Dual Fourier sine series arise in the same manner as dual Fourier cosine series in mixed boundary value problems in rectangular domains. However, for some reason, they do not appear as often and the following example is the only one that we present.
Consider Laplace’s equation on a semi-infinite strip
∂2u
∂x2 +∂2u
∂y2 = 0, 0< x < L, 0< y <∞, (3.2.1)
subject to the boundary conditions
u(0, y) =u(L, y) = 0, 0< y <∞, (3.2.2)
ylim→∞u(x, y)→0, 0< x < L, (3.2.3)
and
u(x,0) =U0(x), 0< x < ,
uy(x,0) = 0, < x < L, (3.2.4) whereL > andU0(x) is a known odd function.
The solution to Equation 3.2.1 and the boundary conditions Equation 3.2.2 and Equation 3.2.3 is
u(x, y) = ∞ n=1
Ane−λnysin(λnx), (3.2.5) whereλn =nπ/L. Substituting Equation 3.2.5 into Equation 3.2.4 yields the dual series of
∞ n=1
Ansin(λnx) =U0(x), 0< x < , (3.2.6)
and ∞
n=1
λnAnsin(λnx) = 0, < x < L. (3.2.7) To solve these dual equations, let us follow Williams14and introducep(x) such that
∞ n=1
λnAnsin(λnx) =
p(x), 0< x < ,
0, < x < L. (3.2.8) Therefore, the Fourier coefficient is given by
λnAn= 2 L
0
p(x) sin(λnx)dx, (3.2.9) or
nπAn =
−
p(x) sin(λnx)dx (3.2.10) if we assume thatp(x) is an odd function. Upon substituting Equation 3.2.10 into Equation 3.2.6,
1 π
∞ n=1
1 nsin
nπx L
−
p(t) sin nπt
L
dt
=U0(x), |x|< . (3.2.11)
14 Williams, W. E., 1964: The solution of dual series and dual integral equations. Proc.
Glasgow Math. Assoc.,6, 123–129.
Let us assume that we can Chebyshev expressp(t) by the expansion p(t) = 1
√2−t2 ∞ m=0
B2m+1T2m+1(t/), (3.2.12) where Tn(ã) is the nth Chebyshev polynomial. Substituting Equation 3.2.12 into Equation 3.2.11 and noting15 that
t
−t
eipτTn(τ /t) dτ
√t2−τ2 =inπJn(pt), (3.2.13) then
∞ n=1
1 n
∞
m=0
(−1)mB2m+1J2m+1
nπ L
sin
nπx
L =U0(x). (3.2.14) Because the Chebyshev polynomial expansion for sin(ax) is
sin(ax) = 2 ∞ k=0
(−1)kJ2k+1(a)T2k+1(x), a >0, (3.2.15) and reexpressingU0(x) as
U0(x) = ∞ k=0
b2k+1T2k+1(x/), (3.2.16) we obtain from Equation 3.2.14 the following set of simultaneous equations
∞ m=0
(−1)m+kB2m+1C2k+1,2m+1=b2k+1, k= 0,1,2, . . . , (3.2.17)
where
C2k+1,2m+1= 2 ∞ n=1
1 nJ2m+1
nπ L
J2k+1
nπ L
. (3.2.18)
Finally, we note that Equation 3.2.10 and Equation 3.2.12 yield nAn=
∞ m=0
(−1)mB2m+1J2m+1
nπ L
. (3.2.19)
15 Gradshteyn, I. S., and I. M. Ryzhik, 1965: Table of Integrals, Series, and Products.
Academic Press, Formula 7.355.1 and 7.355.2.
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
x/L y/L
u(x,y)
Figure 3.2.1: The solutionu(x, y) to the mixed boundary value problem given by Equation 3.2.1 through Equation 3.2.4 whenU0(x) =x/andL/= 3.
For a given U0(x), we compute b2n+1. Equation 3.2.17 then yields B2m+1 while Equation 3.2.19 yieldsAn. Finally,u(x, y) follows from Equation 3.2.5.
Figure 3.2.1 illustrates the solution whenU0(x) =x/andL/= 3.
An alternative method of solving the dual equations Equation 3.2.6 and Equation 3.2.7 involves introducingBn =λnAn,Cn = (−1)n−1Bn,ξ=πx/L, η=π−ξ, andγ=π−π/L. Then, these dual equations become
∞ n=1
nCnsin(nη) =πU0(π−η)/π, 0< η < γ, (3.2.20)
and ∞
n=1
Cnsin(nη) = 0, γ < η < π. (3.2.21) Why did we derive Equation 3.2.20 and Equation 3.2.21? Consider the dual series:
∞ n=1
nansin(nx) =f(x), 0≤x < c, ∞
n=1
ansin(nx) = 0, c < x≤π.
(3.2.22)
To findan, we introduce the Fourier sine series:
y(x) = ∞ n=1
ansin(nx). (3.2.23)
Clearly, to satisfy Equation 3.2.22, y(x) = 0 if c < x ≤π. On the interval [0, c), we assume that
y(x) = sin x
2 c
x
h(t)
cos(x)−cos(t)dt, 0≤x < c, (3.2.24) whereh(x) is unknown. Why have we chosen such an unusual definition for y(x)?
