2.1. TRANSPORT EQUATION 19
Weak solutions. If g is not
C1,
then there is obviously no
C1
solution of
(2). But even in this case formula (3) certainly provides a strong, and in
fact the only reasonable, candidate for a solution. We may thus informally
declare u(x, t) = g(x tb) (x
Rn,
t 0) to be a weak solution of (2), even
should g not be
C1.
This all makes sense even if g and thus u are discontin-
uous. Such a notion, that a nonsmooth or even discontinuous function may
sometimes solve a PDE, will come up again later when we study nonlinear
transport phenomena in §3.4.
2.1.2. Nonhomogeneous problem.
Next let us look at the associated nonhomogeneous problem
(4)
ut + b · Du = f in
Rn
× (0, ∞)
u = g on
Rn
× {t = 0}.
Fix as before (x, t)
Rn
× (0, ∞) and, inspired by the calculation above, set
z(s) := u(x + sb, t + s) for s R. Then
˙(s) z = Du(x + sb, t + s) · b + ut(x + sb, t + s) = f(x + sb, t + s).
Consequently
u(x, t) g(x tb) = z(0) z(−t) =
0
−t
˙(s) z ds
=
0
−t
f(x + sb, t + s) ds
=
t
0
f(x + (s t)b, s) ds,
and so
(5) u(x, t) = g(x tb) +
t
0
f(x + (s t)b, s) ds (x
Rn,
t 0)
solves the initial-value problem (4).
We will later employ this formula to solve the one-dimensional wave
equation, in §2.4.1.
Remark. Observe that we have derived our solutions (3), (5) by in effect
converting the partial differential equations into ordinary differential equa-
tions. This procedure is a special case of the method of characteristics,
developed later in §3.2.
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