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Since F is source free, there is a function g (x, y) g (x, y) with g y = P g y = P and − g x = Q − g x = Q Therefore, F = 〈 g y, − g x 〉 F = 〈 g y, − g x 〉 and div F = g y x − g x y = 0 div F = g y x − g x y = 0 by Clairaut's theorem The converse of Divergence of a SourceFree Vector Field is true on simply connectedA Fact Sheet For Mental Health Care Professionals https//ssagov/disability/professionals/mentalhealthproffactshtm 2/4 MedicaidMore precisely, is there an undirected graph X with a vertex v such that if we construct a graph G 2(X) by taking two copies of X and connecting their v vertices by an edge, and a graph G 1(X) by adding two new vertices to X and Standard Logical Equivalences Ppt Download "¯Œ^ ƒxƒŠ[ƒVƒ‡[ƒg ƒp[ƒ} ƒƒ"ƒY