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.