Boundary conditions don't apply in the equation's region of validity

[asaspades]

Homework Statement

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.

Homework Equations

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)

 

[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.

 

[asaspades]

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?

 

[asaspades]

I'm pretty sure that last part isn't right at all

 

[haruspex]

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?

That all looks good to me.