Mixed Boundary Value Problems Episode 11 docx

Transform Methods 291Finally Selvadurai86solved this problem as a system of integral equations.. Using transform methods or separation of variables, the general solution to Equation 4.4.

where a < 1.

Using transform methods or separation of variables, the general solution

to Equation 4.4.27, Equation 4.4.28, and Equation 4.4.29 is

To solve this set of integral equations, we let A(k) = B(k) + D(k) Then,

Equation 4.4.32 through Equation 4.4.34 can be rewritten

Equation 4.4.36 and Equation 4.4.37 are automatically satisfied if we

define B(k) and D(k) as follows:

d dt

for 0≤ r < a Here we have used Equation 1.4.13 In a similar manner, it is

readily shown that

d dt

 ∞

1

ψ(τ ) d dt

 ∞

1

τ ψ(τ )

0 0.5 1 1.5 2

0

0.5 1 1.5 2

Figure 4.4.1: The solution to Equation 4.4.27 subject to the mixed boundary conditions

given by Equation 4.4.28 through Equation 4.4.30 when a = 0.5.

In a similar manner, we also find that

If we introduce t = ax, τ = ay, Φ(x) = πφ(t)/(2V ), and Ψ(y) = πψ(t)/(2V ),

Equation 4.4.52 and Equation 4.4.53 can be combined to yield

Transform Methods 291

Finally Selvadurai86solved this problem as a system of integral equations.

In the present case we have

(4.4.61)

where b > a.

Using transform methods or separation of variables, the general solution

to Equation 4.4.58, Equation 4.4.59, and Equation 4.4.60 is

To solve this set of integral equations, we introduce two new functions

f (r) and g(r) such that

with permission from Elsevier.

Transform Methods 293Setting

F (s) =

 a s

where we interchanged the order of integration and set s2−r2= y2 Carrying

out the integration in y and simplifying, we finally obtain

Carrying out the integration and differentiation within the wavy brackets, weobtain

Upon solving the dual Fredholm integral equations, Equation 4.4.80 and

Equa-tion 4.4.85, we have F (r) and G(r) Next, we invert EquaEqua-tion 4.4.76 and Equation 4.4.77 to find f (r) and g(r) The Fourier coefficient A(k) follows from Equation 4.4.68 while Equation 4.4.62 yields u(r, z).

u z (r, 0) = 0, b < r < ∞,

(4.4.89)

where b > a > 0.

Using transform methods or separation of variables, the general solution

to Equation 4.4.86, Equation 4.4.87, and Equation 4.4.88 is

Fourier-Bessel coefficient A(k) Can we find some general result that might

assist us in solving these triple Fourier-Bessel equations?

In 1963 Cooke89 studied how to find A(k) governed by the following

of an annular stamp into an elastic half-space Mech Solids, 1(4), 101–103.

We can use Cooke’s results if we set ν = 0 Then, from Equation 4.4.97

through Equation 4.4.100, we have that

A(k) = k

 b a

 b r

t h(t)

√

t2− a2K(η, t) dt, (4.4.104)

 b r

K(η, t)χ(t) dt (4.4.110)

To compute u(r, z), we solve Equation 4.4.110 by replacing the integral

with its representation from the midpoint rule Setting d eta = (b-a) / N,the MATLABcode for computing χ(η) is:

for n = 1:N % rows loop (top to bottom in the matrix)

x = xi(n); b(n) = 1; % right side of the integral equationfor m = 1:N % columns loop (left to right in the matrix)

t = eta(m);

% start setting up Equation 4.4.110

if (n==m) AA(n,m) = (x-a)*(x+a)/(x*x); % first term on left sideelse AA(n,m) = 0; end

Transform Methods 297

% introduce the integral in Equation 4.4.110

temp1 = (x-a)*(x+a)/x; temp2 = (t-a)*(t+a)/t;

if (t == x)

integrand = -a/(2*x*x)+(x*x+a*a)*log((x+a)/(x-a))/(4*x*x*x);else

integrand = temp1*log((x+a)/(x-a))-temp2*log((t+a)/(t-a));integrand = integrand/(2*(x*x-t*t));

Equation 4.4.109 gives g(r) First use the midpoint rule to compute the

integral and put it in F(n) Then compute the derivative to find g(n) The

u(i,j) = u(i,j) + g(m)*integral*eta(m)*d eta;

end; end; end

Figure 4.4.2illustrates this potential when a = 1 and b = 2.

