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198 Mixed Boundary Value Problems −4 −2 0 2 4 0 1 2 −1 −0.5 0 0.5 1 x y u(x,y) Problem 2 and ∞ 0 A(k)sin(kx) dk =0, 1 < |x| < ∞. Step 3:Fredricks 27 showed that dual integral equations of the form ∞ 0 A(k) k sin(kx) dk = f(x), 0 ≤ x<a, and ∞ 0 A(k)sin(kx) dk =0,a<x<∞, have the solution A(k)=2 ∞ n=1 ∞ m=0 (2m +1)B n J 2m+1 (nπ)J 2m+1 (ka), where B n is given by the Fourier sine series f(x)= ∞ n=1 B n sin(nπx/a), 0 ≤ x<a. Clearly f(0) = 0. Use this result to show that A(k)= 4 π ∞ n=1 ∞ m=0 (−1) n+1 n (2m +1)B n J 2m+1 (nπ)J 2m+1 (k). The figure labeled Problem 2 illustrates this solution u(x, y). 27 Fredricks, R. W., 1958: Solution of a pair of integral equations from elastostatics. Proc. Natl. Acad. Sci., 44, 309–312. See also Sneddon, I. N., 1962: Dual integral equations with trigonometrical kernels. Proc. Glasgow Math. Assoc., 5, 147–152. © 2008 by Taylor & Francis Group, LLC Transform Methods 199 3. Following Example 4.1.3, solve Laplace’s equation ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, 0 <x<∞, 0 <y<h, with the boundary conditions u x (0,y)=0, lim x→∞ u(x, y) → 0, 0 <y<h, u y (x, 0) = −g(x), 0 <x<a, u(x, 0) = 0,a<x<∞, (1) and u(x, h)=0, 0 <x<∞. Step 1 :Usingseparation of variables or transform methods, show that the general solution totheproblem is u(x, y)= 2 π ∞ 0 A(k)sinh[k(h −y)] cos(kx) dk. Step 2:Using boundary condition (1), show that A(k)satisfiesthe dual inte- gral equations 2 π ∞ 0 kA(k)[1+M(kh)] sinh(kh)cos(kx) dk = g(x), 0 <x<a, and 2 π ∞ 0 A(k)sinh(kh)cos(kx) dk =0,a<x<∞, where M(kh)=e −kh / sinh(kh). Step 3:Setting sinh(kh)A(k)= a 0 τh(τ)J 0 (kτ)dτ, show that the second integral equation in Step 2 is identically satisfied. Step 4 : Show that the first integral equation in Step 2 leads to the integral equation h(t)+ a 0 τh(τ) ∞ 0 kM(kh)J 0 (kτ)J 0 (kt) dk dτ = t 0 g(x) √ t 2 − x 2 dx with 0 <t<a. Step 5 :Simplify your results in the case g(x)=1andshowthat they are identical with the results given in Example 4.1.3 by Equation 4.1.83 through Equation 4.1.86. © 2008 by Taylor & Francis Group, LLC 200 Mixed Boundary Value Problems 4. Following Example 4.1.2, solve Laplace’s equation 28 ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, −∞ <x<∞, 0 <y<h, with the boundary conditions lim |x|→∞ u(x, y) → 0, 0 <y<h, u y (x, 0) = −p(x)/h, |x| <a, u(x, 0) = 0, |x| >a, (1) and u(x, h)=0, −∞ <x<∞. Step 1 :Usingseparation of variables or transform methods, show that the general solution totheproblem is u(x, y)= 2 π ∞ 0 A(k) e −ky − e ky−2kh 1 − e −2kh cos(kx) dk. Step 2:Using boundary condition (1), show that A(k)satisfiesthe dual inte- gral equations 2 π ∞ 0 k coth(kh)A(k)cos(kx) dk = p(x) h , |x| <a, and 2 π ∞ 0 A(k)cos(kx) dk =0, |x| >a. Step 3:Setting kA(k)= π 2 a 0 g(τ)sin(kτ) dτ, show that the second integral equation in Step 2 is identically satisfied. Step 4: Show that the first integral equation in Step 2 can be rewritten d dx a 0 g(τ) ∞ 0 coth(kh)sin(kτ)sin(kx) dk k dτ = p(x) h , 0 <x<a. 28 Suggested from Singh, B. M., T. B. Moodie, and J. B. Haddow, 1981: Closed-form solutions for finite length crack moving in a strip under anti-plane shear stress. Acta Mech., 38, 99–109. © 2008 by Taylor & Francis Group, LLC Transform Methods 201 −3 −2 −1 0 1 2 3 0 0.5 1 1.5 2 −0.05 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 x y u(x,y)/p 0 Problem 4 Step 5:Usingtherelationship that ∞ 0 coth(kh)sin(kτ)sin(kx) dk k = 1 2 ln tanh(cx)+tanh(cτ) tanh(cx) − tanh(cτ) , where c = π/(2h), show that Step 4 can be written a 0 g(τ)ln tanh(cx)+tanh(cτ) tanh(cx) − tanh(cτ) dτ = 2 h x 0 p(ξ) dξ = F (x), 0 <x<a. Step 6 :Byusingtheresults from Example 1.2.3, show that the solution to the integral equation in Step 5 is g(τ)=− 2c tanh(cτ)sech 2 (cτ) π 2 tanh 2 (ca) −tanh 2 (cτ) a 0 F (x) tanh 2 (ca) − tanh 2 (cx) tanh 2 (cx) −tanh 2 (cτ) dx + 4cF (0) tanh(ca) π 2 sinh(2cτ) tanh 2 (ca) − tanh 2 (cτ) , 0 <τ<a. Step 7:Because F (x)=2p(x)/h with F (0) = 0, show that Step 6 simplifies to g(τ)=− 4c tanh(cτ)sech 2 (cτ) π 2 h tanh 2 (ca) − tanh 2 (cτ) a 0 p(x) tanh 2 (ca) − tanh 2 (cx) tanh 2 (cx) − tanh 2 (cτ) dx, if 0 <τ <a. Step 8:Forthespecial case p(x)=p 0 ,aconstant,showthat g(τ)= 2p 0 sinh(cτ) πh sinh 2 (ca) − sinh 2 (cτ) , 0 <τ <a. The figure entitled Problem 4 illustrates this special solution when a =1and h =2. © 2008 by Taylor & Francis Group, LLC 202 Mixed Boundary Value Problems 4.2 TRIPLE FOURIER INTEGRALS In the previous section we considered the case where the mixed boundary conditions led to two integral equations that are in the form of a Fourier inte- gral. In the present section we take the next step and examine the situation where the mixed boundary condition contains different boundary conditions along three segments. • Example 4.2.1 For our first example, let us solve Laplace’s equation ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, −∞ <x<∞, 0 <y<∞, (4.2.1) subject to the boundary conditions lim |x|→∞ u(x, y) → 0, 0 <y, (4.2.2) lim y→∞ u(x, y) → 0, −∞ <x<∞, (4.2.3) u(x, 0) = −1, −b<x<−a, 1,a<x<b, (4.2.4) and u y (x, 0) = 0, 0 < |x| <a, b<|x| < ∞. (4.2.5) The interesting aspect of this problem is the boundary condition along y =0. Foraportion of the boundary (−b<x<−a and a<x<b), it consists of a Dirichlet condition; otherwise, it is a Neumann condition. If we employ separation of variables or transform methods, the most general solution is u(x, y)= ∞ 0 A(k) k e −ky sin(kx) dk. (4.2.6) Substituting Equation 4.2.6 into the boundary