Tranter1 examined dual trigonometric series of the form
∞ n=1
n−12p
ancos n−12
x
=F(x), 0≤x < c, ∞
n=1
ancos n−12
x
=G(x), c < x≤π,
(3.1.1)
where p = ±1. The most commonly encountered case is when G(x) = 0.
Whenp= 1, he showed that an = 2 π
c 0
h(x) cos n−12
x
dx, (3.1.2)
1 Tranter, C. J., 1960: Dual trigonometrical series. Proc. Glasgow Math. Assoc.,4, 49–57.
where
h(x) = 1
ξ
χ(η)
η2−ξ2dη, (3.1.3)
χ(η) = 4η
π sin(c/2) η
0
F{2 arcsin[xsin(c/2)]}
(η2−x2)[1−x2sin2(c/2)]
dx, (3.1.4)
andξ= sin(x/2) csc(c/2).
Whenp=−1,
an= 2χ(1) sin(c/2)Pn−1[cos(c)]
−2 sin(c/2) 1
0
χ(η)Pn−1
1−2η2sin2(c/2)
dη, (3.1.5) where
χ(η) = 2 π
η 0
x F{2 arcsin[x sin(c/2)]}
η2−x2 dx+C. (3.1.6) Here, C is a constant whose value is determined by substituting Equation 3.1.6 into Equation 3.1.5 and then choosingC so that
∞ n=1
n−12−1
an =F(0). (3.1.7)
•Example 3.1.1
To illustrate Tranter’s solution, let us assume thatF(x) = 1 if 0≤x < c.
From Equation 3.1.6, we have that χ(η) = C; from Equation 3.1.5, an = 2Csin(c/2)Pn−1[cos(c)]. To evaluateC, we substitutean into Equation 3.1.7 and find that
2Csin(c/2) ∞ n=1
n−12−1
Pn−1[cos(c)] = 1. (3.1.8) From the generation formula for Legendre polynomials,
∞ n=1
x2n−2Pn−1[cos(c)] = [1−2x2cos(c) +x4]−1/2. (3.1.9) Integrating Equation 3.1.9 from 0 to 1, we have
∞ n=1
n−12
Pn−1[cos(c)] = 2 1
0
dx
1−2x2cos(c) +x4 (3.1.10)
= π/2
0
dθ
1−cos2(c/2) sin2(θ)
(3.1.11)
=K[cos2(c/2)], (3.1.12)
0.2 0 0.6 0.4
0.8 1 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
x/ π y
u(x,y)
Figure 3.1.1: The solutionu(x, y) to the mixed boundary value problem posed in Example 1.1.1 whenc=π/2.
whereK(ã) denotes the complete elliptic integral2 and sin(θ) = 2x/(1 +x2).
Therefore, 2Csin(c/2)K[cos2(c/2)] = 1, and an =Pn−1[cos(c)]/K[cos2(c/2)]
is the solution to the dual Fourier cosine series Equation 3.0.1 and Equation 3.0.2.
Recall that Equation 3.0.1 and Equation 3.0.2 arose from the separation of variables solution of Equation 1.1.1 through Equation 1.1.4. Therefore, the solution to this particular mixed boundary value problem is
u(x, y) = ∞ n=1
Pn−1[cos(c)]
K[cos2(c/2)]
exp
− n−12
y n−12 cos
n−12 x
. (3.1.13) This solution is illustrated in Fig. 3.1.1 whenc=π/2.
•Example 3.1.2 Let us solve3
∂2u
∂x2 +∂2u
∂y2 = 0, 0< x < π, 0< y < π/2, (3.1.14)
2 See Milne-Thomson, L. M., 1965: Elliptic integrals. Handbook of Mathematical Func- tions, M. Abromowitz and I. A. Stegun, Eds., Dover, 587–626. See Section 17.3.
