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.

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    $\begingroup$ $\partial_t$ holds $x$ constant, but it does not hold $z$ constant. So $\xi$ depends on $t$ through both the explicit $t$ and $z(t,x)$. $\endgroup$ Jul 1, 2022 at 13:51
  • $\begingroup$ 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. $\endgroup$ Jul 1, 2022 at 13:55
  • $\begingroup$ 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 $\endgroup$
    – Mjoseph
    Jul 1, 2022 at 13:57

1 Answer 1


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, $$ \partial_h G = \partial_\xi G\, \partial_h \xi + \partial_\phi G\, \partial_h \phi = \frac{1}{h_*}\partial_\xi G - \frac{\psi}{h^2} \partial_\phi G . $$


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