Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 3 potx

Show that j 1 (z) = sin z z 2 − cos z z , i 0 (z) = sinh z z , k 0 (z) = π 2z exp(−z). Exercise 34.13 Show that as x → ∞, K n (x) ∝ e −x √ x  1 + 4n 2 − 1 8x + (4n 2 − 1)(4n 2 − 9) 128x 2 + ···  . 1654 34.9 Hints Hint 34.2 Hint 34.3 Hint 34.4 Use the generating function e 1 2 z(t−1/t) = ∞  n=−∞ J n (z)t n to show that J n satisfies Bessel’s equation z 2 y  + zy  +  z 2 − n 2  y = 0. Hint 34.6 Use variation of parameters and the Wronskian that was derived in the text. Hint 34.7 Compute the Wronskian of J ν (z) and Y ν (z). Use the relation W [J ν , J −ν ] = − 2 πz sin(πν) Hint 34.8 Derive W [I ν , I −ν ] from the value of W [J ν , J −ν ]. Derive W [I ν , K ν ] from the value of W [I ν , I −ν ]. Hint 34.9 Hint 34.10 1655 Hint 34.11 Hint 34.12 Hint 34.13 Hint 34.14 1656 34.10 Solutions Solution 34.1 Bessel’s equation is L[y] ≡ z 2 y  + zy  +  z 2 − n 2  y = 0. We consider a solution of the form y(z) =  C e 1 2 z(t−1/t) v(t) dt. We substitute the form of the solution into Bessel’s equation.  C L  e 1 2 z(t−1/t)  v(t) dt = 0  C  z 2 1 4  t + 1 t  2 + z 1 2  t − 1 t  2 +  z 2 − n 2   e 1 2 z(t−1/t) v(t) dt = 0 (34.1) By considering d dt t e 1 2 z(t−1/t) =  1 2 x  t + 1 t  + 1  e 1 2 z(t−1/t) d 2 dt 2 t 2 e 1 2 z(t−1/t) =  1 4 x 2  t + 1 t  2 + x  2t + 1 t  + 2  e 1 2 z(t−1/t) we see that L  e 1 2 z(t−1/t)  =  d 2 dt 2 t 2 − 3 d dt t +  1 − n 2   e 1 2 z(t−1/t) . Thus Equation 34.1 becomes  C  d 2 dt 2 t 2 e 1 2 z(t−1/t) −3 d dt t e 1 2 z(t−1/t) +(1 − n 2 ) e 1 2 z(t−1/t)  v(t) dt = 0 1657 We apply integration by parts to move derivatives from the kernel to v(t).  t 2 e 1 2 z(t−1/t) v(t)  C −  t e 1 2 z(t−1/t) v  (t)  C +  −3t e 1 2 z(t−1/t) v(t)  C +  C e 1 2 z(t−1/t)  t 2 v  (t) + 3tv(t) +  1 − n 2  v(t)  dt = 0  e 1 2 z(t−1/t)  (t 2 − 3t)v(t) −tv  (t)   C +  C e 1 2 z(t−1/t)  t 2 v  (t) + 3tv(t) + (1 −n 2 )v(t)  dt = 0 In order that the integral vanish, v(t) must be a solution of the differential equation t 2 v  + 3tv +  1 − n 2  v = 0. This is an Euler equation with the solutions {t n−1 , t −n−1 } for non-zero n and {t −1 , t −1 log t} for n = 0. Consider the case of non-zero n. Since e 1 2 z(t−1/t)  t 2 − 3t  v(t) −tv  (t)  is single-valued and analytic for t = 0 for the functions v(t) = t n−1 and v(t) = t −n−1 , the bound ary term will vanish if C is any closed contour that that does not pass through the origin. Note that the integrand in our solution, e 1 2 z(t−1/t) v(t), is analytic and single-valued except at the origin and infinity where it has essential singularities. Consider a simple closed contour that does not enclose the origin. The integral along such a path would vanish and give us y(z) = 0. This is not an interesting solution. Since