Solve $y''(x)=[a(x^2-1)^2+b]y(x)$

Question

While I was trying to solve a problem, I've found an equation like $y''(x)=[a(x^2-1)^2+b]y(x)$. I've tried everthing I know (like Riccati's algorithm and homogenous treatment, obtaining $u'(x)=a(x^2-1)^2+b-u(x)^2$).

However, none of the methods works (I've also tried Wolfram Mathematica with both expressions). I guess the solution is related to some polynomials (like Hermite polynomials are related to simple quantum oscillator), but I don't know the name of this polynomials (maybe they have never been studied before).

Thanks

PS: I know that $y(x)$ must be symmetric ($y(x)=y(-x)$) and finite $y(\infty)\to 0$. I also know that the "ground state" of this function (is a quantum mechanics problem) is something like two gaussian distribution added together (with $\mu=\pm 1$).

Why this function? I've tried and it doesn't work. Is $A$ a function of $x$ ($A=A(x)$) or just a constant? — MatMorPau22, Apr 20 '18 at 18:33
in that limit $y"(x) \approx \left(-2ax^2 + (a+b)\right)y(x)$ — phdmba7of12, Apr 20 '18 at 20:47
Could you write something about the quantum mechanics problem, to verify that your equation is correct? As it is, the hull curves of the basis solutions appear to be something like $\exp(\pm \sqrt{a}x^3/3)$, where both variants are unlimited in one direction. — Lutz Lehmann, Apr 21 '18 at 09:15
Is the double well potential, trying to solve Schrödinger equation with potential $V(x)=(x^2-1)^2$. — MatMorPau22, Apr 21 '18 at 17:52

score 1 · Answer 1 · answered Aug 30 '18 at 12:56

Hint:

$y''(x)=(a(x^2-1)^2+b)y(x)$

$y''(x)-(ax^4-2ax^2+a+b)y(x)=0$

Let $y(x)=e^{nx^3}u(x)$ ,

Then $y'(x)=e^{nx^3}u'(x)+3nx^2e^{nx^3}u(x)$

$y''(x)=e^{nx^3}u''(x)+3nx^2e^{nx^3}u'(x)+3nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)=e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)$

$\therefore e^{nx^3}u''(x)+6nx^2e^{nx^3}u'(x)+(9n^2x^4+6nx)e^{nx^3}u(x)-(ax^4-2ax^2+a+b)e^{nx^3}u(x)=0$

$u''(x)+6nx^2u'(x)+((9n^2-a)x^4+2ax^2+6nx-a-b)u(x)=0$

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

$u''(x)+2\sqrt ax^2u'(x)+(2ax^2+2\sqrt ax-a-b)u(x)=0$

Let $u(x)=e^{kx}v(x)$ ,

Then $u'(x)=e^{kx}v'(x)+ke^{kx}v(x)$

$u''(x)=e^{kx}v''(x)+ke^{kx}v'(x)+ke^{kx}v'(x)+k^2e^{kx}v(x)=e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)$

$\therefore e^{kx}v''(x)+2ke^{kx}v'(x)+k^2e^{kx}v(x)+2\sqrt ax^2(e^{kx}v'(x)+ke^{kx}v(x))+(2ax^2+2\sqrt ax-a-b)e^{kx}v(x)=0$

$v''(x)+2(\sqrt ax^2+k)v'(x)+((2\sqrt ak+2a)x^2+2\sqrt ax+k^2-a-b)v(x)=0$

Choose $k=-\sqrt a$ , the ODE becomes

$v''(x)+2\sqrt a(x^2-1)v'(x)+(2\sqrt ax-b)v(x)=0$

Which relates to Heun's Triconfluent Equation.

score 0 · Answer 2 · answered Apr 20 '18 at 20:06

0

ran some numerical integration on this ... looks like the "period" of oscillations is decreasing exponentially while the envelope amplitude is exponentially increasing.

answered Apr 20 '18 at 20:06

phdmba7of12

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Solve $y''(x)=[a(x^2-1)^2+b]y(x)$

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