Mixed Boundary Value Problems Episode 3 pps

InChapter 4 we will discuss the numerical procedure used to solve Equation 2.2.25 and Equation 2.2.26 for specific values of L, V1, and V2.. Historical Background 51occurred while solving

Using separation of variables or transform methods, the solution to tion 2.2.1 is

48 Mixed Boundary Value Problems

because r > 1 and we used Equation 1.4.13 A similar demonstration holds for B(k).

We now turn to the solution of Equation 2.2.7 Substituting Equation2.2.11 and Equation 2.2.12 into Equation 2.2.7, we find that

π

d dt

d dt

By using Equation 2.2.8, we can show in a similar manner that

g(t) +

 1

0

K(t, τ )f (τ ) dτ = V2 (2.2.26)

To evaluate u(r, z) in terms of f (t) and g(t), we substitute Equation 2.2.11

and Equation 2.2.12 into Equation 2.2.6 and find that

where we used Equation 1.4.10

Figure 2.2.2illustrates u(r, z) when L = 1 and V1=−V2= 1 InChapter

4 we will discuss the numerical procedure used to solve Equation 2.2.25 and

Equation 2.2.26 for specific values of L, V1, and V2 We used Simpson’s rule

Hafen did not present any numerical computations

The second special case occurs when we set V1 = −V2 = V0 For this

special case, Equation 2.2.25 and Equation 2.2.26 have the solution f (t) =

Historical Background 51occurred while solving the potential problem:

condi-We must next satisfy the mixed boundary value condition given by

Equa-tion 2.3.4 Noting the symmetry about the x-axis, we find that we must only consider that 0 < θ < π Turning to the 0 < θ < π/2 case, we substitute

Equation 2.3.5 and Equation 2.3.6 into Equation 2.3.4 and obtain

Let us now turn to the boundary condition for π/2 < θ < π Substituting

Equation 2.3.5 or Equation 2.3.6 into Equation 2.3.4, we find that

∞



n=1

C n sin(nθ) = V sin(θ) (2.3.9)

In summary then, we solved our mixed boundary value problem, provided

that the C n’s satisfy Equation 2.3.8 and Equation 2.3.9

It was the dual Fourier series given by Equation 2.3.8 and Equation 2.3.9that motivated W M Shepherd12 in the 1930s to study dual Fourier series of

How can we solve Equation 2.3.7 and Equation 2.3.8 by using Shepherd’s

results? The difficulty is the (κ − 1)/(κ + 1) term in Equation 2.3.8 To

12 Shepherd, W M., 1937: On trigonometrical series with mixed conditions Proc

Lon-don Math Soc., Ser 2 , 43, 366–375.

either Equation 2.3.19 or 2.3.22, we equate the coefficients of each harmonicand find that

is the assumption that a large number of small cracks exist in the interior

of the solid body Whether these “Griffith cracks” spread depends upon thedistribution of the stresses about the crack Our interest in computing thisstress field lies in the fact that many fracture problems contain mixed bound-ary conditions

1 + σ V = Σ y − σ(Σ x+ Σy ), (2.4.10)

and

E 2(1 + σ)

τ xy (0, y) = 0, −∞ < y < ∞, (2.4.14)

σ x (0, y) = −p(y), |y| < c, u(0, y) = 0, |y| > c, (2.4.15)

56 Mixed Boundary Value Problems where p(y) is a known even function of y Because

How do we solve the dual integral equations, Equation 2.4.22 and tion 2.4.23? We begin by introducing

Equa-k = ρ/c, g(η) = c

#

π 2η p(cη), y = cη, (2.4.24)

Figure 2.4.1: Most of Ian Naismith Sneddon’s (1919–2000) life involved the University

of Glasgow Entering at age 16, he graduated with undergraduate degrees in mathematics and physics, returned as a lecturer in physics from 1946 to 1951, and finally accepted the Simon Chair in Mathematics in 1956 In addition to his numerous papers, primarily on elasticity, Sneddon published notable texts on elasticity, mixed boundary value problems, and Fourier transforms (Portrait from Godfrey Argent Studio, London.)

