Here we turn to problems in spherical coordinates which lead to dual Fourier-Legendre series. We now present several examples of their solution.
•Example 3.4.1
Let us solve34the axisymmetric Laplace equation within the unit sphere:
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, 0≤r <1, 0< θ < π, (3.4.1) subject to the boundary conditions
θlim→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, 0≤r≤1, (3.4.2)
rlim→0|u(r, θ)|<∞, 0≤θ≤π, (3.4.3)
and
u(1, θ) = 1, 0≤θ < α,
ur(1, θ) =−bu(1, θ), α < θ≤π. (3.4.4) Separation of variables yields the solution
u(r, θ) = ∞ n=0
AnrnPn[cos(θ)]. (3.4.5) Equation 3.4.5 satisfies not only Equation 3.4.1, but also Equation 3.4.2 and Equation 3.4.3. Substituting Equation 3.4.5 into Equation 3.4.4 yields the dual Fourier-Legendre series
∞ n=0
AnPn[cos(θ)] = 1, 0≤θ < α, (3.4.6)
and ∞
n=0
(b+n)AnPn[cos(θ)] = 0, α≤θ < π. (3.4.7) Consider Equation 3.4.6. Using Equation 1.3.4 to eliminate Pn[cos(θ)]
and then interchanging the order of integration and summation, we have θ
0
1
cos(ϕ)−cos(θ) ∞
n=0
Ancos n+12
ϕ
dϕ= π
√2, 0≤ϕ < α.
(3.4.8) Applying the results from Equation 1.2.9 and Equation 1.2.10,
∞ n=0
Ancos n+12
ϕ
= 1
√2 d dϕ
ϕ 0
sin(τ)
cos(τ)−cos(ϕ)dτ
(3.4.9)
=√ 2 d
dϕ
1−cos(ϕ)
(3.4.10)
= cos(ϕ/2). (3.4.11)
34 See Ramachandran, M. P., 1993: A note on the integral equation method to a diffusion- reaction problem. Appl. Math. Lett.,6, 27–30.
In a similar manner, Equation 3.4.7 can be rewritten as π
θ
1
cos(θ)−cos(ϕ) ∞
n=0
(b+n)Ansin n+12
ϕ
dϕ= 0 (3.4.12) forα < θ≤π. Therefore,
∞ n=0
(b+n)Ansin n+12
ϕ
= ∞ n=0
n+12 −γ
Ansin n+12
ϕ
= 0 (3.4.13) withα < ϕ≤πand γ= 12−b. Integrating Equation 3.4.13 with respect to ϕ, we have
∞ n=0
1− γ
n+12
Ancos n+12
ϕ
= 0. (3.4.14)
The constant of integration equals zero because the left side of Equation 3.4.14 vanishes whenϕ=π.
At this point, let us supplement Equation 3.4.11 with ψ(ϕ) =
∞ n=0
Ancos n+12
ϕ
, α < ϕ≤π. (3.4.15)
Therefore, Equation 3.4.14 can be rewritten ψ(ϕ)−γ
∞ n=0
An
cos n+12
ϕ
n+12 = 0. (3.4.16)
From the definition of half-range Fourier series, An= 2
π α
0
cos( ˜ϕ/2) cos n+12
˜ ϕ
dϕ˜+ 2 π
π α
ψ( ˜ϕ) cos n+12
˜ ϕ
dϕ˜ (3.4.17)
=sin(nα)
nπ +sin[(n+ 1)α]
(n+ 1)π +2 π
π α
ψ( ˜ϕ) cos n+12
˜ ϕ
dϕ.˜ (3.4.18) Substituting Equation 3.4.18 into Equation 3.4.16 and interchanging the order of integration and summation,
ψ(ϕ) = 2γ π
α 0
cos( ˜ϕ/2) ∞
n=0
cos n+12
ϕ cos
n+12
˜ ϕ n+12
dϕ˜
+ π
α
ψ( ˜ϕ) ∞
n=0
cos n+12
ϕ cos
n+12
˜ ϕ n+12
dϕ˜
. (3.4.19)
Because35 ∞ n=0
cos n+12
ϕ cos
n+12
˜ ϕ
n+12 =12ln
cos( ˜ϕ/2) + cos(ϕ/2) cos( ˜ϕ/2)−cos(ϕ/2)
, (3.4.20)
Equation 3.4.19 becomes ψ(ϕ) =γ
π α
0
cos( ˜ϕ/2)L( ˜ϕ, ϕ)dϕ˜+γ π
π α
ψ( ˜ϕ)L( ˜ϕ, ϕ)dϕ,˜ (3.4.21) where
L( ˜ϕ, ϕ) = ln
cos( ˜ϕ/2) + cos(ϕ/2) cos( ˜ϕ/2)−cos(ϕ/2)
. (3.4.22)
To simplify Equation 3.4.21, we setψ(ϕ) = cos(ϕ/2) +r(ϕ) and it becomes r(ϕ)−γ
π π
α
r( ˜ϕ)L( ˜ϕ, ϕ)dϕ˜= (2γ−1) cos(ϕ/2), α < ϕ≤π. (3.4.23) To numerically solve Equation 3.4.23, we introduce n nodal points at ϕi = α+ih, i = 0,1, . . . , n−1, where h = (π−α)/n. We do not have to compute r(π) because it equals zero since ψ(π) = 0 from Equation 3.4.15.
Then, Equation 3.4.23 becomes r(ϕi)−γ
π π
α
r(t)L(t, ϕi)dt= (2γ−1) cos(ϕi/2), (3.4.24) where
L(t, ϕ) = ln2 cos[(t+ϕ)/4]+ lncos(t−ϕ)/4 (3.4.25)
−ln2 sin[(t+ϕ)/4]−ln
sin[(t−ϕ)/4]
t−ϕ
−lnt−ϕ.
Why have we written L(t, ϕ) in the form given in Equation 3.4.25? It clearly shows that the integral equation, Equation 3.4.23, contains a weakly singular kernel. Consequently, the finite difference representation of the inte- gral in Equation 3.4.24 consists of two parts. A simple trapezoidal rule is used for the first four terms given in Equation 3.4.25. For the fifth term, we em- ploy a numerical method devised by Atkinson36for kernels with singularities.
