108 Mixed Boundary Value Problems Substituting Equation 3.2.41 into Equation 3.2.35, we obtain the integral equation 1 √ 2  π c h(t)  ∞  n=1 {P n [cos(t)] − P n−1 [cos(t)]}sin(nx)  dt = f(x). (3.2.42) Using the results from Problem 3 in Section 1.3, Equation 3.2.42 simplifies to  π x h(t)  cos(x) −cos(t) dt = −csc  x 2  f(x),c<x≤ π. (3.2.43) From Equation 1.2.11 and Equation 1.2.12, we obtain h(t)= 2 π d dt   π t f(x)cos(x/2)  cos(t) −cos(x) dx  . (3.2.44) Using the results from Equations 3.2.22, 3.2.28, 3.2.34, 3.2.35, 3.2.41, and 3.2.44, the solution to the dual equations            ∞  n=1 nc n sin(ny)=g(π − y), 0 ≤ y<γ, ∞  n=1 c n sin(ny)=f(π − y),γ<y≤ π, (3.2.45) is c n = 1 √ 2  π 0 h(t) {P n−1 [cos(t)] − P n [cos(t)]} dt, (3.2.46) where h(t)= 2 π cot  t 2   t 0 g(π −ξ)sin(ξ/2)  cos(ξ) −cos(t) dξ, 0 ≤ t<γ, (3.2.47) and h(t)=− 2 π d dt   π t f(π −ξ)cos(ξ/2)  cos(t) −cos(ξ) dξ  ,γ<t≤ π. (3.2.48) Therefore, the solution to Equation 3.2.20 and Equation 3.2.21 is C n = 1 √ 2  γ 0 h(t) {P n−1 [cos(t)] − P n [cos(t)]} dt, (3.2.49) where h(t)= 2 L cot  t 2   t 0 U 0 [L(π −ξ)/π]sin(ξ/2)  cos(ξ) −cos(t) dξ, 0 ≤ t<γ. (3.2.50) © 2008 by Taylor & Francis Group, LLC Separation of Variables 109 Consequently, making the back substitution, the dual series            ∞  n=1 b n n sin(nx)=f(x)0≤ x<c, ∞  n=1 b n sin(nx)=g(x).c<x≤ π, (3.2.51) has the solution b n = n √ 2  π 0 k(t) {P n−1 [cos(t)] + P n [cos(t)]} dt, (3.2.52) where k(t)= 2 π d dt   c 0 f(ξ)sin(ξ/2)  cos(ξ) −cos(t) dξ  , 0 ≤ t<c, (3.2.53) and k(t)= 2 π tan  t 2   π t g(ξ)cos(ξ/2)  cos(t) −cos(ξ) dξ, c < t ≤ π. (3.2.54) Using Equation 3.2.51 through Equation 3.2.54, we finally have that A n = 1 √ 2  π/L 0 k(t) {P n−1 [cos(t)] + P n [cos(t)]} dt, (3.2.55) where k(t)= 2 π d dt   t 0 U 0 (Lξ/π)sin(ξ/2)  cos(ξ) −cos(t) dξ  , 0 ≤ t<π/L. (3.2.56) 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. • Example 3.3.1 Let us find 16 the potential forLaplace’s equation in cylindrical coordi- nates: ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<1, 0 <z<∞, (3.3.1) 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. © 2008 by Taylor & Francis Group, LLC 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) and lim z→∞ u(r, z) → 0, 0 ≤ r<1, (3.3.4) where a<1. Separation of variables yields the potential, namely u(r, z)=A 0 + ∞  n=1 A n e −k n z J 0 (k n r), (3.3.5) where k n is the nth positive root of J  0 (k)=−J 1 (k)=0. Equation 3.3.5 satisfies Equation 3.3.1, Equation 3.3.2, and Equation 3.3.4. Substituting Equation 3.3.5 into Equation 3.3.3, we obtain the dual series: ∞  n=1 k n A n J 0 (k n r)=−1, 0 ≤ r<a, (3.3.6) and A 0 + ∞  n=1 A n J 0 (k n r)=0,a<r<1. (3.3.7) Srivastav 17 showed that this dual Dini series has the solution A 0 = −2  a 0 th(t) dt, (3.3.8) and A n = − 2 k n J 2 0 (k n )  a 0 h(t)sin(k n t) dt, (3.3.9) where the unknown function h(t)isgivenbytheregularFredholm integral equation of the second kind: h(t)+  a 0 L(t, η)h(η) dη = t, 0 ≤ t<a, (3.3.10) and L(t, η)= 4 π 2  ∞ 0 K 1 (x) I 1 (x) sinh(tx)sinh(ηx) dx. (3.3.11) 17 Srivastav, R. P., 1961/1962: Dual series