I'm having a problem trying to show that this solution satisfies the wave equation. I discovered this solution given by $$\psi(x,t)=e^{-(ax-bt)^2}$$ but I'm stuck trying to prove that solution satisfied the wave equation given by $$\frac{\partial^2\psi}{\partial t^2} = v^2 \frac{\partial^2\psi}{\partial x^2}$$ Can someone help me?
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2$\begingroup$ Your "solution" is not a wave; there must be $i$ and there must not be a square in the exponential. $\endgroup$– Vladimir KalitvianskiCommented Jan 6, 2021 at 12:44
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1$\begingroup$ hmm, i'm pretty sure this is a wave equation, just like $$Asin(kx-\omega t)$$ $\endgroup$– ThinkQuantumCommented Jan 6, 2021 at 12:50
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1$\begingroup$ Have you tried plugging your wave-pulse into the wave equation? It should satisfy it if $b/a=v$. $\endgroup$– mike stoneCommented Jan 6, 2021 at 12:52
2 Answers
From my answer to Solutions to wave equation in 1+1D:
This solution of one-dimensional wave equation, known as D’Alembert’s solution, can be written in general as $$\psi(x,t) = F(x- vt) + G(x + vt),$$ where $\psi(x,t)$ is the function that satisfies the wave equation, and $F$ and $G$ are any arbitrary functions. (However, the arguments of these functions must be $x-vt$ and $x+vt$ respectively. ) Physically, these represent two waves travelling in opposite directions with the speeds $v$, whose instantaneous "shapes" is given by the functions $F$ and $G$.
I'll leave it to you to show that the $\psi(x,t)$ that you have been provided can be written as (for example) $\psi(x,t) = F(x-vt)$, where you can determine both the function $F$ and the wave velocity $v$ just by inspecting the equation carefully.
Another way to do it, as @mikestone points out in the comment above, is to actually plug the solution into the wave equation, and ask yourself for which combination of $a, b,$ and $v$ the function is a solution.
Plane wave equation is in the form of $\psi = Ae^{i(kx-wt)}$, but if you're only asking if your solution satisfies the given partial differential equation yes it does in case $v=b/a$
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1$\begingroup$ This is really a comment, not an answer. But note that this is a homework-like question, and complete answers are not allowed here. $\endgroup$– garypCommented Jan 6, 2021 at 12:56
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$\begingroup$ Alright, thanks for pointing out. $\endgroup$– MonopoleCommented Jan 6, 2021 at 13:04