Therefore, for a particular alpha and b, we compute dt = (pi-alpha)/N,
35 Parihar, K. S., 1971: Some triple trigonometrical series equations and their applica- tion. Proc. R. Soc. Edinburgh, Ser. A,69, 255–265.
36 Atkinson, K. E., 1967: The numerical solution of Fredholm integral equations of the second kind.SIAM J. Numer. Anal.,4, 337–348. SeeSection 5in particular.
where N is the number of nodal points. The MATLAB code for computing ψ(ϕ) is
for j = 0:N
tt(j+1) = alpha + j*dt;
pphi(j+1) = alpha + j*dt;
end
% Solve the integral equation Equation 3.4.24 to find r(ϕi).
% Note that r(π) = 0.
for n = 0:N-1 % rows loop (top to bottom in the matrix) phi = pphi(n+1); bb(n+1) = (2*gamma-1)*cos(phi/2);
for m = 0:N-1 % columns loop (left to right in the matrix) t = tt(m+1);
% Introduce the first terms from Equation 3.4.24.
if (n==m) AA(n+1,m+1) = 1;
else AA(n+1,m+1) = 0; end
% Compute integral in Equation 3.4.24. Add in the contribution
% from the first four terms of Equation 3.4.25. Use the
% trapezoidal rule. Recall that r(π) = 0. if (m < N)
arg 1 = (t+phi)/4; arg 2 = (t-phi)/4;
AA(n+1,m+1) = AA(n+1,m+1) ...
- 0.5*dt*gamma*log(abs(2*cos(arg 1)))/pi ...
- 0.5*dt*gamma*log(abs(cos(arg 2)))/pi...
+ 0.5*dt*gamma*log(abs(2*sin(arg 1)))/pi;
if (arg 2 == 0)
AA(n+1,m+1) = AA(n+1,m+1) + 0.5*dt*gamma*log(0.25)/pi;
else
AA(n+1,m+1) = AA(n+1,m+1) ...
+ 0.5*dt*gamma*log(abs(sin(arg 2)/(t-phi)))/pi;
end end
if (m > 0)
arg 1 = (t+phi)/4; arg 2 = (t-phi)/4;
AA(n+1,m+1) = AA(n+1,m+1) ...
- 0.5*dt*gamma*log(abs(2*cos(arg 1)))/pi ...
- 0.5*dt*gamma*log(abs(cos(arg 2)))/pi...
+ 0.5*dt*gamma*log(abs(2*sin(arg 1)))/pi;
if (arg 2 == 0)
AA(n+1,m+1) = AA(n+1,m+1) + 0.5*dt*gamma*log(0.25)/pi;
else
AA(n+1,m+1) = AA(n+1,m+1) ...
+ 0.5*dt*gamma*log(abs(sin(arg 2)/(t-phi)))/pi;
end end
% Use Atkinson’s technique to treat the fifth term in
% Equation 3.4.25. See Section 5 of his paper.
if (m > 0)
kk = n+1-m; Psi 0 = -1;
Psi 1 = 0.25*(kk*kk-(kk-1)*(kk-1));
if ( kk ∼= 0 )
Psi 0 = Psi 0 + kk*log(abs(kk));
Psi 1 = Psi 1 - 0.50*kk*kk*log(abs(kk));
end
if ( kk ∼= 1 )
Psi 0 = Psi 0 - (kk-1)*log(abs(kk-1));
Psi 1 = Psi 1 + 0.50*(kk-1)*(kk-1)*log(abs(kk-1));
end
Psi 1 = Psi 1 + kk*Psi 0;
W 0 = Psi 0 - Psi 1; W 1 = Psi 1;
aalpha = 0.5*dt*log(dt) + dt*W 0;
bbeta = 0.5*dt*log(dt) + dt*W 1;
AA(n+1,m ) = AA(n+1,m ) + gamma*aalpha/pi;
AA(n+1,m+1) = AA(n+1,m+1) + gamma*bbeta/pi;
end
end % end of column loop end % end of rows loop
% compute r(ϕi) r = AA\bb
% compute ψ(ϕ) for n = 0:N-1
theta = tt(n+1); psi(n+1) = cos(theta/2) + r(n+1);
end
Onceψ(ϕi) is found withψ(π) = 0, we computeAn via Equation 3.4.18. The MATLAB code to computeAn is
for m = 0:M if ( m == 0 )
A(m+1) = alpha/pi + sin(alpha)/pi;
else
A(m+1) = sin(m*alpha)/(m*pi) + sin((m+1)*alpha)/((m+1)*pi);
end
% This is the n= 0 term for Simpson’s rule.
A(m+1) = A(m+1) + 2*psi( 1 )*cos((m+0.5)*tt( 1 ))*dt/(3*pi);
% Recall that ψ(π) = 0. Therefore, we do not need the n = N
% term in the numerical integration.
for n = 1:N-1
if ( mod(n+1,2) == 0 )
A(m+1) = A(m+1) + 8*psi(n+1)*cos((m+0.5)*tt(n+1))*dt/(3*pi);
else
A(m+1) = A(m+1) + 4*psi(n+1)*cos((m+0.5)*tt(n+1))*dt/(3*pi);
end; end; end
The final solution follows from Equation 3.4.5. The MATLABcode to realize this solution is
for j = 1:41 y = 0.05*(j-21);
for i = 1:41 x = 0.05*(i-21);
u(i,j) = NaN; r = sqrt(x*x + y*y); theta = abs(atan2(y,x));
if (r <= 1) mu = cos(theta);
% Compute the Legendre polynomials for a given theta
% via the recurrence formula Legendre(1) = 1; Legendre(2) = mu;
for m = 2:M
Legendre(m+1) = (2-1/m)*mu*Legendre(m) - (1-1/m)*Legendre(m-1);
end
% For a point within the sphere, find the solution.
u(i,j) = 0; power = 1;
for m = 0:M
u(i,j) = u(i,j) + A(m+1)*Legendre(m+1)*power;
−1
−0.5 0
0.5 1
−1
−0.5 0 0.5 1
−0.2 0 0.2 0.4 0.6 0.8 1
y x
u(r, θ )
Figure 3.4.1: The solutionu(r, θ) to the mixed boundary value problem posed in Example 3.4.1 withα=π/4 andb= 0.25.
power = power*r;
end; end; end; end
Figure 3.4.1 illustrates the solution whenα=π/4 andb= 0.25.