e 1 2 z(t−1/t) v(t), has non-zero residues for v(t) = t n−1 and v(t) = t −n−1 , choosing any simple, positive, closed contour about the origin will give us a non-trivial solution of Bessel’s equation. These solutions are y 1 (t) =  C t n−1 e 1 2 z(t−1/t) dt, y 2 (t) =  C t −n−1 e 1 2 z(t−1/t) dt. Now consider the case n = 0. The two solutions above concide and we have the solution y(t) =  C t −1 e 1 2 z(t−1/t) dt. 1658 Choosing v(t) = t −1 log t would make both the boundary terms and the integrand multi-valued. We do not pursue the possibility of a solution of this form. The solution y 1 (t) and y 2 (t) are not linearly independent. To demonstrate this we make the change of variables t → −1/t in the integral representation of y 1 (t). y 1 (t) =  C t n−1 e 1 2 z(t−1/t) dt =  C (−1/t) n−1 e 1 2 z(−1/t+t) −1 t 2 dt =  C (−1) n t −n−1 e 1 2 z(t−1/t) dt = (−1) n y 2 (t) Thus we see that a solution of Bessel’s equation for integer n is y(t) =  C t −n−1 e 1 2 z(t−1/t) dt where C is any simple, closed contour about the origin. Therefore, the Bessel function of the first kind and order n, J n (z) = 1 ı2π  C t −n−1 e 1 2 z(t−1/t) dt is a solution of Bessel’s equation for integer n. Note that J n (z) is the coefficient of t n in the Laurent series of e 1 2 z(t−1/t) . This establishes the generating function for the Bessel functions. e 1 2 z(t−1/t) = ∞  n=−∞ J n (z)t n 1659 Solution 34.2 The generating function is e z 2 (t−1/t) = ∞  n=−∞ J n (z)t n . In order to show that J n satisfies Bessel’s equation we seek to show that ∞  n=−∞  z 2 J  n (z) + zJ n (z) + (z 2 − n 2 )J n (z)  t n = 0. To get the appropriate terms in the sum we will differentiate the generating function with respect to z and t. First we differentiate it with respect to z. 1 2  t − 1 t  e z 2 (t−1/t) = ∞  n=−∞ J  n (z)t n 1 4  t − 1 t  2 e z 2 (t−1/t) = ∞  n=−∞ J  n (z)t n Now we differentiate with respect to t and multiply by t get the n 2 J n term. z 2  1 + 1 t 2  e z 2 (t−1/t) = ∞  n=−∞ nJ n (z)t n−1 z 2  t + 1 t  e z 2 (t−1/t) = ∞  n=−∞ nJ n (z)t n z 2  1 − 1 t 2  e z 2 (t−1/t) + z 2 4  t + 1 t  2 e z 2 (t−1/t) = ∞  n=−∞ n 2 J n (z)t n−1 z 2  t − 1 t  e z 2 (t−1/t) + z 2 4  t + 1 t  2 e z 2 (t−1/t) = ∞  n=−∞ n 2 J n (z)t n 1660 Now we can evaluate the desired sum. ∞  n=−∞  z 2 J  n (z) + zJ n (z) +  z 2 − n 2  J n (z)  t n =  z 2 4  t − 1 t  2 + z 2  t − 1 t  + z 2 − z 2  t − 1 t  − z 2 4  t + 1 t  2  e z 2 (t−1/t) ∞  n=−∞  z 2 J  n (z) + zJ n (z) +  z 2 − n 2  J n (z)  t n = 0 z 2 J  n (z) + zJ n (z) +  z 2 − n 2  J n (z) = 0 Thus J n satisfies Bessel’s equation. Solution 34.3 J  n = n z J n − J n+1 = 1 2 (J n−1 + J n+1 ) − J n+1 = 1 2 (J n−1 − J n+1 ) J  n = n z J n − J n+1 = n z J n −  2n z J n − J n−1  = J n−1 − n z J n 1661 Solution 34.4 