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u
ξη
=
1
6(
η
−
ξ
)
(
u
η
−
u
ξ
)
The parabolic case is easier, the only characteristic is
y
=
1
2
x
+
K
and so the transformation is
ξ
=
y
−
1
2
x
η
=
x
The last equation is an arbitrary function and one should check the Jacobian. The details
are left to the reader. One can easily show that
A
∗
=
B
∗
=0
Also
C
∗
=1
and the rest of the coeﬃcients are zero. Therefore the equation is
u
ηη
=0
In the elliptic case, one can use the transformation
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.
 Fall '08
 BELL,D

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