Method of characteristics and shock waves

[hunt_mat]

Really it's u=\sigma from whence you can obtain the solution. So the only other case is for x=1, What are the characteristic equations for this case (as I set them up)?

 

[Leb]

Really it's u=\sigma from whence you can obtain the solution. So the only other case is for x=1, What are the characteristic equations for this case (as I set them up)?

I assume you mean the \sigma = 1 which will give the characteristic x = \alpha t + 1, where \alpha = u and in range from 0 to 1. \alpha = x-1/t so u = x-1/t

 

[hunt_mat]

The function u_{0}(0,x) isn't actually defined for x=1 oddly enough and I would expect it to be. The solution seems to indicate that it's 0, so let's go with that. the equations are then:
\frac{dt}{ds}=1\quad t(0)=0,\quad\frac{dx}{ds}=u\quad x(0)=1,\quad\frac{du}{ds}=0\quad u(0)=\alpha, these equations can be solved to yield the equation you said, the equation for x will give you the ranges for the validity of your solution.

 

[Leb]

So where was I wrong ? (I have not really seen x being set to something before (except when playing around with limits)) Did you mean it like sigma (I take it these can be interchanged for t=0) ?

 

[hunt_mat]

You weren't wrong, I just pointed out the usual process that someone would do when solving these characteristics problems.

 

[Leb]

I am still not sure if this all helped (not that your explaining is bad, but me being daft). I am now looking at shocks and weak solutions. I (think I) know how to mechanically get weak solutions, but when it comes to drawing the characteristics for the weak solution - I'm in trouble.

For example here:

solution:

I can sort of understand why it is straight lines before -1 and after 1 (from below). Because I think I should be looking for the gradient (1/c(σ)) and c(σ) = ρ(x,t), so the 1/0 gives and infinite gradient (i.e. a straight line), but what about the non straight lines ? How to I look at
1+t/(1+x) ? I would probably try and set t first, so the -1<x<sqrt would hold (for example, t = 1) and then just pick any x from those values (in this case x is in (-1;1).

I am also not sure where the "fan" (chars from 0 to 1 point to (1,1)) originates (it's not in the weak solution, or at least I cannot see it...

 

[hunt_mat]

The characteristic curves from the system of equations:
\frac{dt}{ds}=1,\quad\frac{dx}{ds}=u,\quad\frac{du}{ds}=...
Are given by dividing the differential equations by each other to obtain:
\frac{dx}{dt}=u
From this you just solve the equation to obtain x=f(t,a) where a is the parameter that decides which characteristic you're on.

 

[Leb]

Maybe you could tell me about how you think (your thought process) when you are drawing the characteristics (especially the sloping ones) ? I have circled the ones that maybe will make life easier for both of us.

Because, however I approach it, I get nonsensical line.

Let's say for the first one from the left, I see that it t=0 at say x=-3/4, so If I plug it into the 1+x/(1+t) and invert it, I get gradient 4 (which seems to be a good approximation, as it is really steep). So σ = -3/4 (the value of sigma will aways be the value of x at t=0, right ?) Using the equation for characteristic x=ρt+σ. What does it mean when the characteristic reaches the path of the shock ? Is it where breaking occurs ?

 

[hunt_mat]

To compute the characteristics as I have mentioned, I would solve a differential equation I wrote down in the previous post of mine, in the region where u is constant, the characteristics are going to be of the form x-ut=\textrm{constant} and will therefore be straight lines, as you have drawn. For the region where 0&lt;x&lt;1 the solution was u=x/(1+t), and therefore you solve the equation:
\frac{dx}{dt}=\frac{x}{1+t}, solve this to get the shape of the characteristics in that region. What do you get?

 

[hunt_mat]

I should add that once we have done this, we can calculate when the shock happens and the path of the shock.

 

[Leb]

It should be x = K (1+t) So a straight line, for some constant K. If I use the range x is in I get that
-1&lt;t&lt;\frac{1}{K}+1.

I think I am wasting your time by now. I think it could be high school maths that's my problem. I have no idea where I get K from and no idea how this leads to shock time and path...I'll just try to waffle something in the exam and solve as much as possible "mechanically" If I have not understood this simple subset of PDE's I never will.

I appreciate you taking time to guide me through this, thank you so much!

 

[hunt_mat]

The K in your equation specifies which characteristic you're on. Now you can calculate the characteristics we're not in a good position to compute the shock position. Now if we integrate between t_{0} and t, we get the equation:
x=\frac{1+t}{1+t_{0}}
Where t_{0} defines the characteristic. Now we expect that neighbouring characteristics to intersect when a shock forms, so:
x(t,t_{0})\approx x(t,t_{0})=\frac{\partial x}{\partial t_{0}}\delta t_{0}
So the point where the shock forms is when \frac{\partial x}{\partial t_{0}}=0, so can you compute this?

 

[hunt_mat]

Having checked the calculation, I realize this is the wrong solution, you should do the same thing for \frac{dx}{dt}=\frac{x-1}{t}, what value of t does this vanish for?

 

[Leb]

So x vanishes for t = 1/C... So the x intercept. If C is the "sigma" this confuses me even further...

 

[hunt_mat]

I think it shows that x=1+\frac{t}{t_{0}}, then:\frac{\partial x}{\partial t_{0}}=-\frac{t}{t_{0}^{2}} which shows that t=0 is the time when the shock forms and that implies that the shock forms when x=1.

 

[Leb]

I did not notice I deleted that part. I got x = 1 - Ct (I did it by differentiating, and equating the logs.

Are we still talking about the same problem as in the picture ? Because it says the shock forms at t=1, x=1.

Anyway, never mind. I understand nothing. Thanks for your help anyway!

 

[hunt_mat]

If you send me a PM, I will send you my characteristics notes.