algebra lineal y otros temas

8/8/2019 Algebra Lineal y Otros Temas

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LECTURE NOTES ON

MATHEMATICAL METHODS

Mihir SenJoseph M. Powers

Department of Aerospace and Mechanical EngineeringUniversity of Notre Dame

Notre Dame, Indiana 46556-5637USA

updatedApril 9, 2003


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Contents

1 Multi-variable calculus 111.1 Implicit functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2 Functional dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.3 Coordinate Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.3.1 Jacobians and Metric Tensors . . . . . . . . . . . . . . . . . . . . . . 19

1.3.2 Covariance and Contravariance . . . . . . . . . . . . . . . . . . . . . 251.4 Maxima and minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.4.1 Derivatives of integral expressions . . . . . . . . . . . . . . . . . . . . 311.4.2 Calculus of variations . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.5 Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2 First-order ordinary differential equations 432.1 Separation of variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.2 Homogeneous equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442.3 Exact equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

2.4 Integrating factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472.5 Bernoulli equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.6 Riccati equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 512.7 Reduction of order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2.7.1 y absent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522.7.2 x absent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2.8 Uniqueness and singular solutions . . . . . . . . . . . . . . . . . . . . . . . . 552.9 Clairaut equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

3 Linear ordinary differential equations 61

3.1 Linearity and linear independence . . . . . . . . . . . . . . . . . . . . . . . . 613.2 Complementary functions for equations with constant coefficients . . . . . . 63

3.2.1 Arbitrary order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 633.2.2 First order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.2.3 Second order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

3.3 Complementary functions for equations with variable coefficients . . . . . . . 663.3.1 One solution to find another . . . . . . . . . . . . . . . . . . . . . . . 663.3.2 Euler equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

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3.4 Particular solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

3.4.1 Method of undetermined coefficients . . . . . . . . . . . . . . . . . . 68

3.4.2 Variation of parameters . . . . . . . . . . . . . . . . . . . . . . . . . 70

3.4.3 Operator D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

3.4.4 Greens functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4 Series solution methods 81

4.1 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.1.1 First-order equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

4.1.2 Second-order equation . . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.1.2.1 Ordinary point . . . . . . . . . . . . . . . . . . . . . . . . . 84

4.1.2.2 Regular singular point . . . . . . . . . . . . . . . . . . . . . 86

4.1.2.3 Irregular singular point . . . . . . . . . . . . . . . . . . . . 89

4.1.3 Higher order equations . . . . . . . . . . . . . . . . . . . . . . . . . . 894.2 Perturbation methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

4.2.1 Algebraic and transcendental equations . . . . . . . . . . . . . . . . . 91

4.2.2 Regular perturbations . . . . . . . . . . . . . . . . . . . . . . . . . . 95

4.2.3 Strained coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

4.2.4 Multiple scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

4.2.5 Harmonic approximation . . . . . . . . . . . . . . . . . . . . . . . . . 104

4.2.6 Boundary layers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

4.2.7 WKB method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

4.2.8 Solutions of the type eS(x) . . . . . . . . . . . . . . . . . . . . . . . . 112

4.2.9 Repeated substitution . . . . . . . . . . . . . . . . . . . . . . . . . . 113Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

5 Special functions 121

5.1 Sturm-Liouville equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

5.1.1 Linear oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

5.1.2 Legendre equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

5.1.3 Chebyshev equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

5.1.4 Hermite equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

5.1.5 Laguerre equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

5.1.6 Bessel equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

5.1.6.1 first and second kind . . . . . . . . . . . . . . . . . . . . . . 132

5.1.6.2 third kind . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

5.1.6.3 modified Bessel functions . . . . . . . . . . . . . . . . . . . 135

5.1.6.4 ber and bei functions . . . . . . . . . . . . . . . . . . . . . . 135

5.2 Representation of arbitrary functions . . . . . . . . . . . . . . . . . . . . . . 135

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140

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6 Vectors and tensors 1436.1 Cartesian index notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1436.2 Cartesian tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

6.2.1 Direction cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1456.2.1.1 Scalars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

6.2.1.2 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1476.2.1.3 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

6.2.2 Matrix representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1486.2.3 Transpose of a tensor, symmetric and anti-symmetric tensors . . . . . 1486.2.4 Dual vector of a tensor . . . . . . . . . . . . . . . . . . . . . . . . . . 1496.2.5 Principal axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

6.3 Algebra of vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1516.3.1 Definition and properties . . . . . . . . . . . . . . . . . . . . . . . . . 1526.3.2 Scalar product (dot product, inner product) . . . . . . . . . . . . . . 1526.3.3 Cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

6.3.4 Scalar triple product . . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.3.5 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

6.4 Calculus of vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1536.4.1 Vector function of single scalar variable . . . . . . . . . . . . . . . . . 1536.4.2 Differential geometry of curves . . . . . . . . . . . . . . . . . . . . . . 155

6.4.2.1 Curves on a plane . . . . . . . . . . . . . . . . . . . . . . . 1566.4.2.2 Curves in 3-dimensional space . . . . . . . . . . . . . . . . . 157

6.5 Line and surface integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1606.5.1 Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1606.5.2 Surface integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

6.6 Differential operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1616.6.1 Gradient of a scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1636.6.2 Divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

6.6.2.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1646.6.2.2 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164

6.6.3 Curl of a vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1646.6.4 Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

6.6.4.1 Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1656.6.4.2 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

6.6.5 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1656.7 Special theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

6.7.1 Path independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1666.7.2 Greens theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1666.7.3 Gausss theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1676.7.4 Greens identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1696.7.5 Stokes theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1696.7.6 Leibnizs Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

6.8 Orthogonal curvilinear coordinates . . . . . . . . . . . . . . . . . . . . . . . 171Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

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7 Linear analysis 1757.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1757.2 Differentiation and integration . . . . . . . . . . . . . . . . . . . . . . . . . . 176

7.2.1 Frechet derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1767.2.2 Riemann integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176

7.2.3 Lebesgue integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1777.3 Vector spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

7.3.1 Normed spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1827.3.2 Inner product spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

7.3.2.1 Hilbert space . . . . . . . . . . . . . . . . . . . . . . . . . . 1917.3.2.2 Non-commutation of the inner product . . . . . . . . . . . . 1927.3.2.3 Minkowski space . . . . . . . . . . . . . . . . . . . . . . . . 1937.3.2.4 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . 1977.3.2.5 Gram-Schmidt procedure . . . . . . . . . . . . . . . . . . . 1987.3.2.6 Representation of a vector . . . . . . . . . . . . . . . . . . . 1997.3.2.7 Parsevals equation, convergence, and completeness . . . . . 205

7.3.3 Reciprocal bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2057.4 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

7.4.1 Linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2097.4.2 Adjoint operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2107.4.3 Inverse operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2137.4.4 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . . 216

7.5 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2257.6 Method of weighted residuals . . . . . . . . . . . . . . . . . . . . . . . . . . 229Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

8 Linear algebra 2458.1 Determinants and rank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2458.2 Matrix algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

8.2.1 Column, row, left and right null spaces . . . . . . . . . . . . . . . . . 2478.2.2 Matrix multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . 2498.2.3 Definitions and properties . . . . . . . . . . . . . . . . . . . . . . . . 250

8.2.3.1 Diagonal matrices . . . . . . . . . . . . . . . . . . . . . . . 2508.2.3.2 Inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2528.2.3.3 Similar matrices . . . . . . . . . . . . . . . . . . . . . . . . 253

8.2.4 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2538.2.4.1 Overconstrained Systems . . . . . . . . . . . . . . . . . . . 253

8.2.4.2 Underconstrained Systems . . . . . . . . . . . . . . . . . . . 2568.2.4.3 Over- and Underconstrained Systems . . . . . . . . . . . . . 2578.2.4.4 Square Systems . . . . . . . . . . . . . . . . . . . . . . . . . 259

8.2.5 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . . 2618.2.6 Complex matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

8.3 Orthogonal and unitary matrices . . . . . . . . . . . . . . . . . . . . . . . . 2668.3.1 Orthogonal matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . 2668.3.2 Unitary matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

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8.4 Matrix decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2688.4.1 L D U decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 2688.4.2 Echelon form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2708.4.3 Q R decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . 2738.4.4 Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

8.4.5 Jordan canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . 2808.4.6 Schur decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . 2828.4.7 Singular value decomposition . . . . . . . . . . . . . . . . . . . . . . 2828.4.8 Hessenberg form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

8.5 Projection matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2858.6 Method of least squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286

