Matematicas avanzadas para ingenieria dennis zill 4ta edicion (solucionario)

www.elsolucionario.net Table of Contents Part I Ordinary Differential Equations First-Order Differential Equations 22 Higher-Order Differential Equations 99 The Laplace Transform 198 Series Solutions of Linear Differential Equations 252 Numerical Solutions of Ordinary Differential Equations 317 Part II Vectors, Matrices, and Vector Calculus Vectors 339 Matrices 373 Vector Calculus 438 Part III Systems of Differential Equations 10 Systems of Linear Differential Equations 551 11 Systems of Nonlinear Differential Equations 604 Part IV Fourier Series and Partial Differential Equations 12 Orthogonal Functions and Fourier Series 634 13 Boundary-Value Problems in Rectangular Coordinates 680 14 Boundary-Value Problems in Other Coordinate Systems 755 15 Integral Transform Method 793 16 Numerical Solutions of Partial Differential Equations 832 www.elsolucionario.net Introduction to Differential Equations www.elsolucionario.net Part V Complex Analysis 17 Functions of a Complex Variable 854 18 Integration in the Complex Plane 877 19 Series and Residues 896 20 Conformal Mappings 919 Appendices 942 Projects 3.7 Road Mirages 944 3.10 The Ballistic Pendulum 946 8.1 Two-Ports in Electrical Circuits 947 8.2 Traffic Flow 948 8.15 Temperature Dependence of Resistivity 949 9.16 Minimal Surfaces 950 14.3 The Hydrogen Atom 952 15.4 The Uncertainity Inequality in Signal Processing 955 15.4 Fraunhofer Diffraction by a Circular Aperture 958 16.2 Instabilities of Numerical Methods 960 www.elsolucionario.net Appendix II Gamma function www.elsolucionario.net Part I Ordinary Differential Equations Introduction to Differential Equations EXERCISES 1.1 Second order; linear Third order; nonlinear because of (dy/dx)4 Fourth order; linear Second order; nonlinear because of cos(r + u) Second order; nonlinear because of (dy/dx)2 or + (dy/dx)2 Second order; nonlinear because of R2 Third order; linear Second order; nonlinear because of x˙ Writing the differential equation in the form x(dy/dx) + y = 1, we see that it is nonlinear in y because of y However, writing it in the form (y − 1)(dx/dy) + x = 0, we see that it is linear in x 10 Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear in v However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear in u 11 From y = e−x/2 we obtain y = − 12 e−x/2 Then 2y + y = −e−x/2 + e−x/2 = 12 From y = − 65 e−20t we obtain dy/dt = 24e−20t , so that dy + 20y = 24e−20t + 20 dt 6 −20t − e 5 13 From y = e3x cos 2x we obtain y = 3e3x cos 2x − 2e3x sin 2x and y y − 6y + 13y = = 24 = 5e3x cos 2x − 12e3x sin 2x, so that 14 From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and y = tan x + cos x ln(sec x + tan x) Then y + y = tan x 15 The domain of the function, found by solving x + ≥ 0, is [−2, ∞) From y = + 2(x + 2)−1/2 we have (y − x)y = (y − x)[1 + (2(x + 2)−1/2 ] = y − x + 2(y − x)(x + 2)−1/2 = y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2 = y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + www.elsolucionario.net Definitions and Terminology www.elsolucionario.net 1.1 Definitions and Terminology An interval of definition for the solution of the differential equation is (−2, ∞) because y is not defined at x = −2 16 Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y {x 5x = π/2 + nπ} or {x x = π/10 + nπ/5} From y = 25 sec2 5x we have = tan 5x is y = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y An interval of definition for the solution of the differential equation is (−π/10, π/10) Another interval is (π/10, 3π/10), and so on 17 The domain of the function is {x − x2 = 0} or {x x = −2 or x = 2} From y = 2x/(4 − x2 )2 we have − x2 = 2xy An interval of definition for the solution of the differential equation is (−2, 2) Other intervals are (−∞, −2) and (2, ∞) √ 18 The function is y = 1/ − sin x , whose domain is obtained from − sin x = or sin x = Thus, the domain is {x x = π/2 + 2nπ} From y = − 12 (1 − sin x)−3/2 (− cos x) we have 2y = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y cos x An interval of definition for the solution of the differential equation is (π/2, 5π/2) Another one is (5π/2, 9π/2), and so on 19 Writing ln(2X −1)−ln(X −1) = t and differentiating implicitly we obtain X dX dX − =1 2X − dt X − dt − 2X − X − dX =1 dt 2X − − 2X + dX =1 (2X − 1)(X − 1) dt -4 dX = −(2X − 1)(X − 1) = (X − 1)(1 − 2X) dt Exponentiating both sides of the implicit solution we obtain -2 t -2 -4 2X − = et X −1 2X − = Xet − et (et − 1) = (et − 2)X X= et − et − Solving et − = we get t = ln Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞) The graph of the solution defined on (−∞, ln 2) is dashed, and the graph of the solution defined on (ln 2, ∞) is solid www.elsolucionario.net y = 2x www.elsolucionario.net 1.1 Definitions and Terminology 20 Implicitly differentiating the solution, we obtain y dy dy −2x2 − 4xy + 2y =0 dx dx −x2 dy − 2xy dx + y dy = 2xy dx + (x2 − y)dy = Using the quadratic formula to solve y − 2x2 y − = for y, we get √ √ y = 2x2 ± 4x4 + /2 = x2 ± x4 + Thus, two explicit solutions √ √ are y1 = x2 + x4 + and y2 = x2 − x4 + Both solutions are defined -4 -2 x -2 on (−∞, ∞) The graph of y1 (x) is solid and the graph of y2 is dashed -4 dP (1 + c1 et ) c1 et − c1 et · c1 et c1 et [(1 + c1 et ) − c1 et ] = = dt + c1 et + c1 et (1 + c1 et ) = x 22 Differentiating y = e−x c1 et + c1 et 1− c1 et = P (1 − P ) + c1 et et dt + c1 e−x we obtain 2 x y = e−x ex − 2xe−x 2 x et dt − 2c1 xe−x = − 2xe−x 2 et dt − 2c1 xe−x 2 Substituting into the differential equation, we have x y + 2xy = − 2xe−x 2 23 From y = c1 e2x + c2 xe2x we obtain x et dt − 2c1 xe−x + 2xe−x et dt + 2c1 xe−x = 2 dy d2 y = (2c1 + c2 )e2x + 2c2 xe2x and = (4c1 + 4c2 )e2x + 4c2 xe2x , so that dx dx2 d2 y dy −4 + 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = dx dx 24 From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain dy = −c1 x−2 + c2 + c3 + c3 ln x + 8x, dx d2 y = 2c1 x−3 + c3 x−1 + 8, dx2 and d3 y = −6c1 x−4 − c3 x−2 , dx3 so that x3 d3 y d2 y dy + y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x + 2x2 − x dx dx dx + (−c3 + c3 )x ln x + (16 − + 4)x2 = 12x2 25 From y = −x2 , x , x for all x, m = and m = Thus y = e2x and y = e3x are solutions 28 (a) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 Then xy + 2y = implies xm(m − 1)xm−2 + 2mxm−1 = [m(m − 1) + 2m]xm−1 = (m2 + m)xm−1 Since xm−1 > for x > 0, m = and m = −1 Thus y = and y = x−1 are solutions (b) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 Then x2 y − 7xy + 15y = implies x2 m(m − 1)xm−2 − 7xmxm−1 + 15xm = [m(m − 1) − 7m + 15]xm = (m2 − 8m + 15)xm = (m − 3)(m − 5)xm = Since xm > for x > 0, m = and m = Thus y = x3 and y = x5 are solutions In Problems 29–32, we substitute y = c into the differential equations and use y = and y = 29 Solving 5c = 10 we see that y = is a constant solution 30 Solving c2 + 2c − = (c + 3)(c − 1) = we see that y = −3 and y = are constant solutions 31 Since 1/(c − 1) = has no solutions, the differential equation has no constant solutions 32 Solving 6c = 10 we see that y = 5/3 is a constant solution 33 From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain dx = −2e−2t + 18e6t dt and dy = 2e−2t + 30e6t dt Then x + 3y = (e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = −2e−2t + 18e6t = dx dt 5x + 3y = 5(e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = 2e−2t + 30e6t = dy dt and 34 From x = cos 2t + sin 2t + 15 et and y = − cos 2t − sin 2t − 15 et we obtain and dx = −2 sin 2t + cos 2t + et dt and dy = sin 2t − cos 2t − et dt d2 x = −4 cos 2t − sin 2t + et dt and d2 y = cos 2t + sin 2t − et dt Then and 1 d2 x 4y + et = 4(− cos 2t − sin 2t − et ) + et = −4 cos 2t − sin 2t + et = 5 dt www.elsolucionario.net = m(m + 1)xm−1 = www.elsolucionario.net 1.1 Definitions and Terminology 1 d2 y 4x − et = 4(cos 2t + sin 2t + et ) − et = cos 2t + sin 2t − et = 5 dt 35 (y )2 + = has no real solutions because (y )2 + is positive for all functions y = φ(x) 36 The only solution of (y )2 + y = is y = 0, since if y = 0, y > and (y )2 + y ≥ y > 37 The first derivative of f (x) = ex is ex The first derivative of f (x) = ekx is kekx The differential equations are y = y and y = ky, respectively 38 Any function of the form y = cex or y = ce−x is its own second derivative The corresponding differential equation is y − y = Functions of the form y = c sin x or y = c cos x have second derivatives that are the negatives of themselves The differential equation is y + y = √ 39 We first note that − y = − sin2 x = cos2 x = | cos x| This prompts us to consider values of x for which cos x < 0, such as x = π In this case = x=π d (sin x) dx = cos x x=π = cos π = −1, x=π but − y |x=π = − sin2 π = √ = Thus, y = sin x will only be a solution of y = − y when cos x > An interval of definition is then (−π/2, π/2) Other intervals are (3π/2, 5π/2), (7π/2, 9π/2), and so on 40 Since the first and second