Advanced Mathematical Methods for Scientists and Engineers Episode 5 Part 1 pdf

Exercise 32.14 Use the cosine transform to solve y − a2 y = on x ≥ with y (0) = b, y(∞) = Use the cosine transform to show that the Green function for the above with b = is G(x, ξ) = − −a|x−ξ| −a(x−ξ) e − e 2a 2a Hint, Solution Exercise 32.15 Use the sine transform to solve y − a2 y = on x ≥ with y(0) = b, y(∞) = Try using the Laplace transform on this problem Why isn’t it as convenient as the Fourier transform? Use the sine transform to show that the Green function for the above with b = is g(x; ξ) = −a(x−ξ) e − e−a|x+ξ| 2a Hint, Solution Exercise 32.16 Find the Green function which solves the equation y + 2µy + (β + µ2 )y = δ(x − ξ), µ > 0, β > 0, in the range −∞ < x < ∞ with boundary conditions y(−∞) = y(∞) = 1574 Use this Green’s function to show that the solution of y + 2µy + (β + µ2 )y = g(x), µ > 0, β > 0, y(−∞) = y(∞) = 0, with g(±∞) = in the limit as µ → is y= x β g(ξ) sin[β(x − ξ)]dξ −∞ You may assume that the interchange of limits is permitted Hint, Solution Exercise 32.17 Using Fourier transforms, find the solution u(x) to the integral equation ∞ −∞ u(ξ) dξ = 2 + a2 ] [(x − ξ) x + b2 < a < b Hint, Solution Exercise 32.18 The Fourer cosine transform is defined by ˆ fc (ω) = π ∞ f (x) cos(ωx) dx From the Fourier theorem show that the inverse cosine transform is given by ∞ ˆ fc (ω) cos(ωx) dω f (x) = 2 Show that the cosine transform of f (x) is ˆ −ω fc (ω) − 1575 f (0) π Use the cosine transform to solve the following boundary value problem y − a2 y = on x > with y (0) = b, y(∞) = Hint, Solution Exercise 32.19 The Fourier sine transform is defined by ˆ fs (ω) = π ∞ f (x) sin(ωx) dx Show that the inverse sine transform is given by ∞ ˆ fs (ω) sin(ωx) dω f (x) = 2 Show that the sine transform of f (x) is ω ˆ f (0) − ω fs (ω) π Use this property to solve the equation y − a2 y = on x > with y(0) = b, y(∞) = Try using the Laplace transform on this problem Why isn’t it as convenient as the Fourier transform? Hint, Solution Exercise 32.20 Show that (Fc [f (x) + f (−x)] − ıFs [f (x) − f (−x)]) where F, Fc and Fs are respectively the Fourier transform, Fourier cosine transform and Fourier sine transform Hint, Solution F[f (x)] = 1576 Exercise 32.21 Find u(x) as the solution to the integral equation: ∞ −∞ u(ξ) dξ = , + a2 (x − ξ) x + b2 < a < b Use Fourier transforms and the inverse transform Justify the choice of any contours used in the complex plane Hint, Solution 1577 32.10 Hints Hint 32.1 H(x + c) − H(x − c) = Hint 32.2 Consider the two cases half plane (ω) < and for |x| < c, for |x| > c (ω) > 0, closing the path of integration with a semi-circle in the lower or upper Hint 32.3 Hint 32.4 Hint 32.5 Hint 32.6 Hint 32.7 Hint 32.8 Hint 32.9 1578 Hint 32.10 Hint 32.11 The left side is the convolution of u(x) and