Mixed Boundary Value Problems Episode 8 ppsx

W., 1958: Solution of a pair of integral equations from elastostatics.. Step 2 : Using boundary condition 1, show that Ak satisfies the dual show that the second integral equation in Ste

−4 −2

4

0 1 2

−1

−0.5 0 0.5 1

x y

The figure labeled Problem 2 illustrates this solution u(x, y).

27 Fredricks, R W., 1958: Solution of a pair of integral equations from elastostatics.

Proc. Natl Acad Sci., 44, 309–312. See also Sneddon, I N., 1962: Dual integral

equations with trigonometrical kernels Proc Glasgow Math Assoc., 5, 147–152.

3 Following Example 4.1.3, solve Laplace’s equation

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

general solution to the problem is

u(x, y) = 2

π

 ∞

0

A(k) sinh[k(h − y)] cos(kx) dk.

Step 2 : Using boundary condition (1), show that A(k) satisfies the dual

show that the second integral equation in Step 2 is identically satisfied

Step 4 : Show that the first integral equation in Step 2 leads to the integral

Step 5 : Simplify your results in the case g(x) = 1 and show that they are

identical with the results given in Example 4.1.3 by Equation 4.1.83 throughEquation 4.1.86

4 Following Example 4.1.2, solve Laplace’s equation28

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

general solution to the problem is

show that the second integral equation in Step 2 is identically satisfied

Step 4 : Show that the first integral equation in Step 2 can be rewritten d

28 Suggested from Singh, B M., T B Moodie, and J B Haddow, 1981: Closed-form

solutions for finite length crack moving in a strip under anti-plane shear stress Acta Mech.,

38, 99–109.

−0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

x y



, where c = π/(2h), show that Step 4 can be written

the integral equation in Step 5 is

tanh2(ca) − tanh2(cτ ) , 0 < τ < a.

Step 7 : Because F  (x) = 2p(x)/h with F (0) = 0, show that Step 6 simplifies

The figure entitled Problem 4 illustrates this special solution when a = 1 and

h = 2.

4.2 TRIPLE FOURIER INTEGRALS

In the previous section we considered the case where the mixed boundaryconditions led to two integral equations that are in the form of a Fourier inte-gral In the present section we take the next step and examine the situationwhere the mixed boundary condition contains different boundary conditionsalong three segments

The interesting aspect of this problem is the boundary condition along y = 0.

For a portion of the boundary (−b < x < −a and a < x < b), it consists of a

Dirichlet condition; otherwise, it is a Neumann condition

If we employ separation of variables or transform methods, the mostgeneral solution is

We must now solve for A(k) which appears in a set of integral equations.

Tranter29showed that triple integral equations of the form given by Equation

4.2.7 through Equation 4.2.9 have the solution

sin

n −1 2



ϕ

= 1, γ < ϕ ≤ π, (4.2.12)

and γ is defined by a = b sin(γ/2), 0 < γ ≤ π If we now introduce the change

of variables θ = π − ϕ and c = π − γ, we find that A n is the solution of thefollowing pair of dual series:

cos

n −1 2

Boundary Value Problems in Potential Theory Wiley, 283 pp.

30 Tranter, C J., 1959: Dual trigonometric series Proc Glasgow Math Assoc., 4,49–57; Sneddon, op cit., Section 5.4.5.

−2 0 2 4

0 0.5 1 1.5 2

−1

−0.5 0 0.5 1

x y

substi- ∞

0

A(k) coth(kπ) sin(kx) dk = 0, 0 < x < a, (4.2.22)

A(k) coth(kπ) sin(kx) dk = 0, b < x < ∞ (4.2.24)

Singh31 showed that the solution to

g[cosh(ξ)] sin(kξ) cosh(ξ/2) dξ

×

 b a

(

cosh(b) − cosh(ξ) cosh(ξ) − cosh(a)

R  (ξ) sinh(ξ/2)

cosh(ξ) − cosh(η) dξ

[cosh(η) − cosh(a)][cosh(b) − cosh(η)] . (4.2.31)

31 Singh, B M., 1973: On triple trigonometrical equations. Glasgow Math J., 14,174–178.

−3

−1 1 3 5

0 0.2 0.4 0.6 0.8 1

−1

−0.5 0 0.5 1

x y/ π

[cosh(η) − cosh(a)][cosh(b) − cosh(η)] . (4.2.32)