An alternative method for solving Equation 3.4.6 and Equation 3.4.7 is to introduce the following integral definition forAn:
An= 2n+ 1 n+b
α 0
h(t) cos
n+1 2
t
dt. (3.4.26)
Integration by parts gives (n+b)An= 2h(α) sin
n+1
2
α
−2 α
0
h(t) sin
n+1 2
t
dt. (3.4.27) Then,
∞ n=0
(n+b)AnPn[cos(θ)] = 2h(α) ∞ n=0
sin
n+1 2
α
Pn[cos(θ)] (3.4.28)
−2 α
0
h(t) ∞
n=0
sin
n+1 2
t
Pn[cos(θ)]
dt
= 0. (3.4.29)
Therefore, our choice for theAn satisfies Equation 3.4.7 identically. Here we used results from Problem 1 at the end of Section 1.3.
Turning to Equation 3.4.6, we have forAn that ∞
n=0
n+12
n+bPn[cos(θ)]
α 0
h(t) cos
n+1 2
t
dt=1
2. (3.4.30) Breaking the summation on the left side of Equation 3.4.30 into two parts, we can rewrite it as
∞ n=0
Pn[cos(θ)]
α 0
h(t) cos
n+1 2
t
dt (3.4.31)
= 1 2 −
∞ n=0
γ
n+bPn[cos(θ)]
α 0
h(τ) cos
n+1 2
τ
dτ.
On the left side of Equation 3.4.31, we interchange the order of integration and summation and employ again the results from Problem 1. Equation 3.4.31 simplifies to
θ 0
h(t)
2[cos(t)−cos(θ)]dt (3.4.32)
= 1 2−∞
n=0
γ
n+bPn[cos(θ)]
α 0
h(τ) cos
n+1 2
τ
dτ.
Upon employing Equation 1.2.9 and Equation 1.2.10, we can solve for h(t) and find that
h(t) = 1 π
d dt
t 0
sin(θ)
2[cos(θ)−cos(t)]dθ
−2 π
α 0
h(τ) ∞
n=0
γ n+bcos
n+1
2
τ
(3.4.33)
× d dt
t 0
sin(θ)Pn[cos(θ)]
2[cos(θ)−cos(t)]dθ
dτ
= 1 π
d dt
*2[1−cos(t)]
+
(3.4.34)
−2 π
α 0
h(τ) ∞
n=0
γ n+bcos
n+1
2
τ
cos
n+1 2
t
dt.
We used results from Problem 2 at the end of Section 1.3 to evaluate the integral within the wavy brackets in Equation 3.4.33. Therefore, we now have the Fredholm integral equation
h(t) + α
0
[K(t−τ) +K(t+τ)]h(τ)dτ = sin(t) π
2[1−cos(t)], (3.4.35)
where
K(z) = 1 π
∞ n=0
γ n+bcos
n+1
2
z
. (3.4.36)
Solving Equation 3.4.35 is straightforward. Having determinedh(t), An fol- lows from Equation 3.4.26. Finally Equation 3.4.5 yieldsu(r, z).
•Example 3.4.2
Let us solve the axisymmetric Laplace equation37
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, 0≤r <∞, 0< θ < π, (3.4.37) subject to the boundary conditions
θlim→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, 0≤r <∞, (3.4.38) lim
r→0|u(r, θ)|<∞, lim
r→∞u(r, θ)→0, 0≤θ≤π, (3.4.39)
and
ur(a−, θ) =ur(a+, θ) =−Ucos(θ), 0≤θ < α,
ur(a−, θ) =ur(a+, θ), α < θ≤π, (3.4.40) wherea− anda+ denote points located just inside and outside ofr=a.
Separation of variables yields the solution u(r, θ) =U a
∞ n=0
An r
a
n
Pn[cos(θ)], 0≤r < a, (3.4.41) and
u(r, θ) =−U a ∞ n=0
n n+ 1An
a r
n+1
Pn[cos(θ)], a < r <∞. (3.4.42) Equation 3.4.41 and Equation 3.4.42 satisfy not only Equation 3.4.37, but also Equation 3.4.38 and Equation 3.4.39. They also yield a continuous value ofur(a, θ) for 0≤θ≤π.
Substituting Equation 3.4.41 and Equation 3.4.42 into Equation 3.4.40 gives the dual Fourier-Legendre series
∞ n=0
nAnPn[cos(θ)] =−cos(θ), 0≤θ < α, (3.4.43)
37 Reprinted fromInt. J. Solids Struct., 38, P. A. Martin, The spherical-cap crack revisited, 4759–4776, c2001, with permission of Elsevier.
and ∞ n=0
(2n+ 1) An
n+ 1Pn[cos(θ)] = 0, α≤θ < π. (3.4.44) Turning to Equation 3.4.44 first, let us introduce the functionh(t) such that
An
n+ 1 = 1 2n+ 1
α 0
h(t) sin n+12
t
dt, h(0) = 0. (3.4.45) Therefore,
∞ n=0
(2n+ 1) An
n+ 1Pn[cos(θ)] = α
0
h(t) ∞
n=0
sin n+12
t
Pn[cos(θ)]
dt= 0.
(3.4.46) This follows from Problem 1 in Section 1.3 since 0≤t≤α < θ < π. Conse- quently, our choice forAn automatically satisfies Equation 3.4.44.
How do we findh(t)? We begin by integrating Equation 3.4.45 by parts.
This yields 2
n+122 An
n+ 1 = α
0
h(t) cos n+12
t
dt−h(α) cos n+12
α . (3.4.47) Next, we rewrite Equation 3.4.43 as
∞ n=0
2
n+122 An
n+ 1Pn[cos(θ)]−1 2
∞ n=0
An
n+ 1Pn[cos(θ)] =−2 cos(θ).
