In this closing section we illustrate a mixed boundary value problem that yields a triple Fourier sine series.
Let us find46 the potential for Laplace’s equation in cylindrical coordi- nates:
∂2u
∂r2 +1 r
∂u
∂r+∂2u
∂z2 = 0, 0≤r < a, 0< z < π, (3.5.1) subject to the boundary conditions
lim
r→0|u(r, z)|<∞, 0< z < π, (3.5.2)
ur(a, z) = 0, 0≤z < α, u(a, z) = 1, α < z < β, ur(a, z) = 0, β < z < π,
(3.5.3) and
u(r,0) =uz(r, π) = 0, 0≤r < a. (3.5.4) Separation of variables yields the potential, namely
u(r, z) = ∞ n=0
Ansin n+12
zI0 n+12
r I1
n+12
a. (3.5.5) Equation 3.5.5 satisfies Equation 3.5.1, Equation 3.5.2, and Equation 3.5.4.
Substituting Equation 3.5.5 into Equation 3.5.3, we obtain the triple Fourier sine series:
∞ n=0
n+12
Ansin n+12
z
= 0, 0< z < α, (3.5.6) ∞
n=0
(1 +Mn)Ansin n+12
z
= 1, α < z < β, (3.5.7)
and ∞
n=0
n+12
Ansin n+12
z
= 0, β < z < π, (3.5.8) where
Mn= I0 n+12
a I1
n+12
a −1. (3.5.9)
46 See Zanadvorov, N. P., V. A. Malinov, and A. V. Charukhchev, 1983: Radial trans- mission distribution in a cylindrical electrooptical shutter with a large aperture. Opt.
Spectrosc. (USSR),54, 212–215.
To solve Equation 3.5.6 through Equation 3.5.8, we first note that ∞
n=0
n+12
Ansin n+12
z
=−d dz
∞
n=0
Ancos n+12
z
. (3.5.10) Following Tranter and Cooke,47we introduce
An= ∞ k=0
Bk
∞
0
J2k+1[xsin(β/2)]J2n+1(x)dx
x. (3.5.11)
The integral in Equation 3.5.11 can be evaluated48in terms of hypergeometric functions. Then,
∞ n=0
Ancos n+12
z
= 1
4√ 2
∞ k=0
Bk
π z
sin(η)
cos(z)−cos(η) (3.5.12)
× ∞
0
J2k+1[xsin(β/2)]J0[xsin(η/2)]dx
dη.
Because49 ∞
0
J2k+1[xsin(β/2)]J0[xsin(η/2)]dx (3.5.13)
=
0, η > β,
csc(β/2)2 1F [k+ 1,−k; 1; sin2(η/2)/sin2(β/2)], η < β, ∞
n=0
Ancos n+12
z
= 0 (3.5.14)
if z > β. Therefore, it follows from Equation 3.5.10 that Equation 3.5.8 is also satisfied. On the other hand, if 0< z < β,
∞ n=0
Ancos n+12
z
= csc(β/2) 4√
2 ∞ k=0
Bk
β z
sin(η)2 1F [k+ 1,−k; 1; sin2(η/2)/sin2(β/2)]
cos(z)−cos(η) dη
(3.5.15)
= 1
4√ 2
∞ k=0
Bk
π y
sin(θ)Pk[cos(θ)]
cos(y)−cos(θ)dθ (3.5.16)
= 1 4
∞ k=0
Bkcos k+12
y
k+12 , 0< y < π, (3.5.17)
47 Tranter, C. J., and J. C. Cooke, 1973: A Fourier-Neumann series and its application to the reduction of triple cosine series.Glasgow Math. J.,14, 198–201.
48 Gradshteyn and Ryzhik, op. cit., Formula 6.574.1
49 Ibid., Formula 6.512.2 withν=n+ 1.
where we substituted sin(θ/2) = sin(η/2)/sin(β/2) and sin(y/2) = sin(z/2) /sin(β/2). We also used Equation 1.3.4 to simplify Equation 3.5.16. Upon substituting Equation 3.5.17 into Equation 3.5.10 and carrying out the differ- entiation, we find that Equation 3.5.6 becomes
∞ k=0
Bksin
(2k+ 1) arcsin
sin(z/2) sin(β/2)
= 0, 0< z < α. (3.5.18) Finally, consider Equation 3.5.7. We can rewrite it
∞ n=0
Ansin n+12
z
= 1−∞
n=0
MnAnsin n+12
z
. (3.5.19)
Substituting Equation 3.5.11, we find that ∞
k=0
Bk
∞
0
J2k+1[xsin(β/2)]
∞
n=0
sin n+12
z
J2n+1(x)
dx
x (3.5.20)
= 1−∞
k=0
Bk
∞
0
J2k+1[xsin(β/2)]
∞
n=0
Mnsin n+12
z
J2n+1(x)
dx x . The summation overn on the left side of Equation 3.5.20 can be replaced50 by sin[xsin(z/2)]/2 so that we now have
∞ k=0
Bk
∞
0
J2k+1[xsin(β/2)] sin[xsin(z/2)]dx
x (3.5.21)
= 2−2 ∞ k=0
Bk
∞
0
J2k+1[xsin(β/2)]
∞
n=0
Mnsin n+12
z
J2n+1(x)
dx x. Evaluating51the integral on the left side of Equation 3.5.21, we finally obtain ∞
k=0
Bk
2k+ 1sin
(2k+ 1) arcsin
sin(z/2) sin(β/2)
(3.5.22)
= 2−2 ∞ k=0
Bk
∞
0
J2k+1[xsin(β/2)]
∞
n=0
Mnsin n+12
z
J2n+1(x)
dx x. In summary, by introducing Equation 3.5.11, we reduced the triple Fourier sine equations, Equation 3.5.6 through Equation 3.5.8, to the dual Fourier sine series
∞ k=0
Bksin k+12
ϕ
= 0, (3.5.23)
