Mixed Boundary Value Problems Episode 15 ppsx

use the differential equation and first two boundary conditions to show thatHere we have assumed that|ux, 0| is bounded by e −x , 0 < so that U+is analytic in the upper half-planek > −, w

use the differential equation and first two boundary conditions to show that

Here we have assumed that|u(x, 0)| is bounded by e −x , 0 < 

so that U+is analytic in the upper half-plane(k) > −, while U −is analytic

in the lower half-plane(k) < 0.

Step 3 : Show that (1) can be rewritten

Note that the right side of this equation is analytic in the lower half-plane

(k) < 0, while the left side is analytic in the upper half-plane (k) > − Step 4 : Use Liouville’s theorem and deduce that

U − (k) =

√

2(1− ki) √ 1 + ki − 1

1− ki . Step 5 : Show that

k − i

e −y √ k2 +1

√

k2+ 1 .

Step 6 : Finish the problem by retracing Step 6 through Step 9 of the previous

problem and show that you recover the same solution Gramberg and van deVen24 found an alternative representation

24 Gramberg, H J J., and A A F van de Ven, 2005: Temperature distribution in a

Newtonian fluid injected between two semi-infinite plates Eur J Mech., Ser B., 24,

767–787.

by evaluating the inverse Fourier transform via contour integration.

3 Use the Wiener-Hopf technique to solve the mixed boundary value problem

u(x, y)e ikx dx,

so that U (k, y) = U+(k, y) + U − (k, y) Here, U+(k, y) is analytic in the

half-space(k) > −, while U − (k, y) is analytic in the half-space (k) < 0 Then

show that the partial differential equation becomes

d2U

dy2 − m2U = 0, 0 < y,

with limy →∞ U (k, y) → 0, where m2= k2+ 1.

Step 2 : Show that the solution to Step 1 is U (k, y) = A(k)e −my.

Step 3 : From the boundary conditions along x = 0, show that

−3

−1 1 3 5

0 1 2 3 4 5

−0.5 0 0.5 1

x y

Problem 3

Note that M+(k) is analytic in the half-space (k) > −, while L − (k) is

analytic in the half-space(k) < 0.

Step 4 : By eliminating A(k) from the equations in Step 3, show that we can

factor the resulting equation as

Note that the left side of the equation is analytic in the upper half-plane

(k) > −, while the right side of the equation is analytic in the lower

half-plane(k) < 0.

Step 5 : Use Liouville’s theorem to show that each side of the equation in Step

4 equals zero Therefore,

−1 0 1 2

−2

−1 0

Chapter 6 Green’s Function

The use of Green’s functions to construct solutions to boundary value lems dates back to nineteenth century electrostatics In this chapter we firstshow how to construct a Green’s function with mixed boundary conditions.Then we will apply integral representations to a mixed boundary value prob-lem when the kernel is a Green’s function In the last section we specialize topotentials

prob-6.1 GREEN’S FUNCTION WITH MIXED BOUNDARY VALUE CONDITIONS

We begin our study of Green’s function methods by examining how we mightconstruct a Green’s function when mixed boundary conditions are present.Consider the rather simple problem1of

1 See Khanzhov, A D., 1966: A mixed heat conduction boundary problem for a

semi-infinite plate J Engng Phys., 11, 370–371.

y →∞ u(x, y) → 0, −∞ < x < ∞, (6.1.3)

u y (x, 0) = 0, 0≤ |x| < a, u(x, 0) = 0, a < |x| < ∞ (6.1.4)

Using Fourier cosine transforms, the partial differential equation and theboundary conditions given by Equation 6.1.2 and Equation 6.1.3 are satisfiedby

The arbitrary constant A(k) will be used to satisfy the final boundary

con-dition, Equation 6.1.4 Direct substitution of Equation 6.1.5 and Equation6.1.6 into this boundary condition yields the dual integral equations

If we introduce x = aρ and η = ka, Equation 6.1.10 and Equation 6.1.11

become the nondimensional integral equations

a 4η e

−bη/a + A(η)e bη/a (6.1.14)