8.6.1 Unweighted least squares . . . . . . . . . . . . . . . . . . . . . . . . . 2868.6.2 Weighted least squares . . . . . . . . . . . . . . . . . . . . . . . . . . 287

8.7 Matrix exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2898.8 Quadratic form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2918.9 Moore-Penrose inverse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296

9 Dynamical systems 3019.1 Paradigm problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

9.1.1 Autonomous example . . . . . . . . . . . . . . . . . . . . . . . . . . . 3019.1.2 Non-autonomous example . . . . . . . . . . . . . . . . . . . . . . . . 305

9.2 General theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3079.3 Iterated maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3089.4 High order scalar differential equations . . . . . . . . . . . . . . . . . . . . . 3119.5 Linear systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

9.5.1 Homogeneous equations with constant A . . . . . . . . . . . . . . . . 3139.5.1.1 n eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . 3149.5.1.2 < n eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . 3159.5.1.3 Summary of method . . . . . . . . . . . . . . . . . . . . . . 3169.5.1.4 Alternative method . . . . . . . . . . . . . . . . . . . . . . . 3169.5.1.5 Fundamental matrix . . . . . . . . . . . . . . . . . . . . . . 319

9.5.2 Inhomogeneous equations . . . . . . . . . . . . . . . . . . . . . . . . 3209.5.2.1 Undetermined coefficients . . . . . . . . . . . . . . . . . . . 3219.5.2.2 Variation of parameters . . . . . . . . . . . . . . . . . . . . 321

9.6 Nonlinear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3229.6.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322

9.6.2 Linear stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3229.6.3 Lyapunov functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3249.6.4 Hamiltonian systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 327

9.7 Fractals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3299.7.1 Cantor set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3309.7.2 Koch curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3309.7.3 Weierstrass function . . . . . . . . . . . . . . . . . . . . . . . . . . . 3319.7.4 Mandelbrot and Julia sets . . . . . . . . . . . . . . . . . . . . . . . . 331

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9.8 Bifurcations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3329.8.1 Pitchfork bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . 3339.8.2 Transcritical bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . 3359.8.3 Saddle-node bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . 3369.8.4 Hopf bifurcation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

9.9 Lorenz equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3389.9.1 Linear stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3389.9.2 Center manifold projection . . . . . . . . . . . . . . . . . . . . . . . . 341

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344

10 Appendix 35310.1 Trigonometric relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35310.2 Routh-Hurwitz criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35410.3 Infinite series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35510.4 Asymptotic expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356

10.4.1 Expansion of integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 356

10.4.2 Integration and differentiation of series . . . . . . . . . . . . . . . . . 35610.5 Limits and continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35610.6 Special functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356

10.6.1 Gamma function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35610.6.2 Beta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35610.6.3 Riemann zeta function . . . . . . . . . . . . . . . . . . . . . . . . . . 35710.6.4 Error function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35710.6.5 Fresnel integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35810.6.6 Sine- and cosine-integral functions . . . . . . . . . . . . . . . . . . . . 35810.6.7 Elliptic integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359

10.6.8 Gausss hypergeometric function . . . . . . . . . . . . . . . . . . . . . 36010.6.9 distribution and Heaviside function . . . . . . . . . . . . . . . . . . 36010.7 Singular integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36110.8 Chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36110.9 Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362

10.9.1 Eulers formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36210.9.2 Polar and Cartesian representations . . . . . . . . . . . . . . . . . . . 36310.9.3 Cauchy-Riemann equations . . . . . . . . . . . . . . . . . . . . . . . 364

Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365

Bibliography 367

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Preface

These are lecture notes for AME 561 Mathematical Methods I, the first of a pair of courseson applied mathematics taught at the Department of Aerospace and Mechanical Engineeringof the University of Notre Dame. Most of the students in this course are beginning graduatestudents in engineering coming from a wide variety of backgrounds. The objective of thecourse is to provide a survey of a variety of topics in applied mathematics, including mul-tidimensional calculus, ordinary differential equations, perturbation methods, vectors and

tensors, linear analysis, and linear algebra, and dynamic systems. The companion course,AME 562, covers complex variables, integral transforms, and partial differential equations.These notes emphasize method and technique over rigor and completeness; the student

should call on textbooks and other reference materials. It should also be remembered thatpractice is essential to the learning process; the student would do well to apply the techniquespresented here by working as many problems as possible.

The notes, along with much information on the course itself, can be found on the worldwide web at http://www.nd.edu/powers/ame.561. At this stage, anyone is free to duplicatethe notes on their own printers.

These notes have appeared in various forms for the past few years; minor changes andadditions have been made and will continue to be made. We would be happy to hear from

you about errors or suggestions for improvement.

Mihir [email protected]://www.nd.edu/msenJoseph M. [email protected]://www.nd.edu/powers

Notre Dame, Indiana; USA

April 9, 2003

Copyright c 2003 by Mihir Sen and Joseph M. Powers.All rights reserved.

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Chapter 1

Multi-variable calculus

see Kaplan, Chapter 2: 2.1-2.22, Chapter 3: 3.9,see Riley, Hobson, Bence, Chapters 4, 19, 20,see Lopez, Chapters 32, 46, 47, 48.

1.1 Implicit functions

We can think of a relation such as f(x1, x2, . . . , xn, y) = 0, also written as f(xi, y) = 0, insome region as an implicit function ofy with respect to the other variables. We cannot havefy

= 0, because then f would not depend on y in this region. In principle, we can write

y = y(x1, x2, . . . , xn) or y = y(xi) (1.1)

if fy

= 0.The derivative

yxi can be determined from f = 0 without explicitly solving for y. First,

from the chain rule, we have

df =f

x1dx1 +

f

x2dx2 + . . . +

f

xidxi + . . . +

f

xndxn +

f

ydy = 0 (1.2)

Differentiating with respect to xi while holding all the other xj , j = i constant, we getf

xi+

f

y

y

xi= 0 (1.3)

so that

yxi

= fxify

(1.4)

which can be found if fy = 0. That is to say, y can be considered a function of xi if fy = 0.Let us now consider the equations

f(x,y,u,v) = 0 (1.5)

g(x,y,u,v) = 0 (1.6)

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Under certain circumstances, we can unravel these equations (either algebraically or numer-ically) to form u = u(x, y), v = v(x, y). The conditions for the existence of such a functionaldependency can be found by differentiation of the original equations, for example:

df =f

x

dx +f

y

dy +f

u

du +f

v

dv = 0

holding y constant and dividing by dx we get

f

x+

f

u

u

x+

f

v

v

x= 0

in the same manner, we get

g

x+

g

u

u

x+

g

v

v

x= 0

fy

+ fu

uy

+ fv

vy

= 0

g

y+

g

u

u

y+

g

v

v

y= 0

Two of the above equations can be solved for ux

and vx

, and two others for uy

and vy

by

using Cramers 1 rule. To solve for ux andvx , we first write two of the previous equations

in matrix form:

fu fvgu

gv

uxvx

= fx gx

(1.7)thus from Cramers rule we have

u

x=

fx fv gx

gv

fu fvgu

gv

(f,g)(x,v)

(f,g)(u,v)

v

x=

fu fxgu

gx

fu fvgu

gv

(f,g)(u,x)

(f,g)(u,v)

In a similar fashion, we can form expressions for uy andvy :

u

y=

fy fvgy

gv

fu fvgu

gv

(f,g)(y,v)

(f,g)(u,v)

v

y=

fu fygu

gy

fu fvgu

gv

(f,g)(u,y)

(f,g)(u,v)

1Gabriel Cramer, 1704-1752, well-travelled Swiss-born mathematician who did enunciate his well knownrule, but was not the first to do so.

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If the Jacobian 2 determinant, defined below, is non-zero, the derivatives exist, and weindeed can form u(x, y) and v(x, y).

(f, g)

(u, v)=

fu

fv

gu

gv

= 0 (1.8)

This is the condition for the implicit to explicit function conversion. Similar conditions holdfor multiple implicit functions fi(x1, . . . , xn, y1, . . . , ym) = 0, i = 1, . . . , m. The derivativesfixj

, i = 1, . . . , m, j = 1, . . . , n exist in some region if the determinant of the matrix fiyj = 0(i, j = 1, . . . , m) in this region.

Example 1.1If

x + y + u6 + u + v = 0

xy + uv = 1

Find u

x.

Note that we have four unknowns in two equations. In principle we could solve for u(x, y) andv(x, y) and then determine all partial derivatives, such as the one desired. In practice this is not alwayspossible; for example, there is no general solution to sixth order equations such as we have here.

