Overview 17 Hafen 10 derived Equation 1.2.20 and Equation 1.2.21 in 1910. • Example 1.2.2 Consider the integral equation of the form x 0 K(x 2 − t 2 )f(t) dt = g(x), 0 <x, (1.2.22) where K(·)isknown. Wecansolveequationsof this type by reducing them to ξ 0 K(ξ −τ)F (τ) dτ = G(ξ), 0 <ξ, (1.2.23) via the substitutions x = √ ξ, t = √ τ, F(τ)=f ( √ τ ) / (2 √ τ ), G(ξ)=g √ ξ . Taking the Laplace transform of both sides of Equation 1.2.23, we have by the convolution theorem L[K(t)]L[F (t)] = L[G(t)]. (1.2.24) Defining L[L(t)] = 1/{sL[K(t)]},wehavebytheconvolution theorem F (ξ)= d dξ ξ 0 L(ξ −τ)G(τ) dτ , (1.2.25) or f(x)=2 d dx x 0 tg(t)L(x 2 − t 2 ) dt . (1.2.26) To illustrate this method, we choose K(t)=cos k √ t / √ t which has the Laplace transform L[K(t)] = √ πe −k 2 /(4s) / √ s.Then, L(t)=L −1 e k 2 /(4s) √ πs = cosh k √ t π √ t . (1.2.27) Therefore, the integral equation x 0 cosh k √ x 2 − t 2 √ x 2 − t 2 f(t) dt = g(x), 0 <x, (1.2.28) has the solution f(x)= 2 π d dx x 0 cosh k √ x 2 − t 2 √ x 2 − t 2 tg(t) dt . (1.2.29) 10 Hafen, M., 1910: Studien ¨uber einige Probleme der Potentialtheorie. Math. Ann., 69, 517–537. © 2008 by Taylor & Francis Group, LLC 18 Mixed Boundary Value Problems In particular, if g(0) = 0, then f(x)= 2x π x 0 cosh k √ x 2 − t 2 √ x 2 − t 2 g (t) dt. (1.2.30) • Example 1.2.3 In 1970, Cooke 11 proved that the solution totheintegral equation 1 0 ln x + t x − t h(t) dt = πf(x), 0 <x<1, (1.2.31) is h(t)=− 2 π d dt 1 t αS(α) √ α 2 − t 2 dα + 2f(0 + ) πt √ 1 − t 2 , (1.2.32) where S(α)= α 0 f (ξ) α 2 − ξ 2 dξ. (1.2.33) We will use this result several times in this book. For example, at the begin- ning of Chapter 4, we must solve the integral equation − 1 π 1 0 g(t)ln tanh(βx)+tanh(βt) tanh(βx) − tanh(βt) dt = x h , 0 ≤ x<1, (1.2.34) where 2hβ = π.HowdoesEquation 1.2.31 help us here? If we introduce the variables tanh(βt)=tanh(β)T and tanh(βx)=tanh(β)X,thenEqua- tion 1.2.34 transforms into an integral equation of the form Equation 1.2.31. Substituting back into the original variables, we find that h(t)= 1 π 2 d dt 1 t tanh(βα)S(α) cosh 2 (βα) tanh 2 (βα) − tanh 2 (βt) dα , (1.2.35) where S(α)= α 0 dξ tanh 2 (βα) −tanh 2 (βξ) . (1.2.36) Another useful result derived by Cooke is that the solution to the integral equation 1 0 g(y)log |x 2 − y 2 | y 2 dy = πf(x), 0 <x<1, (1.2.37) 11 Cooke, J. C., 1970: The solution of some integral equations and their connection with dual integral equations and series. Glasgow Math. J., 11, 9–20. © 2008 by Taylor & Francis Group, LLC Overview 19 is g(y)= 2 π 1 − y 2 1 0 