conditions given by Equation 4.2.4 and Equation 4.2.5, we obtain the following set of integral equations: ∞ 0 A(k)sin(kx) dk =0, 0 ≤ x<a, (4.2.7) ∞ 0 A(k) k sin(kx) dk =1,a<x<b, (4.2.8) and ∞ 0 A(k)sin(kx) dk =0,b<x<∞. (4.2.9) © 2008 by Taylor & Francis Group, LLC Transform Methods 203 We must now solve for A(k)whichappears in a set of integral equations. Tranter 29 showed that triple integral equations of the form given by Equation 4.2.7 through Equation 4.2.9 have the solution A(k)=2 ∞ n=1 (−1) n−1 A n J 2n−1 (bk), (4.2.10) where the constants A n are the solution of the dualseriesrelationship ∞ n=1 (−1) n−1 A n sin n − 1 2 ϕ =0, 0 ≤ ϕ<γ, (4.2.11) ∞ n=1 (−1) n−1 A n n − 1 2 sin n − 1 2 ϕ =1,γ<ϕ≤ π, (4.2.12) and γ is defined by a = b sin(γ/2), 0 <γ≤ π.Ifwenowintroduce the change of variables θ = π − ϕ and c = π − γ,wefind that A n is the solution of the following pair of dual series: ∞ n=1 A n n − 1 2 cos n − 1 2 θ =1, 0 <θ<c, (4.2.13) and ∞ n=1 A n cos n − 1 2 θ =0,c<θ≤ π. (4.2.14) Consequently, we have reduced three integral equations to two dual trigono- metric series. Tranter 30 also analyzed dual trigonometric series of the form given by Equation 4.2.13 and Equation 4.2.14 and showed that in our partic- ular case A n = P n−1 [cos(c)] K(a/b) , (4.2.15) where K(·)denotes the complete elliptic integral and P n (·)istheLegendre polynomial of order n. Substituting Equation 4.2.15 into Equation 4.2.10 with a = b cos(c/2), we obtain A(k)= 2 K(a/b) ∞ n=1 (−1) n−1 P n−1 [cos(c)]J 2n−1 (bk). (4.2.16) 29 Tranter, C. J., 1960: Some triple integral equations. Proc. Glasgow Math. Assoc., 4, 200–203. A more accessible analysis is given in Section 6.4 of Sneddon, I. N., 1966: Mixed Boundary Value Problems in Potential Theory.Wiley, 283 pp. 30 Tranter, C. J., 1959: Dual trigonometric series. Proc. Glasgow Math. Assoc., 4, 49–57; Sneddon, op. cit., Section 5.4.5. © 2008 by Taylor & Francis Group, LLC 204 Mixed Boundary Value Problems −4 −2 0 2 4 0 0.5 1 1.5 2 −1 −0.5 0 0.5 1 x y u(x,y) Figure 4.2.1:ThesolutionforEquation 4.2.1 through Equation 4.2.5 when a =1and b =2. Figure 4.2.1 illustrates u(x, y)whena =1andb =2. • Example 4.2.2 Let us solve Laplace’s equation ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, −∞ <x<∞, 0 <y<π, (4.2.17) with the boundary conditions lim |x|→∞ u(x, y) → 0, 0 <y<π, (4.2.18) u y (x, 0) = 0, |x| <a, u(x, 0) = sgn(x),a<|x| <b, u y (x, 0) = 0,b<|x|, (4.2.19) and u(x, π)=0, −∞ <x<∞, (4.2.20) where a<b. Aquickcheck shows that u(x, y)= ∞ 0 A(k) sinh[k(π −y)] sinh(kπ sin(kx) dk k (4.2.21) satisfies Equation 4.2.17, Equation 4.2.18 and Equation 4.2.20. Upon substi- tuting Equation 4.2.21 into Equation 4.2.19, we obtain three integral equa- tions: ∞ 0 A(k)coth(kπ)sin(kx) dk =0, 0 <x<a, (4.2.22) © 