# Question on partial differential equation using equation in fluid dynamics

In fluid dynamics, the one-dimensional continuity equation is given by $$\frac{\partial \rho}{\partial t}+\frac{\partial}{\partial x}(\rho u)=0$$ where $\rho =\rho(x,t)$ is the density of the fluid and $u=u(x,t)$ is the speed of the fluid. Show that if $\rho=f(x+ct)$, where c is a constant, then $$\frac{\partial}{\partial x}[(u+c)\rho]=0.$$

$\frac{\partial \rho}{\partial t}+\frac{\partial}{\partial x}(\rho u)=0$

$\frac{\partial}{\partial t}[\rho(x,t)]+ \frac{\partial}{\partial x}[\rho(x,t)u(x,t)]=0$

$\frac{\partial}{\partial t}[f(x+ct)]+ \frac{\partial}{\partial x}[f(x+ct)u(x,t)]=0$

$cf'(x+ct)+f(x+ct)[\frac{\partial}{\partial x}u(x,t)]+u(x,t)[\frac{\partial}{\partial x}f(x+ct)]=0$

$f'(x+ct)[c+u(x,t)]+\frac{\partial}{\partial x}u(x,t)f(x+ct)=0$

Then, I get a bit lost to prove the required part.

$$f'(x+ct)[c+u(x,t)]+\frac{\partial}{\partial x}u(x,t)f(x+ct)=0$$
By the product rule, the above line implies $$\frac{\partial}{\partial x}[f(x+ct)(c+u(x,t))]=0.$$ That is $$\frac{\partial}{\partial x}[(u+c)\rho] =0.$$