Quasilinear PDE $u_t + (u^2)_x = 0$ cauchy problem

Question

The problem I am trying to solve is: \begin{equation}\label{eq:3.1} \begin{cases} \partial_t u + \partial_x(u^2)=0 & x\in \mathbb{R}, t \in (0,\infty]\\ u(x,0)= \begin{cases} 0 & x\leq 0\\ x & 0<x\leq 1\\ 1 & x>1 \end{cases} \end{cases} \end{equation}

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What I have done is:

We will try to reduce the problem to ODEs on a curve $x(t)$ on the $(t,x)$ plane. The equation can be compared with the canonical form, \begin{equation} a\frac{\partial u}{\partial x} +b\frac{\partial }{\partial t} = c, \end{equation} where $a = 2u$, $b= 1$ and $c=0$. From the Lagrange-Charpit equations, we have, \begin{align}\label{eq:3.2} &\frac{dx}{a}=\frac{dt}{b}=\frac{du}{c} & \text{ substituting we have,}\nonumber\\ \implies &\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}& \end{align} using second and the third ratio from the equation we have, \begin{align}\label{eq:3.3} &\frac{du}{dt}=0 & \text{integrating we have,} \nonumber\\ \implies&u=B,& \end{align} where $B$ is a arbitrary constant. Using the initial conditions, \begin{equation}\label{eq:3.4} u(x,0)= \begin{cases} 0 & x\leq 0\\ x & 0<x\leq 1\\ 1 & x>1 \end{cases} \end{equation} where the characteristic curve $x(t)$, passes through $(c,0)$. By substitution we have, \begin{equation} B= \begin{cases} 0 & x\leq 0\\ c & 0<x\leq 1\\ 1 & x>1. \end{cases} \end{equation}
Therefore solution can be written as \begin{equation}\label{eq:3.5} u= \begin{cases} 0 & x\leq 0\\ c & 0<x\leq 1\\ 1 & x>1. \end{cases} \end{equation} using first and the second ratios from the equation we have, \begin{align}\label{eq:3.6} &\frac{dx}{dt}=2u & \text{substituting we have,} \nonumber\\ \implies&\frac{dx}{dt}= \begin{cases} 0 & x\leq 0\\ 2c & 0<x\leq 1\\ 2 & x>1. \end{cases} &\text{integrating we have,}\nonumber\\ \implies&x= \begin{cases} B & x\leq 0\\ 2ct+B & 0<x\leq 1\\ 2t+B & x>1. \end{cases} &\nonumber\\ \end{align} where $B$ is a arbitrary constant. Using the initial conditions , and that the characteristic curve $x(t)$ passes through $(c,0)$ we have, \begin{equation} x= \begin{cases} c & x\leq 0\\ 2ct+c & 0<x\leq 1\\ 2t+c & x>1. \end{cases} \end{equation} Therefore $u$ becomes, \begin{equation} u(x,t)= \begin{cases} 0 & x\leq 0\\ \frac{x}{2t+1} & 0<x\leq 1\\ 1 & x>1. \end{cases} \end{equation} I think I am missing something. The solution should have $t$ dependence in the intervals. Thanks.

Answer 1

The main part that you did seems correct. Except the limits at the end. $$\begin{equation} \frac{\partial u}{\partial x} +2u\frac{\partial u}{\partial t} = 0 \end{equation}$$ Your Charpit-lagrange characteristic ODEs are correct : $$\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}$$ A first characteristic equation comes from $du=0$ : $$u=c_1$$ A second characteristic equation comes from $\frac{dx}{2c_1}=\frac{dt}{1}$ : $$x-2c_1t=c_2$$ The general solution of the PDE expressed on implicite form $c_1=F(c_2)$ is : $$u=F(x-2ut)$$ where $F$ is an arbitrary function to be determined according to the initial condition. \begin{equation} u(x,0)=F(x)= \begin{cases} 0 & x\leq 0\\ x & 0<x\leq 1\\ 1 & x>1 \end{cases} \end{equation} So, the function $F$ is determined any variable $\chi$ :

\begin{equation} F(\chi)= \begin{cases} 0 & \chi\leq 0\\ \chi & 0<\chi\leq 1\\ 1 & \chi>1 \end{cases} \end{equation}

We put this function $F(\chi)$ into the above general solution where $\chi=x-2ut$ \begin{equation} u=F(x-2ut)= \begin{cases} 0 & x-2ut\leq 0\\ x-2ut & 0<x-2ut\leq 1\\ 1 & x-2ut>1 \end{cases} \end{equation}

Case $u=0$ and $x-2ut\leq 0\quad\to\quad x\leq 0$ .

Case $u=x-2ut$ and $0<x-2ut\leq 1\quad\to\quad u=\frac{x}{1+2t}$ and $0<x\leq 1+2t$

Case $u=1$ and $x-2ut>1 \quad\to\quad x>1+2t$

The solution is :

\begin{equation} u(x,t)= \begin{cases} 0 & x\leq 0\\ \frac{x}{1+2t} & 0<x\leq 1+2t\\ 1 & x>1+2t \end{cases} \end{equation}

Answer 2

This PDE is very similar to Burgers equation, and the solution $u(x,t)$ deduced from the method of characteristics reads $u = f(x-2u t)$ in implicit form, where $f = u(\cdot, t=0)$. Following the steps in the linked post (see also the comments section), we find $$ u(x,t) = \left\lbrace \begin{aligned} &0 & & x\leq 0\\ &\tfrac{x}{1+2t} & & 0< x\leq 1+2t\\ &1 & & x> 1+2t \end{aligned}\right. $$