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We set α 2 = U − 1 U . β is then β = α 1 − α 2 = (U − 1)/U 1 − (U − 1)/U) = (U − 1)U U − (U − 1) = (U − 1)U. The solution for η becomes αβ 2 sech 2 αx − βt 2 where β = α 1 − α 2 . 2. u tt − u xx − 3 2 u 2 xx − u xxxx = 0 We make the substitution u(x, t) = z(X), X = x −Ut. 2214 (U 2 − 1)z − 3 2 z 2 − z = 0 (U 2 − 1)z − 3 2 z 2 − z = 0 (U 2 − 1)z − 3 2 z 2 − z = 0 We multiply by z and integrate. 1 2 (U 2 − 1)z 2 − 1 2 z 3 − 1 2 (z ) 2 = 0 (z ) 2 = (U 2 − 1)z 2 − z 3 z = (U 2 − 1) sech 2 1 2 √ U 2 − 1X u(x, t) = (U 2 − 1) sech 2 1 2 √ U 2 − 1x − U √ U 2 − 1t The linearized equation is u tt − u xx − u xxxx = 0. Substituting u = e −αx+βt into this equation yields β 2 − α 2 − α 4 = 0 β 2 = α 2 (α 2 + 1). We set α = √ U 2 − 1. 2215 β is then β 2 = α 2 (α 2 + 1) = (U 2 − 1)U 2 β = U √ U 2 − 1. The solution for u becomes u(x, t) = α 2 sech 2 αx − βt 2 where β 2 = α 2 (α 2 + 1). 3. φ tt − φ xx + 2φ x φ xt + φ xx φ t − φ xxxx We make the substitution φ(x, t) = z(X), X = x −Ut. (U 2 − 1)z − 2Uz z − Uz z − z = 0 (U 2 − 1)z − 3Uz z − z = 0 (U 2 − 1)z − 3 2 (z ) 2 − z = 0 Multiply by z and integrate. 1 2 (U 2 − 1)(z ) 2 − 1 2 (z ) 3 − 1 2 (z ) 2 = 0 (z ) 2 = (U 2 − 1)(z ) 2 − (z ) 3 z = (U 2 − 1) sech 2 1 2 √ U 2 − 1X φ x (x, t) = (U 2 − 1) sech 2 1 2 √ U 2 − 1x − U √ U 2 − 1t . 2216 The linearized equation is φ tt − φ xx − φ xxxx Substituting φ = e −αx+βt into this equation yields β 2 = α 2 (α 2 + 1). The solution for φ x becomes φ x = α 2 sech 2 αx − βt 2 where β 2 = α 2 (α 2 + 1). 4. u t + 30u 2 u 1 + 20u 1 u 2 + 10uu 3 + u 5 = 0 We make the substitution u(x, t) = z(X), X = x −Ut. −Uz + 30z 2 z + 20z z + 10zz + z (5) = 0 Note that (zz ) = z z + zz . −Uz + 30z 2 z + 10z z + 10(zz ) + z (5) = 0 −Uz + 10z 3 + 5(z ) 2 + 10zz + z (4) = 0 Multiply by z and integrate. − 1 2 Uz 2 + 5 2 z 4 + 5z(z ) 2 − 1 2 (z ) 2 + z z = 0 2217 Assume that (z ) 2 = P (z). Differentiating this relation, 2z z = P (z)z z = 1 2 P (z) z = 1 2 P (z)z z z = 1 2 P (z)P (z). Substituting this expressions into the differential equation for z, − 1 2 Uz 2 + 5 2 z 4 + 5zP (z) − 1 2 1 4 (P (z)) 2 + 1 2 P (z)P (z) = 0 4Uz 2 + 20z 4 + 40zP (z) −(P (z)) 2 + 4P (z)P (z) = 0 Substituting P (z) = az 3 + bz 2 yields (20 + 40a + 15a 2 )z 4 + (40b + 20ab)z 3 + (4b 2 + 4U)z 2 = 0 This equation is satisfied by b 2 = U, a = −2. Thus we have (z ) 2 = √ Uz 2 − 2z 3 z = √ U 2 sech 2 1 2 U 1/4 X u(x, t) = √ U 2 sech 2 1 2 (U 1/4 x − U 5/4 t) 2218 The linearized equation is u t + u 5 = 0. Substituting u = e −αx+βt into this equation yields β − α 5 = 0. We set α = U 1/4 . The solution for u(x, t) becomes α 2 2 sech 2 αx − βt 2 where β = α 5 . 