# Thread: shocks and strict inequalities

1. ## shocks and strict inequalities

I'm just doing some practice exam questions and the question asks me to solve
$\displaystyle \frac{\partial u}{\partial t}+u^2\frac{\partial u}{\partial x} = 0$

with
$\displaystyle u(x,0) = \left\{ \begin{array}{cc} 1, &x<0\\0, &x \geq 0 \end{array} \right.$

I am expected to use the method of characteristics. The calculation was very straightforward and I ended up with
$\displaystyle u(x,t) = \left\{ \begin{array}{ll} 1,&x<\frac13 t\\0,&x>\frac13 t \end{array} \right.$

What I'm not sure about is the value of $\displaystyle u$ on the shock: does the fact that the initial conditions used $\displaystyle \geq$ instead of > mean that $\displaystyle u=0$ on the shock? or do I need to add to my answer that $\displaystyle u(0,0)=0$

I'm just doing some practice exam questions and the question asks me to solve
$\displaystyle \frac{\partial u}{\partial t}+u^2\frac{\partial u}{\partial x} = 0$

with
$\displaystyle u(x,0) = \left\{ \begin{array}{cc} 1, &x<0\\0, &x \geq 0 \end{array} \right.$

I am expected to use the method of characteristics. The calculation was very straightforward and I ended up with
$\displaystyle u(x,t) = \left\{ \begin{array}{ll} 1,&x<\frac13 t\\0,&x>\frac13 t \end{array} \right.$

What I'm not sure about is the value of $\displaystyle u$ on the shock: does the fact that the initial conditions used $\displaystyle \geq$ instead of > mean that $\displaystyle u=0$ on the shock? or do I need to add to my answer that $\displaystyle u(0,0)=0$
Yes, it does. Your solution must satisfy your initial condition and to do that you must have:
$\displaystyle u(x,t) = \left\{ \begin{array}{ll} 1,&x<\frac13 t\\0,&x\ge\frac13 t \end{array} \right.$