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?

Question

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?

anonymous · Answer

Those derivatives should all be with respect to y also.

anonymous · Answer

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?

anonymous · Answer

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.

anonymous · Answer

I suppose it would also have an imaginary velocity.

anonymous · Answer

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

anonymous · Answer

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 :)

anonymous · Answer

Thanks Jemurray3!

anonymous · Answer

No problem :)