pp. 38-41 Sobolev

We now substantiate that last statement by making a *nonlinear*
change of variables to
transform

into canonical form, if **L** is parabolic or hyperbolic. (The case of
elliptic form is complicated.)
Putting and , and supposing
everything is smooth, and
moreover that the curves const and const intersect
transversally, then we
proceed. We have

where

and

To be in canonical form we have to have . If the equation
is of parabolic type, we
must have either or . The canonical form of the
quadratic form depends on the sign of

**Detail**: the matrix is which has
characteristic polynomial

which has roots . But this is

always real, of course, and . If ,
and have opposite sign and the equation is
hyperbolic, etc. Consider
the parabolic (non degenerate) case where or .
Assume one of **A**
or **C** is not zero (otherwise degenerate). If , then if **B
= C = 0** we are already
in canonical form. Put , which is
because .
Then

Remember, . If we can solve this first order linear PDE, we can find . We may take almost arbitrarily, so long as which is equivalent to the curve const intersecting const tangentially:

implies the curves are tangential .

The hyperbolic case, p. 42.

Wed Feb 4 14:46:04 EST 1998