4
$\begingroup$

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. enter image description here 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}

$\endgroup$

closed as off-topic by Sofia, Kyle Kanos, John Rennie, Rob Jeffries, JamalS Mar 7 '15 at 9:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, John Rennie, Rob Jeffries, JamalS
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I'm a new user so probably I did a mistake. My question is not related to any homework-like question. My question concerns just the solution of the wave equation in a case where the potential is a generic function of the position. I set the question tag to electromagnetism. $\endgroup$ – Upax Mar 7 '15 at 14:07
  • 1
    $\begingroup$ Hi Upax. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Mar 7 '15 at 14:30
  • $\begingroup$ I modified the tags accordingly to your suggestion $\endgroup$ – Upax Mar 8 '15 at 12:11
  • $\begingroup$ Related meta question: meta.physics.stackexchange.com/q/6596/2451 $\endgroup$ – Qmechanic Mar 14 '15 at 12:39
1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The policy of our site is to give hints, but actually the OP should solve his/her home-work. $\endgroup$ – Sofia Mar 6 '15 at 0:58
  • $\begingroup$ Sorry I only saw the (homework-and-exercises) tag after reading your comment. As harm to the OP has been already possibly done, I'll keep the answer posted so that other viewers can benefit from it. $\endgroup$ – Ali Moh Mar 6 '15 at 1:11
  • $\begingroup$ I suppose you are referring to the student badge. Since I add my first question yestearday I got this badge, but I'm not a student. Concerning the solution I don't have a problem in computing the integration constant. As a have a little bit of time I'll ad my computation. On the other hand I'm not sure about the picture B and the direction of the propagating waves. $\endgroup$ – Upax Mar 6 '15 at 15:55
  • $\begingroup$ @Upax no, the student badge is something completely different. This question qualifies as "homework-like" because you're solving a problem for educational value. Besides that, it's not at all clear what you're actually asking. You've posted a problem and what is apparently a solution to the problem, but I don't actually see a question in there. $\endgroup$ – David Z Mar 7 '15 at 12:17
  • 1
    $\begingroup$ @Ali Moh: Tip: In the future please make the judgement whether a post is HW&EX (or not) based on the substance of the post rather than the tags. Tags might (or might not) have been applied correctly. $\endgroup$ – Qmechanic Mar 7 '15 at 14:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.