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?

1. anonymous

Those derivatives should all be with respect to y also.

2. anonymous

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

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

I suppose it would also have an imaginary velocity.

5. anonymous

Mhmm, it's not a wave as defined by the wave equation.

6. anonymous

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

Thanks Jemurray3!

8. anonymous

No problem :)