x √ 1 − x 2 x 2 − y 2 f (x) dx + C 1 − y 2 , (1.2.38) or g(y)= 2 πy d dy τS(τ) τ 2 − y 2 dτ + C 1 − y 2 , (1.2.39) where S(τ)=τ d dτ τ 0 f(x) √ τ 2 − x 2 dx = τ 0 xf (x) √ τ 2 − x 2 dx. (1.2.40) If f(x)isaconstant,the equation has no solution. We will use Equation 1.2.39 and Equation 1.2.40 in Chapter 6. • Example 1.2.4 Many solutions to dual integrals hinge on various improper integrals that contain Bessel functions. For example, consider the known result 12 that ∞ 0 J ν (βx) J µ α √ x 2 + z 2 (x 2 + z 2 ) µ x ν+1 dx (1.2.41) = β ν α µ α 2 − β 2 z µ−ν−1 J µ−ν−1 z α 2 − β 2 H(α − β), if (µ) > (ν) > −1. Akhiezer 13 used Equation 1.2.41 along with the result 14 that the integral equation g(x)= x 0 f(t) √ x 2 − t 2 ik −p J −p ik x 2 − t 2 dt (1.2.42) has the solution f(x)= d dx x 0 tg(t) √ x 2 − t 2 k −q J −q k x 2 − t 2 dt , (1.2.43) 12 Gradshteyn, I. S., and I. M. Ryzhik, 1965: Table of Integrals, Series, and Products. Academic Press, Formula 6.596.6. 13 Akhiezer, N. I., 1954: On some coupled integral equations (in Russian). Dokl. Akad. Nauk USSR, 98, 333–336. 14 Polyanin, A. D., and A. V. Manzhirov, 1998: Handbook of Integral Equations.CRC Press, Formula 1.8.71. © 2008 by Taylor & Francis Group, LLC 20 Mixed Boundary Value Problems where p>0, q>0, and p + q =1. Let us use these results to find the solution to the dual integral equations ∞ α C(k)J ν (kx) dk = f (x), 0 <x<1, (1.2.44) and ∞ 0 C(k)J −ν (kx) dk =0, 1 <x<∞, (1.2.45) where α ≥ 0and0<ν 2 < 1. We begin by introducing C(k)=k 1−ν 1 0 h(t)J 0 t k 2 − α 2 dt. (1.2.46) Then, ∞ 0 C(k)J −ν (kx) dk = 1 0 h(t) ∞ 0 k 1−ν J −ν (kx)J 0 t k 2 − α 2 dk dt (1.2.47) = 1 0 h(t) x ν √ t 2 − x 2 iα ν−1 J ν−1 iα t 2 − x 2 H(t − x) dt (1.2.48) =0, (1.2.49) because 0 ≤ t ≤ 1 <x<∞.Therefore, the introduction of the integral defi- nition for C(k)resultsinEquation 1.2.45 being satisfied identically. Turning now to Equation 1.2.44, 1 0 h(t) ∞ α k 1−ν J ν (kx)J 0 t k 2 − α 2 dk dt = f(x)(1.2.50) for 0 <x<1. Now, if we introduce η 2 = k 2 − α 2 , ∞ α k 1−ν J ν (kx)J 0 t k 2 − α 2 dk = ∞ 0 J ν x η 2 + α 2 J 0 (tη) η (η 2 + α 2 ) ν dη (1.2.51) = 1 x ν √ x 2 − t 2 −α ν−1 J ν−1 −α x 2 − t 2 H(x − t), (1.2.52) upon applying Equation 1.2.41. Therefore, Equation 1.2.50 simplifies to x 0 h(t) x ν √ x 2 − t 2 −α ν−1 J ν−1 −α x 2 − t 2 dt = f(x)(1.2.53) © 2008 by Taylor & Francis Group, LLC Overview 21 with 0 <x<1. Therefore, the solution to the dual integral equations, Equation 1.2.44 and Equation 1.2.45, consists of Equation 1.2.46 and h(t)= d dt t 0 f(x)x ν+1 √ t 2 − x 2 iα −ν J −ν iα t 2 − x 