relations. II. Dual relations involving Dini series. Proc. R. Soc. Edinburgh, Ser. A, 66, 161–172. © 2008 by Taylor & Francis Group, LLC Separation of Variables 111 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 r z u(r,z) 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 Asimilar problem 18 to the previous one arises during the solution of Laplace’s equation in cylindrical coordinates: ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<1, 0 <z<∞, (3.3.12) subject to the boundary conditions lim r→0 |u(r, z)| < ∞,u r (1,z)=0, 0 <z<∞, (3.3.13)  u(r, 0) = 1, 0 ≤ r<a, u z (r, 0) = 0,a<r<1, (3.3.14) and lim z→∞ |u z (r, z)| < ∞, 0 ≤ r<1, (3.3.15) where a<1. Separation of variablesgives u(r, z)=A 0 z + ∞  n=1 A n e −k n z J 0 (k n r) k n , (3.3.16) 18 SeeHunter, A., and A. Williams, 1969: Heat flow across metallic joints – The con- striction alleviation factor. Int. J. Heat Mass Transfer, 12, 524–526. © 2008 by Taylor & Francis Group, LLC 112 Mixed Boundary Value Problems where k n is the nth positive root of J  0 (k)=−J 1 (k)=0. Equation 3.3.16 satisfies Equation 3.3.12, Equation 3.3.13, and Equation 3.3.15. Substituting Equation 3.3.16 into Equation 3.3.14, we obtain the dual series: ∞  n=1 A n J 0 (k n r) k n =1, 0 ≤ r<a, (3.3.17) and A 0 − ∞  n=1 A n J 0 (k n r)=0,a<r<1. (3.3.18) Srivastav 19 has given the solution to the dual Fourier-Bessel series αa 0 + ∞  n=1 a n J 0 (k n r) k n = f(r), 0 ≤ r<a, (3.3.19) and a 0 + ∞  n=1 a n J 0 (k n r)=0,a<r<1. (3.3.20) Then, a 0 =2  a 0 h(t) dt, (3.3.21) and a n = 2 J 2 0 (k n )  a 0 h(t)cos(k n t) dt, (3.3.22) where the function h(t)isgivenbythe integral equation h(t) −  a 0 K(t, τ)h(τ) dτ = x(t), 0 <t<a, (3.3.23) x(t)= 2 π d dt   t 0 rf(r) √ t 2 − r 2 dr  , (3.3.24) and K(t, τ)= 4 π (1 − α)+ 4 π 2  ∞ 0 K 1 (ξ) ξI 1 (ξ) [2I 1 (ξ) −ξ cosh(τξ)cosh(tξ)] dξ. (3.3.25) 19 Ibid. See also Sneddon, op. cit., Equation 5.3.27 through Equation 5.3.35. © 2008 by Taylor & Francis Group, LLC Separation of Variables 113 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1 −1 −0.5 0 0.5 1 1.5 2 r z u(r,z) 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 =0andf(r)=1. Figure 3.3.2 illustrates the solution when a = 1 2 . • Example 3.3.3 Let us solve Laplace equation 20 ∂ 2 u ∂r 2 + 1 r ∂u ∂r + ∂ 2 u ∂z 2 =0, 0 ≤ r<a, 0 <z<∞, (3.3.26) subject to the boundary conditions lim r→0 |u(r, z)| < ∞,u(a, z)=0, 0 <z<∞, (3.3.27)  u z (r, 0) = 1, 0 ≤ r<1, u r (r, 0) = 0, 1 <r<a, (3.3.28) and lim z→∞ u(r, z) → 0, 0 ≤ r<a, (3.3.29) where a>1. The method of separation of variables yields the product solution u(r, z)= ∞  n=1 A n J 0 (k n r)e −k n z , (3.3.30) 20 See Sherwood, J. D., and H. A. Stone, 1997: Added mass of a disc accelerating within apipe.Phys. Fluids, 9, 3141–3148. © 2008 by Taylor & Francis Group, LLC 114 Mixed Boundary Value Problems where k n is the nth root of J 1 (ka)=0. Equation 3.3.30 satisfies not only Laplace’s equation, but also the boundary conditions given by Equation 3.3.27 andEquation 3.3.29. Substituting Equation 3.3.30 into Equation 3.3.28, we obtain the dual series ∞  n=1 A n k