I am trying to approximate the solution to the equation
$f''(x)+(E-U(x))f(x)=0$
where $U(x) = \begin{cases} \frac{U_0}{m}x-U_0 & \text{for $-m<x<0$} \\ \frac{-U_0}{m}x-U_0 & \text{for $0<x<m$} \\ 0 & \text{otherwise} \end{cases}$
using the WKB method.
Assuming that the solution is bounded everywhere, I would then like to find the possible negative values of E.
My attempt (and plans after I got stuck):
When $E>U(x)$, WKB gives
$f\approx \frac{A}{(E-U)^\frac{1}{4}}cos(\int_a^x\sqrt {E-U(x_1)}dx_1-\phi)$
for arbitrary $a$ and constants $A$ and $\phi$, and when $E<U(x)$, WKB gives
$f\approx A\frac{1}{(E-U)^\frac{1}{4}}e^{\int_a^x\sqrt {E-U(x_1)}dx_1}+B\frac{1}{(E-U)^\frac{1}{4}}e^{-\int_a^x\sqrt {E-U(x_1)}dx_1}$
for arbitrary $a$ and constants $A$ and $B$.
I want negative E so when $x<-m$ or $x>m$, then $E<U$ and so the first formula applies but when $-m<x<m$, I'm not sure which formula applies as depending on the value of $E$, $E$ could be more or less than $U$ (??).
Anyway, I'm trying to find a formula for $f$ for each of $x<-m$, $-m<x<0$, $0<x<m$ and $x>m$. Then, how do I combine these? Can $f$ be approximated by a piecewise function? Or do I have to look at continuity between the sections?
Then, for the values of E: I apply the restriction that $f$ is bounded everywhere and hopefully that will lead to a formula for E.
Any ideas or critiques of my ideas?