The linearly independent homogeneous solutions are J 1/2 and J −1/2 . The Wronskian is W [J 1/2 , J −1/2 ] = − 2 πz sin(π/2) = − 2 πz . Using variation of parameters, a particular solution is y p = −J 1/2 (z)  z ζJ −1/2 (ζ) −2/πζ dζ + J −1/2 (z)  z ζJ 1/2 (ζ) −2/πζ dζ = π 2 J 1/2 (z)  z ζ 2 J −1/2 (ζ) dζ − π 2 J −1/2 (z)  z ζ 2 J 1/2 (ζ) dζ. Thus the general solution is y = c 1 J 1/2 (z) + c 2 J −1/2 (z) + π 2 J 1/2 (z)  z ζ 2 J −1/2 (ζ) dζ − π 2 J −1/2 (z)  z ζ 2 J 1/2 (ζ) dζ. We could substitute J 1/2 (z) =  2 πz  1/2 sin z and J −1/2 =  2 πz  1/2 cos z into the solution, but we cannot evaluate the integrals in terms of elementary functions. (You can write the solution in terms of Fresnel integrals.) 1662 [...]... ∂x3 ∂ξi 2 The gradient, divergence, etc., follow a1 ∂u a2 ∂u a3 ∂u + + h1 ∂ξ1 h2 ∂ξ2 h3 ∂ 3 ∂ ∂ ∂ (h2 h3 v1 ) + (h3 h1 v2 ) + (h1 h2 v3 ) ∂ξ1 ∂ξ2 ∂ 3 h2 h3 ∂u ∂ h3 h1 ∂u ∂ h1 h2 ∂u + + h1 ∂ξ1 ∂ξ2 h2 ∂ξ2 ∂ 3 h3 ∂ 3 u= 1 h1 h2 h3 1 ∂ 2 u= h1 h2 h3 ∂ξ1 ·v = 1681 35 . 1 Exercises Exercise 35 . 1 Find the Laplacian in cylindrical coordinates (r, θ, z) x = r cos θ, y = r sin θ, z Hint, Solution Exercise 35 . 2... Hint, Solution 1682 z = r cos φ 35 . 2 Hints Hint 35 . 1 Hint 35 . 2 16 83 35 . 3 Solutions Solution 35 . 1 h1 = (cos θ)2 + (sin θ)2 + 0 = 1 h2 = (−r sin θ)2 + (r cos θ)2 + 0 = r h3 = 2 u= 1 r √ ∂ ∂r 0 + 0 + 12 = 1 ∂u ∂ 1 ∂u ∂ ∂u + + r ∂r ∂θ r ∂θ ∂z ∂z 2 2 1 ∂ ∂u 1∂ u ∂ u 2 u= r + 2 2 + 2 r ∂r ∂r r ∂θ ∂z r Solution 35 . 2 h1 = h2 = u= (r cos φ cos θ)2 + (r cos φ sin θ)2 + (−r sin φ)2 = r h3 = 2 (sin φ cos θ)2 + (sin... then have ξ = F (x, y) Upon solving for ξ and ψ we divide Equation 36 .2 by β(ξ, ψ) to obtain the canonical form Note that we could have solved for ξy /ξx in Equation 36 .4 dx ξy b− =− = dy ξx √ b2 − ac c This form is useful if a vanishes Another canonical form for hyperbolic equations is uσσ − uτ τ = K(σ, τ, u, uσ , uτ ) 1687 (36 .5) We can transform Equation 36 .3 to this form with the change of variables... √ + c2 xY1/p for p = 0, ±1 2 p/2 x p for p = 0 The Airy equation y + xy = 0 is the case p = 3 The general solution of the Airy equation is √ y(x) = c1 xJ1 /3 2 3/ 2 x 3 √ + c2 xJ−1 /3 2 3/ 2 x 3 Solution 34 .11 Consider J1/2 (z) We start with the series expansion ∞ z (−1)m J1/2 (z) = m!Γ(1/2 + m + 1) 2 m=0 Use the identity Γ(n + 1/2) = 1/2+2m (1) (3) ···(2n−1) √ π 2n ∞ = (−1)m 2m+1 √ m!