28 1. Simple Examples of Propagation
In the limit as 0, one finds that both ∂tu and ∂xu are O(1/) while
u = O(1) is negligibly small in comparison. Dropping this small term leads
to the simplified equation for an approximation v ,
∂tv + ∂xv = 0 , v
= a(x) cos(x/) .
The solution is
v (t, x) = a(x t) cos
(x t)/
which misses the exponential decay. It is not a good approximation except
for t 1. The two large terms compensate so that the small term is not
negligible compared to their sum.
1.6. The law of reflection
Consider the wave equation u = 0 in the half-space R−
:= {x1 0}.
At {x1 = 0} a boundary condition is required. The condition encodes the
physics of the interaction with the boundary.
Since the differential equation is of second order, one might guess that
two boundary conditions are needed as for the Cauchy problem. An anal-
ogy with the Dirichlet problem for the Laplace equation suggests that one
condition is required.
A more revealing analysis concerns the case of dimension d = 1. D’Alem-
bert’s formula shows that at all points of space-time, the solution consists
of the sum of two waves, one moving toward the boundary and the other
toward the interior. The waves approaching the boundary will propagate to
the edge of the domain. At the boundary one does not know what values
to give to the waves which move into the domain. The boundary condition
must give the value of the incoming wave in terms of the outgoing wave.
That is one boundary condition.
= (∂t ∂x)(∂t + ∂x) = (∂t + ∂x)(∂t ∂x)
shows that (∂t ∂x)(ut + ux) = 0, so ut + ux is transported to the left.
Similarly, ut ux moves to the right. Thus from the initial conditions,
ut ux is determined everywhere in x 0, including the boundary x = 0.
The boundary condition at {x = 0} must determine ut +ux. The conclusion
is that half of the information needed to find all the first derivatives is already
available and one needs only one boundary condition.
Consider the Dirichlet condition,
(1.6.1) u(t, x)
= 0 .
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