(k)k2s (k

2 + q2)g4

B =1

(k)k2s (k

2 + q2)g3,

C = 2

(k)k2s (p1g4 p2g3) ,

(105)

Hence, after setting z = 0 in (86a) and (86b), the surface displacement field due to a vertical point force at(0, 0, 0) becomes

Uzr(r, 0) =Fz

2

0

k

(k2 + q2) (g4 g3) + 2 (p1g4 p2g3) q

(k)J1 (kr) kdk (106a)

Uzz (r, 0) = Fz

2

0

(k2 q2) (p1g4 p2g3)

(k)J0 (kr) kdk, (106b)

which take the same form as those found by Philippacopoulos [14] for the Lambs problem of a poroelastichalf-space. This supports the fact that the frequency domain solutions derived in the previous section are likelyto be correct.

6.2 Limiting case 2: elastodynamics

Now we check the solution form as , f, m, b, and Mapproach zero. Noting, in this limiting case, that

p1 = p =

k2 k2p , Re p1 0, (107a)

p2 = 0, (107b)

q =

k2 k2s , Re q 0, (107c)

(k) =k2 + q2

2 4k2pq, (107d)

AI = 1, (107e)

AII = 0, (107f)


19/23


where kp and ks are wave numbers of P and S waves, respectively, in the elastic half-space and (107d) isthe conventional Rayleigh function for an elastic material. Additionally, all terms associated with the slowdilatational wave are eliminated. Thus, (84) reduces to

A =

1

(k)k

2

+ q2

k

2

+ q2

e

ph

2k2

e

qh,

B = 0,

C =2

(k)p

k2 + q2

eph 2k2eqh

.

(108)

By substitution of (108) into (86), one has

Uzr(r, 0) =Fz

2

0

k

2pqeph +k2 + q2

eqh

(k)

J1(kr)kdk, (109a)

Uzz (r, 0) =Fz

2

0

p

k2 + q2

eph 2k2eqh

(k) J0(kr)kdk, (109b)which are consistent with those found originally by Pekeris [11]. Checking this limiting case again confirmsthe correctness of the solutions presented in Sect. 5.

7 Numerical results

Explicit expressions for the displacement fields and excess pore pressure due to buried sources in a poroelastichalf-space have been obtained. These solutions are expressed in terms of semi-infinite Hankel-type integralswith respect to horizontal wavenumber, that is, integrals of the form

Im (r; ) =

0

F(k; )Jm (kr) kdk, m = 0, 1, 2, (110)

which must, in general, be evaluated numerically for a given frequency. It should be noted that the kernelF(k; ) has singularities in the form of branch points at k = kp1, k = kp2, and k = ks and poles atk= kr, where kr are the roots of the Rayleigh function defined in (85). Although the real kaxis is free fromany singularities as the viscous dissipation effects are incorporated, the presence of complex poles near thereal kaxis may introduce sharp peaks in the integrand. In our numerical analysis, the material dissipation isincorporated to remove the poles by giving the Lame constants an imaginary part

= (1 + i2) , = (1 + i2) . (111)

In addition, the complete integral is formally obtained by summing the partial integrations, but direct sum-mations are feasible only for rapidly convergent integrals, and the result may be very slowly convergent andeven algebraically divergent. Moreover, the difficulty in evaluating the integrals of the type given by (110)

also stems from the oscillatory nature of both the Bessel functions and the kernel F(k;). In view of the abovefeatures of the integrand, we have followed Chave [23] and used a high-order adaptive algorithm combinedwith continued faction expansions for accelerating the convergence of the truncated integral to evaluate suchintegrals. Numerical tests on various types of irregularly oscillatory functions including those algebraicallydivergent types have been displayedby Chave [23] to demonstrate thefeatures andcapabilities of thealgorithm.Also, this novel quadrature algorithm has been successfully applied to evaluate the responses of a multilayeredelastic half-space [24]. Alternative quadrature schemes have been investigatedby Bouchon andAki[25], Apseland Luco [26], Hisada [27,28], etc.