The two equations are rewritten as

f(x,y,u,v) = x + y + u6 + u + v = 0

g(x,y,u,v) = xy + uv 1 = 0Using the formula developed above to solve for the desired derivative, we get

u

x =

f

xfv

gx

gv

fu fvg

ugv

Substituting, we get

u

x=

1 1y u 6u5 + 1 1v u

=y u

u(6u5 + 1) v

Note when

v = 6u6 + u

that the relevant Jacobian is zero; at such points we can determine neither ux noruy ; thus we cannot

form u(x, y).

At points where the relevant Jacobian (f,g)(u,v) = 0, (which includes nearly all of the (x, y) plane) givena local value of (x, y), we can use algebra to find a corresponding u and v, which may be multivalued,and use the formula developed to find the local value of the partial derivative.

2Carl Gustav Jacob Jacobi, 1804-1851, German/Prussian mathematician who used these determinants,which were first studied by Cauchy, in his work on partial differential equations.

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1.2 Functional dependence

Let u = u(x, y) and v = v(x, y). If we can write u = g(v) or v = h(u), then u and v are saidto be functionally dependent. If functional dependence between u and v exists, then we canconsider f(u, v) = 0. So,

f

u

u

x+

f

v

v

x= 0, (1.9)

f

u

u

y+

f

v

v

y= 0, (1.10)

ux

vx

uy

vy

fufv

=

00

(1.11)

Since the right hand side is zero, and we desire a non-trivial solution, the determinant of thecoefficient matrix, must be zero for functional dependency, i.e.

ux vxuy

vy

= 0. (1.12)Note, since det A = det AT, that this is equivalent to ux uyv

xvy

= (u, v)(x, y) = 0. (1.13)That is the Jacobian must be zero.

Example 1.2Determine if

u = y + z

v = x + 2z2

w = x 4yz 2y2

are functionally dependent.The determinant of the resulting coefficient matrix, by extension to three functions of three vari-

ables, is

(u, v, w)

(x,y,z)=

ux

uy

uz

vx

vy

vz

wx

wy

wz

=

ux

vx

wx

uy

vy

wy

uz

vz

wz

=

0 1 11 0 4(y + z)1 4z 4y

= (1)(4y (4)(y + z)) + (1)(4z)= 4y 4y 4z + 4z= 0

So, u,v,w are functionally dependent. In fact w = v 2u2.

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Example 1.3Let

x + y + z = 0

x2 + y2 + z2 + 2xz = 1

Can x and y be considered as functions of z?If x = x(z) and y = y(z), then dxdz and

dydz must exist. If we take

f(x, y, z) = x + y + z = 0

g(x, y, z) = x2 + y2 + z2 + 2xz 1 = 0df =

f

zdz +

f

xdx +

f

ydy = 0

dg =g

zdz +

g

xdx +

g

ydy = 0

f

z+

f

x

dx

dz+

f

y

dy

dz= 0

g

z+

g

x

dx

dz+

g

y

dy

dz= 0 f

xfy

gx

gy

dxdzdydz

=

fzgz

then the solution matrixdxdz

, dydz

Tcan be obtained by Cramers rule.

dxdz

=

fz

fy

gz gy fx fygx

gy

= 1 1(2z + 2x) 2y 1 12x + 2z 2y

= 2y + 2z + 2x

2y 2x 2z = 1

dy

dz=

fx fzgx gz

fx fygx

gy

=

1 12x + 2z (2z + 2x) 1 12x + 2z 2y

=

0

2y 2x 2z

Note here that in the expression for dxdz that the numerator and denominator cancel; there is no special

condition defined by the Jacobian determinant of the denominator being zero. In the second, dydz = 0 ify x z = 0, in which case this formula cannot give us the derivative.

Now in fact, it is easily shown by algebraic manipulations (which for more general functions arenot possible) that

x(z) = z

2

2

y(z) =

2

2

Note that in fact y x z = 0, so the Jacobian determinant (f,g)(x,y) = 0; thus, the above expressionfor dydz is indeterminant. However, we see from the explicit expression y =

2

2 that in fact,dydz = 0.

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-1

0

1

x

-0.5

0

0.5

y

-1

-0.5

0

0.5

1

z

-1

0

1

x

-0.5

0

0.5

y

-1

-0

0

0

1

-1

-0.5

0

0.5

1x

-2

-1

0

1

2

y

-1

-0.5

0

0.5

1

z

-1

-0.5

0

0.5

1x

-2

-1

0

1

2

y

-1

.5

0

5

1

Figure 1.1: Surfaces ofx + y + z = 0 and x2 + y2 + z2 + 2xz = 1, and their loci of intersection

The two original functions and their loci of intersection are plotted in Figure 1.1.It is seen that the surface represented by the quadratic function is a open cylindrical tube, and thatrepresented by the linear function is a plane. Note that planes and cylinders may or may not intersect.If they intersect, it is most likely that the intersection will be a closed arc. However, when the planeis aligned with the axis of the cylinder, the intersection will be two non-intersecting lines; such is thecase in this example.

Lets see how slightly altering the equation for the plane removes the degeneracy. Take now

5x + y + z = 0

x2 + y2 + z2 + 2xz = 1

Can x and y be considered as functions of z?If x = x(z) and y = y(z), then dxdz and

dydz must exist. If we take

f(x, y, z) = 5x + y + z = 0g(x, y, z) = x2 + y2 + z2 + 2xz 1 = 0

then the solution matrixdxdz ,

dydz

Tis found as before:

dx

dz=

fz fygz gy fx fyg

xgy

=

1 1(2z + 2x) 2y 5 12x + 2z 2y

=

2y + 2z + 2x10y 2x 2z

dydz =fx fzg

x g

z fx fygx

gy

=

5 12x + 2z

(2z + 2x) 5 12x + 2z 2y

= 8x 8z10y 2x 2zThe two original functions and their loci of intersection are plotted in Figure 1.2.Straightforward algebra in this case shows that an explicit dependency exists:

x(z) =6z 213 8z2

26

y(z) =4z 5213 8z2

26

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-0.20 0.2

-1

-0.5

0

0.5

1

-1

0

1

z

0

x

-1

-0.5

0

0.5

1

y

-1

0

1

-1

-0.5

0

0.5

1

x

-2

-1

0

1

2

y

-1

-0.5

0

0.5

1

z

-1

-0.5

0

0.5

1

x

-2

-1

0

1

2

y

-1

-0.5

0

0.5

1

z

Figure 1.2: Surfaces of 5x+ y +z = 0 and x2 +y2 +z2 +2xz = 1, and their loci of intersection

These curves represent the projection of the curve of intersection on the x z and y z planes,respectively. In both cases, the projections are ellipses.

1.3 Coordinate Transformations

Many problems are formulated in three-dimensional Cartesian 3 space. However, many ofthese problems, especially those involving curved geometrical bodies, are better posed in anon-Cartesian, curvilinear coordinate system. As such, one needs techniques to transformfrom one coordinate system to another.

For this section, we will take Cartesian coordinates to be represented by (1, 2, 3). Herethe superscript is an index and does not represent a power of . We will denote this pointby i, where i = 1, 2, 3. Since the space is Cartesian, we have the usual Euclidean 4 formulafor arc length s:

(ds)2

= d12 + d22 + d32 (1.14)(ds)2 =

3i=1

didi didi (1.15)

Here we have adopted the summation convention that when an index appears twice, asummation from 1 to 3 is understood.

3Rene Descartes, 1596-1650, French mathematician and philosopher.4Euclid of Alexandria, 325 B.C.- 265 B.C., Greek geometer.

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Now let us map a point from a point in (1, 2, 3) space to a point in a more convenient(x1, x2, x3) space. This mapping is achieved by defining the following functional dependen-cies:

x1 = x1(1, 2, 3) (1.16)

x2 = x2(1, 2, 3) (1.17)

x3 = x3(1, 2, 3) (1.18)

Taking derivatives can tell us whether the inverse exists.

dx1 =x1

1d1 +

x1

2d2 +

x1

3d3 =

x1

jdj (1.19)

dx2 = x21

d1 + x22

d2 + x23

d3 = x2j

dj (1.20)

dx3 =x3

1d1 +

x3

2d2 +

x3

3d3 =

x3

jdj (1.21)

dx1dx2dx3

=

x1

1x1

2x1

3

x2

1x2

2x2

3

x3

1x3

2x3

3

d1d2

d3

(1.22)

dxi =xi

jdj (1.23)

In order for the inverse to exist we must have a non-zero Jacobian for the transformation,i.e.

