含三次项的微分方程解析求解咨询：变换优化与超几何方程转化可行性

Hi there! Let's break down your problem step by step—first looking for better coordinate transformations to avoid the cubic term, then exploring whether we can still map the equation to a solvable hypergeometric (or related) form even with the cubic term present.

一、尝试更优的变量变换：从原始势能化简入手 #

Your initial transformation $z = e^{-\delta x}$ makes sense given the exponential terms, but let's start by simplifying the original potential in the ODE first, since messy coefficients often lead to complicated transformed equations.

First, rewrite the potential term in your original ODE by combining the fractional terms inside the $D$-bracket:
$$
1 - e^{-\delta x} + 2e^{-\delta x}\frac{\delta}{1 - e^{-\delta x}} = \frac{(1 - e^{-\delta x})^2 + 2\delta e^{-\delta x}}{1 - e^{-\delta x}} = \frac{1 - 2e^{-\delta x} + e^{-2\delta x} + 2\delta e^{-\delta x}}{1 - e^{-\delta x}}
$$
This suggests a more natural variable might be $t = 1 - e^{-\delta x}$ (instead of $z = e^{-\delta x}$), since it directly aligns with the denominator terms in the potential. Let's try this substitution:

• $dt/dx = \delta e^{-\delta x} = \delta(1 - t)$, so $dx/dt = 1/[\delta(1 - t)]$
• Compute the second derivative:
  $$
  y''(x) = \delta^2(1 - t)^2 y''(t) - \delta^2(1 - t) y'(t)
  $$
  Substitute back into the original ODE and divide through by $\delta^2(1 - t)^2$, you'll get a new ODE in $t$:
  $$
  y''(t) - \frac{1}{1 - t}y'(t) + \left[ \frac{E + 2\delta D - Dt}{\delta^2(1 - t)^2} - \frac{2D}{\delta t(1 - t)^2} - \frac{C}{\delta^2 t^2(1 - t)^2} \right] y(t) = 0
  $$
  Notice that the cubic term in $z$ translates to a linear term in $t$ here—we still end up with a polynomial numerator of degree 3, but the structure aligns better with the denominator terms, making subsequent function transformations easier.

Another quick win: absorb constants into the variable to simplify notation first. Let $t = \delta x$, so the original ODE becomes:
$$
y''(t) + \left[ \frac{E}{\delta^2} - \frac{D}{\delta^2}\left(1 - e^{-t} + \frac{2e^{-t}}{1 - e^{-t}}\right) - \frac{C}{\delta^2(1 - e^{-t})2} \right] y(t) = 0
$$
This removes $\delta$ from the exponential terms, cleaning up the algebra for any further substitutions.

二、保留三次项时的解析解：超几何/广义超几何方程转化 #

Even with the cubic term, we can still map your ODE to known solvable forms. Let's start with the transformed ODE in $z$ you derived, rewritten in standard linear ODE form:
$$
z^2(1-z)2 y''(z) + z(1-z)^2 y'(z) + \left[ A + Bz + Cz^2 + Kz^3 \right] y(z) = 0
$$
where $A = -D + E - C\delta^2$, $B = 3D - 2E - 2D\delta$, $C = -3D + E + 2D\delta$, and $K = D$ (renamed to avoid notation conflict).

Step 1: Eliminate the first derivative term #

First, use a Liouville transformation to get rid of the $y'(z)$ term—this is a standard trick for linear ODEs. For your equation, the coefficient of $y'(z)$ is $1/z$, so we choose $y(z) = z^{-1/2} u(z)$ (since $\int \frac{1}{2z} dz = \frac{1}{2}\ln z$, so $e^{-\int \frac{P(z)}{2}dz} = z^{-1/2}$).

Substituting this into the ODE eliminates the first derivative, leading to:
$$
u''(z) + \left[ \frac{A + Bz + Cz^2 + Kz^3}{z2(1-z)^2} + \frac{1}{4z^2} \right] u(z) = 0
$$
The new coefficient simplifies to a cubic numerator over $z^2(1-z)2$, which is closer to the structure of generalized hypergeometric equations (like $_pF_q$), which handle polynomial numerators in their coefficients.

Step 2: Match to special function forms #

Your ODE has two finite regular singularities ($z=0$ and $z=1$) and one irregular singularity at $z=\infty$ (as $z\to\infty$, the dominant term is $y'' + K z y = 0$, an Airy-type equation).

For the finite regular singularities, we can use Frobenius series to build solutions, but the irregular singularity at infinity means the full solution will be a combination of:

1. Generalized hypergeometric functions (to handle the finite regular singularities)
2. Confluent hypergeometric functions or Airy functions (to handle the irregular singularity at infinity)

If you factor the cubic numerator into $K(z - r_1)(z - r_2)(z - r_3)$ (using cubic root formulas for roots $r_1, r_2, r_3$), you can split the coefficient into partial fractions. This might let you express the solution as a product of hypergeometric functions, though the algebra will be tedious.

Step 3: Context for small cubic terms #

If your constant $K$ (original $D$) is small, you could use perturbation theory around the solvable $K=0$ case—but this is a backup, since you're asking for exact analytic solutions.

三、关键总结 #

• Avoiding the cubic term: There's no obvious coordinate transformation that eliminates the cubic polynomial entirely—it comes directly from the $D$-dependent part of your original potential, so it's inherent unless $D=0$. However, substitutions like $t=1-e^{-\delta x}$ or constant-absorbing variable shifts can simplify the equation's structure for further work.
• Analytic solution with cubic term: Yes, it's possible! By using Liouville transformations to eliminate the first derivative, then leveraging generalized hypergeometric functions (combined with Airy/confluent hypergeometric functions for the irregular singularity), you can express the solution in terms of known special functions. The exact form depends on your constant values, but the path forward is:
  1. Simplify the ODE with function transformations to remove the first derivative.
  2. Use partial fractions or Frobenius series to match the equation to a known special function form.

备注：内容来源于stack exchange，提问作者ETorre