2219 Part VIII Appendices 2220 Appendix A Greek Letters The following table shows the greek letters, (some of them have two typeset variants), and thei r corresponding Roman letters. Name Roman Lower Upper alpha a α beta b β chi c χ delta d δ ∆ epsilon e epsilon (variant) e ε phi f φ Φ phi (variant) f ϕ gamma g γ Γ eta h η iota i ι kappa k κ lambda l λ Λ mu m µ 2221 nu n ν omicron o o pi p π Π pi (variant) p theta q θ Θ theta (variant) q ϑ rho r ρ rho (variant) r sigma s σ Σ sigma (variant) s ς tau t τ upsilon u υ Υ omega w ω Ω xi x ξ Ξ psi y ψ Ψ zeta z ζ 2222 [...]... Dirac delta function Fourier transform Fourier cosine transform Fourier sine transform ∞ Euler’s constant, γ = 0 e−x Log x dx Gamma function Heaviside function Hankel function of the first kind and order ν Hankel function of the second kind and order ν √ ı ≡ −1 Bessel function of the first kind and order ν Modified Bessel function of the first kind and order ν Laplace transform 2223 N Nν (x) R R+ R− o(z)... n 16 1 6 = n6 945 31π 6 (−1)n+1 = n6 30240 2240 Appendix H Table of Taylor Series ∞ (1 − z)−1 = zn |z| < 1 (n + 1)z n |z| < 1 n=0 ∞ −2 (1 − z) = n=0 ∞ α (1 + z) = n=0 ∞ z e = n=0 α n z n |z| < 1 zn n! |z| < ∞ ∞ log(1 − z) = − n=1 zn n |z| < 1 2241 log ∞ 1+z 1−z =2 n=1 ∞ |z| < ∞ (−1)n z 2n+1 (2n + 1)! |z| < ∞ n=0 ∞ sin z = n=0 tan z = z + z 3 2z 5 17z 7 + + + ··· 3 15 315 π z3 1 · 3z 5 1 · 3 · 5z 7. .. half plane 22 37 Appendix G Table of Sums ∞ rn = r , 1−r rn = r − rN +1 1−r n=1 N n=1 b for |r| < 1 n= (a + b)(b + 1 − a) 2 n= N (N + 1) 2 n=a N n=1 b n2 = n=a b(b + 1)(2b + 1) − a(a − 1)(2a − 1) 6 2238 N n2 = n=1 ∞ n=1 ∞ n=1 ∞ n=1 ∞ n=1 ∞ n=1 ∞ n=1 ∞ n=1 N (N + 1)(2N + 1) 6 (−1)n+1 = log(2) n 1 π2 = n2 6 π2 (−1)n+1 = n2 12 1 = ζ(3) n3 (−1)n+1 3ζ(3) = 3 n 4 1 π4 = n4 90 (−1)n+1 7 4 = n4 72 0 2239 ∞ n=1... 7 − z+ + + + ··· 2 2·3 2·4·5 2·4 6 7 sin−1 z = z + ∞ tan −1 |z| < 1 (−1)n z 2n (2n)! cos z = cos−1 z = z 2n−1 2n − 1 z= n=1 ∞ cosh z = n=0 ∞ sinh z = n=0 |z| < π 2 |z| < 1 z3 1 · 3z 5 1 · 3 · 5z 7 + + + ··· 2·3 2·4·5 2·4 6 7 |z| < 1 (−1)n+1 z 2n−1 2n − 1 |z| < 1 z 2n (2n)! |z| < ∞ z 2n+1 (2n + 1)! |z| < ∞ 2242 tanh z = z − ∞ Jν (z) = n=0 ∞ Iν (z) = n=0 z 3 2z 5 17z 7 + − + ··· 3 15 315 (−1)n z n!Γ(ν... x0 + ıy0 and the partial derivatives of u and v are continuous and satisfy the Cauchy-Riemann equations ux = v y , uy = −vx , then f (z0 ) exists Residues If f (z) has the Laurent expansion ∞ an z n , f (z) = n=−∞ then the residue of f (z) at z = z0 is Res(f (z), z0 ) = a−1 2225 Residue Theorem Let C be a positively oriented, simple, closed contour If f (z) is analytic in and on C except for