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Selected Solutions Manual for Introduction to Partial Differential Equations by Peter J. Olver

Note: Answers not provided in this manual are given at

the end of Chapters/Text Book. 1 / 4

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.

1.2. (a) (i) ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 = 0, (ii)u xx +u yy = 0; ⋆(c) (i) ∂u ∂t = ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 , (ii)u t =u xx +u yy .

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.

1.5. (a) ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 =e x cosy−e x cosy= 0; defined and C ∞ on all ofR 2 .⋆(c) ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 = 6x−6x= 0; defined and C ∞ on all ofR 2 .⋆(d) ∂ 2 u ∂x 2 + ∂ 2 u ∂y 2 = 2y 2 −2x 2 (x 2 +y 2 ) 2 + 2x 2 −2y 2 (x 2 +y 2 ) 2 = 0; defined and C ∞ onR 2

\ {0}.

1.7.u= log h c(x−a) 2 +c(y−b) 2 i , fora, b, carbitrary constants.⋆1.8. (a)c

+c 1 x+c 2 y+c 3 z+c 4 (x 2 −y 2

• +c

5 (x 2 −z 2

• +c

6 xy+c 7 xz+c 8 y z, wherec

, . . . , c 8 are arbitrary constants.

1.10. (a) ∂ 2 u ∂t 2 −4 ∂ 2 u ∂x 2 = 8−8 = 0;⋆(c) ∂ 2 u ∂t 2 −4 ∂ 2 u ∂x 2 =−4 sin 2tcosx+ 4 sin 2tcosx= 0.

1.11. (a)c

+c 1 t+c 2 x+c 3 (t 2 +x 2

• +c

4 tx, wherec

, . . . , c 4 are arbitrary constants.⋆1.13.u=a+ b r =a+ b q x 2 +y 2 +z 2 , wherea, bare arbitrary constants.c2020 Peter J. Olver 2 / 4

2Chapter 1: Selected Solutions

1.15. Example: (b)u

2 x +u 2 y +u 2 = 0 — the only real solution isu≡0.⋆1.16. When (x, y)6= (0,0), a direct computation shows that ∂u ∂x = y(x 4

• 4x

2 y 2 −y 4 ) (x 2 +y 2 ) 2 , ∂u ∂y = x(x 4 −4x 2 y 2 −y 4 ) (x 2 +y 2 ) 2 , while, from the definition of partial derivative, ∂u ∂x (0,0) = lim h→0 u(h,0)−u(0,0) h = 0, ∂u ∂y (0,0) = lim k→0 u(0, k)−u(0,0) k = 0.Thus, ∂ 2 u ∂x ∂y (0,0) = lim h→0 u y (h,0)−u y (0,0) h = 1, ∂ 2 u ∂y ∂x (0,0) = lim k→0 u x (0, k)−u x (0,0) k

=−1.

This does not contradict the equality of mixed partials because the theorem requires continuity, while ∂ 2 u ∂x ∂y = ∂ 2 u ∂x ∂y = x 6

• 9x

4 y 2

• 9x

2 y 4 −y 4 (x 2 +y 2 ) 3 ,(x, y)6= (0,0), is not continuous at (x, y) = (0,0). Indeed, lim h→0 ∂ 2 u ∂x ∂y (h,0) = 16=−1 = lim k→0 ∂ 2 u ∂x ∂y (0, k).

1.17. (a) homogeneous linear; (d) nonlinear;⋆(f) inhomogeneous linear.

1.19. (a) (i) ∂ 2 u ∂t 2 =−4 cos(x−2t) = 4 ∂ 2 u ∂x 2 .⋆1.20. (a) cos(x−2t) + 1 4 cosx−5 sin(x−2t)− 5 4 sinx.

1.21. (a)∂ x [cf+dg] = ∂ ∂x [cf(x) +dg(x)] =c ∂f ∂x +d ∂g ∂x =c∂ x [f] +d∂ x [g]. The same proof works for∂ y . (b) Linearity requiresd= 0, whilea, b, ccan be arbitrary functions ofx, y.

⋆1.23. Using standard vector calculus identities:

(b)∇ ×(f+g) =∇ ×f+∇ ×g,∇ ×(cf) =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)[cu] =L[cu]−M[cu] =cL[u]−cM[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)[cu] =f L[cu] =f cL[u] =c(f L)[u].

1.27. (b)u(x) = 1 6 e x sinx+c 1 e 2x/5 cos 4 5 x+c 2 e 2x/5 sin 4 5 x.

1.28. (b)u(x) =− 1 9 x− 1 10 sinx+c 1 e 3x +c 2 e −3x , ⋆(d)u(x) = 1 6 x e x − 1 18 e x + 1 4 e −x +c 1 e x +c 2 e −2x .c2020 Peter J. Olver 3 / 4

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), wherefis an arbitrary C 1 function.

2.1.3. (a)u(t, x) =f(t);⋆(c)u(t, x) =tx− 1 2 t 2 +f(x); (e)u(t, x) =e −t x f(t).

2.1.5.u(t, x, y) =f(x, y) wherefis an arbitrary C 1 function of two variables. This is valid pro- vided each sliceD a,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 int, withxas a parameter. Solving this ordinary differential equation by standard methods, [20,23], the solution to the initial value problem is u(t, x) = f(x) tf(x) + 1 . Thus, iff(x)>0, then the denominator does not vanish fort≥0, and, moreover, goes to∞ast→ ∞. Therefore,u(t, x)→0 ast→ ∞.(b) Iff(x)<0, then the denominator in the preceding solution formula vanishes when t=τ=−1/f(x). Moreover, fort < τ, the numerator is negative, while the denominator is positive, and so lim t→τ − u(t, x) =−∞.(c) The solution is defined for 0< t < t ⋆ , wheret ⋆ =−1/minf(x). In particular, if minf(x) =−∞, then the solution is not defined for allx∈Rforanyt >0.♦2.1.9. It suffices to show that, given two points (t 1 , x),(t 2 , x)∈D, thenu(t 1 , x) =u(t 2 , x). By the assumption, (t, x)∈Dfort 1 ≤t≤t 2 , and sou(t, x) is defined and continuously differentiable at such points. Thus, by the Fundamental Theorem of Calculus, u(t 2 , x)−u(t 1 , x) = Z t2 t1 ∂u ∂t (s, x)ds= 0.Q.E.D.

2.2.2. (a)u(t, x) =e −(x+3t) 2 t= 1t= 2t= 3 c2020 Peter J. Olver

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