Complex Variables & Integral Transforms

Transcription

1 Complex Variables & Integral Transforms Notes taken by J.Pearson, from a S4 course at the U.Manchester. Lecture delivered by Dr.W.Parnell July 9, 007 Contents 1 Complex Variables General Relations & Definitions Hyperbolics, Sinusoidals & Logarithms Complex Functions Continuity & Differentiability Cauchy-Riemann Equations Laurent Series Singularities Complex Integration Integration by Parameterisation Cauchy s Theorem Cauchy s Integral Formula Cauchy s Residue Theorem Finding Residues Evaluation of Real Integrals Jordan s Lemma Integral Transforms 8.1 Laplace Transform Bromwich Contour Fourier Transforms

2 .3.1 Derivation of Fourier Transforms Summary of Fourier Transform

3 1 Complex Variables 1.1 General Relations & Definitions z = x + iy = re iθ (1) e iθ = cos θ + i sin θ () z n = (re iθ ) n = r n (cos nθ + i sin nθ) (3) z = x iy (4) z = zz (5) e iπ + 1 = 0 (6) Principle value range: range within which functions are well defined/single valued. Open: in a set of complex numbers, if one can draw an arbitrarily small circle around any point, with the circle still in the set, then the set is open. Connected: if two points in a set can be connected with a continuous curve, then the set is connected. If the set is not-empty and open and connected, then the set is a domain Hyperbolics, Sinusoidals & Logarithms cosh z = ez + e z (7) sinh z = ez e z (8) cosh iz = cos z (9) sinh iz = i sin z (10) ln z = ln(re iθ ) = ln r + iθ (11) 1.1. Complex Functions So that R{f(z)} = u(x, y) and I{f(z)} = v(x, y) f(z) = u(x, y) + iv(x, y) (1) Continuity & Differentiability A complex function f(z) is continuous if f(z) f(z 0 ) 0 (13) as z z 0 in any manner. 3

4 A complex function f(z) is differentiable if in any manner. lim z z 0 ( ) f(z) f(z0 ) 0 (14) z z 0 1. Cauchy-Riemann Equations The complex function f(z) = u(x, y) + iv(x, y) is regular (analytic) if and only if u x = v y u = v y x (15) (16) hold. The functions u(x, y) and v(x, y) are known as conjugate functions, and are harmonic: u = v = 0 (17) 1.3 Laurent Series Useful to know the following common expansions: 1 1 z z = 1 + z + z + z (18) = z + 1 z + 1 z (19) the first of which is convergent on z < 1, and the second on z > 1. Now, a n z n converges on z < R, and b n z n on z > R 1. Hence: an z n + b n z n (0) converges on R 1 < z < R. We have: a n = 1 πi b n = 1 πi So, generally: is a Laurent expansion about the point z = z 0. f(z) dz n = 0, 1,,... (1) (z z 0 ) n+1 (z z 0 ) n 1 f(z) dz n = 1,, 3,... () f(z) = A n (z z 0 ) n (3) 4

5 1.3.1 Singularities If, in the Laurent expansion, all coefficients of the 1 z terms (i.e. the b n n ) are zero, then z = z 0 is a removable singularity. If, for n > k, b n = 0, then there is a pole of order k at z = z 0. k = 1 is known as a simple pole, with k = being a double pole. If there is an infinite number of b n s, then an essential singularity at z = z 0. These have all been examples of isolated singularities A non-isolated is one in which in any small region surrounding one singularity, there is always another. For example f(z) = 1 has an essential singularity at z = 1 sin 1 nπ. z A branch point singularity has an example of the logarithm: it is only uniquely defined in the cut plane, and is singular everywhere on the branch point. f(z) has a simple pole at z = z 0 if f(z 0 ) = 0 and f (z 0 ) 0. f(z) has a double pole at z = z 0 if f(z 0 ) = f (z 0 ) = 0 and f (z 0 ) 0. g(z) has a pole of order k at z = z 0, and f(z) has a zero of order k. Where g(z) = 1 f(z). 1.4 Complex Integration Integration by Parameterisation Here, we parameterise z in terms of some function of t: f(z) dz = t=β t=α f{z(t)} dz dz (4) dt 1.4. Cauchy s Theorem If f(z) is regular inside some closed contour, then: f(z) dz = 0 (5) Cauchy s Integral Formula If f(z) is regular in some domain D, and is some closed Jordan contour in D; and z 0 inside, then: f(z 0 ) = 1 f(z) dz (6) πi z z 0 5