Then Equation 6.1.12 and Equation 6.1.13 become

 ∞

0

ηB(η)J −1(ρη) dη = h(ρ) =

#2

Upon using B(η) to find A(k), substituting A(k) into Equation 6.1.5 and

Equation 6.1.6, and then evaluating the integrals, we find that

u(x, y) = 1

2π

1

Efimov and Vorob’ev3 found the Green’s function for the

three-dimen-sional Laplace equation in the half-space z ≥ 0 with the boundary conditions

3 Efimov, A B., and V N Vorob’ev, 1975: A mixed boundary value problem for the

Laplace equation J Engng Phys., 26, 664–666.

then the Green’s function is

6.2 INTEGRAL REPRESENTATIONS INVOLVING GREEN’S FUNCTIONS

Green’s functions have long been used to create integral representations forboundary value problems Here we illustrate how this technique is used in thecase of mixed boundary value problems As before, we will face an integralequation that must solved, usually numerically

where f (x) is given by the integral equation

tanh[πx/(2L)] + tanh[πξ/(2L)] tanh[πx/(2L)] − tanh[πξ/(2L)] dξ (6.2.8)

for 0≤ x ≤ 1 From Example 1.2.3, we have that

dη





dξ (6.2.10)

for 0 ≤ x ≤ 1 Consequently, the portion of the potential u1(x, y) due to

h e (η) can be computed from

cosh(π |x + ξ|/L) − cos(πy/L) dξ, (6.2.11) where f1(ξ) is given by Equation 6.2.10.

Turning now to finding that portion of f (ξ), f2(ξ), due to h o (x), Equation

Integrating Equation 6.2.12 from 0 to x,

 1

x

sinh(πξ/L)

sinh2[πξ/(2L)] − sinh2[πx/(2L)]

where A is an undetermined constant and 0 < x < 1 Integrating Equation 6.2.14 with respect to x, we obtain

d dχ

 1

χ

sinh(πξ/L)

sinh2[πξ/(2L)] − sinh2[πχ/(2L)]

d dχ

 1

χ

sinh(πξ/L)

sinh2[πξ/(2L)] − sinh2[πχ/(2L)]

−2

−1 0 1 2 3

0 0.02 0.04 0.06 0.08 0.1

Figure 6.2.1: The solution to Laplace’s equation with the boundary conditions given by

Equation 6.2.2 through Equation 6.2.4 when h(x) = (1 − x)2and L = 0.1.

The potential due to h(x) equals the sum of u1(x, y) and u2(x, y) Figure 6.2.1 illustrates this solution when h(x) = (1 − x)2 and L = 0.1.

Yang et al.4 solved this problem when they replaced the boundary

condi-tion u(x, L) = 0 with u y (x, L) = 0 In this case the Green’s function becomes g(x, y |ξ, η) = 2

π e

−π|x−ξ|/(2L)sinπy

2L

sin

4 Taken with permission from Yang, F., V Prasad, and I Kao, 1999: The thermal

constriction resistance of a strip contact spot on a thin film J Phys D: Appl Phys., 32,

930–936 Published by IOP Publishing Ltd.

d dχ

 1

χ

sinh(πξ/L)

sinh2[πξ/(2L)] − sinh2[πχ/(2L)]

d dχ

 1

χ

sinh(πξ/L)

sinh2[πξ/(2L)] − sinh2[πχ/(2L)]

−2

−1 0 1 2 3

0 0.02 0.04 0.06 0.08 0.1





The potential due to h(x) equals the sum of u1(x, y) and u2(x, y) We have

illustrated this solution in Figure 6.2.2 when h(x) = (1 − x)2 and L = 0.1.

5 Taken with permission from Lal, B., 1978: A note on mixed boundary value problems

in electrostatics Z Angew Math Mech., 58, 56–58 To see how to solve this problem

using conformal mapping, see Homentcovschi, D., 1980: On the mixed boundary value

problem for harmonic functions in plane domains J Appl Math Phys., 31, 352–366.

We also require that both u(r, nπ/2) and u θ (r, nπ/2) are continuous if r > 1.

From the theory of Green’s function,

−1 0 1 2

−2

−1 0 1 2

electro-• Example 6.2.3: The method of Yang and Yao

Building upon Example 1.1.4 and Example 6.2.1, Yang and Yao6

de-veloped a general method for finding the potential u(x, y) governed by the

nondimensional partial differential equation:

6 Yang, F.-Q., and R Yao, 1996: The solution for mixed boundary value problems of

two-dimensional potential theory Indian J Pure Appl Math., 27, 313–322.

