Characteristic curves of the advection equation

Question

I'm trying to understand Pg. 2 of this paper. Why do the characteristics of the equation $$ \partial_tu + (1-\rho)^2\partial_\rho u = 0 $$ satisfy the differential equation $$ \dfrac{d\rho}{dt} = -(1-\rho(t))^2 $$ rather than $d\rho/dt = (1-\rho(t))^2$? Why is there a negative sign? I was under the impression that in general, the characteristics of an equation $a\partial_xu+b\partial_yu = 0$ satisfy $dy/dx = b/a$.

Answer 1

I believe that it is a typo. Please see Figure 1 of that paper (attached below).

As time $t$ increases, $\rho$ increases, so its derivative must be positive.

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Answer 2

$$\partial_tu + (1-\rho)^2\partial_\rho u = 0 $$ The characteristic system of ODEs (Charpit-Lagrange) is : $$\frac{dt}{1}=\frac{d\rho}{(1-\rho)^2}=\frac{du}{0}$$ A first characteristic equation, coming from $du=0$, is : $$u=c_1$$ A second characteristic equation, coming from $\frac{dt}{1}=\frac{d\rho}{(1-\rho)^2}$ , is : $$t+\frac{1}{\rho-1}=c_2$$ The general solution of the PDE on the form of implicit equation $c_1=F(c_2)$ is : $$\boxed{u=F\left(t+\frac{1}{\rho-1}\right)}$$ $F$ is an arbitrary function, to be determined according to some boundary conditions.

So, you are right to point out that $\frac{d\rho}{dt} = (1-\rho(t))^2$ on the characteristic curves.