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Solutions to Exercises

(a) The equation is parabolic. The required transformation is y = t, x = s−t .
3
(3x+y)y 4 y 5
(b) u(x, y) = − + yφ(3x + y) + ψ(3x + y).
324 540
(3x+y)y 4 y 5 y 1 y
(c) u(x, y) = − + y cos(x + ) − cos(x + ) + sin(x + y/3).
324 540 3 3 3

Exercise 80. We obtain:
1
(a) Writing w(s, t) = u(x, y), the canonical form is w + w = 0.
st
t
4
(b) Using W := w , the general solution is u(x, y) = f(y − 4x)e −y/4 + g(y), for
t
2
arbitrary functions f, g ∈ C (R).
(c) u(x, y) = (−y/2 + 4x)e −y/4 .

Exercise 83. Considering the equation we deduce
(a) The equation is hyperbolic for q > 0, i.e. for y > 1. The equation is elliptic for
q < 0, i.e. for y < −1. The equation is parabolic for q = 0, i.e. for |y| ≤ 1.
0 2
(b) The characteristics equation is (y ) − 2y + (1 − q) = 0; its roots are y 0 =
√ 1,2
−1 ± q.
0
(a) The hyperbolic regime y > 1. There are two real roots y 1,2 = 1 ± 1. The
solutions of the ODEs are y = constant, y = 2x+constant. Hence the
1
2
new variables are s(x, y) = y and t(x, y) = y − 2x.
0
(b) The elliptic regime y < −1. The two roots are imaginary: y 1,2 = 1 ± i.
0
Choose one of them, y = 1 + i, to obtain y = (1 + i)x + constant. The
new variables are s(x, y) = y − x, t(x, y) = x.

(c) The parabolic regime |y| ≤ 1. There is a single real root y = 1; The
solution of the resulting ODE is y = x + constant. The new variables are
s(x, y) = x, t(x, y) = x − y.

Exercise 92. u = g[(l − x)u ]x where l is the length of the chain.
x
tt

Exercise 93. u = k/cρ)u −(µP/cρA)(u−T ), where P is the perimeter of the
0
t
xx
crosssection, A is its area, and µ is the conductance across the contact surface.
Exercise 94. u = ku xx + V u .
x
t

Exercise 95. u = ku xx − V u .
t
x

Exercise 99. We show
2
(a) c = v ± α where α = T/ρA;
(b) u(x, t) = a cos[k(x − vt)] cos(kαt) − (va/α) sin[k(x − vt)] sin(kαt).

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