Mixed Boundary Value Problems Episode 4 doc

First, the solution for the µ = 0 case, Equation 2.6.16, creates a divergent integral when it is substituted back into dual equations, Equation 2.6.8 and Equation 2.6.9.Second, the gover

(1 + aλ0k)A(k) sin(kr) dk = 0, a < r < ∞ (2.6.38)

We can rewrite Equation 2.6.38 as

πak J3(ak)

− 2λ0π

78 Mixed Boundary Value Problems The second term in Equation 2.6.41 equals aλ0kA(k) Therefore,

(1 + aλ0k)A(k) = 2ωa2

#2

aξA(ξ) sin(ξr) dξ sin(kr) dr (2.6.42)

Interchanging the order of integration, we finally have that

(1 + aλ0k)A(k) = 2ωa2

#2

One of the intriguing aspects of Equation 2.6.43 is that the unknown is the

Hankel transform A(k) In general, the integral equations that we will see will involve an unknown which is related to A(k) via an integral definition.

See Equation 2.5.61, Equation 2.5.62 and Equation 2.5.65 In Goodrich’spaper he solved the integral equation as a variational problem and found anapproximate solution by the optimization of suitable solutions However, we

shall shortly outline an alternative method for any value of λ0

In 1978 Shail27 reexamined Goodrich’s paper for two reasons First, the

solution for the µ = 0 case, Equation 2.6.16, creates a divergent integral when

it is substituted back into dual equations, Equation 2.6.8 and Equation 2.6.9.Second, the governing Fredholm integral equation is over an infinite range and

appears to be unsuitable for asymptotic solution as λ0→ 0 or λ0→ ∞ Shail’s analysis for the µ = 0 case begins by noting that

27 Taken from Shail, R., 1978: The torque on a rotating disk in the surface of a liquid

with an adsorbed film J Engng Math., 12, 59–76 with kind permission from Springer

Science and Business Media.

If we use Equation 2.6.47 in place of Equation 2.6.4, then the solution tion 2.6.17 follows directly.

Equa-Shail also examined the general case and solved it using the method

of complementary representations for generalized axially symmetric tial functions This method is very complicated and we introduce a greatlysimplified version of an analysis first done by Chakrabarti.28

poten-We begin by introducing the function g(x) such that

From tables,29 the integral within the square brackets of the first integral in

Equation 2.6.53 can be evaluated and Equation 2.6.53 simplifies to

28 Chakrabarti, A., 1989: On some dual integral equations involving Bessel functions of

order one Indian J Pure Appl Math., 20, 483–492.

29 Gradshteyn and Ryzhik, op cit., Formula 6.671.1.

80 Mixed Boundary Value Problems

for 0≤ r < a Using Equation 1.2.13 and Equation 1.2.14, we can solve for

Equation 2.6.57 is identical to Chakrabarti’s equations (50) and (52) Once

we solve Equation 2.6.57 numerically, its values of g(x) can be substituted into Equation 2.6.48 Finally the solution u(r, z) follows from Equation 2.6.5.

The numerical solution of Equation 2.6.57 is nontrivial due to the nature

of the integration over k To solve it, we use a spectral method If we take g(τ ) to be an odd function over ( −a, a), we have that

0 0.5 1 1.5 2

0 0.5 1 1.5 2

0 0.2 0.4 0.6 0.8 1

r/az/a

0 0.5 1 1.5 2

−2

−1

0

1 2 3 4

Chapter 3 Separation of Variables

Separation of variables is the most commonly used technique for solvingboundary value problems In the case of mixed boundary value problemsthey lead to dual or higher-numbered Fourier series which yield the Fouriercoefficients In this chapter we examine dual Fourier series in Section 3.1 andSection 3.2, while dual Fourier-Bessel series are treated in Section 3.3 anddual Fourier-Legendre series in Section 3.4 Finally Section 3.5 treats tripleFourier series

In Example 1.1.1 we showed that the method of separation of variablesled to the dual cosine series:

cos

n −1 2

equations The general form of these dual series can be written

3.1 DUAL FOURIER COSINE SERIES

Tranter1 examined dual trigonometric series of the form

Separation of Variables 85where

dθ



1− cos2(c/2) sin2(θ) (3.1.11)

= K[cos2(c/2)], (3.1.12)

0 0.2 0.4 0.6 0.8 1

0

0.5 1 1.5 2

Figure 3.1.1: The solution u(x, y) to the mixed boundary value problem posed in Example

1.1.1 when c = π/2.

where K( ·) denotes the complete elliptic integral2 and sin(θ) = 2x/(1 + x2).

Therefore, 2C sin(c/2)K[cos2(c/2)] = 1, and a

n = P n −1 [cos(c)]/K[cos2(c/2)]

is the solution to the dual Fourier cosine series Equation 3.0.1 and Equation3.0.2

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, thesolution to this particular mixed boundary value problem is



y

n −1 2

cos

n −1 2

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.

