Instructors Solution Manual Introduction to Partial Differential Equations 1st Edition By Peter J. O

INSTRUCTORS SOLUTION MANUAL

Selected Solutions Manual for Instructors for

Introduction to Partial Differential Equations by

Peter J. Olver Undergraduate Texts in Mathematics Springer, 2014 ISBN 978–3–319–02098–3

c 2020 Peter J. Olver

Table of Contents Chapter 1. What Are Partial Differential Equations? . . . . . . . . . . 1 Chapter 2. Linear and Nonlinear Waves . . . . . . . . . . . . . . . . 3 Chapter 3. Fourier Series . . . . . . . . . . . . . . . . . . . . . 14 Chapter 4. Separation of Variables . . . . . . . . . . . . . . . . . 29 Chapter 5. Finite Differences

. . . . . . . . . . . . . . . . . . . 43

Chapter 6. Generalized Functions and Green’s Functions . . . . . . . 60 Chapter 7. Fourier Transforms . . . . . . . . . . . . . . . . . . . 71 Chapter 8. Linear and Nonlinear Evolution Equations

. . . . . . . . 77

Chapter 9. A General Framework for Linear Partial Differential Equations . . . . . . 87 Chapter 10. Finite Elements and Weak Solutions . . . . . . . . . . . 99 Chapter 11. Dynamics of Planar Media . . . . . . . . . . . . . . . 110 Chapter 12. Partial Differential Equations in Space . . . . . . . . . . 126

c 2020

Peter J. Olver

c 2020

Peter J. Olver

Selected Solutions to Chapter 1: What Are Partial Differential Equations?

Note: Solutions marked with a ⋆ do not appear in the Student Solutions Manual.

1.1. (a) Ordinary differential equation, equilibrium, order = 1; (c) partial differential equation, dynamic, order = 2; (e) partial differential equation, equilibrium, order = 2;

⋆ (g) partial differential equation, equilibrium, order = 2; ⋆ (i) partial differential equation, dynamic, order = 3; ⋆ (k) partial differential equation, dynamic, order = 4. ∂2u ∂2u + = 0, (ii) uxx + uyy = 0; ∂x2 ∂y2 ∂2u ∂2u ⋆ (c) (i) ∂u = + , (ii) ut = uxx + uyy . ∂t ∂x2 ∂y2

1.2. (a) (i)

1.4. (a) independent variables: x, y; dependent variables: u, v; order = 1;

⋆ (b) independent variables: x, y; dependent variables: u, v; order = 2; ⋆ (d) independent variables: t, x, y; dependent variables: u, v, p; order = 1. ∂2u ∂2u + = ex cos y − ex cos y = 0; defined and C∞ on all of R 2 . ∂x2 ∂y2 ∂2u ∂2u ⋆ (c) ∂x + = 6 x − 6 x = 0; defined and C∞ on all of R 2 .

2 ∂y2 ∂2u ∂2u 2 y2 − 2 x2 2 x2 − 2 y2 ⋆ (d) ∂x + = + = 0; defined and C∞ on R 2 \ {0}.

2 ∂y2 (x2 + y2 )2 (x2 + y2 )2

1.5. (a)

i

h

1.7. u = log c (x − a)2 + c (y − b)2 , for a, b, c arbitrary constants.

⋆

1.8. (a) c0 + c1 x + c2 y + c3 z + c4 (x2 − y2 ) + c5 (x2 − z 2 ) + c6 x y + c7 x z + c8 y z, where c0 , . . . , c8 are arbitrary constants.

1.10. (a)

∂2u ∂2u − 4 = 8 − 8 = 0; ∂t2 ∂x2

2

2

⋆ (c) ∂∂t2u − 4 ∂∂xu2 = −4 sin 2 t cos x + 4 sin 2 t cos x = 0.

1.11. (a) c0 + c1 t + c2 x + c3 (t2 + x2 ) + c4 t x, where c0 , . . . , c4 are arbitrary constants.

⋆ 1.13. u = a + rb = a + q 2

b

x + y2 + z 2

, where a, b are arbitrary constants.

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Peter J. Olver

2

Chapter 1: Selected Solutions

1.15. Example: (b) u2x + u2y + u2 = 0 — the only real solution is u ≡ 0.

⋆ 1.16. When (x, y) 6= (0, 0), a direct computation shows that

∂u y (x4 + 4 x2 y2 − y4 ) ∂u x (x4 − 4 x2 y2 − y4 ) = , = , ∂x (x2 + y2 )2 ∂y (x2 + y2 )2 while, from the definition of partial derivative, ∂u ∂u u(h, 0) − u(0, 0) u(0, k) − u(0, 0) (0, 0) = lim = 0, (0, 0) = lim = 0. ∂x h ∂y k h→0 k→0 Thus, 2 uy (h, 0) − uy (0, 0) u (0, k) − ux (0, 0) ∂2u ∂ u (0, 0) = lim = 1, (0, 0) = lim x = −1. ∂x ∂y h ∂y ∂x k h→0 k→0 This does not contradict the equality of mixed partials because the theorem requires continuity, while ∂2u x6 + 9 x4 y2 + 9 x2 y4 − y4 ∂2u = = , (x, y) 6= (0, 0), ∂x ∂y ∂x ∂y (x2 + y2 )3 is not continuous at (x, y) = (0, 0). Indeed, ∂2u ∂2u (h, 0) = 1 6= −1 = lim (0, k). h → 0 ∂x ∂y k → 0 ∂x ∂y lim

