THÔNG TIN TÀI LIỆU
258 Mixed Boundary Value Problems lim z→∞ u(r, z) → 0, 0 ≤ r<∞, (4.3.329) and αu z (r, 0) − βu(r, 0) = −f(r), 0 ≤ r<1, u(r, 0) = 0, 1 <r<∞. (4.3.330) Both α and β are nonzero. In line with previous examples the solution that satisfies Equation 4.3.327 through Equation 4.3.329 is u(r, z)= ∞ 0 A(k)e −kz J 0 (kr) dk. (4.3.331) Substituting u(r, z)intothe mixed boundary condition, ∞ 0 (αk + β)A(k)J 0 (kr) dk = f(r), 0 ≤ r<1, (4.3.332) and ∞ 0 A(k)J 0 (kr) dk =0, 1 ≤ r<∞. (4.3.333) At this point we introduce an integral definition for A(k), A(k)= 1 0 h(t)sin(kt) dt, h(0) = 0. (4.3.334) The demonstration that this definition of A(k)satisfiesEquation 4.3.333 is left as part of Problem 6. Turning to Equation 4.3.332, we substitute A(k) into Equation 4.3.332 and find that α 1 0 h(t) ∞ 0 k sin(kt)J 0 (kr) dk dt (4.3.335) + β 1 0 h(t) ∞ 0 sin(kt)J 0 (kr) dk dt = f(r), 0 ≤ r<1. At this point, we would normally manipulate Equation 4.3.335 into a Fredholm integral equation. This is left as an exercise in Problem 6. Here we introduce an alternative method developed by Gladwell et al. 62 The derivation begins by showing that x 0 t ∞ 0 kA(k)J 0 (kt) dk dt √ x 2 − t 2 = ∞ 0 A(k)sin(kx) dk. (4.3.336) 62 TakenfromGladwell, G. M. L., J. R. Barber, and Z. Olesiak, 1983: Thermal problems with radiation boundary conditions. Quart. J. Mech.Appl. Math., 36, 387–401 by permission of Oxford University Press; see also Lemczyk, T. F., and M. M. Yovanovich, 1988: Thermal constriction resistance with convective boundary conditions–1. Half-space contacts. Int. J. Heat Mass Transfer, 31, 1861–1872. © 2008 by Taylor & Francis Group, LLC Transform Methods 259 This follows from interchanging the order of integration and applying Equa- tion 1.4.9. Next, we view the quantity within the square brackets on the left side of Equation 4.3.336 as the unknown in an integral equation of the Abel type. From Equation 1.2.13 and Equation 1.2.14, we have that t ∞ 0 kA(k)J 0 (kt) dk = 2 π d dt t 0 ∞ 0 A(k)sin(kτ) dk τdτ √ t 2 − τ 2 (4.3.337) = 2 π d dt ∞ 0 A(k) t 0 τ sin(kτ) √ t 2 − τ 2 dτ dk (4.3.338) = d dt t ∞ 0 A(k)J 1 (kt) dk (4.3.339) = d dt t 1 0 h(τ) ∞ 0 sin(kτ)J 1 (kt) dk dτ (4.3.340) = d dt t 0 τh(τ) √ t 2 − τ 2 dτ . (4.3.341) If we divide the left side of Equation 4.3.341 by t,wehavethefirst term on theleft side of Equation 4.3.335. The second term can be evaluated from integral tables. 63 Consequently Equation 4.3.335 becomes α r d dr r 0 th(t) √ r 2 − t 2 dt + β 1 r h(t) √ t 2 − r 2 dt = f(r), 0 ≤ r<1. (4.3.342) To solve Equation 4.3.342, let h(t)=ψ(θ)= ∞ n=0 A n cos[(2n +1)θ], 0 <θ<π/2, (4.3.343) and sin(θ)ψ(θ)= ∞ n=0 B n cos[(2n +1)θ], 0 <θ<π/2, (4.3.344) where t =cos(θ). If r =cos(ϕ), then r 0 th(t) √ r 2 − t 2 dt = ∞ n=0 A n π ϕ sin(2θ)cos[(2n +1)θ] 2[cos(2ϕ) − cos(2θ)] dθ (4.3.345) = π 8 ∞ n=0 A n {P n+1 [cos(2ϕ)] −P n−1 [cos(2ϕ)]} (4.3.346) 63 Gradshteyn and Ryzhik, op. cit., Formula 4.671.1. © 2008 by Taylor & Francis Group, LLC 260 Mixed Boundary Value Problems from Equation 1.3.4 and Equation 1.3.5, where the prime denotes that when- ever P −1 (·)occurs, then it is replaced by P 0 (·). Similarly, 1 r h(t) √ r 2 − t 2 dt =2 ∞ n=0 B n ϕ 0 cos[(2n +1)θ] 2[cos(2ϕ) −cos(2θ)] dθ (4.3.347) = π 2 ∞ n=0 B n P n [cos(2ϕ)]. (4.3.348) Because P n (x) − P