A problem that is similar to the Reissner-Sagoci problem involves finding the steady-state velocity field within a laminar, infinitely deep fluid that is driven by a slowly rotating disc of radius ain contact with the free surface.
The disc rotates at the angular velocityω. The Navier-Stokes equations for the angular componentu(r, z) of the fluid’s velocity reduce to
∂2u
∂r2 +1 r
∂u
∂r − u r2 +∂2u
∂z2 = 0, 0≤r <∞, 0< z <∞. (2.6.1) At infinity the velocity must tend to zero which yields the boundary conditions
rlim→0|u(r, z)|<∞, lim
r→∞u(r, z)→0, 0< z <∞, (2.6.2) and
zlim→∞u(r, z)→0, 0≤r <∞. (2.6.3) At the interface, the mixed boundary condition is
u(r,0) =ωr, 0≤r < a,
àuz(r,0) +ηuzz(r,0) = 0, a < r <∞. (2.6.4) Goodrich23was the first to attack this problem. Using Hankel transforms, the solution to Equation 2.6.1 through Equation 2.6.3 is
u(r, z) = ∞
0
A(k)J1(kr)e−kzdk. (2.6.5)
23 Taken with permission from Goodrich, F. C., 1969: The theory of absolute surface shear viscosity. I.Proc. Roy. Soc. London, Ser. A,310, 359–372.
Substituting Equation 2.6.5 into Equation 2.6.4, we have that ∞
0
A(k)J1(kr)dk=ωr, 0≤r < a, (2.6.6)
and ∞
0
(ηk2−àk)A(k)J1(kr)dk= 0, a < r <∞. (2.6.7) Before Goodrich tackled the general problem, he considered the following special cases.
à= 0
In this special case, Equation 2.6.6 and Equation 2.6.7 simplify to ∞
0
A(k)J1(kr)dk=ωr, 0≤r < a, (2.6.8)
and ∞
0
k2A(k)J1(kr)dk= 0, a < r <∞. (2.6.9) Now, multiplying Equation 2.6.8 byrand differentiating with respect tor,
∞
0
A(k)d
dr[rJ1(kr)]dk= 2ωr, 0≤r < a. (2.6.10) In the case of Equation 2.6.9, integrating both sides with respect to r, we obtain
∞
0
k2A(k) ∞
r
J1(kξ)dξ
dk= 0, a < r <∞. (2.6.11) From the theory of Bessel functions,24
d
dr[rJ1(kr)] =krJ0(kr), (2.6.12)
and ∞
r
J1(kξ)dξ= J0(kr)
k . (2.6.13)
Substituting Equation 2.6.12 and Equation 2.6.13 into Equation 2.6.10 and Equation 2.6.11, respectively, they become
∞
0
kA(k)J0(kr)dk= 2ω, 0≤r < a, (2.6.14)
24 Gradshteyn and Ryzhik, op. cit., Formulas 8.472.3 and 8.473.4 withz=kr.
0 0.5 1 1.5
2 0
0.5 1 1.5 2 0
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
r/a z/a
u(r,z)/(ω a)
Figure 2.6.1: The solution to Laplace’s equation, Equation 2.6.1, with the boundary conditions given by Equation 2.6.2 through Equation 2.6.4 whenà= 0.
and ∞
0
kA(k)J0(kr)dk= 0, a < r <∞. (2.6.15) Taking the inverse Hankel transform given by Equation 2.6.14 and Equation 2.6.15, we have that
A(k) = 2ω a
0
rJ0(kr)dr= 2ωaJ1(ak)
k . (2.6.16)
Therefore,
u(r, z) = 2ωa ∞
0
J1(ξ)J1(ξr/a)e−ξz/adξ
ξ . (2.6.17)
Figure 2.6.1 illustrates this solution.
