On the operator equation A1XA2 − B1XB2 = Q when A1, A2, B1, B2, Q may be all unbounded

AbstractLet A1,B1ϵL(V1K) with A2θ, B2θϵV2 and Qϵ L(V2, H), where V1, V2, H, K are appropriate normed spaces. It is shown that there exists a unique solution Xϵ L(V2, H) of the equation, A1XA2φ − B1XB2φ = Qφ (for all θ V2), provided (i) V2 is spanned by an orthonormal basis {bi}∞i = 1 where bi, is an eigenvector of both A2 and B2 belonging to eigenvalues λi and [;μi] respectively, (ii) for all i, λi − μiBi: H→H has a continuous inverse defined on its range, with ∑i∥(λiA1 − μiB1)−1∥2 < ∞, and (iii) some other natural conditions are satisfied. A Galerkin method of approach is utilized and examples are given. A necessary and sufficient condition is arrived at for the finite dimensional case