n J 0 (k n r)=−1, 0 ≤ r<1, (3.3.31) and ∞  n=1 A n k n J 1 (k n r)=0, 1 ≤ r<a. (3.3.32) We b egin our solution of these dual equations by applying the identity 21 ∞  n=1 J ν+2m+1−p (ζ n )J ν (ζ n r) ζ 1−p n J 2 ν+1 (ζ n a) =0, 1 <r<a, (3.3.33) where |p|≤ 1 2 , ν>p− 1, m =0, 1, 2, ,andζ n denotes the nth root of J ν (ζa)=0. Bydirection substitution it is easily seen that Equation 3.3.32 is satisfied if ν =1and k 2−p n J 2 2 (k n a)A n = ∞  m=0 C m J 2m+2−p (k n ). (3.3.34) Here p is still a free parameter. Substituting Equation 3.3.34 into Equation 3.3.31, ∞  n=1 ∞  m=0 C m J 2m+2−p (k n )J 0 (k n r) k 1−p n J 2 2 (k n a) = −1, 0 ≤ r<1. (3.3.35) 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  ∞ 0 η 1−k J ν+2m+k (η)J ν (rη) dη = Γ(ν + m +1)r ν (1 − r 2 ) k−1 2 k−1 Γ(ν +1)Γ(m + k) P (k+ν,ν+1) m  r 2  (3.3.36) 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. © 2008 by Taylor & Francis Group, LLC Separation of Variables 115 if 0 ≤ r<1; this integral equals 0 if 1 <r<∞.HereP (a,b) m (x)= 2 F 1 (−m, a+ m; b; x)istheJacobi polynomial. If we view Equation 3.3.36 as a Hankel transform of η −k J ν+2m+k (η), then its inverse is η −k J ν+2m+k (η) =  1 0 Γ(ν + m +1)r 1+ν  1 − r 2  k−1 2 k−1 Γ(ν +1)Γ(m + k) J ν (rη)P (k+ν,ν+1) m  r 2  dr. (3.3.37) We also have from the orthogonality condition 23 of Jacobi polynomials that  1 0 r 2ν+1  1 − r 2  k−1 P (k+ν,ν+1) m  r 2  dr = Γ(ν +1)Γ(k) 2Γ(ν + k +1) δ 0m , (3.3.38) where δ nm is the Kronecker delta. Multiplying both sides of Equation 3.3.35 by r  1 − r 2  −p P (1−p,1) m (r 2 )andapplying Equation 3.3.38, we obtain − Γ(1 − p) 2Γ(2 −p) δ 0j = ∞  n=1 ∞  m=0 C m J 2m+2−p (k n )J 2j+1−p (k n )Γ(j +1− p) 2 p Γ(j +1)k 2−2p n J 2 2 (k n a) ; (3.3.39) or ∞  m=0 A jm C m = B j ,j=0, 1, 2, , (3.3.40) where A jm = ∞  n=1 J 2m+2−p (k n )J 2j+1−p (k n ) k 2−2p n J 2 2 (k n a) , (3.3.41) and B j =  −2 p−1 /Γ(2 − p),j=0, 0, otherwise. (3.3.42) For a given p,wecan solve Equation 3.3.40 after we truncate the infinite num- ber of equations to just M .Foragivenk n we solve the truncated Equation 3.3.40, which yields C m for m =0, 1, 2, ,M.ThenEquation3.3.34 gives A n .Finally, the potential u(r, z)follows from Equation 3.3.30. Figure 3.3.3 illustrates this solution when a =2andp =0.5. • Example 3.3.4 In the previous example we solved Laplace’s equation over a semi-infinite right cylinder. Here, let us solve Laplace’s equation 24 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 SeeGalceran,J., J. Cecilia, E. Companys, J. Salvador, and J. Puy, 2000: Analytical expressions for feedback currents at the scanning electrochemical microscope. J. Phys. Chem., Ser. B, 104, 7993-8000. © 2008 by Taylor & Francis Group, LLC Separation of Variables 117 and ∞  n=1 A n J 0 (k n r)=0, 1 ≤ r<a. (3.3.49) Let us reexpress A n as follows: A n = 1 √ k n J 2 1 (k n a) ∞  m=0 B m J 2m+ 1 2 (k m ). (3.3.50) Substituting Equation 3.3.50 into Equation 3.3.49, we have 25 that ∞  n=1 A n J 0 (k n r)= ∞  m=o B m  ∞  n=1 J 2m+ 1 2 (k m )J 0 (k n r) √ k n J 2 1 (k n a)  =0 (3.3.51) for 1 <r≤ a.Therefore, Equation 3.3.49 is satisfied