shocks and strict inequalities

**badgerigar** · June 14th 2009, 06:43 PM

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

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

I am expected to use the method of characteristics. The calculation was very straightforward and I ended up with
$ 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 $u$ on the shock: does the fact that the initial conditions used $\geq$ instead of > mean that $u=0$ on the shock? or do I need to add to my answer that $u(0,0)=0$

**HallsofIvy** · June 19th 2009, 06:21 AM

Originally Posted by badgerigar

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

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

I am expected to use the method of characteristics. The calculation was very straightforward and I ended up with
$ 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 $u$ on the shock: does the fact that the initial conditions used $\geq$ instead of > mean that $u=0$ on the shock? or do I need to add to my answer that $u(0,0)=0$

Yes, it does. Your solution must satisfy your initial condition and to do that you must have:
$ u(x,t) = \left\{ \begin{array}{ll} 1,&x<\frac13 t\\0,&x\ge\frac13 t \end{array} \right.$

Thread: shocks and strict inequalities

LinkBack

Thread Tools

Display

shocks and strict inequalities

Similar Math Help Forum Discussions

inequalities

validity of convergent sequences with strict inequalities

Strict

inequalities

Search Tags