η= 0
In this case Equation 2.6.6 and Equation 2.6.7 become ∞
0
A(k)J1(kr)dk=ωr, 0≤r < a, (2.6.18)
and ∞
0
kA(k)J1(kr)dk= 0, a < r <∞. (2.6.19) If we now introduce a functiong(r) such that
∞
0
kA(k)J1(kr)dk=g(r), 0≤r < a, (2.6.20)
then
A(k) = a
0
ξ g(ξ)J1(kξ)dξ. (2.6.21)
Upon substituting Equation 2.6.21 into Equation 2.6.18 and interchang- ing the order of integration,
∞
0
ξ g(ξ) ∞
0
J1(kξ)J1(kr)dk
dξ=ωr. (2.6.22) Because25
∞
0
Jν(kξ)Jν(kx)dk= 2(ξx)−ν π
min(ξ,x) 0
s2ν
(ξ2−s2)(x2−s2)ds, (2.6.23) Equation 2.6.22 can be rewritten
2 π
r 0
g(ξ) ξ
0
s2
(r2−s2)(ξ2−s2)ds
dξ (2.6.24)
+2 π
∞
r
g(ξ) r
0
s2
(r2−s2)(ξ2−s2)ds
dξ=ωr2. Interchanging the order of integration,
2 π
r 0
f(s)
√r2−s2ds=ωr2, 0≤r < a, (2.6.25) where we set
f(s) =s2 a
s
g(ξ)
ξ2−s2dξ. (2.6.26)
From Equation 1.2.13 and Equation 1.2.14, f(s) =ω d
ds s
0
r3
√s2−r2dr
= 2ωs2. (2.6.27) Substituting the results from Equation 2.6.27 into Equation 2.6.26 we find that
g(ξ) =−2 π
d dξ
a ξ
2ωτ
τ2−ξ2dτ
= 4ωξ
π
a2−ξ2, 0≤ξ < a, (2.6.28)
25 Cooke, J. C., 1963: Triple integral equations. Quart. J. Mech. Appl. Math., 16, 193–203. See Appendix 1.
0 0.5 1 1.5 2 0
0.5 1
1.5 2 0
0.2 0.4 0.6 0.8 1
z/a r/a
u(r,z)/(ω a)
Figure 2.6.2: This is similar toFigure 2.6.1except thatà= 0 andη= 0.
after using Equation 1.2.15 and Equation 1.2.16. Therefore,26 A(k) = 4ω
π a
0
ξ2
a2−ξ2J1(kξ)dξ= 2ωa2
# 2 πkaJ3
2(ka). (2.6.29) The solutionu(r, z) follows from Equation 2.6.5. Figure 2.6.2 illustrates this solution.
à= 0, η = 0
In this general case Equation 2.6.6 and Equation 2.6.7 become ∞
0
A(k)J1(kr)dk=ωr, 0≤r < a, (2.6.30)
and ∞
0
k(1 +aλ0k)A(k)J1(kr)dk= 0, a < r <∞, (2.6.31) whereλ0=−η/(aà). Let us turn to Equation 2.6.30 first. Multiplying Equa- tion 2.6.30 by (2/π)1/2r2/√
x2−r2and integrating the resulting equation over rfrom 0 tox≤a, we find that
∞
0
A(k)
#2 π
x 0
r2J1(kr)
√x2−r2 dr
dk=ω
#2 π
x 0
r3
√x2−r2dr, (2.6.32)
26 Gradshteyn and Ryzhik, op. cit., Formula 6.567.1 withν= 1 andà=−12.
or
x3/2 ∞
0
A(k)J3 2(kx)
√k dk= 2√ 2ωx3 3√
π (2.6.33)
after using integral tables. Differentiating both sides of Equation 2.6.33 with respect tox,
∞
0
√k A(k)J1
2(kx)dk= 2ω√
√ 2x
π , 0≤x < a. (2.6.34) We now turn to Equation 2.6.31. Multiplying this equation by (2x/π)1/2 /√
r2−x2 and integrating the resulting equation overrfromx > ato∞, we find that
∞
0
k(1 +aλ0k)A(k)
#2 π
∞
x
√x J1(kr)
√r2−x2 dr
dk= 0; (2.6.35)
or ∞
0
√k(1 +aλ0k)A(k)J1
2(kx)dk= 0, a < x <∞. (2.6.36) Let us now replacexwithrin Equation 2.6.34 and Equation 2.6.36 and expressJ1
2(z) =
2/(πz) sin(z). This yields ∞
0
A(k) sin(kr)dk= 2ωr, 0≤r < a, (2.6.37)
and ∞
0
(1 +aλ0k)A(k) sin(kr)dk= 0, a < r <∞. (2.6.38) We can rewrite Equation 2.6.38 as
∞
0
A(k) sin(kr)dk=−λ0 ∞
0
akA(k) sin(kr)dk, a < r <∞. (2.6.39) From the theory of Fourier integrals,
A(k) = 4ω π
a 0
rsin(kr)dr
−2λ0 π
∞
a
∞
0
akA(k) sin(kr)dk
sin(kr)dr (2.6.40)
= 2ωa2
# 2 πakJ3
2(ak)
−2λ0 π
∞
0
∞
0
aξA(ξ) sin(ξr)dξ
sin(kr)dr +2λ0
π a
0
∞
0
aξA(ξ) sin(ξr)dξ
sin(kr)dr. (2.6.41)
The second term in Equation 2.6.41 equalsaλ0kA(k). Therefore, (1 +aλ0k)A(k) = 2ωa2
# 2 πakJ3
2(ak) +2λ0
π a
0
∞
0
aξA(ξ) sin(ξr)dξ
sin(kr)dr. (2.6.42) Interchanging the order of integration, we finally have that
(1 +aλ0k)A(k) = 2ωa2
# 2 πakJ3
2(ak) +λ0 a
0
aξA(ξ)K(k, ξ)dξ, (2.6.43) where
K(y, ξ) = 2 π
a 0
sin(ry) sin(rξ)dr = 1 π
sin[a(y−ξ)]
y−ξ −sin[a(y+ξ)]
y+ξ
. (2.6.44) One of the intriguing aspects of Equation 2.6.43 is that the unknown is the Hankel transformA(k). In general, the integral equations that we will see will involve an unknown which is related to A(k) via an integral definition.
