Mixed boundary value problems often appear in elasticity problems. An early example involved finding the distribution of stress within a semi-infinite elastic medium when a load is applied to the surface z = 0. Reissner and Sagoci15 used separation of variables and spheroidal coordinates. In 1947 Sneddon16 resolved the static (time-independent) problems applying Hankel transforms. This is the approach that we will highlight here.
Ifu(r, z) denotes the circumferential displacement, the mathematical the- ory of elasticity yields the governing equation
∂2u
∂r2 +1 r
∂u
∂r − u r2 +∂2u
∂z2 = 0, 0≤r <∞, 0< z <∞, (2.5.1)
14 See Section 12 in Sneddon, I. N., 1995:Fourier Transforms. Dover, 542pp.
15 Reissner, E., and H. F. Sagoci, 1944: Forced torsional oscillation of an elastic half- space. J. Appl. Phys.,15, 652–654; Sagoci, H. F., 1944: Forced torsional oscillation of an elastic half-space. II.J. Appl. Phys.,15, 655–662.
16 Sneddon, I. N., 1947: Note on a boundary value problem of Reissner and Sagoci. J.
Appl. Phys.,18, 130–132; Rahimian, M., A. K. Ghorbani-Tanha, and M. Eskandari-Ghadi, 2006: The Reissner-Sagoci problem for a transversely isotropic half-space. Int. J. Numer.
Anal. Methods Geomech.,30, 1063–1074.
subject to the boundary conditions
rlim→0|u(r, z)|<∞, lim
r→∞u(r, z)→0, 0< z <∞, (2.5.2)
zlim→∞u(r, z)→0, 0≤r <∞, (2.5.3)
and
u(r,0) =r, 0≤r≤a,
uz(r,0) = 0, a≤r <∞. (2.5.4) Hankel transforms are used to solve Equation 2.5.1 via
U(k, z) = ∞
0
r u(r, z)J1(kr)dr (2.5.5) which transforms the governing partial differential equation into the ordinary differential equation
d2U(k, z)
dz2 −k2U(k, z) = 0, 0< z <∞. (2.5.6) Taking Equation 2.5.3 into account,
U(k, z) =A(k)e−kz. (2.5.7) Therefore, the solution to Equation 2.5.1, Equation 2.5.2, and Equation 2.5.3 is
u(r, z) = ∞
0
k A(k)e−kzJ1(kr)dk. (2.5.8) Upon substituting Equation 2.5.8 into Equation 2.5.4, we have that
∞
0
k A(k)J1(kr)dk=r, 0≤r < a, (2.5.9)
and ∞
0
k2A(k)J1(kr)dk= 0, a < r <∞. (2.5.10) The dual integral equations, Equation 2.5.9 and Equation 2.5.10, can be solved using the Busbridge results, Equation 2.4.28 through Equation 2.4.30. This yields
A(k) = 4a πk2
sin(ak)
ak −cos(ak)
, (2.5.11)
and
u(r, z) = 4a π
∞
0
sin(ak)
ak −cos(ak)
e−kzJ1(kr)dk
k . (2.5.12)
0 0.5
1 1.5
2
0 0.2 0.4 0.6 0.8 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
z/a r/a
u(r,z)/a
Figure 2.5.1: The solutionu(r, z)/ato the mixed boundary value problem governed by Equation 2.5.1 through Equation 2.5.4.
We can evaluate the integral in Equation 2.5.12 and it simplifies to u(r, z) = 2a2
rπ
λRsin(ψ+ϕ)−2Rcos(ϕ) +r2
a2arctan
Rsin(ϕ) +λsin(ψ) Rcos(ϕ) +λcos(ψ)
, (2.5.13) whereλ2sin(2ψ) tan(ψ) = 2,λ2= 1 +z2/a2,ztan(ψ) =a,
R4= r2
a2 +z2 a2 −1
2
+4z2
a2 , and 2z
a cot(2ϕ) = r2 a2 +z2
a2 −1. (2.5.14) We illustrate Equation 2.5.13 in Figure 2.5.1.