(3.4.48) Then using Equation 3.4.45, Equation 3.4.47, the results from Problem 1 in Section 1.3 and the fact that
∞ n=0
Pn[cos(θ)]sin n+12 n+12 =
t 0
H(θ−τ)
2 cos(τ)−2 cos(θ)dτ, (3.4.49)
we find that θ
0
h(t)
2 cos(t)−2 cos(θ)dt−1 4
θ 0
1
2 cos(t)−cos(θ) α
t
h(τ)dτ
dt
=−2 cos(θ). (3.4.50)
Finally, we employ Equation 1.2.11 and Equation 1.2.12 and find the inhomo- geneous Fredholm equation of the second kind:
h(t)−1 4
α t
h(τ)dτ =−4
πcos(3t/2), h(0) = 0. (3.4.51)
−2
−1 0 1 2
−2
−1 0
1 2
−0.8
−0.7
−0.6
−0.5
−0.4
−0.3
−0.2
−0.1 0
y/a x/a
u(x,y)/(Ua)
Figure 3.4.2: The solutionu(r, θ) to the mixed boundary value problem posed by Equation 3.4.37 through Equation 3.4.40 whenα=π/4.
To solve Equation 3.4.51, we rewrite it as
h(t) +1 4
t 0
h(τ)dτ =M1 4 −4
πcos(3t/2), (3.4.52) whereM1="α
0 h(t)dt, an unknown constant. Taking the Laplace transform of Equation 3.4.52,
s2+14
H(s) =M1 4 −4
π s2
s2+94, (3.4.53) or
h(t) = M1
2 +1 π
sin(t/2)− 3
πsin(3t/2). (3.4.54) From the definition ofM1, M1=−(4/π) sin(α) sin(α/2). Therefore,
h(t) = sec(α/2) cos(3α/2) sin(t/2)/π−3 sin(3t/2)/π. (3.4.55) Figure 3.4.2 illustrates the solution whenα=π/4.
•Example 3.4.3 Let us solve38
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, 0≤r <∞, 0< θ < π, (3.4.56) subject to the boundary conditions
rlim→0|u(r, θ)|<∞, lim
r→∞u(r, θ)→0, 0< θ < π, (3.4.57) and
u(a−, θ) =u(a+, θ) = 1, 0≤θ < α, u(a−, θ) =u(a+, θ), ur(a−, θ) =ur(a+, θ), α < θ < π−α,
u(a−, θ) =u(a+, θ) = (−1)m, π−α < θ < π,
(3.4.58)
where m = 0 or 1 and 0 < α < π/2. The solution must be finite at the θ= 0, π.
Separation of variables yields the solution u(r, θ) =
∞ n=0
A2n+m
r a
2n+m
P2n+m[cos(θ)], 0≤r < a, (3.4.59)
and u(r, θ) =
∞ n=0
A2n+m
a r
2n+m+1
P2n+m[cos(θ)], a < r <∞. (3.4.60) We have written the solution in this form so that we can take advantage of symmetry and limitθ between 0 and π/2 rather than 0< θ < π. Equation 3.4.59 and Equation 3.4.60 satisfy not only Equation 3.4.56, but also Equation 3.4.57. Substituting Equation 3.4.59 and Equation 3.4.60 into Equation 3.4.58 yields the dual Fourier-Legendre series
∞ n=0
A2n+mP2n+m[cos(θ)] = 1, 0< θ < α, (3.4.61) and
∞ n=0
2n+m+12
A2n+mP2n+m[cos(θ)] = 0, α < θ < π/2. (3.4.62)
38 Minkov, I. M., 1963: Electrostatic field of a sectional spherical capacitor. Sov. Tech.
Phys.,7, 1041–1043.
At this point, we introduce A2n+m=
α 0
g(t) cos
2n+m+12 t
dt (3.4.63)
=g(α)sin
2n+m+12 α 2n+m+12 −
α 0
g(t)sin
2n+m+12 t 2n+m+12 dt.
(3.4.64) Now,
∞ n=0
2n+m+12
A2n+mP2n+m[cos(θ)]
=g(α) ∞ n=0
P2n+m[cos(θ)] sin
2n+m+12 α
(3.4.65)
− α
0
g(t) ∞
n=0
P2n+m[cos(θ)] sin
2n+m+12 t
dt= 0.
Equation 3.4.65 follows from Problem 4 at the end of Section 1.3 as well as the facts that 0≤t≤α < θ < π/2. Therefore, our choice forA2n+msatisfies Equation 3.4.62 identically.
Turning to Equation 3.4.61, α
0
g(t) ∞
n=0
P2n+m[cos(θ)] cos
2n+m+12 t
dt= 1. (3.4.66)
Again, using the results from Problem 4 at the end of Section 1.3, we have θ
0
g(t)
2[cos(t)−cos(θ)]dt= 2−(−1)m α
0
g(τ)
2[cos(τ)−cos(θ)]dτ, (3.4.67)
where 0≤θ≤α. Applying Equation 1.2.9 and Equation 1.2.10,
g(t) = 4 π
d dt
t 0
sin(θ)
2[cos(θ)−cos(t)]dθ
−(−1)m 2π
α 0
K(t, τ)g(τ)dτ, (3.4.68) where
K(t, τ) = 2d dt
t 0
sin(θ) cos(τ) + cos(θ)
cos(θ)−cos(t)dθ
(3.4.69)
= sec[(t+τ)/2] + sec[(t−τ)/2]. (3.4.70)
−2
−1 0
1
2
−2
−1 0 1 2 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
x/a y/a
u(x,y)
Figure 3.4.3: The solution u(x, y) to the mixed boundary value problem governed by Equation 3.4.56 through Equation 3.4.58 whenα=π/4 andm= 0.
Evaluating the first integral in Equation 3.4.68, the integral equation that governsg(t) is
g(t) +(−1)m 2π
α 0
K(t, τ)g(τ)dτ = 4
πcos(t/2), 0≤t≤α. (3.4.71) Figure 3.4.3 illustrates our solution whenα=π/4 andm= 0.
