robustify RBF by always resorting to least squares

            M[:,j]=Phi(d/r)

        op=M

        del M

        try:

            ab=np.linalg.solve(op,rhs)

        except np.linalg.LinAlgError:

            #print('Bad conditioning of RBF operator, using least square solution')

            ab=np.linalg.lstsq(op,rhs,cond=10**(9))

        ab=np.linalg.lstsq(op,rhs,rcond=10**(9))[0]

        del op

        alpha=ab

        ds=np.zeros((nb_nodes,mesh.GetDimensionality()))