2 dx (1.2.54) if 0 <ν<1. This solution also applies to the dual integral equation: ∞ α C(k)J 1+ν (kx)kdk= −x ν d dx x −ν f(x) , 0 <x<1, (1.2.55) and ∞ 0 C(k)J −1−ν (kx)kdk=0, 1 <x<∞, (1.2.56) when −1 <ν<0. In a similar manner, let us show the dual integral equation ∞ α S(k)J 0 (kx)kdk= f(x), 0 <x<1, (1.2.57) and ∞ 0 S(k)J 0 (kx)(k 2 − α 2 ) p kdk=0, 1 <x<∞, (1.2.58) where α ≥ 0and0<p 2 < 1. We begin by introducing S(k)= 1 0 h(t) √ k 2 − α 2 p J −p t k 2 − α 2 dt. (1.2.59) It is straightforward to show that this choice for S(k)satisfiesEquation 1.2.58 identically. When we perform an analysis similar to Equation 1.2.50 through Equation 1.2.53, we find that f(x)= x 0 h(t) t p √ x 2 − t 2 α p−1 J p−1 α x 2 − t 2 dt (1.2.60) if 0 <p<1; or f(x)= 1 x d dx x 0 h(t) t p √ x 2 − t 2 α p J p α x 2 − t 2 dt (1.2.61) if −1 <p<0. To obtain Equation 1.2.61, we integrated Equation 1.2.57 with respect to x which yields 1 x x 0 f(ξ)ξdξ= ∞ α S(k)J 1 (kx) dx. (1.2.62) © 2008 by Taylor & Francis Group, LLC 22 Mixed Boundary Value Problems Applying Equation 1.2.42 and Equation 1.2.43, we have that h(t)=t p d dt ∞ 0 xf(x) √ t 2 − x 2 iα −p J −p iα t 2 − α 2 dx (1.2.63) for 0 <p<1; and h(t)=t 1+p t 0 xf(x) √ t 2 − x 2 iα −1−p J −1−p iα t 2 − α 2 dx (1.2.64) for −1 <p<0. Therefore, the solution to the dual integral equations Equa- tion 1.2.57 and Equation 1.2.58 has the solution Equation 1.2.59 along with Equation 1.2.63 or Equation 1.2.64 depending on the value of p. 1.3 LEGENDRE POLYNOMIALS In this book we will encounter special functions whose properties will be repeatedly used to derive important results. This section focuses on Legendre polynomials. Legendre polynomials 15 are defined by the power series: P n (x)= m k=0 (−1) k (2n − 2k)! 2 n k!(n − k)!(n − 2k)! x n−2k , (1.3.1) where m = n/2, or m =(n−1)/2, depending upon which is an integer. Figure 1.3.1 illustrates the first four Legendre polynomials. Legendre polynomials were originally developed to satisfy the differential equation (1 − x 2 ) d 2 y dx 2 − 2x dy dx + n(n +1)y =0, (1.3.2) or d dx (1 − x 2 ) dy dx + n(n +1)y =0, (1.3.3) that arose in the separation-of-variables solution of partial differential equa- tions in spherical coordinates. Several of their properties are given in Table 1.3.1. 