Days ago I saw this:
... (1) we have $$\frac{dx}{u}=\frac{dt}{1}=\frac{du}{0}$$
i.e.
$$\frac{u}{dx}=\frac{1}{dt}=\frac{0}{du}$$
Why are both equalities equivalent ?
I can see that $\frac{dx}{u}=\frac{dt}{1}\iff\frac{u}{dx}=\frac{1}{dt}$ but then you have to swap again with the other expression $\frac{du}{0}$ to have $\frac{0}{du}$ but if you do this then you will modify this $\frac{dx}{u}=\frac{dt}{1}\iff\frac{u}{dx}=\frac{1}{dt}$ again, i.e. you wouldn't end up having $\frac{u}{dx}=\frac{1}{dt}=\frac{0}{du}$.
Edit
Consider the equation $u_t+uu_x=0$, then we have (1)
$$\frac{dx}{u}=\frac{dt}{1}=\frac{du}{0}=ds$$
I wouldn't say that is is a trick. May be a more convenient word is a short-cut, i.e. a synoptic way to write : $$\frac{dx}{u}=ds$$ $$\frac{dt}{1}=ds$$ $$\frac{du}{0}=ds$$
On a formal manner:
$$\frac{1}{ds}=\frac{u}{dx}$$ $$\frac{1}{ds}=\frac{1}{dt}$$ $$\frac{1}{ds}=\frac{0}{du}$$
and so :
$$\frac{1}{ds}=\frac{u}{dx}=\frac{1}{dt}=\frac{0}{du}$$
All this cannot be understood independently from the theory of the method of characteristics, which can be found in textbooks and which leads to more extended relationships.
The above symbolic is useful to simplify the writing in some cases. For example, see section 2.2 in http://www1.maths.leeds.ac.uk/~kersale/Teach/M3414/Notes/m3414_2.pdf
EXAMPLE : $$u_t+uu_x=0$$ $\frac{dx}{u}=\frac{dt}{1}=\frac{du}{0}$
A first family of characteristic curves comes from $\quad \frac{du}{0}=$finite$\quad\implies du=0$. $$u=c_1$$ A second family of characteristic curves comes from $\quad \frac{dx}{c_1}=\frac{dt}{1}$ $$c_1t-x=c_2$$ The general solution of the PDE, expressed on the form of implicit equation, is : $$\Phi\left(u\:,\: u\,t-x \right)=0$$ where $\Phi$ is an arbitrary function of two variables, to be determined according to the boundary and/or initial condition.
Equivalently : $$u=F(u\,t-x) \tag 1$$ where $F$ is an arbitrary function, to be determined according to the boundary and/or initial condition.
Without specified condition one cannot go further.
With some well posed condition, the function $F$ can be determined. Then in some cases (not always) Eq.$(1)$ can be solved for $u$ and the solution $u(x,t)$ is explicitly obtained. In other cases, the solution remains on the form of an implicit equation.
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