(x1, x2, x3)

(1, 2, 3)= 0 (1.24)

It can then be inferred that the inverse transformation exists.

1 = 1(x1, x2, x3) (1.25)2 = 2(x1, x2, x3) (1.26)

3 = 3(x1, x2, x3) (1.27)

Likewise then,

di = i

xjdxj (1.28)

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1.3.1 Jacobians and Metric Tensors

Defining 5 the Jacobian matrix J, which we associate with the inverse transformation, thatis the transformation from non-Cartesian to Cartesian coordinates, to be

J = ixj

=

1

x1

1

x2

1

x3

2

x12

x22

x33

x13

x23

x3

(1.29)we can rewrite di in Gibbs 6 vector notation as

d = J dx (1.30)

Now for Euclidean spaces, distance must be independent of coordinate systems, so werequire

(ds)2 = didi = i

xkdxk

i

xldxl =

i

xk

i

xldxkdxl (1.31)

In Gibbs vector notation this becomes

(ds)2 = dT d (1.32)= (J dx)T (J dx) (1.33)= dxT JT J dx (1.34)

If we define the metric tensor, gkl or G, as follows:

gkl = i

xk

i

xl(1.35)

G = JT J (1.36)

then we have, equivalently in both index and Gibbs notation,

(ds)2 = gkl dxkdxl (1.37)

(ds)2 = dxT G dx (1.38)

Now gkl can be represented as a matrix. If we define

g = det(gkl) , (1.39)

5The definition we adopt is that used in most texts, including Kaplan. A few, e.g. Aris, define theJacobian determinant in terms of the transpose of the Jacobian matrix, which is not problematic since thetwo are the same. Extending this, an argument can be made that a better definition of the Jacobian matrixwould be the transpose of the traditional Jacobian matrix. This is because when one considers that thedifferential operator acts first, the Jacobian matrix is really xj

i, and the alternative definition is more

consistent with traditional matrix notation, which would have the first row as x1 1, x1

2, x1 3. As long

as one realizes the implications of the notation, however, the convention adopted ultimately does not matter.6Josiah Willard Gibbs, 1839-1903, prolific American physicist and mathematician with a lifetime affiliation

with Yale University.

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it can be shown that the ratio of volumes of differential elements in one space to that of theother is given by

d1 d2 d3 =

g dx1 dx2 dx3 (1.40)

We also require dependent variables and all derivatives to take on the same values atcorresponding points in each space, e.g. ifS [S = f(1, 2, 3) = h(x1, x2, x3)] is a dependentvariable defined at (1, 2, 3), and (1, 2, 3) maps into (x1, x2, x3)), we require f(1, 2, 3) =h(x1, x2, x3))

The chain rule lets us transform derivatives to other spaces

( S1S2

S3 ) = (

Sx1

Sx2

Sx3 )

x1

1x1

2x1

3

x2

1x2

2x2

3

x3

1x3

2x3

3

(1.41)

S

i =S

xjxj

i (1.42)

This can also be inverted, given that g = 0, to find Sx1 , Sx2 , Sx3T. The fact that the gradientoperator required the use of row vectors in conjunction with the Jacobian matrix, while thetransformation of distance, earlier in this section, required the use of column vectors is offundamental importance, and will be examined further in an upcoming section where wedistinguish between what are known as covariant and contravariant vectors.

Example 1.4Transform the Cartesian equation

S

1 S = 12 + 22

under under the following:

1. Cartesian to affine coordinates.

Consider the following linear non-orthogonal transformation (those of this type are known as affine):

x1 = 41 + 22

x2 = 31 + 22

x3 = 3

This is a linear system of three equations in three unknowns; using standard techniques of linear algebraallows us to solve for 1, 2, 3 in terms of x1, x2, x3; that is we find the inverse transformation, whichis

1 = x1 x2

2 = 32

x1 + 2x2

3 = x3

Lines of constant x1 and x2 in the 1, 2 plane are plotted in Figure 1.3.

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-4 -2 2 4

-4

-3

-2

-1

1

2

3

4

1

2

Figure 1.3: Lines of constant x1 and x2 in the 1, 2 plane for affine transformation of exampleproblem.

The appropriate Jacobian matrix for the inverse transformation is

J = i

xj=

(1, 2, 3)

(x1, x2, x3)=

1

x11

x21

x3

2

x12

x22

x3

3

x13

x23

x3

J = 1 1 0 32 2 00 0 1

The determinant of the Jacobian matrix is

2 32

=1

2

So a unique transformation always exists, since the Jacobian determinant is never zero.The metric tensor is

gkl = i

xk i

xl=

1

xk 1

xl+

2

xk 2

xl+

3

xk 3

xl

for example for k = 1, l = 1 we get

g11 = i

x1 i

x1=

1

x1 1

x1+

2

x1 2

x1+

3

x1 3

x1

g11 = (1)(1) +

3

2

3

2

+ (0)(0) =

13

4

repeating this operation, we find the complete metric tensor is

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gkl =

134 4 04 5 0

0 0 1

g = det (gkl) =65

4 16 = 1

4

This is equivalent to the calculation in Gibbs notation:

G = JT J

G =

1 32 01 2 0

0 0 1

1 1 032 2 0

0 0 1

G =

134 4 04 5 0

0 0 1

Distance in the transformed system is given by

(ds)2

= gkl dxk dxl

(ds)2 = dxT G dx

(ds)2

= ( dx1 dx2 dx3 )

134 4 04 5 0

0 0 1

dx1dx2

dx3

(ds)2

= ( dx1 dx2 dx3 )

134 dx1 4 dx24 dx1 + 5 dx2

dx3

(ds)2

=13

4

dx1

2+ 5

dx2

2+

dx32 8 dx1 dx2

Algebraic manipulation reveals that this can be rewritten as follows:

(ds)2

= 8.22 0.627 dx1 0.779 dx2

2+ 0.0304 0.779 dx1 + 0.627 dx2

2+ dx3

2

Note:

The Jacobian matrix J is not symmetric. The metric tensor G = JT J is symmetric. The fact that the metric tensor has non-zero off-diagonal elements is a consequence of the transfor-

mation being non-orthogonal.

The distance is guaranteed to be positive. This will be true for all affine transformations in ordinarythree-dimensional Euclidean space. In the generalized space-time continuum suggested by the theoryof relativity, the generalized distance may in fact be negative; this generalized distance ds for an

infintesmial change in space and time is given by ds2 =

d12

+

d22

+

d32 d42, where the

first three coordinates are the ordinary Cartesian space coordinates and the fourth is d4

2

= (c dt)2,

where c is the speed of light.Also we have the volume ratio of differential elements as

d1 d2 d3 =1

2dx1 dx2 dx3

Now

S

1=

S

x1x1

1+

S

x2x2

1+

S

x3x3

1

= 4S

x1+ 3

S

x2

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-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

2

1

1x = 1

1x = 2

1x = 3

2x = 0

2x = /4

2x = /2

2x = 3/2

2x =

2x = 5/4

x = 3/4

2x = 7/4

Figure 1.4: Lines of constant x1

and x2

in the 1

, 2

plane for cylindrical transformation ofexample problem.

So the transformed equation becomes

4S

x1+ 3

S

x2 S = x1 x22 +3

2x1 + 2 x2

24

S

x1+ 3

S

x2 S = 13

4

x12 8 x1x2 + 5 x22

2. Cartesian to cylindrical coordinates.

The transformations are

x1 = +

(1)2

+ (2)2

x2 = tan1

2

1

x3 = 3

Note this system of equations is non-linear. For such systems, we cannot always find an explicit algebraicexpression for the inverse transformation. In this case, some straightforward algebraic and trigonometricmanipulation reveals that we can find an explicit representation of the inverse transformation, which is

1 = x1 cos x2

2 = x1 sin x2

3 = x3

Lines of constant x1 and x2 in the 1, 2 plane are plotted in Figure 1.4. Notice that the lines ofconstant x1 are orthogonal to lines of constant x2 in the Cartesian 1, 2 plane. For general transfor-mations, this will not be the case.

The appropriate Jacobian matrix for the inverse transformation is

J = i

xj=

(1, 2, 3)

(x1, x2, x3)=

1

x11

x21

x3

2

x12

x22

x3

3

x13

x23

x3

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J =

cos x2 x1 sin x2 0sin x2 x1 cos x2 0

0 0 1

The determinant of the Jacobian matrix is

x1 cos2 x2 + x1 sin2 x2 = x1.

