Mixed Boundary Value Problems Episode 5 pptx

• Example 3.3.2 A similar problem18 to the previous one arises during the solution of Laplace’s equation in cylindrical coordinates: 18 See Hunter, A., and A.. Equation 3.3.30 satisfies n

108 Mixed Boundary Value Problems

Substituting Equation 3.2.41 into Equation 3.2.35, we obtain the integralequation

f (x), c < x ≤ π (3.2.43)

From Equation 1.2.11 and Equation 1.2.12, we obtain

h(t) = 2π

d dt

 π t

and

h(t) = −2

π

d dt

 π t

f (π − ξ) cos(ξ/2) cos(t) − cos(ξ) dξ

Consequently, making the back substitution, the dual series

3.3 DUAL FOURIER-BESSEL SERIES

Dual Fourier-Bessel series arise during mixed boundary value problems

in cylindrical coordinates where the radial dimension is of finite extent Here

we show a few examples

16 Originally solved by Borodachev, N M., and F N Borodacheva, 1967: Considering

the effect of the walls for an impact of a circular disk on liquid Mech Solids, 2(1), 118.

110 Mixed Boundary Value Problems

subject to the boundary conditions

lim

r →0 |u(r, z)| < ∞, u r (1, z) = 0, 0 < z < ∞, (3.3.2)

u z (r, 0) = 1, 0≤ r < a, u(r, 0) = 0, a < r < 1, (3.3.3)

where the unknown function h(t) is given by the regular Fredholm integral

equation of the second kind:

17 Srivastav, R P., 1961/1962: Dual series relations II Dual relations involving Dini

series Proc R Soc Edinburgh, Ser A, 66, 161–172.

0 0.2 0.4 0.6 0.8 1

0 0.05 0.1 0.15 0.2

−0.4

−0.3

−0.2

−0.1 0 0.1

rz

Figure 3.3.1: The solution to Laplace’s equation subject to the boundary conditions given

by Equation 3.3.2, Equation 3.3.3, and Equation 3.3.4 when a = 0.5.

Figure 3.3.1 illustrates this solution when a = 0.5.

• Example 3.3.2

A similar problem18 to the previous one arises during the solution of

Laplace’s equation in cylindrical coordinates:

18 See Hunter, A., and A Williams, 1969: Heat flow across metallic joints – The

con-striction alleviation factor Int J Heat Mass Transfer , 12, 524–526.

112 Mixed Boundary Value Problems where k n is the nth positive root of J 

d dt

0 0.2

0.4 0.6

0.8 10

0.2 0.4 0.6 0.8 1

−1

−0.5 0 0.5 1 1.5 2

rz

Figure 3.3.2: The solution to Laplace’s equation subject to the boundary conditions given

by Equation 3.3.13 through Equation 3.3.15 when a =1

2.

Equation 3.3.19 through Equation 3.3.25 provide the answer to our problem

if we set α = 0 and f (r) = 1 Figure 3.3.2 illustrates the solution when a = 1

20 See Sherwood, J D., and H A Stone, 1997: Added mass of a disc accelerating within

a pipe Phys Fluids, 9, 3141–3148.

114 Mixed Boundary Value Problems

where k n is the nth root of J1(ka) = 0 Equation 3.3.30 satisfies not only

Laplace’s equation, but also the boundary conditions given by Equation 3.3.27and Equation 3.3.29 Substituting Equation 3.3.30 into Equation 3.3.28, weobtain the dual series

2, ν > p − 1, m = 0, 1, 2, , and ζ n denotes the nth root of

J ν (ζa) = 0 By direction substitution it is easily seen that Equation 3.3.32 is satisfied if ν = 1 and

Our remaining task is to compute C m Although Equation 3.3.35 holds

for any r between 0 and 1, it would be better if we did not have to deal with

its presence It can be eliminated as follows: From Sneddon’s book,22

21 Tranter, C J., 1959: On the analogies between some series containing Bessel functions

and certain special cases of the Weber-Schafheitlin integral Quart J Math., Ser 2 , 10,

110–114.

