# Shock-wave solution for PDE $u_t+(u-1)u_x=2$

I want to solve the following PDE initial value problem

$$u_t+(u-1)u_x=2$$

and

$$u (x,0)=\begin{cases} 1 & \text{for } x <0,\\ 1-x & \text{for } 0

However, I find that I have intersecting characteristics between $$x=t^2$$ and $$x=t^2-t+1$$.

How would I apply the shockwave method in this case since the PDE is given? Is it possible to solve this PDE as is?

• Did you consider reducing this to Burgers' equation for the function $u-1$? – Michał Miśkiewicz Dec 3 '16 at 16:41

## 2 Answers

Following the comments, we transform this problem by setting $$v=u-1$$ as v_t + v v_x = 2, \qquad v(x,0) = \left\lbrace \begin{aligned} &0 &&\text{for } x<0\\ &{-x} &&\text{for } 0 The breaking time for the Burgers equation above is $$t_b = -1/\inf v_x(x,0) = 1$$. Hence, there is indeed a shock formation. Before the shock, the solution is given by the method of characteristics. Here, the characteristic curves are $$x = v(x_0,0) t + t^2 + x_0$$ along which $$v = v(x_0,0)+2t$$. Hence, for $$t<1$$, the solution reads v(x,t) = \left\lbrace \begin{aligned} &2t &&\text{for } x When characteristic curves intersect, the Rankine-Hugoniot condition gives the shock speed $$\dot x_s(t) = \frac{1}{2}(2t - 1 + 2t)$$ with $$x_s(1) = 1$$. Therefore, for $$t\geq 1$$, the solution reads v(x,t) = \left\lbrace \begin{aligned} &2t &&\text{for } x< \tfrac{1}{2}(4t-1)\\ &{-1}+2t &&\text{for } \tfrac{1}{2}(4t-1) To recover $$u$$, add $$1$$ to the values of $$v$$ above.

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$

$\dfrac{du}{ds}=2$ , letting $u(0)=u_0$ , we have $u=u_0+2s=u_0+2t$

$\dfrac{dx}{ds}=u-1=u_0+2s-1$ , letting $x(0)=f(u_0)$ , we have $x=s^2+(u_0-1)s+f(u_0)=t^2+(u-2t-1)t+f(u-2t)=(u-1)t-t^2+f(u-2t)$ , i.e. $u=2t+F(x+t^2-(u-1)t)$

$u(x,0)=\begin{cases}1&\text{when}~x<0\\1-x&\text{when}~0<x<1\\0&\text{when}~1<x\end{cases}$ :

$\therefore u=\begin{cases}1&\text{when}~x+t^2-(u-1)t<0\\2t+1-x-t^2+(u-1)t&\text{when}~0<x+t^2-(u-1)t<1\\0&\text{when}~1<x+t^2-(u-1)t\end{cases}$

Hence $u=\begin{cases}1&\text{when}~x<-t^2\\\dfrac{x+t^2-t-1}{t-1}&\text{when}~0<x+t^2-\dfrac{(x+t^2-t-1)t}{t-1}+t<1\\0&\text{when}~x>1-t^2-t\end{cases}$

• I don't really understand how you went from $\frac{dx}{ds}=u(s)-1$ to $\frac{dx}{ds}=u(0)-1$. Wouldn't this mean that $u$ is constant (along the characteristic) which it is not? – Louis Dec 4 '16 at 18:26
• @Louis There is indeed no reason for having time-independent values of $u$ for some $x$. Some terms $2t$ are missing. – Harry49 Feb 19 at 11:48