15 Legendre, A. M., 1785: Sur l’attraction des sph´ero¨ıdes homog`enes. M´em. math. phys. pr´esent´es `al’Acad. Sci. pars divers savants, 10, 411–434. The best reference on Legendre polynomials is Hobson, E. W., 1965: The Theory of Spherical and Ellipsoidal Harmonics. Chelsea Publishing Co., 500 pp. © 2008 by Taylor & Francis Group, LLC Overview 25 They began by using the formula 18 √ 2 ∞ n=0 cos n + 1 2 x P n [cos(η)] = 1/ cos(x) −cos(η), 0 ≤ x<η<π, 0, 0 <η<x<π. (1.3.11) Now, 19 1 2 zJ 0 [z sin(η/2)] = ∞ n=0 (2n +1)P n [cos(η)]J 2n+1 (z). (1.3.12) Here the hypergeometric function in Watson’s formula is replaced with Leg- endre polynomials. Multiplying Equation 1.3.12 by sin(η)/ cos(x) −cos(η), integrating with respect to η from x to π and using the results from Problem 2, we obtain Equation 1.3.10. In a similar manner, we also have ∞ n=0 J 2n+1 (z)sin n + 1 2 x = z 4 √ 2 x 0 sin(η)J 0 [z sin(η/2)] cos(η) − cos(x) dη. (1.3.13) Problems 1. Using Equation 1.3.4 and Equation 1.3.5, show that the following general- ized Fourier series hold: H(θ − t) 2cos(t) − 2cos(θ) = ∞ n=0 P n [cos(θ)] cos n + 1 2 t , 0 ≤ t, θ ≤ π, if we use the eigenfunction y n (x)=cos n + 1 2 x ,0<x<π,andr(x)=1; and H(t − θ) 2cos(θ) − 2cos(t) = ∞ n=0 P n [cos(θ)] sin n + 1 2 t , 0 ≤ θ, t ≤ π, if we use the eigenfunction y n (x)=sin n + 1 2 x ,0<x<π,andr(x)=1. 2. The series given in Problem 1 are also expansions in Legendre polynomials. In that light, show that t 0 P n [cos(θ)] sin(θ) 2cos(θ) − 2cos(t) dθ = sin n + 1 2 t n + 1 2 , 18 Gradshteyn and Ryzhik, op. cit., Formula 8.927. 19 Watson, G. N., 1966: ATreatiseonthe Theory of Bessel Functions.Cambridge University Press, 804 pp. See Equation (3) in Section 5.21. © 2008 by Taylor & Francis Group, LLC 26 Mixed Boundary Value Problems and π t P n [cos(θ)] sin(θ) 2cos(t) −2cos(θ) dθ = cos n + 1 2 t n + 1 2 , where 0 <t<π. 3. Using 1 √ 2 ∞ n=0 {P n−1 [cos(t)] − P n [cos(t)]}sin(nx) = 1 √ 2 ∞ n=0 P n [cos(t)] {sin[(n +1)x] − sin(nx)} and the results from Problem 1, show that 1 √ 2 ∞ n=1 {P n−1 [cos(t)] − P n [cos(t)]}sin(nx)= sin(x/2)H(t − x) cos(x) −cos(t) , and 1 √ 2 ∞ n=1 {P n−1 [cos(t)] + P n [cos(t)]}sin(nx)= cos(x/2)H(x − t) cos(t) − cos(x) , provided 0 <x,t<π. 4. The generating function for Legendre polynomials is 1 − 2ξ cos(θ)+ξ 2 −1/2 = ∞ n=0 P n [cos(θ)]ξ n , |ξ| < 1. Setting ξ = h,thenξ = −h,andfinally adding and subtracting the resulting equations, show 20 that 2 ∞ n=0 P 2n+m [cos(θ)] h 2n+m = 1 − 2h cos(θ)+h 2 −1/2 +(−1) m 1+2h cos(θ)+h 2 −1/2 for m =0and1. 20 Forits use, see Minkov, I. M., 1963: Electrostatic field of a sectional spherical capac- itor. Sov. Tech. Phys., 7, 1041–1043. © 2008 by Taylor & Francis Group, LLC Overview 27 Multiplying the previous equation by √ h,setting h = e it ,andthen sep- arating the real and imaginary parts, show that 2 ∞ n=0 P 2n+m [cos(θ)] sin 2n + m + 1 2 t = H(t − θ) 2[cos(θ) − cos(t)] , and 2 ∞ n=0 P 2n+m [cos(θ)] cos 2n + m + 1 2 t = H(θ − t) 2[cos(t) − cos(θ)] + (−1) m 2[cos(θ)+cos(t)] , where 0 <t,θ<π/2. 