22 Sneddon, op cit., Equation 2.1.33 and Equation 2.1.34.

if 0≤ r < 1; this integral equals 0 if 1 < r < ∞ Here P m (a,b) (x) =2F1(−m, a+ m; b; x) is the Jacobi polynomial If we view Equation 3.3.36 as a Hankel transform of η −k J ν

where δ nm is the Kronecker delta Multiplying both sides of Equation 3.3.35

num-3.3.40, which yields C m for m = 0, 1, 2, , M Then Equation 3.3.34 gives

A n Finally, the potential u(r, z) follows from Equation 3.3.30. Figure 3.3.3

illustrates this solution when a = 2 and p = 0.5.

• Example 3.3.4

In the previous example we solved Laplace’s equation over a semi-infiniteright cylinder Here, let us solve Laplace’s equation24 when the cylinder has

23 See page 83 in Magnus, W., and F Oberhettinger, 1954: Formulas and Theorems for

the Functions of Mathematical Physics Chelsea Publ Co., 172 pp.

24 See Galceran, J., J Cecilia, E Companys, J Salvador, and J Puy, 2000: Analyticalexpressions for feedback currents at the scanning electrochemical microscope. J Phys.

Chem., Ser B , 104, 7993-8000.

Next, we substitute Equation 3.3.50 into Equation 3.3.48, multiply both

sides of the resulting equation by r F2 1

Equation 3.3.55 is now solved to yield B m Next, we compute A nfrom

Equa-tion 3.3.50 Finally u(r, z) follows from EquaEqua-tion 3.3.47.

Figure 3.3.4 illustrates the solution to Equation 3.3.43 through

Equa-tion 3.3.46 when a = 2 and b = 1 As suggested by Galceran et al.,26 the

25 Tranter, op cit.

26 Galceran, Cecilia, Companys, Salvador, and Puy, op cit.

Figure 3.3.5: Schematic of a hollow cylinder containing discs at z = 0 and z = h.

u(r, 0 − ) = u(r, 0+), u(r, h − ) = u(r, h+), 0≤ r < a, (3.3.60)

slightly above or below h, respectively.

Separation of variables yields the potential, namely

120 Mixed Boundary Value Problems

where k n is the nth positive root of J0(ka) = 0 Equation 3.3.63 through

Equation 3.3.65 satisfy Equation 3.3.58, Equation 3.3.59, Equation 3.3.60,and Equation 3.3.62 Substituting Equation 3.3.63 through Equation 3.3.65into Equation 3.3.61, we obtain the following system of simultaneous dualseries equations:

where g(t) and h(t) are unknown functions.

Taken together, Equation 3.3.68 through Equation 3.3.71 are a Bessel series over the interval 0≤ r < a From Equation 1.4.16 and Equation

122 Mixed Boundary Value Problems

It is readily shown30that

 ∞

0

K0(aη)

I0(aη) I0(rη) cosh(tη) cos(hη) dη.

Substituting Equation 1.4.14, Equation 3.3.82, and Equation 3.3.83 into tion 3.3.80 and Equation 3.3.81, we have that

0 0.5 1 1.5 2

−2

−1 0 1 2 3 4

−1

−0.5 0 0.5 1

rz

Figure 3.3.6: The electrostatic potential within an infinitely long, grounded, and hollow

cylinder of radius 2 when two discs with potential−1 and 1 are placed at z = 0 and z = 2,

to evaluate the potential for any given r and z Figure 3.3.6 illustrates the electrostatic potential when F (r) = 1, G(r) = −1, and a = h = 2.

• Example 3.3.6: Electrostatic problem

In electrostatics the potential due to a point charge located at r = 0 and

z = h in the upper half-plane z > 0 above a grounded plane z = 0 is

124 Mixed Boundary Value Problems

(0,h) z

r (1,0)

Figure 3.3.7: Schematic of the spatial domain for which we are finding the potential in

Example 3.3.6.

this hole See Figure 3.3.7 Let us find the potential31 in this case.

The potential in this new configuration is

when 0≤ r ≤ 1 and −∞ < z ≤ 0 The integral32in Equation 3.3.89 vanishes

when z = 0 and 1 ≤ r < ∞ Here g(t) is an odd real-valued function and

k n denotes the nth root of J0(k) = 0 To determine g(t) and the Fourier

coefficients A n, the potential and its normal derivative must be continuous

across the aperture z = 0 and 0 ≤ r ≤ 1 Mathematically these conditions

31 Taken from Shail, R., and B A Packham, 1986: Some potential problems associated

with the sedimentation of a small particle into a semi-infinite fluid-filled pore IMA J Appl.

