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 ) u± = 0 . Example 1.1.1 implies that there are smooth F± so that (1.1.6) u± = 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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