First, since the potential is unbounded from below at the left side, the particle has continuous spectrum. This means that what you just have to do is compute the coefficients $c_i$, taking $E$ (real, never complex!) as input parameter.
One of these coefficients, $c_3$, is in fact arbitrary, because it only influences normalization, not smoothness of wavefunction. Thus, to compute the connection coefficients you have to find $c_1,c_2,c_4,c_5$.
And finally, your equation is a simple system of $4$ algebraic equations with $4$ unknowns. If I take $c_3=1$ and use arbitrary letters to denote all those Airy functions and their derivatives, I get
$$\left\{\begin{align}
c_4A+c_5B&=c_1C+c_2D,\\
c_4G+c_5H&=c_1I+c_2J,\\
c_1K+c_2L&=M,\\
c_1N+c_2O&=P.
\end{align}\right.$$
Solve it for $c_i$ and you've solved almost the whole of your problem. Now if you need normalized wavefunctions, you have to use some scheme for normalization of unbounded states, e.g. so called "normalization by Dirac delta", discussed e.g. in [1].
References:
- L.D. Landau & E.M. Lifshitz, Quantum Mechanics. Non-Relativistic Theory, $\S5$.
Disclaimer: In this answer, we will just derive a rough semiclassical estimate for the threshold between the existence of zero and one bound state. Of course, one should keep in mind that the semiclassical WKB method is not reliable$^1$ for predicting the ground state. We leave it to others to perform a full numerical analysis of the problem using Airy Functions.
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$\uparrow$ Fig.1: Potential $V(x)$ as a function of position $x$ in OP's example.
First let us include the metaplectic correction/Maslov index. The turning point at an infinitely hard wall and an inclined potential wall have Maslov index $2$ and $1$, respectively, cf. e.g. this Phys.SE post. In total $3$. We should then adjust the Bohr-Sommerfeld quantization rule with a fraction $\frac{3}{4}$.
$$ \int_{x_-}^{x_+} \! \frac{dx}{\pi} k(x)~\simeq~n+\frac{3}{4},\qquad n~\in~\mathbb{N}_0,\tag{1} $$
where
$$ k(x)~:=~\frac{\sqrt{2m(E-V(x))}}{\hbar}, \qquad
V(x)~:=~-V_0 \frac{L-x}{L}. \tag{2} $$
At the threshold, we can assume $n=0$ and $E=0$. The limiting values of the turning points are $x_-=0$ and $x_+=L$. Straightforward algebra yields that the
threshold between the existence of zero and one bound state is
$$V_0~\simeq~\frac{81}{128} \frac{\pi^2\hbar^2}{mL^2} \tag{3} .$$
$^1$ For comparison, the WKB approximation for the threshold of the corresponding square well problem yields
$$V_0~\simeq~\frac{\pi^2\hbar^2}{2m L^2} \tag{4} ,$$
while the exact quantum mechanical result is
$$V_0~=~\frac{\pi^2\hbar^2}{8m L^2} \tag{5} ,$$
cf. e.g. Alonso & Finn, Quantum and Statistical Physics, Vol 3, p. 77-78. Not impressive!
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$\uparrow$ Fig.2: Corresponding square well potential as a function of position $x$. Each of the 2 infinitely hard walls has Maslow index 2.
Best Answer
Since $k_E \propto \hat{p}^2$ and $\hat{p}$ is Hermitian you may see that this makes $k_E$ positive semidefinite, that is all of its eigenvalues are larger or equal to 0. In other words when you measure this operator you will always get results which are larger or greater than zero. This "contradiction" is resolved by the fact that the potential is a function of the coordinate and the momentum (and hence kinetic energy) does not commute with functions of coordinate. This in turn means that you can't simultaneously measure both $k_E$, $E$, and potential.