So a unique transformation fails to exist when x1 = 0.The metric tensor is

gkl = i

xk i

xl=

1

xk 1

xl+

2

xk 2

xl+

3

xk 3

xl

for example for k = 1, l = 1 we get

g11 = i

x1 i

x1=

1

x1 1

x1+

2

x1 2

x1+

3

x1 3

x1

g11 = cos

2

x

2

+ sin

2

x

2

+ 0 = 1

repeating this operation, we find the complete metric tensor is

gkl =

1 0 00 x12 0

0 0 1

g = det (gkl) = (x1)2

This is equivalent to the calculation in Gibbs notation:

G = JT J

G = cos x2 sin x2 0x1 sin x2 x1 cos x2 0

0 0 1

cos x2 x1 sin x2 0sin x2 x1 cos x2 00 0 1

G =

1 0 00 x12 0

0 0 1

Distance in the transformed system is given by

(ds)2

= gkl dxk dxl

(ds)2

= dxT G dx

(ds)

2

= [ dx

1

dx

2

dx

3

]1 0 0

0 x12 00 0 1dx1

dx

2

dx3

(ds)2 = ( dx1 dx2 dx3 )

dx1x12 dx2

dx3

(ds)2

=

dx12

+

x1dx22

+

dx32

Note:

The fact that the metric tensor is diagonal can be attributed to the transformation being orthogonal.

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Since the product of any matrix with its transpose is guaranteed to yield a symmetric matrix, themetric tensor is always symmetric.

Also we have the volume ratio of differential elements as

d1 d2 d3 = x1 dx1 dx2 dx3

Now

S

1=

S

x1x1

1+

S

x2x2

1+

S

x3x3

1

=S

x11

(1)2

+ (2)2

Sx2

2

(1)2

+ (2)2

= cos x2S

x1 sin x

2

x1S

x2

So the transformed equation becomes

cos x2S

x1 sin x2

x1

S

x2 S = x12

1.3.2 Covariance and Contravariance

Quantities known as contravariant vectors transform according to

ui

=

xi

xj uj

(1.43)

Quantities known as covariant vectors transform according to

ui =xj

xiuj (1.44)

Here we have considered general transformations from one non-Cartesian coordinate system(x1, x2, x3) to another (x1, x2, x3).

Example 1.5Lets say (x, y, z) is a normal Cartesian system and define the transformation

x = x y = y z = z

Now we can assign velocities in both the unbarred and barred systems:

ux =dx

dtuy =

dy

dtuz =

dz

dt

ux =dx

dtuy =

dy

dtuz =

dz

dt

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ux =x

x

dx

dtuy =

y

y

dy

dtuz =

z

z

dz

dt

ux = ux uy = uy uz = uz

ux =x

xux uy =

y

yuy uz =

z

zuz

This suggests the velocity vector is contravariant.Now consider a vector which is the gradient of a function f(x, y, z). For example, let

f(x, y, z) = x + y2 + z3

ux =f

xuy =

f

yuz =

f

z

ux = 1 uy = 2y uz = 3z2

In the new coordinates

f x

,

y

,

z

=

x

+

y2

2+

z3

3

so

f(x, y, z) =x

+y2

2

+z3

3

Now

ux =f

xuy =

f

yuz =

f

z

ux =1

uy =

2y

2uz =

3z2

3

In terms of x, y, z, we have

ux =1

uy =

2y

uz =

3z2

So it is clear here that, in contrast to the velocity vector,

ux =1

ux uy =

1

uy uz =

1

uz

Somewhat more generally we find for this case that

ux =x

xux uy =

y

yuy uz =

z

zuz,

which suggests the gradient vector is covariant.

Contravariant tensors transform according to

vij =xi

xkxj

xlvkl

Covariant tensors transform according to

vij =xk

xixl

xjvkl

Mixed tensors transform according to

vij =xi

xkxl

xjvkl

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The idea of covariant and contravariant derivatives play an important role in mathemat-ical physics, namely in that the equations should be formulated such that they are invariantunder coordinate transformations. This is not particularly difficult for Cartesian systems,but for non-orthogonal systems, one cannot use differentiation in the ordinary sense butmust instead use the notion of covariant and contravariant derivatives, depending on the

problem. The role of these terms was especially important in the development of the theoryof relativity.

Consider a contravariant vector ui defined in xi which has corresponding components Ui

in the Cartesian i. Take wij and Wij to represent the covariant spatial derivative of u

i andUi, respectively. Lets use the chain rule and definitions of tensorial quantities to arrive ata formula for covariant differentiation.

From the definition of contravariance

Ui = i

xlul (1.45)

Take the derivative in Cartesian space and then use the chain rule:

Wij =Ui

j=

Ui

xkxk

j(1.46)

=

xk

i

xlul

xk

j(1.47)

=

2i

xkxlul +

i

xlul

xk

xk

j(1.48)

Wpq =

2p

xkxlul +

p

xlul

xk

xk

q(1.49)

From the definition of a mixed tensor

wij = Wpq

xi

pq

xj(1.50)

=

2p

xkxlul +

p

xlul

xk

xk

qxi

pq

xj(1.51)

=2p

xkxlxk

qxi

p q

xjul +

p

xlxk

qxi

pq

xjul

xk(1.52)

=2p

xkxlxk

xjxi

pul +

xi

xlxk

xjul

xk(1.53)

=

2p

xkxl k

j

xi

pul

+ i

lk

j

ul

xk (1.54)

=2p

xjxlxi

pul +

ui

xj(1.55)

Here we have used the identity that xi

xj= ij , where

ij is the Kronecker

7 delta, ij = 1, i =j,ij = 0, i = j. We define the Christoffel8 symbols ijl as follows:

7Leopold Kronecker, 1823-1891, German/Prussian mathematician.8Elwin Bruno Christoffel, 1829-1900, German mathematician.

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ijl =2p

xjxlxi

p(1.56)

and use the term j to represent the covariant derivative. Thus we have found the covariantderivative of a contravariant vector ui is as follows:

jui = wij =

ui

xj+ ijlu

l (1.57)

Example 1.6Find u in cylindrical coordinates. The transformations are

x1 = +

(1)

2+ (2)

2

x2 = tan1

2

1

x3 = 3

The inverse transformation is

1 = x1 cos x2

2 = x1 sin x2

3 = x3

This corresponds to finding

iui = wii =

ui

xi+ iilu

l

Now for i = j

iilul =

2p

xixlxi

pul

=21

xixlxi

1ul +

22

xixlxi

2ul +

23

xixlxi

3ul

noting that all second partials of 3 are zero,

=21

xixlxi

1ul +

22

xixlxi

2ul

=21

x1xlx1

1ul +

21

x2xlx2

1ul +

21

x3xlx3

1ul

+22

x1xlx1

2ul +

22

x2xlx2

2ul +

22

x3xlx3

2ul

noting that partials of x3 with respect to 1 and 2 are zero,

=21

x1xlx1

1ul +

21

x2xlx2

1ul

+22

x1xlx1

2ul +

22

x2xlx2

2ul

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=21

x1x1x1

1u1 +

21

x1x2x1

1u2 +

21

x1x3x1

1u3

+21

x2x1x2

1u1 +

21

x2x2x2

1u2 +

21

x2x3x2

1u3

+22

x1x1x1

2u1 +

22

x1x2x1

2u2 +

22

x1x3x1

2u3

+22

x2x1x2

2u1 +

22

x2x2x2

2u2 +

22

x2x3x2

2u3

again removing the x3 variation

=21

x1x1x1

1u1 +

21

x1x2x1

1u2

+21

x2x1x2

1u1 +

21

x2x2x2

1u2

+22

x1x1x1

2u1 +

22

x1x2x1

2u2

+ 22

x2x1x2 2

u1 + 22

x2x2x2 2

u2

substituting for the partial derivatives

= 0u1 sin x2 cos x2u2

sin x2 sin x2

x1

u1 x1 cos x2

sin x2x1

u2

+0u1 + cos x2 sin x2u2

+cos x2

cos x2

x1

u1 x1 sin x2

cos x2

x1

u2

= u1

x1

So in cylindrical coordinates

u = u1

x1+

u2

x2+

u3

x3+

u1

x1

Note: In standard cylindrical notation, x1 = r, x2 = , x3 = z. Considering u to be a velocity vector,we get

u = r

dr

dt

+

d

dt

+

z

dz

dt

+

1

r

dr

dt

u = 1r

r

r dr

dt

+ 1

r

r d

dt

+

z

dzdt

u = 1

r

r(rur) +

1

r

u

+uzz

Here we have also used the more traditional u = rddt = x

1u2, along with ur = u1, uz = u

3. Forpractical purposes, this insures that ur, u, uz all have the same dimensions.

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We summarize some useful identities, all of which can be proved, as well as some othercommon notation, as follows

gkl = i

xk i

xl(1.58)

g = det(gij) (1.59)gikg

kj = ji (1.60)

ui = gijuj (1.61)

ui = gijuj (1.62)

u v = uivi = uivi = gijujvi = gijujvi (1.63)u v = ijkgjmgknumvn = ijkujvk (1.64)

ijk =2p

xjxkxi

p=

1

2gip

gpjxk

+gpkxj

gjkxp

(1.65)

u = ju

i = ui,j =ui

xj+ ijlu

l (1.66)

u = iui = ui,i =ui

xi+ iilu

l =1

g

xi

g ui

(1.67)

u = ujxi

uixj

(1.68)