identically with this definition of A n . Next, we substitute Equation 3.3.50 into Equation 3.3.48, multiply both sides of the resulting equation by rF 21  −s, s + 1 2 , 1,r 2  dr/ √ 1 − r 2 and in- tegrate between r =0andr =1.Wefind that ∞  m=0 C m,s B m = √ 2Γ(s +1) Γ  s + 1 2   1 0 r √ 1 − r 2 F 21  −s, s + 1 2 , 1,r 2  dr (3.3.52) =   2/π, s =0, 0,s>0, (3.3.53) where C m,s = ∞  n=1 coth(k n b)J 2m+ 1 2 (k n )J 2s+ 1 2 (k n ) k 2 n J 2 1 (k n a) (3.3.54) and s =0, 1, 2, Weused J 2s+ 1 2 (k n ) √ k n = √ 2Γ(s +1) Γ  s + 1 2   1 0 r √ 1 − r 2 F 21  −s, s + 1 2 , 1,r 2  J 0 (k n r) dr. (3.3.55) Equation 3.3.55 is now solved to yield B m .Next,wecompute A n from Equa- tion 3.3.50. Finally u(r, z)follows from Equation 3.3.47. Figure 3.3.4 illustrates the solution to Equation 3.3.43 through Equa- tion 3.3.46 when a =2andb =1. Assuggested by Galceran et al., 26 the 25 Tranter, op. cit. 26 Galceran, Cecilia, Companys, Salvador, and Puy, op. cit. © 2008 by Taylor & Francis Group, LLC Separation of Variables 119 z r z = h z = 0 1 a Figure 3.3.5:Schematicof a hollow cylinder containing discs at z =0andz = h. u(r, 0 − )=u(r, 0 + ),u(r, h − )=u(r, h + ), 0 ≤ r<a, (3.3.60)  u(r, 0) = G(r),u(r, h)=F (r), 0 ≤ r<1, u z (r, 0 − )=u z (r, 0 + ),u z (r, h − )=u z (r, h + ), 1 <r<a, (3.3.61) and lim |z|→∞ u(r, z) → 0, 0 ≤ r<a. (3.3.62) Here, G(r)andF (r)denotetheprescribed potential on the discs at z =0 and z = h,respectively. The parameters h + and h − denote points that are slightly above or below h,respectively. Separation of variables yields the potential, namely u(r, z)= ∞  n=0 A n e −k n (z−h) J 0 (k n r) k n ,h≤ z<∞, (3.3.63) u(r, z)= ∞  n=0  A n sinh(k n z) sinh(k n h) + B n sinh[k n (h −z)] sinh(k n h)  J 0 (k n r) k n , 0 ≤ z ≤ h, (3.3.64) and u(r, z)= ∞  n=0 B n e k n z J 0 (k n r) k n , −∞ <z≤ 0, (3.3.65) © 2008 by Taylor & Francis Group, LLC [...]... Group, LLC Separation of Variables 133 1 u(r,θ ) 0.8 0.6 0.4 0.2 0 −0.2 1 1 0 .5 0 .5 0 0 −0 .5 −0 .5 y x −1 −1 Figure 3.4.1: The solution u(r, θ) to the mixed boundary value problem posed in Example 3.4.1 with α = π/4 and b = 0. 25 power = power*r; 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... (t2 0 1 rJ0 (km r) r 1 rJ0 (km r) r (t2 + sin(kn t) √ dt dr t2 − r 2 dt √ h2 ) Gradshteyn and Ryzhik, op cit., Formula 2. 252 , Point II © 2008 by Taylor & Francis Group, LLC h2 ) t2 − r 2 dr (3.3.99) 126 Mixed Boundary Value Problems 10 9 8 u(r,z) 7 6 5 4 3 2 1 0 0 2 0 .5 1 1 0 r 1 .5 −1 2 −2 z Figure 3.3.8: The electrostatic potential when a unit point charge is placed at r = 0 and z = h for the structure... Separation of Variables - 0 .5* dt*gamma*log(abs(2*cos(arg 1)))/pi - 0 .5* dt*gamma*log(abs(cos(arg 2)))/pi + 0 .5* dt*gamma*log(abs(2*sin(arg 1)))/pi; if (arg 2 == 0) AA(n+1,m+1) = AA(n+1,m+1) + 0 .5* dt*gamma*log(0. 25) /pi; else AA(n+1,m+1) = AA(n+1,m+1) + 0 .5* dt*gamma*log(abs(sin(arg 2)/(t-phi)))/pi; end end % Use Atkinson’s technique to treat the fifth term in % Equation 3.4. 25 See Section 5 of his paper if (m...120 Mixed Boundary Value Problems where kn 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. 