n+2 (x) 1 − x 2 = (2n +3) 2(n +1)(n +2) n k=0 1+(−1) n+k (2k +1)P k (x) (4.3.349) and using the derivative rule for Legendre polynomials, we find that 1 r d dr P n+1 (2r 2 − 1) −P n−1 (2r 2 − 1) =4(2n +1)P n (2r 2 − 1). (4.3.350) Therefore, Equation 4.3.342 becomes απ 2 ∞ n=0 (2n +1)A n P n (2r 2 − 1) + βπ 2 ∞ n=0 B n P n (2r 2 − 1) = f(r). (4.3.351) If we reexpress f(r)astheFourier-Legendreexpansion f(r)= π 2 ∞ n=0 (2n +1)C n P n (2r 2 − 1), (4.3.352) then αA n + β 2n +1 B n = C n ,n=0, 1, 2, (4.3.353) In addition to Equation 4.3.353, we also have from Equation 4.3.344 that B n = ∞ m=0 c m,n A n , (4.3.354) where c m,n = 2 π π 0 sin(θ)cos[(2m +1)θ]cos[(2n +1)θ] dθ (4.3.355) = 1 π 1 2n +2m +3 − 1 2n +2m +1 + 1 2n − 2m +1 − 1 2n − 2m − 1 . (4.3.356) © 2008 by Taylor & Francis Group, LLC Transform Methods 261 We can now solve for A n via Equation 4.3.353 and Equation 4.3.354 after we truncate the set of infinite equations to a finite number. Then h(t)follows from Equation 4.3.343. Finally, A(k)iscomputed from Equation 4.3.334 while u(r, z)isobtained from Equation 4.3.331. Problems 1. Solve the potential problem 64 ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, subject to the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ |u(r, z)| < ∞, 0 <z<∞, lim z→∞ u(r, z) → u ∞ , 0 ≤ r<∞, and u(r, 0) = u ∞ − ∆u, 0 ≤ r<a, u z (r, 0) = 0,a≤ r<∞, where u ∞ and ∆u are constants. Step 1: Show that u(r, z)=u ∞ − ∞ 0 A(k)e −kz J 0 (kr) dk satisfies the partial differential equation and the boundary conditions as r → 0, r →∞,andz →∞. Step 2: Show that ∞ 0 kA(k)J 0 (kr) dk =0,a<r<∞. Step 3:Usingtherelationship 65 ∞ 0 sin(ka)J 0 (kr) dk = (a 2 − r 2 ) − 1 2 ,r<a, 0,r>a, 64 See Fleischmann, M., and S. Pons, 1987: The behavior of microdisk and microring electrodes. J. Electroanal. Chem., 222, 107–115; Gupta, S. C., 1957: Slow broad side motion of a flat plate in a viscous liquid. Z. Angew. Math. Phys., 8, 257–261. 65 Gradshteyn and Ryzhik, op. cit., Formula 6.671.7. © 2008 by Taylor & Francis Group, LLC 262 Mixed Boundary Value Problems show that kA(k)=C sin(ka). Step 4:Usingtherelationship 66 ∞ 0 sin(ka)J 0 (kr) dk k = π/2,r≤ a, arcsin(a/r),r≥ a, show that u(r, z)=u ∞ − 2∆u π ∞ 0 e −kz sin(ka)J 0 (kr) dk k = u ∞ − 2∆u π arcsin 2a (r − a) 2 + z 2 + (r + a) 2 + z 2 . Wiley and Webster 67 used this solution in an improved design for a circular electrosurgical dispersive electrode. 2. Solve Laplace’s equation 68 1 r ∂ ∂r r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, with the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, lim z→∞ u(r, z) → 0, 0 ≤ r<∞, and u z (r, 0) = 1, 0 ≤ r<1, u(r, 0) = 0, 1 <r<∞. (1) See Problem 5 for a generalization of this problem. Step 1 :Usingseparation of variables or transform methods, show that the general solution totheproblem is u(r, z)= ∞ 0 kA(k)e −kz J 0 (kr) dk. 66 Ibid., Formula 6.693.1 with ν =0. 67 Wiley, J. D., and J. G. Webster, 1982: Analysis and control of the current distribution under circular dispersive electrode. IEEE Trans. Biomed. Engng., BME-29, 381–385. 68 See Yang, F Q., and J. C. M. Li, 1995: Impression and diffusion creep of anisotropic media. J. Appl. Phys., 77, 110–117. This problem also appears while finding the tempera- ture field in a paper by Florence, A. L., and J. N. Goodier, 1963: The linear thermoelastic problem of uniform heat flow disturbed by a penny-shaped insulated crack. Int. J. Engng. Sci., 1, 533–540. © 2008 by Taylor & Francis Group, LLC Transform Methods 263 0 0.5 1 1.5 2 0 0.5 1 1.5 2 −0.7 −0.6 −0.5 −0.4 −0.3 −0.2 −0.1 0 r z u(r,z) Problem 2 Step 2:Using boundary condition (1), show that A(k)satisfiesthe dual inte- gral equations ∞ 0 k 2 A(k)J 0 (kr) dk = −1, 0 ≤ r<1, and ∞ 0 kA(k)J 0 (kr) dk =0, 1 <r<∞. Step 3:UsingEquation 1.4.13 and the result 69 from integral tables that ∞ 0 J ν (ax)sin(bx) dx x = sin[ν arcsin(b/a)]/ν, b ≤ a, a ν sin(νπ/2) ν b + √ b 2 − a 2 ν b ≥ a, if (ν) > −1, show that A(k)=− 2 kπ 1 0 t sin(kt) dt = − 2 π sin(k) k 3 − cos(k) k 2 satisfies both integral equations given in Step 2. Step 4: Show that the solution to this problem is u(r, z)=− 2 π ∞ 0 sin(k) k 2 − cos(k) k e −kz J 0 (kr) dk. In particular, show that u(r, 0) = −2 √ 1 − r 2 /π if 0 ≤ r<1. The figure labeled Problem 2 illustrates the solution u(r, z). 69 Gradshteyn and Ryzhik, op. cit., Formula 6.693.1. © 2008 by Taylor & Francis Group, LLC 264 Mixed Boundary Value Problems 3. Solve the potential problem 70 1 r ∂ ∂r r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, subject to the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, lim z→∞ u(r, z) → 0, 0 ≤ r<∞, and u(r, 0) = U 0 ,r<a, u z (r, 0) = 0,r>a. Step 1:Byusingeither separation of variables or transform methods, show that the general solution to partial differential equation is u(r, z)= 2 π ∞ 0 A(k)e −kz J 0 (kr) dk. Note that this solution also satisfies the first three boundary conditions. Step 2:Setting A(k)= a 0 f(ξ)cos(kξ) dξ, show that u z (r, 0) = 0 if r>a. Step 3: Show that 2 π ∞ 0 J 0 (kr) a 0 f(ξ)cos(kξ) dξ dk = U 0 , 0 <r<a. Step 4:Byreplacing r by η in the integral equation given in Step 3, multiplying the resulting equation by ηdη/ r 2 − η 2 ,integratingfrom0tor,andtaking the derivative with respect to r,showthat 2 π d dr r 0 η r 2 − η 2 ∞ 0 J 0 (kη) a 0 f(ξ)cos(kξ) dξ dk dη = U 0 . Step 5:Byinterchanging the order of the η and k integrations and using the relationship that r 0 ηJ 0 (kη) r 2 − η 2 dη = sin(kr) k , 70 See Laporte, O., and R. G. Fowler, 1967: Weber’s mixed boundary value problem in electrodynamics. J. Math. Phys., 8 , 518–522. © 2008 by Taylor & Francis Group, LLC Transform Methods 265 show that the integral equation in Step 4 simplifies to 2 π d dr a 0 f(ξ) ∞ 0 sin(kr)cos(kξ) dk k dξ = U 0 . Step 6:Usetheresults from Step 5 to show that if f(ξ)=U 0 ,thenA(k)= U 0 sin(ka)/k,and u(r, z)= 2U 0 π ∞ 0 sin(ka) k e −kz J 0 (kr) dk = 2U 0 π arcsin 2a (r − a) 2 + z 2 + (r + a) 2 + z 2 . 4. Solve Laplace’s equation 71 ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, subject to the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, lim z→∞ u(r, z) → 0, 0 ≤ r<∞, and u z (r, 0) = 1, 0 ≤ r<a, u(r, 0) = 0,a<r<∞. Step 1: Show that u(r, z)= ∞ 0 kA(k)e −kz J 0 (kr) dk satisfies the partial differential equation and the boundary conditions provided that A(k)satisfiesthedual integral equations ∞ 0 k 2 A(k)J 0 (kr) dk = −1, 0 ≤ r<a, (1) and ∞ 0 kA(k)J 0 (kr) dk =0,a≤ r<∞. (2) 71 See Shindo, Y., 1986: The linear magnetoelastic problem of a uniform current flow disturbed by a penny-shaped crack in a constant axial magnetic field. Eng. Fract. Mech., 23, 977–982. © 2008 by Taylor & Francis Group, LLC 266 Mixed Boundary Value Problems 0 0.5 1 1.5 2 0 0.5 1 1.5 2 −0.2 0 0.2 0.4 0.6 0.8 1 1.2 a u(r,z) Problem 4 Step 2:Byintroducing kA(k)= a 0 h(t)sin(kt) dt, h(0) = 0, show that Equation (2) is automatically satisfied. Step 3:ByusingEquation (1), show that h(t)=−2t/π and kA(k)=− 2 πk 2 [sin(ka) − kacos(ka)] . Step 4: Show that the solution to the problem is u(r, z)=− 2 π ∞ 0 [sin(ka) − kacos(ka)] e −kz J 0 (kr) dk k 2 . The figure labeled Problem 4 illustrates this solution u(r, z). 5. Solve Laplace’s equation 72 ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, subject to the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, 72 See Riffert, H., 1980: Pulsating X-ray sources: The oblique dipole configuration. Astrophys. Space Sci., 71, 195–201. © 2008 by Taylor & Francis Group, LLC 268 Mixed Boundary Value Problems 6. Solve the partial differential equation 73 ∂ 2 u ∂r 2 + 1 r ∂u ∂r − u r 2 + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, subject to the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, lim z→∞ u(r, z) → 0, 0 ≤ r<∞, and u z (r, 0) = 0, 0 ≤ r<a, u(r, 0) = a/r, a < r < ∞. Step 1: Show that u(r, z)= ∞ 0 A(k)e −kz J 1 (kr) dk satisfies the partial differential equation and the boundary conditions provided that A(k)satisfiesthedual integral equations ∞ 0 kA(k)J 1 (kr) dk =0, 0 ≤ r<a, and ∞ 0 A(k)J 1 (kr) dk = a/r, a ≤ r<∞. Step 2: Show that A(k)= sin(ka) k satisfies the integral equations. The figure labeled Problem 6 illustrates this solution. 7. A generalization of a problem originally suggested by Popova 74 was given by Kuz’min 75 who solved ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, 73 Suggested by a problem solved in Appendix A of Raynolds, J. E., B. A. Munk, J. B. Pryor, and R. J. Marhefka, 2003: Ohmic loss in frequency-selective surfaces. J. Appl. Phys., 93, 5346–5358. 74 Popova,A.P., 1973: Nonstationary mixed problem of thermal conductivity for the half-space. J. Engng. Phys., 25, 934–935. 75 Kuz’min, Yu. N., 1966: Some axially symmetric problems in heat flow with mixed boundary conditions. Sov. Tech. Phys., 11, 169–173. © 2008 by Taylor & Francis Group, LLC [...]... in cylindrical coordinates: ∂ 2 u 1 ∂u ∂ 2 u + 2 = 0, + ∂r2 r ∂r ∂z 0 ≤ r < ∞, −h < z < ∞, subject to the boundary conditions lim |u(r, z)| < ∞, r→0 lim u(r, z) → 0, r→∞ u(r, −h) = 0, 82 Ehrenmark, op cit © 2008 by Taylor & Francis Group, LLC −h < z < ∞, 0 ≤ r < ∞, (1) (2) 282 Mixed Boundary Value Problems u(r, 0− ) = u(r, 0+ ) = 1, u(r, 0− ) = u(r, 0+ ), uz (r, 0− ) = uz (r, 0+ ), 0 ≤ r < 1, 1 < r . & Francis Group, LLC 264 Mixed Boundary Value Problems 3. Solve the potential problem 70 1 r ∂ ∂r r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, subject to the boundary conditions lim r→0 |u(r,. Group, LLC 268 Mixed Boundary Value Problems 6. Solve the partial differential equation 73 ∂ 2 u ∂r 2 + 1 r ∂u ∂r − u r 2 + ∂ 2 u ∂z 2 =0, 0 ≤ r<∞, 0 <z<∞, subject to the boundary conditions lim r→0 |u(r,. and Ryzhik, op. cit., Formula 6.671.7. © 2008 by Taylor & Francis Group, LLC 262 Mixed Boundary Value Problems show that kA(k)=C sin(ka). Step 4:Usingtherelationship 66 ∞ 0 sin(ka)J 0 (kr) dk k = π/2,r≤
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