18 2. FOUR IMPORTANT LINEAR PDE
2.1. TRANSPORT EQUATION
One of the simplest partial differential equations is the transport equation
with constant coefficients. This is the PDE
(1) ut + b · Du = 0 in
Rn
× (0, ∞),
where b is a fixed vector in
Rn,
b = (b1,... , bn), and u :
Rn
× [0, ∞) → R
is the unknown, u = u(x, t). Here x = (x1,... , xn) ∈
Rn
denotes a typical
point in space, and t ≥ 0 denotes a typical time. We write Du = Dxu =
(ux1 , . . . , uxn ) for the gradient of u with respect to the spatial variables x.
Which functions u solve (1)? To answer, let us suppose for the moment
we are given some smooth solution u and try to compute it. To do so, we
first must recognize that the partial differential equation (1) asserts that a
particular directional derivative of u vanishes. We exploit this insight by
fixing any point (x, t) ∈
Rn
× (0, ∞) and defining
z(s) := u(x + sb, t + s) (s ∈ R).
We then calculate
˙(s) z = Du(x + sb, t + s) · b + ut(x + sb, t + s) = 0
·
=
d
ds
,
the second equality holding owing to (1). Thus z(·) is a constant function of
s, and consequently for each point (x, t), u is constant on the line through
(x, t) with the direction (b, 1) ∈
Rn+1.
Hence if we know the value of u at
any point on each such line, we know its value everywhere in
Rn
× (0, ∞).
2.1.1. Initial-value problem.
For definiteness therefore, let us consider the initial-value problem
(2)
ut + b · Du = 0 in
Rn
× (0, ∞)
u = g on
Rn
× {t = 0}.
Here b ∈
Rn
and g :
Rn
→ R are known, and the problem is to compute
u. Given (x, t) as above, the line through (x, t) with direction (b, 1) is
represented parametrically by (x + sb, t + s) (s ∈ R). This line hits the
plane Γ :=
Rn
× {t = 0} when s = −t, at the point (x − tb, 0). Since u is
constant on the line and u(x − tb, 0) = g(x − tb), we deduce
(3) u(x, t) = g(x − tb) (x ∈
Rn,t
≥ 0).
So, if (2) has a sufficiently regular solution u, it must certainly be given
by (3). And conversely, it is easy to check directly that if g is
C1,
then u
defined by (3) is indeed a solution of (2).