•Example 3.4.4 Let us solve
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, 0≤r <∞, 0< θ < π, (3.4.72) subject to the boundary conditions
lim
θ→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, 0< r <∞, (3.4.73)
rlim→0|u(r, θ)|<∞, lim
r→∞u(r, θ)→0, 0< θ < π, (3.4.74)
and
u(1−, θ) =u(1+, θ) = 1, 0≤θ < α, ur(1−, θ) =ur(1+, θ), α < θ < β, u(1−, θ) =u(1+, θ) = 0, β < θ < π.
(3.4.75)
Separation of variables yields the solution u(r, θ) =
∞ n=0
AnrnPn[cos(θ)], 0≤r <1, (3.4.76) and
u(r, θ) = ∞ n=0
Anr−n−1Pn[cos(θ)], 1< r <∞. (3.4.77) Equation 3.4.76 and Equation 3.4.77 satisfy not only Equation 3.4.72, but also Equation 3.4.73 and Equation 3.4.74. Substitution of Equation 3.4.76 and Equation 3.4.77 into Equation 3.4.75 yields the triple Fourier-Legendre series
∞ n=0
AnPn[cos(θ)] = 1, 0≤θ < α, (3.4.78) ∞
n=0
(2n+ 1)AnPn[cos(θ)] = 0, α < θ < β, (3.4.79)
and ∞
n=0
AnPn[cos(θ)] = 0, β < θ≤π. (3.4.80) How do we determineAn? Recently Singh et al.39 solved the triple series equation
∞ n=0
AnPn[cos(θ)] =f1(θ), 0≤θ < α, (3.4.81) ∞
n=0
(2n+ 1)AnPn[cos(θ)] =f2(θ), α < θ < β, (3.4.82)
and ∞
n=0
AnPn[cos(θ)] =f3(θ), β < θ≤π. (3.4.83) They showed thatAn is given by
An =12 α
0
g1(η) sin(η)Pn[cos(η)]dη+ β
α
f2(η) sin(η)Pn[cos(η)]dη +
π β
g3(η) sin(η)Pn[cos(η)]dη
, (3.4.84)
39 Results quoted with permission from Singh, B. M., R. S. Dhaliwal, and J. Rokne, 2002:
The elementary solution of triple series equations involving series of Legendre polynomials and their application to an electrostatic problem.Z. Angew. Math. Mech.,82, 497–503.
where
sin(x)g1(x) =−1 π
d dx
α x
G1(η) sin(η) cos(x)−cos(η)dη
, 0< x < α, (3.4.85)
sin(x)g3(x) = 1 π
d dx
x β
G3(η) sin(η) cos(η)−cos(x)dη
, β < x < π, (3.4.86)
G1(x) + 2
πcos(x/2) π
β
sin(η/2)G3(η)
cos(x)−cos(η)dη= d dx
x 0
F1(θ) sin(θ) cos(θ)−cos(x)dθ
(3.4.87) for 0< x < α,
G3(x) =−2
πsin(x/2) α
0
cos(η/2)G1(η)
cos(η)−cos(x)dη− d dx
π x
F3(θ) sin(θ) cos(x)−cos(θ)dθ
(3.4.88) forβ < x < π,
F1(θ) = 2f1(θ)− β
α
f2(η) sin(η)K(η, θ)dη, 0< θ < α, (3.4.89)
F3(θ) = 2f3(θ)− β
α
f2(η) sin(η)K(η, θ)dη, β < θ < π, (3.4.90) and
K(η, θ) = ∞ n=0
Pn[cos(θ)]Pn[cos(η)]. (3.4.91) Let us apply these results to our problem. Becausef1(θ) = 1 andf2(θ) = f3(θ) = 0,F1(θ) = 2 andF3(θ) = 0. Therefore,
An =12 α
0
g1(η) sin(η)Pn[cos(η)]dη+ π
β
g3(η) sin(η)Pn[cos(η)]dη
, (3.4.92) where
G1(x) +2
πcos(x/2) π
β
sin(η/2)G3(η)
cos(x)−cos(η)dη= 2 sin(x)
1−cos(x), (3.4.93)
for 0< x < α, and G3(x) =−2
πsin(x/2) α
0
cos(η/2)G1(η)
cos(η)−cos(x)dη (3.4.94)
−2
−1 0
1 2
−2
−1 0 1 2
−0.2 0 0.2 0.4 0.6 0.8 1 1.2
y x
u(x,y)
Figure 3.4.4: The solution u(x, y) to the mixed boundary value problem governed by Equation 3.4.72 and Equation 3.4.75 whenα=π/4 andβ= 3π/4.
with Equation 3.4.85 and Equation 3.4.86. Figure 3.4.4 illustrates our solution whenα=π/4 andβ = 3π/4.
A special case of particular interest occurs whenβ→π. Here, Equation 3.4.81 through Equation 3.4.83 reduce to
∞ n=0
AnPn[cos(θ)] =f1(θ), 0≤θ < α, (3.4.95)
and ∞
n=0
(2n+ 1)AnPn[cos(θ)] = 0, α < θ≤π. (3.4.96) From Equation 3.4.92, we have that
An= 12 α
0
g1(x) sin(x)Pn[cos(x)]dx (3.4.97)
= 1
√2π α
0
g1(x) sin(x) x
0
cos n+12
η cos(η)−cos(x)dη
dx (3.4.98)
= 1
√2π α
0
α η
g1(x) sin(x) cos(η)−cos(x)dx
cos
n+12 η
dη (3.4.99)
= 1
√2π α
0
G1(η) cos n+12
η
dη, (3.4.100)
−2
−1 0
1 2
−2
−1 0 1 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
y x
u(x,y)
Figure 3.4.5: The solution u(x, y) to the mixed boundary value problem governed by Equation 3.4.72 through Equation 3.4.74 and Equation 3.4.104 whenα=π/4.
where we used the Mehler integral representation of Pn[cos(x)] and inter- changed the order of integration in Equation 3.4.98. We also used the fact that
G1(η) = α
η
g1(x) sin(x)
cos(η)−cos(x)dx (3.4.101) which follows from Equation 3.4.85, Equation 1.2.11, and Equation 1.2.12.