We can generalize17the original Reissner-Sagoci equation so that it now reads
∂2u
∂r2 +1 r
∂u
∂r − u r2 +∂2u
∂z2 +K∂u
∂z +ω2u= 0, 0≤r <∞, 0< z <∞, (2.5.15)
17 See Chakraborty, S., D. S. Ray and A. Chakravarty, 1996: A dynamical problem of Reissner-Sagoci type for a non-homogeneous elastic half-space. Indian J. Pure Appl.
Math.,27, 795–806.
subject to the boundary conditions lim
r→0|u(r, z)|<∞, lim
r→∞u(r, z)→0, 0< z <∞, (2.5.16)
zlim→∞u(r, z)→0, 0≤r <∞, (2.5.17)
and
u(r,0) =r, 0≤r≤1,
uz(r,0) = 0, 1≤r <∞. (2.5.18) If we use Hankel transforms, the solution to Equation 2.5.15 through Equation 2.5.17 is
u(r, z) = ∞
0
A(k)e−κ(k)zJ1(kr)dk, (2.5.19) where κ(k) = K/2 +√
k2+a2 and a2 = K2/4 −ω2. Upon substituting Equation 2.5.19 into Equation 2.5.18, we have that
∞
0
A(k)J1(kr)dk=r, 0≤r <1, (2.5.20)
and ∞
0
κ(k)A(k)J1(kr)dk= 0, 1< r <∞; (2.5.21)
or ∞
0
B(k)[1 +M(k)]J1(kr)dk
k = 2r, 0≤r <1, (2.5.22)
and ∞
0
B(k)J1(kr)dk= 0, 1< r <∞, (2.5.23) whereB(k) = 2κ(k)A(k) andM(k) =k/κ(k)−1.
To solve the dual integral equations, Equation 2.5.22 and Equation 2.5.23, we set
B(k) =k 1
0
h(t) sin(kt)dt. (2.5.24)
We have done this because ∞
0
B(k)J1(kr)dk= 1
0
h(t) ∞
0
ksin(kt)J1(kr)dk
dt (2.5.25)
=− 1
0
h(t)d dr
∞
0
sin(kt)J0(kr)dk
dt= 0, (2.5.26) where we used Equation 1.4.13 and 0≤t ≤1 < r. Consequently our choice forB(k) satisfies Equation 2.5.23 identically.
Turning to Equation 2.5.22, we now substitute Equation 2.5.24 and in- terchange the order of integration:
1
0
h(t) ∞
0
sin(kt)J1(kr)dk
dt (2.5.27)
+ 1
0
h(τ) ∞
0
M(k) sin(kτ)J1(kr)dk
dτ = 2r.
Using Equation 1.4.13 again, Equation 2.5.27 simplifies to t
0
t h(t)
√r2−t2dt+
1
0
h(τ) ∞
0
M(k) sin(kτ)rJ1(kr)dk
dτ = 2r2. (2.5.28) Applying Equation 1.2.13 and Equation 1.2.14, we have
t h(t) = 4 π
d dt
t 0
η3 t2−η2dη
(2.5.29)
− 2 π
1
0
h(τ) ∞
0
M(k) sin(kτ)d dt
t 0
η2J1(kη) t2−η2dη
dk
dτ.
Now
d dt
t 0
ξ3 t2−ξ2dξ
= 2t2, (2.5.30)
and
d dt
t 0
ξ2J1(kξ) t2−ξ2dξ
=tsin(kt) (2.5.31)
after using integral tables.18 Substituting Equation 2.5.30 and Equation 2.5.31 into Equation 2.5.29, we finally obtain
h(t) + 2 π
1
0 h(τ) ∞
0 M(k) sin(kt) sin(kτ)dk
dτ =8t
π. (2.5.32) Equation 2.5.32 must be solved numerically. We examine this in detail in Section 4.3. Once h(t) is computed, B(k) and A(k) follow from Equation 2.5.24. Finally Equation 2.5.19 gives the solutionu(r, z). We illustrate this solution inFigure 2.5.2.
