# General equations for transmitted and reflected waves

1. Jan 10, 2015

### ELB27

1. The problem statement, all variables and given/known data
Suppose you send an incident wave of specified shape, $g_I(z-v_1t)$, down string number 1. It gives rise to a reflected wave, $h_R(z+v_1t)$, and a transmitted wave, $g_T(z-v_2t)$. By imposing the appropriate boundary conditions [see below], find $h_R$ and $g_T$.

2. Relevant equations
The appropriate boundary conditions as given in the book:
$f(0^-,t)=f(0^+,t)$ and $\frac{\partial f}{\partial z}|_{z=0^-} = \frac{\partial f}{\partial z}|_{z=0^+}$.

3. The attempt at a solution
The net displacement of the strings is $f=g_I(z-v_1t)+h_R(z+v_1t)$ for $(z<0)$ and $f=g_T(z-v_2t)$ for $(z>0)$. Using the first boundary condition: $g_I(-v_1t)+h_R(v_1t)=g_T(-v_2t)$. For the second condition, I think that the definition of a derivative as a limit is needed. Thus, $$\lim_{\Delta z→0}\frac{g_I(-v_1t)-g_I(-v_1t-\Delta z) + h_R(v_1t)-h_R(v_1t-\Delta z)}{\Delta z} = \lim_{\Delta z→0}\frac{g_T(-v_2t-\Delta z)-g_T(-v_2t)}{\Delta z}$$
Rearranging and plugging in the first boundary condition, $$\lim_{\Delta z→0}\frac{2g_T(-v_2t)-g_I(-v_1t-\Delta z)-h_R(v_1t-\Delta z)-g_T(-v_2t-\Delta z)}{\Delta z}=0$$
Now I'm stuck. I tried to evaluate the last equation at different times and using the first constraint to attempt to eliminate either $h_R$ or $g_T$. For instance, evaluating at $t=\frac{\Delta z}{v_1}$ and noting that $h_R(0) = g_T(0)-g_I(0)$ by the first constraint I got: $$\lim_{\Delta z→0}\frac{2g_T(\frac{v_2}{v_1}t)-g_T(0)-g_T(-\frac{v_2}{v_1}\Delta z-\Delta z)+g_I(0)-g_I(-2\Delta z)}{\Delta z} = 0$$
I don't see where this leads me and the same happens when I try to evaluate at different times. Any ideas on how to progress from here?

Any suggestions/comments will be greatly appreciated!

2. Jan 10, 2015

### gabbagabbahey

Hi ELB27,

Hint 1 if $a(u)=b(u+1)$, then $a(u-\Delta z)=b(u-\Delta z + 1)$ (can you see how to use this to eliminate $g_{T}$?)

Hint 2 $\frac{d}{du}a(w) = \frac{dw}{du}\frac{d}{dw}a(w)$

Using some clever substitutions and applying the above derivative rule you should be able to arrive at a simple relationship between $g'_{I}(u)$ and $h'_{R}(u)$

Cheers,
Gabba

3. Jan 11, 2015

### ELB27

Thank you very much for the hints.
I'm afraid that I still don't see the solution you were suggesting. Also, what is the justification for hint no.1? Why is it true?
However, your two hints gave me another idea. Using your hint no.2 (the chain rule) $\frac{\partial g_T}{\partial z} = \frac{d g_T}{d (z-v_2t)}\frac{\partial (z-v_2t)}{\partial z} = \frac{d g_T}{d (z-v_2t)}$ and $\frac{\partial g_T}{\partial t} = \frac{d g_T}{d (z-v_2t)}\frac{\partial (z-v_2t)}{\partial t} = -v_2\frac{d g_T}{d (z-v_2t)}$. Thus, $\frac{\partial g_T}{\partial z} = -\frac{1}{v_2}\frac{\partial g_T}{\partial t}$. Acquiring a similar expression for $g_I$ and $h_R$, and noting that $\frac{\partial g(z-vt)}{\partial t}|_{z=0} = \frac{\partial g(-vt)}{\partial t}$ since the derivative is with respect to $t$, the 2nd condition can be rewritten as $$-\frac{1}{v_2}\frac{\partial g_T(-v_2t)}{\partial t}=-\frac{1}{v_1}\frac{\partial g_I(-v_1t)}{\partial t}+\frac{1}{v_1}\frac{\partial h_R(v_1t)}{\partial t}$$Integrating with respect to time, $$-\frac{1}{v_2}g_T(-v_2t)=-\frac{1}{v_1}g_I(-v_1t)+\frac{1}{v_1}h_R(v_1t) + c$$ where $c$ is a constant of integration. Now, by the first condition, $g_T(-v_2t) = g_I(-v_1t)+h_R(v_1t)$. Using these two equations, we can isolate both $h_R$ and $g_T$. To wit: $h_R(v_1t)=\frac{v_2-v_1}{v_2+v_1}g_I(-v_1t) + c$ and $g_T(-v_2t) = \frac{2v_2}{v_2+v_1}g_I(-v_1t)+d$. Now your first hint comes into play. If I understood it correctly, I can just add $+z$ into the brackets of each function and get the final solution.
I'm not sure about this last step as there is also a constant in front (although it doesn't depend on $t$). Also, I would like to solve this problem with the original method.

Last edited: Jan 11, 2015
4. Jan 11, 2015

### gabbagabbahey

There's more than one way to skin a cat

Well, $a$ and $b$ are functions (although I realize now I could have made that clearer in the Hint), so it is just a substitution $u \rightarrow u +\Delta z$ (specifically, it can be thought of as a "shift" of the $u$ axis)

If it is not intuitively obvious still, consider an example like the functions $\cos$ and $\sin$ which are related by $\cos\left(x \right) = \sin \left( \frac{\pi}{2} - x\right)$, and compare the plots of $\cos\left(x + a\right)$ and $\sin \left( \frac{\pi}{2} - (x +a)\right)$ for some random values of $a$

Not sure why you have two different constants here.

Well, you can't "just add $+z$ into the brackets of each function and get the final solution", but you can make a substitution (carefully) to express the relationships between the functions in a different form if desired. Are you expected to provide your final answer in a specific format?
If not, I would personally write it in the form $h_{R}(u) = ...$ and $g_T(u) = ...$

5. Jan 12, 2015

### ELB27

I understand now, thanks.
You're right. After keeping track of the changes to the original $c$ I realized that in both cases the constant equals to $c'=\left(-\frac{v_2}{v_1+v_2}\right)c$.
Well, in the case of $h_R$, I want $u=z+v_1t$. In order to get it, I would make the substitution $t→t+\frac{z}{v_1}$ and get $$h_R(z+v_1t)=h_R(u)=\frac{v_2-v_1}{v_2+v_1}g_I(-v_1t-z)+c'=\frac{v_2-v_1}{v_2+v_1}g_I(-u)+c'$$ In the case of $g_T$, I want $u=z-v_2t$. To get it, I would make the substitution $t→t-\frac{z}{v_2}$ to get $$g_T(z-v_2t)=g_T(u)=\frac{2v_2}{v_2+v_1}g_I(\frac{v_1}{v_2}z-v_1t)+c' = \frac{2v_2}{v_2+v_1}g_I(\frac{v_1}{v_2}u)+c'$$ I have to say that I feel more confident in these substitutions - are they correct?