Mixed Boundary Value Problems Episode 7 doc

Next, we numerically integrate Equation 4.1.54 to find Ak.. At this point,8 we can also show how to solve Equation 4.1.46, Equation Closed-form solutions for finite length crack moving in

x = xx(n); b(n) = x; % the right side of Equation 4.1.16

psi 0 = psi 0 + k*log(abs(k));

psi 1 = psi 1 - 0.5*k*k*log(abs(k));

end

if (k ∼= 1)

psi 0 = psi 0 - (k-1)*log(abs(k-1));

psi 1 = psi 1 + 0.5*(k-1)*(k-1)*log(abs(k-1));

end

psi 1 = psi 1 + k*psi 0;

alpha = 0.5*dx*log(dx) + dx*(psi 0-psi 1);

beta = 0.5*dx*log(dx) + dx*psi 1;

AA(n,m-1) = AA(n,m-1) - alpha/pi;

AA(n, m ) = AA(n, m ) - beta/pi;

end; end

f = AA\b’ % Compute f(x) from Equation 4.1.16

Having found f (x) from Equation 4.1.16, g(t) follows from Equation 4.1.15.

% **************************************************************

% Compute A(k) from Equation 4.1.8.

% Use Simpson’s rule

% **************************************************************for n = 1:N

−1 0 1 2

0 1 2 3 4

Figure 4.1.1: The solution to Equation 4.1.1 subject to the mixed boundary conditions

given by Equation 4.1.2 through Equation 4.1.4 when h = 1.

end

% **************************************************************

% Compute the solution u(x, y) for a given x and y.

% Use Simpson’s rule

% **************************************************************u(i,j) = 0;

% integral contribution from wavenumber

% between k = 0 and k = K max

Using separation of variables or transform methods, the general

solu-tion to Equasolu-tion 4.1.18, Equasolu-tion 4.1.19 and Equasolu-tion 4.1.20 is u(x, y) =

u+(x, y) + u − (x, y), where

u+(x, y) =

 ∞0

A(k)e −ky cos(kx) dk

and

u − (x, y) =

 ∞0

B(k)e −ky sin(kx) dk

The idea here is that the solution consists of two parts: an even portion

denoted by u+(x, y) and an odd portion denoted by u − (x, y) In a similar

for mixed boundary conditions Sov Tech Phys., 11, 996–999.

g 

+(t)

 ∞0

cos(kx)J0(k) dk dt

−

 10

g 

+(t)

 ∞0

cos(kx)J0(kt) dk dt = 0. (4.1.27)

Thus, our choice for A(k) identically satisfies Equation 4.1.25 This follows

from Equation 1.4.14 since|x| > 1 and 0 ≤ t ≤ 1.

Next, we integrate Equation 4.1.26 by parts and find that

A(k) = −k

 10

g+(τ )

 ∞0

g+(τ )

 ∞0

g+(τ )

 ∞0

g+(τ )

 ∞0

F (π/2, ·), and E(·) = E(π/2, ·).

Turning to B(k), we substitute Equation 4.1.23 into Equation 4.1.21 and

t g 

− (t)J1(kt) dt = k

 10

tg 

− (t)

 ∞0

τ g − (τ )

 ∞0

sin(kx)J0(kτ ) dk dτ. (4.1.41)

−1 0 1 2

0 0.5 1 1.5 2

Figure 4.1.2: The solution to Equation 4.1.18 subject to the mixed boundary conditions

given by Equation 4.1.19, Equation 4.1.20, and Equation 4.1.21 when f+(x) = 0, f − (x) = x and h = 1.

sin(kx)J0(kτ ) dk √ dx

x2− t2 (4.1.44)

= hτ

 10

u y (x, h) = 0, −∞ < x < ∞ (4.1.49)

Using separation of variables or transform methods, the general solution

to Equation 4.1.46, Equation 4.1.47 and Equation 4.1.49 is

u(x, y) = 2

π

 ∞0

A(k) cosh[k(y − h)] cos(kx) dk (4.1.50)

Direct substitution of Equation 4.1.50 into Equation 4.1.48 yields the dualintegral equations:

A(k) cosh(kh) cos(kx) dk = 0, |x| > 1 (4.1.52)

We begin our solution of these dual integral equations by noting that for

|x| < 1,

u(x, 0) = 2

π

 ∞0

A(k) cosh(kh) cos(kx) dk (4.1.53)

with u(1, 0) = 0 Because Equation 4.1.53 is the Fourier cosine representation

of u(x, 0),

A(k) cosh(kh) =

 10

u(ξ, 0) cos(kξ) dξ = −1

k

 10

h(ξ) sin(kξ) dξ, (4.1.54)

diffusion I Straight punch J Appl Phys., 74, 4382–4389.

since u(1, 0) = 0 and h(ξ) = u ξ (ξ, 0). Substituting Equation 4.1.54 intoEquation 4.1.51,

tanh(kh) sin(kξ) sin(kx) dk

tanh(kh) {cos[k(x − ξ)] − cos[k(x + ξ)]} dk

4β , α, β > 0. (4.1.60)

