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Solutions to Exercises

2
(c) The initial curve (s, 0, s ) is a characteristic curve (see the characteristic equa-
2
2
tions). Thus, there exist infinitely many solutions: u(x, y) = x +ky ∀k ∈ R.

Exercise 49. The Jacobian satisfies J ≡ 0. Since u ≡ 0 is a solution of the problem,
there exist infinitely many solutions. To compute other solutions, define a new
1
2
Cauchy problem such as (y + u)u + yu = 0, u(x, 1) = x − . Now the Jacobian
x
y
2
1
1 2t
1
satisfies J ≡ 1. The parametric form of the solution is x(t, s) = (s− )t+ e +s− ,
2
2
2
t
1
y(t, s) = e , u(t, s) = s − . It is convenient in this case to express the solution as
2 2
a graph of the form x(y, u) = y + u ln y + u.
2

Exercise 50. u(x, y) = y 1−y x/y−y .
x−y 2

Exercise 57. u = (x − ct)/[1 + t(x − ct)].

Exercise 61. General solution has form: (a) z = cy/b + f(bx − ay) (b 6= 0).
2
2
(b) z = e cx/a f(bx − ay) (a 6= 0). (c) z = f(x + y ). (d) z = e −x 2 f(y − x).
−1
(e) z = xf(xy). (f) z −1 = x −1 + f(x −1 − y ).

2
Exercise 62. z = (x + y) .

√
x y + f(y). (c) z =
Exercise 63. (a) z = x + yf(x − y) + I. (b) z = 1 2 2
2
xy + f(y/x).

y
Exercise 68. a) hyperbolic; ξ = xy; η = ; u ξη = 0;
x
y
b) parabolic; ξ = y tan ; η = y (arbitrary function); u ηη = 0;
2
c) elliptic; ξ = y + x; η = x; u + u ηη = 0.
ξξ

Exercise 71. (a) D = 3, so it’s hyperbolic;.(b) parabolic.

Exercise 74. α = 1, β = 4, γ = 148.

Exercise 76. Respectively, |k| < 2, = 2, > 2.

2
2
Exercise 77. Respectively in, on, ot outside the unit circle x + y = 1.

Exercise 78. We establish:

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