Pragabhava
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 Jan 14 comment How to find the following derivative? @GitGud Edited. By the way, if you use \arcsin x, you produce $$\arcsin x.$$ Jan 14 revised How to find the following derivative? added 484 characters in body Jan 14 answered How to find the following derivative? Jan 13 comment Wave equation solution properties @rlgordonma I'd say that the fact that $v_{\alpha \beta} = 0$ implies that $v(\alpha,\beta) = F(\beta) + G(\alpha)$ which, in turn, means that $u(x,t) = F(x-t) + G(x+t)$ (this can be ensured because the transformation $(x,t) \to (\alpha,\beta)$ is invertible), and given that the wave equation with initial conditions has a unique solution (provided $f$ and $g$ are in some space, details left to the OP), this has to be it. Finally, using the initial conditions, one can determine $F$ and $G$ in terms of $f$ and $g$, and derive d'Alambert solution. Jan 13 comment Wave equation solution properties For completeness: Question no. 2 hasn't been answered. An argument of unicity must be performed in order to ensure that every solution has the desired form. Jan 9 awarded Revival Dec 21 awarded Revival Dec 10 comment Consider the wave equation I believe you need to solve using the reflection principle, which means your soltution is valid in the triangular region given by T = \left\{(x,t) \in \mathbb{R}\,\big|\, t > 0\,,\,t < x+a,\,t