= ,i = xi

(1.69)

2 = = gij,ij = xj

gij

xi

+ jjkg

ik

xi(1.70)

=1

g

xj

g gij

xi (1.71)T = Tij,k =

Tij

xk+ ilkT

lj + jlkTil (1.72)

T = Tij,j =Tij

xj+ iljT

lj + jljTil =

1g

xj

g Tij

+ ijkTjk (1.73)

=1

g

xj

g Tkj

i

xk

(1.74)

1.4 Maxima and minima

Consider the real function f(x), where x [a, b]. Extrema are at x = xm, where f(xm) = 0,if xm [a, b]. It is a local minimum, a local maximum, or an inflection point according towhether f(xm) is positive, negative or zero, respectively.

Now consider a function of two variables f(x, y), with x [a, b], y [c, d]. A necessarycondition for an extremum is

f

x(xm, ym) =

f

y(xm, ym) = 0 (1.75)

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where xm [a, b], ym [c, d]. Next we find the Hessian 9 matrix (Hildebrand 356)

H =

2fx2

2fxy

2fxy

2fy2

(1.76)

and its determinant D = det H. It can be shown thatf is a maximum if

2fx2 < 0 and D < 0

f is a minimum if 2fx2

> 0 and D < 0f is a saddle ifD > 0Higher order derivatives must be considered if D = 0.

Example 1.7

f = x2 y2

Equating partial derivatives with respect to x and to y to zero, we get

2x = 0

2y = 0

This gives x = 0, y = 0. For these values we find that

D = 2 00 2

= 4

Since D > 0, the point (0,0) is a saddle point.

1.4.1 Derivatives of integral expressions

Often functions are expressed in terms of integrals. For example

y(x) =

b(x)a(x)

f(x, t) dt

Here t is a dummy variable of integration. Leibnizs 10 rule tells us how to take derivativesof functions in integral form:

y(x) =

b(x)a(x)

f(x, t) dt (1.77)

dy(x)

dx= f(x, b(x))

db(x)

dx f(x, a(x))da(x)

dx+

b(x)a(x)

f(x, t)

xdt (1.78)

9Ludwig Otto Hesse, 1811-1874, German mathematician, studied under Jacobi.10Gottfried Wilhelm von Leibniz, 1646-1716, German mathematician and philosopher of great influence;

co-inventor with Sir Isaac Newton, 1643-1727, of the calculus.

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Inverting this arrangement in a special case, we note if

y(x) = y(xo) +

xx0

f(t) dt (1.79)

then (1.80)

dy(x)dx

= f(x) dxdx

f(x0) dxodx

+b(x)a(x)

f(t)x

dt (1.81)

dy(x)

dx= f(x) (1.82)

Note that the integral expression naturally includes the initial condition that when x = x0,y = y(x0). This needs to be expressed separately for the differential version of the equation.

Example 1.8Find dydx if

y(x) =

x2x

(x + 1)t2 dt (1.83)

Using Leibnizs rule we get

dy(x)

dx= [(x + 1)x4](2x) [(x + 1)x2](1) +

x2x

t2 dt (1.84)

= 2x6 + 2x5 x3 x2 +

t3

3

x2x

(1.85)

= 2x6 + 2x5 x3 x2 + x6

3 x

3

3(1.86)

= 7x63

+ 2x5 4x33

x2 (1.87)(1.88)

In this case, but not all, we can achieve the same result from explicit formulation of y(x):

y(x) = (x + 1)

x2x

t2 dt (1.89)

= (x + 1)

t3

3

x2x

(1.90)

= (x + 1)x6

3 x

3

3 (1.91)y(x) =

x7

3+

x6

3 x

4

3 x

3

3(1.92)

dy(x)

dx=

7x6

3+ 2x5 4x

3

3 x2 (1.93)

So the two methods give identical results.

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1.4.2 Calculus of variations

(See Hildebrand, p. 360)The problem is to find the function y(x), with x [x1, x2], and boundary conditions

y(x1) = y1, y(x2) = y2, such that the integral

I =x2x1

f(x,y,y) dx (1.94)

is an extremum. If y(x) is the desired solution, let Y(x) = y(x) + h(x), where h(x1) =h(x2) = 0. Thus Y(x) also satisfies the boundary conditions; also Y

(x) = y(x) + h(x).We can write

I() =

x2x1

f(x,Y,Y) dx

Taking dId , utilizing Leibnizs formula, we get

dI

d=

x2

x1fx

x

+

f

Y

Y

+

f

Y

Y

dxEvaluating, we find

dI

d=

x2x1

f

x0 +

f

Yh(x) +

f

Y h(x)

dx

Since I is an extremum at = 0, we have dI/d = 0 for = 0. This gives

0 =

x2x1

f

Yh(x) +

f

Yh(x)

=0

dx

Also when = 0, we have Y = y, Y = y, so

0 =

x2x1

f

yh(x) +

f

y h(x)

dx

Look at the second term in this integral. Since from integration by parts we getx2x1

f

y h(x) dx =

x2x1

f

y dh

=f

y h(x)

x2

x1

x2x1

d

dx

f

y

h(x) dx

The first term above is zero because of our conditions on h(x1) and h(x2). Thus substitutinginto the original equation we havex2x1

fy

ddx

f

y

h(x) dx = 0 (1.95)

The equality holds for all h(x), so that we must have

f

y d

dx

f

y

= 0 (1.96)

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called the Euler11 equation.While this is, in general, the preferred form of the Euler equation, its explicit dependency

on the two end conditions is better displayed by considering a slightly different form. Byexpanding the total derivative term, that is

d

dx

f

y (x,y,y)

=

2f

y x+

2f

y ydy

dx+

2f

y y dy

dx(1.97)

=2f

y x+

2f

y yy +

2f

y y y (1.98)

the Euler equation after slight rearrangement becomes

2f

y y y +

2f

y yy +

2f

y x f

y= 0 (1.99)

fyyd2y

dx2+ fyy

dy

dx+ (fyx

fy) = 0 (1.100)

This is a clearly second order differential equation for fyy = 0, and in general, non-linear.If fyy is always non-zero, the problem is said to be regular. If fyy = 0 at any point, theequation is no longer second order, and the problem is said to be singular at such points.Note that satisfaction of two boundary conditions becomes problematic for equations lessthan second order.

There are several special cases of the function f.

1. f = f(x, y)

The Euler equation is

fy

= 0 (1.101)

which is easily solved:f(x, y) = A(x) (1.102)

which, knowing f, is then solved for y(x).

2. f = f(x, y)

The Euler equation isd

dx

f

y

= 0 (1.103)

which yields

f

y = A (1.104)

f(x, y) = Ay + B(x) (1.105)

Again, knowing f, the equation is solved for y and then integrated to find y(x).11Leonhard Euler, 1707-1783, prolific Swiss mathematician, born in Basel, died in St. Petersburg.

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3. f = f(y, y)


f

y d

dx

f

y (y, y)

= 0 (1.106)

fy

2fyy

dydx

+ 2fy y

dydx

= 0 (1.107)

f

y

2f

yydy

dx

2f

y y d2y

dx2= 0 (1.108)

Multiply by y to get

y

f

y

2f

yydy

dx

2f

y y d2y

dx2

= 0 (1.109)

f

yy +

f

y y y

2f

yydy

dx

2f

y y d2y

dx2 f

y y = 0 (1.110)

(1.111)

d

dx

f y f

y

= 0 (1.112)

which can be integrated. Thus

f(y, y) y fy

= A (1.113)

which is effectively a first order ordinary differential equation which is solved. Anotherintegration constant arises. This along with A are determined by the two end pointconditions.

