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

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SOLUTIONS 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 WeakSolutions . . . . . . . . . . . 99

Chapter 11. Dynamics of Planar Media . . . . . . . . . . . . . . . 110

Chapter 12. Partial Differential Equations in Space . . . . . . . . . . 126




c 2020 Peter J. Olver

, SelectedSolutionsto
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.
u
1.2. (a) (i ) 2u ∂2 = 0, (ii ) u +u
+ = 0;
∂x2 ∂y 2 xx yy
∂u ∂ 2 2
u +u .
⋆ (c) (i ) = u ∂
+
, (ii ) u = u
∂t ∂x2 ∂y2 t xx 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.

∂ 2 2u
1.5. (a) u ∂ = ex cos y −ex cos y = 0; defined and C∞ on all of R 2 .
2 + 2
∂x2 ∂y
2
∂ u
⋆ (c) u ∂
+
= 6 x − 6 x = 0; defined and C∞ on all of R 2 .
∂x2 2 ∂y2 2 2 2 2 2
∂ u ∂ u 2y − 2x 2x − 2 y
⋆ (d)
∂x2
+
∂y 2
= ∞
2
\ {0}.
2 + y2 ) 2 + 2 + y2 )2 = 0; defined and C on R
h
(x i
(x
1.7. u = log c (x − a)2 + c (y − b)2 , for a, b, c arbitrary constants.
+ c1 x + c2 y + c3 z + c4 (x2 − y 2) + c5 (x2 − z 2) + c6 xy + c7 xz + c8 yz,
⋆ 1.8. (a) c 0
where c0, . . . , c8 are arbitrary constants.
∂ u ∂ u
1.10. (a) 2
u = 8 − 8 = 0; ⋆ (c)
2
u = −4 sin 2 t cos x + 4 sin 2 t cos x = 0.
∂ ∂
2 2
−4 −4
∂t2 ∂x2 ∂t2 ∂x2
2 2
1.11. (a) c0 + c1t + c 2x + c3(t + x ) + c 4 tx, where c0, . . . , c4 are arbitrary constants.
b b , where a, b are arbitrary constants.
⋆ 1.13. u = a + = a +
q
r x2 + y 2 + z 2




c 2020 Peter J. Olver

, 2 Chapter 1: Selected Solutions


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

⋆ 1.16. When (x, y) 6= (0, 0), a direct4 computation
2 2 4
shows that
4 2 2 4
∂u y (x + 4 x y − y ) ∂u x(x − 4 x y − y )
= , = ,
∂x 2 2
(x + y ) 2 ∂y (x2 + y2) 2
while, from the definition of partial derivative,
∂u
(0, 0) = lim u(h, 0) − u(0, 0) ∂u u(0, k) − u(0, 0)
∂ = 0, (0, 0) = lim = 0.
h→ 0 h ∂y k→ 0 k
x
Thus,
∂2 u ux(0, k) − ux(0, 0) = −1.
∂ 2u uy(h, 0) − uy(0, 0) = 1,
(0, 0) = lim (0, 0) = lim
∂x ∂y h→ 0 h ∂y ∂x k→ 0 k
This does not contradict the equality of mixed partials because the theorem
requires continuity, while
2
∂2u = ∂ u x6 + 9 x 4 y 2 + 9 x 2 y 4 − y 4 , (x, y) 6= (0, 0),
= 2 + y 2 )3
∂x ∂y ∂x ∂y (x
is not continuous at (x, y) = (0, 0). Indeed,
2
lim ∂ u (h, 0) = 1 6= −1 = lim ∂2 u
(0, k).
h → 0 ∂x ∂y k → 0 ∂x ∂y




1.17. (a) homogeneous linear; (d) nonlinear; ⋆
(f ) inhomogeneous linear.
∂ 2u ∂ u2
1.19. (a) (i ) = −4 cos(x − 2 t) = 4 .
∂t2 ∂x2
⋆ 1.20. (a) cos(x − 2 t) + 1 cos x − 5 sin(x − 2 t) − 5 sin x.
4 4
∂ ∂f ∂g
1.21. (a) ∂x[ cf + dg ] = [ cf (x) + dg(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.
⋆ 1.23. 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)[ 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 ex sin x + c1 e2 x/5 cos 45x + c2 e2 x/5 sin 45x.
16 1 sin x + c e3 x + c
1.28. (b) u(x) = − x − e− 3 x ,
9 10 1 2
⋆(d) u(x) = 1 x ex − 1 ex + 1 e − x + c ex + c e− 2 x .
6 18 4 1 2




c 2020 Peter J. Olver

, SelectedSolutionsto
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) = tx − 1 t22 + f(x); (e) u(t, x) = e− tx 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 pro-
vided 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) =
tf (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

u(t2, x) − u(t1, x) = (s, x) ds = 0. Q.E.D.
t 1 ∂t




2.2.2. (a) u(t, x) = e − (x+3 t) 2




t= 1 t= 2 t= 3




c 2020 Peter J. Olver

, 4 Chapter 2: Selected Solutions



⋆(c) u(t, x) = e − t/2 tan−1(x − t)




t= 1 t= 2 t= 3
2.2.3. (b) Characteristic lines: x = 5 t + c; general solution: u(t,x) = f(x −5t);
x




t



⋆(d) Characteristic lines: x = −4 t + c; general solution: u(t, x) = e −t f (x + 4 t);
x




t



⋆ 2.2.4. u(t, x) = t + e − (x−2 t) .
2


♦ 2.2.6. By the chain rule
∂v ∂u ∂v ∂u , x) ,
(t, x) = (t − t , x) , (t, x) = (t − t
0
∂t ∂t ∂x ∂x
and hence
0
∂v ∂v ∂u ∂u , x) = 0.
(t, x) + c (t, x) = (t − t , x) + c (t − t
∂t ∂x 0
∂t ∂x

