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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