S Manko, 2004: Magnetic fields of pulsars surrounded by accretion disks of finite extension.

Astron Astrophys., 422, 587–590.

300 Mixed Boundary Value Problems

We begin our solution of Equation 4.4.117 through Equation 4.4.119 byintroducing

µ2(ξ) sin(kξ) dξ. (4.4.120)

Note that µ1(ξ) is defined over the interval [0, a] while µ2(ξ) is defined over

the interval [a, b] Substituting Equation 4.4.120 into Equation 4.4.119 and

interchanging the order of integration, we find that

µ2(ξ)

 ∞

0

sin(kξ)J0(kr) dk dξ (4.4.121)

and Equation 4.4.119 is satisfied because the integrals within the square

brack-ets vanish according to Equation 1.4.13 since r > b > a In a similar manner,

substituting Equation 4.4.120 into Equation 4.4.117 yields

µ2(ξ)



ξ2− r2dξ = K, 0≤ r < a, (4.4.122)

again by using Equation 1.4.13 and noting that r < a < b.

To solve Equation 4.4.118, we first note that

µ 

2(ξ) cos(kξ) dξ

by integrating the second integral in Equation 4.4.120 by parts ing Equation 4.4.123 into Equation 4.4.118 and interchanging the order ofintegration,

where we used tables91 to simplify the integral on the right side of Equation

4.4.125 and d[z n J n (z)] = z n J n −1 (z) dz, n = 1, 2, Using Equation 1.4.13

and Equation 4.4.126, Equation 4.4.124 becomes

where C is an arbitrary constant.

Because we can write Equation 4.4.129 as an integral equation of theAbel type:

 a r

302 Mixed Boundary Value Problems Carrying out the τ integration and then taking the r derivative,

not singular We obtained it from Equation 4.4.138 by using the relationship

which follows from Equation 4.4.133 in the limit of r → a.

The potential is computed as follows: For a specific C  , we find y(x)

from Equation 4.4.139 The corresponding value of G follows from Equation 4.4.140 By varying C  , we can compute the y(x) for a desired G We com-

pute the function µ1(ξ) from Equation 4.4.135 Finally, combining Equation

4.4.116 and Equation 4.4.120, we have that

0 1 2 3 4

0 0.5 1 1.5 2

−0.2

−0.1

0 0.1

2 u(r,z)/A

Figure 4.4.3: The solution to Equation 4.4.112 subject to the mixed boundary conditions

We evaluate numerically the integrals inside of the square brackets (except

for the case when z = 0 where there is an exact expression) and then we compute the ξ integration Figure 4.4.3 illustrates the solution for b/a = 2 and G = 0.5.

u z (r, 0) = 0, b < r < ∞,

(4.4.145)

by the rotation of a ring-shaped punch Mech Solids, 1(1), 63–66.

304 Mixed Boundary Value Problems

where b > a > 0.

Using transform methods or separation of variables, the general solution

to Equation 4.4.142, Equation 4.4.143, and Equation 4.4.144 is

We can use Cooke’s results if we set ν = 1 Then, from Equation 4.4.97

through Equation 4.4.100, we have that

A(k) = k

 b a

 b r

0 0.5 1 1.5 2 2.5 3

0 0.5 1 1.5 2 0

Figure 4.4.4: The solution to Equation 4.4.142 subject to the mixed boundary conditions

given by Equation 4.4.143 through Equation 4.4.145 when a = 1 and b = 2.

In the special case when t = η, we employ L’Hospital rule and find that

 b r

K(η, t)χ(t) dt (4.4.159)

The evaluation of u(r, z) begins by solving Equation 4.4.159 by replacing

the integral with its representation from the midpoint rule With those values

of χ(η), Equation 4.4.158 gives g(r) Finally, combining Equation 4.4.146 and

Equation 4.4.158, we find that

u(r, z) =

 b a

306 Mixed Boundary Value Problems

Using transform methods or separation of variables, the general solution

to Equation 4.4.161, Equation 4.4.162, and Equation 4.4.163 is

ξF (ξ) J1(kξ)

annular disk in a viscous fluid J Phys Soc Japan, 54, 3337–3341.