3 Taken from Whiteman, J. R., 1968: Treatment of singularities in a harmonic mixed boundary value problem by dual series methods. Quart. J. Mech. Appl. Math.,21, 41–50 with permission of Oxford University Press.
subject to the boundary conditions
ux(0, y) = 0, u(π, y) = 1, 0< y < π/2, (3.1.15) u(x,0) = 12, 0≤x < π/2,
uy(x,0) = 0, π/2< x≤π, (3.1.16) and
uy(x, π/2) = 0, 0< x < π. (3.1.17) Using separation of variables, a solution to Equation 3.1.14 which also satisfies Equation 3.1.15 and Equation 3.1.17 is
u(x, y) = 1−∞
n=0
Bn
n+12 cosh
n+12 y−π2 sinh
n+12
π/2 cos n+12
x
. (3.1.18) If we then substitute Equation 3.1.18 into the mixed boundary condition Equation 3.1.17, we obtain the dual series
∞ n=0
Bn
2n+ 1coth n+12
π/2 cos
n+12 x
= 14, 0≤x < π/2, (3.1.19)
and ∞
n=0
Bncos n+12
x
= 0, π/2< x≤π. (3.1.20) The remaining challenge is to solve this dual series.
Fortunately, in the 1960s Tranter4 showed that the dual trigonometrical series
∞ n=0
An 2n+ 1cos
n+12 x
=f(x), 0< x < c, (3.1.21)
and ∞
n=0
Ancos n+12
x
= 0, c < x < π, (3.1.22) has the solution
An=A0Pn[cos(c)]− c
0
F(θ)Pn[cos(θ)] sin(θ)dθ, n= 1,2,3, . . . , (3.1.23) where
F(θ) = 2√ 2 π
θ 0
f(x) sin(x)
cos(x)−cos(θ)dx, (3.1.24)
4 Tranter, C. J., 1964: An improved method for dual trigonometrical series. Proc.
Glasgow Math. Assoc.,6, 136–140.
andA0 is found by substituting Equation 3.1.23 into Equation 3.1.21.
Can we apply Tranter’s results, Equation 3.1.21 through Equation 3.1.24, to solve Equation 3.1.19 and Equation 3.1.20? We begin by rewriting these equations as follows:
∞ n=0
Bn
2n+ 1cos n+12
x
= 1 4+
∞ m=0
Bm
2m+ 1
$1−coth m+12
π/2%
×cos m+12
x
, 0≤x < π/2, (3.1.25)
and ∞
n=0
Bncos n+12
x
= 0, π/2< x≤π. (3.1.26) By inspection, we set
f(x) =1 4 +
∞ m=0
Bm 2m+ 1
$1−coth m+12
π/2% cos
m+12 x
. (3.1.27) Substituting Equation 3.1.27 into Equation 3.1.24,
F(θ) =−
√2 π
θ 0
∞ m=0
Bm
$1−coth m+12
π/2%
×sin(x) sin m+12
x
cos(x)−cos(θ) dx (3.1.28)
=−
√2 π
∞ m=0
Bm$
1−coth m+12
π/2%
× θ
0
sin(x) sin m+12
x
cos(x)−cos(θ) dx (3.1.29)
= 1 π
∞ m=0
Bm
$1−coth m+12
π/2%
× θ
0
cos m+32
x
−cos m−12
x
2[cos(x)−cos(θ)] dx (3.1.30)
= 1 2
∞ m=0
Bm
$1−coth m+12
π/2%
{Pm+1[cos(θ)]−Pm−1[cos(θ)]}, (3.1.31) where we used Mehler formula, Equation 1.3.4. Substituting the results from Equation 3.1.31 into Equation 3.1.23, we have that
Bn=B0Pn(0) +1 2
∞ m=0
Bm
$1−coth m+12
π/2%
×
[Pm+1(0)−Pm−1(0)]Pn(0) + (2m+ 1) 1
0
Pn(t)Pm(t)dt
, (3.1.32)
because π/2
0 {Pm+1[cos(θ)]−Pm−1[cos(θ)]}Pn[cos(θ)] sin(θ)dθ
