$t$ derivative of Kirchhoff's solution

Question

I know that the solution to the PDE \begin{align*} u_{tt} - \Delta u = 0, \quad \mathbb{R}^3\times[0, \infty)\\ u(x, 0) = 0, \quad x \in \mathbb{R}^3\\ u_t(x, 0) = g(x), \quad x \in \mathbb{R}^3 \end{align*} is $$u(x,t) = \mathrel{\int\!\!\!\!\!\!-}_{\partial B(0,1)}t g(x + tw)dS(w).$$

My question is how is this found: $$ u_t(x,t) = \mathrel{\int\!\!\!\!\!\!-}_{\partial B(0,1)}[g(x + tw) + t \nabla g(x + tw)\cdot w] dS(w). $$ I can tell that the first term in the integral is from the product rule, but I do not understand how apparently $\frac{\partial}{\partial t} g(x + tw) = \nabla g(x + tw)\cdot w$. Is the gradient with respect to $x$? Is this an application of the chain rule and I just don't see it?

Answer 1

Yes, this is just the multivariate chain rule, and the gradient is wrt $x$. You can set $h(t)=x+tw$, then $g(x+tw)=g\circ h(t)$ so $$ \frac{d}{dt} g\circ h(t) = (\nabla g)(h(t))\cdot \frac{dh}{dt}(t) = (\nabla g)(x+tw)\cdot w$$

Answer 2

This equality follows from the notion of directional derivative: given a function $f: \mathbb{R}^n \to \mathbb{R}$ and $x,w \in \mathbb{R}^n$, we have $$ \frac{\operatorname{d}}{\operatorname{d} t} \, f(x + tw) = \nabla f(x) \cdot w$$