See Equation 2.5.61, Equation 2.5.62 and Equation 2.5.65. In Goodrich’s paper he solved the integral equation as a variational problem and found an approximate solution by the optimization of suitable solutions. However, we shall shortly outline an alternative method for any value ofλ0.
In 1978 Shail27 reexamined Goodrich’s paper for two reasons. First, the solution for theà= 0 case, Equation 2.6.16, creates a divergent integral when it is substituted back into dual equations, Equation 2.6.8 and Equation 2.6.9.
Second, the governing Fredholm integral equation is over an infinite range and appears to be unsuitable for asymptotic solution asλ0→0 orλ0→ ∞.
Shail’s analysis for theà= 0 case begins by noting that
∂2u(r,0)
∂r2 +1 r
∂u(r,0)
∂r −u(r,0)
r2 = 0, a≤r <∞ (2.6.45) from Equation 2.6.1 and Equation 2.6.4. The general solution to Equation 2.6.45 is
u(r,0) =Cr+D/r, a < r <∞. (2.6.46) The values ofCandD follow from the limits thatu(r,0)→0 asr→ ∞and continuity atu(r,0) atr=a. This yields
u(r,0) =
ωr, 0≤r < a,
ωa2/r, a < r <∞. (2.6.47)
27 Taken from Shail, R., 1978: The torque on a rotating disk in the surface of a liquid with an adsorbed film. J. Engng. Math.,12, 59–76 with kind permission from Springer Science and Business Media.
If we use Equation 2.6.47 in place of Equation 2.6.4, then the solution Equa- tion 2.6.17 follows directly.
Shail also examined the general case and solved it using the method of complementary representations for generalized axially symmetric poten- tial functions. This method is very complicated and we introduce a greatly simplified version of an analysis first done by Chakrabarti.28
We begin by introducing the functiong(x) such that (1 +aλ0k)A(k) = 2
π a
0
g(x) sin(kx)dx. (2.6.48) Turning to Equation 2.6.31 first, direct substitution yields
∞
0
k(1 +aλ0k)A(k)J1(kr)dk
= a
0
g(x) ∞
0
ksin(kx)J1(kr)dk
dx (2.6.49)
=− a
0
g(x)d dr
∞
0
sin(kx)J0(kr)dk
dx (2.6.50)
= 0, (2.6.51)
because 0 ≤ x ≤ a < r < ∞. Thus, our choice of A(k) satisfies Equation 2.6.31 identically.
Turning to Equation 2.6.30 next, direct substitution gives 2
π a
0
g(x) ∞
0
sin(kx)J1(kr) 1 +aλ0k dk
dx=ωr, 0≤r < a; (2.6.52) or
a 0
g(x) ∞
0
sin(kx)J1(kr)dk
dx (2.6.53)
+ a
0
g(x) ∞
0
1−aλ0k
1 +aλ0ksin(kx)J1(kr)dk
dx=ωπr, 0≤r < a.
From tables,29 the integral within the square brackets of the first integral in Equation 2.6.53 can be evaluated and Equation 2.6.53 simplifies to
r 0
x g(x)
√r2−x2dx+ a
0
g(τ) ∞
0
1−aλ0k
1 +aλ0ksin(kτ)r J1(kr)dk
dτ =ωπr2 (2.6.54)
28 Chakrabarti, A., 1989: On some dual integral equations involving Bessel functions of order one.Indian J. Pure Appl. Math.,20, 483–492.