e−ax Hint 32.12 Hint 32.13 Hint 32.14 Hint 32.15 Hint 32.16 Hint 32.17 Hint 32.18 Hint 32.19 Hint 32.20 1579 Hint 32.21 1580 32.11 Solutions Solution 32.1 F[H(x + c) − H(x − c)] = = = = ∞ (H(x + c) − H(x − c)) e−ıωx dx 2π −∞ c e−ıωx dx 2π −c c e−ıωx 2π −ıω −c eıωc e−ıωc − 2π −ıω −ıω F[H(x + c) − H(x − c)] = sin(cω) πω Solution 32.2 F If 1 = x2 + c2 2π = 2π ∞ −∞ ∞ −∞ e−ıωx dx x2 + c2 e−ıωx dx (x − ıc)(x + ıc) (ω) < then we close the path of integration with a semi-circle in the upper half plane F 1 = 2πi Res + c2 x 2π e−ıωx , x = ıc (x − ıc)(x + ıc) 1581 = cω e 2c If ω > then we close the path of integration in the lower half plane F 1 = − 2πi Res + c2 x 2π e−ıωx , −ıc (x − ıc)(x + ıc) = −cω e 2c Thus we have that F x2 1 −c|ω| e , = +c 2c for (c) = Solution 32.3 ∞ y sin(ωx) dx π ∞ ω ∞ = y sin(ωx) − y cos(ωx) dx π π 0 = −ω yc (ω) ˆ ∞ Fs [y ] = y sin(ωx) dx π ∞ ω ∞ = y sin(ωx) − y cos(ωx) dx π π 0 ∞ ω ω2 ∞ = − y cos(ωx) − y sin(ωx) dx π π 0 ω = −ω ys (ω) + y(0) ˆ π Fs [y ] = 1582 We take the Fourier transform, utilizing the convolution theorem e−a|ω| e−b|ω| = 2a 2b −(b−a)|ω| ae u(ω) = ˆ 2πb a 2(b − a) u(x) = 2πb x + (b − a)2 2πˆ(ω) u a(b − a) πb(x2 + (b − a)2 ) u(x) = Solution 32.18 ˆ Note that fc (ω) is an even function We compute the inverse Fourier cosine transform −1 ˆ f (x) = Fc fc (ω) ∞ ˆ fc (ω) eıωx dω = −∞ ∞ ˆ fc (ω)(cos(ωx) + ı sin(ωx)) dω = −∞ ∞ ˆ fc (ω) cos(ωx) dω = −∞ ∞ ˆ fc (ω) cos(ωx) dω =2 1598 ∞ y cos(ωx) dx π ω ∞ ∞ y sin(ωx) dx = [y cos(ωx)]0 + π π ω2 ∞ ω = − y (0) + [y sin(ωx)]∞ − y cos(ωx) dx π π π Fc [y ] = Fc [y ] = −ω yc (ω) − ˆ y (0) π We take the Fourier cosine transform of the differential equation b − a2 y (ω) = ˆ π b y (ω) = − ˆ π(ω + a2 ) −ω y (ω) − ˆ Now we take the inverse Fourier cosine transform We use the fact that y (ω) is an even function ˆ b + a2 ) b = F −1 − + a2 ) π(ω b eıωx , ω = ıa = − ı2π Res + a2 π ω eıωx = −ı2b lim , for x ≥ ω→ıa ω + ıa −1 y(x) = Fc − π(ω b y(x) = − e−ax a 1599 Solution 32.19 Suppose f (x) is an odd function The Fourier transform of f (x) is 2π = 2π ı =− π ∞ f (x) e−ıωx dx F[f (x)] = −∞ ∞ f (x)(cos(ωx) − ı sin(ωx)) dx −∞ ∞ f (x) sin(ωx) dx ˆ ˆ Note that f (ω) = F[f (x)] is an odd function of ω The inverse Fourier transform of f (ω) is ∞ ˆ F −1 [f (ω)] = ˆ f (ω) eıωx dω −∞ ∞ ˆ f (ω) sin(ωx) dω = 2ı Thus we have that ∞ − f (x) = 2ı ∞ =2 π ı π ∞ f (x) sin(ωx) dx sin(ωx) dω ∞ f (x) sin(ωx) dx sin(ωx) dω This gives us the Fourier sine transform pair ∞ ˆ fs (ω) sin(ωx) dω, f (x) = 1600 ˆ fs (ω) = π ∞ f (x) sin(ωx) dx ∞ y sin(ωx) dx π ∞ ω ∞ = y sin(ωx) − y cos(ωx) dx π π 0 ∞ ω ω2 ∞ y sin(ωx) dx = − y cos(ωx) − π π 0 Fs [y ] = Fs [y ] = −ω ys (ω) + ˆ ω y(0) π We take the Fourier sine transform of the differential equation bω − a2 y (ω) = ˆ π bω y (ω) = ˆ + a2 ) π(ω −ω y (ω) + ˆ Now we take the inverse Fourier sine transform We use the fact that