6 1.5 Cauchy s Residue Theorem If f(z) is regular everywhere inside a closed contour, except at a finite number of isolated singularities z k. Then: f(z) dz = πi Res{f(z); z = z k } (7) Where: z k inside Res{f(z); z = z 0 } = b 1 (8) 1 Thus, the residue of f(z) at a singlarity at z = z 0, is the coefficient of the expansion. z z 0 term in the Laurent Finding Residues We can find residues by a number of methods: Let: f(z) = φ(z) (z z 0 ) m (9) Where f(z) has a pole of order m at z = z 0. If φ(z 0 ) 0, and φ(z 0 ) is regular we have the following cases: For m = 1 simple pole: For m = double pole: For m = m pole order m: Res{f(z); z = z 0 } = φ(z 0 ) (30) Res{f(z); z = z 0 } = Res{f(z); z = z 0 } = 1 (m 1)! ( ) φ (31) z z=z 0 ( m 1 ) φ z m 1 (3) z=z 0 Another method: Let f(z) = p(z) q(z) where both p(z) & q(z) are regular at z = z 0. q(z 0 ) is a zero of order m; and if p(z 0 ) 0, then: Res{f(z); z = z 0 } = p(z 0) q (z 0 ) (33) (34) 6

7 And, if p(z 0 ) = 0, and p (z 0 ) 0 & q ( 0 ) 0: Res{f(z); z = z 0 } = p (z 0 ) q (z 0 ) (35) 1.6 Evaluation of Real Integrals Use to solve integrals of the form: Now, z = re iθ = cos θ + i sin θ. Thus: π 0 f(cos θ, sin θ) dθ (36) cos θ = 1 ( z + 1 ) z sin θ = 1 ( z 1 ) i z dθ = dz iz Thus, the integral can be changed to a complex integral about the unit circle C g(z) dz, which can be solved. (37) (38) (39) 1.7 Jordan s Lemma Let R be the semi-circular contour extending from R +R, and 1 the portion of the real line joining these two points. Thus, the closed semi-circular contour is = R + 1. Then, for the integral: f(z)e imz dz (40) That is, the integral:: f(z)e imz dz = = f(z)e imz dz + f(z)e imz dz (41) R 1 R f(z)e imz dz + f(z)e imz dz (4) R R If the following conditions hold: f(z) only has simple poles in the finite part of the upper-half-plane; lim z f(z) = 0; m > 0; Jordan s Lemma states that: lim R f(z)e imz dz R = 0 (43) (44) 7

8 So that the integral becomes: lim R f(z)e imz dz = lim = = R 1 f(z)e imz dz (45) f(z)e imz dz (46) f(x)e imx dx (47) We use this in the evaluation of integrals of the form f(x) dx, by working backwards: f(x) dx = f(z)e imz dz = 1 Res{f(z)e imz ; z = z k } (48) πi Which is therefore evalulated by Cauchy s Residue theorem. Remembering that it is the sum over residues enclosed by the countour: i.e. over all residues due to all singularities in the upper-halfplane. k Integral Transforms.1 Laplace Transform f(p) = L{f(t)} = The transform is a linear operator: 0 f(t)e pt dt (49) L{f 1 (t) + f (t)} = L{f 1 (t)} + L{f (t)} (50) Some common transforms: f(t) f(p) = L{f(t)} e αt 1 p α t n n! p n+1 dy dt d y dt e iωt py(p) y(0) p y(p) y (0) py(0) p+iω p +ω. Bromwich Contour f(t) = L 1 {f(p)} = 1 c+i f(p)e pt dp (51) πi c i 8

9 This inversion integral is directly analogous to Jordan s Lemma. c is chosen to the right of any singularities of fe pt. Only works if f has a finite number of isolated poles in the left-half-plane, lim p f = 0 and m > 0. That is: f(t) = 1 πi πi z k in LHP fe pt ; z = z k (5).3 Fourier Transforms.3.1 Derivation of Fourier Transforms Suppose we expand f(x) on some range λ x λ as a Fourier series: f(x) = a 0 + a n cos(k n x) + b n sin(k n x) (53) n=1 Where: k n πn λ a 0 = 1 λ a n = λ b n = λ λ λ λ (54) f(t) dt (55) f(t) cos(k n t) dt (56) f(t) sin(k n t) dt (57) (58) Now, using the identity cos[k n (t x)] = cos(k n t) cos(k n x) + sin(k n t) sin(k n x), we get: f(x) = 1 λ λ f(t) dt + λ n=1 λ f(t) cos[k n (t x)] dt (59) Notice: cos[k n (t x)] = 1 [ e ikn(t x) + e ikn(t x)] (60) k n πn λ k n = π( n) = k n λ (61) k n = k n (6) Thus, by changing the summation limits to n : f(x) = 1 λ n= λ f(t)e ikn(t x) dt (63) 9

10 Now, to tidy up; we define h π λ k n = nh: f(x) = 1 π g(nh) = 1 π π π n= n= π h f(t)e inh(t x) dt (64) π hg(nh) (65) f(t)e inh(t x) dt (67) Now, as h 0: lim hg(nh) = g(k) dk (68) h 0 Thus: Where: f(x) = 1 π = 1 π = 1 π F (k) g(k) dk (69) f(t)e ik(t x) dtdk (70) F (k)e ikx dk (71) f(x)e ikx dx (7).3. Summary of Fourier Transform Transform: Inverse: F (k) = f(x)e ikx dx (73) f(x) = 1 F (k)e ikx dk (74) π 10