Mixed Boundary Value Problems Episode 9 pptx

Substituting for Ak from Equation 4.3.118 into Equation 4.3.119, we findafter interchanging the order of integration that where the contour Γ consists of the real axis from the origin to

Substituting for A(k) from Equation 4.3.118 into Equation 4.3.119, we find

after interchanging the order of integration that

where the contour Γ consists of the real axis from the origin to R, an arc in

the first quadrant|z| = R, 0 ≤ θ ≤ π/2, and imaginary axis from iR to the origin As R → ∞, we find that

0 5 10 15 20

0 5 10 15 20

u z

Figure 4.3.4: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.3.108 through Equation 4.3.111 with a = 2.

for a < t < ∞ Figure 4.3.4 illustrates the solution when a = 2.

ground level: Three-dimensional turbulent diffusion without convection J Geophys Res.,

72, 5631–5639 c1967 American Geophysical Union Reproduced/modified by permission

of American Geophysical Union.

Using transform methods or separation of variables, the general solution

to Equation 4.3.126, Equation 4.3.127, and Equation 4.3.128 is

1969: Unsteady torsional oscillations of an elastic half-space Mech Solids, 4(1), 79–83.

Setting x = r/a, ξ = ka, and g(ξ) =

ξ2+ (κa)2A(ξ) in Equation 4.3.143

and Equation 4.3.144, we find that

By comparing our problem with the canonical form given by Equation 4.3.26

through Equation 4.3.27, then ν = 1 and G(ξ) =

where L1(·) denotes a modified Struve function of the first kind Vasudevaiah

and Majhi46showed how to evaluate the integral in Equation 4.3.151.

As in the previous examples, we must solve for h(x) numerically Then g(ξ) is computed from Equation 4.3.147. Finally, Equation 4.3.142 gives

u(r, z). Figure 4.3.6illustrates this solution when κa = 1.

coaxial disks Indian J Pure Appl Math., 12, 1027–1042.

Substituting Equation 4.3.157 into Equation 4.3.156, we have that

k cos{k[τ − r sin(θ)]} − cos{k[a − r sin(θ)]}

+ cos{k[t + r sin(θ)]} − cos{k[a + r sin(θ)]}

dk dθ (4.3.165)

We used the integral definition of J0(kr) to obtain Equation 4.3.164.

Consider now the integral

0 0.5 1 1.5 2

−1

−0.5 0 0.5 1

Figure 4.3.7: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.3.153 through Equation 4.3.156 with a = 1 and α = 0.1.

Applying these results to Equation 4.3.163, we have

In a similar manner,49 we can solve

Using transform methods or separation of variables, the general solution

to Equation 4.3.181, Equation 4.3.182, and Equation 4.3.183 is

since the integral within the square brackets vanishes in Equation 4.3.190when 0≤ t ≤ a < r.

Turning Equation 4.3.186, we substitute Equation 4.3.188 into it Thisyields

We used the integral definition of J1(kr) to obtain Equation 4.3.193.

Consider now the integral

2

2

0 0.5 1 1.5 2

−1

−0.5 0 0.5 1

Figure 4.3.8: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.3.181 through Equation 4.3.184 with a = 1 and α = 0.1.

Using transform methods or separation of variables, the general solution

to Equation 4.3.206 through Equation 4.3.210 is

k2+ α2 with (λ) > 0 Substituting Equation 4.3.212 and

Equation 4.3.213 into Equation 4.3.211, we have that

circular disk parallel to an infinite plane wall Fluid Dyn Res., 34, 77–97.

To solve the dual integral equations, Equation 4.3.214 and Equation4.3.215, we introduce

d dr

Figure 4.3.9: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.3.206 through Equation 4.3.211 with h = 1 and α = 1.

and

d dr

In the previous examples, the boundary condition was u(r, 0) = 0 or

u z (r, 0) = 0 for 0 < a < r < ∞ In this example we consider the other situation where u(r, 0) = 0 applies when 0 < r < 1 In particular, we find the

Applied Mathematics Reprinted with permission.

subject to the boundary conditions

Using transform methods or separation of variables, the general solution

to Equation 4.3.225, Equation 4.3.226, and Equation 4.3.227 is

from Equation 1.4.14 with 0 < r < 1 ≤ t < ∞ Turning to Equation 4.3.231,

the substitution of Equation 4.3.232 yields

We now simplify Equation 4.3.234 in two ways In the first term we

integrate by parts the integral within the square brackets and apply Equation

1.4.13 We then replace J0(kr) by its integral representation.54 This gives

0 M (η) cos(ξη/h) dη Viewing Equation 4.3.236 as an integral

equation of the Abel type, Equation 1.2.15 and Equation 1.2.16 yield

Figure 4.3.10: The solution u(r, z) to the mixed boundary value problem governed by

Equation 4.3.225 through Equation 4.3.228 with a = h = 2.

Figure 4.3.10 illustrates this solution when a = h = 2.

All of the coefficients in Equation 4.3.245 are nonzero

Using transform methods or separation of variables, the general solution

to Equation 4.3.242, Equation 4.3.243, and Equation 4.3.244 is

u(r, z) =

 ∞

0

A(k)J0(kr)e −kz dk. (4.3.246)

boundary conditions Sov Tech Phys., 11, 169–173.

Substituting Equation 4.3.246 into Equation 4.3.245, we have that

Upon applying Equation 1.4.13 to the integral within the square brackets onthe left side of Equation 4.3.256,

R(τ, t, k) = sin[k(t + τ )] si[k(t + τ )] + cos[k(t + τ )] ci[k(t + τ )]

+ sin[k |t − τ|] si[k|t − τ|] + cos[k|t − τ|] ci[k|t − τ|], (4.3.264)

λ = δ/γ and si( ·) and ci(·) are the sine and cosine integrals.