2008 by Taylor & Francis Group, LLC Transform Methods 205 ∞ 0 A(k)sin(kx) dk k =1,a<x<b, (4.2.23) and ∞ 0 A(k)coth(kπ)sin(kx) dk =0,b<x<∞. (4.2.24) Singh 31 showed that the solution to ∞ 0 A(k)coth(kπ)sin(kη) dk = F 1 (η), 0 <η<a, (4.2.25) ∞ 0 A(k)sin(kη) dk k = F 2 (η),a<η<b, (4.2.26) and ∞ 0 A(k)coth(kπ)sin(kη) dk =0 b<η<∞ (4.2.27) is coth(kπ)A(k)= 2 π a 0 F 1 (ξ)sin(kξ) dξ + 2 π b a g[cosh(ξ)] sin(kξ)cosh(ξ/2) dξ + 2 π ∞ b F 3 (ξ)sin(kξ) dξ, (4.2.28) where R(η)= b a g[cosh(ξ)] cosh(ξ/2) ln sinh(ξ/2) + sinh(η/2) sinh(ξ/2) −sinh(η/2) dξ, a < η <b, (4.2.29) or R(η)=πF 2 (η) − a 0 F 1 (ξ)ln sinh(ξ/2) + sinh(η/2) sinh(ξ/2) −sinh(η/2) dξ − ∞ b F 3 (ξ)ln sinh(ξ/2) + sinh(η/2) sinh(ξ/2) −sinh(η/2) dξ, (4.2.30) and g[cosh(η)] = − 2 π 2 cosh(η) − cosh(a) cosh(b) −cosh(η) × b a cosh(b) −cosh(ξ) cosh(ξ) − cosh(a) R (ξ)sinh(ξ/2) cosh(ξ) − cosh(η) dξ + C [cosh(η) − cosh(a)][cosh(b) −cosh(η)] . (4.2.31) 31 Singh, B. M., 1973: On triple trigonometrical equations. Glasgow Math. J., 14, 174–178. © 2008 by Taylor & Francis Group, LLC 206 Mixed Boundary Value Problems −5 −3 −1 1 3 5 0 0.2 0.4 0.6 0.8 1 −1 −0.5 0 0.5 1 x y/π u(x,y) Figure 4.2.2:ThesolutionforEquation 4.2.17 through Equation 4.2.20 when a =1and b =2. In the present case, F 1 (η)=F 3 (η)=0andF 2 (η)=1. Therefore, g[cosh(η)] = C [cosh(η) − cosh(a)][cosh(b) −cosh(η)] . (4.2.32) From Equation 4.2.29 and Equation 4.2.30, we find that b a g[cosh(ξ)] cosh(ξ/2) ln sinh(ξ/2) + sinh(η/2) sinh(ξ/2) −sinh(η/2) dξ = π, a < η < b. (4.2.33) Substituting Equation 4.2.32 into Equation 4.2.33 and evaluating the integral, we obtain C = sinh(b/2) K[sinh(a/2)/ sinh(b/2)] , (4.2.34) where K(·)denotes the complete elliptic integral. We then introduce this value of C into Equation 4.2.32 and find that coth(kπ)A(k)= 2 π b a g[cosh(ξ)] sin(kξ)cosh(ξ/2) dξ. (4.2.35) Finally, the potential u(x, y)follows from Equation 4.2.21. Figure 4.2.2 illus- trates the present example. • Example 4.2.3 Ageneralization 32 of the previous example is ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, −∞ <x<∞, 0 <y<h, (4.2.36) 32 See Singh, B. M., and R. S. Dhaliwal, 1984: Closed form solutions to dynamic punch problems by integral transform method. Z. Angew. Math. Mech., 64, 31–34. © 2008 by Taylor & Francis Group, LLC Transform Methods 207 with the boundary conditions lim |x|→∞ u(x, y) → 0, 0 <y<h, (4.2.37) u y (x, 0) = 0, |x| <a, u(x, 0) = f(x),a<|x| <b, u y (x, 0) = 0,b<|x|, (4.2.38) and u(x, h)=0, −∞ <x<∞, (4.2.39) where a<b. Aquickcheck shows that u(x, y)= 2 π ∞ 0 A(k) e −ky − e ky−2kh 1+e −2kh cos(kx) dk (4.2.40) satisfies Equation 4.2.36, Equation 4.2.37 and Equation 4.2.39. Upon substi- tuting Equation 4.2.40 into Equation 4.2.38, we obtain three integral equa- tions: 2 π ∞ 0 kA(k)cos(kx) dk =0, 0 <x<a, (4.2.41) 2 π ∞ 0 tanh(kh)A(k)cos(kx) dk = f(x),a<x<b, (4.2.42) and 2 π ∞ 0 kA(k)cos(kx) dk =0,b<x<∞. (4.2.43) Let us introduce 2 π ∞ 0 kA(k)cos(kx) dk = g(x)sinh(cx),a<x<b, (4.2.44) where g(x)isanunknown function and c = π/(2h). Using Fourier’s inversion theorem, kA(k)= b a g(τ)sinh(cτ)cos(kτ) dτ. (4.2.45) If we substitute Equation 4.2.45 into Equation 4.2.42, interchange the order of integration in the resulting equation, and use ∞ 0 cos(kx)cos(kτ)tanh(kh) dk k = 1 2 ln cosh(cx)+cosh(cτ ) cosh(cx) − cosh(cτ) , (4.2.46) we find that g(τ)isgivenbythe integral equation b a g(τ)sinh(cτ)ln cosh(cx)+cosh(cτ ) cosh(cx) − cosh(cτ) dτ = πf(x),a<x<b. (4.2.47) © 2008 by Taylor & Francis Group, LLC [...]... ] dτ h[r sin(θ)] + 0 41 dθ = 1 0 Gradshteyn and Ryzhik, op cit., Formula 8. 411.1 with n = 0 © 20 08 by Taylor & Francis Group, LLC (4.3.106) 226 Mixed Boundary Value Problems 1 u(r,z) 0 .8 0.6 0.4 0.2 0 2 2 1.5 1.5 1 1 0.5 z 0.5 0 r 0 Figure 4.3.3: The solution u(r, z) to the mixed boundary value problem governed by Equation 4.3 .86 through Equation 4.3.90 with a = b = 1 Equation 4.3.106 is satisfied if...2 08 Mixed Boundary Value Problems 1.4 1.2 u(x,y)/f0 1 0 .8 0.6 0.4 0.2 0 −4 0 −2 0.5 0 1 y 2 1.5 2 4 x Figure 4.2.3: The solution to Equation 4.2.36 subject to the mixed boundary conditions given by Equation 4.2.37, Equation 4.2. 38, and Equation 4.2.39 when a = 1, b = 2, h = 2 and f (x) = f0 Taking the derivative with... equation, Equation 4.3 .81 , then evaluate A(k) using values of h(t) via Equation 4.3.75, and finally employ Equation 4.3.71 • Example 4.3.4 Let us solve40 ∂ 2 u 1 ∂u ∂ 2 u + 2 = 0, + ∂r2 r ∂r ∂z 0 ≤ r < ∞, 0 < z < ∞, (4.3 .86 ) 0 < z < ∞, (4.3 .87 ) subject to the boundary conditions lim |u(r, z)| < ∞, lim u(r, z) → 0, r→∞ r→0 lim u(r, z) → 1, z→∞ u(r, 0) = 0, 0 ≤ r < ∞, 0 ≤ r < ∞, (4.3 .88 ) (4.3 .89 ) 40 See Lebedev,... when they arise in mixed boundary value problems • Example 4.3.2 Let us solve36 ∂ 2 u 1 ∂u ∂ 2 u + 2 = 0, + ∂r2 r ∂r ∂z 0 ≤ r < ∞, 0 < z < a, (4.3. 28) 35 Cooke, J C., 1956: A solution of Tranter’s dual integral equations problem Quart J Mech Appl Math., 9, 103–110 36 See Leong, M S., S C Choo, and K H Tay, 1976: The resistance of an infinite slab with a disc electrode as a mixed boundary value problem Solid-State... contact Solid-State Electron., 36, 143–146 © 20 08 by Taylor & Francis Group, LLC 214 Mixed Boundary Value Problems subject to the boundary conditions lim |u(r, z)| < ∞, lim u(r, z) → 0, r→∞ r→0 0 < z < a, 0 ≤ r < 1, 1 ≤ r < ∞, u(r, 0) = 1, uz (r, 0) = 0, (4.3.29) (4.3.30) and 0 ≤ r < ∞ u(r, a) = 0, (4.3.31) Using Hankel transforms, the solution to Equation 4.3. 