1.17. (a) homogeneous linear; 2

1.19. (a) (i)

(d) nonlinear;

⋆ (f ) inhomogeneous linear.

2

∂ u ∂ u = −4 cos(x − 2 t) = 4 2 .

2 ∂t ∂x

⋆ 1.20. (a) cos(x − 2 t) + 41 cos x − 5 sin(x − 2 t) − 54 sin x.

∂ ∂f ∂g [ c f (x) + d g(x) ] = c +d = c ∂x [ f ] + d ∂x [ g ]. The same proof ∂x ∂x ∂x works for ∂y . (b) Linearity requires d = 0, while a, b, c can be arbitrary functions of x, y.

1.21. (a) ∂x [ c f + d g ] =

⋆ 1.23. Using standard vector calculus identities: (b) ∇ × (f + g) = ∇ × f + ∇ × g, ∇ × (c f ) = c ∇ × f .

1.24. (a) (L − M )[ u + v ] = L[ u + v ] − M [ u + v ] = L[ u ] + M [ u ] − L[ v ] − M [ v ] = (L − M )[ u ] + (L − M )[ v ],

(L − M )[ c u ] = L[ c u ] − M [ c u ] = c L[ u ] − c M [ u ] = c (L − M )[ u ];

⋆ (c) (f L)[ u + v ] = f L[ u + v ] = f L[ u ] + f L[ v ] = (f L)[ u ] + (f L)[ v ], (f L)[ c u ] = f L[ c u ] = f c L[ u ] = c (f L)[ u ].

1.27. (b) u(x) = 16 ex sin x + c1 e2 x/5 cos 45 x + c2 e2 x/5 sin 54 x.

1 sin x + c1 e3 x + c2 e− 3 x , 1.28. (b) u(x) = − 19 x − 10

⋆ (d) u(x) = 16 x ex − 181 ex + 14 e− x + c1 ex + c2 e− 2 x .

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Peter J. Olver

Selected Solutions to Chapter 2: Linear and Nonlinear Waves Note: Solutions marked with a ⋆ do not appear in the Student Solutions Manual.

⋆

2.1.1. u(t, x) = t x + f (x), where f is an arbitrary C1 function.

2.1.3. (a) u(t, x) = f (t);

⋆ (c) u(t, x) = t x − 12 t2 + f (x); (e) u(t, x) = e− t x f (t).

2.1.5. u(t, x, y) = f (x, y) where f is an arbitrary C1 function of two variables. This is valid provided each slice Da,b = D ∩ { (t, a, b) | t ∈ R }, for fixed (a, b) ∈ R 2 , is either empty or a connected interval.

⋆ ♥ 2.1.8. (a) The partial differential equation is really an autonomous first-order ordinary differen-

tial equation in t, with x as a parameter. Solving this ordinary differential equation by standard methods, [ 20, 23 ], the solution to the initial value problem is f (x) . Thus, if f (x) > 0, then the denominator does not vanish for t ≥ 0, u(t, x) = t f (x) + 1 and, moreover, goes to ∞ as t → ∞. Therefore, u(t, x) → 0 as t → ∞.

(b) If f (x) < 0, then the denominator in the preceding solution formula vanishes when t = τ = −1/f (x). Moreover, for t < τ , the numerator is negative, while the denominator is positive, and so lim u(t, x) = − ∞. t → τ−

(c) The solution is defined for 0 < t < t⋆ , where t⋆ = −1/ min f (x). In particular, if min f (x) = − ∞, then the solution is not defined for all x ∈ R for any t > 0. ♦ 2.1.9. It suffices to show that, given two points (t1 , x), (t2 , x) ∈ D, then u(t1 , x) = u(t2 , x). By the assumption, (t, x) ∈ D for t1 ≤ t ≤ t2 , and so u(t, x) is defined and continuously differentiable at such points. Thus, by the Fundamental Theorem of Calculus, Z t 2 ∂u (s, x) ds = 0.

Q.E .D. u(t2 , x) − u(t1 , x) = t1 ∂t

2

2.2.2. (a) u(t, x) = e− (x+3 t)

t=1

t=2

t=3

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Peter J. Olver