Here is the outline of a proof that L [see Equation (3)] is not selfadjoint if b(x)=a′(x). Observe first that if b(x)=a′(x) and we can choose u and v so that
v(x)u′(x)−u(x)v′(x)=b(x)−a′(x),
then the integral in Equation (10) reduces to
∫αβ[b(x)−a′(x)]2dx=0.
The only problem is that u and v have to satisfy the boundary conditions (3'). To modify the proof,
(a) Observe that
v(x)u′(x)−u(x)v′(x)=v2(x)(vu)′.
(b) Choose u, v, and constants c1,c3,…,ck+1 (with k to be determined) so that
(vu)′=c1[b(x)−a′(x)]+c3[b(x)−a′(x)]3+c5[b(x)−a′(x)]5+⋯,u and v satisfy the boundary conditions (3'), and the integral in Equation (10) is =0.