= 1
0
[Pm+1(t)−Pm−1(t)]Pn(t)dt (3.1.33)
=Pn(t)Pm+1(t)1
0−Pn(t)Pm−1(t)1
0− 1
0
Pm+1(t)−Pm −1(t)
Pn(t)dt (3.1.34)
=Pn(0)Pm−1(0)−Pn(0)Pm+1(0)−(2m+ 1) 1
0
Pn(t)Pm(t)dt, (3.1.35) andPm(1) = 1. Equation 3.1.32 can be expressed in the succinct form of
Bn=B0Pn(0) +C(n,& 0)B0+ ∞ m=1
&
C(n, m)Bm, (3.1.36) where
&
C(n,0) = [coth(π/4)−1]Pn(0) + 1
0
Pn(t)dt, (3.1.37) and
&
C(n, m) =
1−coth m+12
π/2
[Pm+1(0)−Pm−1(0)]Pn(0) + (2m+ 1)
1
0
Pn(t)Pm(t)dt
. (3.1.38) The coefficientsBm/B0 form≥1 are found by solving the linear equa- tions
∞ m=1
C(n, m)Bm
B0 =D(n), (3.1.39)
where
C(n, m) =
C(n, m),˜ n=m,
C(n, m)˜ −1, n=m, (3.1.40)
and
D(n) =−Pn(0)−C(n,˜ 0). (3.1.41) Having found these Bm’s, we use Equation 3.1.19 with x=π/4 to compute B0 via
B0= 1 4
' ∞ n=0
Bn/B0 2n+ 1 cosh
n+12 π/2
cos n+12
x
. (3.1.42)
0 0.2
0.4 0.6
0.8 1
0 0.1 0.2 0.3 0.4 0.5 0.4 0.5 0.6 0.7 0.8 0.9 1
x/ π y/ π
u(x,y)
Figure 3.1.2: The solutionu(x, y) to the mixed boundary value problem given by Equation 3.1.14 through Equation 3.1.17.
Finally, u(x, y) follows from Equation 3.1.18. We illustrate this solution in Figure 3.1.2.
•Example 3.1.3 Let us solve
∂2u
∂x2 +∂2u
∂y2 = 0, −b < x < b, 0< y < h, (3.1.43) subject to the boundary conditions
u(x,0) =u(x, h) = 0, u(x, c−) =u(x, c+), −b < x < b, (3.1.44)
u(x, c) = 1, |x|< w,
1uy(x, c−) =2uy(x, c+), w <|x|< b, (3.1.45) and
u(−b, y) =u(b, y) = 0, 0< y < h. (3.1.46) Figure 3.1.3illustrates the geometry for this problem.
If we use separation of variables, the solution to Equation 3.1.43 is u(x, y) =
∞ n=1
An
ekn(y−c)−e−kn(y+c)
cos(knx), 0< y < c, (3.1.47)
y
h ε ε
w b x
−b −w
2
1
c
Figure 3.1.3: Schematic of the geometry of Example 3.1.3.
and
u(x, y) = ∞ n=1
Bn
ekn(y+c−2h)−ekn(c−y)
, c < y < h, (3.1.48)
wherekn= n−12
π/b. Equation 3.1.47 and Equation 3.1.48 satisfy not only Equation 3.1.43, but also Equation 3.1.46 and u(x,0) =u(x, h) = 0. Using the fact thatu(x, c−) =u(x, c+) for|x|< band Equation 3.1.47 and Equation 3.1.48, we find that
Bn =−An
1−e−2knc
1−e−2kn(h−c). (3.1.49) Finally, we substitute Equation 3.1.47 and Equation 3.1.48 into Equation 3.1.45. The results can be written
∞ n=1
1 +Hn
n−12 Cncos n−12
θ
=f(θ), 0< θ < d, (3.1.50)
and ∞
n=1
Cncos n−12
θ
= 0, d < θ < π, (3.1.51) where
Cn = n−12
An
1−e−2kn(h−c)−2knc+κ
e−2knh−e−2kn(h−c)
1−e−2kn(h−c) , (3.1.52) Hn=−e−2knc+e−2kn(h−c)−2e−2knh+κ
e−2knc−e−2kn(h−c) 1−e−2knh+κ
e−2knc−e−2kn(h−c) , (3.1.53) d=πw/b,θ=πx/b, andκ= (1−2)/(1+2). Heref(θ) = 1.
