Solution of the wave equation I am dealing with the solution of the wave equation in two different cases represented in figure by Case A and Case B. The two figures were obtained by shining a slab with an incident wave coming from the right (Case A) and from the left (Case B). I verified the solution for Case A. It appears to be correct. Thus I used the same methodology to derive the solution for Case B. However I suspect the solution for Case B is wrong, even if I am not able to understand why.
Independently of the correct/wrong result of the wave equation in case B, I would like to know if there is a way to derive the solution of the wave equation in Case B from that of Case A by applying a kind of symmetry. Please note that $\eta(x)$ does not necessarely any symmetry with respect to  $x$.
This is my solution. Given the wave equation:
\begin{equation}
\frac{d^2y(x)}{dx^2}+k^2 \left(1+\eta(x)\right) y(x)=0
\end{equation}
Its solution can be found by using the Green function method. Details of the method are proviced for Case B. When applied to Case A the wave equation has the solution: 
\begin{equation}
y(x)=c_1 e^{-j k x}+c_2 e^{j k x}+\frac{j k}{2} \int_0^Le^{j k |x-t|} \eta(t) y(t) dt
\end{equation}
where $j^2 = -1$ and $c_1$ and $c_2$ are the integration constant. In order to determine the constants we have to consider the boundary conditions. 
When refferring to the case A in figure it is possible to distinguis an incident wave and a reflected wave on the right hand side, and a transmitted wave on the left hand side. In this case the following boundatry conditions hold:
\begin{align}
y(L)=1+R_L\\
y'(L)=-jk(2-y(L))\\
y(0)=T_L\\
y'(0)=-jky(0)
\end{align}
By applying these boundary condition while distinguish the two cases $t-x>0$ (left side of the picture) and $t-x<0$ (right side of the picture) it is possibler to find a differential equation in $T_L$. 
Now I would like to repeat the same kind of computation on the second picture (Case B). In this picture the incident wave is coming from the left side, while the transmitted wave is on the right side. In the case A a wave propagating in the same direction of the x axis has a positive exponential term. I have a doubt concerning figure B. In this case a wave travelling in the same direction of the x axis has a dependance on $-x$. Is this correct? I suppose that it's just a question of conventions. The boundary condition for the case in figure B are of course switched:
\begin{align}
y(0)=1+R_L\\
y'(0)=-jk(2-y(0))\\
y(L)=T_L\\
y'(L)=-jky(L)
\end{align}
Concerning the case A, I first differentiate the general solution:
\begin{equation}
y'(x)=-jk c_1 e^{-jk x}+jk c_2 e^{+jk x}+\frac{j k}{2} \int_0^L\frac{d}{dx}\left(e^{j k |x-t|} \right) \eta(t) y(t) dt
\end{equation}
Now I considered two different cases: whel $t-x>0$ (left) and $t-x<0$ (right).
