. O A ] 1 4 Ju l 1 99 9 An Algebraic Treatment of Totally Linear Partial Differential Equations

We construct the field generated by n algebraically independent elements, and show that the linear space of derivations over this field is faithfully represented by the linear space of the n − th fold Cartesian product of this field acting through inner product on the gradient of this field. We prove also that functional independence of a set in this field is equivalent to linear independence of the gradient set in the vector space of Cartesian product. It is also shown that every subfield S of A which is generated by (n − 1) functionally independent elements defines a one-dimensional space of derivations, such that each member L of the latter subspace has S as its kernel. Each coset of the multiplicative subgroup S defines a non-homogeneous differential operator L + q whose kernel coincides with this coset. We prove also that every element in A defines a coset of the subgroup Ker(L + q) in the additive group A, on which L + q is constant. 1. Introduction In a recent work [4] the relations between the solutions of three types of partial differential equations involving first order differential operators were obtained. It was shown that if L is a homogeneous differential operator of class C 0 on a manifold M , q is C 0 function on M then all operators of the form L + q are isomorphic to each other when acting on appropriate Hilbert spaces of functions on M. Neat relations between the solutions of the totally linear partial differential equations (i) Lφ = 0, (ii) (L + q)ψ = 0, (iii) (L + q)χ = b, where b is a C 0 function, were derived. More precisely, it was shown that if η ∈ Ker(L + q), is any particular solution of (ii) then the set of solutions of (ii) is given by Ker(L + q) = ηKerL. If ζ is any solution of (iii) then the set of solutions of (iii) is given by ζ + Ker(L + q). The set of solutions of (i) is a sub-algebra of C 1 (M), whereas the set of solutions of (ii), Ker(L + q), is a sub-space of the vector space C 1 (M). The algebraic features of these results encourages us to seek an algebraic approach to this problem. Following up this line of thoughts-we first formulate the problem in a field A generated …