Mixed Boundary Value Problems Episode 6 docx

Substituting Equation 3.4.59 and Equation 3.4.60 into Equation 3.4.58yields the dual Fourier-Legendre series... Evaluating the first integral in Equation 3.4.68, the integral equation tha

138 Mixed Boundary Value Problems

We have written the solution in this form so that we can take advantage of

symmetry and limit θ 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 Equation3.4.57 Substituting Equation 3.4.59 and Equation 3.4.60 into Equation 3.4.58yields the dual Fourier-Legendre series

At this point, we introduce

140 Mixed Boundary Value Problems

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 when α = π/4 and m = 0.

Evaluating the first integral in Equation 3.4.68, the integral equation that

Separation of variables yields the solution

142 Mixed Boundary Value Problems

where

sin(x)g1(x) = −1

π

d dx

 α x

sin(η/2)G3(η) cos(x) − cos(η) dη =

d dx

 π x

f2(η) sin(η)K(η, θ) dη, 0 < θ < α, (3.4.89)

F3(θ) = 2f3(θ) −

 β α

−1 0 1 2

−2

−1 0 1 2

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 when α = π/4 and β = 3π/4.

with Equation 3.4.85 and Equation 3.4.86 Figure 3.4.4 illustrates our solution

when α = π/4 and β = 3π/4.

A special case of particular interest occurs when β → π Here, Equation

3.4.81 through Equation 3.4.83 reduce to

g1(x) sin(x)



cos(η) − cos(x) dx

cos

n +1 2



η

144 Mixed Boundary Value Problems

−2

−1 0 1 2

−2

−1 0 1 2

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 when α = π/4.

where we used the Mehler integral representation of P n [cos(x)] and

inter-changed the order of integration in Equation 3.4.98 We also used the factthat

G1(η) =

 α η

Subse-included in the table

To illustrate these results, we apply them to a case examined by Collins.42

40 Sneddon, op cit., Section 5.6.

41 Boridy, E., 1987: Solution of some electrostatic potential problems involving sphericalconductors: 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.

Substituting Equation 3.4.102 into Equation 3.4.100 and carrying out theintegration, we find that

Before we solve our original problem, let us find the solution to a simplerone when we replace Equation 3.4.108 with

Let us return to our original problem We can view the introduction of

the aperture between θ0< θ ≤ π as a perturbation on the solution given by

Equation 3.4.110 Therefore, we write the solution as

Separation of Variables 147and

−1 0 1 2

−2

−1 0 1 2

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 when a = 1, b = 2 and θ0= π/2.

Subtracting Equation 3.4.119 from Equation 3.4.113, we obtain the followingdual equations:

2n+1, then we can immediately use the resultsfromTable 3.4.1and find that

From the nature of the boundary conditions, we anticipate that C0 = C2 =

C4= = 0 Upon substituting Equation 3.4.127 into Equation 3.4.126,

To find C1, C3, C5, , let us set C n = A n + B n with B n = (−1) n+1A n.

Therefore, Equation 3.4.128 and Equation 3.4.129 can be rewritten

44 See Dryden, J R., and F W Zok, 2004: Effective conductivity of partially sintered

solids J Appl Phys., 95, 156–160.

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 Equation3.4.131 Using the integral representation of Legendre polynomials, Equation1.3.4, we can rewrite Equation 3.4.130 as

3.4.135 the t-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 when t = π Consequently,

Separation of Variables 151

−1

−0.5 0 0.5 1

−1

−0.5 0 0.5 1

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 when α = π/4 and β = π/2.

Applying the orthogonality properties of cos

n +1 2

 α

0

h(ξ) cos

n +1 2



ξ

where δ nm is the Kronecker delta Upon setting B n = (−1) n+1A n and

sub-stituting Equation 3.4.138 into Equation 3.4.136, we obtain

+ π sin

2ξ + cos ξ2ln

tan2 ξ

2 . (3.4.143)

Once h(t) is computed numerically from Equation 3.4.141, we can find A n

from Equation 3.4.139 Finally, C 2n+1 = 2A 2n+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.

Step 2 : Returning to the original problem, show that the solution to the

partial differential equation plus the first two boundary conditions is

2n+1

A n

b r

2n+1

A n P n [cos(θ)] = 0, θ0< θ ≤ π.

Separation of Variables 153

−2

−1 0 1 2

−2

−1 0 1 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.

where b − and b+ denote points slightly inside and outside of r = b,

Step 2 : Returning to the original problem, show that the solution to the

partial differential equation plus the first two boundary conditions is

2n+1

A n

b r

Separation of Variables 155

−2

−1 0 1 2

−2

−1 0 1 2

tively, and 0 < α < π The parameter C2 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.

