I know that if we have $f:\mathbb{R}^2\to \mathbb{R}$, defined as $f(x(t),y(t))$ for some
$x(t):\mathbb{R}\to\mathbb{R}$ and $y(t):\mathbb{R}\to\mathbb{R}$.
Then,
$$\frac{d}{dt}f(x(t),y(t))=\frac{\partial f}{\partial x}.\frac{dx}{dt}+\frac{\partial f}{\partial y}.\frac{dy}{dt}$$
And we have the corresponding result for any $n^{th}$ dimension.
But for :
$z(s)=u(x_1+x_2s,x_2-x_1s,t+s)$.
Where
$u(x_1,x_2,t):\mathbb{R^3}\to\mathbb{R}$
and for a fixed $(x_1,x_2,t)\in\mathbb{R^3}$
Can you tell me what will be the equation when you differentiate $z(s)$ with respect to $s$ (as in the example I did before).
(This question originally came when going over the transport equation on PDE)
Thank you
It's the same. In fact, we have $z(s)=u(y_1(s),y_2(s),y_3(s))$ with $y_1(s)=x_1+x_2s,\,y_2(s)=x_2-x_1s$ and $y_3(s)=t+s$. A renaming of the arguments of $u$ is necessary as their names were the same as the constants forming $z(s)$ (note that when you write "for a fixed $(x_1,x_2,t)$", you are referring to two different things, but we understand that $x_1$, $x_2$, $t$ in $u(x_1,x_2,t)$ cannot be constants. Anyway, I'd like to see the formulas in the context of your question)
$$\frac{d}{dt}u(y_1(s),y_2(s),y_3(s))=\frac{\partial u}{\partial y_1}\frac{dy_1}{ds}+\frac{\partial u}{\partial y_2}\frac{dy_2}{ds}+\frac{\partial u}{\partial y_3}\frac{dy_3}{ds}=$$
$$=\frac{\partial u}{\partial y_1}x_2+\frac{\partial u}{\partial y_2}(-x_1)+\frac{\partial u}{\partial y_3}$$