1.7. Snell’s law of refraction 37 We analyze the transmission condition that imposes continuity of u and ∂1u across {x1 = 0}. Seek solutions of (1.7.1) satisfying the transmission condition (1.7.2) u(t, 0−,x ) = u(t, 0+,x ) , ∂1u(t, 0−,x ) = ∂1u(t, 0+,x ) . Denote by square brackets the jump u (t, x ) := u(t, 0+,x ) u(t, 0−,x ) . The transmission condition is then u = 0 , ∂1u = 0 . For solutions that are smooth on both sides of the boundary {x1 = 0}, the transmission condition (1.7.2) can be differentiated in t or x2,...,xd to find (1.7.3) β t,x u = 0 , β t,x ∂1u = 0 . The partial differential equations then imply that in x1 0 and x1 0, respectively, one has ∂2u ∂x2 1 = ∂2u ∂t2 d j=2 ∂2u ∂x2 j , ∂2u ∂x2 1 = 1 c2 ∂2u ∂t2 d j=2 ∂2u ∂x2 j . Therefore at the boundary ∂2u ∂x2 1 = 1 1 c2 ∂2u ∂t2 , the second derivative ∂2u 1 is expected to be discontinuous at {x1 = 0}. The physical conditions for Maxwell’s equations at an air-water or air- glass interface can be analyzed in the same way. In that case, the dielectric constant is discontinuous at the interface. Define γ(x) := 1 when x1 0, c−2 when x1 0 , e(t, x) := γ ut 2 + |∇xu|2 2 . From (1.7.1) it follows that solutions suitably small at infinity satisfy ∂t x1 0 e dx = ut(t, 0−,x ) ∂1u(t, 0+,x ) dx , ∂t x1 0 e dx = ut(t, 0+,x ) ∂1u(t, 0+,x ) dx . The transmission condition guarantees that the terms on the right compen- sate exactly so ∂t R3 e dx = 0 .
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