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Lecture 3, Section 2

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.


Robert Corless
Wed Feb 4 14:46:04 EST 1998