My question deals with the continuity equation and the changing of the derivative function.
The continuity equation for one-dimensional flow requires that
$$ \frac{\partial(\rho v)}{\partial x} = - \phi \frac{\partial \rho}{\partial t} $$
From some math steps that are performed (not shown here), the change in pressure with respect to time is found to be:
$$ \frac{\partial P}{\partial t} = \frac{g \sqrt{c+1}}{\sqrt{c+g}} \cdot P'_o $$
where $$g \equiv \left(1-\frac{x}{L}\right)$$
Since $ \rho = PM/(RT) $ and $g = 1 - x/L $, then we can substitute our relations back into our continuity equation as: $$ \frac{\partial(\rho v)}{\partial g} = \frac{ \phi LMP'_o \sqrt{c+1}}{RT} \cdot \frac{g}{\sqrt{c+g}}$$
As can be seen, our LHS changed: $$ \frac{\partial(\rho v)}{\partial x} \rightarrow \frac{\partial(\rho v)}{\partial g} $$ with a corresponding change in the numerator on the RHS by multiplying by L.
My questions is, What are the steps mathematically to show that this is true? I feel that it is simple algebra saying that $$\frac{\partial(\rho v)}{\partial g} = \frac{\partial(\rho v)}{\partial \big(\frac{L-x}{L}\big)} $$
therefore: $$\frac{\partial(\rho v)}{\partial ({L-x})} = \frac{ \phi LMP'_o \sqrt{c+1}}{RT} \cdot \frac{g}{\sqrt{c+g}} $$
And we can leave it written as $\frac{\partial(\rho v)}{\partial g} = \frac{ \phi LMP'_o \sqrt{c+1}}{RT} \cdot \frac{g}{\sqrt{c+g}} $ ?
I hope that someone can clear this up for me. Thanks.
This is a direct application of the chain rule. Seeing $g$ as a function of $x$, the following is true :
$$ \frac{\partial}{\partial x} = \frac{\partial g}{\partial x}\frac{\partial}{\partial g}. $$
Since $g=1-\frac{x}{L}$, we find
$$ \frac{\partial}{\partial x} = -\frac{1}{L}\frac{\partial}{\partial g}, $$
or
$$ \frac{\partial}{\partial g} = -L\frac{\partial}{\partial x}. $$
Plugging this result in your previous calculations gives
$$ \begin{align} \frac{\partial(\rho v)}{\partial g} &= -L\frac{\partial(\rho v)}{\partial x} \\ &= L\phi \frac{\partial}{\partial t}\left(\frac{PM}{RT}\right) \\ &= \frac{L\phi M}{RT}\frac{g\sqrt{c+1}}{\sqrt{c+g}}P_0'. \end{align} $$