I was trying to apply the WKB method to analyze the approximate eigenvalue condition for the Schrodinger like equation
\begin{equation} \varepsilon^{4} y^{(4)} = (E-V(x))y, \quad y(\pm \infty) = 0, V(x) \to \infty \text{ as } |x| \to \infty \end{equation}
To do so, I considered the more general equation \begin{equation} \varepsilon^{4}y^{(4)} = Q(x)y, \quad y(\pm \infty) = 0, Q(\pm \infty) = -\infty \end{equation} where $Q(x)$ is assumed to have only 2 turning points $A, B$ ($A < B$) and that $Q(x) > 0$ for $A < x < B$, and $Q(x) < 0$ for $x < A$ or $x > B$.
I seem to have encountered a problem in matching the coefficients for the solution in the region $A < x < B$, where the solution consists of exponential terms and sinusoidal/cosinusoidal terms (as $Q(x) > 0$).
I wonder if the WKB method is suitable for tackling this kind of problem and if my approach to it (applying the usual method of analyzing two turning points in WKB) is correct, and if so, how should the coefficient matching be done?
Edit 1 : If my calculations are correct, here's what I found :
On the region $x > B$, the WKB approximation gives
\begin{align*} y_{I_{B}}(x) \sim [|Q(x)|]^{-3/8}\Bigg[&b_{3}\exp\left(\dfrac{-1}{\varepsilon\sqrt{2}}\kappa(x)\right)\cos\left(\dfrac{1}{\varepsilon\sqrt{2}}\kappa(x)\right) \\ + &b_{4}\exp\left(\dfrac{-1}{\varepsilon\sqrt{2}}\kappa(x)\right)\sin\left(\dfrac{1}{\varepsilon\sqrt{2}}\kappa(x)\right)\Bigg] \end{align*}
Where $\kappa(x) = \int_{B}^{x}|Q(t)|^{1/4}dt$.
Similarly, on the region $x < A$, the WKB approximation is
\begin{align*} y_{I_{A}}(x) \sim [|Q(x)|]^{-3/8}\Bigg[&a_{3}\exp\left(\dfrac{-1}{\varepsilon\sqrt{2}}\gamma(x)\right)\cos\left(\dfrac{1}{\varepsilon\sqrt{2}}\gamma(x)\right) \\ + &a_{4}\exp\left(\dfrac{-1}{\varepsilon\sqrt{2}}\gamma(x)\right)\sin\left(\dfrac{1}{\varepsilon\sqrt{2}}\gamma(x)\right)\Bigg] \end{align*}
where $\gamma(x) = \int_{x}^{A}|Q(t)|^{1/4}dt$.
On the region $A < x < B$, we might have 2 representations for $y(x)$, they are \begin{align*} y_{IIIB}(x) \sim [Q(x)]^{-3/8}\Bigg[&l_{1}\exp\left(-\frac{1}{\varepsilon}(-\kappa(x))\right) + l_{2}\exp\left(\frac{1}{\varepsilon}(-\kappa(x))\right) \\ + &l_{3}\cos\left(\frac{1}{\varepsilon}(-\kappa(x))\right) + l_{4}\sin\left(\frac{1}{\varepsilon}(-\kappa(x))\right)\Bigg] \\ y_{IIIA}(x) \sim [Q(x)]^{-3/8}\Bigg[&k_{1}\exp\left(-\frac{1}{\varepsilon}(-\gamma(x))\right) + k_{2}\exp\left(\frac{1}{\varepsilon}(-\gamma(x))\right) \\ &+ k_{3}\cos\left(\frac{1}{\varepsilon}(-\gamma(x))\right) + k_{4}\sin\left(\frac{1}{\varepsilon}(-\gamma(x))\right)\Bigg] \end{align*}
Now, I have the connection formulas for connecting $b_{3},b_{4}$ with $l_{1},l_{2},l_{3},l_{4}$ and for connecting $a_{3},a_{4}$ with $k_{1},k_{2},k_{3},k_{4}$ through analyzing the fourth order Airy like equation \begin{align*} y^{(4)} = xy \end{align*} which is the valid approximation for $x$ close to (either one of) the turning points.
Using those solutions, essentially what I have found is that $l_{1},l_{2},l_{3},l_{4}$ and $b_{3},b_{4}$ are connected through a matrix, that is, if $l = (l_{1},l_{2},l_{3},l_{4})$ and $b = (b_{3},b_{4})$, then
\begin{align*} l = Mb \end{align*}
for some matrix $M$.