derivatives of sin t and cos t involve sin t and cos t, it is plausible that a linear combination of these functions, A sin t + B cos t, could be a solution of the differential equation Using y = A cos t − B sin t and y = −A sin t − B cos t and substituting into the differential equation we get y + 2y + 4y = −A sin t − B cos t + 2A cos t − 2B sin t + 4A sin t + 4B cos t = (3A − 2B) sin t + (2A + 3B) cos t = sin t Thus 3A − 2B = and 2A + 3B = Solving these simultaneous equations we find A = particular solution is y = 15 13 sin t − 10 13 15 13 and B = − 10 13 A cos t 41 One solution is given by the upper portion of the graph with domain approximately (0, 2.6) The other solution is given by the lower portion of the graph, also with domain approximately (0, 2.6) 42 One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second quadrant together with the lower part of the graph in the first quadrant A second solution, with domain approximately (0, 1.6) is the upper part of the graph in the first quadrant The third solution, with domain (0, ∞), is the part of the graph in the fourth quadrant 43 Differentiating (x3 + y )/xy = 3c we obtain xy(3x2 + 3y y ) − (x3 + y )(xy + y) =0 x2 y 3x3 y + 3xy y − x4 y − x3 y − xy y − y = (3xy − x4 − xy )y = −3x3 y + x3 y + y y = y − 2x3 y y(y − 2x3 ) = 2xy − x4 x(2y − x3 ) 44 A tangent line will be vertical where y is undefined, or in this case, where x(2y − x3 ) = This gives x = and 2y = x3 Substituting y = x3 /2 into x3 + y = 3xy we get www.elsolucionario.net dy dx www.elsolucionario.net 1.1 Definitions and Terminology 1 x3 + x3 = 3x x 21/3 3 x = 1/3 x2 2 x3 = 22/3 x2 x2 (x − 22/3 ) = Thus, there are vertical tangent lines at x = and x = 22/3 , or at (0, 0) and (22/3 , 21/3 ) Since 22/3 ≈ 1.59, the estimates of the domains in Problem 42 were close √ √ 45 The derivatives of the functions are φ1 (x) = −x/ 25 − x2 and φ2 (x) = x/ 25 − x2 , neither of which is defined at x = ±5 46 To determine if a solution curve passes through (0, 3) we let t = and P = in the equation P = c1 et /(1+c1 et ) P = (−3/2)et −3et = − (3/2)et − 3et passes through the point (0, 3) Similarly, letting t = and P = in the equation for the one-parameter family of solutions gives = c1 /(1 + c1 ) or c1 = + c1 Since this equation has no solution, no solution curve passes through (0, 1) 47 For the first-order differential equation integrate f (x) For the second-order differential equation integrate twice In the latter case we get y = ( f (x)dx)dx + c1 x + c2 48 Solving for y using the quadratic formula we obtain the two differential equations y = + + 3x6 x and y = 2−2 x + 3x6 , so the differential equation cannot be put in the form dy/dx = f (x, y) 49 The differential equation yy − xy = has normal form dy/dx = x These are not equivalent because y = is a solution of the first differential equation but not a solution of the second 50 Differentiating we get y = c1 + 3c2 x2 and y = 6c2 x Then c2 = y /6x and c1 = y − xy /2, so y= y − xy y 6x x+ x3 = xy − x2 y and the differential equation is x2 y − 3xy + 3y = 51 (a) Since e−x is positive for all values of x, dy/dx > for all x, and a solution, y(x), of the differential equation must be increasing on any interval 2 dy dy (b) lim = lim e−x = and lim = lim e−x = Since dy/dx approaches as x approaches −∞ x→−∞ dx x→−∞ x→∞ dx x→∞ and ∞, the solution curve has horizontal asymptotes to the left and to the right (c) To test concavity we consider the second derivative d2 y d = dx2 dx dy dx = 2 d e−x = −2xe−x dx Since the second derivative is positive for x < and negative for x > 0, the solution curve is concave up on (−∞, 0) and concave down on (0, ∞) www.elsolucionario.net This gives = c1 /(1 + c1 ) or c1 = − 32 Thus, the solution curve www.elsolucionario.net 1.1 (d) Definitions and Terminology y x 52 (a) The derivative of a constant solution y = c is 0, so solving − c = we see that c = and so y = is a constant solution (b) A solution is increasing where dy/dx = 5−y > or y < A solution is decreasing where dy/dx = 5−y < or y > (b) A solution is increasing where dy/dx = y(a − by) = by(a/b − y) > or < y < a/b A solution is decreasing where dy/dx = by(a/b − y) < or y < or y > a/b (c) Using implicit differentiation we compute d2 y = y(−by ) + y (a − by) = y (a − 2by) dx2 Solving d2 y/dx2 = we obtain y = a/2b Since d2 y/dx2 > for < y < a/2b and d2 y/dx2 < for a/2b < y < a/b, the graph of y = φ(x) has a point of inflection at y = a/2b (d) y y=aêb y=0 x 54 (a) If