58 Mixed Boundary Value Problems

In the present case, α = 1, ν = −1

2 and Equation 2.4.30 simplifies to

F (ρ) =

#

2ρ π

A simple illustration of this solution occurs if p(y) = p0 for all y Then,

P (y) = p0cJ1(ck) and u(0, y) = 2(1 − ν2)

c2− y2/E In this case, the crack

has the shape of an ellipse with semi-axes of 2(1− ν2)p0/E and c.

2.5 THE BOUNDARY VALUE PROBLEM OF REISSNER AND SAGOCI

Mixed boundary value problems often appear in elasticity problems Anearly example involved finding the distribution of stress within a semi-infinite

elastic medium when a load is applied to the surface z = 0 Reissner and

Sagoci15 used separation of variables and spheroidal coordinates In 1947

Sneddon16 resolved the static (time-independent) problems applying Hankel

transforms This is the approach that we will highlight here

If u(r, z) denotes the circumferential displacement, the mathematical

the-ory of elasticity yields the governing equation

14 See Section 12 in Sneddon, I N., 1995: Fourier Transforms Dover, 542pp.

15 Reissner, E., and H F Sagoci, 1944: Forced torsional oscillation of an elastic

half-space J Appl Phys., 15, 652–654; Sagoci, H F., 1944: Forced torsional oscillation of an elastic half-space II J Appl Phys., 15, 655–662.

16 Sneddon, I N., 1947: Note on a boundary value problem of Reissner and Sagoci J.

Appl Phys., 18, 130–132; Rahimian, M., A K Ghorbani-Tanha, and M Eskandari-Ghadi,

2006: The Reissner-Sagoci problem for a transversely isotropic half-space Int J Numer.

Anal Methods Geomech., 30, 1063–1074.

subject to the boundary conditions

60 Mixed Boundary Value Problems

0 0.5 1 1.5 2

0 0.2 0.4 0.6 0.8 1

Figure 2.5.1: The solution u(r, z)/a to the mixed boundary value problem governed by

Equation 2.5.1 through Equation 2.5.4.

We can evaluate the integral in Equation 2.5.12 and it simplifies to

We illustrate Equation 2.5.13 in Figure 2.5.1.

We can generalize17the original Reissner-Sagoci equation so that it now

17 See Chakraborty, S., D S Ray and A Chakravarty, 1996: A dynamical problem

of Reissner-Sagoci type for a non-homogeneous elastic half-space Indian J Pure Appl.

Math., 27, 795–806.

subject to the boundary conditions

k2+ a2 and a2 = K2/4 − ω2 Upon substituting

Equation 2.5.19 into Equation 2.5.18, we have that

where B(k) = 2κ(k)A(k) and M (k) = k/κ(k) − 1.

To solve the dual integral equations, Equation 2.5.22 and Equation 2.5.23,

62 Mixed Boundary Value Problems

Turning to Equation 2.5.22, we now substitute Equation 2.5.24 and terchange the order of integration:

after using integral tables.18 Substituting Equation 2.5.30 and Equation 2.5.31

into Equation 2.5.29, we finally obtain

Equation 2.5.32 must be solved numerically We examine this in detail in

Section 4.3 Once h(t) is computed, B(k) and A(k) follow from Equation 2.5.24 Finally Equation 2.5.19 gives the solution u(r, z) We illustrate this

solution inFigure 2.5.2

18 Gradshteyn, I S., and I M Ryzhik, 1965: Table of Integrals, Series, and Products.

Academic Press, Formula 6.567.1 with ν = 1 and µ = −1

2.

Upon substituting Equation 2.5.37 into Equation 2.5.36, we obtain the tripleintegral equations:

(· · ·) dτ dt +

 a

0

 b a

 t a

(· · ·) dτ dt +

 ∞

b

 b a

for 0 < r < a Then, applying Equation 2.5.48 and interchanging the order

of integration in the second integral, we obtain

τ2f2(τ )

√

t2− τ2dτ, b < t < ∞ (2.5.53)

If we regard the right side of Equation 2.5.51 as a known function of r, then it

is an integral equation of the Abel type From Equation 1.2.13 and Equation1.2.14, its solution is