For $t-x>0$ I can write:
\begin{equation}
y'(x)=-jk c_1 e^{-jk x}+jk c_2 e^{+jk x}-jk U(x)
\end{equation}
where
\begin{equation}
U(x)= \int_0^L\frac{d}{dx}\left(e^{-j k (x-t)} \right) \eta(t) y(t) dt
\end{equation}
Then
\begin{equation}
y'(x)=-jk \left( c_1 e^{-jk x}- c_2 e^{+jk x} + U(x) \right)
\end{equation}
While when $t-x<0$ I have:
\begin{equation}
y'(x)=-jk \left( c_1 e^{-jk x}- c_2 e^{+jk x}- K(x) \right)
\end{equation}
where
\begin{equation}
K(x)= \int_0^L\frac{d}{dx}\left(e^{j k (x-t)} \right) \eta(t) y(t) dt
\end{equation}
The boundary conditions for in $x=0$ correspond to the case $t-x>0$, while the boundary conditions in $x=L$ correspond to the case $t-x>0$. For $x=0$ we have
\begin{align}
y(0)=c_1+c_2+U(0)\\
y'(0)=-jk(c_1-c_2+U(0))\\
\end{align}
So that $c_2=0$. By repeating the same computation in $x=L$ it is possible to determine that $c_1=e^{jkL}$. Then the solution is:
\begin{equation}
y(x)=e^{j k(L- x)}+\frac{j k}{2} \int_0^Le^{j k |x-t|} \eta(t) y(t) dt
\end{equation}
Now the case in figure B. This is where I'm not sure. According to the picture $e^{-jkx}$ is the forward propagating wave. The Green functions are for $x>L$:
\begin{equation}
\begin{cases}
y_1(x)=C_1 e^{-jkx}\\
y_1(L)=A'
\end{cases}
\end{equation}
While for $x<0$:
\begin{equation}
\begin{cases}
y_2(x)=c_2 e^{jkx}\\
y_2(0)=B'
\end{cases}
\end{equation}
Then
\begin{equation}
\begin{cases}
y_1(L)=c_1 e^{-jkL}=A' & \Rightarrow y_1(x)=A' e^{jk(L-x)}\\
y_2(0)=c_2=B' & \Rightarrow y_2(x)=B' e^{jkx}
\end{cases}
\end{equation}
The Wronskian is
\begin{equation}
W=
\begin{vmatrix}
A' e^{jk(L-x)} & B' e^{jkx}\\
-jk A' e^{jk(L-x)} & jk B' e^{jkx}\\
\end{vmatrix}
=2 jk A'B' e^{jkL}
\end{equation}
So that the Green function is
\begin{equation}
G(x,t)=
\begin{cases}
\frac{A' e^{jk(L-x)} B' e^{jkt}}{2 jk A'B' e^{jkL}}\\
\frac{A' e^{jk(L-t)} B' e^{jkx}}{2 jk A'B' e^{jkL}}\\
\end{cases}
\end{equation}
Which reduces to
\begin{equation}
G(x,t)=\frac{1}{2jk} e^{jk|t-x|}
\end{equation}
So that the solution of the differential equation for the case in figure B is
\begin{equation}
y(x)=c_1 e^{-j k x}+c_2 e^{j k x}+\frac{j k}{2} \int_0^Le^{j k |t-x|} \eta(t) y(t) dt
\end{equation}
 A: Apply a parity transformation to your solution (but first please consider the region with non vanishing potential to be $[-L/2, +L/2]$ otherwise it will be notationally not convenient to shift by $L/2$ the final solutions, but it is up to you)
$$ y(x) = c_1 e^{-jkx} + c_2 e^{+jkx} + \frac{jk}{2}\int_{-L/2}^{L/2} dt e^{jk|x-t|}\eta(t)y(t)$$
gives
$$ y_P(x)\equiv y(-x) = c_1 e^{+jkx} + c_2 e^{-jkx} + \frac{jk}{2}\int_{-L/2}^{L/2} dt e^{jk|-x-t|}\eta(t)y(t)$$
Now change variable of integration
\begin{align*} y_P(x) &= c_1 e^{+jkx} + c_2 e^{-jkx} - \frac{jk}{2}\int_{+L/2}^{-L/2} dt e^{jk|-x+t|}\eta(-t)y(-t)\\
&= c_1 e^{+jkx} + c_2 e^{-jkx} + \frac{jk}{2}\int_{-L/2}^{L/2} dt e^{jk|x-t|}\eta_P(t)y_P(t)
\end{align*}
So we can see that $y(-x)$ satisfies the same equation as $y(x)$ given that we identify $c_1 \leftrightarrow c_2$, and we use the parity transformed potential $\eta(-x)$. This straightforwardly implies that for an asymmetric potential you wont get the same reflection/transmission.
Notice that had you used the range $[0,L]$ instead, the new $c_{1}$ and $c_2$ would also differ by phases $e^{\pm jk L/2}$, and well as a not aesthetic shift for the additional integral term.