1.4 BESSEL FUNCTIONS The solution to the classic differential equation r 2 d 2 y dr 2 + r dy dr +(λ 2 r 2 − n 2 )y =0, (1.4.1) commonly known as Bessel’s equation of order n with a parameter λ,is y(r)=c 1 J n (λr)+c 2 Y n (λr), (1.4.2) where J n (·)andY n (·)arenth order Bessel functions of the first and second kind, respectively. Figure 1.4.1 illustrates J 0 (x), J 1 (x), J 2 (x), and J 3 (x)while in Figure 1.4.2 Y 0 (x), Y 1 (x), Y 2 (x), and Y 3 (x)aregraphed. Bessel functions have been exhaustively studied and a vast literature now exists on them. 21 The Bessel function J n (z)isanentirefunction, has no complex zeros, and has an infinite number of real zeros symmetrically located with respect to the point z =0,whichis itself a zero if n>0. All of the zeros are simple, except the point z =0,whichisazero of order n if n>0. On the other hand, Y n (z) is analytic in the complex plane with a branch cut along the segment (−∞, 0] andbecomes infinite as z → 0. Considerable insight into the nature of Bessel functions is gained from their asymptotic expansions. These expansions are J n (z) ∼ 2 πz 1/2 cos z − 1 2 nπ − 1 4 π , |arg(z)|≤π −, |z|→∞, (1.4.3) 21 The standard reference isWatson,op. cit. © 2008 by Taylor & Francis Group, LLC Overview 29 0.0 1.0 2.0 3.0 x 0.0 1.0 2.0 3.0 4.0 5.0 1 I I I 2 I 0 3 Figure 1.4.3:Thefirst four modified Bessel functions of the first kind over 0 ≤ x ≤ 3. in the complex z-plane provided that we introduce a branch cut along the segment (−∞, 0]. As z → 0, K ν (z)becomes infinite. Figure 1.4.3 illustrates I 0 (x), I 1 (x), I 2 (x), and I 3 (x)while in Figure 1.4.4 K 0 (x), K 1 (x), K 2 (x), and K 3 (x)aregraphed. Turningtothezeros,I ν (z)haszeros that are purely imaginary for ν> −1. On the other hand, K ν (z)hasnozeros in the region |arg(z)|≤π/2. In the remaining portion of the cut z-plane, it has a finite number of zeros. The modified Bessel functions also have asymptotic representations: I ν (z) ∼ e z √ 2πz + e −z±π(ν+ 1 2 ) √ 2πz , |arg(z)|≤π − , |z|→∞, (1.4.7) and K n (z) ∼ πe −z √ 2πz , |arg(z)|≤π − , |z|→∞, (1.4.8) where wechosetheplus sign if (z) > 0, and the minus sign if (z) < 0. Note that K n (z)decrease exponentially as x →∞,while I n (z)increases exponentially as x →∞and x →−∞. Having introduced Bessel functions, we now turn to some of their useful properties. Repeatedly in the following chapters, we will