Math., 37, 37–66 with permission of Oxford University Press.

32 See Section 5.10 in Green, A E., and W Zerna, 1992: Theoretical Elasticity NewYork: Dover, 457 pp.

r2+ (z + it)2= ξe iη/2, 

r2+ (z − it)2= ξe −iη/2 , (3.3.94)

with ξ2cos(η) = r2+ z2− t2, ξ2sin(η) = 2zt, ξ ≥ 0, and 0 ≤ η ≤ π On the

other hand, Equation 3.3.92 gives

t2− r2dt.

(3.3.98)

The left side of Equation 3.3.98 is a Fourier-Bessel expansion Multiplying

both sides of this equation by rJ0(k m r) and integrating with respect to r from

t2− r2 dr. (3.3.99)

33 Gradshteyn and Ryzhik, op cit., Formula 2.252, Point II.

126 Mixed Boundary Value Problems

0

0.5 1 1.5

2 −2

−1 0 1 2 0

Figure 3.3.8: The electrostatic potential when a unit point charge is placed at r = 0 and

z = h for the structure illustrated inFigure 3.3.7

Interchanging the order of integration in Equation 3.3.99 and evaluating the

inner r integrals, we finally achieve

where m = 1, 2, 3, To find A m, we truncate the infinite number of

equa-tions given by Equation 3.3.100 to a finite number, say M As M increases the A m ’s with a smaller m increase in accuracy The procedure is stopped when the leading A m’s are sufficiently accurate for the evaluation of Equation3.3.89, Equation 3.3.90 and Equation 3.3.97 Figure 3.3.8 illustrates this so-

lution when h = 1 and the first 300 terms have been retained when M = 600.

3.4 DUAL FOURIER-LEGENDRE SERIES

Here we turn to problems in spherical coordinates which lead to dualFourier-Legendre series We now present several examples of their solution

Consider Equation 3.4.6 Using Equation 1.3.4 to eliminate P n [cos(θ)]

and then interchanging the order of integration and summation, we have

34 See Ramachandran, M P., 1993: A note on the integral equation method to a

diffusion-reaction problem Appl Math Lett., 6, 27–30.

128 Mixed Boundary Value Problems

In a similar manner, Equation 3.4.7 can be rewritten as

A ncos

n +1 2



ϕ

n +1 2



˜

ϕ

d ˜ ϕ + 2π

 π α

ψ( ˜ ϕ) cos

n +1 2

ψ( ˜ ϕ) cos

n +1 2



˜

ϕ

d ˜ ϕ (3.4.18)

Substituting Equation 3.4.18 into Equation 3.4.16 and interchanging the order

of integration and summation,



ϕcos

n +1 2



˜

ϕ

n +1 2



d ˜ ϕ

+

 π α



ϕcos

n +1 2



˜

ϕ

n +1 2



d ˜ ϕ (3.4.19)

n +1 2



˜

ϕ

n +1 2

=1

2ln

cos( ˜ϕ/2) + cos(ϕ/2)

To numerically solve Equation 3.4.23, we introduce n nodal points at

ϕ i = α + ih, i = 0, 1, , n − 1, where h = (π − α)/n We do not have to compute r(π) because it equals zero since ψ(π) = 0 from Equation 3.4.15.

Then, Equation 3.4.23 becomes

r(ϕ i)− γ

π

 π α

Why have we written L(t, ϕ) in the form given in Equation 3.4.25? It

clearly shows that the integral equation, Equation 3.4.23, contains a weaklysingular kernel Consequently, the finite difference representation of the inte-gral in Equation 3.4.24 consists of two parts A simple trapezoidal rule is usedfor the first four terms given in Equation 3.4.25 For the fifth term, we em-ploy a numerical method devised by Atkinson36for kernels with singularities.

Therefore, for a particular alpha and b, we compute dt = (pi-alpha)/N,

35 Parihar, K S., 1971: Some triple trigonometrical series equations and their

applica-tion Proc R Soc Edinburgh, Ser A, 69, 255–265.