0
Moreover, v(t0, x) = u(0,x) = f(x). Q.E.D.
⋆ 2.2.9. (a) | u(t, x) | = | f(x − ct) | e− at ≤ M e− at → 0 as t → ∞ since a > 0.
(b) For example, if c ≥ a, then the solution u(t, x) = e(c−a) t−x −6 → 0 as t → ∞.




c 2020 Peter J. Olver

, Chapter 2: Selected Solutions 5
1 1
⋆ ♥ 2.2.11. (a) u(t, x) = , where h(ξ) is an arbitrary C function.
t + h(x − t)
(b) The solution to the initial value problem is
f(x − t)
u(t, x) = .
1 + t f(x − t)
If f(x) ≥ 0, then the denominator does not vanish for t ≥ 0, and hence the solution
y is strictly
exists for t > 0. Moreover, for fixed t > 0, the function g(y) =
1 + ty
M
increasing for y ≥ 0. Therefore, 0 ≤ u(t, x) ≤ −→ 0 as t → ∞.
(c) Using the preceding solution formula, if 1f(x) + M< t 0, then, at the point x = x − 1/f(x),
⋆
the solution u(t, x⋆ ) → −∞ as t → τ = −1/f(x).
(d) If m = min f(x) < 0, then, by part (c), the minimal blow-up time is τ⋆ = −1/m.
(
f(x − ct), x ≥ c t,
2.2.14. (a) u(t, x) = g(t − x/c), x ≤ c t, defines a classical C1 solution provided the
compatibility conditions g(0) = f(0), g′(0) = −cf(0), hold.
(b) The initial condition affects the solution for x ≥ c t, whereas the boundary
condition affects the solution for x ≤ c t. Apart from the compatibility condition
along the characteristic line x = ct, they do not affect each other.



2.2.17. (a) u(t, x) =
1 e− 2 t
= .
(b) (xet )2 + 1 2
x +e − 2 t
t = 0: t = 1:




t = 2: t = 3:




(
1, x = 0,
(c) The limit is discontinuous: lim u(t, x) = 0, otherwise.
t→ ∞

0, x < −1, f f(−1), x < 1,
⋆ 2.2.18. (a) lim
(−1), x = −1, (b) lim f(1), x = 1,
t→ ∞ u(t, x) = t→ − ∞
u(t, x) =
f(1), x > −1. 0, x > 1.




c 2020 Peter J. Olver

, 6 Chapter 2: Selected Solutions

2.2.20. (a) The characteristic curves are given by x = tan(t + k) for k ∈ R.
x




t




(b) The general solution is u(t, x) = g(tan−1 x − t), where g(ξ) is an arbitrary C1 function of
the characteristic variable.
(c) The solution is u(t, x) = f tan(tan−1 x − t) . Observe that the solution is not defined
for x < tan t − 1 π for 0 < t < π, nor at any value of x after t ≥ π. As t increases up
to π, the wave moves
2 rapidly off to +∞ at an ever accelerating rate, and the solution
effectively disappears.
⋆ 2.2.22. If u(t, x) = v(β(x) − t), then, by the chain rule, ut = −v′(β(x) − t), while
ux = v′(β(x) − t) β′(x) = v′ (β(x) − t)/c(x), and hence ut + c(x) ux = 0. Q.E.D.
⋆ 2.2.24. If | c(x) | ≤ c⋆, then each characteristic curve exists for all t ∈ R; indeed, if | dx/dt | ≤ c⋆,
then | x(t) − x(0) | ≤ c⋆ | t |, and hence, by the existence and uniqueness theorem for
first-order ordinary differential equations, the solution x(t) is uniquely defined for all
t.
Thus, the characteristic curve through a point (t,x) ∈ R 2 intersects the x axis at,
say, the point x = x⋆. Since the solution must be constant along the characteristic,
its value u(t,x) equals the initial value f(x⋆).
⋆ ♥ 2.2.25. (a) The characteristic curves are solutions to the first-order ordinary differential equation
dx
= c(x). They cannot cross each other because of uniqueness of solutions to the
dt
initial value problem. Indeed, uniqueness says that, since c(x) ∈ C1, there is
exactly one solution x = g(t) to the initial value problem x(t0) = x0, and hence
exactly one characteristic curve going through any point (t0, x 0).
dx
(b) If x(t) = x⋆ is constant, = 0 = c(x )⋆solves the characteristic ordinary differential
dt
then
equation, and hence, by uniqueness, its graph, which is a horizontal line, is the
characteristic curve going through the point (t, x⋆).
(e) Suppose x = g(t) is not monotone, so g′(t1) = c(g(t1)) > 0 and g′(t 2) = c(g(t2)) < 0,
say. Then, by continuity, 0 = g′(t 0) = c(g(t0)) for some t0 between t1 and t2, and
hence x0 = g(t0) is a fixed point. But, this would violate uniqueness: the non-
horizontal char- acteristic curve x = g(t) would cross the horizontal characteristic
line at the fixed point
(t0, x0) = (t0, x(t0)). We conclude that each non-horizontal characteristic curve
must be the graph of a monotone function.
(f ) Because the individual points on the wave follow the monotone characteristic curves
x(t), and hence the wave speed c(x(t)) has the same sign at all times t.


c 2020 Peter J. Olver

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