A soil (coarse sand) is considered in the following numerical examples. The material parameters, takenfrom literature [29], are given as follows: = 1.29 108 Pa, = 9.79 107 Pa, M = 2.50 109 Pa, =1.884 103 kg/m3, f = 1.0 10

3 kg/m3, m = 3.646 103 kg/m3, = 0.981, b = 1.185 108 Ns/m4.In the examples that follow, the damping term takes the value = 5 %, the magnitudes of vertical andhorizontal point force are Fx = Fz = 2 10

6 N, and the pore fluid is injected at a rate of = 2 m3/s. It


20/23

36 P. Zheng et al.

1.2

1.0

0.8

0.6

0.4

0.2

0.0-4 -2 0 2 4

h=3.0 m

h=2.0 m

h=1.0 m

Verticaldisplacement

Reuz

/m10-2

+x/m-x/m

=1.0Hz

(a)

0.20

0.15

0.10

0.05

0.00

-0.05

-0.10

-0.15

-0.20-4 -2 0 2 4

VerticaldisplacementReuz

/m10-2

=1.0Hz

+x/m-x/m(b)

h =3.0 m

h =2.0 m

h =1.0 m

0.00

-0.02

-0.04

-0.06

-0.08

-0.10

-0.12

-0.14

-0.16

-0.18-4 -2 0 2 4

Verticaldisplacem

entReuz

/m

h=3.0 m

h=2.0 m

h=1.0 m

=1.0Hz

+x/m-x/m(c)

Fig. 1 Variation of vertical displacement along the xaxis on the free surface with source depth due to a a vertical point force,b a horizontal point force, and c a fluid injection point source

should be mentioned that only the real parts of the amplitudes of interested physical quantities, that is the cor-responding real parts of the quantities after removing the harmonic time factor eit, are shown schematicallybelow.

Figure 1 shows the variation of vertical displacement along the xaxis on the free surface with source depthdue to a vertical point force, a horizontal point force, and a fluid injection point source, respectively. As thesource deepens, the amplitude of vertical displacement decreases as can be seen in Fig. 1. Figure 2 shows thevariation of vertical displacement and pore pressure along the vertical line (2.0 m, 0, 0 10.0m) due to ahorizontal point force located at (0, 0, 0.5m), (0, 0, 1.0m), and (0, 0, 1.5m), respectively. It can be clearly seenfrom Fig. 2b that pore pressure is zero at the surface and tends to zero as z .

Figures 3 and 4 show, respectively, pore pressure and total stress zz along the z axis for field point atr = 1.0 m and r = 0.05 m due to a fluid injection point source at (0, 0, 0.5m). The effects of free surface arealso shown in Figs. 3a and 4a by drawing a comparison with the full-space results. As expected, full-spaceand half-space solutions agree well for large depths. At the free surface, however, the difference between thetwo solutions reaches a maximum. Additionally, it can also be seen from Figs. 3a and 4a that pore pressureand total normal stress is zero at the surface, attains a maximum value at a depth depending upon rand sourcedepth h and tends to zero as z . It is well known that there is a singularity at the point where the source

is located. The behavior near the singularity can be also observed from Figs. 3b and 4b, respectively, for smallvalues ofr.

The variation of vertical displacement, pore pressure versus excitation frequency is shown in Fig. 5 forthe field point at (0.5 m, 0, 0.25m) and (1.0m, 0, 0.25 m) due to a vertical point force at (0, 0, 0.5m). Thefrequency change is in the range of 0150Hz. As the excitation frequency increases, the amplitude of ver-tical displacement decreases as can be seen in Fig. 5a. As the loading frequency 0, the pore pressureapproaches zero, as shown in Fig. 5b. On the other hand, at high frequencies, the pore pressure reaches its peakvalue and remains nearly unchanged, especially for the relative far field point (1.0m, 0, 0.25m).