y (ω) is an odd function ˆ bω + a2 ) bω = −ıF −1 π(ω + a2 ) b ω eıωx , ω = ıa = −ı ı2π Res + a2 π ω ıωx ωe = 2b lim ω→ıa ω + ıa −ax = be for x ≥ −1 y(x) = Fs π(ω y(x) = b e−ax 1601 Now we solve the differential equation with the Laplace transform y − a2 y = s2 y (s) − sy(0) − y (0) − a2 y (s) = ˆ ˆ We don’t know the value of y (0), so we treat it as an unknown constant bs + y (0) s − a2 y (0) y(x) = b cosh(ax) + sinh(ax) a y (s) = ˆ In order to satisfy the boundary condition at infinity we must choose y (0) = −ab y(x) = b e−ax We see that solving the differential equation with the Laplace transform is not as convenient, because the boundary condition at infinity is not automatically satisfied We had to find a value of y (0) so that y(∞) = Solution 32.20 The Fourier, Fourier cosine and Fourier sine transforms are defined: 2π F[f (x)]c = π F[f (x)]s = π ∞ f (x) e−ıωx dx, F[f (x)] = −∞ ∞ f (x) cos(ωx) dx, ∞ f (x) sin(ωx) dx We start with the right side of the identity and apply the usual tricks of integral calculus to reduce the expression to the left side (Fc [f (x) + f (−x)] − ıFs [f (x) − f (−x)]) 1602 2π 2π 2π ∞ ∞ ∞ f (−x) cos(ωx) dx − ı f (x) cos(ωx) dx + −∞ ∞ ∞ f (x) sin(ωx) dx − ı ∞ f (x) cos(ωx) dx − ı −∞ 2π f (x) sin(ωx) dx − ı f (x) sin(ωx) dx −∞ ∞ f (x) sin(−ωx) dx 0 f (x) cos(ωx) dx + −∞ ∞ 0 f (−x) sin(ωx) dx f (x) cos(−ωx) dx − ı f (x) cos(ωx) dx − ∞ f (x) sin(ωx) dx + ı ∞ f (x) cos(ωx) dx − ı −∞ f (x) sin(ωx) dx −∞ 2π ∞ f (x) e−ıωx dx −∞ F[f (x)] Solution 32.21 We take the Fourier transform of the integral equation, noting that the left side is the convolution of u(x) and 2πˆ(ω)F u We find the Fourier transform of f (x) = even, real-valued function F x2 +c2 1 = + c2 x 2π x2 x2 +a2 1 =F 2 +a x + b2 ˆ Note that since f (x) is an even, real-valued function, f (ω) is an ∞ −∞ x2 e−ıωx dx + c2 1603 For x > we close the path of integration in the upper half plane and apply Jordan’s Lemma to evaluate the integral in terms of the residues ı2π Res 2π e−ıωıc =ı 2ıc −cω e = 2c = e−ıωx , x = ıc (x − ıc)(x + ıc) ˆ Since f (ω) is an even function, we have F x2 1 −c|ω| e = +c 2c Our equation for u(ω) becomes, ˆ −b|ω| −a|ω| e e = 2a 2b a −(b−a)|ω| e u(ω) = ˆ 2πb 2πˆ(ω) u We take the inverse Fourier transform using the transform pair we derived above u(x) = a 2(b − a) + (b − a)2 2πb x u(x) = a(b − a) πb(x2 + (b − a)2 ) 1604 Chapter 33 The Gamma Function 33.1 Euler’s Formula For non-negative, integral n the factorial function is n! = n(n − 1) · · · (1), with 0! = We would like to extend the factorial function so it is defined for all complex numbers Consider the function Γ(z) defined by Euler’s formula ∞ e−t tz−1 dt Γ(z) = (Here we take the principal value of tz−1 ) The integral converges for (z) > If be at least as singular as 1/t at t = and thus the integral will diverge 1605 (z) ≤ then the integrand