50 Ibid., Formula 8.514.6.
51 Ibid., Formula 6.693.1.
and ∞ k=0
Bk k+12 sin
k+12 ϕ
= 4−4 ∞ k=0
BkFk(ϕ), (3.5.24) where
Fk(ϕ) = ∞ n=0
Mnsin{(2n+ 1) arcsin[sin(ϕ/2) sin(β/2)]}
× ∞
0
J2k+1[xsin(β/2)]J2n+1(x)dx
x, (3.5.25) andϕ= 2 arcsin[sin(z/2)/sin(β/2)].
Our final task is to computeBk. To this end, let us introduce Bk =APk[cos(ϕ0)] +
π ϕ0
f(τ) d dτ
$Pk[cos(τ)]%
dτ, (3.5.26)
whereAis a free parameter andϕ0= 2 arcsin[sin(α/2)/sin(β/2)]. Substitut- ing Equation 3.5.26 into Equation 3.5.23, we obtain
∞ k=0
Bksin k+12
ϕ
=A ∞ k=0
sin k+12
ϕ
Pk[cos(ϕ0)] (3.5.27)
+ π
ϕ0
f(τ) d dτ
∞
k=0
sin k+12
ϕ
Pk[cos(τ)]
dτ,
=A H(ϕ−ϕ0)
2 cos(ϕ0)−2 cos(ϕ) (3.5.28)
+ π
ϕ0
f(τ) d dτ
H(ϕ−τ) 2 cos(τ)−2 cos(ϕ)
dτ,
where we used results from Problem 1 at the end of Section 1.3. Because ϕ < ϕ0, both Heaviside functions in Equation 3.5.28 equal zero and our choice forBk satisfies Equation 3.5.23.
Turning to Equation 3.5.24, we take its derivative with respect toϕand obtain
∞ k=0
Bkcos k+12
ϕ
=−4 ∞ k=0
BkFk(ϕ). (3.5.29) Next, we substitute forBk and find that
A ∞ k=0
cos k+12
ϕ
Pk[cos(ϕ0)]
+ π
ϕ0
f(t)d dt
∞
k=0
cos k+12
ϕ
Pk[cos(t)]
dt (3.5.30)
=−4A ∞ k=0
Fk(ϕ)Pk[cos(ϕ0)]−4 π
ϕ0
f(t)d dt
∞
k=0
Fk(ϕ)Pk[cos(t)]
dt.
0 0.2
0.4 0.6
0.8 1
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
z/ π r/a
u(r,z)
Figure 3.5.1: The solution u(r, z) to the mixed boundary value problem governed by Equation 3.5.1 through Equation 3.5.3 whena=π,α=π/3, andβ= 2π/3.
If we now integrate the second term in Equation 3.5.30 by parts and again introduce the results from Problem 1 from Section 1.3, we derive
A H(ϕ0−ϕ)
2 cos(ϕ)−2 cos(ϕ0)−f(ϕ0) ∞ k=0
cos k+12
ϕ
Pk[cos(ϕ0)]
− π
ϕ0
f(t)H(t−ϕ)
2 cos(ϕ)−2 cos(t)dt+ π
ϕ0
f(t)d dt
∞
k=0
4Fk(ϕ)Pk[cos(t)]
dt
=−A ∞ k=0
4Fk(ϕ)Pk[cos(ϕ0)]. (3.5.31)
The first two terms in Equation 3.5.31 vanish while the limits of integra- tion for the integral in the third term run fromϕtoπ. Finally, let us multiply Equation 3.5.31 by sin(ϕ)dϕ/
2 cos(τ)−2 cos(ϕ) and then integrate fromτ toπ. We find then that
− π
τ
sin(ϕ) 2 cos(τ)−2 cos(ϕ)
π ϕ
f(t)
2 cos(ϕ)−2 cos(t)dt
dϕ
+ π
ϕ0
f(t)d dt
π τ
sin(ϕ) 2 cos(τ)−2 cos(ϕ)
∞ k=0
4Fk(ϕ)Pk[cos(t)]dϕ
dt
=−A π
τ
sin(ϕ) 2 cos(τ)−2 cos(ϕ)
∞ k=0
4Fk(ϕ)Pk[cos(ϕ0)]dϕ. (3.5.32)
Using results given by Equation 1.2.11 and Equation 1.2.12, the first term in Equation 3.5.32 equalsf(τ) and Equation 3.5.32 becomes
f(τ) + 2 π
π ϕ0
f(t)dL(τ, t)
dt dt=−2A
π L(τ, ϕ0), ϕ0< τ < π, (3.5.33) where
L(τ, t) = π
τ
sin(ϕ) 2 cos(τ)−2 cos(ϕ)
∞ k=0
4Fk(ϕ)Pk[cos(t)]dϕ. (3.5.34)
It is clear from Equation 3.5.33 thatf(τ) is proportional to A. Conse- quently, bothBk andAn also are proportional to A. Therefore, A must be chosen to thatu(r, z) = 1 forα < z < β. Figure 3.5.1illustrates this solution whena=π,α=π/3 andβ = 2π/3. It is better to use
Bk= [A−f(ϕ0)]Pk[cos(ϕ0)] +f(π)(−1)k− π
ϕ0
f(τ)Pk[cos(τ)]dτ (3.5.35) rather than Equation 3.5.26 so that we avoid large values of the derivative of the Legendre polynomials for largeknear τ=π.