Turning to Equation 2.5.44, we employ Equation 2.5.46 and Equation2.5.47 and find that

if b < r < ∞ If we now apply Equation 2.5.49, interchange the order of

inte-gration in the second integral and use Equation 2.5.52 and Equation 2.5.53,

Once we find X1(at1) and X2(bt1) via Equation 2.5.62 and Equation

2.5.63, we can find f1(r) and f2(r) from

f1(r) = −2

π

d dr

 a r

t F1(t)

√

t2− r2dt , 0 < r < a, (2.5.64)

Historical Background 67

0 0.5 1 1.5 2 2.5 3

0

0.2

0.4 0.6 0.8 1

Figure 2.5.3: The solution u(r, z)/a to the mixed boundary value problem governed by

Equation 2.5.33 through Equation 2.5.36 when c =1.

and

f2(r) = πr22dr d

 r b

2.

In the previous examples of the Reissner-Sagoci problem, we solved it in

the half-space z > 0 Here we solve this problem21within a cylinder of radius

b when the shear modulus of the material varies as µ0z α, where 0 ≤ α < 1.

Mathematically the problem is

∂u

∂z = 0, 0≤ r < b, 0 < z < ∞, (2.5.67)

21 Reprinted from Int J Engng Sci., 8, M K Kassir, The Reissner-Sagoci problem

for a non-homogeneous solid, 875–885, c1970, with permission from Elsevier.

subject to the boundary conditions

Using separation of variables, the solution to Equation 2.5.67 throughEquation 2.5.69 is

22 Sneddon, I N., and R P Srivastav, 1966: Dual series relations I Dual relations

involving Fourier-Bessel series Proc R Soc Edinburgh, Ser A, 66, 150–160.

Historical Background 69

0 0.5 1 1.5 2

0 0.5 1 1.5 2

Figure 2.5.4: The solution u(r, z)/ω to the mixed boundary value problem governed by

Equation 2.5.33 through Equation 2.5.36 when a = 1, b = 2 and α =1



rf (r) dr (t2− r2)p , (2.5.78)

I1(by) I 1−p (ty)I 1−p (τ y) y dy. (2.5.79)

To illustrate our results, we choose f (r) = ωr Then

subject to the boundary conditions

Historical Background 71or

0 0.5 1 1.5

0.2 0.4 0.6 0.8 1

0 0.2 0.4 0.6 0.8 1

z r

Problem 1

Step 8 : Taking the limit as α → 0, show that you recover the solution given

by Equation 2.5.11 and Equation 2.5.12 with a = 1 and q(r) = r In the figure labeled Problem 1, we illustrate the solution when α = 1 and q(r) = r.

2.6 STEADY ROTATION OF A CIRCULAR DISC

A problem that is similar to the Reissner-Sagoci problem involves findingthe steady-state velocity field within a laminar, infinitely deep fluid that is

driven by a slowly rotating disc of radius a in contact with the free surface The disc rotates at the angular velocity ω The Navier-Stokes equations for the angular component u(r, z) of the fluid’s velocity reduce to

Goodrich23was the first to attack this problem Using Hankel transforms,

the solution to Equation 2.6.1 through Equation 2.6.3 is

u(r, z) =

 ∞

0

A(k)J1(kr)e −kz dk. (2.6.5)

23 Taken with permission from Goodrich, F C., 1969: The theory of absolute surface

shear viscosity I Proc Roy Soc London, Ser A, 310, 359–372.

0 0.5 1 1.5

0.5 1 1.5 2 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

z/ar/a

Figure 2.6.1: The solution to Laplace’s equation, Equation 2.6.1, with the boundary

conditions given by Equation 2.6.2 through Equation 2.6.4 when µ = 0.

Historical Background 75then

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

after using Equation 1.2.15 and Equation 1.2.16 Therefore,26

Figure 2.5.4: The solution u(r, z)/ω to the mixed boundary value problem governed by

Equation 2.5 .33 through Equation 2.5 .36 when a = 1, b = and α =1... 1.5 2.5 3< /small>

0

0.2

0.4 0.6 0.8 1

Figure 2.5 .3: The solution u(r, z)/a to the mixed boundary value problem...

23< /small> Taken with permission from Goodrich, F C., 1969: The theory of absolute surface

shear viscosity I Proc Roy Soc London, Ser A, 31 0, 35 9? ?37 2.