Kiyono and Shimasaki5 developed a method for computing Cn. They showed that
Cn= 2 π
d 0
g(ϕ) cos n−12
ϕ
dϕ, (3.1.54)
where
g(ϕ) = 1 π
d dϕ
ϕ 0
h(η)
cos(η)−cos(ϕ)dη
. (3.1.55)
The functionh(η) is given by the integral equation
h(η) + d
0
K(η, ξ)h(ξ)dξ=− d dη
d η
sin(θ)f(θ) cos(η)−cos(θ)dθ
(3.1.56)
for 0< η < d, where
K(η, ξ) =K0(η, ξ) + sin(η) ∞ n=1
HnI(η, n−1)I(ξ, n−1), (3.1.57)
K0(η, ξ) = 2 π2
sin(η) cos(η)−cos(ξ)
(cos(η)−cos(d) cos(ξ)−cos(d)ln
(
1 + cos(η) cos(η)−cos(d)
− (
cos(ξ)−cos(d) cos(η)−cos(d)ln
(
1 + cos(ξ) cos(ξ)−cos(d)
,(3.1.58)
I(ξ, n) =
√2 cos n+12
d
π
n+12
[cos(ξ)−cos(d)]
+
n+12Rn(ξ), (3.1.59)
R0(ξ) = 1− 2 πarcsin
cos(d/2) cos(ξ/2)
, (3.1.60)
R1(ξ) = cos(ξ)R0(ξ) +2√ 2
π cos(d/2)
cos(ξ)−cos(d), (3.1.61) and
(n+ 1)Rn+1(ξ)−(2n+ 1) cos(ξ)Rn(ξ) +nRn−1(ξ)
= 2√ 2 π cos
n+12
d cos(ξ)−cos(d), n≥1.(3.1.62)
5 Kiyono, T., and M. Shimasaki, 1971: On the solution of Laplace’s equation by certain dual series equations. SIAM J. Appl. Math.,21, 245–257.
−1
−0.5 0 0.5 1
0 0.2 0.4 0.6 0.8 1 0
0.2 0.4 0.6 0.8 1
y/b x/b
u(x,y)
Figure 3.1.4: The solutionu(x, y) to the mixed boundary value problem given by Equation 3.1.43 through Equation 3.1.46.
Figure 3.1.4 illustrates the solutionu(x, y) whenb =h= 3, c=w= 1, 1= 1 and2= 2.
•Example 3.1.4 Let us solve6
∂2u
∂x2 +∂2u
∂y2 = 0, 0≤x≤π, −∞< y <∞, (3.1.63) subject to the boundary conditions
ux(0, y) =ux(π, y) = 0, −∞< y <∞, (3.1.64)
|ylim|→∞uy(x, y)→1, 0≤x≤π, (3.1.65) uy(x,0−) =uy(x,0+), 0≤x≤c, (3.1.66)
and
u(x,0−) =u(x,0+) = 0, 0≤x < c,
uy(x,0−) =uy(x,0+) = 0, c < x≤π. (3.1.67)
6 Reprinted fromInt. J. Heat Mass Transfer,19, J. Dundurs and C. Panek, Heat con- duction between bodies with wavy surfaces, 731–736, c1976, with permission of Elsevier.
If we use separation of variables, the solution to Equation 3.1.63 is u(x, y) =y+A0+
∞ n=1
Ane−nycos(nx), 0< y <∞, (3.1.68)
and
u(x, y) =y−A0− ∞ n=1
Anenycos(nx), −∞< y <0. (3.1.69) Equation 3.1.68 and Equation 3.1.69 satisfy not only Equation 3.1.63, but also Equation 3.1.64 through Equation 3.1.66. Finally, we substitute Equation 3.1.68 and Equation 3.1.69 into Equation 3.1.67. This results in
A0+ ∞ n=1
Ancos(nx) = 0, 0≤x < c, (3.1.70)
and ∞
n=1
nAncos(nx) = 1, c < x≤π. (3.1.71) To solve Equation 3.1.70 and Equation 3.1.71, we first substitute x = π−ξ, c=π−γ,A0=a0/2 andAn= (−1)nan. We then obtain
∞ n=1
nancos(nξ) = 1, 0≤ξ < γ, (3.1.72) and
12a0+ ∞ n=1
ancos(nξ) = 0, γ < ξ≤π. (3.1.73) Recently Sbragaglia and Prosperetti7also solved this dual series when it arose during their study of the effects of surface deformation on a type of superhy- drophobic surface.