y = c is a constant solution then y = 0, but c2 + is never for any real value of c (b) Since y = y + > for all x where a solution y = φ(x) is defined, any solution must be increasing on any interval on which it is defined Thus it cannot have any relative extrema (c) Using implicit differentiation we compute d2 y/dx2 = 2yy = 2y(y + 4) Setting d2 y/dx2 = we see that y = corresponds to the only possible point of inflection Since d2 y/dx2 < for y < and d2 y/dx2 > for y > 0, there is a point of inflection where y = www.elsolucionario.net 53 (a) The derivative of a constant solution is 0, so solving y(a − by) = we see that y = and y = a/b are constant solutions www.elsolucionario.net 20.3 Linear Fractional Transformations (w − w1 )(w2 − w3 ) maps w1 , w2 , w3 to 0, 1, ∞ and so S maps 0, 1, ∞ to w1 , w2 , w3 Therefore, (w − w3 )(w2 − w1 ) (w − 0)(i − 2) 2z z= and so w = maps 0, 1, ∞ to 0, i, (w − 2)(i − 0) z − − 2i 11 S(w) = 12 As in Exercise 11, z = − i (w − − i)(−1 + i) 2z − and, solving for w, w = maps 0, 1, ∞ to + i, 0, (w − + i)(−1 − i) (1 + i)z − + i 13 Using the cross-ratio formula (7), (w − i)(1) (z + 1)(−1) = w(1) (z − 1)(1) i z−1 maps −1, 0, to i, ∞, z 14 Using the cross-ratio formula (7), (1)(−i − 1) (z + 1)(−1) = (w − 1)(1) (z − 1)(1) (2 + i)z − i maps −1, 0, to ∞, −i, z+1 15 Using the cross-ratio formula (7), and so w = S(w) = (w + 1)(−3) (z − 1)(2i) = = T (z) (w − 3)(1) (z + i)(i − 1) We can solve for w to obtain w=3 (1 + i)z + (1 − i) (−3 + 5i)z − − 5i Alternatively we can apply the matrix method to compute w = S −1 (T (z)) 16 Using the cross-ratio formula (7), S(w) = (w − i)(−i − 1) (z − 1)(2i) = = T (z) (w − 1)(−2i) (z + i)(i − 1) We can solve for w to obtain w= (1 + 2i)z − i iz + − 2i Alternatively we can apply the matrix method to compute w = S −1 (T (z)) w+2 17 From Example 2, z = maps the annular region < |w| < onto the w−1 region R and the circle |w| = corresponds to the line x = −1/2 Solving for z+2 w, w = maps R onto the annulus and the transferred boundary conditions z−1 are shown in the figure to the right The solution to this new Dirichlet problem is U = loge r/ loge and so u = U (f (z)) = z+2 loge loge z−1 is the solution to the Dirichlet problem in Figure 20.37 The level curves are the images of the level curves of U , |w| = r for < r < under the mapping z = w+2 Since these circles not pass through the pole at w+1 w = 1, the images are circles 927 www.elsolucionario.net and so w = www.elsolucionario.net 20.3 Linear Fractional Transformations z+1 maps −1, 1, to 0, 1, ∞ and maps each of the z two circles in R to lines since both circles pass through the pole at z = Since 18 The mapping T (z) = T ( 12 + 12 i) = − i and T (1) = 1, the circle |z − 12 | = 12 is mapped onto the line u = Likewise, the circle |z + 12 | = 12 is mapped onto the line u = The transferred boundary conditions are shown in the figure and U (u, v) = u is the solution The solution to the Dirichlet problem in Figure 20.38 is u = U (T (z)) = Re z+1 z = 1 x + 2 x2 + y The level curves u = c are the circles with centers on the x-axis which pass through the origin The level curve u = 12 , however, is the vertical line x = (w − 0)(2) (z − 1)(2i) = (w + 1)(1) (z + i)(i − 1) Solving for w, w = i 1−z = T (z) Since T (0) = i, the image of the disk |z| ≤ is the upper half-plane v ≥ 1+z 20 The linear fractional transformation that sends 1, i, −1 to ∞, i, is T (z) = maps 1, i, −1 to ∞, −1, 1+z and so f (z) = 1−z 1+z 1−z The upper semi-circle is mapped by T to the positive imaginary axis, and the real i = + i, the image of R under T is the first 5 quadrant w = z12 doubles the size of the opening so that the image under f is the upper half-plane v ≥ See the figures below interval [−1, 1] is mapped to the positive real axis Since T a2 T1 (z) + b2 21 T2 (T1 (z)) = = c2 T1 (z) + d2 a1 z + b1 + b2 a1 a2 z + a2 b1 + b2 c1 z + b2 d1 (a1 a2 + c1 b2 )z + (b1 a2 + d1 b2 ) c1 z + d1 = = a1 z + b1 a1 c2 z + b1 c2 + c1 d2 z + d1 d2 (a1 c2 + c1 d2 )z + (b1 c2 + d1 d2 ) c2 + d2 c1 z + d1 a2 22 |w − w1 | = λ|w − w2 | =⇒ (u − u1 )2 + (v − v1 )2 = λ2 [(u − u2 )2 + (v − v2 )2 ] The latter equation may be put in the form Au2 + Bv + Cu + Dv + F = where A = B = − λ2 If λ = 1, the line Cu + Dv + F = results If λ > and λ = 1, then the equation defines a circle 928 www.elsolucionario.net 19 The linear fractional transformation that sends 1, i, −i to 0, 1, −1 satisfies the cross-ratio equation www.elsolucionario.net 20.4 Schwarz-Christoffel