65 into Equation 3.3.61, we obtain the following system of simultaneous dual... terms of Equation 3.4. 25 Use the % trapezoidal rule Recall that r(π) = 0 if (m < N) arg 1 = (t+phi)/4; arg 2 = (t-phi)/4; AA(n+1,m+1) = AA(n+1,m+1) - 0 .5* dt*gamma*log(abs(2*cos(arg 1)))/pi - 0 .5* dt*gamma*log(abs(cos(arg 2)))/pi + 0 .5* dt*gamma*log(abs(2*sin(arg 1)))/pi; if (arg 2 == 0) AA(n+1,m+1) = AA(n+1,m+1) + 0 .5* dt*gamma*log(0. 25) /pi; else AA(n+1,m+1) = AA(n+1,m+1) + 0 .5* dt*gamma*log(abs(sin(arg... cos(3t/2), π h(0) = 0 (3.4 .51 ) Separation of Variables 137 0 u(x,y)/(Ua) −0.1 −0.2 −0.3 −0.4 −0 .5 −0.6 −0.7 −0.8 −2 −2 −1 −1 0 0 1 y/a 1 x/a 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 (t) + t 1 4 h(τ ) dτ = 0 4 M1 − cos(3t/2), 4 π (3.4 .52 ) α where M1 = 0 h(t) dt,... continuous value of ur (a, θ) for 0 ≤ θ ≤ π Substituting Equation 3.4.41 and Equation 3.4.42 into Equation 3.4.40 gives the dual Fourier-Legendre series ∞ nAn Pn [cos(θ)] = − cos(θ), 0 ≤ θ < α, (3.4.43) n=0 37 Reprinted from Int J Solids Struct., 38, P A Martin, The spherical-cap crack revisited, 4 759 –4776, c 2001, with permission of Elsevier © 2008 by Taylor & Francis Group, LLC 136 Mixed Boundary Value Problems. .. and ∞ 1 h(t) 0 J0 (kn r) cos(kn t) dt 2 a2 kn J1 (kn a) n=0 ∞ 1 g(t) + 0 29 e−kn h J0 (kn r) cos(kn t) 2 a2 kn J1 (kn a) n=0 Gradshteyn and Ryzhik, op cit., Formula 6 .55 2.4 © 2008 by Taylor & Francis Group, LLC 122 Mixed Boundary Value Problems It is readily shown30 that 2 a2 ∞ ∞ J0 (kn r) cos(kn t) = 2 kn J1 (kn a) n=0 − J0 (rη) cos(tη) dη 0 ∞ 2 π 0 K0 (aη) I0 (rη) cosh(tη) dη (3.3.82) I0 (aη) In a... & Francis Group, LLC 131 132 Mixed Boundary Value Problems Once ψ(ϕi ) is found with ψ(π) = 0, we compute An via Equation 3.4.18 The MATLAB code to compute An is for m = 0:M if ( m == 0 ) A(m+1) = alpha/pi + sin(alpha)/pi; else A(m+1) = sin(m*alpha)/(m*pi) + sin((m+1)*alpha)/((m+1)*pi); end % 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... integration for n = 1:N-1 if ( mod(n+1,2) == 0 ) A(m+1) = A(m+1) + 8*psi(n+1)*cos((m+0 .5) *tt(n+1))*dt/(3*pi); else A(m+1) = A(m+1) + 4*psi(n+1)*cos((m+0 .5) *tt(n+1))*dt/(3*pi); end; end; end The final solution follows from Equation 3.4 .5 The MATLAB code to realize this solution is for j = 1:41 y = 0. 05* (j-21); for i = 1:41 x = 0. 05* (i-21); u(i,j) = NaN; r = sqrt(x*x + y*y); theta = abs(atan2(y,x)); if (r . and Ryzhik, op. cit., Formula 2. 252 , Point II. © 2008 by Taylor & Francis Group, LLC 126 Mixed Boundary Value Problems 0 0 .5 1 1 .5 2 −2 −1 0 1 2 0 1 2 3 4 5 6 7 8 9 10 z r u(r,z) Figure 3.3.8:Theelectrostatic. (3.3. 65) © 2008 by Taylor & Francis Group, LLC 120 Mixed Boundary Value Problems where k n is the nth positive root of J 0 (ka)=0. Equation 3.3.63 through Equation 3.3. 65 satisfy Equation 3.3 .58 ,. G(r). (3.3.81) 29 Gradshteyn and Ryzhik, op. cit., Formula 6 .55 2.4. © 2008 by Taylor & Francis Group, LLC 122 Mixed Boundary Value Problems It is readily shown 30 that 2 a 2 ∞  n=0 J 0 (k n r)cos(k n t) k n J 2 1 (k n a) =  ∞ 0 J 0 (rη)cos(tη)