Equation 3.1.72 and Equation 3.1.73 are an example of Equation 3.0.6.
For the special casep = 1, Sneddon8 showed that the solution to Equation 3.0.6 is
a0= 2 π
π
√2 c
0
h(t)dt+ π
c
g(t)dt
, (3.1.74)
7 Sbragaglia, M., and A. Prosperetti, 2007: A note on the effective slip properties for microchannel flows with ultrahydrophobic surfaces.Phys. Fluids,19, Art. No. 043603.
8 See Section 5.4.3 in Sneddon, I. N., 1966:Mixed Boundary Value Problems in Poten- tial Theory.North Holland, 283 pp.
and an= 2
π π
2√ 2
c 0
h(t){Pn[cos(t)] +Pn−1[cos(t)]} dt+ π
c
g(t) cos(nt)dt
, (3.1.75) wheren= 1,2,3, . . .,
h(t) = 2 π
d dt
t 0
sin(x/2) cos(x)−cos(t)dx
× x
0
f(ξ)dξ−12αa0x+ ∞ n=1
bnsin(nx)
, (3.1.76) and
bn = 2 π
π c
g(ξ) cos(nξ)dξ. (3.1.77)
Applying these results to Equation 3.1.72 and Equation 3.1.73, we find that h(t) = 2
π d dt
t 0
ξsin(ξ/2) cos(ξ)−cos(t)dξ
=
√2 sin(t) 1 + cos(t) =√
2 tan(t/2).
(3.1.78) Therefore,
a0= 2
√2 γ
0
h(t)dt=−4 ln[cos(γ/2)], (3.1.79) and
an = 1
√2 γ
0
h(t){Pn[cos(θ)] +Pn−1[cos(t)]}dt (3.1.80)
= 1
n{Pn−1[cos(γ)]−Pn[cos(γ)]}. (3.1.81) Returning to the original Fourier coefficient,
A0=−2 ln[sin(c/2)], (3.1.82) and
An=−1
n{Pn[cos(c)] +Pn−1[cos(c)]}. (3.1.83) Figure 3.1.5illustrates the solution whenc=12.
•Example 3.1.5 Let us solve9
∂2u
∂x2 +∂2u
∂y2 = 0, 0≤x≤L, 0< y < h, (3.1.84)
9 Adapted from Westmann, R. A., and W. H. Yang, 1967: Stress analysis of cracked rectangular beams.J. Appl. Mech.,34, 693–701.
0 0.2
0.4 0.6
0.8 1
−1
−0.5 0 0.5 1
−3
−2
−1 0 1 2 3
x/ π y
u(x,y)
Figure 3.1.5: The solutionu(x, y) to the mixed boundary value problem given by Equation 3.1.63 through Equation 3.1.67.
subject to the boundary conditions
ux(0, y) =ux(L, y) = 0, 0< y < h, (3.1.85) uy(x, h) =S, 0< x < L, (3.1.86)
and
uy(x,0) = 0, 0< x < a,
u(x,0) = 0, a < x < L, (3.1.87) whereL > a.