Substituting Equation 4.1.59 into Equation 4.1.57 and integrating, we obtainthe integral equation

 10

h(ξ) ln

sinh(βx) + sinh(βξ) sinh(βx) − sinh(βξ)



 dξ = πAx (4.1.61)

If we define

γ = sinh(βξ) sinh(β) and γ0=sinh(βx)

Figure 4.1.3: The solution to Equation 4.1.46 subject to the mixed boundary conditions

given by Equation 4.1.47, Equation 4.1.48, and Equation 4.1.49 when h = 2.

where

g(γ) = sinh(β) cosh(βξ) h

 1

ξ

sinh(2βτ )

sinh2(βτ ) − sinh2(βξ)

×



 τ0

dη

sinh2(βτ ) − sinh2(βη)

×



 τ0

dη

sinh2(βτ ) − sinh2(βη)



 dτ (4.1.66)

To compute u(x, y), we first evaluate u(x, 0) Next, we numerically integrate Equation 4.1.54 to find A(k) Finally, we employ Equation 4.1.50 Figure 4.1.3 illustrates this solution when h = 2.

At this point,8 we can also show how to solve Equation 4.1.46, Equation

Closed-form solutions for finite length crack moving in a strip under anti-plane shear stress Acta

Mech., 38, 99–109.

4.1.47, and Equation 4.1.48, while modifying Equation 4.1.49 to read

u(x, h) = 0, −∞ < x < ∞ (4.1.67)

We begin once again using separation of variables or transform methods

to find the general solution to Equation 4.1.46, Equation 4.1.47 and Equation4.1.67 This now gives

u(x, y) = 2

π

 ∞0

A(k) sinh[k(y − h)] cos(kx) dk (4.1.68)

Direct substitution of Equation 4.1.68 into Equation 4.1.48 yields the dualintegral equations:

A(k) sinh(kh) cos(kx) dk = 0, |x| > 1 (4.1.70)

To solve Equation 4.1.69 and Equation 4.1.70, we introduce

A(k) sinh(kh) =

 10

coth(kh) sin(kξ) sin(kx) dk

coth(kh) {cos[k(x − ξ)] − cos[k(x + ξ)]} dk

 dξ = πAx (4.1.78)

If we define

γ = tanh(βξ) tanh(β) and γ0=tanh(βx) tanh(β) , (4.1.79)

we find that Equation 4.1.78 becomes

dη

tanh2(βτ ) − tanh2(βη)

Konishi and Atsumi10 have given an alternative method of attackingthe problem given by Equation 4.1.46 through Equation 4.1.48 and Equation4.1.67 For clarity let us restate the problem:

flow disturbed by a two-dimensional crack in a strip Int J Engng Sci., 11, 1–7.

To solve these dual equations, we set

cos(kx)J0(kt) dk dt = 0,

(4.1.91)

where we used integral tables11 with 0≤ t ≤ a < |x| < ∞.

Turning to Equation 4.1.80, we have

 a

0

h(t)

 ∞0

Now, from integral tables,12

and

u y (x, h) = 0, −∞ < x < ∞, (4.1.103)

where p(x) is an even function.

We begin by applying Fourier cosine transforms to solve Equation 4.1.100.This yields the solution

u(x, y) = 2

π

 ∞0

Equation 4.1.104 satisfies not only Equation 4.1.100, but also Equation 4.1.101and Equation 4.1.103 Substituting Equation 4.1.104 into Equation 4.1.102,

we obtain

2

π

 ∞0

sin(kτ ) cos(kx) dk

k dτ = 0,(4.1.109)

where the integral14 within the square brackets vanishes since |x| > a and

0≤ τ ≤ a Thus, Equation 4.1.107 satisfies Equation 4.1.105 identically.

We now turn our attention to Equation 4.1.106 Substituting Equation4.1.107 into Equation 4.1.106, we have that

 ∞

0 tanh(kh)

 a

0 g(τ ) sin(kτ ) dτ cos(kx) dk = p(x), (4.1.110)or

sinh(cx) + sinh(cτ ) sinh(cx) − sinh(cτ), (4.1.112)

where c = π/(2h) Therefore, substituting Equation 4.1.112 into Equation

−2

−1 0 1 2 3

0 0.5 1 1.5 2

−0.2

0 0.2

0.4

0.6

0.8

1 1.2

x y

Figure 4.1.5: The solution to Equation 4.1.100 subject to the mixed boundary conditions

given by Equation 4.1.101, Equation 4.1.102, and Equation 4.1.103 when a = 1 and h = 2.

