# Partial Derivatives involving Change of Variables

I am currently reading Elementary Fluid Dynamics by Acheson and following along $$3.9$$. In this section, the author uses a change of variables: $$z=3c-2c_0$$ (where $$c_0$$ is a constant and $$z,c$$ are functions of $$x,t$$) to transform the equation: $$c_t + (2c_0-3c)c_x=0 \quad\implies\quad z_t-zz_x=0$$ which is simply the advection equation with $$z$$ as the speed and hence the solution is: $$z = F(x+zt)$$ All of this I understand just fine. However, immediately after the author verifies this is the solution by actually differentiating and substituting this form for $$z$$ into the PDE. This is where I am a little confused. Why is it that: $$\frac{\partial}{\partial t}(F) = F'(\xi)\frac{\partial \xi}{\partial t} + F'(\xi)\frac{\partial \xi}{\partial z}\frac{\partial z}{\partial t}$$ where $$\xi = x+zt$$. It seems to me that either of these terms on their own would be equal to $$\frac{\partial}{\partial t}(F)$$ but I don't quite see why both are necessary. This is something that I keep running into and it always confuses me. Note my confusion is not with the chain rule, I understand that just fine and see how either term individually would be found, it's the fact that both terms are needed that confuses especially considering $$F$$ is a function of only one variable.

Example 2: Another similar example that I can follow along but can't quite justify to myself comes from a fluid dynamics paper. In the paper, the film height $$h=h(x,y,t)$$ and the height-averaged per volume solute concentration $$\phi = \phi(x,y,t)$$ are reformulated as: $$\xi = h/h_* \qquad\text{and}\qquad \psi = h\phi$$ Later on in the paper, the author takes the partial derivative with respect to $$h$$ of a function given as $$G=G(\xi,\phi)$$. Hence, in this case it makes sense to me that this partial derivative will be of two terms, namely: $$\frac{\partial}{\partial h}G = \frac{\partial G}{\partial \xi}\frac{\partial\xi}{\partial h} + \frac{\partial G}{\partial \phi}\frac{\partial \phi}{\partial h} = \frac{1}{h_*}G_\xi -\frac{\psi}{h^2} G_\phi$$ but I can't quite reconcile these two examples. In the second case, $$G$$ is a function of two variables both of which depend on $$h$$. Hence, it seems intuitive that both terms will be needed. However, in the first case, we only have a function of one variable.

• $\partial_t$ holds $x$ constant, but it does not hold $z$ constant. So $\xi$ depends on $t$ through both the explicit $t$ and $z(t,x)$. Jul 1, 2022 at 13:51
• Oh, I think I see the problem here. When he writes $\partial_t \xi$, he means the derivative of $\xi$ holding $x$ and $z$ constant. But for $\partial_t F$ and $\partial_t z$, he means the derivative holding $x$ (but not $z$) constant. Jul 1, 2022 at 13:55
• And that's just because $z=z(x,t)$? I don't think I quite understand why you hold it constant in the first case and not the second Jul 1, 2022 at 13:57

For the implicit solution $$z = F(\xi) ,\qquad \xi = x+zt$$ where $$z = z(x,t)$$, we note that all the above quantities are functions of $$(x,t)$$. Thus, partial differentiation with respect to $$t$$ gives directly $$\partial_t z = F'(\xi)\, \partial_t \xi , \qquad \partial_t \xi = 0 + (\partial_t z)\, t + z ,$$ according to the chain rule and the product rule. This was the easy computation, which is pretty much straightforward to follow. Now, let us view $$\xi$$ as a function of $$(x,t,z)$$. It follows that time-domain partial differentiation takes the following form $$\partial_t \xi = \partial_t\xi + \partial_z \xi\, \partial_t z = z + t\, \partial_t z ,$$ which is a total derivative in time. Note that both approaches give the exact same result. Hope this observation / explanation helps.
NB. This identity is then used to write $$\partial_t z = \frac{zF'(\xi)}{1-tF'(\xi)}$$ and similar computations are performed for $$\partial_x z$$ to end the proof.
The second example involves the computation of a total derivative. In fact, we have set $$G = G(\xi,\phi)$$ where $$\xi$$ and $$\phi$$ may be expressed in terms of $$h$$. Similarly to above, $$\partial_h G = \partial_\xi G\, \partial_h \xi + \partial_\phi G\, \partial_h \phi = \frac{1}{h_*}\partial_\xi G - \frac{\psi}{h^2} \partial_\phi G .$$