I have been given the Eulerian density function for a one-dimensional flow in the region $x\ge0$,$t\ge0$ to be: $$\rho(x,t)=(t+1)e^{-(t+1)x}$$ and have used the given fact that $v(0,t)=0$ and the conservation of mass equation to deduce that the Eulerian velocity function is given by $$v(x,t)=\frac{-x}{1+t}$$
I need to find the (Lagrange) particle trajectories $x(X,t)$, where $x(X,0)=X$. In these terms, the relation between Lagrange and Eulerian is: $$x(X(x,t),t)=x$$ I have done questions like this a lot the other way round (finding eulerian functions using lagrange) but I have no idea how to work backwards and find $x$ from $v$ and $\rho$. Is this more complicated?? i.e involving constructing a pde from the given information...? I'm really stuck, so any help would be greatly appreciated!
If we start with the definition of the conservation of mass, given by $$ \frac{\partial \rho}{\partial t} = -\nabla\cdot\left(\rho \mathbf{v}\right). $$ and $$ \rho(x,t) = (t+1)\mathrm{e}^{-(t+1)x},\\ \mathbf{v} = (v,0,0). $$ Using the fact that $\rho$ is a scalar function, we can expand as follows: $$ \frac{\partial \rho}{\partial t} = -\rho \nabla\cdot\mathbf{v}-\mathbf{v}\cdot\nabla\rho. $$ inserting the Euler density as leads to $$ \frac{\rho}{t+1} -x\rho = -\rho\frac{\partial v}{\partial x} - v(t+1)\rho $$ you can check this for yourself (I left it in terms of $\rho$'s instead of writing exponentials everywhere :) ) Which reduces to $$ \frac{\partial v}{\partial x} +(t+1)v = x -\frac{1}{t+1}. $$ we can solve for v using the integrating factor $$ v\mathrm{e}^{-(t+1)x} = \int x\mathrm{e}^{-(t+1)x} - \frac{1}{t+1}\mathrm{e}^{-(t+1)x}dx +\lambda $$ where $\lamda$ is a constant. We can solve the above easily to lead to
$$ v = -\frac{(t+1)x+1}{\left(t+1\right)^{2}} +\frac{1}{\left(t+1\right)^{2}} + \lambda\mathrm{e}^{(t+1)x} = -\frac{x}{t+1} + \lambda\mathrm{e}^{(t+1)x} $$ using the i.c $v(0,t) = 0$ we find that $\lambda=0$.