How to solve this Sturm Liouville problem?

Question

$\dfrac{d^2\phi}{dx^2} + (\lambda - x^4)\phi = 0$

Would really appreciate a solution or a significant hint because I could find anything that's helpful in my textbook. Thanks!

Answer 1

Hint:

Let $\phi=e^{ax^3}y$ ,

Then $\dfrac{d\phi}{dx}=e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}y$

$\dfrac{d^2\phi}{dx^2}=e^{ax^3}\dfrac{d^2y}{dx^2}+3ax^2e^{ax^3}\dfrac{dy}{dx}+3ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y=e^{ax^3}\dfrac{d^2y}{dx^2}+6ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y$

$\therefore e^{ax^3}\dfrac{d^2y}{dx^2}+6ax^2e^{ax^3}\dfrac{dy}{dx}+(9a^2x^4+6ax)e^{ax^3}y+(\lambda-x^4)e^{ax^3}y=0$

$\dfrac{d^2y}{dx^2}+6ax^2\dfrac{dy}{dx}+((9a^2-1)x^4+6ax+\lambda)y=0$

Choose $9a^2-1=0$ , i.e. $a=\dfrac{1}{3}$ , the ODE becomes

$\dfrac{d^2y}{dx^2}+2x^2\dfrac{dy}{dx}+(2x+\lambda)y=0$

Let $t=bx$ ,

Then $b^2\dfrac{d^2y}{dt^2}+\dfrac{2t^2}{b}\dfrac{dy}{dt}+\left(\dfrac{2t}{b}+\lambda\right)y=0$

$\dfrac{d^2y}{dt^2}+\dfrac{2t^2}{b^3}\dfrac{dy}{dt}+\left(\dfrac{2t}{b^3}+\dfrac{\lambda}{b^2}\right)y=0$

Choose $b^3=2$ , i.e. $b=\sqrt[3]2$ , the ODE becomes

$\dfrac{d^2y}{dt^2}+t^2\dfrac{dy}{dt}+\left(t+\dfrac{\lambda}{\sqrt[3]4}\right)y=0$

Which relates to Heun's Triconfluent Equation.