Separation of Variables 87subject to the boundary conditions

cosh

n +1 2



π/2cos

n +1 2



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



n +1 2

4 Tranter, C J., 1964: An improved method for dual trigonometrical series. Proc.

Glasgow Math Assoc., 6, 136–140.

and A0 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 theseequations as follows:

π/2cos

n +1 2



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

subject to the boundary conditions

u(x, 0) = u(x, h) = 0, u(x, c − ) = u(x, c+), −b < x < b, (3.1.44)

Figure 3.1.3illustrates the geometry for this problem

If we use separation of variables, the solution to Equation 3.1.43 is

92 Mixed Boundary Value Problems

Kiyono and Shimasaki5 developed a method for computing C

n Theyshowed that

(

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)



d

π

n +1 2



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.

−0.5 0 0.5 1

0 0.2 0.4 0.6 0.8 1

6 Reprinted from Int 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.

94 Mixed Boundary Value Problems

If we use separation of variables, the solution to Equation 3.1.63 is

To solve Equation 3.1.70 and Equation 3.1.71, we first substitute x =

π − ξ, c = π − γ, A0= a0/2 and A n= (−1) n a n We then obtain

Recently Sbragaglia and Prosperetti7also solved this dual series when it arose

during their study of the effects of surface deformation on a type of drophobic surface

superhy-Equation 3.1.72 and superhy-Equation 3.1.73 are an example of superhy-Equation 3.0.6

For the special case p = 1, Sneddon8 showed that the solution to Equation

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.

9 Adapted from Westmann, R A., and W H Yang, 1967: Stress analysis of cracked

rectangular beams J Appl Mech., 34, 693–701.

Separation of Variables 97

To solve the dual equations, Equation 3.1.89 and Equation 3.1.90, let us

assume that u(x, 0) is known for 0 < x < L Then

h(t)

√

t2− x2dt dx =

π 2L

order of integration and summation,

If we integrate the function csc(πz)e iπz J0(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 + 1

Therefore, Equation 3.1.99 can be rewritten and made nondimensional with

respect to a so that it becomes

nπax

L

− a L

Figure 3.1.6illustrates the solution when a/L = 0.5 and h/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.

100 Mixed Boundary Value Problems

Equation 3.1.110 satisfies not only Equation 3.1.106, but also Equation 3.1.107and Equation 3.1.108 Upon substituting Equation 3.1.110 into Equation3.1.109, we find the dual series

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 A n’s The potential then follows from Equation3.1.110 Figure 3.1.7illustrates this solution when ϕ = π/3 and N = 100.

r2

∂2u

∂θ2 = 0, 0≤ r < ∞, 0 < θ < 2π,

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

u(r, θ)/V when κ = 6.

3.2 DUAL FOURIER SINE SERIES

Dual Fourier sine series arise in the same manner as dual Fourier cosineseries in mixed boundary value problems in rectangular domains However,for some reason, they do not appear as often and the following example is theonly 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)

Separation of Variables 103subject to the boundary conditions

The solution to Equation 3.2.1 and the boundary conditions Equation3.2.2 and Equation 3.2.3 is

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 express p(t) by the expansion

where T n(·) is the nth Chebyshev polynomial Substituting Equation 3.2.12

into Equation 3.2.11 and noting15 that

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.

Separation of Variables 105

0 0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

Figure 3.2.1: The solution u(x, y) to the mixed boundary value problem given by Equation

3.2.1 through Equation 3.2.4 when U0(x) = x/ and L/ = 3.

For a given U0(x), we compute b 2n+1 Equation 3.2.17 then yields B 2m+1

while Equation 3.2.19 yields A n Finally, u(x, y) follows from Equation 3.2.5 Figure 3.2.1 illustrates the solution when U0(x) = x/ and L/ = 3.

An alternative method of solving the dual equations Equation 3.2.6 and

Equation 3.2.7 involves introducing B n = λ n A n , C n = (−1) n −1 B n , ξ = πx/L,

η = π − ξ, and γ = π − π/L Then, these dual equations become

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

x2

 c x

h(t)



cos(x) − cos(t) dt, 0≤ x < c, (3.2.24) where h(x) is unknown Why have we chosen such an unusual definition for y(x)?

Recall that we seek an a nthat satisfies Equation 3.2.22 From the theory

of Fourier series, we know that

Turning now to the first equation in Equation 3.2.22, we eliminate n

inside of the summation by integrating both sides:

 x

0

f (ξ) dξ (3.2.32)

 x c

over the range c < x ≤ π, where h(t) is unknown From the properties of

Fourier sine series,

sin

x2

 x c

 x c

h(t) {P n [cos(t)] − P n −1 [cos(t)] } dt (3.2.41)