29 Gradshteyn and Ryzhik, op. cit., Formula 6.671.1.
for 0≤r < a. Using Equation 1.2.13 and Equation 1.2.14, we can solve for x g(x) and find that
x g(x) + 2 π
a 0
g(τ) ∞
0
1−aλ0k
1 +aλ0ksin(kτ) d dx
x 0
ξ2J1(kξ) x2−ξ2dξ
dk
dτ
= 2ω d dx
x 0
ξ3 x2−ξ2dξ
, 0≤x < a. (2.6.55)
From integral tables,30 d ds
s 0
ξ2J1(kξ) s2−ξ2dξ
=ssin(ks), (2.6.56)
and Equation 2.6.55 becomes g(x) + 2
π a
0
g(τ) ∞
0
1−aλ0k
1 +aλ0ksin(kτ) sin(kx)dk
dτ = 4ωx, 0≤x < a.
(2.6.57) Equation 2.6.57 is identical to Chakrabarti’s equations (50) and (52). Once we solve Equation 2.6.57 numerically, its values of g(x) can be substituted into Equation 2.6.48. Finally the solutionu(r, z) follows from Equation 2.6.5.
The numerical solution of Equation 2.6.57 is nontrivial due to the nature of the integration overk. To solve it, we use a spectral method. If we take g(τ) to be an odd function over (−a, a), we have that
g(τ) = ∞ n=1
Ansin nπτ
a , (2.6.58)
sin(kτ) = 2 ∞ n=1
(−1)nnπ
k2a2−n2π2sin(ka) sin nπτ
a , (2.6.59)
and
x=−2a ∞ n=1
(−1)n nπ sin
nπx
a . (2.6.60)
Substitution of Equation 2.6.58 through Equation 2.6.60 into Equation 2.6.57 gives the infinite set of equations
Am+ ∞ n=1
HmnAn=Cm, m= 1,2,3, . . . , (2.6.61)
30 Ibid., Formula 6.567.1 withν= 1 andà=−12.
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/a z/a
u(r,z)/(aω)
Figure 2.6.3: This is similar toFigure 2.6.1except thatà, η= 0 andλ0= 5.
where
Hmn= 4a(−1)n+mnmπ ∞
0
1−aλ0k 1 +aλ0k
sin2(ka)
(k2a2−n2π2)(k2a2−m2π2)dk, (2.6.62) and
Cm= 8ωa
mπ(−1)m+1. (2.6.63)
The system of equations is then truncated to, say, N spectral components and the system is inverted to yield Am form= 1,2, . . . , N. Next, A(k) can be found via
(1 +aλ0k)A(ak)
a = 2
N n=1
n(−1)n
k2a2−n2π2sin(ka)An. (2.6.64) The larger the value ofN, the greater the accuracy. Finally,
u(r, z) = ∞
0
A(ξ)
a e−ξz/aJ1(ξr/a)dξ. (2.6.65) Figure 2.6.3 illustrates this solution whenλ0= 5.
0 0.5 1 1.5 2
−2
−1 0
1 2
3 4
−1
−0.5 0
r z
u(r,z)
Chapter 3 Separation of Variables
Separation of variables is the most commonly used technique for solving boundary value problems. In the case of mixed boundary value problems they lead to dual or higher-numbered Fourier series which yield the Fourier coefficients. In this chapter we examine dual Fourier series in Section 3.1 and Section 3.2, while dual Fourier-Bessel series are treated in Section 3.3 and dual Fourier-Legendre series in Section 3.4. Finally Section 3.5 treats triple Fourier series.
In Example 1.1.1 we showed that the method of separation of variables led to the dual cosine series:
∞ n=1
an n−12 cos
n−12 x
= 1, 0≤x < c, (3.0.1)
and ∞
n=1
ancos n−12
x
= 0, c < x≤π. (3.0.2) Equations 3.0.1 and 3.0.2 are examples of a larger class of dual trigonometric
83
equations. The general form of these dual series can be written
∞ n=1
npansin(nx) =f(x), 0≤x < c, ∞
n=1
ansin(nx) =g(x), c < x≤π,
(3.0.3)
∞ n=1
n−12p
ancos n−12
x
=f(x), 0≤x < c, ∞
n=1
ancos n−12
x
=g(x), c < x≤π,
(3.0.4)
∞ n=1
n−12p
ansin n−12
x
=f(x), 0≤x < c, ∞
n=1
ansin n−12
x
=g(x), c≤x≤π,
(3.0.5)
and
12αa0+ ∞ n=1
npancos(nx) =f(x), 0≤x < c,
12a0+ ∞ n=1
ancos(nx) =g(x), c < x≤π,
(3.0.6)
where −1 ≤ p ≤ 1. Comparing Equation 3.0.1 and Equation 3.0.2 with Equation 3.0.4, they are identical if we set p = −1. The purpose of this chapter is to focus on those mixed value problems that lead to these dual equations and solve them.