Transformations EXERCISES 20.4 Schwarz-Christoffel Transformations arg f (t) = − 12 Arg(t − 1) = −π/2, t < Since f (1) = 0, the image is the first quadrant 0, t>1 arg f (t) = − Arg(t + 1) = −π/3, 0, t < −1 In (2), α1 = 2π/3 and since f (−1) = 0, the image is the wedge t > −1  t < −1  0, 1 arg f (t) = − Arg(t + 1) + Arg(t − 1) = π/2, −1 < t <  2 0, t>1 and α1 = π/2 and α2 = 3π/2 Since f (−1) = 0, the image of the upper half-plane is the region shown in the figure   −5π/4, arg f (t) = − Arg(t + 1) − Arg(t − 1) = −3π/4,  0, and α1 = π/2 and α2 = π/4 Since f (−1) = 0, the t < −1 −1 < t < t>1 image of the upper half-plane is the region shown in the figure Since α1 = α2 = α3 = π/2, αi /π − = −1/2 and so f (z) = A(z + 1)−1/2 z −1/2 (z − 1)−1/2 for some constant A Since α1 = π/3 and α2 = π/2, α1 /π − = −2/3 and α2 /π − = −1/2 and so f (z) = A(z + 1)−2/3 z −1/2 for some constant A Since α1 = α2 = 2π/3, αi /π − = −1/3 and so f (z) = A(z + 1)−1/3 z −1/3 for some constant A Since α1 = 3π/2 and α2 = π/4, α1 /π − = 1/2 and α2 /π − = −3/4 Therefore, f (z) = A(z + 1)1/2 z −3/4 for some constant A Since α1 = α2 = π/2, f (z) = A(z + 1)−1/2 (z − 1)−1/2 = A/(z − 1)1/2 Therefore, f (z) = A cosh−1 z + B But f (−1) = πi and f (1) = Since cosh−1 = 0, B = Since cosh−1 (−1) = πi, πi = A(πi) and so A = Hence f (z) = cosh−1 z 10 Since α1 = 3π/2 = α2 , f (z) = A(z + 1)1/2 (z − 1)1/2 = A(z − 1)1/2 Therefore, f (z) = A z(z − 1)1/2 − Ln(z + (z − 1)1/2 + B 2 but f (−1) = −ai and f (1) = It follows that = f (1) = B, −ai = f (−1) = A − πi +B and so B = and A = 4a/π Therefore, f (z) = 4a z(z − 1)1/2 4a z(z − 1)1/2 1 − Ln(z + (z − 1)1/2 ) + = − cosh−1 z + π 2 π 2 11 f (z) = A(z + 1)(α1 /π)−1 z (α2 /π)−1 (z − 1)(α3 /π)−1 from (3) Since f (−1) = πi, α1 → π as w1 → ∞ in the horizontal direction Likewise α2 → and α3 → π This suggests we examine f (z) = Az −1 = A/z Therefore, 929 www.elsolucionario.net ≤ Arg w ≤ 2π/3 www.elsolucionario.net 20.4 Schwarz-Christoffel Transformations f (z) = ALn z + B But f (−1) = πi and f (1) = It follows that A = and B = so that f (z) = Ln z We verified in Example 1, Section 20.1 that f (z) = Ln z maps the upper half-plane y ≥ to the horizontal strip ≤ v ≤ π 12 From (3), f (z) = Az −3/4 (z − 1)(α2 /π)−1 But α2 → π as θ → This suggests that we examine f (z) = Az −3/4 Therefore, f (z) = A1 z 1/4 + B1 But f (0) = and f (1) = so that B1 = and A1 = Hence f (z) = z 1/4 and we recognize that this power function maps the upper half-plane onto the wedge ≤ Arg w ≤ π/4 13 From (3), f (z) = A(z + 1)(α1 /π)−1 z (α2 /π)−1 (z − 1)(α3 /π)−1 But as u1 → 0, α1 → π/2, α2 → 2π, and α3 → π/2 This suggests that we examine z f (z) = A(z + 1)−1/2 z(z − 1)−1/2 = A (z − 1)1/2 Therefore, f (z) = A(z − 1)1/2 + B But f (−1) = f (1) = and f (0) = This implies that B = and = Ai 1/2 or A = a Therefore, f (z) = a(z − 1)1/2 By expressing f (z) as the composite of z1 = z , z2 = z1 − 1, z3 = z2 and w = az3 we can show that the image of the upper half-plane is R www.elsolucionario.net 14 If w(t) = u(t) + iv(t), then w (t) = u (t) + iv (t) and so dv v (t) = u (t) du tan(arg w (t)) = If arg w (t) is constant, then dv/du = m or v = mu + b for some constants m and b EXERCISES 20.5 Poisson Integral Formulas Using (3) with x0 = −1, x1 = 0, x2 = and u1 = −1 and u1 = 1, u = − Arg π z z+1 + Arg π z−1 z Using (3) with x0 = −2, x1 = 0, x2 = and u1 = and u2 = 1, u= Arg π z z+2 + Arg π z−1 z The harmonic function u1 = [π − Arg(z − 1)] = π 5, 0, x>1 , x Therefore u = u1 + u2 is the solution to the given Dirichlet problem The harmonic function u1 = Arg(z + 2) = π 1, 0, x < −2 , x > −2 and u2 = − Arg π z+1 z+2 + Arg π z+1 z−1 satisfies all boundary conditions except that u2 = for x < −2 Therefore u = u1 + u2 is the solution to the given Dirichlet problem 930 www.elsolucionario.net 20.5 Poisson Integral Formulas By Theorem 20.5, u(x, y) = y π t2 dt (x − t)2 + y If we let s = x − t, this integral can be expressed in terms of the natural log and inverse tangent Using Maple we obtain x−1 y y − x2 tan−1 y y π u= − tan−1 x y + x ln (x − 1)2 + y +1 x2 + y From Theorem 20.5, u(x, y) = y π ∞ −∞ cos t y dt = (x − t)2 + y π ∞ −∞ cos(x − s) ds s2 + y u(x, y) = y cos x π ∞ −∞ cos s y sin x ds + + y2 π s2 ∞ −∞ sin s y cos x ds = + y2 π s2 πe−y y = e−y cos x, y > The mapping f (z) = z maps R onto the upper half-plane R The corresponding boundary value problem in R is shown in the figure From (3), U= Arg(w + 1) + Arg π π w−1 w is the solution in R