If we use separation of variables, the solution to Equation 3.1.84 is u(x, y) =Sy+
∞ n=0
Ancosh[nπ(h−y)/L]
cosh(nπh/L) cos nπx
L . (3.1.88)
Equation 3.1.88 satisfies not only Equation 3.1.84, but also Equation 3.1.85 and Equation 3.1.86. Upon substituting Equation 3.1.88 into Equation 3.1.87, we find the dual series
∞ n=0
nπ
L Antanh nπh
L
cos nπx
L =S, 0< x < a, (3.1.89)
and ∞
n=0
Ancos nπx
L = 0, a < x < L. (3.1.90)
To solve the dual equations, Equation 3.1.89 and Equation 3.1.90, let us assume thatu(x,0) is known for 0< x < L. Then
u(x,0) = ∞ n=0
Ancos nπx
L . (3.1.91)
From the theory of Fourier series we have that A0= 1
L L
0
u(x,0)dx, (3.1.92)
and
An= 2 L
L 0
u(x,0) cos nπx
L dx. (3.1.93)
Let us assume that we can expressu(x,0) as u(x,0) =
a x
√h(t)
t2−x2dt, 0< x < a. (3.1.94) Upon substituting Equation 3.1.94 into Equation 3.1.92 and Equation 3.1.93,
A0= 1 L
a 0
a x
√h(t) t2−x2dt
dx= π 2L
a 0
h(t)dt, (3.1.95) and
An= 2 L
a 0
a x
√h(t)
t2−x2dt
cos nπx
L dx (3.1.96)
= π L
a 0
h(t)J0 nπt
L
dt (3.1.97)
forn= 1,2,3, . . .. Next, we substitute Equation 3.1.95 and Equation 3.1.97 into Equation 3.1.89,
∞ n=1
nπ L
a 0
h(t)J0 nπt
L
dt
tanh nπh
L
cos nπx
L = LS
π . (3.1.98) Integrating both sides of Equation 3.1.98 from 0 toxand interchanging the order of integration and summation,
a 0
h(t) ∞
n=1
tanh nπh
L
J0 nπt
L
sin nπx
L dt= LSx
π . (3.1.99) If we integrate the function csc(πz)eiπzJ0(πtz/L) sin(πxz/L) around a contour which consists of 1) the positive real axis, 2) the positive imaginary
axis, and 3) the arc in the first quadrant of the circle|z|=N+12, Sneddon and Srivastav10showed that
∞ n=1
J0 nπt
L
sin nπx
L =L H(x−t) π√
x2−t2 −1 π
∞
0
e−ξsinh(ξx/L)I0(ξt/L)
sinh(ξ) dξ.
(3.1.100) Therefore, Equation 3.1.99 can be rewritten and made nondimensional with respect toaso that it becomes
x 0
√h(t)
x2−t2dt=Sax− 1
0
K(x, η)h(η)dη, 0< x <1, (3.1.101) where
K(x, η) =aπ L
∞ n=1
tanh
nπh L
−1
J0 nπaη
L sin
nπax L
− a L
∞
0
e−ξsinh(ξax/L)I0(ξat/L)
sinh(ξ) dξ. (3.1.102)
Equation 3.1.101 is an integral equation of the Abel type. Applying Equation 1.2.14 and Equation 1.2.15, we find that
h(t) =Sat+ 1
0
L(t, η)h(η)dη, 0≤t≤1, (3.1.103) where
L(t, η) =−2 π
d dt
t 0
x K(x, η)
√t2−x2 dx
(3.1.104)
= aπ
L
2∞ n=1
nt
1−tanh nπh
L
J0 nπaη
L J0 nπat
L
+ a2 L2
∞
0
ξte−ξI0(ξat/L)I0(ξaη/L)
sinh(ξ) dξ. (3.1.105)
Figure 3.1.6illustrates the solution when a/L= 0.5 andh/L= 1. Keer and Sve11applied this technique to the biharmonic equation. Later, Sezgin12used this method to solve a coupled set of Laplace-like equations.
10 Sneddon, I. N., and R. P. Srivastav, 1964: Dual series relationships.Proc. Roy. Soc.
Edinburgh, Ser. A,66, 150–191.
11 Keer, L. M., and C. Sve, 1970: On the bending of cracked plates. Int. J. Solids Struct.,6, 1545–1559.
12 Sezgin, M., 1987: Magnetohydrodynamic flow in a rectangular duct. Int. J. Numer.
Meth. Fluids,7, 697–718.
0 0.5
1 1.5
2
0 0.5 1 1.5 2
−0.5 0 0.5 1 1.5 2 2.5 3
y/a x/a
u(x,y)/(aS)
Figure 3.1.6: The solutionu(x, y) to the mixed boundary value problem given by Equation 3.1.84 through Equation 3.1.87.
