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).

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