1.6. The law of reflection 35 The method of images computes the reflection. Define v to be the reversed mirror image solution, v (t, x1,x2,...,xd) := −u (t, −x1,x2,...,xd) . The solution of the Dirichlet problem is then equal to the restriction of u + v to {x1 0}. Then v = ei(˜ −|ξ|t)/ h(˜−tξ/|ξ|)+h.o.t. = ei(x˜−|˜|t)/ (−Rh)(x−t˜ |˜|)+h.o.t . To leading order, u + v is equal to (1.6.6) ei(xξ−t|ξ|)/ h(x tξ/|ξ|) ei(x˜−|ξ|t)/ (Rh)(x |˜|) . The wave represented by u has a leading term which moves with ve- locity ξ/|ξ|. The wave corresponding to v has a leading term with velocity ˜ |˜| that comes from ξ/|ξ| by reversing the first component. At the bound- ary x1 = 0, the tangential components of ξ/|ξ| and ˜ |˜| are equal and their normal components are opposite. The directions are related by the standard law that the angle of incidence equals the angle of reflection. The amplitude of the reflected wave v on the reflected ray is equal to −1 times the ampli- tude of the incoming wave u on the incoming wave. This is summarized by the statement that the reflection coefficient is equal to −1. Suppose that t,x is a point on the boundary and O is a neighborhood of size large compared to the wavelength and small compared to the scale on which h varies. Then, on O, the solution is approximately equal to ei(xξ−t|ξ|)/ h(x tξ/|ξ|) ei(x˜−t|˜|)/ ˜(x |˜|) . This recovers the reflected plane waves of §1.6.2. An observer on such an in- termediate scale sees the structure of the plane waves. Thus, even though the plane waves are completely nonlocal, the asymptotic solutions of geometric optics shows that they predict the local behavior at points of reflection. The method of images also solves the Neumann boundary value problem in a half-space using an even mirror reflection in x1 = 0. It shows that for the Neumann condition, the reflection coefficient is equal to 1. Proposition 1.6.4. i. If u C∞(R1+d) is an even solution of u = 0, then its restriction to {x1 0} is a smooth solution of u = 0 satisfying the Neumann boundary condition (1.6.7) ∂1u|x 1 =0 = 0. ii. Conversely, if u C∞({x1 0}) is a smooth solution of u = 0 satisfying (1.6.8), then the even extension of u to R1+d is a smooth even solution of u = 0.
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