1.1. The method of characteristics 3 Example 1.1.1. If c R is constant, then u C∞([0,T] × R) satisfies (1.1.2) ∂tu + c ∂xu = 0 if and only if there is an f C∞(R) so that u = f(x ct). The function f is uniquely determined. Proof. For constant c the characteristics along which u is constant are the lines (t, x + ct). Therefore, u(t, x) = u(0,x ct) proving the result with f(x) := u(0,x). This shows that the Cauchy problem consisting of (1.1.2) together with the initial condition u|t=0 = f is uniquely solvable with solution f(x ct). The solutions are waves translating rigidly with velocity equal to c. Exercise 1.1.1. Find an explicit solution formula for the solution of the Cauchy problem ∂tu + c ∂xu + z(t, x)u = 0, u|t=0 = g, where z C∞. Example 1.1.2 (D’Alembert’s formula). If c R \ 0, then u C∞([0,T] × R) satisfies (1.1.3) utt c2 uxx = 0 if and only if there are smooth f, g C∞(R) so that (1.1.4) u = f(x ct) + g(x + ct) . The set of all pairs ˜ ˜ so that this is so is of the form ˜ = f + b , ˜ = g b with b C. Proof. Factor ∂2 t c2∂2 x = (∂t c∂x) (∂t + c∂x) = (∂t + c∂x) (∂t c∂x) . Conclude that u+ := ∂tu c ∂xu and u− := ∂tu + c ∂xu satisfy (1.1.5) ( ∂t ± c ∂x ) = 0 . Example 1.1.1 implies that there are smooth so that (1.1.6) = F±(x ct) . In order for (1.1.4) and (1.1.6) to hold, one must have (1.1.7) F+ = (∂t c∂x)u = (1 + c2)f , F− = (∂t + c∂x)u = (1 + c2)g .
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