Euler equation derivation

Question

I am trying to follow the solution for i) , but i'm stuck on the parts I have underlined in green and orange.

Answer 1

For the green part:

Using $(1)$ $\frac{\partial\rho}{\partial t}=-\nabla\cdot(\rho \underline{u})$, Thus:

$\frac{d}{dt}\int_V\rho dxdydz=\int_V\frac{\partial\rho}{\partial t} dxdydz=-\int_V\nabla\cdot(\rho \underline{u})dxdydz$

$=-\int_{\partial V}(\rho \underline{u})\cdot n\,dxdydz=-[FLUX]|_{\partial V}=-([Flux]^{A_x^+}_{A_x^-}+[Flux]^{A_y^+}_{A_y^-}+[Flux]^{A_z^+}_{A_z^-})=([Flux]^{A_x^-}_{A_x^+}+[Flux]^{A_y^-}_{A_y^+}+[Flux]^{A_z^-}_{A_z^+})$

$=\int_{A_x^{\mp}}(\rho \underline{u})\cdot n_{A_x}\,dydz+\int_{A_y^{\mp}}(\rho \underline{u})\cdot n_{A_y}\,dxdz+\int_{A_z^{\mp}}(\rho \underline{u})\cdot n_{A_z}\,dxdy$

Taking into account that $\underline{u}=(u,v,w)$, and on $A^{\mp x}$, $\, n=(\mp 1,0,0)$, on $A^{\mp y}$, $\, n=( 0,\mp 1,0)$, and on $A^{\mp z}$, $\, n=(0,0,\mp 1)$ Now we compare the integrands, we get:

$\frac{\partial \rho}{\partial t}\,\Delta x\Delta y\Delta z=\{\rho u|_{(x-\frac{\Delta x}{2},y,z,t)}-\rho u|_{(x+\frac{\Delta x}{2},y,z,t)}\}\Delta y\Delta z$

$~~~~~~~~~~~~~~~~~~~~~~~+\{\rho v|_{(x,y-\frac{\Delta y}{2},z,t)}-\rho v|_{(x,y+\frac{\Delta y}{2},z,t)}\}\Delta x\Delta z$

$~~~~~~~~~~~~~~~~~~~~~~~+\{\rho w|_{(x,y,z-\frac{\Delta z}{2},t)}-\rho w|_{(x,y,z+\frac{\Delta z}{2},t)}\}\Delta x\Delta y$

For the orange part let's take the first term, as it has done in the notes, and then you can do the others. We will expand centering at $(x,y,z,t)$

Here: $\rho u|_{(x-\frac{\Delta x}{2},y,z,t)}=\rho(x-\frac{\Delta x}{2},y,z,t)u(x-\frac{\Delta x}{2},y,z,t)$, since we are only considering a change in the $x$ variable, we can generalise to $\rho(x-\frac{\Delta x}{2},y,z,t)u(x-\frac{\Delta x}{2},y,z,t)=G(x-\frac{\Delta x}{2})$

By taylors expansion $G(x-\frac{\Delta x}{2})=G(x)-\frac{\Delta x}{2} \frac{\partial}{\partial x}G(x)+\frac{\Delta x^2}{8} \frac{\partial^2}{\partial x^2}G(x)\pm...$

and for $\rho u|_{(x+\frac{\Delta x}{2},y,z,t)}=\rho(x+\frac{\Delta x}{2},y,z,t)u(x+\frac{\Delta x}{2},y,z,t)=G(x+\Delta x)$

So again: $G(x+\frac{\Delta x}{2})=G(x)+\frac{\Delta x}{2} \frac{\partial}{\partial x}G(x)+\frac{\Delta x^2}{8} \frac{\partial^2}{\partial x^2}G(x)\pm...$