18 Gradshteyn, I. S., and I. M. Ryzhik, 1965: Table of Integrals, Series, and Products.
Academic Press, Formula 6.567.1 withν= 1 andà=−12.
0 0.5
1 1.5
2
0 0.2 0.4 0.6 0.8 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
z r
u(r,z)
Figure 2.5.2: The solution u(r, z) to the mixed boundary value problem governed by Equation 2.5.15 through Equation 2.5.18.
In 1989 Singh et al.19 extended the classical Reissner-Sagoci problem so that it now reads
∂2u
∂r2 +1 r
∂u
∂r − u r2 +∂2u
∂z2 = 0, 0≤r <∞, 0< z <∞, (2.5.33) subject to the boundary conditions
rlim→0|u(r, z)|<∞, lim
r→∞u(r, z)→0, 0< z <∞, (2.5.34)
zlim→∞u(r, z)→0, 0≤r <∞, (2.5.35)
and
u(r,0) =r, 0≤r < a, uz(r,0) = 0, a < r < b, u(r,0) = 0, b < r <∞.
(2.5.36)
As we showed earlier, the solution to Equation 2.5.33, Equation 2.5.34 and Equation 2.5.35 is
u(r, z) = ∞
0
A(k)e−kzJ1(kr)dk. (2.5.37)
19 Singh, B. M., H. T. Danyluk, and A. P. S. Selvadurai, 1989: The Reissner-Sagoci problem for a half-space with a surface constraint. Z. Angew. Math. Phys.,40, 762–768.
This work was extended in Singh, B. M., H. T. Danyluk, J. Vrbik, J. Rokne, and R. S.
Dhaliwal, 2003: The Reissner-Sagoci problem for a non-homogeneous half-space with a surface constraint.Meccanica,38, 453–465.
Upon substituting Equation 2.5.37 into Equation 2.5.36, we obtain the triple integral equations:
∞
0
A(k)J1(kr)dk=r, 0≤r < a, (2.5.38) ∞
0
k A(k)J1(kr)dk= 0, a < r < b, (2.5.39)
and ∞
0
A(k)J1(kr)dk= 0, b < r <∞. (2.5.40) Let us now solve this set of triple integral equations by assuming that
∞
0
k A(k)J1(kr)dk=
f1(r), 0< r < a,
f2(r), b < r <∞. (2.5.41) Taking the inverse of the Hankel transform, we obtain from Equation 2.5.39 and Equation 2.5.41 that
A(k) = a
0
rf1(r)J1(kr)dr+ ∞
b
rf2(r)J1(kr)dr. (2.5.42) Substituting Equation 2.5.42 into Equation 2.5.38 and Equation 2.5.40, we find that
a
0 τ f1(τ)L(r, τ)dτ + ∞
b
τ f2(τ)L(r, τ)dτ =r, 0< r < a, (2.5.43) and
a 0
τ f1(τ)L(r, τ)dτ+ ∞
b
τ f2(τ)L(r, τ)dτ = 0, b < r <∞, (2.5.44) where
L(r, τ) = ∞
0
J1(kr)J1(kτ)dk. (2.5.45) At this point, we introduce several results by Cooke,20namely that
L(r, τ) = 2 πrτ
min(r,τ) 0
t2
(r2−t2)(τ2−t2)dt (2.5.46)
= 2rτ π
∞
max(r,τ)
dt t2
(t2−r2)(t2−τ2), (2.5.47)