2u

∂z2 = 0, 0≤ r < ∞, 0 < z < 1, (4.3.265)

with permission from Elsevier.

subject to the boundary conditions

The interesting aspect of this example is the mixture of boundary conditions

along the boundary z = 1 For 0 < r < a, we have a Dirichlet boundary condition that becomes a Robin boundary condition when a < r < ∞.

Applying Hankel transforms, the solution to Equation 4.3.265 and theboundary conditions given by Equation 4.3.266 is

g(a) = 1 and G(λ) = 1 + λ coth(λ)/σ.

with axial symmetry Quart J Mech Appl Math., 3, 411–419.

Figure 4.3.11: Educated at Queen’s College, Oxford, Clement John Tranter, CBE, (1909–

1991) excelled both as a researcher and educator, primarily at the Military College of Science at Woolrich and then Shrivenham His mathematical papers fall into two camps: (a) the solution of boundary value problems by classical and transform methods and (b) the solution of dual integral equations and series He is equally well known for a series

of popular textbooks on integral transforms and Bessel functions (Portrait provided by kind permission of the Defense College of Management and Technology Library’s Heritage Centre.)

What is the value of κ here? Clearly, we would like our solution to be valid for a wide range of σ Because G(λ) → 1 as σ → ∞, a reasonable choice

where B mn depends only on a and σ Multiplying Equation 4.3.275 by dk/k ×

J 2p−1 (ka) and integrating

Equa-by P m −1 (ξ) dξ, integrating between −1 and 1, and using the orthogonality

properties of the Legendre polynomial, we have

Table 4.3.1: The Convergence of the Coefficients A n Given by Equation

4.3.287 Where S mnHas Nonzero Values for 1≤ m, n ≤ N

Thus, we reduced the problem to the solution of an infinite number of linear

equations that yield A n Selecting some maximum value for n and m, say

N , each term in the matrix S mn, 1 ≤ m, n ≤ N, is evaluated numerically for a given value of a and σ By inverting Equation 4.3.287, we obtain the coefficients A n for n = 1, , N Because we solved a truncated version

of Equation 4.3.287, they will only be approximate To find more accurate

values, we can increase N by 1 and again invert Equation 4.3.287 In addition

to the new A N+1, the previous coefficients will become more accurate We

can repeat this process of increasing N until the coefficients converge to their correct values This is illustrated in Table 4.3.1 when σ = a = 1.

Once we have computed the coefficients A n necessary for the desired

ac-curacy, we use Equation 4.3.274 to find A(k) and then obtain u(r, z) from

Equation 4.3.268 via numerical integration Figure 4.3.12illustrates the

solu-tion when σ = 1 and a = 2.

u2(r, z) =

 ∞

0

A(k) tanh(kb/a)e kz/a J0(kr/a) dk k . (4.3.296)

Equation 4.3.295 satisfies not only Equation 4.3.288, but also Equation 4.3.290and Equation 4.3.292 Similarly, Equation 4.3.296 satisfies not only Equation4.3.289, but also Equation 4.3.291 and Equation 4.3.293 Substituting Equa-tion 4.3.295 and Equation 4.3.296 into Equation 4.3.294, we obtain the dualintegral equations

then direct substitution of Equation 4.3.299 into Equation 4.3.298 shows that

it is satisfied identically We next substitute Equation 4.3.299 into Equation4.3.297 and interchange the order of integration This yields

t2− r2/a2, integrate from 0 to at, differentiate with respect

to t, and use Equation 4.3.301, we obtain the following integral equation that gives f (t):

0 0.5 1

1.5 2

0

0.5 1 1.5 2

u 1

Figure 4.3.13: The solution u1(r, z) to the mixed boundary value problem governed by

if 0 < t < 1.

At this point we must solve Equation 4.3.303 numerically to compute

f (t) Before we do that, there are two limiting cases of interest When  = 0,

we have the same problem that we solved in Section 2.2 on the disc capacitor

The second limit is 0 2(r, z) → 0 and u1(r, z) is given by

the solution to Example 4.3.2 Figure 4.3.13 shows the solution somewhere

between these two limits with  = 30.

• Example 4.3.14

During their study of a circular disk in a Brinkman medium, Feng et

al.61 solved a system of mixed boundary value problems. We join their

problem midway in progress where they derived the following governing partialdifferential equations and boundary conditions:

in a Brinkman medium Phys Fluids, 10, 2137–2146.

Let us now turn to the equation involving p(r, z) in the mixed boundary

condition Equation 4.3.311 Now,

0 0.5 1 1.5 2

−1

−0.5 0 0.5 1

u 1

0 0.5 1 1.5 2

−1

−0.5 0 0.5 1

u 2

Figure 4.3.14: The solution to the mixed boundary value problem governed by Equation

4.3.304 through Equation 4.3.311 when γ = 1.

where f (r) is an arbitrary function of r The mixed boundary condition,

Equation 4.3.311, then gives

Equation 4.3.313 and Equation 4.3.314 yields

Equation 4.3.316, respectively Finally, Equation 4.3.313 and Equation 4.3.314

yield u(r, z). Figure 4.3.14illustrates the solution when γ = 1

then direct substitution of Equation 4.3. 299 into... (4.3. 296 )

Equation 4.3. 295 satisfies not only Equation 4.3.288, but also Equation 4.3. 290 and Equation 4.3. 292 Similarly, Equation 4.3. 296 satisfies not only Equation4.3.2 89, but... system of mixed boundary value problems. We join their

problem midway in progress where they derived the following governing partialdifferential equations and boundary conditions: