318 Mixed Boundary Value Problems 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 r z u(r,z) Problem 1 The figure labeled Problem 1 illustrates this solution when a =0.5. 4.5 JOINT TRANSFORM METHODS In the previous sections we sought to separate problems according to whether the kernels of dual or triple integral equations contained trigonomet- rical or Bessel functions. Such clear-cut lines of demarcation are not always possible and we conclude with examples where the analysis includes both Fourier and Hankel transforms as well as Fourier and Fourier-Bessel series. • Example 4.5.1 In Section 4.1 we solved Laplace’s equation on an infinite strip. See Equation 4.1.84 through Equation 4.1.100. Here, we again solve 101 Laplace’s equation but on a semi-infinite domain which contains two regions with dif- ferent properties: ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =0, 0 <x<∞, 0 <y<L, (4.5.1) subject to the boundary conditions u x (0,y)=0, 0 <y<L, (4.5.2) lim x→∞ u(x, y) → 0, 0 <y<L, (4.5.3) u(x, L)=0, 0 <x<∞, (4.5.4) 101 See Shindo, Y., and A. Atsumi, 1975: Thermal stresses in a laminate composite with infinite row of parallel cracks normal to the interfaces. Int. J. Engng. Sci., 13, 25–42. © 2008 by Taylor & Francis Group, LLC Transform Methods 319 u(h − ,y)=u(h + ,y), 1 u x (h − ,y)= 2 u x (h + ,y), 0 <y<L, (4.5.5) and u y (x, 0) = −1, 0 <x<a, u(x, 0) = 0,a<x<∞, (4.5.6) where h>a.Theeffect of these two different regions introduces an interfacial condition, Equation 4.5.5. The solution to Equation 4.5.1 to Equation 4.5.4 is u(x, y)= ∞ 0 A(k) sinh[k(L − y)] sinh(kL) cos(kx) dk + ∞ n=1 A n cosh nπx L sin nπy L (4.5.7) for 0 <x<h;and u(x, y)= ∞ n=1 B n exp − nπx L sin nπy L (4.5.8) for h<x<∞.Aninteresting aspect of this problem is that Equation 4.5.7 contains both a Fourier cosine transform and a Fourier sine series. Substi- tuting Equation 4.5.7 and Equation 4.5.8 into Equation 4.5.6 yields the dual integral equations ∞ 0 kA(k)[1+M(kL)] cos(kx) dk− ∞ n=1 nπ L A n cosh nπx L =1, 0 <x<a, (4.5.9) and ∞ 0 A(k)cos(kx) dk =0,a<x<∞, (4.5.10) where M (µ)=e −µ / sinh(µ). To solve this set of dual integral equations, we introduce A(k)= a 0 g(t)J 0 (kt) dt. (4.5.11) We chose this definition for A(k)because ∞ 0 A(k)cos(kx) dk = a 0 g(t) ∞ 0 cos(kx)J 0 (kt) dk dt =0, (4.5.12) where weusedEquation 1.4.14. Note that 0 ≤ t ≤ x<∞. Turning toEquation4.5.9, we first integrate it with respect to x and obtain ∞ 0 A(k)[1 + M(kL)] sin(kx) dk − ∞ n=1 A n sinh nπx L = x, 0 <x<a; (4.5.13) © 2008 by Taylor & Francis Group, LLC 320 Mixed Boundary Value Problems or a 0 g(t) ∞ 0 sin(kx)J 0 (kt) dk dt + a 0 g(t) ∞ 0 M(kL)sin(kx)J 0 (kt) dk dt − ∞ n=1 A n sinh nπx L = x, 0 <x<a. (4.5.14) Using Equation 1.4.13, the first term in Equation 4.5.14 can be simplified and we find that x 0 g(t) √ x 2 − t 2 dt + a 0 g(τ) ∞ 0 M(kL)sin(kx)J 0 (kτ) dk dτ (4.5.15) − ∞ n=1 A n sinh nπx L = x, 0 <x<a. Solving for g(t)inthefirst term by using Equation 1.2.13 and Equation 1.2.14, g(x)+ 2 π a 0 g(τ) ∞ 0 M(kL) d dx x 0 t sin(kt) √ x 2 − t 2 dt J 0 (kτ) dk dτ − 2 π ∞ n=1 A n d dx x 0 t sinh(nπx/L) √ x 2 − t 2 dt = 2τ 0 π d dx t 2 √ x 2 − t 2 dt (4.5.16) when 0 <x<a.UsingEquation 4.1.98, Equation 4.1.99, and the fact 102 that x 0 t sinh(nπt) √ x 2 − t 2 dt = πx 2 I 1 nπx L , (4.5.17) we finally obtain g(x)+x a 0 g(τ) ∞ 0 kM(kL)J 0 (kx)J 0 (kτ) dk dτ − ∞ n=1 A n nπx L I 0 nπx L = x (4.5.18) for 0 <x<a. Before we can solve Equation 4.5.18, we must eliminate A n from this equation. To that end, we apply the interfacial condition Equation 4.5.5 and find that −cosh nπh L A n +exp − nπh L B n = 2 L ∞ 0 A(k) nπ/L k 2 + n 2 π 2 /L 2 cos(kh) dk (4.5.19) 102 Gradshteyn and Ryzhik, op. cit., Formula 3.365.1 and Formula 3.389.3. © 2008 by Taylor & Francis Group, LLC Transform Methods 321 and sinh nπh L A n + 2 1 exp − nπh L B n = 2 L ∞ 0 A(k) k sin(kh) k 2 + n 2 π 2 /L 2 dk. (4.5.20) Solving for A n and B n , A n = 2(1 − 2 / 1 ) L[sinh(nπh/L)+( 2 / 1 )cosh(nπh/L)] F (nπ/L), (4.5.21) and B n = 2[sinh(nπh/L)+cosh(nπh/L)] L[sinh(nπh/L)+( 2 / 1 )cosh(nπh/L)] F (nπ/L), (4.5.22) where F (nπ/L)= ∞ 0 A(k) k sin(kh) k 2 + n 2 π 2 /L 2 dk (4.5.23) = ∞ 0 A(k) nπL cos(kh) k 2 L 2 + n 2 π 2 dk (4.5.24) = π 2 e −nπh/L a 0 g(t)I 0 nπt L dt. (4.5.25) Finally, we can bring everything into a nondimensional form by introduc- ing ξ = x/a, η = τ/a, ρ = L/a, κ = a/h,andG(ξ)=h(aξ)/(a √ ξ ). This gives G(ξ)+ 1 0 G(η)K(ξ, η) dη = ξ, (4.5.26) where K(ξ,η)= √ ξη ρ 2 ∞ 0 µM(µ)J 0 (ξµ/ρ)J 0 (ηµ/ρ) dµ (4.5.27) − π 2 ∞ n=1 n(1 − 2 / 1 )e −nπ/(κρ) sinh[nπ/(κρ)] + 2 cosh[nπ/(κρ)]/ 1 I 0 nπξ ρ I 0 nπη ρ . Once G(ξ)iscomputed via Equation 4.5.26, we find A n , B n and A(k)from Equation 4.5.21, Equation 4.5.22, and Equation 4.5.11, respectively. The solution u(x, y)follows from Equation 4.5.7 and Equation 4.5.8. Figure 4.5.1 illustrates this solution when h/a = 1 2 and a/L =1. • Example 4.5.2 Let us solve 103 ∂ 2 u ∂r 2 + 1 r ∂u ∂r − u r 2 + ∂ 2 u ∂z 2 =0, 0 ≤ r<a, −∞ <z<∞, (4.5.28) 103 Takenwithpermission from Kim, M U., 1981: Slow rotation of a disk in a fluid-filled circular cylinder. J. Phys. Soc. Japan, 50, 4063–4067. © 2008 by Taylor & Francis Group, LLC Transform Methods 323 The interesting aspect of this problem is that the dual integral equations contain both Fourier and Fourier-Bessel transforms. We begin our solution of Equation 4.5.33 and Equation 4.5.34 by intro- ducing A(k)= 1 0 h(t)sin(kt) dt. (4.5.35) If we substitute Equation 4.5.35 into Equation 4.5.34, we can show that this definition of A(k)satisfiesEquation 4.5.34 identically. On the other hand, Equation 4.5.33 yields 1 0 h(t) ∞ 0 sin(kt)J 1 (kt) dk dt = r − ∞ 0 B(k)I 1 (kr) dk. (4.5.36) Because ∞ 0 sin(kt)J 1 (kr) dk = tH(r − t) r √ r 2 − t 2 , (4.5.37) Equation 4.5.36 becomes r 0 th(t) √ r 2 − t 2 dt = r 2 − r ∞ 0 B(k)I 1 (kr) dk. (4.5.38) We now use results from Chapter 1, namely Equation 1.2.7 with α = 1 2 ,that the solution to Equation 4.5.38 is h(t)= 2 πt d dt t 0 ξ 3 − ∞ 0 B(k)ξ 2 I 1 (kξ) dk dξ t 2 − ξ 2 . (4.5.39) Using the relationship that t 0 ξ 2 I 1 (kξ) t 2 − ξ 2 dξ = kt cosh(kt) − sinh(kt) k 2 , (4.5.40) we can simplify Equation 4.5.39 to h(t)= 2 π 2t − ∞ 0 B(k)sinh(kt) dk . (4.5.41) To evaluate B(k), we first substitute Equation 4.5.32 into the boundary condition given by Equation 4.5.31. This yields ∞ 0 B(k)I 1 (ka)cos(kz) dk = − ∞ 0 A(η)J 1 (ηa)e −η|z| dη. (4.5.42) © 2008 by Taylor & Francis Group, LLC 324 Mixed Boundary Value Problems Recognizing that the left side of Equation 4.5.42 is the Fourier cosine trans- form representation of the right side, B(k)isgivenby I 1 (ka)B(k)=− 2 π ∞ 0 ∞ 0 J 1 (ηa)e −ηz 1 0 h(τ)sin(ητ) dτ dη cos(kz) dz (4.5.43) = − 2 π 1 0 h(τ) ∞ 0 sin(ητ)J 1 (ηa) ∞ 0 e −ηz cos(kz) dz dη dτ (4.5.44) = − 2 π 1 0 h(τ) ∞ 0 sin(ητ)J 1 (ηa) η η 2 + k 2 dη dτ (4.5.45) = − 2 π 1 0 h(τ)sinh(kτ)K 1 (ka) dτ, (4.5.46) where weusedintegraltables 104 for the η integration. Substituting Equation 4.5.46 into Equation 4.5.41, we obtain h(t)= 2 π 2t + 2 π 1 0 h(τ) ∞ 0 K 1 (ka) I 1 (ka) sinh(kt)sinh(kτ) dk dτ ; (4.5.47) or h(t) − 4 π 2 1 0 h(τ) ∞ 0 K 1 (ka) I 1 (ka) sinh(kt)sinh(kτ) dk dτ = 4t π , (4.5.48) for 0 <t<1. To compute u(r, z), we first solve Equation 4.5.48 to find h(t). Next, we use Simpson’s rule to evaluate A(k)andB(k)fromEquation 4.5.35 and Equation 4.5.46, respectively. Finally, Equation 4.5.32 gives u(r, z). Figure 4.5.2 illustrates the solution when a =2. • Example 4.5.3 Given a<1anddenoting a nonnegative integer by n,letussolve 105 ∂ 2 u ∂r 2 + 1 r ∂u ∂r − n 2 r 2 u + ∂ 2 u ∂z 2 =0, 0 ≤ r<1, 0 ≤ z<∞, (4.5.49) subject to the boundary conditions lim r→0 |u(r, z)| < ∞,u(1,z)=g(z), 0 ≤ z<∞, (4.5.50) 104 Gradshteyn and Ryzhik, op. cit., Formula 6.718. 105 See Rusia, K. C., 1968: On certain asymmetric mixed boundary value problems of an electrified circular disc situated inside a coaxial infinite hollow cylinder. Indian J. Pure Appl. Phys., 6, 44–46. © 2008 by Taylor & Francis Group, LLC 326 Mixed Boundary Value Problems We begin our solution of Equation 4.5.54 and Equation 4.5.55 by intro- ducing A(k)= √ k a 0 h(t)t − ( n− 1 2 ) J n+ 1 2 (kt) dt. (4.5.56) If we substitute Equation 4.5.56 into Equation 4.5.55 and interchange the order of integration, ∞ 0 A(k)J n (kr) dk = a 0 h(t)t − ( n− 1 2 ) ∞ 0 √ kJ n (kr)J n+ 1 2 (kt) dk dt. (4.5.57) From tables 106 and noting that r>t,thevalueofthe integral within the square brackets equals zero and this choice of A(k)satisfiesEquation 4.5.55. If we integrate Equation 4.5.56 by parts and assuming that h(t)t − ( n− 1 2 ) tends to zero as t → 0, A(k)=− h(a)J n− 1 2 (ka) √ ka n− 1 2 + 1 √ k a 0 h (t)t − ( n− 1 2 ) J n− 1 2 (kt) dt. (4.5.58) Substituting Equation 4.5.58 into Equation 4.5.54, interchanging the order of integration, and carrying out the k-integration, − h(a) a n− 1 2 ∞ 0 √ kJ n− 1 2 (ka)J n (kr) dk + a 0 h (t) t n− 1 2 ∞ 0 √ kJ n− 1 2 (kt)J n (kr) dk dt = ∞ 0 kB(k)I n (kr) dk − f(r). (4.5.59) In Equation 4.5.59, the first integral on the left side vanishes while the integral inside the square brackets can be evaluated using tables. The end result is r 0 h (t) √ r 2 − t 2 dt = π 2 r n ∞ 0 kB(k)I n (kr) dk − f(r) , 0 ≤ r<a. (4.5.60) Applying the results from Equation 1.2.13 and Equation 1.2.14, we find for h(t)that h(t)= 2 π t 0 rdr √ t 2 − r 2 π 2 r n ∞ 0 kB(k)I n (kr) dk − f(r) (4.5.61) = 2 π t 0 r n+1 √ t 2 − r 2 ∞ 0 kB(k)I n (kr) dk − f(r) dr (4.5.62) 106 Gradshteyn and Ryzhik, op. cit., Formula 6.575.1. Note that this formula has a typo; the condition should read (ν +1) > (µ) > −1. See p. 100 in Magnus, W., F. Oberhettinger, and R. P. Soni, 1966: Formulas and Theorems for the Special Functions of Mathematical Physics. Springer-Verlag, 508 pp. © 2008 by Taylor & Francis Group, LLC Transform Methods 327 = − 2 π t 0 r n+1 f(r) √ t 2 − r 2 dr + 2 π ∞ 0 kB(k) t 0 r n+1 I n (kr) √ t 2 − r 2 dr dk (4.5.63) = − 2 π t 0 r n+1 f(r) √ t 2 − r 2 dr + t n+ 1 2 ∞ 0 √ kB(k)I n+ 1 2 (kt) dk, (4.5.64) where weused 107 t 0 r n+1 I n (kr) √ t 2 − r 2 dr = π 2 t n+ 1 2 √ k I n+ 1 2 (kt), (n) > −1. (4.5.65) Turning to the last boundary condition, the second part of Equation 4.5.50, ∞ 0 A(k)J n (k)e −kz dk + ∞ 0 B(k)I n (k)sin(kz) dk = g(z), 0 <z<∞. (4.5.66) Noting that Equation 4.5.66 is a Fourier sine transform with B(k)asthe Fourier coefficient, I n (k)B(k)= 2 π ∞ 0 g(z)sin(kz) dz − 2 π ∞ 0 ∞ 0 A(t)J n (t)e −tz dt sin(kz) dz (4.5.67) = 2 π ∞ 0 g(z)sin(kz) dz − 2k π ∞ 0 A(t)J n (t) t 2 + k 2 dt. (4.5.68) Substituting for A(k)theexpression given by Equation 4.5.56, I n (k)B(k)= 2 π ∞ 0 g(z)sin(kz) dz − 2k π ∞ 0 J n (t) t 2 + k 2 √ t a 0 h(ξ) ξ n− 1 2 J n+ 1 2 (tξ) dξ dt (4.5.69) = 2 π ∞ 0 g(z)sin(kz) dz − 2k π a 0 h(ξ) ξ n− 1 2 ∞ 0 √ t t 2 + k 2 J n (t)J n+ 1 2 (ξt) dt dξ (4.5.70) = 2 π ∞ 0 g(z)sin(kz) dz − 2 π √ kK n (k) a 0 h(ξ) ξ n− 1 2 I n+ 1 2 (kξ) dξ. (4.5.71) 107 Sneddon [Sneddon, op. cit., p. 30.] showed how Sonine’s first integral can be rewritten as t 0 ξ µ+1 (t 2 − ξ 2 ) ν J µ (yξ) dξ =2 ν t µ+ν+1 y −ν−1 Γ(ν +1)J ν+ν+1 (ty). Equation 4.5.65 follows by setting ξ = r, µ = n, ν = − 1 2 ,andy = ik. © 2008 by Taylor & Francis Group, LLC 328 Mixed Boundary Value Problems Substituting Equation 4.5.71 into Equation 4.5.64 and interchanging the order of integration, h(t)=− 2 π t 0 r n+1 f(r) √ t 2 − r 2 dr + t n+ 1 2 ∞ 0 √ kI n+ 1 2 (kt) I n (k) dk (4.5.72) × 2 π ∞ 0 g(z)sin(kz) dz − 2 π √ kK n (k) a 0 h(ξ) ξ n− 1 2 I n+ 1 2 (kξ) dξ = − 2 π t 0 r n+1 f(r) √ t 2 − r 2 dr + 2 π t n+ 1 2 ∞ 0 √ kI n+ 1 2 (kt) I n (k) ∞ 0 g(z)sin(kz) dz dk − 2 π t n+ 1 2 ∞ 0 h(ξ) ξ n− 1 2 ∞ 0 kK n (k)I n+ 1 2 (kt)I n+ 1 2 (kξ) I n (k) dk dξ, (4.5.73) where 0 <t<a. From Equation 4.5.73 we can compute h(t), and consequently A(k)and B(k). The potential u(r, z)thenfollows from u(r, z)= ∞ 0 A(k)e −kz J n (kr) dk + ∞ 0 B(k)I n (kr)sin(kz) dk (4.5.74) = a 0 h(t) t n− 1 2 ∞ 0 √ ke −kz J n+ 1 2 (kt)J n (kr) dk dt + 2 π ∞ 0 I n (kr) I n (k) ∞ 0 g(t)sin(kt) dt sin(kz) dk (4.5.75) − 2 π a 0 h(t) t n− 1 2 ∞ 0 √ k K n (k) I n (k) I n (kr)I n+ 1 2 (kt)sin(kz) dk dt. Figure 4.5.3 illustrates Equation 4.5.75 when a =0.5, n =1,g(z)=0and f(r)=1. We can also use this technique to solve Equation 4.5.49 through Equation 4.5.52 when Equation 4.5.50 reads lim r→0 |u(r, z)| < ∞,u r (1,z)=g(z), 0 ≤ z<∞. (4.5.76) The analysis is identical to our earlier problem except for computing B(k). Equation 4.5.71 now becomes I n (k)B(k)= 2 πk ∞ 0 g(z)sin(kz) dz − 2 π √ kK n (k) a 0 h(ξ) ξ n− 1 2 I n+ 1 2 (kξ) dξ (4.5.77) © 2008 by Taylor & Francis Group, LLC 330 Mixed Boundary Value Problems 0 0.2 0.4 0.6 0.8 1 0 0.1 0.2 0.3 0.4 −0.25 −0.2 −0.15 −0.1 −0.05 0 r z u(r,z) Figure 4.5.4:Thesolutionu(r, z)tothe mixed boundary value problem governed by Equation 4.5.49, Equation 4.5.76, Equation 4.5.51 and Equation 4.5.52 with a =0.8, n =1, g(z)=0andf(r)=1. subject to the boundary conditions lim r→0 |u(r, z)| < ∞, lim r→∞ u(r, z) → 0, 0 <z<∞, (4.5.80) lim z→∞ u(r, z) → 0, 0 ≤ r<∞, (4.5.81) u(1 − ,z)=u(1 + ,z), 0 <z<∞, (4.5.82) 1 ∂ ∂r u(r, z) r r=1 − = 2 ∂ ∂r u(r, z) r r=1 + , 0 <z<∞, (4.5.83) and u z (r, 0) = f(r), 0 ≤ r<c, u(r, 0) = 0,c<r<∞, (4.5.84) where c<1. Using separation of variables, the general solution to Equation 4.5.79 is u(r, z)= ∞ 0 A(k)e −kz J 1 (kr) dk + ∞ 0 B(k)I 1 (kr)sin(kz) dk, 0 ≤ r<1, (4.5.85) and u(r, z)= ∞ 0 C(k)K 1 (kr)sin(kz) dk, 1 <r<∞, (4.5.86) © 2008 by Taylor & Francis Group, LLC [...]... (4.5 .124 ) and Vθ (r, 0, s) = V (r, π, s) = 0, 0 ≤ r < ∞ (4.5 .125 ) We next express the solution to Equation 4.5 .123 as the Fourier series ∞ V (r, θ, s) = Vn (r, s) cos n + 1 2 θ (4.5 .126 ) n=0 Note that Equation 4.5 .126 satisfies the boundary condition given by Equation 4.5 .125 Each Fourier coefficient Vn (r, s) is governed by d2 Vn 1 dVn − + dr2 r dr n+ 1 2 r2 2 + s 2U0 (−1)n Vn = 2 a πa2 n + 1 2 (4.5 .127 )... to the boundary conditions u(1, z) = 0, lim u(r, z) → 0, r→∞ lim u(r, z) → 0, z→∞ 0 ≤ z < ∞, 1 ≤ r < ∞, 116 See Rusia, K C., 1967: Some asymmetric mixed boundary value problems for a half-space with a cylindrical cavity Indian J Pure Appl Phys., 5, 419–421 Rusia’s contribution was to simplify the solution of this problem first posed by Narain, P., 1965: A note on an asymmetric mixed boundary value problem... involving products of Bessel functions Philos Trans R Soc London, Ser A, 247, 529–551 © 2008 by Taylor & Francis Group, LLC 332 Mixed Boundary Value Problems 0 u(r,z) −0.05 −0.1 −0.15 −0.2 −0.25 0 0.2 0.4 0.6 z 0 0.5 0.8 1 1 1.5 2 r Figure 4.5.5: The solution u(r, z) to the mixed boundary value problem governed by Equation 4.5.79 through Equation 4.5.84 with β = 0 and c = 0.9 Solving this integral equation... during the manufacture of p-n junctions: ∂2u ∂2u ∂u = a2 + 2 ∂t ∂x2 ∂y , −∞ < x < ∞, 0 < y < ∞, 0 < t, (4.5.110) subject to the mixed boundary value conditions lim |u(x, y, t)| < ∞, |x|→∞ © 2008 by Taylor & Francis Group, LLC 0 < y < ∞, 0 < t, (4.5.111) 336 Mixed Boundary Value Problems uθ (r, 0, t) = 0, 0 ≤ r < ∞, u(r, π, t) = U0 , 0 < t, (4.5.117) and 0 ≤ r < ∞, u(r, θ, 0) = 0, 0 < θ < π (4.5.118) Equation... π, (4.5.119) subject to the boundary conditions lim |U (r, θ, s)| < ∞, lim |U (r, θ, s)| < ∞, 0 < θ < π, r→∞ r→0 and Uθ (r, 0, s) = 0, U (r, π, s) = U0 , s 0 ≤ r < ∞ (4.5 .120 ) (4.5 .121 ) Let U0 + V (r, θ, s) s Then, Equation 4.5.119 through Equation 4.5 .121 become U (r, θ, s) = ∂ 2 V 1 ∂V 1 ∂ 2V s U0 + 2 + − 2V = 2 , 2 2 ∂r r ∂r r ∂θ a a 0 ≤ r < ∞, (4.5 .122 ) 0 < θ < π, (4.5 .123 ) with lim |V (r, θ, s)|... cosh(kτ ) cosh(kt) dk I0 (kc) 340 Mixed Boundary Value Problems 1.2 1 u(r,z) 0.8 0.6 0.4 0.2 0 0 0 0.5 0.5 1 1 1.5 1.5 z 2 2 r Problem 1 The figure labeled Problem 1 illustrates the solution when f (r) = 1, a = 1 and c = 2 2 Given a > 1 and denoting a non-negative integer by n, let us solve115 ∂ 2 u 1 ∂u n2 ∂ 2u − 2 u + 2 = 0, + ∂r2 r ∂r r ∂z 0 ≤ r < a, 0 ≤ z < ∞, subject to the boundary conditions lim |u(r,... Sommerfeld, A., 1896: Mathematische Theorie der Diffraction Math Ann., 47, 317–374 6 Kaup, S N., 1950: Wiener-Hopf techniques and mixed boundary value problems Comm Pure Appl Math., 3, 411–426; Clemmow, P C., 1951: A method for the exact solution of a class of two-dimensional diffraction problems Proc R Soc London, Ser A, 205, 286–308 See Noble, B., 1958: Methods Based on the Wiener-Hopf Technique for the Solution... Step 8 : Using the results from Step 5 to eliminate B(k) in Step 7, show that h(t) is given by the integral equation 1 h(t) = χ(t) + h(ξ)K(t, ξ) dξ, 0 © 2008 by Taylor & Francis Group, LLC 342 Mixed Boundary Value Problems 1 u(r,z) 0.8 0.6 0.4 0.2 0 0 0 0.2 0.5 0.4 1 0.6 r 1.5 z 0.8 2 Problem 2 where 2√ tη π K(t, η) = and χ(t) = − ∞ 0 ξKn (ξa) I 1 (tξ)In− 1 (ηξ) dξ, 2 In (ξa) n− 2 t n+1 2 −n d r f (r)... Solving Equation 4.5 .127 by Hankel transforms via ∞ Vn (r, s) = © 2008 by Taylor & Francis Group, LLC 0 A(k)Jn+ 1 (kr) k dk, 2 (4.5 .128 ) Transform Methods 337 we find that U (r, θ, s) = U0 2 1 − s π ∞ (−1)n n=0 cos n + 1 θ dk 2 Jn+ 1 (kr) 2 k2 2 s+a k (4.5 .129 ) Straightforward inversion yields the final solution: u(r, θ, t) = U0 1 − 2 π ∞ (−1)n cos n + 1 2 ∞ θ 2 2 e−a 0 n=0 k t Now 112 ∞ Γ n+1 dk 2 4 =... (kr) dk , a < r < ∞ Step 6 : Solve the integral equation in Step 5 and show that h (t) = 2 d π dt ∞ t © 2008 by Taylor & Francis Group, LLC √ r1−n f (r) + r 2 − t2 ∞ 0 kB(k)Kn (kr) dk dr , 344 Mixed Boundary Value Problems or h(t) = 2 π = 2 π = 2 π ∞ t ∞ t ∞ t Hint: kB(k)Kn (kr) dk dr 0 r1−n f (r) √ dr + r 2 − t2 r1−n f (r) √ dr + r 2 − t2 ∞ ∞ 2 π ∞ t ∞ r1−n √ f (r) + r 2 − t2 0 ∞ kB(k) 0 t r1−n Kn (kr) . Group, LLC 330 Mixed Boundary Value Problems 0 0.2 0.4 0.6 0.8 1 0 0.1 0.2 0.3 0.4 −0.25 −0.2 −0.15 −0.1 −0.05 0 r z u(r,z) Figure 4.5.4:Thesolutionu(r, z)tothe mixed boundary value problem governed. Francis Group, LLC 332 Mixed Boundary Value Problems 0 0.5 1 1.5 2 0 0.2 0.4 0.6 0.8 1 −0.25 −0.2 −0.15 −0.1 −0.05 0 r z u(r,z) Figure 4.5.5:Thesolutionu(r, z)tothe mixed boundary value problem governed. (4.5.110) subject to the mixed boundary value conditions lim |x|→∞ |u(x, y, t)| < ∞, 0 <y<∞, 0 <t, (4.5.111) © 2008 by Taylor & Francis Group, LLC 336 Mixed Boundary Value Problems u θ (r,