Using the results from Example 1.2.3, the solution to the integral equationEquation 4.1.113 is

sinh2(ca) − sinh2(cτ ) , 0 < τ < a. (4.1.114)

Because F  (x) = 2p(x) with F (0) = 0, Equation 4.1.114 simplifies to

problems Theoret Appl Fract Mech., 26, 211–217.

with the boundary conditions

=−

 10

g(t) d dx

 10

h  (τ )J

0(kτ ) dτ cos(kx) dk (4.1.144)

=

 10

We now turn our attention to Equation 4.1.142 Substituting Equation

4.1.143 into Equation 4.1.142, we have, after integrating with respect to x,

 ∞

0

 10

0 0.5 1 1.5

−0.5 0 0.5 1

−0.5

0 0.5

1 1.5

y x

Figure 4.1.7: The solution to Equation 4.1.133 subject to the mixed boundary conditions

given by Equation 4.1.134 through Equation 4.1.138 when q(x) = h = l.

after integrating with respect to x Using integral tables,18 Equation 4.1.150simplifies to

To compute the potential, we first find h  (t) via Equation 4.1.151 Then

C(k) or A(k) follows from 4.1.143 Finally, we employ Equation 4.1.139 and Equation 4.1.140 Figure 4.1.7 illustrates this solution when q(x) = h = 1.

• Example 4.1.7

In this problem we find the solution to Laplace’s equation19in the upper

half-plane z > 0 into which we insert a semi-circular cylinder of radius a that

has a potential of 1 SeeFigure 4.1.8 Mathematically this problem is

lens consisting of two semi-infinite cylinders Sov Tech Phys., 7, 501–506.

if z < b, lim r →∞ u(r, z) → 0, and |u(r, z)| < ∞ as z → ∞ Thus the first

integrals are particular solutions to our problem while the second integralsare homogeneous solutions

Upon substituting Equation 4.1.157 and Equation 4.1.158 into Equation4.1.156, we obtain the dual integral equations

[1− cos(kz)] cos(kb) g(k) (π2/4 − k2b2)

dk

and g(k) = 2aI0(ka)K0(ka) To solve these dual integral equations, we

intro-duce a function ψ(t) such that

A(k) =

 b

0

t ψ(t)J1(kt) dt (4.1.162)

If we substitute Equation 4.1.162 into Equation 4.1.160, interchange the order

of integration, and then use integral tables,20we can show that this choice for

A(k) satisfies Equation 4.1.160 identically.

Upon substituting Equation 4.1.162 into Equation 4.1.159,

 ∞0

[1− cos(kz)] cos(kb) g(k) (π2/4 − k2b2)

dk

k2.

If we now interchange the order of integration in Equation 4.1.165 and useintegral tables,21

1− cos(kz)

kz h(k) k J1(kt) dk dt

= π2

 ∞0

[1− cos(kz)] cos(kb)

k2z g(k)(π2/4 − k2b2)dk (4.1.166)Because22

1− cos(x)

 π/2 0

h(k)J1[kz sin(θ)]J1(kt) k dk dt dθ

=π2

 π/2 0

 ∞0

cos(kb)J1(kx) g(k)(π2/4 − k2b2)

From the form of the boundary conditions along x = 0 and y = 0,

we anticipate that we should use Fourier cosine transforms Therefore, thesolution to Equation 4.1.172 through Equation 4.1.174 is

k J1(k) cos(kx) dk

−

 10

k cos(kx)J0(kt) dk dt

+

 10

t h(t)

 ∞0

k e −kx J

0(kt) dk dt = f (x), (4.1.184)

0 0.5 1 1.5 2

0 0.5 1 1.5 2

Figure 4.1.10: The solution to Equation 4.1.172 subject to the mixed boundary conditions

given by Equation 4.1.173 through Equation 4.1.176.

sin(kx)J0(kt) dk dt



+

 10

t h(t)

 ∞0

t h(t) (x2+ t2)3/2 dt = f (x), 0≤ x < 1 (4.1.186)

Using Equation 1.2.13 and Equation 1.2.14, we can solve for h(t) and find

that

h(t) +

 10

In the case when f (x) = 1, Equation 4.1.187 simplifies to

h(t) +

 10

K(t, τ )h(τ ) dτ = 1 (4.1.189)

Figure 4.1.10 illustrates the solution

general solution to the problem is

u(x, y) =

 ∞0

A(k)e −ky cos(kx) dk.

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

inte-gral equations

 ∞0

tJ0(kt) dt = − πJ1(k)

2k

satisfies both integral equations given in Step 2

Transient singular stresses of a finite crack in an elastic conductor under electromagnetic

force (in Japanese) Nihon Kikai Gakkai Rombunshu (Trans Japan Soc Mech Engrs.),

Ser A, 56, 278–282.

x y

J1(k)e −ky cos(kx) dk k .

In particular, verify that u(x, 0) = − √1− x2 if |x| < 1 The figure labeled Problem 1 illustrates the solution u(x, y).

2 Solve Laplace’s equation

problem is

u(x, y) =

 ∞0

A(k)e −ky sin(kx)

k dk.

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

inte-gral equations

 ∞0

A(k)

k sin(kx) dk = x, 0≤ |x| < 1,

problems Theoret Appl Fract Mech., 26, 211–2 17.