20 Cooke, J. C., 1963: Triple integral equations problems.Quart. J. Mech. Appl. Math., 16, 193–201.
b a
min(r,τ) 0
(ã ã ã)dt dτ = r
0
b t
(ã ã ã)dτ dt+ a
0
b a
(ã ã ã)dτ dt, (2.5.48) and
b a
∞
max(r,τ)
(ã ã ã)dt dτ= b
r
t a
(ã ã ã)dτ dt+ ∞
b
b a
(ã ã ã)dτ dt. (2.5.49) Why have we introduced Equation 2.5.46 through Equation 2.5.49? Ap- plying Equation 2.5.46, we can rewrite Equation 2.5.43 as
a 0
f1(τ)
min(r,τ) 0
t2
(r2−t2)(τ2−t2)dt
dτ (2.5.50)
+r2 ∞
b
τ2f2(τ) ∞
τ
dt t2
(t2−r2)(t2−τ2)
dτ = πr2 2 for 0< r < a. Then, applying Equation 2.5.48 and interchanging the order of integration in the second integral, we obtain
r 0
t2F1(t)
√r2−t2dt=πr2 2 −r2
∞
b
F2(t) t2√
t2−r2dt, 0< r < a, (2.5.51) where
F1(t) = a
t
f1(τ)
√τ2−t2dτ, 0< t < a, (2.5.52) and
F2(t) = t
b
τ2f2(τ)
√t2−τ2dτ, b < t <∞. (2.5.53) If we regard the right side of Equation 2.5.51 as a known function ofr, then it is an integral equation of the Abel type. From Equation 1.2.13 and Equation 1.2.14, its solution is
t F1(t) = 2t−1 π
∞
b
2ηt
η2−t2 −log η−t
η+t
F2(η)dη
η2, 0< t < a, (2.5.54) where we used the following results:
d dt
t 0
r3
√t2−r2dr
= 2t2, (2.5.55)
and d dt
t 0
r3
(t2−r2)(η2−r2)dr
= t 2
2ηt
η2−t2 −log η−t
η+t
. (2.5.56)
Turning to Equation 2.5.44, we employ Equation 2.5.46 and Equation 2.5.47 and find that
∞
b
τ2f2(τ) ∞
max(r,τ)
dt t2
(t2−r2)(t2−τ2)
dτ (2.5.57)
+ 1 r2
a 0
f1(τ) τ
0
t2
(r2−t2)(τ2−t2)dt
dτ = 0, ifb < r <∞. If we now apply Equation 2.5.49, interchange the order of inte- gration in the second integral and use Equation 2.5.52 and Equation 2.5.53, we find that
∞
r
F2(t) t2√
t2−r2dt=−1 r2
a 0
t2F1(t)
√r2−t2dt, b < r <∞. (2.5.58) Solving this integral equation of the Abel type yields
F2(t) t2 = 2
π d dt
∞
t
dr r√
r2−t2 a
0
η2F1(η) r2−η2dη
, b < t <∞. (2.5.59) Because
d dt
∞
t
dr r
(r2−t2)(r2−η2)
= 1
2ηt2log t−η
t+η
− 1
t(t2−η2), (2.5.60) F2(t)
t = 1 π
a 0
ηF1(η) 1
tlog t−η
t+η
− 2η t2−η2
dη, b < t <∞. (2.5.61) Setting ηF1(η) = 2aX1(η), F2(η)/η = 2aX2(η), c =a/b, and introduc- ing the variables η = bη1 and t = at1, we can rewrite Equation 2.5.54 and Equation 2.5.59 as
X1(at1) =t1− 1 π
∞
1
2ct1 η21−c2t21− 1
η1log
1−ct1/η1 1 +ct1/η1
X2(bη1)dη1, (2.5.62) when 0< t1<1; and
X2(bt1) = 1 π
1
0
c t1log
1−cη1/t1 1 +cη1/t1
− 2c2η1 t21−c2η12
X1(aη1)dη1, (2.5.63) when 1< t1<∞.
