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) ∂t,x
β
u = 0 , ∂t,x
β
∂1u = 0 .
The partial differential equations then imply that in x1 0 and x1 0,
respectively, one has
∂2u
∂x1
2
=
∂2u
∂t2

d
j=2
∂2u
∂xj
2
,
∂2u
∂x1
2
=
1
c2
∂2u
∂t2

d
j=2
∂2u
∂xj2
.
Therefore at the boundary
∂2u
∂x1
2
= 1
1
c2
∂2u
∂t2
,
the second derivative ∂1
2u
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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