Then, from Equation 4.4.167,

 b τ

(· · ·) dξ dτ +

 a

0

 b a

 b ξ

308 Mixed Boundary Value Problems

0 0.5 1 1.5 2 2.5 3

0

0.5 1 1.5 2

Figure 4.4.5: The solution to Equation 4.4.161 subject to the mixed boundary conditions

given by Equation 4.4.162 through Equation 4.4.164 when b/a = 2

F (ξ) and A(k) follow from Equation 4.4.176 and Equation 4.4.170, tively Finally Equation 4.4.165 provides u(r, z) Figure 4.4.5 illustrates this solution when b/a = 2.

for infinite slab Appl Sci Res., 16, 228–240 with kind permission from Springer Science

and Business Media.

Using Hankel transforms, the solution to Equation 4.4.179 and Equation4.4.180 is

u(r, z) =

 ∞

0

[A(k) cosh(kz) + B(k) sinh(kz)] J0(kr) dk (4.4.182)

Substituting Equation 4.4.182 into the mixed boundary conditions, Equation4.4.181, we find that

1 0

1 0

ψ2(t) sin(kt) dt.

(4.4.190)

310 Mixed Boundary Value Problems

Introducing Equation 4.4.187 and Equation 4.4.189 into Equation 4.4.183 and

Equation 4.4.184, multiplying the resulting equations by dη/

0 0.5 1 1.5 2

0

0.5 1 1.5 2

Figure 4.4.6: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.4.179 through Equation 4.4.181.

K2(η, t) = G2(η + t) − G2(η − t) = 2

 ∞

0

cos(ηξ/h) sin(tξ/h) cosh(ξ) dξ, (4.4.198)

K3(η, t) = G2(η + t) + G2(η − t) = 2

 ∞

0

sin(ηξ/h) cos(tξ/h) cosh(ξ) dξ, (4.4.199)

We compute u(r, z) as follows: First, we find ψ1(η) and ψ2(η) via

Equa-tion 4.4.195 and EquaEqua-tion 4.4.196 Next, A(k) and B(k) follow from EquaEqua-tion

4.4.187 and Equation 4.4.189, respectively Finally, Equation 4.4.182 yields

u(r, z) Figure 4.4.6 illustrates the solution when f (r) = 0, g(r) = 1, and

an aperture Sov Tech Phys., 17, 473–476.

312 Mixed Boundary Value Problems

subject to the boundary conditions

u(r, h) = V, 0 < r < a, u(r, h − ) = u(r, h+), a < r < ∞, (4.4.207) where b > a.

Using Hankel transforms, the solution to Equation 4.4.201 through tion 4.4.204 and Equation 4.4.206 is

Equation 4.4.217 and Equation 4.4.219 are satisfied identically if we

in-troduce a φ(t) and ψ(t) such that

Substituting for B(k) and C(k) in Equation 4.4.216 and interchanging the

order of integration, we have that

314 Mixed Boundary Value Problems

t2− r2, integrating from 0 to t, and

using Equation 1.4.9, Equation 4.4.230 transforms into

0 1 2 3 4

−2

−1 0 1 2

Figure 4.4.7: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.4.201 through Equation 4.4.207.

Upon substituting Equation 4.4.221, Equation 4.4.232 becomes

In summary, once we find φ(t) and ψ(t) by solving the simultaneous

inte-gral equations, Equation 4.4.227 and Equation 4.4.236, respectively, we can

compute A(k) and B(k) via Equation 4.4.222, Equation 4.4.223, and tion 4.4.215 Finally u(r, z) follows from Equation 4.4.208 through Equation 4.4.210 Figure 4.4.7 illustrates the solution when a = 1, b = 2, and h = 1.

pressure-driven extrusion through an annular hole in a wall J Fluid Mech., 231, 51–71.

316 Mixed Boundary Value Problems

subject to the boundary conditions

u z (r, 0) = 0, 1 < r < ∞.

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

general solution to the partial differential equation and the first two boundaryconditions is

312 Mixed Boundary Value Problems< /i>

subject to the boundary conditions

u(r, h) = V, < r < a, u(r, h −... class="text_page_counter">Trang 28

316 Mixed Boundary Value Problems< /i>

subject to the boundary conditions

u z (r, 0) = 0, < r <... class="text_page_counter">Trang 18

306 Mixed Boundary Value Problems< /i>

Using transform methods or separation of variables, the general solution