Does the nonlinear WKB equation always have eigenfunctions?

[eljose]

SE equation in (1d) $$a\frac{d^{2}\phi}{dx^{2}}+V(x)\phi=E_{n}\phi$$ (1)

WKB approach $$\phi=exp(iW/\hbar)$$ (2)

$$W=[2m(E_{n}-V)]^{1/2}$$

then from (2) we can obtain the potential and introduce it in (1) to get a differential NOn-linear equation...

$$a\frac{d^{2}\phi}{dx^{2}}+(b/phi)^{2}(\frac{d\phi}{dx})^{2}+c\phi=0$$ we could use this to prove that always the eigenfunction of the Hamiltonian $$\phi$$ will exist...

 

[eljose]

$$a\frac{d^{2}\phi}{dx^2}+V(x)\phi-E_{n}\phi=0$$

$$\phi=exp(iW/\hbar)$$

combinign both and substituting the potential of V as a function of $$\phi$$ we could form the equation:

$$a\frac{d^{2}\phi}{dx}+(b/\phi)^{2}(\frac{d\phi}{dx})^{2}+c\phi=0$$ or if we call $$a=-\hbar^{2}/2m$$

$$a(\frac{d^{2}\phi}{dx}+(1\phi)(\frac{d\phi}{dx})^{2})=0$$

EDIT:sorry it would be...$$a(\frac{d^{2}\phi}{dx}+(1/\phi)(\frac{d\phi}{dx})^{2})=0$$

from this equation and applying existence theorem we would get that the eigenfunctions $$\phi$$ exist so the potential V(x) will always exist for the WKB case...