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General equations for transmitted and reflected waves
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[QUOTE="ELB27, post: 4972710, member: 516388"] 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 [tex]-\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}[/tex]Integrating with respect to time, [tex]-\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[/tex] 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. Thanks in advance! [/QUOTE]
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General equations for transmitted and reflected waves
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