(1) (3) · · · (2m... difference of Equation 34 .3 and Equation 34 .4 by ı2 gives us the other identity ∞ (1 − (−1)n ) Jn (z)ın = sin z −ı n=1 ∞ Jn (z)ın−1 = sin z 2 n=1 odd n ∞ (−1)(n−1)/2 Jn (z) = sin z 2 n=1 odd n ∞ (−1)n J2n+1 (z) = sin z 2 n=0 1668 4 We substitute −t for t in the generating function ∞ e − 1 z(t−1/t) 2 Jn (z)(−t)n = (34 .5) n=−∞ We take the product of Equation 34 .2 and Equation 34 .5 to obtain the final identity... , uψ ) (36 .3) 1686 We require that the uξξ and uψψ terms vanish That is α = γ = 0 in Equation 36 .2 This gives us two constraints on ξ and ψ 2 2 aξx + 2bξx ξy + cξy = 0, √ ξx −b + b2 − ac = , ξy a √ b − b2 − ac ξx + ξy = 0, a 2 2 aψx + 2bψx ψy + cψy = 0 √ ψx −b − b2 − ac = ψy a √ b + b2 − ac ψx + ψy = 0 a (36 .4) Here we chose the signs in the quadratic formulas to get different solutions for ξ and ψ Now... The series does not converge for any x in the finite complex plane However, if we take only a finite number of terms in the series, it gives a good approximation of Kn (x) for large, positive x At x = 10, the one, two and three term approximations give relative errors of 0.01, 0.0006 and 0.00006, respectively 1679 Part V Partial Differential Equations 1680 Chapter 35 Transforming Equations I’m about two... quasi-linear partial differential equations for the coordinates ξ and ψ We solve these equations with the method of characteristics The characteristic equations for ξ are √ b − b2 − ac d dy = , ξ(x, y(x)) = 0 dx a dx Solving the differential equation for y(x) determines ξ(x, y) We just write the solution for y(x) in the form F (x, y(x)) = const Since the solution of the differential equation for ξ is ξ(x,... = eız (34 .3) (−1)n Jn (z)ın = e−ız (34 .4) J0 (z) + 2 n=1 Next we substitute t = −ı into the generating function ∞ J0 (z) + 2 n=1 1667 Dividing the sum of Equation 34 .3 and Equation 34 .4 by 2 gives us the desired identity ∞ (1 + (−1)n ) Jn (z)ın = cos z J0 (z) + n=1 ∞ Jn (z)ın = cos z J0 (z) + 2 n=2 even n ∞ (−1)n/2 Jn (z) = cos z J0 (z) + 2 n=2 even n ∞ (−1)n J2n (z) = cos z J0 (z) + 2 n=1 3 Dividing... Equation 36 .1 yields an equation in ξ and ψ 2 2 aξx + 2bξx ξy + cξy uξξ + 2 (aξx ψx + b(ξx ψy + ξy ψx ) + cξy ψy ) uξψ 2 2 + aψx + 2bψx ψy + cψy uψψ = H(ξ, ψ, u, uξ , uψ ) α(ξ, ψ)uξξ + β(ξ, ψ)uξψ + γ(ξ, ψ)uψψ = H(ξ, ψ, u, uξ , uψ ) 36 .1.1 (36 .2) Hyperbolic Equations We start with a hyperbolic equation, (b2 −ac > 0) We seek a change of independent variables that will put Equation 36 .1 in the form uξψ . − 2 πz sin(πν) Hint 34 .8 Derive W [I ν , I −ν ] from the value of W [J ν , J −ν ]. Derive W [I ν , K ν ] from the value of W [I ν , I −ν ]. Hint 34 .9 Hint 34 .10 1 655 Hint 34 .11 Hint 34 .12 Hint 34 . 13 Hint 34 .14 1 656 34 .10. c 2 √ xY 1/p  2 p x p/2  for p = 0 The Airy equation y  + xy = 0 is the case p = 3. The general solution of the Airy equation is y(x) = c 1 √ xJ 1 /3  2 3 x 3/ 2  + c 2 √ xJ −1 /3  2 3 x 3/ 2  . Solution 34 .11 Consider. = π 2z exp(−z). Exercise 34 . 13 Show that as x → ∞, K n (x) ∝ e −x √ x  1 + 4n 2 − 1 8x + (4n 2 − 1)(4n 2 − 9) 128x 2 + ···  . 1 654 34 .9 Hints Hint 34 .2 Hint 34 .3 Hint 34 .4 Use the generating