$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ (u(x,t).\nabla)\rho(x,t)$.

Question

We are Studying the behavior of the fluid using Navier Stokes' Equation, and I am new to this lesson, but I am stuck on my below question:

If I have $g(t)=\rho(x(t),t)$, then using Chain rule, we get: $$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+\frac{\partial \rho}{\partial x}(x(t),t)x'(t)$$

So how to continue in order to get the latter is equal to : $$\frac{\partial \rho}{\partial t}(x(t),t)+ (u(x,t).\nabla)\rho(x,t).$$

Remark that $u(x,t)=x'(t)$ is the velocity field of the flow of the fluid (here we are dealing with the Eulerian description), and $\rho:\Omega×[0,T] \rightarrow \mathbb{R}$ is the mass density, $\nabla$ is the nabla function

I am really stuck and I need some help please!!!

Answer 1

Observe that $x(t)=(x_1(t),x_2(t),x_3(t))$ and $u(t,x)=(x'_1(t),x'_2(t),x'_3(t))$ by the chain rules results

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+\frac{\partial \rho}{\partial x_1}(x(t),t)x_1'(t)+\frac{\partial \rho}{\partial x_2}(x(t),t)x_2'(t)+\frac{\partial \rho}{\partial x_3}(x(t),t)x_3'(t)$$

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+(x'_1(t),x'_2(t),x'_3(t)) \cdot \left(\frac{\partial \rho}{\partial x_1}(x(t),t),\frac{\partial \rho}{\partial x_2}(x(t),t),\frac{\partial \rho}{\partial x_3}(x(t),t)\right ) $$

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ u(x,t) \cdot \left(\frac{\partial \rho}{\partial x_1}(x(t),t),\frac{\partial \rho}{\partial x_2}(x(t),t),\frac{\partial \rho}{\partial x_3}(x(t),t)\right )$$

$$g'(t)=\frac{\partial \rho}{\partial t}(x(t),t)+ u(x,t) \cdot \nabla \rho(x(t),t). $$