Explicit solution to IVP of PDE $\rho_t = [\rho (1-\rho)]_x$

Question

When trying to determine the density profile $\rho(t,x)$ of a system of particles I came across the PDE:
$$\frac{\partial \rho}{\partial t}=\frac{\partial}{\partial x}\big(\rho (1-\rho)\big), \qquad\text{for all}\qquad x \in \mathbb{R}, t>0,$$ with the initial condition $\displaystyle \lim _{t \rightarrow 0^{+}}\rho(t,x)=\mathbb{1}_{\{x<0\}}$.

Does this equation have an explicit solution?

I tried the solution of the form $\rho(t,x)=h(x/t)$ which leads to $h(u)=\frac{u+1}{2}$ but the initial condition is not satisfied. By $\mathbb{1}_{\{x<0\}}$, I mean the Heaviside step function (i.e., the characteristic function of $\Bbb R_-^*$).

Answer 1

The PDE $\rho_t + Q(\rho)_x = 0$ with the flux $Q(\rho) = -\rho (1-\rho)$ corresponds actually to the Lighthill-Whitham-Richards (LWR) traffic flow model for cars propagating towards decreasing $x$, where $\rho$ denotes the car density. The initial-value problem $\rho(x,0) = 1$ for $x< 0$ and $\rho(x,0) = 0$ for $x\geq 0$ is a Riemann problem. The situation can be interpreted as a one-way road, which is saturated for $x<0$ and empty for $x\geq 0$. Using the method of characteristics, we obtain the set of lines along which $\rho$ is constant (represented below in the $x$-$t$ plane):

The Lax entropy condition tells that the admissible solution is a shock wave, since characteristic curves intersect. According to the Rankine-Hugoniot condition, the shock propagates at the speed $s = 0$. The solution $$ \rho(x,t) = \left\lbrace \begin{aligned} &1 &&\text{if}\quad x<0 \\ &0 &&\text{if}\quad x\geq 0 \end{aligned}\right. $$ is a static discontinuity located at $x=0$, which is coherent with intuition. Indeed, since the cars are moving towards decreasing $x$, the cars initially located in the saturated part $x<0$ cannot move forward.

The opposite situation where $\rho(x,0) = 0$ for $x\leq 0$ and $\rho(x,0) = 1$ for $x> 0$ may be interpreted as a red light located at $x=0$, which turns green at $t=0$. Cars initially located in the saturated half-line $x>0$ can move forward towards decreasing $x$. The entropy solution is a rarefaction wave (i.e., a self-similar smooth solution) of the form $$ \rho(x,t) = \left\lbrace \begin{aligned} &0 &&\text{if}\quad x\leq -t \\ &\tfrac{1}{2}(1+x/t) &&\text{if}\quad {-t} \leq x\leq t \\ &1 &&\text{if}\quad x> t \end{aligned}\right. $$

Answer 2

$$\frac{\partial \rho}{\partial t}-\frac{\partial}{\partial x}\big(\rho (1-\rho)\big)=0\qquad;\qquad \rho(0,x)=\mathbb{1}$$ As already pointed out in comments, an obvious solution is : $$\rho(x,t)=1$$ If one cannot see that at first sight, one can solve analytically the PDE.

The Charpit-Lagrange system of ODEs is : $$\frac{dx}{1}=\frac{dt}{-\rho (1-\rho)}=\frac{d\rho}{0}$$ A first characteristic equation is : $$\rho=c_1$$ A second characteristic equation comes from $\frac{dx}{1}=\frac{dt}{-c_1 (1-c_1)}$ : $$x+c_1(1-c_1)t=c_2$$ The general solution of the PDE is expressed on the form of the implicit equation : $$\rho=F\left(x+\rho (1-\rho)t\right)$$ where $F$ is any function (to be determined according to the boundary condition).

Boundary condition : $\rho(0,x)=1=F\left(x+1 (1-1)t\right)=F(x)$

The function $F$ is determined : $F(x)=1$ any value of $x$.

We put it into the general solution, which leads to the particular solution satisfying the boundary condition : $\rho=F\left(x+\rho (1-\rho)t\right)=1$ $$\rho(x,t)=1$$ Of course, the general solving only to get such an obvious solution is like using a sledgehammer to kill a fly.