1.5. A cautionary example in geometric optics 27 The sin t|ξ|/|ξ| factor is ≤ |t|. For small, the distance of e1/ to the support of χ is ≥ C/. Therefore, χ(ξ) sin t|ξ| |ξ| 1 ξ − e1/ s L∞(Rd) ξ ≤ Cs |t| s , 0 ≤ 1 . It follows that χ(ξ) sin t|ξ| |ξ| 1 ξ − e1/ s ks(ξ − e1/) L2(Rd) ≤ Cs |t| s ks L2(Rd) , with s arbitrarily large. The small frequency contribution is negligible in the limit → 0. It is removed with a cutoff as above, and then the analysis away from ξ = 0 proceeds by decomposition into plane wave as in the case with ut(0) = 0. It yields left and right moving waves with the same phases as before. Exercise 1.4.3. Solve the Cauchy problem for the anisotropic wave equa- tion, utt = uxx + 4uyy with initial data given by u (0,x) = γ(x) eixξ/ , u t (0,x) = 0 , γ ∈ s Hs(Rd) . Find the leading term in the approximate solution to u+. In particular, find the velocity of propagation as a function of ξ. Discussion. The velocity is equal to the group velocity from §1.3. 1.5. A cautionary example in geometric optics A typical science text discussion of a mathematics problem involves simplify- ing the underlying equations. The usual criterion applied is to ignore terms which are small compared to other terms in the equation. It is striking that in many of the problems treated under the rubric of geometric optics, such an approach can lead to completely inaccurate results. It is an example of an area where more careful mathematical consideration is not only useful but necessary. Consider the initial value problems ∂tu + ∂xu + u = 0 , u t=0 = a(x) cos(x/) in the limit → 0. The function a is assumed to be smooth and to vanish rapidly as |x| → ∞, so the initial value has the form of a wave packet. The initial value problem is uniquely solvable, and the solution depends continuously on the data. The exact solution of the general problem ∂tu + ∂xu + u = 0 , u t=0 = f(x) , is u(t, x) = e−t f(x − t), so the exact solution u is u (t, x) = e−t a(x − t) cos((x − t)/) .

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