−1 0 1 2

−2

−1 0 1 2

Problem 3

Step 1 : Show that the solution to the differential equation and first two

boundary conditions are

Separation of Variables 157

3.5 TRIPLE FOURIER SINE SERIES

In this closing section we illustrate a mixed boundary value problem thatyields a triple Fourier sine series

Let us find46 the potential for Laplace’s equation in cylindrical



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 Fouriersine series:



a

I1

n +1 2

To solve Equation 3.5.6 through Equation 3.5.8, we first note that



z

=− d dz

sin(η) F2 1[k + 1, −k; 1; sin2(η/2)/ sin2(β/2)]



y

k +1 2

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 with ν = n + 1.

Finally, consider Equation 3.5.7 We can rewrite it

by sin[x sin(z/2)]/2 so that we now have

k + 1 2

and ϕ = 2 arcsin[sin(z/2)/ sin(β/2)].

Our final task is to compute B k To this end, let us introduce

B k = AP k [cos(ϕ0)] +

 π

ϕ0

f (τ ) d dτ

$

P k [cos(τ )]%

where A is a free parameter and ϕ0= 2 arcsin[sin(α/2)/ sin(β/2)]

Substitut-ing Equation 3.5.26 into Equation 3.5.23, we obtain

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 for B k satisfies Equation 3.5.23

Turning to Equation 3.5.24, we take its derivative with respect to ϕ and

Separation of Variables 161

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

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 when a = π, α = π/3, and β = 2π/3.

If we now integrate the second term in Equation 3.5.30 by parts and againintroduce the results from Problem 1 from Section 1.3, we derive

The first two terms in Equation 3.5.31 vanish while the limits of

integra-tion for the integral in the third term run from ϕ to π Finally, let us multiply Equation 3.5.31 by sin(ϕ) dϕ/

2 cos(τ ) − 2 cos(ϕ) and then integrate from τ

Using results given by Equation 1.2.11 and Equation 1.2.12, the first term in

Equation 3.5.32 equals f (τ ) and Equation 3.5.32 becomes

Conse-when a = π, α = π/3 and β = 2π/3 It is better to use

rather than Equation 3.5.26 so that we avoid large values of the derivative of

the Legendre polynomials for large k near τ = π.

−4 −2

4 0

Chapter 4 Transform Methods

In Example 1.1.2 we showed that applying a Fourier cosine transformleads to the dual integral equations:

Before we proceed to our study of dual and triple integral equations, let

us finish Example 1.1.2 We begin by introducing

Referring back to Equation 1.1.15, we see that u(x, 0) is the solution to tion 1.1.11 along the boundary y = 0 Next, for convenience, let us define

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 of g(x),

The integral within the square brackets in Equation 4.0.7 can be evaluated1

exactly and the integral equation simplifies to

tanh(βx) + tanh(βξ) tanh(βx) − tanh(βξ) dξ = x h , 0≤ x < 1, (4.0.8)

where β = π/(2h) The results from Example 1.2.3 can be employed to solve Equation 4.0.8 after substituting x  = tanh(βx)/ tanh(β) This yields

g(ξ) = 1

h2

d dξ

Transform Methods 165

−2

−1

0 1

2

0 0.5

1

0.25 0.75

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 when h = 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, tion 1.1.13, and Equation 1.1.14 by numerically integrating Equation 1.1.17.Figure 4.0.1 illustrates this solution

Equa-4.1 DUAL FOURIER INTEGRALS

A common technique in solving boundary value problems in rectangularcoordinates involves Fourier transforms In the case of mixed boundary valueproblems, this leads to sets of integral equations In this section we focus oncommonly occuring cases of dual integral equations

2 See Iossel, Yu Ya., and R A Pavlovskii, 1966: A plane steady heat conduction

problem J Engng Phys., 10, 163–166.

subject to the boundary conditions

if we assume that g(0) = 0 Turning first to Equation 4.1.7, if we substitute

Equation 4.1.8 into Equation 4.1.7 and interchange the order of integration,

h . (4.1.10)

x n =

n −1

2



∆x, n = 1, 2, · · · , N, the MATLAB code to find f(x) isR

3 Gradshteyn and Ryzhik, op cit., Formula 6.671.1 and Formula 6.693.7.

4 Atkinson, K E., 1967: The numerical solution of Fredholm integral equations of the

second kind SIAM J Numer Anal., 4, 337–348 SeeSection 5 in particular.

A common technique in solving boundary value problems in rectangularcoordinates involves Fourier transforms In the case of mixed boundary valueproblems, this leads to sets of integral... Variables 161

0 0.2 0.4 0 .6 0.8 1

0 0.2 0.4 0 .6 0.8 1

Figure 3.5.1: The solution u(r, z) to the mixed boundary value problem...

2 See Iossel, Yu Ya., and R A Pavlovskii, 1 966 : A plane steady heat conduction

problem J Engng Phys., 10, 163 – 166 .