Once we find X1(at1) and X2(bt1) via Equation 2.5.62 and Equation 2.5.63, we can findf1(r) andf2(r) from
f1(r) =−2 π
d dr
a r
t F1(t)
√t2−r2dt
, 0< r < a, (2.5.64)
0 0.5 1 1.5 2 2.5 3 0
0.2 0.4
0.6 0.8
1 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
z/a r/a
u(r,z)/a
Figure 2.5.3: The solutionu(r, z)/ato the mixed boundary value problem governed by Equation 2.5.33 through Equation 2.5.36 whenc=12.
and
f2(r) = 2 πr2
d dr
r b
t F2(t)
√r2−t2dt
, b < r <∞. (2.5.65) Substituting Equation 2.5.64 and Equation 2.5.65 into Equation 2.5.42 and integrating by parts, we find that
A(k) =4ka3 π
1
0
1
t
X1(aτ)
√τ2−t2dτ
t J0(akt)dt +4ka3
c2π ∞
1
t 1
τ2X2(bτ)
√t2−τ2 dτ
J2(akt/c)
t dt. (2.5.66) Finally, Equation 2.5.37 gives the solutionu(r, z). Figure 2.5.3 illustrates this solution whenc=12.
In the previous examples of the Reissner-Sagoci problem, we solved it in the half-spacez >0. Here we solve this problem21within a cylinder of radius b when the shear modulus of the material varies as à0zα, where 0 ≤α <1.
Mathematically the problem is
∂2u
∂r2 +1 r
∂u
∂r − u r2 +∂2u
∂z2 +α z
∂u
∂z = 0, 0≤r < b, 0< z <∞, (2.5.67)
21 Reprinted fromInt. J. Engng. Sci.,8, M. K. Kassir, The Reissner-Sagoci problem for a non-homogeneous solid, 875–885, c1970, with permission from Elsevier.
subject to the boundary conditions
rlim→0|u(r, z)|<∞, u(b, z) = 0, 0< z <∞, (2.5.68)
zlim→∞u(r, z)→0, 0≤r < b, (2.5.69)
and
u(r,0) =f(r), 0≤r < a, zαuz(r, z)
z=0
= 0, a < r≤b, (2.5.70) whereb > a.
Using separation of variables, the solution to Equation 2.5.67 through Equation 2.5.69 is
u(r, z) =zp ∞ n=1
k−npAn(kn)Kp(knz)J1(knr), (2.5.71)
where 2p= 1−α, 0< p ≤ 12. Here kn denotes the nth root ofJ1(kb) = 0 andn= 1,2,3, . . .. Upon substituting Equation 2.5.71 into Equation 2.5.70, we obtain the dual series:
∞ n=1
kn−2pAnJ1(knr) = 21−p
Γ(p)f(r), 0≤r≤a, (2.5.72)
and ∞
n=1
AnJ1(knr) = 0, a < r≤b. (2.5.73) Sneddon and Srivastav22 studied dual Fourier-Bessel series of the form
∞ n=1
k−npAnJν(knρ) =f(ρ), 0< ρ <1, (2.5.74)
and ∞
n=1
AnJν(knρ) =f(ρ), 1< ρ < a. (2.5.75) Applying here the results from their Section 4, we have that
An= 21−pΓ(1−p)kpn b2J22(knb)
a 0
t1−pJ1−p(knt)g(t)dt. (2.5.76)
22 Sneddon, I. N., and R. P. Srivastav, 1966: Dual series relations. I. Dual relations involving Fourier-Bessel series.Proc. R. Soc. Edinburgh, Ser. A,66, 150–160.
0 0.5 1 1.5 2
0 0.5 1 1.5 2 0
0.2 0.4 0.6 0.8 1
r z
u(r,z)/ ω
Figure 2.5.4: The solution u(r, z)/ω to the mixed boundary value problem governed by Equation 2.5.33 through Equation 2.5.36 whena= 1,b= 2 andα=14.