encounter them in improper integrals. Here, we list some of the most common ones: 22 t 0 xJ 0 (kx) √ t 2 − x 2 dx = sin(kt) k ,k>0, (1.4.9) ∞ 0 e −αx J ν (βx) x ν dx = (2β) ν Γ ν + 1 2 √ π (α 2 + β 2 ) ν+ 1 2 , (ν) > − 1 2 , (α) > |(β)|, (1.4.10) 22 Gradshteyn and Ryzhik, op. cit., Formulas 6.554.2, 6.623.1, 6.623.2, 6.671.1, 6.671.2, and 6.699.8. © 2008 by Taylor & Francis Group, LLC [...]... (µn x/b) 2 √ , = 2 2 − x2 b n=1 µn J0 (µn ) a 0 ≤ x < b, where a < b and µn is the nth positive root of J0 (µ) = −J1 (µ) = 0 11 Given the definite integral27 a 0 cos(cx) J0 b a2 − x2 √ sin a b2 + c2 √ dx = , b 2 + c2 0 < b, show that √ cosh b t2 − x2 2 √ H(t − x) = 2 a t2 − x2 ∞ sin t 2 − b2 J0 (µk x) k 2 2 − b2 J1 (µk a) k k=1 , where 0 < x < a and µk is the kth positive root of J0 (µa) = 0 12 Using... once more, we find that Ak = = 2 2 µ4 J1 (µk ) k 2 2 J1 (µk ) µ3 J1 (µk ) − 2 2 J2 (µk ) k k J1 (µk ) 2J2 (µk ) − µk 2 k (1.4. 32) (1.4.33) However, from the first line of Table 1.4.1 with n = 1, J1 (µk ) = 1 µk [J2 (µk ) + J0 (µk )] , 2 or J2 (µk ) = 2J1 (µk ) , µk (1.4.34) (1.4.35) because J0 (µk ) = 0 Therefore, Ak = and x2 = 2 ∞ k=1 2( 2 − 4)J1 (µk ) k , 2 µ3 J1 (µk ) k ( 2 − 4)J0 (µk x) k , µ3 J1 (µk... L) (1.4.17) 2 On the other hand, if Jn (µk L) = 0, then Ck = © 20 08 by Taylor & Francis Group, LLC 2 L2 − n2 2 k Jn (µk L) 2 2 k (1.4.18) 32 Mixed Boundary Value Problems Finally, if µk Jn (µk L) = −hJn (µk L), 2 L2 − n2 + h2 L2 2 k Jn (µk L) 2 2 k Ck = (1.4.19) All of the preceding results must be slightly modified when n = 0 and the boundary condition is J0 (µk L) = 0 or µk J1 (µk L) = 0 For this... (µk ) 0 However, from the third line of Table 1.4.1, d 2 x J2 (x) = x2 J1 (x), dx (1.4 .24 ) if n = 2 Therefore, Equation 1.4 .23 becomes Ak = 2x2 J2 (x) 2 µ3 J2 (µk ) k µk = 0 2 , µk J2 (µk ) (1.4 .25 ) and the resulting expansion is ∞ x =2 k=1 © 20 08 by Taylor & Francis Group, LLC J1 (µk x) , µk J2 (µk ) 0 ≤ x < 1 (1.4 .26 ) Overview 33 1 1 term f(x) 2 terms f(x) 1 0.5 0.5 0 0 0 1 0.5 1 0 3 terms f(x) 0.5... ρ J0 (kρ) cos(kξ) 0 0 ∞ c 2 r 2 − 2 0 f (ξ) 0 0 Because r r 0 r2 − 0 ρ r2 η J0 (kη) 0 r cos(kξ) 1 + e2ka πU0 2 = 2 dρ dk ρ J0 (kρ) r 2 − 2 dρ dk dξ dρ (2. 1.14) sin(kr) , k (2. 1.15) − 2 dη = dξ Equation 2. 1.14 simplifies to ∞ c f (ξ) dk dξ k cos(kξ) sin(kr) dk 1 + e2ka k cos(kξ) sin(kr) 0 0 ∞ c 2 f (ξ) 0 0 dξ = πU0 r 2 (2. 1.16) Upon taking the derivative of Equation 2. 