From Equation 4.2.29 and Equation 4.2.30, we find that

 b

a

g[cosh(ξ)] cosh(ξ/2) ln

sinh(ξ/2) + sinh(η/2) sinh(ξ/2) − sinh(η/2)

coth(kπ)A(k) = 2

π

 b a

g[cosh(ξ)] sin(kξ) cosh(ξ/2) dξ (4.2.35)

Finally, the potential u(x, y) follows from Equation 4.2.21 Figure 4.2.2

illus-trates the present example

32 See Singh, B M., and R S Dhaliwal, 1984: Closed form solutions to dynamic punch

problems by integral transform method Z Angew Math Mech., 64, 31–34.

with the boundary conditions

kA(k) cos(kx) dk = g(x) sinh(cx), a < x < b, (4.2.44)

where g(x) is an unknown function and c = π/(2h) Using Fourier’s inversion

theorem,

kA(k) =

 b a

g(τ ) sinh(cτ ) cos(kτ ) dτ (4.2.45)

If we substitute Equation 4.2.45 into Equation 4.2.42, interchange the order

of integration in the resulting equation, and use

−2 0 2 4

0 0.5 1 1.5 2

Figure 4.2.3: The solution to Equation 4.2.36 subject to the mixed boundary conditions

given by Equation 4.2.37, Equation 4.2.38, and Equation 4.2.39 when a = 1, b = 2, h = 2 and f (x) = f0

Taking the derivative with respect to x of Equation 4.1.47, we find that

×

 b a

c cosh(cx)f  (x)

cosh(2cx) − cosh(2cτ)

(

cosh(2cb) − cosh(2cx) cosh(2cx) − cosh(2ca) dx

[cosh(2cτ ) − cosh(2ca)][cosh(2cb) − cosh(2cτ)] . (4.2.49) The constant B is found by substituting Equation 4.2.49 into Equation 4.2.47 and solving for B Figure 4.2.3 illustrates this special case when a = 1, b = 2,

u y (x, 0) = 0, b < |x|.

(4.2.50)

In the present case, the solution is given by

kA(k) sin(kx) dk = ϕ(x) cosh(cx), a < x < b, (4.2.54)

where ϕ(x) is an unknown function and c = π/(2h) Using Fourier’s inversion

theorem,

kA(k) =

 b a

ϕ(τ ) cosh(cτ ) sin(kτ ) dτ (4.2.55)

If we substitute Equation 4.2.55 into Equation 4.2.53, interchange the order

of integration in the resulting equation, and simplifying, ϕ(τ ) is given by the



 dτ = πf(x), a < x < b.

(4.2.56)

Finally, taking the derivative with respect to x of Equation 4.2.56 and solving

the resulting equation, we find that

ϕ(τ ) = −4

π

(

cosh(2cτ ) − cosh(2ca) cosh(2cb) − cosh(2cτ)

×

 b a

c sinh(cx)f  (x)

cosh(2cx) − cosh(2ca)

(

cosh(2cb) − cosh(2cx) cosh(2cx) − cosh(2ca) dx

[cosh(2cτ ) − cosh(2ca)][cosh(2cb) − cosh(2cτ)] . (4.2.57) The constant B is found by substituting Equation 4.2.57 into Equation 4.2.56 and solving for B. Figure 4.2.4illustrates this special case when a = 1, b = 2,

h = 2 and f (x) = f0.

Using Hankel transforms, the solution to Equation 4.3.1 is

in-2.4.30 Noting that ν = 0 and α = −1 and associating his f(y) with k2A(k),

the solution to Equation 4.3.6 and Equation 4.3.7 is

An alternative form33of expressing the solution to Equation 4.3.1 through

Equation 4.3.4 follows by rewriting Equation 4.3.5 as

d dt

Recently, Fu et al.34 showed that if f (r) is a smooth function in 0 ≤ r < 1,

so that it can be experienced as the Maclaurin series

34 Fu, G., T Cao, and L Cao, 2005: On the evaluation of the dopant concentration

of a three-dimensional steady-state constant-source diffusion problem Mater Lett., 59,

3018–3020.

where G(k) is a known function of k. In 1956, Cooke35 proved that the

solution to Equation 4.3.24 and Equation 4.3.25 is

A(k) = 2

β Γ(ν + 1) Γ(ν − β + 1) k 1+β

Here, a, α and β are at our disposal as long as 0 < (β) < 1 and −1 <

(ν − β) Cooke suggested that we choose a and β so that G(k) closely approximates a −1 k −2β The following examples illustrate the use of Cooke’s

method for solving Equation 4.3.24 and Equation 4.3.25 when they arise inmixed boundary value problems

35 Cooke, J C., 1956: A solution of Tranter’s dual integral equations problem Quart.

J Mech Appl Math., 9, 103–110.

