Recent content by Mrx

Okay great! thank you for your help. Do you know what this equation ectually represents? I can't find anything about the physical meaning, or a derivation of this equation. I tried searching on conveyor belt equation but got nothing.

Thanks for your reply. The equation you got for X(x) is the same I tried to solve. Let's do it again: \lambda^2+2VX'\lambda + (V^2-c^2)X''=0 has solution X(x)=\exp\left(r_{\pm} x\right) With r=\frac{V\lambda \pm c\sqrt{\lambda^2}}{V^2-c^2} Now you can distinguish three cases...

After seperating variables and deviding by XT: \frac{T''}{T} + 2V\frac{X'T'}{XT} + (V^2-c^2) \frac{X''}{X}=0 differentiation wrt to t: \frac{d}{dt}\left(\frac{T''}{T}\right) = -2V\frac{d}{dt}\left(\frac{X'T'}{XT}\right) (1) differentiating wrt to x: (V^2...

@dirk_mec1 How and what expression do you get for the separation constant. I know that usually this is done by the boundary conditions, but I don't see how that works here, because application of the boundary conditions imply a trivial solution .