Sneddon40 was the first to derive the solution to the dual series of Equa- tion 3.4.95 and Equation 3.4.96;Table 3.4.1 summarizes the results. Subse- quently Boridy41derived several additional solutions and they have also been included in the table.
To illustrate these results, we apply them to a case examined by Collins.42 Forf1(θ) = 1,
G1(η) = 2 sin(η)
1−cos(η) = 2√
2 cos(η/2). (3.4.102)
40 Sneddon, op. cit., Section 5.6.
41 Boridy, E., 1987: Solution of some electrostatic potential problems involving spherical conductors: A dual series approach. IEEE Trans. Electromagn. Compat., EMC-29, 132–140. c1987 IEEE.
42 Collins, W. D., 1961: On some dual series equations and their application to elec- trostatic problems for spheroidal caps. Proc. Cambridge Philosoph. Soc.,57, 367–384.
Table 3.4.1: The Solution (given in the right column) of Dual Series Involv- ing Legendre Polynomials (given in the left column). Taken from Boridy, E., 1987: Solution of some electrostatic poten- tial problems involving spherical conductors: A dual series approach. IEEE Trans. Electromagn. Compat., EMC-29, 132–140. c1987 IEEE.
∞ n=0
AnPn[cos(θ)] =F(θ), An =
√2 π
θ0 0
f(ξ) cos
n+1 2
ξ
dξ 0≤θ < θ0
∞ n=0
(2n+ 1)AnPn[cos(θ)] = 0, f(ξ) = d dξ
ξ 0
F(θ) sin(θ) cos(θ)−cos(ξ)dθ
θ0< θ≤π ∞
n=0
AnPn[cos(θ)] = 0, An=− 1
√2π π
θ0
f(ξ) cos
n+1 2
ξ
dξ 0≤θ < θ0
∞ n=0
(2n+ 1)AnPn[cos(θ)] =F(θ), f(ξ) = π
ξ
F(θ) sin(θ) cos(ξ)−cos(θ)dθ θ0< θ≤π
∞ n=0
(2n+ 1)AnPn[cos(θ)] =F(θ), An=(−1)n+1
√2π π
π−θ0
f(ξ) cos
n+1 2
ξ
dξ 0≤θ < θ0
∞ n=0
AnPn[cos(θ)] = 0, f(ξ) = π
ξ
F(π−θ) sin(θ) cos(ξ)−cos(θ)dθ θ0< θ≤π
∞ n=0
(2n+ 1)AnPn[cos(θ)] = 0, An =(−1)n√ 2 π
π−θ0 0
f(ξ) cos
n+1 2
ξ
dξ 0≤θ < θ0
∞ n=0
AnPn[cos(θ)] =F(θ), f(ξ) = d dξ
ξ 0
F(π−θ) sin(θ) cos(θ)−cos(ξ)dθ
θ0< θ≤π
Substituting Equation 3.4.102 into Equation 3.4.100 and carrying out the integration, we find that
An= sin(nα)
nπ +sin[(n+ 1)α]
(n+ 1)π . (3.4.103)
Figure 3.4.5 illustrates the solution to Equation 3.4.72 through Equation 3.4.74 and
u(1−, θ) =u(1+, θ) = 1, 0≤θ < π/4,
ur(1−, θ) =ur(1+, θ), π/4< θ < π. (3.4.104)
•Example 3.4.5 Let us solve43
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, 0≤r < b, 0≤θ≤π, (3.4.105) subject to the boundary conditions that
lim
θ→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, 0≤r < b, (3.4.106)
rlim→0|u(r, θ)|<∞, u(b, θ) = 0, 0≤θ≤π, (3.4.107)
and
ur(a−, θ) =ur(a+, θ), 0≤θ < θ0,
u(a, θ) =V0, θ0< θ≤π. (3.4.108) Before we solve our original problem, let us find the solution to a simpler one when we replace Equation 3.4.108 with
u(a, θ) =V0, 0≤θ≤π. (3.4.109) The solution to this new problem is
u(r, θ) =V0, 0≤r≤a, u(r, θ) = aV0
b−a b
r−1
, a≤r≤b.
(3.4.110)
Let us return to our original problem. We can view the introduction of the aperture betweenθ0< θ ≤π as a perturbation on the solution given by Equation 3.4.110. Therefore, we write the solution as
u(r, θ) =V0+ ∞ n=0
1−a
b
2n+1 An
r a
n
Pn[cos(θ)], 0≤r≤a, (3.4.111)
43 Boridy, op. cit.
and
u(r, θ) = aV0 b−a
b r−1
+
∞ n=0
An
a r
n+1
−a b
2n+1r a
n
Pn[cos(θ)]
(3.4.112) fora≤r≤b. The coefficients in Equation 3.4.111 and Equation 3.4.112 were chosen so that Equation 3.4.107 is satisfied andu(r, θ) is continuous atr=a.
Turning to the mixed boundary condition, direct substitution yields ∞
n=0
(2n+ 1)AnPn[cos(θ)] =− bV0
b−a, 0≤θ < θ0, (3.4.113) and
∞ n=0
1−a
b
2n+1
AnPn[cos(θ)] = 0, θ0< θ≤π. (3.4.114)
At this point, we would like to use the results given inTable 3.4.1but Equation 3.4.114 is not in the correct form. To circumvent this difficulty, let us set x=a/b <1. Then we can rewrite Equation 3.4.114 as
∞ n=0
AnPn[cos(θ)] = ∞ n=0
Anx2n+1Pn[cos(θ)], θ0< θ≤π. (3.4.115)
Setting ξ= cos(θ), let us integrate Equation 3.4.115 from −1 to ξ. We then
have ∞
n=0
An
ξ
−1
Pn(ξ)dξ = ∞ n=0
Anx2n+1 ξ
−1
Pn(ξ)dξ. (3.4.116) However, because
1
−1
Pn(ξ)dξ= 2δn0, (3.4.117) whereδij is the Kronecker delta,
2A0− ∞ n=0
An
1
ξ
Pn(ξ)dξ= 2A0x− ∞ n=0
Anx2n+1 1
ξ
Pn(ξ)dξ. (3.4.118) If we differentiate Equation 3.4.118 with respect of x, then differentiate it with respect ofξ, and finally multiply byx, we obtain
∞ n=0
(2n+ 1)Anx2n+1Pn[cos(θ)] = 0, 0≤θ < θ0. (3.4.119)
−2
−1 0
1 2
−2
−1 0 1 2 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
y x
u(x,y)
Figure 3.4.6: The solution u(x, y) to the mixed boundary value problem governed by Equation 3.4.105 through Equation 3.4.108 whena= 1,b= 2 andθ0=π/2.
