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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



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