isolated... lim R→∞ then R max |f (z)| z∈CR n ∞ f (x) dx = −ı2π −∞ =0 Res (f (z), bj ) j=1 22 36 Integrals from 0 to ∞ Let f (z) be analytic except for isolated singularities, none of which lie on the positive real axis, [0, ∞) Let z1 , , zn be the singularities of f (z) If f (z) z α as z → 0 for some α > −1 and f (z) z β as z → ∞ for some β < −1 then n ∞ f (x) dx = − 0 ∞ 0 1 f (x) log dx = − 2 n n 2 Res f (z)... function that is analytic and single valued in the disk |z − z0 | < R ∞ f (z) = n=0 f (n) (z0 ) (z − z0 )n n! The series converges for |z − z0 | < R 22 26 Laurent Series Let f (z) be a function that is analytic and single valued in the annulus r < |z − z0 | < R In this annulus f (z) has the convergent series, ∞ cn (z − z0 )n , f (z) = n=−∞ where cn = 1 ı2π f (z) dz (z − z0 )n+1 and the path of integration... ··· 3 15 315 (−1)n z n!Γ(ν + n + 1) 2 π 2 ν+2n 1 z n!Γ(ν + n + 1) 2 |z| < ν+2n |z| < ∞ |z| < ∞ 2243 Appendix I Continuous Transforms I.1 Properties of Laplace Transforms Let f (t) be piecewise continuous and of exponential order α Unless otherwise noted, the transform is defined for s > 0 To reduce clutter, it is understood that the Heaviside function H(t) multiplies the original function in the following... Integrals u dv dx = uv − dx v du dx dx f (x) dx = log f (x) f (x) f (x) 2 f (x) xα dx = dx = xα+1 α+1 f (x) for α == −1 1 dx = log x x eax dx = eax a 2232 abx a dx = b log a bx for a > 0 log x dx = x log x − x x2 1 1 x dx = arctan 2 +a a a 1 dx = 2 − a2 x √ √ x a2 1 2a 1 2a log a−x a+x log x−a x+a for x2 < a2 for x2 > a2 1 x x dx = arcsin = − arccos 2 |a| |a| −x √ 1 dx = log(x + x2 ± a2 ) x 2 ± a2 √ 1 x2 −... −1 then ∞ Res (f (z) log z, zk ) k=1 z α as z → 0 for some α > −1 and z a f (z) ∞ ı2π x f (x) log x dx = 1 − eı2πa a 0 z β as z → ∞ for some n ı2π x f (x) dx = 1 − eı2πa a 0 Res (f (z) log z, zk ) k=1 Res (z a f (z), zk ) k=1 n π2a Res (z f (z) log z, zk ) , + 2 sin (πa) k=1 n a Res (z a f (z), zk ) k=1 Fourier Integrals Let f (z) be analytic except for isolated singularities, none of which lie on . kind and order ν H (2) ν (x) Hankel function of the second kind and order ν ı ı ≡ √ −1 J ν (x) Bessel function of the first kind and order ν K ν (x) Modified Bessel function of the first kind and. u(x, y) + ıv(x, y) is defined in some neighborhood of z 0 = x 0 + ıy 0 and the partial d erivatives of u and v are continuous and satisfy the Cauchy-Riemann equations u x = v y , u y = −v x , then. of complex numbers δ(x) Dirac delta function F[·] Fourier transform F c [·] Fourier cosine transform F s [·] Fourier sine transform γ Euler’s constant, γ = ∞ 0 e −x Log x dx Γ(ν) Gamma function H(x)