Their analysis begins by noting that we can express u(x, y) in terms of the Green’s function g(x, y |ξ, η) by the integral

where r = exp[ −π(|x−ξ|−iy)/L ] Substituting Equation 6.2.46 into Equation

6.2.41 and using Equation 6.2.39, we find that

At this point, we specialize according to whether h(x) is an even or odd

function Because any function can be written as the sum of an even and odd

function, we can first rewrite h(x) as a sum of an even function h e (x) and an odd function h o (x) Then we find the potentials for the corresponding h e (x) and h o (x) The potential for the given h(x) then equals their sum by the

principle of linear superposition

7 Duffy, D G., 2001: Green’s Functions with Applications Chapman & Hall/CRC, 443

pp See Section 5.2

and find that

2L sinh(πx/L)

d dx

 1

x

sinh(πξ/L)

sinh2[πξ/(2L)] − sinh2[πx/(2L)]

where A is a constant that is determined by f (0+) Substituting these results

into Equation 6.2.41, we obtain the final result







sinh2[π |x − ξ|/(2L)] + sin2[πy/(2L)]

In the case when h(x) is an odd function, Equation 6.2.47 can be rewritten

sinh[πx/(2L)] + sinh[πξ/(2L)] sinh[πx/(2L)] − sinh[πξ/(2L)] dξ (6.2.54)

for 0≤ x ≤ 1 Again, applying the methods from Example 6.2.1, we find that

f (x) = 1

2L

d dx

  1

x

sinh(πξ/L)

sinh2[πξ/(2L)] − sinh2[πx/(2L)]

× ξ

0

h  (χ)

sinh2[πξ/(2L)] − sinh2[πχ/(2L)] dχ

Substituting Equation 6.2.55 into Equation 6.2.41, the potential is







sinh2[π |x − ξ|/(2L)] + sin2[πy/(2L)]

The Green’s function is now governed by

nπy

L

cos

where r = exp[ −π(|x − ξ| − iy)/L ] Upon substituting Equation 6.2.65 into

Equation 6.2.41 and using Equation 6.2.39, we obtain an integral equation for

[cosh(πχ/L) − 1][cosh(π/L) − cosh(πχ/L)] dχ. (6.2.69)

Substituting Equation 6.2.69 into Equation 6.2.41, we find that

tanh2[π/(2L)] − tanh2[πx/(2L)]

Substituting Equation 6.2.65 and Equation 6.2.73 into Equation 6.2.41, weobtain the final results that



f(ξ)dξ (6.2.74)

if h(x) is an odd function.

• Example 6.2.4: The method of Clements and Love

In 1974 Clements and Love8 published a method for finding

axisymm-metric potentials Mathematically, this problem is given by

(6.2.78)

Clements and Love referred to this problem as a “Neumann problem” because

of the boundary condition between a < r < b.

Clements and Love’s method expresses the potential in terms of a Green’sfunction:

where σ(ρ) is presently unknown The quantity inside of the square brackets

is the free-space Green’s function Clements and Love then proved that σ(ρ)

is given by

ρ σ(ρ) = −2

π

d dρ

 a ρ

 ρ b

8 Clements, D L., and E R Love, 1974: Potential problems involving an annulus Proc.

Cambridge Phil Soc., 76, 313–325. 1974 Cambridge Philosophical Society Reprintedc with the permission of Cambridge University Press.

where f1(ξ) and f2(ξ) are found from the coupled integral equations

Clements and Love also considered the case when the mixed boundary

condition along z = 0 reads







u z (r, 0) = −σ1(r), 0 < r < a, u(r, 0) = U0(r), a < r < b,

u z (r, 0) = −σ2(r), b < r < ∞.

(6.2.87)

Clements and Love referred to this problem as a “Dirichlet problem” because

of the boundary condition between a < r < b.

The potential is given by Equation 6.2.79 once again In the present case,

To evaluate Equation 6.2.89 and Equation 6.2.90, we must first compute the

quantities U3(r) and U4(r) via

d dr

 b r

s

√

s2− r2

d ds

d dr

subject to the boundary conditions

(6.2.103)

From the nature of the boundary conditions, we use Equation 6.2.79

through Equation 6.2.86 We begin by computing g1(r) and g2(r) ing U1(ξ) = 2

At this point, we must turn to numerical methods to compute u(r, z).