8 Conclusions

The preceding developments lead to the following conclusions:


21/23


10

8

6

4

2

0

Depthz/m

Vertical displacement Reuz/m10

-2

h=1.5 m

h=1.0 m

h=0.5 m

=1.0Hz

(a)

10

8

6

4

2

0-0.08 -0.06 -0.04 -0.02 0.00 0.02 0.04 0.06 0.0 0.5 1.0 1.5 2.0 2.5

Depthz/m

h=1.5 m

h=1.0 m

h=0.5 m

Pore pressure Rep m2/N10

4(b)

=1.0Hz

Fig. 2 Variation of vertical displacement and pore pressure along the vertical line (2.0m, 0, 0 10.0m) due to a horizontal pointforce located at (0, 0, 0.5m), (0, 0, 1.0m), and (0, 0, 1.5m), respectively

5

4

3

2

1

0

Depthz/m

full-space

half-space

=1.0Hz,h=0.5m

r=1.0m

(a) Pore pressure Repm2/N10

7

5

4

3

2

1

0-0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 0 20 40 60 80 100 120

Depthz/m

Pore pressure Repm2/N10

7

(b)

=1.0Hz,h=0.5m

r=0.05m

Fig. 3 Depth profiles of pore pressure for field point at a r = 1.0 m and b r = 0.05 m due to a fluid injection point source at(0, 0, 0.5m)

5

4

3

2

1

0

-1.0 -0.8 -0.6 -0.4 -0. 2 0.0 0.2 0.4

Depthz/m

full-space

half-space

=1.0Hz,h=0.5m

Stress Rezz

m2/N10

7

r=1.0m

(a)

5

4

3

2

1

0-1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 0.4 -30 -25 -20 -15 -10 -5 0

Depthz/m

Stress Rezz

m2/N10

7

=1.0Hz,h=0.5m

r=0.05m

(b)

Fig. 4 Depth profiles of total stress for field point at a r = 1.0 m and b r = 0.05 m due to a fluid injection point source at(0, 0, 0.5m)


22/23

38 P. Zheng et al.

0.2

0.4

0.6

0.8

1.0

1.2

1.4

VerticaldisplacementReuz

/m10-2

Excitation frequency /Hz

r=1.0 m

r=0.5 m

h =0.5m

(a)

0 20 40 60 80 100 120 140 0 20 40 60 80 100 120 140

0

2

4

6

8

10

12

PorepressureRepm

2/N105

Excitation frequency /Hz

r=1.0 m

r=0.5 m

h =0.5m

(b)

Fig. 5 a Vertical displacement and b pore pressure of the field point at (0.5m, 0, 0.25m) and (1.0m, 0, 0.25m) versus excitationfrequency due to a vertical point force at (0, 0, 0.5 m)

(i) Representing displacement fields in terms of six scalar potentials (three for the solid phase and threefor the fluid, respectively), we have derived the 3D general solutions in the frequency domain for thehomogeneous wave equations.

(ii) Thefundamental solutionsof thefull space dueto a point load andfluid source, respectively, areobtainedin the frequency domain.

(iii) Explicit analytical expressions for the displacement components and excess pore pressure due to avertical point force, a horizontal point force, and a fluid injection point source, respectively, in a poro-elastic half-space, that is Greens functions fora poroelastic half-space, have been obtained in frequencydomain by using mirror-image technique.

(iv) In two limiting cases, the solutions are shown to reduce to known counterparts of elastodynamics andthose of Lambs problem, thus ensuring the validity of our result.

(v) The solution for an arbitrary time-dependent source can be Fourier synthesized from the solutions pre-sented in this study. Based on the dynamic Greens functions derived in this study, a BEM (the boundaryelement method) formulation can be developed to study the wave propagation problems in a poroelastichalf-space.

Acknowledgments This work is financially supported by the National Natural Science Foundation of China through Grant No.11172268.

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