1.2. Examples of propagation of singularities using progressing waves 13 The rays are the integral curves of (1.2.3) ∂t ± ∂x . Structures are rigidly transported at speeds ±1. There is an energy law. If u is a smooth solution whose support intersects each time slab a t b in a compact set, one has d dt R u2 t + u2 x dx = ∂t(ut 2 + u2) x dx = 2ut(utt uxx) + ∂x(2ut ux) dx = 0 , since the first summand vanishes and the second is the x derivative of a function vanishing outside a compact set. The fundamental solution that solves (1.2.3) together with the initial values (1.2.4) u(0,x) = 0 , ut(0,x) = δ(x) , is given by the explicit formula (1.2.5) u(t, x) = sgn t 2 χ[−t,t] = 1 2 h(x t) h(x + t) , where h denotes Heaviside’s function, the characteristic function of ]0, ∞[. Exercise 1.2.1. i. Derive (1.2.5) by solving the initial value problem using the Fourier transform in x. Hint. You will likely decompose an expression regular at ξ = 0 into two that are not. Use a principal value to justify this step. ii. The proof of D’Alembert’s formula (1.2.2) shows that every distribu- tion solution of (1.2.1) is given by (1.2.2) for f, g distributions on R. Derive (1.2.5) by finding the f, g that yield the solution of (1.2.4). Hint. You will need to find the solutions of du/dx = δ(x). The singularities of the solution (1.2.5) lie on the characteristic curves through (0, 0). This is a consequence of Theorem 1.1.4. In fact, define v as the solution of vtt vxx = 0, v(0,x) = 0, vt(0,x) = x2 + /2, x+ := max{x, 0} . Introduce V := (v1,v2,v), v1 := ∂tv ∂xv, v2 := ∂tv + ∂xv to find ∂tV + ⎝0 1 0 0 1 0 0 0 −1 ∂xV + ⎝0 0 0 1 0 0⎠ 0 0 0 V = 0 .
Previous Page Next Page