Using MATLAB, we begin our calculations at a given radius r and heightR

z by solving the coupled integral equations, Equation 6.2.83 and Equation6.2.84 We do this by introducing N nodal point in the region 0 ≤ ξ ≤ a

such that xi = (n− 1) ∗ dr 1, where n = 1,2, .,N and dr 1 = a/(N-1).

Similarly, for 1≤ ξ < ∞, we introduce M nodal points such that xi = 1)*dr 2, where m = 1,2, .,M and dr 2 is the resolution of the grid Thus,

1+(m-Equation 6.2.83 and 1+(m-Equation 6.2.84 yield N+M equations which we express

in matrix notation as Af = b, where N equations arise from Equation 6.2.83

and M equations are due to Equation 6.2.84 Because we will evaluate theintegrals using Simpson’s rule, both N and M must be odd integers

The MATLABcode that approximates Equation 6.2.83 is

A = zeros(N+M,N+M); % zero out the array A

end; end; end

The MATLABcode that approximates Equation 6.2.84 is

end; end; end

Solving these (N + M ) × (N + M) equations, we find f which holds f1(ξ)

in its first N elements, while f2(ξ) is given in the remaining M elements:

f = A\b;

Given f, we now solve for σ(ρ) This is a two-step procedure First,

we evaluate the bracketed terms in Equation 6.2.80 or Equation 6.2.82 For

accurate computations in Equation 6.2.80, we note that ξ dξ/

ξ2− ρ2 =

d

ξ2− ρ2

Equation 6.2.82 employs a similar trick Then, we evaluate

the integral using the trapezoidal rule Finally, σ(ρ) follows from simple finite

bracket 1(m) = bracket 1(m) + f1*sqrt(xi end*xi end-t*t)

- f1*sqrt(xi begin*xi begin-t*t);

end;end

bracket 1(N) = 0;

for n = 1:N-1

bracket 2(m) = bracket 2(m) - f2*sqrt(t*t-xi end*xi end)

+ f2*sqrt(t*t-xi begin*xi begin);

With σ(ρ) we are ready to compute Equation 6.2.79 There are two steps.

First, we find the Green’s function via Simpson’s rule; it is called green here.Then we evaluate the outside integral using the midpoint rule The finalsolution is called u(i,j)

green = green + dphi/(3*denom);

end; end; end

if (k < N)

u(i,j) = u(i,j) + sigma(k)*green*rho(k)*dr 1;

else

0 0.5 1

1.5 2

0

0.5 1 1.5 2

Figure 6.2.4: The solution to Laplace’s equation with the boundary conditions given by

Equation 6.2.101 through Equation 6.2.103.

u(i,j) = u(i,j) + sigma(k)*green*rho(k)*dr 2;

• Example 6.3.1

In this example let us find the solution to Laplace’s equation in three

dimen-sions when the boundary condition on the z = 0 plane is

R2= ρ2+ r2− 2ρr cos(ϑ − θ) + z2. (6.3.3)

Here R gives the distance from the general point (r, ϑ, z) to the point (r, θ, 0)

on the disk Because

the boundary condition u z (r, ϑ, 0) = 0 is satisfied if r > 1 because R does not

vanish outside of the unit circle Kanwal and Sachdeva10 found expressions

for σ(ρ) in the special case of

 ρ

0

J −1(kx)

√ x

We can invert Equation 6.3.12 and find that

S(x) = 1

2π

d dx

2π +

(1− γ)x 2π √

Figure 6.3.1illustrates the potential u(x, y, z) when z = 0 and z = 0.2 Here

we also selected α = 0.5 and λ = 2 During the numerical evaluation of Equation 6.3.2 the integration with respect to ρ was done first and the time derivative in t σ(t) was eliminated by an integration by parts.

• Example 6.3.2: Fabrikant’s method

In the previous example we found the solution to Laplace’s equation in three

dimensions in the half-space z ≥ 0 with the mixed boundary condition given

by Equation 6.3.1 During the 1980s Fabrikant11 generalized this mixed

boundary value problem to read

11 Fabrikant, V I., 1986: A new approach to some problems in potential theory Z.

Angew Math Mech., 66, 363–368 Quoted with permission.