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

In summary, the equation for the displacement of a tight string attached to a boundary at x=0 is e^{−ikx−i\omega t} + re^{ikx−i\omega t}, where x>0 and the reflection coefficient r is given by |r|e^{i\phi}.
  • #1
asaspades
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Homework Statement


A tight string lies along the positive x-axis when unperturbed. Its displacement from the x-axis is denoted by [itex]y(x, t)[/itex]. It is attached to a boundary at [itex]x = 0[/itex]. The condition at the boundary is
[tex]y+\alpha \frac{\partial y}{\partial x} =0[/tex] where [itex]\alpha[/itex] is a constant.

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

Homework Equations





The Attempt at a Solution


Since the boundary condition applies at [itex]x=0[/itex] and the equation given is only valid for [itex]x>0[/itex] 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 [itex]x=0[/itex], 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)
 
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  • #2
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
Am I doing this correctly.

Just assuming it works at [itex]x=0[/itex] gives [tex]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}[/tex] so [tex]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}[/tex] and dividing by [itex]e^{-i \omega t}[/itex] (or letting [itex]t=0[/itex], since the equation must hold then as well as any time) [tex]1+|r|e^{i\phi} = \alpha ik - \alpha ik|r|e^{i\phi} \\ |r|e^{i\phi}=\frac{\alpha ik -1}{\alpha ik +1}[/tex] and so [tex]|r|=|r||e^{i\phi}|=\left| \frac{\alpha ik -1}{\alpha ik +1} \right|=\frac{|\alpha ik -1|}{|\alpha ik +1|} = 1. [/tex] Now since [tex]\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}[/tex] we have [tex]\phi = \tan^{-1} \left\{ \frac{2\alpha k}{\alpha^2 k^2 -1} \right\}.[/tex] Is that right?
 
  • #4
I'm pretty sure that last part isn't right at all
 
  • #5
asaspades said:
Am I doing this correctly.

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

That all looks good to me.
 

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

1. What are boundary conditions in an equation's region of validity?

Boundary conditions refer to the constraints or limitations placed on a mathematical equation in a specific region where it is considered to be valid. These conditions help to define the behavior of the equation and restrict its solutions to a specific range.

2. Why don't boundary conditions apply in the equation's region of validity?

Boundary conditions are only applicable in specific regions where the equation is considered to be valid. Outside of this region, the equation may not accurately represent the behavior of the system and therefore the boundary conditions do not apply.

3. How do boundary conditions affect the solutions of an equation?

Boundary conditions play a crucial role in determining the solutions of an equation in its region of validity. They help to define the behavior of the equation and restrict its solutions to a specific range, making it more accurate and relevant to the system being studied.

4. Can boundary conditions be modified or changed?

Yes, in some cases, boundary conditions can be modified or changed to better fit the system being studied. However, this should be done carefully and with a thorough understanding of the implications on the solutions of the equation.

5. What happens if boundary conditions are not considered in an equation's region of validity?

If boundary conditions are not considered, the solutions of the equation may not accurately represent the behavior of the system being studied. This can lead to inaccurate predictions and conclusions, highlighting the importance of including boundary conditions in the region of validity.

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