Recall that we seek ananthat satisfies Equation 3.2.22. From the theory of Fourier series, we know that
an= 2 π
c 0
y(x) sin(nx)dx (3.2.25)
= 1 π
c 0
h(t) t
0
cos n−12
x
−cos n+12
x cos(x)−cos(t) dx
dt. (3.2.26)
We now simplify Equation 3.2.26 by applying Mehler integral, Equation 1.3.4:
Pn[cos(x)] =
√2 π
x 0
cos n+12
ξ
cos(ξ)−cos(x)dξ, (3.2.27) and find that
an = 1
√2 c
0
h(t){Pn−1[cos(t)]−Pn[cos(t)]} dt. (3.2.28) Turning now to the first equation in Equation 3.2.22, we eliminate n inside of the summation by integrating both sides:
∞ n=1
an[1−cos(nx)] = x
0
f(ξ)dξ, (3.2.29)
or c
0
h(t)
√1 2
∞ n=1
[1−cos(nx)]{Pn−1[cos(t)]−Pn[cos(t)]}
dt= x
0
f(ξ)dξ.
(3.2.30) Using the results from Example 1.3.1, we can simplify Equation 3.2.30 to
sin x
2 c
0
h(t) H(x−t)
cos(t)−cos(x)dt= x
0
f(ξ)dξ, (3.2.31)
or x
0
h(t)
cos(t)−cos(x)dt= csc x
2 x
0
f(ξ)dξ. (3.2.32)
Finally, using the results given in Equation 1.2.9 and Equation 1.2.10, the solution to the integral equation Equation 3.2.32 is
h(t) = 2 π
d dt
t 0
cos(x/2) cos(x)−cos(t)
x 0
f(ξ)dξ
dx
, (3.2.33) or
h(t) = 2 πcot
t 2
t 0
sin(x/2)f(x)
cos(x)−cos(t)dx. (3.2.34) Consider now the set of dual equation similar to Equation 3.2.22 is
∞ n=1
nansin(nx) = 0, 0≤x < c, ∞
n=1
ansin(nx) =f(x), c < x≤π.
(3.2.35)
To findan, we begin by introducing a functiong(x), defined by g(x) =
∞ n=1
nansin(nx), c < x≤π. (3.2.36)
Next, we assume thatg(x) can be represented by g(x) =−d
dx
sin x
2 x
c
h(t)
cos(t)−cos(x)dt
(3.2.37)
over the range c < x ≤ π, where h(t) is unknown. From the properties of Fourier sine series,
nan= 2 π
π c
g(x) sin(nx)dx (3.2.38)
=−2 π
π c
d dx
sin
x 2
x
c
h(t)
cos(t)−cos(x)dt
sin(nx)dx (3.2.39)
= 2n π
π c
sin x
2 x
c
h(t)
cos(t)−cos(x)dt
cos(nx)dx (3.2.40) by integration by parts. Interchanging the order of integration and using Equation 1.3.5, we have
an = 1
√2 π
c
h(t){Pn[cos(t)]−Pn−1[cos(t)]} dt. (3.2.41)
Substituting Equation 3.2.41 into Equation 3.2.35, we obtain the integral equation
√1 2
π c
h(t) ∞
n=1
{Pn[cos(t)]−Pn−1[cos(t)]}sin(nx)
dt=f(x). (3.2.42) Using the results from Problem 3 in Section 1.3, Equation 3.2.42 simplifies to
π x
h(t)
cos(x)−cos(t)dt=−csc x
2 f(x), c < x≤π. (3.2.43) From Equation 1.2.11 and Equation 1.2.12, we obtain
h(t) = 2 π
d dt
π t
f(x) cos(x/2) cos(t)−cos(x)dx
. (3.2.44)
Using the results from Equations 3.2.22, 3.2.28, 3.2.34, 3.2.35, 3.2.41, and 3.2.44, the solution to the dual equations
∞ n=1
ncnsin(ny) =g(π−y), 0≤y < γ, ∞
n=1
cnsin(ny) =f(π−y), γ < y≤π,
(3.2.45)
is
cn= 1
√2 π
0
h(t){Pn−1[cos(t)]−Pn[cos(t)]} dt, (3.2.46) where
h(t) = 2 πcot
t 2
t 0
g(π−ξ) sin(ξ/2)
cos(ξ)−cos(t)dξ, 0≤t < γ, (3.2.47)
and
h(t) =−2 π
d dt
π t
f(π−ξ) cos(ξ/2) cos(t)−cos(ξ) dξ
, γ < t≤π. (3.2.48) Therefore, the solution to Equation 3.2.20 and Equation 3.2.21 is
Cn= 1
√2 γ
0
h(t){Pn−1[cos(t)]−Pn[cos(t)]} dt, (3.2.49) where
h(t) = 2 Lcot
t 2
t 0
U0[L(π−ξ)/π] sin(ξ/2)
cos(ξ)−cos(t) dξ, 0≤t < γ. (3.2.50)
Consequently, making the back substitution, the dual series
∞ n=1
bn
n sin(nx) =f(x) 0≤x < c, ∞
n=1
bnsin(nx) =g(x). c < x≤π,
(3.2.51)
has the solution bn= n
√2 π
0
k(t){Pn−1[cos(t)] +Pn[cos(t)]} dt, (3.2.52) where
k(t) = 2 π
d dt
c 0
f(ξ) sin(ξ/2) cos(ξ)−cos(t)dξ
, 0≤t < c, (3.2.53) and
k(t) = 2 πtan
t 2
π
t
g(ξ) cos(ξ/2)
cos(t)−cos(ξ)dξ, c < t≤π. (3.2.54)
Using Equation 3.2.51 through Equation 3.2.54, we finally have that An = 1
√2 π/L
0
k(t){Pn−1[cos(t)] +Pn[cos(t)]} dt, (3.2.55) where
k(t) = 2 π
d dt
t 0
U0(Lξ/π) sin(ξ/2) cos(ξ)−cos(t) dξ
, 0≤t < π/L. (3.2.56)