will Difference Equation Using integration by parts, ∞ e−t tz dt Γ(z + 1) = = −e −t z ∞ ∞ Since − e−t ztz−1 dt − t (z) > the first term vanishes ∞ e−t tz−1 dt =z = zΓ(z) Thus Γ(z) satisfies the difference equation Γ(z + 1) = zΓ(z) For general z it is not possible to express the integral in terms of elementary functions However, we can evaluate the integral for some z The value z = looks particularly simple to ∞ e−t dt = − e−t Γ(1) = ∞ = 0 Using the difference equation we can find the value of Γ(n) for any positive, integral n Γ(1) = Γ(2) = Γ(3) = (2)(1) = Γ(4) = (3)(2)(1) = ··· = ··· Γ(n + 1) = n! 1606 Thus the Gamma function, Γ(z), extends the factorial function to all complex z in the right half-plane For nonnegative, integral n we have Γ(n + 1) = n! Analyticity The derivative of Γ(z) is ∞ e−t tz−1 log t dt Γ (z) = Since this integral converges for 33.2 (z) > 0, Γ(z) is analytic in that domain Hankel’s Formula We would like to find the analytic continuation of the Gamma function into the left half-plane We accomplish this with Hankel’s formula et tz−1 dt Γ(z) = ı2 sin(πz) C Here C is the contour starting at −∞ below the real axis, enclosing the origin and returning to −∞ above the real axis A graph of this contour is shown in Figure 33.1 Again we use the principle value of tz−1 so there is a branch cut on the negative real axis The integral in Hankel’s formula converges for all complex z For non-positive, integral z the integral does not vanish Thus because of the sine term the Gamma function has simple poles at z = 0, −1, −2, For positive, integral z, the integrand is entire and thus the integral vanishes Using L’Hospital’s rule you can show that the points, z = 1, 2, 3, are removable singularities and the Gamma function is analytic at these points Since the only zeroes of sin(πz) occur for integral z, Γ(z) is analytic in the entire plane except for the points, z = 0, −1, −2, 1607 Figure 33.1: The Hankel Contour Difference Equation Using integration by parts we can derive the difference equation from Hankel’s formula et tz dt ı2 sin(π(z + 1)) C −∞+ı0 et tz = − −ı2 sin(πz) −∞−ı0 et tz−1 dt = z ı2 sin(πz) C = zΓ(z) Γ(z + 1) = Evaluating Γ(1), Γ(1) = lim z→1 et tz−1 dt ı2 sin(πz) C 1608 et ztz−1 dt C Both the numerator and denominator vanish Using L’Hospital’s rule, et tz−1 log t dt = lim z→1 ı2π cos(πz) t e log t dt = C ı2π C Let Cr be the circle of radius r starting at −π radians and going to π radians ı2π = ı2π = ı2π −∞ −r et [log(−t) + πi] dt et log t dt + et [log(−t) − πi] dt + = −∞ −∞ −r Cr −∞ −r −r −∞ Cr et log t dt et ı2π dt + −r et log t dt et [log(−t) + πi] dt + et [− log(−t) + πi] dt + Cr The integral on Cr vanishes as r → ı2π ı2π = −∞ et dt = Thus we obtain the same value as with Euler’s formula It can be shown that Hankel’s formula is the analytic continuation of the Gamma