Subtracting Equation 3.4.119 from Equation 3.4.113, we obtain the following dual equations:
∞ n=0
(2n+ 1)An
1−a b
2n+1
Pn[cos(θ)] =− bV0
b−a, 0≤θ < θ0, (3.4.120) and
∞ n=0
An
1−a b
2n+1
Pn[cos(θ)] = 0, θ0< θ≤π. (3.4.121)
If we set Cn = An
1−a
b
2n+1
, then we can immediately use the results fromTable 3.4.1and find that
An=− bV0
π(b−a)
1−a
b
2n+1
sin(nθ0)
n −sin[(n+ 1)θ0] n+ 1
. (3.4.122)
Figure 3.4.6 illustrates our solution whena= 1,b= 2 andθ0=π/2.
•Example 3.4.6
A problem44 that is similar to Example 3.4.3 consists of solving
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, 0≤r <1, 0< θ < π, (3.4.123) subject to the boundary conditions
lim
θ→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, 0≤r <1, (3.4.124)
rlim→0|u(r, θ)|<∞, 0< θ < π, (3.4.125)
and
u(1, θ) = cos(θ), 0≤θ < α, ur(1, θ) = 0, α < θ < β, u(1, θ) = cos(θ), β < θ≤π.
(3.4.126) Separation of variables yields the solution
u(r, θ) =rP1[cos(θ)]−∞
n=1
2n+ 1
n CnrnPn[cos(θ)]. (3.4.127) From the nature of the boundary conditions, we anticipate thatC0 =C2 = C4=. . .= 0. Upon substituting Equation 3.4.127 into Equation 3.4.126,
∞ n=1
Cn(1 +Hn)Pn[cos(θ)] = 0, 0≤θ < α, β < θ≤π, (3.4.128) and
∞ n=1
Cn(2n+ 1)Pn[cos(θ)] =P1[cos(θ)], α < θ < β, (3.4.129) whereHn = 1/(2n).
To findC1, C3, C5, . . ., let us set Cn =An+Bn with Bn = (−1)n+1An. Therefore, Equation 3.4.128 and Equation 3.4.129 can be rewritten
∞ n=0
(An+Bn)(1 +Hn)Pn[cos(θ)] = 0, 0≤θ < α, (3.4.130) ∞
n=0
An(2n+ 1)Pn[cos(θ)]−12P1[cos(θ)] = 0, α < θ≤π, (3.4.131)
44 See Dryden, J. R., and F. W. Zok, 2004: Effective conductivity of partially sintered solids.J. Appl. Phys.,95, 156–160.
∞ n=0
Bn(2n+ 1)Pn[cos(θ)]−12P1[cos(θ)] = 0, 0≤θ < β, (3.4.132) and
∞ n=0
(An+Bn)(1 +Hn)Pn[cos(θ)] = 0, β < θ≤π. (3.4.133) In this formulation, A0 and B0 are nonzero although A0+B0 = 0. For convenience we introduce H0 ≡1 so that no difficulty arises in solving this system of equations.
Due to symmetry, we must only solve Equation 3.4.130 and Equation 3.4.131. Using the integral representation of Legendre polynomials, Equation 1.3.4, we can rewrite Equation 3.4.130 as
θ 0
∞
n=0
(An+Bn)(1 +Hn) cos n+12
t dt
cos(t)−cos(θ)= 0 (3.4.134) after interchanging the order of integration and summation. In a similar manner, we can use Equation 1.3.5 to write Equation 3.4.131 as
π θ
d dt
∞
n=0
Ancos n+12
t
−16cos 3t
2
dt
cos(θ)−cos(t) = 0.
(3.4.135) Equation 3.4.134 and Equation 3.4.135 are integral equations of the Abel type; see Equation 1.2.9 and Equation 1.2.12. In the case of Equation 3.4.134 the quantity within the wavy brackets must vanish. In the case of Equation 3.4.135 thet-derivative of the quantity within the wavy brackets must vanish.
Actually the quantity within the wavy brackets also vanishes because this quantity equals zero since it vanishes whent=π. Consequently,
∞ n=0
(An+Bn)(1 +Hn) cos n+12
t
= 0, 0≤t < α, (3.4.136)
and ∞ n=0
Ancos n+12
t
−16cos 3t
2
= 0, α < t≤π. (3.4.137)
To findAn, let us introduce an unknown functionh(t) such that ∞
n=0
Ancos n+12
t
=
cos(3t/2)/6−h(t)/2, 0≤t < α,
cos(3t/2)/6, α < t≤π. (3.4.138)
−1
−0.5 0
0.5 1
−1
−0.5 0 0.5 1
−1
−0.5 0 0.5 1
x y
u(x,y)
Figure 3.4.4: The solution u(x, y) to the mixed boundary value problem governed by Equation 3.4.123 through Equation 3.4.126 whenα=π/4 andβ=π/2.
Applying the orthogonality properties of cos n+12
t
over [0, π], An= 16δ1n− 1
π α
0
h(ξ) cos n+12
ξ
dξ, (3.4.139)
where δnm is the Kronecker delta. Upon settingBn = (−1)n+1An and sub- stituting Equation 3.4.138 into Equation 3.4.136, we obtain
h(t) + 2 ∞ n=0
An{(−1)n−Hn[1−(−1)n]}cos n+12
t
= 13cos(3t/2).
