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anonymous

  • 4 years ago

Is \(y=Ae^{-x-t^2}\) a wave? I'm testing it by using the wave equation:\[\frac{\partial ^2}{\partial x^2}=\frac{1}{u^2} \frac{\partial ^2}{\partial t^2}\]So I currently have\[\frac{\partial ^2}{\partial x^2}=Ae^{-x-t^2}\]and\[\frac{\partial ^2}{\partial t^2}=Ae^{-x-t^2} [-2+4t^2]\]Normally, a wave has the form \(kx-\omega t\), and the velocity can be calculated from that, but this one has a \(t^2\), which confuses me. Any thoughts?

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  1. anonymous
    • 4 years ago
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    Those derivatives should all be with respect to y also.

  2. anonymous
    • 4 years ago
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    They are derivatives of y, but not with respect to y. In the wave equation, u is a constant, i.e. the speed of propagation of the wave. Do your results suggest that your function satisfies the wave equation?

  3. anonymous
    • 4 years ago
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    Aye, I meant derivatives of y. Early morning head I suppose. If the two sides were equal, the velocity squared would need to be \(-2+4t^2\), which is not constant, so I suppose it cannot be a wave.

  4. anonymous
    • 4 years ago
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    I suppose it would also have an imaginary velocity.

  5. anonymous
    • 4 years ago
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    Mhmm, it's not a wave as defined by the wave equation.

  6. anonymous
    • 4 years ago
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    I shouldn't say "as defined by the wave equation".... it suffices to say that the function you are given does not satisfy the one-dimensional wave equation :)

  7. anonymous
    • 4 years ago
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    Thanks Jemurray3!

  8. anonymous
    • 4 years ago
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    No problem :)

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