40 1. Simple Examples of Propagation

From section 1.6.3, the leading amplitude d0 must be constant on the

rays t → (t , x + c t η/|η|). To determine d0, it suﬃces to know the values

d0(t,

0+,x

) at the interface. One could choose d0 to guarantee the continuity

of u or of ∂1u, but not both. One cannot construct a good approximate

solution consisting of just an incident and transmitted wave.

Add to the recipe a reflected wave. Seek a reflected wave in x1 ≥ 0 in

the form

R ∼

ei(xζ+tσ)/

b(, t, x) , b(, t, x) ∼ b0(t, x) + b1(t, x) + · · · .

In order that the reflected wave oscillate with the same phase as the incident

wave in the interface x1 = 0, one must have ζ = ξ and σ = −1. To satisfy

the wave equation in x1 0 requires

σ2

=

|ζ|2.

Together these imply

ζ1

2

= ξ1.

2

To have propagation away from the boundary requires ζ1 = −ξ1 so

ζ =

˜.

ξ Therefore,

(1.7.7) R ∼

ei(x˜−t)/

ξ

b(, t, x) , b(, t, x) ∼ b0(t, x)+ b1(t, x)+ · · · .

Summarizing, seek

v

=

I + R in x1 0,

T in x1 0.

The continuity required at x1 = 0 forces

(1.7.8)

ei(x ξ −t)/

(

a(, t, 0,x ) + b(, t, 0,x )

)

=

ei(x ξ −t)/

d(, t, 0,x ) .

The continuity of u and ∂1u hold if and only if at x1 = 0 one has

(1.7.9) a + b = d , and

iξ1

a + ∂1a −

iξ1

b + ∂1b =

iη1

d + ∂1d .

The first of these relations yields

(1.7.10) aj + bj − dj

x1=0

= 0, j = 0, 1, 2,....

The second relation in (1.7.9) is expanded in powers of . The coeﬃcients

of

j

must match for all all j ≥ −1. The leading order is

−1

and yields

(1.7.11)

(

a0 − b0 − (η1/ξ1)d0

)

x1=0

= 0 .

Since a0 is known, the j = 0 equation from (1.7.10) together with (1.7.11)

is a system of two linear equations for the two unknown b0,d0,

−1 1

1 η1/ξ1

b0

d0

=

a0

a0

.

Since the matrix is invertible, this determines the values of b0 and d0 at

x1 = 0.

The amplitude b0 (resp., d0) is constant on rays with velocity

˜

ξ (resp.,

cη/|η|). Thus the leading amplitudes are determined throughout the half-

spaces on which they are defined.