(3.4.140) Finally, substituting forAn, we obtain an integral equation that governsh(t):
h(t) = cos(3t/2) + 1 π
α 0
K(τ, t)h(τ)dτ, 0≤t < α, (3.4.141) whereK(τ, t) =G(t−τ) +G(t+τ) and
G(ξ) = ∞ n=0
{(−1)n−Hn[1−(−1)n]}cos n+12
ξ
(3.4.142)
=1 4
2 sec
ξ 2
+πsin ξ
2 + cos
ξ 2
ln
tan2
ξ 2
. (3.4.143) Once h(t) is computed numerically from Equation 3.4.141, we can find An
from Equation 3.4.139. Finally, C2n+1 = 2A2n+1 and u(r, θ) follows from Equation 3.4.127. Figure 3.4.4 illustrates the solution when α = π/4 and β=π/2.
Problems 1. Solve Laplace’s equation
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, a < r <∞, 0≤θ≤π, subject to the boundary conditions that
lim
θ→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, a < r <∞, (1) u(a, θ) =V0, lim
r→∞u(r, θ)→0, 0≤θ≤π, (2)
and
ur(b−, θ) =ur(b+, θ), 0≤θ < θ0,
u(b, θ) = 0, θ0< θ≤π, (3)
where b− and b+ denote points slightly inside and outside of r =b, respec- tively, and 0< θ0< π.
Step 1: First solve the simpler problem when we replace Equation (3) with u(b, θ) = 0 for 0≤θ≤πand show that
u(r, θ) = aV0 b−a
b r−1
, a < r < b, u(r, θ) = 0, b < r <∞.
Step 2: Returning to the original problem, show that the solution to the partial differential equation plus the first two boundary conditions is
u(r, θ) = aV0 b−a
b r−1
+
∞ n=0
An
r b
n−a b
2n+1 b r
n+1
Pn[cos(θ)]
fora≤r≤b, and u(r, θ) =
∞ n=0
1−a
b
2n+1 An
b r
n+1
Pn[cos(θ)], b≤r <∞.
Step 3: Using Equation (3), show thatAn is given by the dual series:
∞ n=0
(2n+ 1)AnPn[cos(θ)] = aV0
b−a, 0≤θ < θ0,
and ∞
n=0
1−a
b
2n+1
AnPn[cos(θ)] = 0, θ0< θ≤π.
−2
−1 0
1 2
−2
−1 0 1 2
−0.2 0 0.2 0.4 0.6 0.8 1 1.2
x y u(x,y)/V 0
Problem 1
Step 4: Following the analysis given by Equation 3.4.115 through Equation 3.4.121, show that
An= aV0
π(b−a)
1−a
b
2n+1
sin(nθ0)
n −sin[(n+ 1)θ0] n+ 1
.
The figure labeled Problem 1 illustrates this solution whena= 1,b= 2 and θ0=π/2.
2. A problem similar to the previous one involves finding the electrostatic potential when an uniform external electric field is applied along the z-axis.
In this case,
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, a < r <∞, 0≤θ≤π, subject to the boundary conditions that
lim
θ→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, a < r <∞, (1) u(a, θ) =V0, lim
r→∞u(r, θ)→E0rcos(θ), 0≤θ≤π, (2)
and
ur(b−, θ) =ur(b+, θ), 0≤θ < θ0,
u(b, θ) = 0, θ0< θ≤π, (3)
where b− and b+ denote points slightly inside and outside of r =b, respec- tively, and 0< θ0< π.
Step 1: First solve the simpler problem when we replace Equation (3) with u(b, θ) = 0 for 0≤θ≤πand show that
u(r, θ) = aV0 b−a
b r−1
, a < r < b, u(r, θ) =E0rcos(θ)−E0b3cos(θ)/r2, b < r <∞.
Step 2: Returning to the original problem, show that the solution to the partial differential equation plus the first two boundary conditions is
u(r, θ) = aV0 b−a
b r−1
+
∞ n=0
An r
b
n−a b
2n+1 b r
n+1
Pn[cos(θ)]
fora≤r≤b, and
u(r, θ) =E0rcos(θ)−E0b3
r2cos(θ)+
∞ n=0
1−a
b
2n+1 An
b r
n+1
Pn[cos(θ)]
forb≤r <∞.
Step 3: Using the third boundary condition, show that An is given by the dual series:
∞ n=0
(2n+ 1)AnPn[cos(θ)] = aV0
b−a+ 3E0bcos(θ), 0≤θ < θ0,
and ∞
n=0
1−a
b
2n+1
AnPn[cos(θ)] = 0, θ0< θ≤π.
Step 4: Following the analysis given by Equation 3.4.115 through Equation 3.4.121, show that
An = aV0
π(b−a)
1−a
b
2n+1
sin(nθ0)
n −sin[(n+ 1)θ0] n+ 1
− E0b π
1−a
b
2n+1
sin[(n−1)θ0]
n−1 −sin[(n+ 2)θ0] n+ 2
.
−2
−1 0
1 2
−2
−1 0 1 2
−1.5
−1
−0.5 0 0.5 1 1.5
y x u(x,y)/V 0
Problem 2
The figure labeled Problem 2 illustrates this solution whena= 0.7,b= 1.4, θ0=π/2 andV0=bE0.
3. Solve45 Laplace equation
∂
∂r
r2∂u
∂r
+ 1
sin(θ)
∂
∂θ
sin(θ)∂u
∂θ
= 0, 0≤r <∞, 0≤θ≤π, subject to the boundary conditions that
lim
θ→0|u(r, θ)|<∞, lim
θ→π|u(r, θ)|<∞, 0≤r <∞, (1)
rlim→0|u(r, θ)|<∞, lim
r→∞|u(r, θ)|<∞, 0≤θ≤π, (2)
and
u(a−, θ) =u(a+, θ) =C1+C2cos(θ), 0≤θ < α,
ur(a−, θ) =ur(a+, θ), α < θ≤π, (3) where a− and a+ denote points slightly inside and outside of r=a, respec- tively, and 0< α < π. The parameterC2 is nonzero.
45 Taken with permission from Casey, K. F., 1985: Quasi-static electric- and magnetic- field penetration of a spherical shield through a circular aperture.IEEE Trans. Electromag.
Compat.,EMC-27, 13–17. c1985 IEEE.