Therefore u = U (f (z)) = Arg(z + 1) + Arg π π z2 − z2 is the solution to the original Dirichlet problem in R Using H-4 with a = 3, f (z) = cos(πz/3) maps R onto the upper half-plane R The corresponding Dirichlet problem in R is shown in the figure From (3), U= 1 [π − Arg(w − 1)] + Arg(w + 1) π π is the solution in R Therefore [Arg(cos(πz/3) + 1) − Arg(cos(πz/3) − 1)] π is the solution to the original Dirichlet problem in R 1−z Using H-1, f (z) = i maps R onto the upper half-plane R The corresponding 1+z u = U (f (z)) = + Dirichlet problem in R is shown in the figure From (3), U = − Arg π w w+1 + Arg π w−1 w = Arg π w−1 w − Arg w w+1 The harmonic function u = U (f (z)) may be simplified to u= Arg π (1 − i)z − (1 + i) 1−z and is the solution to the original Dirichlet problem in R 931 − Arg 1−z −(1 + i)z + − i www.elsolucionario.net letting s = x − t But cos(x − s) = cos x cos x + sin x sin s It follows that www.elsolucionario.net 20.5 Poisson Integral Formulas 10 Using H-5, f (z) = 1+z 1−z maps R onto the upper half-plane R The corre- sponding Dirichlet problem in R is shown in the figure From (3), U= Arg π w w+1 + 1 [π − Arg(w − 1)] = − Arg(w − 1) + Arg π π w w w+1 The harmonic function u = U (f (z)) may be simplified to 11 From Theorem 20.6, u(x, y) = 2π π −π Arg π 4z (1 − z)2 + Arg π (1 + z)2 2(1 + z ) t2 − |z|2 dt Therefore, π |eit − z|2 u(0, 0) = 2π π −π t2 dt = π To estimate u(0.5, 0) and u(−0.5, 0) we must use a numerical integration method With the aid of Simpson’s Rule, u(0.5, 0) = 0.1516 and u(−0.5, 0) = 0.5693 12 From Theorem 20.6, u(x, y) = 2π π e−|t| −π u(0, 0) = 2π − |z|2 dt Therefore, |eit − z|2 π −π e−|t| dt = π π e−t dt = (1 − e−π ) π With the aid of Simpson’s Rule, u(0.5, 0) = 0.5128 and u(−0.5, 0) = 0.1623 13 u(0.0) = 2π π −π u(eit ) dt The latter integral is just the average value of u(eit ) for −π ≤ t ≤ π iθ 14 For u(e ) = cos 2θ, the Fourier series solution (6) reduces to u(r, θ) = r2 cos 2θ = Re(z ) or u(x, y) = x2 − y The corresponding system of level curves is shown in the figure 15 For u(eiθ ) = sin θ + cos θ, the Fourier series solution (6) reduces to u(r, θ) = r sin θ + r cos θ or u(x, y) = y + x The corresponding system of level curves is shown in the figure 932 www.elsolucionario.net u=1− www.elsolucionario.net 20.6 Applications EXERCISES 20.6 Applications g(z) = cos θ0 −i sin θ0 = e−iθ0 is analytic everywhere and so div F = and curl F = by Theorem 20.7 A complex potential is G(z) = e−iθ0 z and φ(x, y) = Re(G(z)) = x cos θ0 + y sin θ0 The equipotential lines (corresponding to θ0 = π/6) are shown in the figure i z φ(x, y) = Re www.elsolucionario.net g(z) = −y + xi = i(x + iy) = iz is analytic everywhere and so div F = and curl F = by Theorem 20.7 A complex potential is G(z) = 2i z and = −xy The equipotential lines −xy = c are shown in the figure and are hyperbolas x y − i = is analytic for z = and so div F = and 2 +y x +y z curl F = by Theorem 20.7 A complex potential is G(z) = Lnz and g(z) = x2 φ(x, y) = Re(G(z)) = loge (x2 + y ) The equipotential lines φ(x, y) = c are circles x2 + y = e2c and are shown in the figure g(z) = x2 − y − 2xyi = (x2 + y )2 z is analytic for z = and so div F = and curl F = by Theorem 20.7 A complex potential is G(z) = − φ(x, y) = Re − The equipotential lines − 2c z =− x2 and z x + y2 x = c can be writen as x2 + y x+ 2c + y2 = for c = See the figure The mapping f (z) = z maps the wedge ≤ Arg z ≤ π/4 to the upper half-plane R and U = Arg w is the π solution to the corresponding Dirichlet problem in R Therefore, Arg z = Arg z π π 4 is the potential in the wedge A complex potential is G(z) = Ln z and, since φ(r, θ) = θ, the equipotential π π φ(x, y) = U (z ) = 933 www.elsolucionario.net 20.6 Applications lines are the rays θ = π c Finally F = G (z) = 4 = π z¯ π x y , x2 + y x2 + y maps the original region R to the strip − 12 ≤ v ≤ (see z Example 2, Section 20.1) The boundary conditions transfer as shown in the figure The function f (z) = U = −2v is the solution in the horizontal strip and so φ(x, y) = −2Im z = x2 2y + y2 2y = c may be written as x2 + y is the potential in the original region R The equipotential lines c φ(x, y) = Re = 2i z c for c = and are circles If c = 0, we obtain the line y = Note that and so G(z) = 2i is a complex potential The corresponding vector field is z F = G (z) = Using H-5, w = 1−z 1+z = z−1 z+1 2i = z¯2 −4xy 2(x2 − y ) , (x2 + y )2 (x2 + y )2 maps R onto the upper half-plane R and U = Arg w is the