4 1. Simple Examples of Propagation Thus if are primitives of that vanish at the origin, then one must have f = G+ (1 + c2) + C+, g = G+ (1 + c2) + C− , C+ + C− = u(0, 0) . Reversing the process shows that if G+, f, g are defined as above, then ˜ := f(x ct) + g(x + ct) yields a solution of D’Alembert’s equation with (∂t c ∂x)˜ = so (∂t c ∂x)(u ˜) = 0 . Adding and subtracting this pair of equations shows that ∇t,x(u ˜) = 0. Since u(0, 0) = ˜(0, 0), it follows by connectedness of [0,T] × R that u = ˜, and the proof is complete. For speeds c(t, x) that are not bounded, it is possible that characteristics escape to infinity with interesting consequences. Example 1.1.3 (Nonuniqueness in the Cauchy problem). Consider c(t, x):= x2. The characteristic through (0,x0) is the solution of x = x2, x(0) = x0 . Then, 1 = x x2 = d dt −1 x . Integrating from t = 0 yields −1 x(t) −1 x0 = t and, therefore, x(t) = x0 1 x0 t . Through each point t, x with t 0 there is a unique characteristic tracing backward to t = 0. Therefore, given initial data u(0,x) = g(x), the solution u(t, x) is uniquely determined in t 0 by requiring u to be constant on characteristics.
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