Mixed Boundary Value Problems Episode 12 ppsx

On the other hand, To evaluate Bk, we first substitute Equation 4.5.32 into the boundary condition given by Equation 4.5.31... 324 Mixed Boundary Value ProblemsRecognizing that the left s

318 Mixed Boundary Value Problems

0 0.5 1 1.5 2

0 0.2 0.4 0.6 0.8 1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

r z

Problem 1

The figure labeled Problem 1 illustrates this solution when a = 0.5.

4.5 JOINT TRANSFORM METHODS

In the previous sections we sought to separate problems according towhether the kernels of dual or triple integral equations contained trigonomet-rical or Bessel functions Such clear-cut lines of demarcation are not alwayspossible and we conclude with examples where the analysis includes bothFourier and Hankel transforms as well as Fourier and Fourier-Bessel series

101 See Shindo, Y., and A Atsumi, 1975: Thermal stresses in a laminate composite with

infinite row of parallel cracks normal to the interfaces Int J Engng Sci., 13, 25–42.

nπy

L

(4.5.8)

for h < x < ∞ An interesting aspect of this problem is that Equation 4.5.7

contains both a Fourier cosine transform and a Fourier sine series tuting Equation 4.5.7 and Equation 4.5.8 into Equation 4.5.6 yields the dualintegral equations

where we used Equation 1.4.14 Note that 0≤ t ≤ x < ∞.

Turning to Equation 4.5.9, we first integrate it with respect to x and

320 Mixed Boundary Value Problems

Transform Methods 321and

Finally, we can bring everything into a nondimensional form by

introduc-ing ξ = x/a, η = τ /a, ρ = L/a, κ = a/h, and G(ξ) = h(aξ)/(a √

I0 nπη ρ .

Once G(ξ) is computed via Equation 4.5.26, we find A n , B n and A(k) from

Equation 4.5.21, Equation 4.5.22, and Equation 4.5.11, respectively The

solution u(x, y) follows from Equation 4.5.7 and Equation 4.5.8. Figure 4.5.1

illustrates this solution when h/a =1

103 Taken with permission from Kim, M.-U., 1981: Slow rotation of a disk in a fluid-filled

circular cylinder J Phys Soc Japan, 50, 4063–4067.

If we substitute Equation 4.5.35 into Equation 4.5.34, we can show that this

definition of A(k) satisfies Equation 4.5.34 identically On the other hand,

To evaluate B(k), we first substitute Equation 4.5.32 into the boundary

condition given by Equation 4.5.31 This yields

324 Mixed Boundary Value Problems

Recognizing that the left side of Equation 4.5.42 is the Fourier cosine

trans-form representation of the right side, B(k) is given by

where we used integral tables104for the η integration Substituting Equation

4.5.46 into Equation 4.5.41, we obtain

h(t) = 2

π

2t + 2π

Figure 4.5.2illustrates the solution when a = 2.

104 Gradshteyn and Ryzhik, op cit., Formula 6.718.

105 See Rusia, K C., 1968: On certain asymmetric mixed boundary value problems of

an electrified circular disc situated inside a coaxial infinite hollow cylinder Indian J Pure

Appl Phys., 6, 44–46.

326 Mixed Boundary Value Problems

We begin our solution of Equation 4.5.54 and Equation 4.5.55 by ducing

From tables106 and noting that r > t, the value of the integral within the

square brackets equals zero and this choice of A(k) satisfies Equation 4.5.55.

If we integrate Equation 4.5.56 by parts and assuming that h(t)t −(n −1)

Substituting Equation 4.5.58 into Equation 4.5.54, interchanging the order of

integration, and carrying out the k-integration,

106 Gradshteyn and Ryzhik, op. cit., Formula 6.575.1. Note that this formula has a

typo; the condition should read(ν + 1) > (µ) > −1 See p 100 in Magnus, W., F.

Oberhettinger, and R P Soni, 1966: Formulas and Theorems for the Special Functions of

Mathematical Physics Springer-Verlag, 508 pp.

328 Mixed Boundary Value Problems

Substituting Equation 4.5.71 into Equation 4.5.64 and interchanging the order

The analysis is identical to our earlier problem except for computing B(k).

