AIDJEX Bulletin #40 - Polar Science Center - University of Washington

q1 = -A2
q2 =
A1
Therefore, the vectors
and e2 are
R2 = [11]
(74)
Making use of the zero terms in coefficients, equations along stress characteristics
are found by substituting for R and sa, a = 1,2 into equation
-a
(66) neglecting the advection. They become
A similar analysis of equations along a ueZocity characteristic given
by a = 3,4 requires that eigenvectors
-a
R
and (68) with
Then, since ca'is nonsingular, we require that
be determined from equations (67)
det ca = 0 det ?& # 0 (76)
R = O
-a -
(77)
But S can be found from (67) by solving the linear algebraic equations. It
-a
can be shown that
where the first component has arbitrarily been chosen to be unity. Again
making use of zero coefficients and neglecting advection, the governing
equations (66) become
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