36 Atkinson, K E., 1967: The numerical solution of Fredholm integral equations of the

second kind SIAM J Numer Anal., 4, 337–348 SeeSection 5 in particular.

130 Mixed Boundary Value Problems

where N is the number of nodal points The MATLAB code for computing

for n = 0:N-1 % rows loop (top to bottom in the matrix)

phi = pphi(n+1); bb(n+1) = (2*gamma-1)*cos(phi/2);

for m = 0:N-1 % columns loop (left to right in the matrix)

t = tt(m+1);

% Introduce the first terms from Equation 3.4.24

if (n==m) AA(n+1,m+1) = 1;

else AA(n+1,m+1) = 0; end

% Compute integral in Equation 3.4.24 Add in the contribution

% from the first four terms of Equation 3.4.25 Use the

% trapezoidal rule Recall that r(π) = 0.

end

if (m > 0)

arg 1 = (t+phi)/4; arg 2 = (t-phi)/4;

AA(n+1,m+1) = AA(n+1,m+1)

% Use Atkinson’s technique to treat the fifth term in

% Equation 3.4.25 See Section 5 of his paper

if (m > 0)

kk = n+1-m; Psi 0 = -1;

Psi 1 = 0.25*(kk*kk-(kk-1)*(kk-1));

if ( kk ∼= 0 )

Psi 0 = Psi 0 + kk*log(abs(kk));

Psi 1 = Psi 1 - 0.50*kk*kk*log(abs(kk));

end

if ( kk ∼= 1 )

Psi 0 = Psi 0 - (kk-1)*log(abs(kk-1));

Psi 1 = Psi 1 + 0.50*(kk-1)*(kk-1)*log(abs(kk-1));

end

Psi 1 = Psi 1 + kk*Psi 0;

W 0 = Psi 0 - Psi 1; W 1 = Psi 1;

aalpha = 0.5*dt*log(dt) + dt*W 0;

bbeta = 0.5*dt*log(dt) + dt*W 1;

AA(n+1,m ) = AA(n+1,m ) + gamma*aalpha/pi;

AA(n+1,m+1) = AA(n+1,m+1) + gamma*bbeta/pi;

end

end % end of column loop

end % end of rows loop

132 Mixed Boundary Value Problems

Once ψ(ϕ i ) is found with ψ(π) = 0, we compute A n via Equation 3.4.18 The

MATLAB code to compute A n is

% This is the n = 0 term for Simpson’s rule.

A(m+1) = A(m+1) + 2*psi( 1 )*cos((m+0.5)*tt( 1 ))*dt/(3*pi);

% Recall that ψ(π) = 0. Therefore, we do not need the n = N

% term in the numerical integration

% Compute the Legendre polynomials for a given theta

% via the recurrence formula

Legendre(1) = 1; Legendre(2) = mu;

−0.5 0 0.5 1

−1

−0.5 0 0.5 1

end; end; end; end

Figure 3.4.1 illustrates the solution when α = π/4 and b = 0.25.

An alternative method for solving Equation 3.4.6 and Equation 3.4.7 is

to introduce the following integral definition for A n:

134 Mixed Boundary Value Problems Turning to Equation 3.4.6, we have for A n that

× d dt

Solving Equation 3.4.35 is straightforward Having determined h(t), A n

fol-lows from Equation 3.4.26 Finally Equation 3.4.5 yields u(r, z).

Separation of variables yields the solution

136 Mixed Boundary Value Problems



n +1 2

h(τ ) dτ = −4

π cos(3t/2), h(0) = 0. (3.4.51)

−1 0 1 2

−2

−1 0 1 2

Figure 3.4.2: The solution u(r, θ) to the mixed boundary value problem posed by Equation

3.4.37 through Equation 3.4.40 when α = π/4.

To solve Equation 3.4.51, we rewrite it as



H(s) = M1

4 −4π

s2

s2+9 4

h(t) = sec(α/2) cos(3α/2) sin(t/2)/π − 3 sin(3t/2)/π (3.4.55)

Figure 3.4.2 illustrates the solution when α = π/4.