−4 −2 0 2 4 0
1 2
−1
−0.5 0 0.5
x y
u(x,y)
Chapter 4 Transform Methods
In Example 1.1.2 we showed that applying a Fourier cosine transform leads to the dual integral equations:
−2 π
∞
0
kcoth(kh)A(k) cos(kx)dk= 1/h, 0≤x <1, (4.0.1)
and 2
π ∞
0
A(k) cos(kx)dk= 0, 1< x <∞. (4.0.2) The purpose of this chapter is to illustrate how these dual integral equations are solved. In Sections 4.1 and 4.2 we focus on Fourier-type of integrals while Sections 4.3 and 4.4 treat Fourier-Bessel integrals. Finally Section 4.5 deals with situations where we have a mixture of Fourier series and transforms, Fourier and Fourier-Bessel transforms and Fourier series and Laplace trans- forms.
Before we proceed to our study of dual and triple integral equations, let us finish Example 1.1.2. We begin by introducing
u(x,0) = 2 π
∞
0
A(k) cos(kx)dk. (4.0.3)
163
Referring back to Equation 1.1.15, we see thatu(x,0) is the solution to Equa- tion 1.1.11 along the boundaryy= 0. Next, for convenience, let us define
g(x) =−du(x,0)
dx = 2
π ∞
0
kA(k) sin(kx)dk. (4.0.4) From Equation 4.0.2,u(x,0) is nonzero only if 0< x <1. Consequently,g(x) is nonzero only between 0< x <1. Taking the Fourier sine transform ofg(x),
kA(k) = 1
0
g(x) sin(kx)dx. (4.0.5)
If we integrate Equation 4.0.1 with respect tox, we have that
−2 π
∞
0
coth(kh)A(k) sin(kx)dk= x
h, 0≤x <1. (4.0.6) Substituting Equation 4.0.5 into Equation 4.0.6, we have the integral equation
−2 π
1
0
g(ξ) ∞
0
coth(kh) sin(kξ) sin(kx)dk k
dξ= x
h. 0≤x <1.
(4.0.7) The integral within the square brackets in Equation 4.0.7 can be evaluated1 exactly and the integral equation simplifies to
−1 π
1
0 g(ξ) ln
tanh(βx) + tanh(βξ) tanh(βx)−tanh(βξ)
dξ=x
h, 0≤x <1, (4.0.8) whereβ=π/(2h). The results from Example 1.2.3 can be employed to solve Equation 4.0.8 after substitutingx= tanh(βx)/tanh(β). This yields
g(ξ) = 1 h2
d dξ
1
ξ
tanh(βx) cosh2(βx)
tanh2(βx)−tanh2(βξ)
×
x 0
dτ
tanh2(βx)−tanh2(βτ)
dx
(4.0.9)
= tanh(βξ)
h2cosh2(βξ)
tanh2(β)−tanh2(βξ)
× 1
0
tanh2(β)−tanh2(βx)
tanh2(βx)−tanh2(βξ) dx (4.0.10)
=− πtanh(βξ)
2h2βcosh(β)
tanh2(β)−tanh2(βξ)
. (4.0.11)
1 Gradshteyn, I. S., and I. M. Ryzhik, 1965: Table of Integrals, Series, and Products.
Academic Press, Formula 4.116.3.
−2
−1 0
1
2
0 0.5
1
0.25 0.75
−0.8
−0.6
−0.4
−0.2 0 0.2
y x
u(x,y)
Figure 4.0.1: The solution to Equation 1.1.11 subject to the mixed boundary conditions given by Equation 1.1.12, Equation 1.1.13, and Equation 1.1.14 whenh= 1.
Substituting Equation 4.0.11 into Equation 4.0.5,A(k) follows via numerical integration. Finally, we can use this A(k) to find the solution to Equation 1.1.11 subject to the boundary conditions given by Equation 1.1.12, Equa- tion 1.1.13, and Equation 1.1.14 by numerically integrating Equation 1.1.17.
Figure 4.0.1 illustrates this solution.