solution π to the corresponding Dirichlet problem in R Therefore, µ= Arg π z−1 z+1 In R the equipotential lines are rays θ = θ0 The inverse transformation is z = S(T (w)) where T (w) = w1/2 and S(w) = 1+w T maps the ray θ = θ0 to the ray θ = θ0 /2 and S maps θ = θ0 /2 to an arc of a circle since 1−w S(0) = and S(∞) = −1 and S is a linear fractional transformation √ √ 3+ 7−3 z−a Using C-1 with b = and c = 4, we have a = and r0 = , so T (z) = maps R onto 2 az − the annular region R defined by r0 ≤ |w| ≤ U= loge |w| and so loge r0 φ= The solution to the corresponding Dirichlet problem in R is z−a loge loge r0 az − The level curves of U are circles |w| = r where r0 < r < 1, and the equipotential lines φ(x, y) = c are the images −w + a √ of these circles under the inverse transformation T −1 (w) = T −1 has a pole at w = = −aw + a 3+ (≈ 0.38) and r0 < < All circles |w| = r are mapped onto circles in the z-plane with the exception of a |w| = which is mapped onto a line a (a) ψ(x, y) = Im(z ) = 4xy(x2 − y ) and so ψ(x, y) = when y = x and y = 934 www.elsolucionario.net x2 + y + www.elsolucionario.net 20.6 Applications (b) V = G (z) = 4z = 4(x3 − 3xy , y − 3x2 y) (c) In polar coordinates r4 sin 4θ = c or r = (c csc 4θ)1/4 , for < θ < π/4, are the streamlines See the figure 10 (a) Since G(reiθ ) = r2/3 ei2θ/3 , ψ(r, θ) = Im(G(reiθ )) = r2/3 sin 2θ Note that ψ = on the boundary where θ = and θ = 3π/2 (b) V = G (z) = 23 z −1/3 Therefore, letting z = reiθ , V = −1/3 (cos(θ/3), sin(θ/3)) for < θ < 3π/2 r 11 (a) ψ(x, y) = Im(sin z) = cos x sinh y and ψ(x, y) = when x = ±π/2 or when y = (b) V = G (z) = cos z = (cos x cosh y, sin x sinh y) (c) cos x sinh y = c =⇒ y = sinh−1 (c sec x) and the streamlines are shown in the figure 12 (a) The image of R under w = i sin−1 z is the horizontal strip (see E-6) −π/2 ≤ v ≤ π/2 and ψ(x, y) = Im(i sin−1 z) = π/2, x≥1 −π/2, x ≤ −1 Each piece of boundary is therefore a streamline (b) V = G (z) = i −i = (1 − z )1/2 (1 − z¯2 )1/2 (c) The streamlines are the images of the lines v = b, −π/2 < b < π/2 under z = −i sin w and are therefore hyperbolas See Example 2, Section 20.2, and the figure Note that at z = 0, v = −i and the flow is downward 13 (a) If z = reiθ , G(reiθ ) = r2 e2iθ + −2iθ e and so r2 ψ(r, θ) = Im(G(reiθ )) = r2 − r2 sin 2θ Note that ψ = when r = or when either θ = or θ = π/2 Therefore ψ = or the boundary of R (b) V = G (z) = (2z − 2z −3 ) and so in polar coordinates V = 2re−iθ − 3iθ e = 2(r cos θ − r−3 cos 3θ, −r sin θ − r−3 sin 3θ) r3 (c) In rectangular coordinates, the streamlines are ψ(x, y) = 2xy − (x2 = c + y )2 935 www.elsolucionario.net (c) r2/3 sin(2θ/3) = c implies that r = [c csc(2θ/3)]2/3 for < θ < 3π/2 The streamlines are shown in the figure www.elsolucionario.net 20.6 Applications 14 (a) ψ(x, y) = Im(ez ) = ex sin y and so ψ = when y = or π Therefore ψ = on the boundary of R (b) V = G (z) = ez = (ex cos y − ex sin y) (c) The streamlines are ex sin y = c and so x = loge (c csc y) for < y < π See the figure f (t) = πi − [loge |t + 1| + loge |t − 1| + iArg(t + 1) + iArg(t − 1)]  t < −1  0, and so Im(f (t)) = π/2, −1 < t < Hence Im(G(z)) = ψ(x, y) = on the boundary of R  π, t>1 1 (b) x = − [loge |t + + ic| + loge |t − + ic|], y = π − [Arg(t + + ic) + Arg(t − + ic)] for c > 2 (c) 16 (a) For f (z) = (z − 1)1/2 , f (t) = |t2 − 1|1/2 cos and so f (t) = |t2 − 1|1/2 , i|t − 1| 1/2 (b) x = |(t + ic)2 − 1|1/2 cos |t| > , |t| < 1 Arg(t2 − 1) + i|t2 − 1|1/2 sin Arg(t2 − 1) Hence Im(G(z)) = on the boundary of R Arg((t + ic)2 − 1) , y = |(t + ic)2 − 1|1/2 sin Arg((t + ic)2 − 1) for c > (c) [(z − 1)1/2 + cosh−1 z], π 1 f (t) = (t2 − 1)1/2 + cosh−1 t = (t2 − 1)1/2 + Ln(t + (t2 − 1)1/2 ) π π 1, t < −1 and so Im(f (t)) = and Re(f (t)) = 0, for −1 < t < Hence Im(G(z)) = ψ(x, y) = on the 0, t > boundary of R 1/2 (b) x = Re (t + ic)2 − + cosh−1 (t + ic) , π 17 (a) For f (z) = y = Im π (t + ic)2 − 1/2 + cosh−1 (t + ic) for c > 936 www.elsolucionario.net 15 (a) For f (z) = πi − 12 [Ln(z + 1) + Ln(z − 1)] www.elsolucionario.net 20.6 Applications (c) (z + 1)1/2 − , (z + 1)1/2 + 18 (a) For f (z) = 2(z + 1)1/2 + Ln f (t) = 2(t + 1)1/2 + Ln (t + 1)1/2 − (t + 1)1/2 + If we write (t + 1)1/2 = |t + 1|1/2 e(i/2)Arg(t+1) , we may conclude that 0, t>0 π, −1 < t < and Re(f (t)) = for t < −1 www.elsolucionario.net Im(f (t)) = Therefore Im(G(z)) = ψ(x, y) = on the boundary of R (b) x = Re 2(t + ic + 1)1/2 + Ln (t + ic + 1)1/2 − (t + ic + 1)1/2 + y = Im 2(t + ic + 1)1/2 + Ln (t + ic + 1)1/2 − (t + ic + 1)1/2 + for c > (c) 19 In Example 5, V = (2x, −2y) and so the only stagnation point occurs at z = In Example 6, V = − 1/¯ z2 and so if V = 0, z¯2 = Therefore z = 1, −1 are the only stagnation points 20 (a) ψ(x, y) = Im(G(z)) = kArg(z − x1 ) and so if ψ(x, y) = c, Arg(z − x1 ) is constant This implies that the streamlines are rays with vertex at z = x1 k k k = = (z − x1 ) z − x1 z − x1 |z − x1 |2 The direction of the flow is determined by the sign of k, and if k < the flow is directed towards z = x1 (b) V = G (z) = 21 f (z) = z maps the first quadrant onto the upper half-plane and f (ξ0 ) = f (1) = Therefore G(z) = Ln(z − 1) is the complex potential, and so ψ(x, y) = Arg(z − 1) = tan−1 2xy x2 − y − is the streamline function where tan−1 is chosen to be between and π If ψ(x, y) = c, then x2 +Bxy −y −1 = where B = −2 cot c Each hyperbola in the family passes through (1, 0) and the boundary of the first quadrant satisfies ψ(x, y) = 937 www.elsolucionario.net 20.6 Applications 22 (a) From E-5, f (z) = ez maps the horizontal strip < y < π onto the upper half-plane and f (ξ0 ) = f (0) = Therefore G(z) = kLn(ez − 1) is a complex potential To construct a sink at ξ0 = 0, we must have k < (b) ψ(x, y) = Im(kLn(ez − 1)) = kArg(ez − 1) = k tan−1 ex sin y ex cos y − where tan−1 is chosen to be between and π If we set k = −1, then the streamlines ψ(x, y) = c, −π < c < 0, satisfy ex [B cos y − sin y] = B where B = − tan c, and so B B cos y − sin y x = loge Note that B will vary over all real values The streamlines are also the images of rays through w = under the inverse transformation z = Ln w See the figure z−1 z+1 23 ψ = Im(G(z)) = kArg(z − 1) − kArg(z + 1) = kArg ψ(x, y) = k tan−1 2y x2 + y − where tan−1 is chosen to be between and π Level curves ψ(x, y) = c can be put in the form x2 + y − 2By = or x2 + (y − B)2 = + B Each member of the family passes through (1, 0) and (−1, 0) 24 (a) V = a + ib = z¯ ax − by bx + ay , x2 + y x2 + y and since (x (t), y (t)) = V, the path of the particle satisfies dx ax − by , = dt x + y2 dy bx + ay = dt x + y2 (b) Switching to polar coordinates, dr = dt r dθ = dt r Therefore (c) x dx dy +y dt dt −y = dx dy +x dt dt r = ax2 − bxy bxy + ay + r2 r2 r2 = a r −axy + by bx2 + axy + r r2 = b r2 dr a = r and so r = ceaθ/b dθ b dr a dθ b = < if and only if a < 0, and in this case r is decreasing and the curve spirals inward = 1 If A is real and we require that Im(f (t)) = for t < −1, then = Aπi + Im(B) If Im(f (t)) = π for t > 1, then π = Im(B) All three equations are satisfied by letting B = πi and A = −1 Therefore f (z) = πi − [Ln(z + 1) + Ln(z − 1)] 18 (a) From Theorem 20.5, u(x, y) = y π ∞ −∞ sin t y dt = (x − t)2 + y π ∞ −∞ sin(x − s) ] ds (letting s = x − t) s2 + y But sin(x − s) = sin x cos s − cos x sin s We now proceed as in the solution to Problem 6, Section 20.5 to show that u(x, y) = e−y sin x (b) For u(eiθ ) = sin θ, the Fourier Series solution (6) in Section 20.5 reduces to u(r, θ) = r sin θ 940 www.elsolucionario.net (b) Since www.elsolucionario.net CHAPTER 20 REVIEW EXERCISES 19 If f (w) = w + ew + 1, G(z) = f −1 (z) maps R onto the strip ≤ v ≤ π and the transferred boundary conditions are shown in the figure to the right The solution for the strip is U = v/π and so the solution in R is φ(x, y) = U (G(z)) = 1 Im(G(z)) = ψ(x, y) π π Therefore the equipotential lines φ(x, y) = c are the streamlines ψ(x, y) = cπ of the flow in Figure 20.72 √ 20 G(reiθ ) = −r1/2 sin(θ/2) + i[r1/2 cos(θ/2) − 1] and so ψ(r, θ) = r cos(θ/2) = If ψ(r, θ) = 0, r cos2 (θ/2) = or r cos θ + r = (using cos2 (θ/2) = (1 + cos θ)/2.) In www.elsolucionario.net rectangular coordinates, x + x2 + y = or y = − 4x Therefore the boundary of R is a streamline To sketch the streamlines, note that ψ(r, θ) = c implies that r = (c + 1)2 sec2 (θ/2) See the figure to the right 941 ... “outside” the nullcline parabola are increasing When y > 12 x2 , 17 When y < 2x , y = x2 − 2y is negative and the portions of the solution curves “inside” the nullcline parabola are decreasing... www.elsolucionario.net 2.2 X X 2α α α −2/α Separable Variables α/2 t 1/α t For X0 > α, X(t) increases without bound up to t = 1/α For t > 1/α, X(t) increases but X → α as EXERCISES 2.2 Separable Variables In many... x y = sin sin−1 x + = x cos + − x2 sin = + 3 2 Setting x = and y = −1 Separable Variables www.elsolucionario.net 2.2 Separable Variables Using partial fractions, we obtain 1 − y−1 y dy = ln |x|