function into the left half-plane 33.3 Gauss’ Formula Gauss defined the Gamma function as an infinite product This form is useful in deriving some of its properties We can obtain the product form from Euler’s formula First recall that e −t = lim n→∞ t 1− n 1609 n Substituting this into Euler’s formula, ∞ e−t tz−1 dt Γ(z) = n 1− = lim n→∞ t n n tz−1 dt With the substitution τ = t/n, (1 − τ )n nz−1 τ z−1 n dτ = lim n→∞ = lim nz n→∞ (1 − τ )n τ z−1 dτ Let n be an integer Using integration by parts we can evaluate the integral 1 (1 − τ )n τ z τz − −n(1 − τ )n−1 dτ z z 0 n (1 − τ )n−1 τ z dτ = z n(n − 1) = (1 − τ )n−2 τ z+1 dτ z(z + 1) n(n − 1) · · · (1) τ z+n−1 dτ = z(z + 1) · · · (z + n − 1) (1 − τ )n τ z−1 dτ = n(n − 1) · · · (1) τ z+n z(z + 1) · · · (z + n − 1) z + n n! = z(z + 1) · · · (z + n) = 1610 Thus we have that n! n→∞ z(z + 1) · · · (z + n) (1)(2) · · · (n) = lim nz n→∞ (z + 1)(z + 2) · · · (z + n) z 1 nz = lim z n→∞ (1 + z)(1 + z/2) · · · (1 + z/n) 1 2z 3z · · · nz = lim z n→∞ (1 + z)(1 + z/2) · · · (1 + z/n) 1z 2z · · · (n − 1)z Γ(z) = lim nz Since limn→∞ (n+1)z nz = we can multiply by that factor = 1 2z 3z · · · (n + 1)z lim z n→∞ (1 + z)(1 + z/2) · · · (1 + z/n) 1z 2z · · · nz ∞ 1 (n + 1)z = z n=1 + z/n nz Thus we have Gauss’ formula for the Gamma function ∞ Γ(z) = z n=1 1+ n z 1+ z n −1 We derived this formula from Euler’s formula which is valid only in the left half-plane However, the product formula is valid for all z except z = 0, −1, −2, 33.4 Weierstrass’ Formula 1611 The Euler-Mascheroni Constant Before deriving Weierstrass’ product formula for the Gamma function we will need to define the Euler-Mascheroni constant γ = lim n→∞ 1+ 1 + + ··· + n − log n = 0.5772 · · · In deriving the Euler product formula, we had the equation n! z(z + 1) · · · (z + n) z −1 z −1 z −1 z = lim z −1 + 1+ ··· + n n→∞ n z z z −z log n e 1+ ··· + = lim z + n→∞ Γ(z) n z −z z −z/2 z −z/n e e e = lim z + 1+ ··· + exp n→∞ n Γ(z) = lim nz n→∞ 1+ 1 + · · · + − log n z n Weierstrass’ formula for the Gamma function is then ∞ = z eγz Γ(z) n=1 1+ z −z/n e n Since the product is uniformly convergent, 1/Γ(z) is an entire function Since 1/Γ(z) has no singularities, we see that Γ(z) has no zeros 1612 ... 32 .10 Hint 32 .11 The left side is the convolution of u(x) and e−ax Hint 32 .12 Hint 32 .13 Hint 32 .14 Hint 32. 15 Hint 32 .16 Hint 32 .17 Hint 32 .18 Hint 32 .19 Hint 32.20 15 7 9 Hint 32. 21 15 8 0 32 .11 ... integral 1 (1 − τ )n τ z τz − −n (1 − τ )n? ?1 dτ z z 0 n (1 − τ )n? ?1 τ z dτ = z n(n − 1) = (1 − τ )n−2 τ z +1 dτ z(z + 1) n(n − 1) · · · (1) τ z+n? ?1 dτ = z(z + 1) · · · (z + n − 1) (1 − τ )n τ z? ?1 dτ... factor = 1 2z 3z · · · (n + 1) z lim z n→∞ (1 + z) (1 + z/2) · · · (1 + z/n) 1z 2z · · · nz ∞ 1 (n + 1) z = z n =1 + z/n nz Thus we have Gauss’ formula for the Gamma function ∞ Γ(z) = z n =1 1+ n z 1+ z