Example 1.9Find the curve of minimum length between the points (x1, y1) and (x2, y2).If y(x) is the curve, then y(x1) = y1 and y(x2) = y2. The length of the curve is

L =

x2x1

1 + (y)2 dx


d

dx

y

1 + (y)2

= 0

which can be integrated to give y1 + (y)2

= C

Solving for y we get

y = A =

C2

1 C2from which

y = Ax + B

The constants A and B are obtained from the boundary conditions y(x1) = y1 and y(x2) = y2.

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-1 -0.5 0 0.5 1 1.5 2x

0.5

1

1.5

2

2.5

3

-1

0

1

2x

-2

0

2y

-2

0

2

z

-1

0

1

2x

-2

0

2y

-2

0

2

curve withendpoints at(-1, 3.09), (2, 2.26)which minimizessurface area of bodyof revolution corresponding

surface ofrevolution

.

.

Figure 1.5: Body of revolution of minimum surface area for (x1, y1) = (

1, 3.08616) and

(x2, y2) = (2, 2.25525)

Example 1.10Find the curve through the points (x1, y1) and (x2, y2), such that the surface area of the body of

revolution by rotating the curve around the x-axis is a minimum.We wish to minimize

I =

x2x1

y

1 + (y)2 dx

Here f(y, y) = y1 + (y)2. So the Euler equation reduces tof(y, y) y f

y = A

y

1 + y2 yy y

1 + y2= A

y(1 + y2) yy2 = A

1 + y2

y = A

1 + y2

y =

yA

2 1

y(x) = A coshx B

AThis is a catenary. The constants A and B are determined from the boundary conditions y(x1) = y1and y(x2) = y2. In general this requires a trial and error solution of simultaneous algebraic equations.If (x1, y1) = (1, 3.08616) and (x2, y2) = ( 2, 2.25525), one finds solution of the resulting algebraicequations gives A = 2, B = 1.

For these conditions, the curve y(x) along with the resulting body of revolution of minimum surfacearea are plotted in Figure 1.5.

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1.5 Lagrange multipliers

Suppose we have to determine the extremum off(x1, x2, . . . , xm) subject to the n constraints

gi(x1, x2, . . . , xm) = 0, i = 1, 2, . . . , n (1.114)

Definef = f 1g1 2g2 . . . ngn (1.115)

where the i (i = 1, 2, , n) are unknown constants called Lagrange 12 multipliers. To getthe extremum off, we equate to zero its derivative with respect to x1, x2, . . . , xm. Thus wehave

f

xi= 0, i = 1, . . . , m (1.116)

gi = 0, i = 1, . . . , n (1.117)

which are (m+n) equations that can be solved for xi (i = 1, 2, . . . , m) and i (i = 1, 2, . . . , n).

Example 1.11Extremize f = x2 + y2 subject to the constraint 5x2 6xy + 5y2 = 8.Let

f = x2 + y2 (5x2 6xy + 5y2 8)from which

2x 10x + 6y = 02y + 6x 10y = 05x2 6xy + 5y2 = 8

From the first equation

=2x

10x 6ywhich, when substituted into the second, gives

x = y

The last equation gives the extrema to be at (x, y) = (

2,

2), (2, 2), ( 12

, 12

), ( 12

, 12

).

The first two sets give f = 4 (maximum) and the last two f = 1 (minimum).The function to be maximized along with the constraint function and its image are plotted in Figure

1.6.

A similar technique can be used for the extremization of a functional with constraint.We wish to find the function y(x), with x [x1, x2], and y(x1) = y1, y(x2) = y2, such thatthe integral

I =

x2x1

f(x,y,y) dx (1.118)

12Joseph-Louis Lagrange, 1736-1813, Italian-born French mathematician.

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-10

1

x

-1

0

1y

0

1

2

3

4

f(x,y)

-10

1

-1

0

1y

0

1

2

3

4

-10

12

x

-2

-1012y

0

2

4

6

8

f(x,y)

-10

12

xy

0

2

4

6

8

f(x,y)

constraintfunction

constrainedfunction

unconstrainedfunction

Figure 1.6: Unconstrained function f(x, y) along with constrained function and constraintfunction (image of constrained function)

is an extremum, and satisfies the constraint

g = 0 (1.119)

DefineI = I g (1.120)

and continue as before.

Example 1.12Extremize I, where

I =

a0

y

1 + (y)2 dx

with y(0) = y(a) = 0, and subject to the constrainta0

1 + (y)2 dx =

That is find the maximum surface area of a body of revolution which has a constant length. Let

g = a

01 + (y)2 dx = 0

Then let

I = I g =a

0

y

1 + (y)2 dx a

0

1 + (y)2 dx +

=

a0

(y )

1 + (y)2 dx +

=

a0

(y )

1 + (y)2 +

a

dx

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-0.2

0

0.2

y

-0.2

0

0.2

z

0

0.25

0.5

0.751

x

0.2 0.4 0.6 0.8 1x

-0.3-0.25

-0.2

-0.15

-0.1

-0.05

y

Figure 1.7: Curve of length = 5/4 with y(0) = y(1) = 0 whose surface area of correspondingbody of revolution (also shown) is maximum.

With f = (y )1 + (y)2 +a , we have the Euler equation

f

y d

dx

f

y

= 0

Integrating from an earlier developed relationship when f = f(y, y), and absorbing a into a constantA, we have

(y )

1 + (y)2 y(y ) y

1 + (y)2= A

from which(y )(1 + (y)2) (y)2(y ) = A

1 + (y)2

(y ) 1 + (y)2 (y)2 = A1 + (y)2y

= A1 + (y)2

y =

y

A

2 1

y = + A coshx B

A

Here A,B, have to be numerically determined from the three conditions y(0) = y(a) = 0, g = 0.If we take the case where a = 1, = 5/4, we find that A = 0.422752, B = 12 , = 0.754549. For

these values, the curve of interest, along with the surface of revolution, is plotted in Figure 1.7.

Problems

1. Ifz3 + zx + x4y = 3y,

(a) find a general expression forz

x

y

,z

y

x

,

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13. Show that the functions

u =x + y

x yv =

xy

(x y)2

are functionally dependent.

14. Find the point on the curve of intersection of z xy = 10 and x + y + z = 1, that is closest to theorigin.

15. Find a function y(x) with y(0) = 1, y(1) = 0 that extremizes the integral

I =

10

1 +

dydx

2y

dx

Plot y(x) for this function.

16. For elliptic cylindrical coordinates

1 = cosh x1 cos x2

2 = sinh x1 sin x2

3 = x3

Find the Jacobian matrix J and the metric tensor G. Find the inverse transformation. Plot lines ofconstant x1 and x2 in the 1 and 2 plane.

17. For the elliptic coordinate system of the previous problem, fine u where u is an arbitrary vector.18. Find the covariant derivative of the contravariant velocity vector in cylindrical coordinates.

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Chapter 2

First-order ordinary differentialequations

see Lopez, Chapters 1-3,

see Riley, Hobson, and Bence, Chapter 12,see Bender and Orszag, 1.6.

A first-order ordinary differential equation is of the form

F(x,y,y) = 0 (2.1)

where y = dydx

.

2.1 Separation of variables

Equation (2.1) is separable if it can be written in the formP(x)dx = Q(y)dy (2.2)

which can then be integrated.

Example 2.1Solve

yy =8x + 1

y, with y(1) = 5.

Separating variablesy2dy = 8xdx + dx.

Integrating, we have y3

3= 4x2 + x + C.

The initial condition gives C = 1403 , so that the solution isy3 = 12x2 + 3x 140.

The solution is plotted in Figure 2.1.

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-10 -5 5 10x

-5

-2.5

2.5

5

7.5

10

y

Figure 2.1: y(x) which solves yy = (8x + 1)/y with y(1) = 5.

2.2 Homogeneous equations

An equation is homogeneous if it can be written in the form

y = fy

x

. (2.3)

Defining

u =y

x

(2.4)

we gety = ux,

from whichy = u + xu.

Substituting in equation (2.3) and separating variables, we have

u + xu = f(u) (2.5)

u + xdu

dx= f(u) (2.6)

xdudx

= f(u) u (2.7)du

f(u) u =dx

x(2.8)

which can be integrated.Equations of the form

y = f

ax + by + c

dx + ey + h

(2.9)

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can be similarly integrated.

Example 2.2Solve

xy = 3y +y2

x, with y(1) = 4.

This can be written as

y = 3y

x

+y

x

2.

Letting u = y/x, we getdu

2u + u2=

dx

x.

Since1

2u + u2=

1

2u 1

4 + 2u

both sides can be integrated to give

1

2 (ln |u| ln |2 + u|) = ln |x| + C.The initial condition gives C = 12 ln

23 , so that the solution can be reduced to y2x + y

= 23 x2.This can be solved explicitly for y(x) where we find for each case of the absolute value. The first case

y(x) =43

x3

1 23 x2

is seen to satisfy the condition at x = 1. The second case is discarded as it does not satisfy the condition

at x = 1.The solution is plotted in Figure 2.2.

