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Example 1. .10
 |
(1.15) |
with the initial condition
The characteristic equations are
and
and rearranging we obtain
Thus,
On using
, (1.15) is now written as
where the constant
is actually constant along the characteristic
defined by the value of
. Thus,
is really a function of
and we write
since
and the function
is determined by the initial
conditions. At
, we have
and
so
that
. Replacing
by the above expression
involving
and
, we obtain the final solution
Example 1. .11Consider the initial-boundary-value problem
 |
(1.16) |
with
Characteristic equations are
 |
(1.17) |
and
and (1.16) gives
Hence,
since
is constant along the characteristic and dx/dr is the
derivative of
along the characteristic curve. Therefore, we have
Thus, substituting for
into
we obtain the
implicit solution
where
is determined by the initial condition, namely,
,
and
. Thus,
and so
Squaring both sides, we can manipulate this solution into an explicit
solution for
in terms of
and
.
and so the final solution is
We can easily check that this is the solution to (1.16) by
calculating the partial derivatives,
Therefore,
Next: Partial Differential Equations of
Up: Method of Characteristics
Previous: Initial-boundary-value Problems
Prof. Alan Hood
2000-10-30