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I want to solve the following PDE initial value problem

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

and

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

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?

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    $\begingroup$ Did you consider reducing this to Burgers' equation for the function $u-1$? $\endgroup$ – Michał Miśkiewicz Dec 3 '16 at 16:41
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Following the comments, we transform this problem by setting $v=u-1$ as $$ v_t + v v_x = 2, \qquad v(x,0) = \left\lbrace \begin{aligned} &0 &&\text{for } x<0\\ &{-x} &&\text{for } 0<x<1\\ &{-1} &&\text{for } 1<x\\ \end{aligned}\right. $$ 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 $$ v(x,t) = \left\lbrace \begin{aligned} &2t &&\text{for } x<t^2\\ &\tfrac{t^2-x}{1-t} + 2t &&\text{for } t^2<x<t^2-t+1\\ &{-1}+2t &&\text{for } t^2-t+1<x\\ \end{aligned}\right. $$ 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 $$ v(x,t) = \left\lbrace \begin{aligned} &2t &&\text{for } x< \tfrac{1}{2}(4t-1)\\ &{-1}+2t &&\text{for } \tfrac{1}{2}(4t-1)<x\\ \end{aligned}\right. $$ To recover $u$, add $1$ to the values of $v$ above.

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

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  • $\begingroup$ 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? $\endgroup$ – Louis Dec 4 '16 at 18:26
  • $\begingroup$ @Louis There is indeed no reason for having time-independent values of $u$ for some $x$. Some terms $2t$ are missing. $\endgroup$ – Harry49 Feb 19 at 11:48

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