36 See Leong, M S., S C Choo, and K H Tay, 1976: The resistance of an infinite

slab with a disc electrode as a mixed boundary value problem Solid-State Electron., 19,

397–401 See also Belmont, B., and M Shur, 1993: Spreading resistance of a round ohmic

contact Solid-State Electron., 36, 143–146.

subject to the boundary conditions

To derive Equation 4.3.38, we used the relation that

J

− 12 (z) =

#2

Because f (t) is an even function of t, we can rewrite Equation 4.3.40 in the

more compact form of

Equation 4.3.43 cannot be solved analytically and we must employ ical techniques Let us use MATLABand show how this is done We introduce

numer-nodal points at t j = (j − N/2)∆t, where j = 0, 1, , N, and ∆t = 2/N.

Therefore, the first thing that we do in the MATLAB code is to compute the

AA(n+1,m+1) = AA(n+1,m+1) + 4*coeff*dt / (3*pi);

end % end of inside logic loop

else

AA(n+1,m+1) = AA(n+1,m+1) + 2*coeff*dt / (3*pi);

end % end of outside logic loop

end % end of m loop

end % end of n loop

for n = 0:N2

t2(n+1) = n*dt; f2(n+1) = f(n+N2+1);

We are now ready to compute the solution u(r, z) For a given r and z, the

solution u is computed as follows:

Note that because the contribution from k = 0 is zero, we simply did not

consider that case Figure 4.3.1 illustrates u(r, z) when a = 1 Schwarzbek

u(r, a) = 0, 0≤ r < ∞, (4.3.46)

u z (r, 0) = 1/a, 0≤ r < 1, u(r, 0) = 0, 1≤ r < ∞ (4.3.47)

Using Hankel functions, the solution to Equation 4.3.44 is

Our solution of the dual integral equations, Equation 4.3.49 and Equation

4.3.50, begins by introducing the undetermined function h(t) defined by

where q(k) = 1 − coth(ka) Because

0 0.5 1 1.5 2

0 0.2 0.4 0.6 0.8 1

Figure 4.3.2: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.3.44 through Equation 4.3.47 with a = 1.

As in the previous example, we must solve for h(x) numerically The MATLAB

code is very similar with the exception that the kernel in Equation 4.3.43 isreplaced with a numerical integration of "∞

0 q(k) sin(kx) sin(kt) dk by using

Simpson’s rule This integral is easily evaluated due to the nature of q(k) Figure 4.3.2 illustrates the solution u(r, z) when a = 1.

Let us now generalize our results We now wish to solve

The solution to Equation 4.3.67 through Equation 4.3.70 is

Substituting Equation 4.3.76 into Equation 4.3.77 and interchanging the order

of integration, we have that

Applying Equation 1.2.13 and Equation 1.2.14,

40 See Lebedev, N N., 1957: The electrostatic field of an immersion electron lens formed

by two diaphragms Sov Tech Phys., 2, 1943–1950.

and

u z (r, b − ) = u z (r, b+), 0≤ r < a, u(r, b) = 1, a < r < ∞, (4.3.90) where b − and b+ denote points located slightly below and above the point

z = b > 0.

Using transform methods or separation of variables, the general solution

to Equation 4.3.86 through Equation 4.3.89 is

We can rewrite Equation 4.3.98 as

1− e −2kb cos{k[t − r sin(θ)]} − cos{k[a − r sin(θ)]}

+ cos{k[t + r sin(θ)]} − cos{k[t + r sin(θ)]}

0 0.5 1 1.5 2

0 0.5 1 1.5 2

Figure 4.3.3: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.3.86 through Equation 4.3.90 with a = b = 1.

42 See Srivastav, R P., and P Narain, 1966: Stress distribution due to pressurized

exterior crack in an infinite isotropic elastic medium with coaxial cylindrical cavity Int J.

Engng Sci., 4, 689–697.

Using transform methods or separation of variables, the general solution

to Equation 4.3.108, Equation 4.3.109, and Equation 4.3.110 is

xf (x)[J0(xy)Y1(y) − Y0(xy)J1(y)] dx (4.3.116)

has the solution

f (x) =

 ∞

0

yF (y) J0(xy)Y1(y) − Y0(xy)J1(y)

Y2(y) + J2(y) dy. (4.3.117)

Therefore, from Equation 4.3.113 and Equation 4.3.115,

43 Titchmarsh, E C., 1946: Eigenfunction Expansions Associated with Second Order

Differential Equations Part I Oxford, 203 pp See Section 4.10.