Let $(x,\tau)\in \mathbb{R}\times\mathbb{R^+}$. and we consider the following term: $$\frac{du_1}{d\tau}-\phi(\beta).\frac{du_1}{dx}-u_1\frac{d\phi(\beta)}{dx}$$ with : $\beta=x+\tau.\phi(\beta)$ ($\beta$ verifies this implicit equation).
I can't figure out why in this article : https://www.ams.org/journals/qam/1986-44-02/S0033-569X-1986-0856182-5/S0033-569X-1986-0856182-5.pdf
They author managed to prove that: $$\frac{du_1}{d\tau}-\phi(\beta).\frac{du_1}{dx}-u_1\frac{d\phi(\beta)}{dx}=\frac{\partial}{\partial\tau}[(1-\tau.\phi'(\beta)u_1]$$ It says that we need to move from the variables $(x,\tau)$ to $(\beta, \tau)$.
My idea for differentiation:
$\frac{du_1}{d\tau}=\frac{\partial u_1}{d\tau}+\frac{\partial u_1}{dx}\frac{d x}{d\tau}$
$\frac{du_1}{dx}=\frac{\partial u_1}{dx}+\frac{\partial u_1}{d\tau}\frac{d \tau}{dx}$
then $\frac{du_1}{d\tau}-\phi(\beta).\frac{du_1}{dx}-u_1\frac{d\phi(\beta)}{dx}=\frac{\partial u_1}{\partial \tau}(1-\phi(\beta).\frac{d\tau}{dx})+\frac{\partial u_1}{\partial x}(\frac{dx}{d\tau}-\phi(\beta))+u_1(\frac{\phi'(\beta)}{(1-\tau\phi'(\beta))}+\phi'(\beta)\frac{1}{1-\tau \phi'(\beta)}\frac{\partial x}{\partial \tau}) $
but using these derivation I could not come up to the conclusion
As I indicated, it is best to rename $\tau$ in one of the coordinate systems to $t$. I'll point out why later. So, consider $(x, t) \in \Bbb R \times \Bbb R^+$. Given functions $\phi : \Bbb R \to \Bbb R$ and $\beta : \Bbb R \times \Bbb R^+ \to \Bbb R$ which satisfy $$\beta(x, t) = x +t\phi(\beta(x,t))$$
From this formula, we can calculate
$$\frac{\partial \beta}{\partial x} = 1 + t\phi'(\beta)\frac{\partial \beta}{\partial x}\\\frac{\partial \beta}{\partial x} = \frac 1{1-t\phi'(\beta)}$$ and $$\frac{\partial \beta}{\partial t}= \phi(\beta) + t\phi'(\beta)\frac{\partial \beta}{\partial t}\\\frac{\partial \beta}{\partial t}= \frac {\phi(\beta)}{1-t\phi'(\beta)} = \phi(\beta)\frac{\partial \beta}{\partial x}$$
Now we define a second coordinate system $(\beta, \tau)$ on $\Bbb R \times \Bbb R^+$ with $\tau = t$, and therefore $$\frac{\partial \tau}{\partial x} = 0, \quad \frac{\partial \tau}{\partial t} = 1$$. We then express $(x, t)$ in terms of $(\beta, \tau)$: $$t = \tau\\x = \beta - \tau\phi(\beta)$$ giving $$\frac{\partial t}{\partial \beta} = 0, \quad \frac{\partial t}{\partial \tau} = 1$$ and $$\frac{\partial x}{\partial \beta} = 1 - \phi'(\beta),\quad \frac{\partial x}{\partial \tau} = -\phi(\beta)$$
Consider a map $u : \Bbb R \times \Bbb R^+ \to \Bbb R$ satisfying $$\frac{\partial u}{\partial t}-\phi(\beta)\frac{\partial u}{\partial x}-u\frac{\partial \phi(\beta)}{\partial x}=\frac{\partial^2 \phi(\beta)}{\partial x^2}$$
By the chain rule, $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial x} + \frac{\partial u}{\partial \tau}\frac{\partial \tau}{\partial x} = \frac 1{1-t\phi'(\beta)}\frac{\partial u}{\partial \beta} + 0$$ $$\frac{\partial u}{\partial t} = \frac{\partial u}{\partial \beta}\frac{\partial \beta}{\partial t} + \frac{\partial u}{\partial \tau}\frac{\partial \tau}{\partial t} = \frac {\phi(\beta)}{1-t\phi'(\beta)}\frac{\partial u}{\partial \beta} + \frac{\partial u}{\partial \tau}$$
Notice that last equation. Even though $t = \tau$, $$\frac{\partial u}{\partial t} \ne \frac{\partial u}{\partial \tau}$$ This is why is was useful to use different variables for the same value in the $(x, t)$ and $(\beta, \tau)$ coordinate systems. Otherwise that statement would have been a very confusing $\frac{\partial u}{\partial \tau} \ne \frac{\partial u}{\partial \tau}$. And it is also a good demonstration that partial derivatives are not the same as ordinary derivatives.
How is it that $\frac{\partial u}{\partial t} \ne \frac{\partial u}{\partial \tau}$ even though $t = \tau$? Because $\frac{\partial u}{\partial t}$ is the derivative of $u$ along a curve with the value of $x$ constant., but $\frac{\partial u}{\partial \tau}$ is the derivative of $u$ along a curve with the value of $\beta$ constant. In this case, holding $x$ constant and holding $\beta$ constant give you different curves, which approach the place where the derivative is being taken in different directions. And in general, $u$ need not behave the same way in these two different directions.
What is "partial" about the partial derivative is the notation. It pretends to involve just two variables, $u$ and $\tau$. But in fact it depends on the entire coordinate system of which $\tau$ is a part. The use of the $\partial$ symbol instead of $d$ is a reminder of this. Whenever you see it, you should remember that dependence on the coordinate system.
Now I have a kitten recovering from a certain operation who is demanding my attention, so you can take it from here. But one more thing to notice: The author transformed the entire equation, not just one side. You can expect some rearrangement of what is on each side.