1.16 with respect to r, we... 1.4.17, 1 2 x3 J0 (µk x) dx (1.4 .28 ) Ak = 2 J1 (µk ) 0 If we let t = µk x, the integration Equation 1.4 .28 becomes Ak = µk 2 2 µ4 J1 (µk ) k 0 t3 J0 (t) dt (1.4 .29 ) We now let u = t2 and dv = tJ0 (t) dt so that integration by parts results in 2 µk 3 4 J 2 (µ ) t J1 (t) 0 − 2 µk 1 k 2 = 4 2 µ3 J1 (µk ) − 2 µk J1 (µk ) k µk Ak = © 20 08 by Taylor & Francis Group, LLC 0 µk 0 t2 J1 (t) dt (1.4.30) t2 J1 (t)... condition along z = ±a and r > c Turning to the boundary condition for 0 ≤ r < c, we have that ∞ c tanh(ka) 0 © 20 08 by Taylor & Francis Group, LLC 0 f (ξ) cos(kξ) dξ J0 (kr) dk = πU0 , 2 (2. 1. 12) 44 Mixed Boundary Value Problems or ∞ c 1− f (ξ) 0 0 2 1 + e2ak cos(kξ)J0 (kρ) dk dξ = πU0 2 (2. 1.13) We now multiply both sides of Equation 2. 1.13 by ρ dρ/ r2 − 2 and integrate from 0 to r Interchanging the... the potential problem 1 ∂ ∂u r r ∂r ∂r + ∂2u = 0, ∂z 2 0 ≤ r < ∞, −∞ < z < ∞, (2. 2.1) 5 Kirchhoff, G., 1877: Zur Theorie des Kondensators Monatsb Deutsch Akad Wiss Berlin, 144–1 62 6 Carlson, G T., and B L Illman, 1994: The circular disk parallel plate capacitor Am J Phys., 62, 1099–1105 © 20 08 by Taylor & Francis Group, LLC 46 Mixed Boundary Value Problems Figure 2. 2.1: Gustav Robert Kirchhoff’s (1 824 –1887)... 1967: Weber’s mixed boundary value problem in electrodynamics J Math Phys., 8, 518– 522 © 20 08 by Taylor & Francis Group, LLC Historical Background 43 variables or transform methods, the solution to Equation 2. 1.1 and Equation 2. 1 .2 can be written u(r, z) = ∞ 2 π A(k) 0 sinh(kz) J0 (kr) dk cosh(ka) (2. 1.4) Substituting Equation 2. 1.4 into Equation 2. 1.3, we obtain the dual integral equations 2 π ∞ tanh(ka)A(k)J0... Francis Group, LLC 2 π ∞ sin(kc) 0 sinh(kz) dk J0 (kr) cosh(ka) k (2. 1 .20 ) Historical Background 45 1.5 u(r,z)/U 0 1 0.5 0 −0.5 −1 −1.5 1 0.5 2 0 1.5 1 −0.5 z/a 0.5 −1 0 r/c Figure 2. 1 .2: The solution u(r, z)/U0 to the mixed boundary value problem governed by Equation 2. 1.1 through Equation 2. 1.3 when c/a = 2 In Chapter 4, we discuss how to numerically solve Equation 2. 1.19 for a finite value of c, so . J 0 b a 2 − x 2 dx = sin a √ b 2 + c 2 √ b 2 + c 2 , 0 <b, show that cosh b √ t 2 − x 2 √ t 2 − x 2 H(t − x)= 2 a 2 ∞ k=1 sin t µ 2 k − b 2 J 0 (µ k x) µ 2 k − b 2 J 2 1 (µ k a) , where. L.IfJ n (µ k L)=0, then C k = 1 2 L 2 J 2 n+1 (µ k L). (1.4.17) On the other hand, if J n (µ k L)=0,then C k = µ 2 k L 2 − n 2 2µ 2 k J 2 n (µ k L). (1.4.18) © 20 08 by Taylor & Francis Group, LLC 32 Mixed Boundary Value. 1.4.1, d dx x 2 J 2 (x) = x 2 J 1 (x), (1.4 .24 ) if n =2. Therefore, Equation 1.4 .23 becomes A k = 2x 2 J 2 (x) µ 3 k J 2 2 (µ k ) µ k 0 = 2 µ k J 2 (µ k ) , (1.4 .25 ) and the resulting