The function g(t) is determined via the Fredholm integral equation of the second kind
g(t) =h(t) + 2
πsin(πp)tp a
0
τ1−pg(τ)L(t, τ)dτ, (2.5.77) where
h(t) = 21+psin(πp) πΓ(1−p) t2p−1
1
0
d dr
rf(r) dr
(t2−r2)p, (2.5.78) and
L(t, τ) = ∞
0
K1(by)
I1(by)I1−p(ty)I1−p(τ y)y dy. (2.5.79) To illustrate our results, we choosef(r) =ωr. Then
h(t) = 21+pωsin(πp)
πΓ(2−p) t. (2.5.80)
Figure 2.5.4 illustrates the case whena= 1, b= 2 and α= 14. Problems
1. Solve Helmholtz’s equation
∂2u
∂r2 +1 r
∂u
∂r +∂2u
∂z2 −
α2+ 1 r2
u= 0, 0≤r <∞, 0< z <∞,
subject to the boundary conditions
rlim→0|u(r, z)|<∞, lim
r→∞u(r, z)→0, 0< z <∞,
zlim→∞u(r, z)→0, 0≤r <∞,
and
u(r,0) =q(r), 0≤r <1, uz(r,0) = 0, 1< r <∞.
Step 1: Show that the solution to the differential equation plus the first three boundary conditions is
u(r, z) = ∞
0
A(k)e−z
√k2+α2J1(kr)dk.
Step 2: Using the last boundary condition, show that we obtain the dual integral equations
∞
0
A(k)J1(kr)dk=q(r), 0≤r <1,
and ∞
0
k2+α2A(k)J1(kr)dk= 0, 1< r <∞. Step 3: If√
k2+α2A(k) =kB(k), then the dual integral equations become ∞
0
√ k
k2+α2B(k)J1(kr)dk=q(r), 0≤r <1,
and ∞
0
kB(k)J1(kr)dk= 0, 1< r <∞.
Step 4: Consider the first integral equation in Step 3. By multiplying both sides of this equation bydr/√
t2−r2, integrating from 0 totand using t
0
J1(kr)
√t2−r2dr=1−cos(kt)
kt ,
show that ∞
0
√ k
k2+α2B(k)
1−cos(kt) kt
dk=
t 0
√q(r)
t2−r2dr, 0≤t <1,
or ∞
0
√ k
k2+α2B(k) sin(kt)dk= d dt
t 0
t q(r)
√t2−r2dr
, 0≤t <1.
Step 5: Consider next the second integral equation in Step 3. By multiplying both sides bydr/√
r2−t2, integrating fromt to∞and using ∞
t
J1(kr)
√r2−t2dr= sin(kt) kt ,
show that ∞
0
B(k) sin(kt)dk= 0, 1≤t <∞.
Step 6: If we introduce
g(t) = ∞
0
B(k) sin(kt)dk,
show that the integral equations in Step 4 and Step 5 become
g(t)− ∞
0
1− k
√k2+α2
B(k) sin(kt)dk= d dt
t 0
t q(r)
√t2−r2dr
, for 0≤t <1, and
g(t) = 0, 1< t <∞. Step 7: Because
B(k) = 2 π
1
0
g(t) sin(kt)dt, show that the functiong(t) is governed by
g(t)−2 π
1
0
g(τ) ∞
0
1− k
√k2+α2
sin(kt) sin(kτ)dk
dτ
= d dt
t 0
t q(r)
√t2−r2 dr
for 0≤t <1. Onceg(t) is computed numerically, thenB(k) andA(k) follow.
Finally, the values ofA(k) are used in the integral solution given in Step 1.
0 0.5 1 1.5
2 0
0.2 0.4 0.6 0.8 1 0
0.2 0.4 0.6 0.8 1
r z
u(r,z)
Problem 1
Step 8: Taking the limit asα→0, show that you recover the solution given by Equation 2.5.11 and Equation 2.5.12 witha= 1 andq(r) =r. In the figure labeled Problem 1, we illustrate the solution whenα= 1 andq(r) =r.