Solution of continuity equation

Question

How can I find exact solution to $$ \rho_t + \frac{d}{dx}(\rho f) = 0, $$ where $\rho(x,0) = 1$ and $f(x,t) = \frac{x}{t-1}$?

Answer 1

The continuity equation can be written as follows: $$ \rho_t+f\rho_x+f_x\rho=0$$ this is a case of the so-called transport PDE equations of first order which have as a general form: $$ u_t+v(t,x)\cdot\nabla u+b(t,x) u=F(t,x) \quad(*)$$ where, $$v(t,x)=(v_1(t,x),v_2(t,x),\cdots,v_n(t,x))^T $$ $$\nabla u=(u_{x_1},u_{x_2},\cdots,u_{x_n})^T $$ Here we have $$ v(t,x)=f(t,x) \mbox{ , } b(t,x)=f_x(t,x) \mbox{ and F(t,x)=0} $$ Then we define the following ODE sytem that is derived from the characteristics method: Find $X(s,t,x)$ such that $$ \begin{cases}\dfrac{\partial X}{\partial s}=v(s,X)\\X(t,t,x)=x \end{cases}$$ But in our case, we shall not resolve this ODE system because $b(t,x)=f_x(t,x)$ does not depend on $x$. the solution of $(*)$ under an intitial condition: $$ u(0,x)=u_0(x)\quad \mbox{ , in our case: }\quad u_0(x)=\rho (0,x)=1$$ \begin{align*} u(t,x)&=\underbrace{u_0(X(0,t,x))\exp\left(-\int_0^t b(\tau,X(\tau,t,x))d\tau\right)}_{Homogeneous\ solution}\\&+\underbrace{\int_0^t\exp\left(-\int_s^tb(\tau,X(\tau,t,x))d\tau\right)f(s,X(s,t,x))ds}_{Particle\ solution}\end{align*} In our case, we don't have a RHS, so we don't have a particle solution. then our solution is: \begin{align*}\rho(x,t)&=1\cdot\exp\left(-\int_0^t\dfrac{1}{\tau-1}d\tau\right)\\ &=\dfrac{1}{|t-1|} \end{align*} So the solution of the given continuity equation is defined by: $$ \rho(t,x)=\begin{cases}\dfrac{1}{t-1}&t>1\\ \dfrac{1}{1-t}&0\leqslant t< 1 \end{cases}$$

Answer 2

Hint For a given number $c$, find the equation satisfied by $\varphi(t) = \rho(c(1-t), t)$.