Plot characteristic curves for Initial value problem

Question

I'm trying to plot the characteristic of the following initial value problem, but I am stuck without a curve after finding the characteristic equation.

IVP: $$u_t + [u(1 − u)]_x = 0 \text{ for } x ∈ \mathbb{R}, t > 0, \\ u(0, x) = x\text{ for }x ∈ \mathbb{R}$$ for sufficiently small $t$.

Characteristic equation that I found $$ξ(t) = (1 - 2ξ_0)t + ξ_0 \\ 1-2ξ_0 = 0, ξ_0 = 1/2 \\ 1-2ξ_0 < 0, ξ_0 < 1/2 \\ 1-2ξ_0 > 0, ξ_0 > 1/2 \\ $$

Is the characteristic equation correct? And if yes, what is the plot for the same?

Thank you

Answer 1

The equation for the characteristics looks good to me, but note that $$ 1 - 2 \xi_0 <0 \Rightarrow \xi_0 > \frac12 \\ 1 - 2 \xi_0 >0 \Rightarrow \xi_0 < \frac12 $$

The picture looks then (plotted up to $t = 0.5$, where the characteristics intersect):

Answer 2

$$u_t+(1-2u)u_x=0$$ Charpit-Lagrange characteristic ODEs : $$\frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{0}$$ A first characteristic equation comes from $du=0$ : $$u=c_1$$ A second characteristic equation comes from $\frac{dt}{1}=\frac{dx}{1-2c_1}$ $$(1-2c_1)t-x=c_2$$ General solution of the PDE on the form of implicit equation $c_1=F\Big((1-2c_1)t-x\Big)$ : $$\boxed{u=F\Big((1-2u)t-x\Big)}$$ $F$ is an arbitrary function.

Condition : $$u(0,x)=x=F\Big((1-2x)0-x\Big)$$ $$x=F(-x)\quad\implies\quad F(X)=-X$$ Now the function $F(X)$ is determined. We put it into the above general solution where $X=(1-2u)t-x$ : $$u=-\Big((1-2u)t-x\Big)=(2u-1)t+x$$ $$\boxed{u(t,x)=\frac{x-t}{1-2t}}$$ This is the solution which satisfies both the PDE and the condition.

Starting from $t=0$ this is valid up to $t=\frac12$ where $u$ tends to infinity except for $(x=\frac12\:,\:u=\frac12)$.

The drawing of $u(x)$ at various values of $t$ in cases of $0<t<\frac12$ is a bundle of straight lines with common point $(x=\frac12\:,\:u=\frac12)$. This is in fact what you are looking for.

The drawing of $u(t)$ at various values of $x$ in cases $x\neq\frac12$is a bundle of hyperbolas with common vertical asymptote at $t=\frac12$.