Equation 4.5.71 now becomes

330 Mixed Boundary Value Problems

0 0.2 0.4 0.6 0.8 1

0 0.1 0.2 0.3 0.4

Figure 4.5.4: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.5.49, Equation 4.5.76, Equation 4.5.51 and Equation 4.5.52 with a = 0.8, n = 1,

Using separation of variables, the general solution to Equation 4.5.79 is

Transform Methods 331where

This choice for A(k) ensures that Equation 4.5.85 satisfies Equation 4.5.84 for

if z > 0 Then, combining Equation 4.5.82 and Equation 4.5.83 with Equation

4.5.85 and Equation 4.5.86, we find that

109 Eason, G., B Noble, and I N Sneddon, 1955: On certain integrals of Lipschitz-Hankel

type involving products of Bessel functions Philos Trans R Soc London, Ser A, 247,

529–551.

332 Mixed Boundary Value Problems

0 0.5

1 1.5

2

0

0.2

0.4 0.6 0.8 1

Figure 4.5.5: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.5.79 through Equation 4.5.84 with β = 0 and c = 0.9.

Solving this integral equation of the Abel type,

G(t) =3

√ 2π

Using separation of variables, the general solution to Equation 4.5.98 is

u(r, z) = 2 r2− r4

2

+

Solving for A(k),

{2I1(k)I0(k) − k[I2

110 See Jeong, J.-T., and S.-R Choi, 2005: Axisymmetric Stokes flow through a circular

orifice in a tube Phys Fluids, 17, Art No 053602.

334 Mixed Boundary Value Problems

Substituting A(k) into Equation 4.5.102 and carrying out the integration over

k, we have

u(r, z) = 2 r2− r4

2

+ 4

The procedure for computing B n is as follows: We truncate Equation

4.5.108 and Equation 4.5.109 to N linear equations with r n = 

n −1 2



∆r with ∆r = 1/N ; we truncate the m summation to M terms Inverting the

N × N system of linear equations, we employ B n in Equation 4.5.106 to find

u(r, z). Figure 4.5.6 illustrates the results when N = 50 and M = 100 when

a = 1

2.

• Example 4.5.6: Change of variables

Consider the following two-dimensional heat conduction problem that

arises during the manufacture of p-n junctions:

336 Mixed Boundary Value Problems



θ

Note that Equation 4.5.126 satisfies the boundary condition given by Equation

4.5.125 Each Fourier coefficient V n (r, s) is governed by

Transform Methods 337

we find that

U (r, θ, s) = U0

1

s − 2π

n + 3 2

where 1F1(a; b; z) is the Kummer function Reexpressing the Kummer

func-tion in terms of Bessel funcfunc-tions,113 Equation 3.5.130 can be written as

n +1 2

Figure 4.5.7illustrates Equation 4.5.132 when a2t = 1 In Example 5.2.1 we

will show how we could have solved this problem via the Wiener-Hopf method

112 Gradshteyn and Ryzhik, op cit., Formula 6.631.

113 Abramowitz, M., and I A Segun, 1968: Handbook of Mathematical Functions Dover

Publications, 1046 pp See Formula 13.3.6.

114 Reprinted from Int J Engng Sci., 6, B R Das, Thermal stresses in a long cylinder

containing a penny-shaped crack, 497–516, c1968, with permission of Elsevier.

Transform Methods 339Equation (2) is satisfied.

Step 4 : From Equation (1), show that g(t) is given by

g(t) + 2π

d dt

Step 6 : Recognizing that B(k) is the Fourier coefficient in a Fourier cosine

transform, show that

340 Mixed Boundary Value Problems

0 0.5 1 1.5 2

0 0.5 1 1.5 2 0

0.2 0.4 0.6 0.8 1 1.2

r z

Step 1 : Using separation of variables or transform methods, show that the

general solution to the problem is

115 Suggested by Sneddon, I N., 1962: Note on an electrified circular disk situated inside

a coaxial infinite hollow cylinder Proc Cambridge Philos Soc., 58, 621–624. 1962c Cambridge Philosophical Society Reprinted with the permission of Cambridge University Press.

