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
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.