How does the solution for the initial value problem change over time?

[evinda]

Hello! (Wave)

We have the problem \begin{align*}&u_t+uu_x=0 \\ &u(x,0)=\begin{cases}2, & x<0 \\ 2-x, & x\in[0,1] \\1, & x>1\end{cases}\end{align*} We have the problems $$\frac{dx}{dt}=u \ \text{ and } \ \frac{du}{dt}=0$$ From the first one we get $x=ct+c_2$ and for $t=0$ we get $x_0=c_2$ and so we get $x_0=x-ct$.

From the second one we get $$\frac{dx}{dt}=u(x,t)=u(x_0,0)=\begin{cases}2, & x_0<0 \\ 2-x_0, & x_0\in[0,1] \\1 , & x_0>1\end{cases}$$ Solving for $x_0$ we get $$x_0=\begin{cases}x-2t, & x<2t \\ \frac{2t-x}{t-1}, & \frac{2t-x}{t-1} \in [0.1] \\ x-t, & x>t\end{cases}$$ So, $u$ is equal to $$u=\begin{cases}2, & x<2t \\ 2-\frac{2t-x}{t-1}, & \frac{x-2}{t-1}\in[0,1] \\1 , & x>t\end{cases}=\begin{cases}2, & x<2t \\ \frac{x-2}{t-1}, & \frac{x-2}{t-1}\in[0,1] \\1 , & x>t\end{cases}$$ For $t<1$ we get $$u=\begin{cases}2, & x<2t \\ \frac{x-2}{t-1}, & 2t\leq x\leq t+1 \\1 , & x>t\end{cases}$$ and for $t\geq 1$ we get $$u=\begin{cases}2, & x<2t \\ \frac{x-2}{t-1}, & t+1\leq x\leq 2t \\1 , & x>t\end{cases}$$ This can be simplified into $$u=\begin{cases}2, & x<2t \\1 , & x>t\end{cases}$$ or not?

According to the solution the case $t\geq 1$ is $$u=\begin{cases}2, & x<\frac{3}{2}t+\frac{1}{2} \\1 , & x>\frac{3}{2}t+\frac{1}{2}\end{cases}$$ But how do we get the inequalities with $\frac{3}{2}t+\frac{1}{2}$ ? (Thinking)Also... why do we have for $t<1$ a classical solution, but for $t>1$ a weak one?

 

[evinda]

Or can we not apply the method I did? (Thinking)

 

[I like Serena]

Hmm... doesn't the solution break down at $x=2$? (Wondering)

The initial value problem gives us solutions up to $x=2$, which is a singularity.
What happens after is effectively another initial value problem, which I think could go anywhere, including the given solution. (Thinking)