# Homework Help: Boundary conditions don't apply in the equation's region of validity

1. Apr 21, 2014

1. The problem statement, all variables and given/known data
A tight string lies along the positive x-axis when unperturbed. Its displacement from the x-axis is denoted by $y(x, t)$. It is attached to a boundary at $x = 0$. The condition at the boundary is
$$y+\alpha \frac{\partial y}{\partial x} =0$$ where $\alpha$ is a constant.

Write the displacement as the sum of an incident wave and reflected wave,
$$y(x, t) = e^{−ikx−i\omega t} + re^{ikx−i\omega t},\qquad x > 0,$$ and compute the reflection coefficient, $r$. Writing $r = |r|e^{i\phi}$, show that $|r| = 1$ and find $\phi$.

2. Relevant equations

3. The attempt at a solution
Since the boundary condition applies at $x=0$ and the equation given is only valid for $x>0$ I can't use that, so what equation should I use?

(If you just apply the condition that the incident and reflected wave are equal at $x=0$, since there is no transmission, you get what I believe is the desired result, but how would one go about this problem with the method it wants)

2. Apr 21, 2014

### haruspex

The equation is only given for x > 0 because the derivative is not defined at x= 0. If we take the derivative at 0 as meaning the one sided derivative then the equation must also be valid at 0; if not there would be a discontinuity in the string. Just take it as valid at x = 0 and see where it leads.

3. Apr 21, 2014

Am I doing this correctly.

Just assuming it works at $x=0$ gives $$e^{−ikx−i\omega t} + |r|e^{ikx−i\omega t +i\phi} = \alpha ike^{−ikx−i\omega t} - \alpha ikre^{ikx−i\omega t +i\phi}$$ so $$e^{−i\omega t} + |r|e^{−i\omega t +i\phi} = \alpha ike^{−i\omega t} - \alpha ik|r|e^{−i\omega t +i\phi}$$ and dividing by $e^{-i \omega t}$ (or letting $t=0$, since the equation must hold then as well as any time) $$1+|r|e^{i\phi} = \alpha ik - \alpha ik|r|e^{i\phi} \\ |r|e^{i\phi}=\frac{\alpha ik -1}{\alpha ik +1}$$ and so $$|r|=|r||e^{i\phi}|=\left| \frac{\alpha ik -1}{\alpha ik +1} \right|=\frac{|\alpha ik -1|}{|\alpha ik +1|} = 1.$$ Now since $$\frac{\alpha ik -1}{\alpha ik +1} = \frac{\alpha k +i}{\alpha k -i} = \frac{\alpha^2 k^2 +2\alpha ki -1}{\alpha^2 k^2 +1} = \frac{\alpha^2 k^2 -1}{\alpha^2k^2 +1}+\frac{2\alpha ki}{\alpha^2k^2 +1}$$ we have $$\phi = \tan^{-1} \left\{ \frac{2\alpha k}{\alpha^2 k^2 -1} \right\}.$$ Is that right?

4. Apr 21, 2014