28 is ∞ u(r, z) = A(k) 0 sinh[k(a − z)] J0 (kr)... Sov Tech Phys., 2, 1943–1950 © 20 08 by Taylor & Francis Group, LLC 224 Mixed Boundary Value Problems and uz (r, b− ) = uz (r, b+ ), u(r, b) = 1, 0 ≤ r < a, a < r < ∞, (4.3.90) where b− and b+ denote points located slightly below and above the point z = b > 0 Using transform methods or separation of variables, the general solution to Equation 4.3 .86 through Equation 4.3 .89 is ∞ z − b u(r, z) = A(k) 0 sinh(kz)... ∂u ∂ 2 u + 2 = 0, + ∂r2 r ∂r ∂z 0 ≤ r < ∞, 0 < z < h, (4.3.67) 0 < z < h, (4.3. 68) subject to the boundary conditions lim |u(r, z)| < ∞, r→0 lim u(r, z) → 0, r→∞ u(r, h) = 0, and uz (r, 0) = −g(r), u(r, 0) = 0, © 20 08 by Taylor & Francis Group, LLC 0 ≤ r < ∞, 0 ≤ r < a, a ≤ r < ∞ (4.3.69) (4.3.70) 222 Mixed Boundary Value Problems The solution to Equation 4.3.67 through Equation 4.3.70 is ∞ u(r, z) =... Integrating Equation 4.3 .80 with respect to t, we obtain the integral equation h(t) = 2 π t rg(r) 1 √ dr − 2 − r2 π t 0 a K(t, τ )h(τ ) dτ, (4.3 .81 ) 0 where ∞ t K(t, τ ) = 2 0 ∞ =2 kG(k) sin(kτ )J0 (kr) dk 0 t kG(k) 0 0 ∞ =2 r dr √ t2 − r 2 rJ0 (kr) √ dr sin(kτ ) dk t2 − r 2 G(k) sin(kt) sin(kτ ) dk 0 ∞ = 0 (4.3 .82 ) (4.3 .83 ) (4.3 .84 ) G(k){cos[k(t − τ )] − cos[k(t + τ )]} dk (4.3 .85 ) To compute u(r, z),... dopant concentration of a three-dimensional steady-state constant-source diffusion problem Mater Lett., 59, 30 18 3020 © 20 08 by Taylor & Francis Group, LLC Transform Methods 213 then ∞ h f (n) (0) Γ(1 + n/2) n √ + t π n=1 n! Γ(1/2 + n/2) 2 g(t) = √ π (4.3.23) The solution of mixed boundary value problems in cylindrical coordinates often yields dual Fourier-Bessel integral equations of the form ∞ 0 G(k)A(k)Jν... J0 (kr) dk 2 π © 20 08 by Taylor & Francis Group, LLC ∞ 1 q(k) 0 0 h(t)k sin(kt) dt J0 (kr) dk = 1, (4.3.55) 220 Mixed Boundary Value Problems where q(k) = 1 − coth(ka) Because 1 0 and ∞ 0 ∞ 1 h(t)k sin(kt) dt J0 (kr) dk = 0 h (t) cos(kt) dt, (4.3.56) 0 H(r − t) cos(kt)J0 (kr) dk = √ , r 2 − t2 1 0 1 h(t)k sin(kt) dt = −h(1) cos(k) + (4.3.57) ∞ h (t) 0 cos(kt)J0 (kr) dk dt 0 (4.3. 58) r = 0 h (t) √ dt . by Equation 4.1 .83 through Equation 4.1 .86 . © 20 08 by Taylor & Francis Group, LLC 200 Mixed Boundary Value Problems 4. Following Example 4.1.2, solve Laplace’s equation 28 ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0,. =1and h =2. © 20 08 by Taylor & Francis Group, LLC 202 Mixed Boundary Value Problems 4.2 TRIPLE FOURIER INTEGRALS In the previous section we considered the case where the mixed boundary conditions. trigonometrical equations. Glasgow Math. J., 14, 174–1 78. © 20 08 by Taylor & Francis Group, LLC 206 Mixed Boundary Value Problems −5 −3 −1 1 3 5 0 0.2 0.4 0.6 0 .8 1 −1 −0.5 0 0.5 1 x y/π u(x,y) Figure 4.2.2:ThesolutionforEquation
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