Transform Methods 341

 ∞

0

A(k)J n (kr) dk = 0, 1 < r < a, (2)and

Step 3 : Using Equation (3) and recognizing the B(k) is the coefficient of a

Fourier cosine transform, show that

show that Equation (2) is identically satisfied

Step 5 : Using the results from Step 3 and Step 4 and following Equation

4.5.42 through Equation 4.5.46, show that

d dt

342 Mixed Boundary Value Problems

0 0.5 1 1.5 2

0 0.2 0.4 0.6 0.8

0 0.2 0.4 0.6 0.8 1

r z

116 See Rusia, K C., 1967: Some asymmetric mixed boundary value problems for a

half-space with a cylindrical cavity Indian J Pure Appl Phys., 5, 419–421. Rusia’s contribution was to simplify the solution of this problem first posed by Narain, P., 1965:

A note on an asymmetric mixed boundary value problem for a half space with a cylindrical

cavity Glasgow Math Assoc Proc., 7, 45–47 Earlier Srivastav (Srivastav, R P., 1964:

An axisymmetric mixed boundary value problem for a half-space with a cylindrical cavity.

J Math Mech., 13, 385–393.) solved this problem for the special case n = 0.

Transform Methods 343

u(r, 0) = 0, 1≤ r < a,

u z (r, 0) = −f(r), a < r < ∞.

Step 1 : Using separation of variables or transform methods, show that the

general solution to the problem is

show that Equation (1) is identically satisfied

Step 4 : Integrating by parts A(k) defined in Step 3, show that

344 Mixed Boundary Value Problems

1−n K1

−n (kt).

Step 7 : Recognizing that K n (k)B(k) is the coefficient of a Fourier sine

trans-form given by Equation (3), show that

−1.5

−0.5 0.5 1.5 2.5

Chapter 5 The Wiener-Hopf Technique

In Example 1.1.3 we posed a mixed boundary value problem that we couldnot solve by using conventional Fourier transforms because the solution does

not vanish as we approach infinity at both ends The Wiener-Hopf technique

is a popular method that avoids this problem by defining Fourier transformsover certain regions and then uses function-theoretic analysis to piece togetherthe complete solution

Although Wiener and Hopf1 first devised this method to solve singular

integral equations of the form

f (x) =

 ∞

0

K(x − y)f(y) dy + ϕ(x), 0 < x < ∞, (5.0.1)

that had arisen in Hopf’s 1928 work on the Milne–Schwarzschild equation,2

it has been in boundary value problems that it has found its greatest plicability For example, this technique reduces the problem of diffraction

ap-1 Wiener, N., and E Hopf, 1931: ¨Uber eine Klasse singul¨arer Integralgleichungen Sitz.

Ber Preuss Akad Wiss., Phys.-Math Kl., 696–706.

2 Shore, S N., 2002: The evolution of radiative transfer theory from atmospheres to

nuclear reactors Hist Math., 29, 463–489.

348 Mixed Boundary Value Problems

Figure 5.0.1: One of the great mathematicians of the twentieth century, Norbert Wiener

(1894–1964) graduated from high school at the age of 11 and Tufts at 14 Obtaining a doctorate in mathematical logic at 18, he repeatedly traveled to Europe for further educa- tion His work extends over an extremely wide range from stochastic processes to harmonic analysis to cybernetics (Photo courtesy of the MIT Museum.)

by a semi-infinite plate to the solution of a singular integral equation.3 The

Wiener-Hopf technique4 then yields the classic result given by Sommerfeld.5

Since its original formulation, the Wiener-Hopf technique has undergonesimplification by formulating the problem in terms of dual integral equations.6

The essence of this technique is the process of factorization of the Fourier

transform of the kernel function into the product of two other Fourier forms which are analytic and nonzero in certain half planes

trans-Before we plunge into the use of the Wiener-Hopf technique for solvingpartial differential equations, let us focus our attention on the mechanics ofthe method itself To this end, let us solve the integral equation

6 Kaup, S N., 1950: Wiener-Hopf techniques and mixed boundary value problems.

Comm Pure Appl Math., 3, 411–426; Clemmow, P C., 1951: A method for the exact

solution of a class of two-dimensional diffraction problems Proc R Soc London, Ser A,

205, 286–308 See Noble, B., 1958: Methods Based on the Wiener-Hopf Technique for the

Solution of Partial Differential Equations Pergamon Press, 246 pp.