Methods of Applied Mathematics Lecture Notes

1.3. VECTOR FIELDS AND DIFFERENTIAL FORMS 29The idea is that the vector field is somehow approximated by this linear vectorfield. The u and v represent deviations of the x and y from their values at thispoint. This linear differential equation can be analyzed using the Jordan formof the matrix.It is natural to ask whether at an isolated zero of the vector field coordinatescan be chosen so that the linear equation is exactly equivalent to the originalnon-linear equation. The answer is: usually, but not always. In particular,there is a problem when the eigenvalues satisfy certain linear equations withinteger coefficients, such as λ 1 + λ 2 = 0. This situation includes some of themost interesting cases, such as that of conjugate pure imaginary eigenvalues.The precise statement of the eigenvalue condition is in terms of integer linearcombinations m 1 λ 1 + m 2 λ 2 with m 1 ≥ 0, m 2 ≥ 0, and m 1 + m 2 ≥ 2. Thecondition is that neither λ 1 nor λ 2 may be expressed in this form.How can this condition be violated? One possibility is to have λ 1 = 2λ 1 ,which says λ 1 = 0. Another is to have λ 1 = 2λ 2 . More interesting is λ 1 =2λ 1 + λ 2 , which gives λ 1 + λ 2 = 0. This includes the case of conjugate pureimaginary eigenvalues.Theorem. (Sternberg linearization theorem). Suppose that a vector fieldsatisfies the eigenvalue condition at a zero. Then there is a new coordinatesystem near that zero in which the vector field is a linear vector field.For the straightening out theorem and the Sternberg linearization theorem,see E. Nelson, Topics in Dynamics I: Flows, Princeton University Press, Princeton,NJ, 1969.1.3.5 Quadratic approximation to a function at a criticalpointThe useful way to picture a function h is by its contour lines.Consider a function h such that at a particular point we have dh = 0. Thenthere is a naturally defined quadratic formd 2 h = h xx (dx) 2 + 2h xy dxdy + h yy (dy) 2 . (1.107)The partial derivatives are evaluated at the point. This quadratic form is calledthe Hessian. It is determined by a symmetric matrix[ ]hxx hH =xy. (1.108)h xy h yyIts value on a tangent vector L at the point isd 2 h(L, L) = h xx a 2 + 2h xy ab + h yy b 2 . (1.109)Theorem (Morse lemma) Let h be such that dh vanishes at a point. Let h 0be the value of h at this point. Suppose that the Hessian is non-degenerate atthe point. Then there is a coordinate system u, v such that near the pointh − h 0 = ɛ 1 u 2 + ɛ 2 v 2 , (1.110)