•Example 3.1.6
So far we have only encountered Fourier cosine series in a rectilinear problem. Consider now the following problem in spherical coordinates:13
1 r2
∂
∂r
r2∂u
∂r
+ ∂2u
∂ϕ2 = 0, a < r <∞, 0≤ϕ≤π, (3.1.106) subject to the boundary conditions
ϕlim→0|u(r, ϕ)|<∞, lim
ϕ→π|u(r, ϕ)|<∞, a < r <∞, (3.1.107)
rlim→∞u(r, ϕ) =u0, 0≤ϕ≤π, (3.1.108)
and
u(a, ϕ) = 0, 0< ϕ < ϕ0,
ur(a, ϕ) = 0, ϕ0< ϕ < π. (3.1.109) If we use separation of variables, the solution to Equation 3.1.106 is
u(r, ϕ) =u0+ ∞ n=0
An
en(a−r)
r cos(nϕ). (3.1.110)
13 See Baldo, M., A. Grassi, and A. Raudino, 1989: Modeling the mechanisms of enzyme reactivity by the rototranslational diffusion equation. Phys. Review, Ser. A,39, 3700–
3702.
Equation 3.1.110 satisfies not only Equation 3.1.106, but also Equation 3.1.107 and Equation 3.1.108. Upon substituting Equation 3.1.110 into Equation 3.1.109, we find the dual series
∞ n=0
Ancos(nϕ) =−u0a, 0< ϕ < ϕ0, (3.1.111)
and ∞
n=0
An(1 +na) cos(nϕ) = 0, ϕ0< ϕ < π. (3.1.112) Let us rewrite Equation 3.1.112 as
∞ n=0
Ancos(nϕ) =− ∞ m=0
Ammacos(mϕ), ϕ0< ϕ < π. (3.1.113) Because the left side of both Equation 3.1.111 and Equation 3.1.113 are the same, the Fourier cosine series expresses the function
f(ϕ) =
−u0a, 0< ϕ < ϕ0,
−)∞
m=0Ammacos(mϕ), ϕ0< ϕ < π, (3.1.114) which is given by the right side of Equation 3.1.111 and Equation 3.1.113.
From the theory of Fourier series, A0=−au0ϕ0
π + 1
π ∞ m=1
Amsin(mϕ0), (3.1.115) and
An =−2au0sin(nϕ0)
nπ −n
πAn
π−ϕ0−sin(2nϕ0) 2n
+1 π
∞
m=1 m=n
mAm
sin[(m−n)ϕ0]
m−n +sin[(m+n)ϕ0] m+n
. (3.1.116)
Equation 3.1.115 and Equation 3.1.116 yield an infinite set of equations. If we only retain the first N terms, we can invert these equations and find approximate values for the An’s. The potential then follows from Equation 3.1.110. Figure 3.1.7illustrates this solution whenϕ=π/3 andN= 100.
Problems 1. Solve Laplace’s equation
1 r
∂
∂r
r∂u
∂r
+ 1 r2
∂2u
∂θ2 = 0, 0≤r <∞, 0< θ <2π,
1 1.5
2 2.5
3
0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1
φ / π r u(r, φ )/u 0
Figure 3.1.7: The solutionu(r, ϕ) to the mixed boundary value problem given by Equation 3.1.106 through Equation 3.1.109 whena= 1 andϕ0=π/3.
subject to the boundary conditions
rlim→0|u(r, θ)|<∞, lim
r→∞u(r, θ)→V rcos(θ), 0< θ <2π, u(1−, θ) =u(1+, θ), 0< θ <2π,
and
κur(1−, θ) =ur(1+, θ), −π/2< θ < π/2, u(1, θ) = 0, π/2< θ <3π/2,
where 1− and 1+ denote points slightly inside and outside of the circler= 1, respectively.
Step 1: Use separation of variables and show that the general solution to the problem is
u(r, θ) =V rcos(θ) + ∞ n=0
Dnrncos(nθ), 0< r <1, and
u(r, θ) =V rcos(θ) + ∞ n=0
Dnr−ncos(nθ), 1< r <∞.
Note that this solution satisfies not only the differential equation, but also the first three boundary conditions.
Problem 1
Step 2: Using the final boundary condition and the results given by Equation 2.3.10 through Equation 2.3.16, show that
D0=− V
κ+ 1, D1= 1−2κ
2(κ+ 1)V, D2=− V 2(κ+ 1), and
D2n−1=−D2n= (−1)n+11ã3ã ã ã ã(2n−3)V 2ã4ã ã ã2n ,
where n = 2,3,4, . . .. The figure entitled Problem 1 illustrates the solution u(r, θ)/V whenκ= 6.