2.3 Exact equations

A differential equation is exact if it can be written in the form

dF(x, y) = 0. (2.10)where F(x, y) = 0 is a solution to the differential equation Using the chain rule to expandthe derivative

F

xdx +

F

ydy = 0

So for an equation of the form

P(x, y)dx + Q(x, y)dy = 0 (2.11)

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-6 -4 -2 2 4 6x

-20

-15

-10

-5

5

10

15

20

y

Figure 2.2: y(x) which solves xy = 3y + y2

xwith y(1) = 4

we have an exact differential if

F

x= P(x, y),

F

y= Q(x, y) (2.12)

2F

xy=

P

y,

2F

yx=

Q

x(2.13)

As long as F(x, y) is continuous and differentiable, the mixed second partials are equal, thus,

P

y=

Q

x(2.14)

must hold if F(x, y) is to exist and render the original differential equation to be exact.

Example 2.3Solve

dy

dx=

exy

exy 1exy dx + 1 exy dy = 0P

y= exy

Q

x= exy

Since Py =Qx , the equation is exact. Thus

F

x= P(x, y)

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-2

2

4

6

y

-6 -4 -2 2 4 6x

C = -2

C = -1

C = 0

C = 1

C = 2

Figure 2.3: y(x) which solves y = exp(x y)/(exp(x y) 1)

F

x= exy

F(x, y) = exy + A(y)F

y= exy + dA

dy= Q(x, y) = 1 exy

dA

dy= 1

A(y) = y CF(x, y) = exy + y C = 0

exy + y = C

The solution for various values ofC is plotted in Figure 2.3.

2.4 Integrating factors

Sometimes, an equation of the form (2.11) is not exact, but can be made so by multiplicationby a function u(x, y), where u is called the integrating factor. It is not always obvious that

integrating factors exist; sometimes they do not.

Example 2.4Solve

dy

dx=

2xy

x2 y2Separating variables, we get

(x2 y2) dy = 2xy dx.

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-1.5 -1 -0.5 0.5 1 1.5x

-3

-2

-1

1

2

3y

C = 3

C = 2

C = 1

C = -1

C = -2

C = -3

Figure 2.4: y(x) which solves y(x) = 2xy(x2y2)

This is not exact according to criterion (2.14). It turns out that the integrating factor is y2, so thaton multiplication, we get

2x

ydx

x2

y2 1

dy = 0.

This can be written as

d

x2

y+ y

= 0

which gives

x2 + y2 = Cy.

The solution for various values ofC is plotted in Figure 2.4.

The general first-order linear equation

dy(x)

dx+ P(x) y(x) = Q(x) (2.15)

withy(xo) = yo

can be solved using the integrating factor

exaP(s)ds = e(F(x)F(a)).

We choose a such that

F(a) = 0.

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Multiply by the integrating factor and proceed:exaP(s)ds

dy(x)dx

+

exaP(s)ds

P(x) y(x) =

exaP(s)ds

Q(x) (2.16)

product rule:d

dx exaP(s)dsy(x) = e

xaP(s)ds

Q(x) (2.17)replace x by t:

ddt

etaP(s)dsy(t)

=

etaP(s)ds

Q(t) (2.18)

integrate:

xxo

d

dt

etaP(s)dsy(t)

dt =

xxo

etaP(s)ds

Q(t)dt (2.19)

exaP(s)dsy(x) e

xoa

P(s)dsy(xo) =

xxo

etaP(s)ds

Q(t) dt (2.20)

which yields

y(x) = exaP(s)ds

exoa

P(s)dsyo +

xxo

etaP(s)ds

Q(t)dt

. (2.21)

Example 2.5Solve

y y = e2x; y(0) = yo.Here

P(x) = 1or

P(s) = 1xa

P(s)ds =

xa

(1)ds = s|xa = a xSo

F() = For F(a) = 0, take a = 0. So the integrating factor is

exaP(s)ds = eax = e0x = ex

Multiplying and rearranging, we get

exdy(x)

dx exy(x) = ex

d

dx

exy(x)

= ex

d

dt ety(t)

= et

xxo=0

ddtety(t) dt = x

xo=0

etdt

exy(x) e0y(0) = ex e0exy(x) yo = ex 1

y(x) = ex (yo + ex 1)

y(x) = e2x + (yo 1) ex

The solution for various values ofyo is plotted in Figure 2.5.

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-3 -2 -1 1 2 3x

-3

-2

-1

1

2

3

yy

=

2

oy

=

0

oy

=

-2

o

Figure 2.5: y(x) which solves y y = e2x with y(0) = yo

2.5 Bernoulli equation

Some first-order nonlinear equations also have analytical solutions. An example is theBernoulli 1 equation

y + P(x)y = Q(x)yn. (2.22)

where n = 1. Letu = y1n,

so thaty = u

1

1n .

The derivative is

y =1

1 n

un

1n

u.

Substituting in equation (2.22), we get

1

1 n

un

1n

u + P(x)u

1

1n = Q(x)un

1n .

This can be written asu + (1 n)P(x)u = (1 n)Q(x) (2.23)

which is a first-order linear equation of the form (2.15) and can be solved.

1after one of the members of the prolific Bernoulli family.

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-1 -0.8 - 0.6 - 0.4 - 0.2 0.2 0.4x

-1

-0.5

0.5

1

1.5

2

2.5

3C=

0

C=

2

C= -1

C= -2

C= -2

C= -1C=

-2

Figure 2.6: y(x) which solves y = exp(3x)x y/x + 3 exp(3x)

which is an equation of first order.

Example 2.7Solve

xy + 2y = 4x3.

Let u = y, so that

xdu

dx+ 2u = 4x3.

Multiplying by x

x2du

dx+ 2xu = 4x4

d

dx(x2u) = 4x4.

This can be integrated to give

u =4

5x3 +

C1x2

from which

y =1

5x4 C1

x+ C2

for x= 0.

2.7.2 x absent

Iff(y, y, y) = 0 (2.34)

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let u(x) = y, so that

y =dy

dx=

dy

dy

dy

dx=

du

dyu

The equation becomes

fy,u,ududy = 0 (2.35)which is also an equation of first order. Note however that the independent variable is nowy while the dependent variable is u.

Example 2.8Solve

y 2yy = 0; y(0) = yo, y(0) = yo.Let u = y, so that y = dudx =

dydx

dudy = u

dudy . The equation becomes

udu

dy 2yu = 0

Nowu = 0

satisfies the equation. Thus

dy

dx= 0

y = C

applying one initial condition: y = yo

This satisfies the initial conditions only under special circumstances, i.e. yo = 0. For u = 0,

dudy

= 2y

u = y2 + C1

apply I.C.s: yo = y2o + C1

C1 = yo y2o

dy

dx= y2 + yo y2o

dy

y2 + yo y2o= dx

from which for yo y2o > 0

1yo y2o

tan1 y

yo y2o

= x + C2

1yo y2o

tan1

yoyo y2o

= C2

y(x) =

yo y2o tan

x

yo y2o + tan1

yoyo y2o

The solution for yo = 0, yo = 1 is plotted in Figure 2.7.

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1 2 3 4x

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

y

1 2 3 4x

-1

-0.75

-0.5

-0.25

0.25

0.5

0.75

1

y

Figure 2.8: Two solutions y(x) which satisfy y = 3y2/3 with y(2) = 0

2.9 Clairaut equation

The solution of a Clairaut 3 equation

y = xy + f(y) (2.36)

can be obtained by letting y = u(x), so that

y = xu + f(u). (2.37)

Differentiating with respect to x, we get

y = xu + u +df

duu (2.38)

u = xu + u +df

duu (2.39)

ux + dfdu = 0. (2.40)If we take

u =du

dx= 0, (2.41)

we can integrate to getu = C (2.42)

3Alexis Claude Clairaut, 1713-1765, Parisian/French mathematician.

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(h) y = x+2y52xy+4

(i) y + xy = y

Plot solutions for y(0) = 1, 0, 1 (except for part e).10. Find all solutions of

(x + 1)(y)2

+ (x y)y y = 011. Find an a for which a unique solution of

(y)4 + 8(y)3 + (3a + 16)(y)2 + 12ay + 2a2 = 0, with y(1) = 2

exists. Find the solution.

12. Solve

y 1x2

y2 +1

xy = 1

60

algebra lineal y otros temas

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