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Given Burgers' equation: $u_{t}+\frac{1}{2}(u^{2})_{x}=0$ with the initial condition:

$$u(x,0) =\begin{cases} \displaystyle u_{l},\quad x <0 \\ \displaystyle u_{r},\quad x>0 \end{cases}$$

where $u_{l} < u_{r}$.

I understand how to find the solution:

$$u(x,t) =\begin{cases} \displaystyle u_{l},\quad x < u_{l}t \\ \displaystyle \frac{x}{t},\quad u_{l}t < x< u_{r}t \\ \displaystyle u_{r},\quad x>u_{r}t \end{cases}$$

I am not sure how to find the solution when $\frac{x}{t} = 0$. Would the solution just be this?

$$u(x,t) =\begin{cases} \displaystyle u_{l},\quad 0 < u_{l} \\ \displaystyle 0,\quad u_{l} < 0< u_{r} \\ \displaystyle u_{r},\quad 0>u_{r} \end{cases}$$

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rtob
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1 Answers1

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The solution to the Riemann problem with $u_l < u_r$ is nearly correct. To be correct, the strict inequalities must be replaced by not strict ones. Moreover, one should be aware that $t$ is assumed positive here. For all $t> 0$, the value along the half-line $x/t=0$ is found by setting $x/t=0$ or $x=0$ in the solution: $$ \left. u(x,t) \right|_{x/t=0} = \left\lbrace \begin{aligned} &u_l, & & \quad\text{if}\quad 0\leq u_l \\ &0, & & \quad\text{if}\quad u_l \leq 0\leq u_r \\ &u_r, & & \quad\text{if}\quad u_r \leq 0 \end{aligned}\right. $$

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