Calculate $\Box w$ for $w(t,x):=g(v\cdot x-tc),\: (t,x)\in\Bbb R\times \Bbb R^m$ with $g\in C^2(\Bbb R)$, $c>0$ and $v\in\mathrm S^{m-1}$.
Here $\Box$ is the wave operator, defined by $\Box u(t,x):=\partial^2_t u(t,x)-\Delta_x u(t,x)$ and $\Delta$ is the Laplace operator. What I did was
$$\partial^2_t w(t,x)=c^2 \partial^2_t g(v\cdot x-tc)$$
$$\partial_{x_j} w(t,x)=v_j\partial_{x_j}g(v\cdot x-tc)\implies \partial^2_{x_j} w(t,x)=v_j^2\partial^2_{x_j}g(v\cdot x-tc)$$
Consequently
$$\Box w(t,x)=c^2\partial^2_t g(v\cdot x-tc)-\sum_{j=1}^mv_j^2\partial^2_{x_j}g(v\cdot x-tc)\tag1$$
so I dont get the point of this exercise, so my question is: can be $(1)$ simplified? I did the exercise correctly? I dont used the fact that $v\in\mathrm S^{m-1}$. The negative element of the RHS seems a weighted laplacian but I dont see that it can be simplified in any way.
I would be glad if some one can confirm that this was calculated correctly and if $(1)$ can be simplified further, thank you.
In your second line, you wrote $\partial_{x_j}w(t,x) =v_j\partial_{x_j}g(x\cdot v-tc),$ but that makes no sense because $g\in C^2(\mathbb{R})$. In other words, you are talking about partials of a single variable function.
Other than that, there is nothing wrong with the solution and once you correct it, you will get a slightly simpler expression.