By my analysis, I have also found that $a = (a_{3},a_{4})$ and $k = (k_{1},k_{2},k_{3},k_{4})$ are also connected similarly, in fact, the matrix connecting them are the same, that is \begin{align*} k = Ma \end{align*}
Now, the two representations for $y(x)$ in the middle region are supposed to be the same, in the usual WKB analysis for the second order Schrodinger equation, this gives the eigenvalue condition (WKB quantization condition). I approach to equate the two expressions as follows, I write \begin{align*} \int_{A}^{x}|Q(t)|^{1/4}dt = \int_{A}^{B}|Q(t)|^{1/4} dt - \int_{x}^{B}|Q(t)|^{1/4}dt \end{align*}
and therefore, $y_{IIIA}(x)$ might be rewritten as \begin{align*} y_{IIIA}(x) \sim [Q(x)]^{-3/8}\Bigg[&k_{1}\exp\left(-\int_{A}^{B}\right)\exp\left(\int_{x}^{B}\right) + k_{2}\exp\left(\int_{A}^{B}\right)\exp\left(-\int_{x}^{B}\right) \\ + &\left(k_{3}\cos\left(\int_{A}^{B}\right)+k_{4}\sin\left(\int_{A}^{B}\right)\right)\cos\left(\int_{x}^{B}\right) \\ + &\left(k_{3}\sin\left(\int_{A}^{B}\right)-k_{4}\cos\left(\int_{A}^{B}\right)\right)\sin\left(\int_{x}^{B}\right)\Bigg] \end{align*}
Where I have shortened $\int_{x_{1}}^{x_{2}} = \dfrac{1}{\varepsilon}\int_{x_{1}}^{x_{2}}|Q(t)|^{1/4} dt$ for brevity.
Now $y_{IIIA}(x)$ is in the form of $y_{IIIB}(x), and I deduced that it must be that
\begin{align*} k_{1}\exp\left(-\int_{A}^{B}\right) &= l_{2} \\ k_{2}\exp\left(\int_{A}^{B}\right) &= l_{1} \\ k_{3}\cos\left(\int_{A}^{B}\right)+k_{4}\sin\left(\int_{A}^{B}\right) &= l_{3} \\ k_{3}\sin\left(\int_{A}^{B}\right)-k_{4}\cos\left(\int_{A}^{B}\right) &= l_{4} \end{align*}
From this system of equations, and using the fact that $l = Mb$ and $k = Ma$, I hope to find a condition relating $a$ and $b$, and also the value of $\int_{A}^{B} = \dfrac{1}{\varepsilon}\int_{A}^{B}|Q(t)|^{1/4} dt$ (the WKB quantization condition). It is here that I seem to be lost, from my calculations, I cannot seem to get the relation between $a$ and $b$, and I cannot find a condition on $\int_{A}^{B} = \dfrac{1}{\varepsilon}\int_{A}^{B}|Q(t)|^{1/4} dt$.
It looks like you've essentially got it down to the linear algebra. Let $\alpha = \epsilon^{-1}\int_A^B|Q|^{1/4}\,\mathrm dt$. You have the linear relationship between the coefficients: $$ l = A(\alpha)\kappa, \qquad l = \begin{bmatrix} l_1 \\ l_2 \\ l_3 \\ l_4 \end{bmatrix}, \quad \kappa = \begin{bmatrix} \kappa_1 \\ \kappa_2 \\ \kappa_3 \\ \kappa_4 \end{bmatrix}, \quad A(\alpha) = \begin{bmatrix} 0 &e^\alpha & 0 & 0 \\ e^{-\alpha} & 0 & 0 & 0 \\ 0 & 0 & \cos\alpha & \sin\alpha \\ 0 & 0 & \sin\alpha & -\cos\alpha \end{bmatrix}, $$ as well as the connection formulae $$ \kappa = Ma, \qquad l = Mb. $$ In combination this is $$ AMa - Mb = \mathbf 0. $$ Since $a,b \in \mathbb R^2$, and $M \in \mathbb R^{4\times 2}$ this represents four equations and four unknowns, and may be equivalently written as $$ \begin{bmatrix} AM & -M \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \mathbf 0 $$ where the block matrix $\begin{bmatrix} AM & -M \end{bmatrix} \in \mathbb R^{4\times 4}$ and $\begin{bmatrix} a \\ b \end{bmatrix} \in \mathbb R^4$. For a nontrivial solution, the block matrix must be singular, which gives the equation for $\alpha$ (by setting the determinant to zero).