# Method of characteristics and shock waves

1. Mar 8, 2013

### Leb

Hi all!

I just wanted to ask if anyone is finding the usage of method of characteristics difficult ? I sort of feel, that it is a very simple approach to solving PDE's, but I get easily lost, when for instance, we have to keep switching back and forth between variables and such. When it comes to introducing shocks, I get lost even more (I think my problem is, I am not comfortable with all these substitutions and expressions for time and speed etc).

I sort of get the idea behind it ( the most important part are the ICs and you can get any value for u in x-t plane etc). But when it comes to solving actual problems, I can only do basic ones and with "pretend" understanding (i.e. I think I know what I am doing, but in reality I only know the steps).

2. Mar 11, 2013

### Ben Niehoff

What source are you learning from? I've found that most sources on the method of characteristics are terrible at explaining what's going on.

3. Mar 11, 2013

### Leb

My university lecture course and a few notes online (I only find them somewhat useful). I got the intuition from this guy here but when it comes to more complex problems my intuition brakes down. As I said, I don't know why, but it really LOOKS as a very easy thing and that's annoying me greatly. Maybe I just need to practice more to get getter basics...

Last edited by a moderator: Sep 25, 2014
4. Mar 28, 2013

### hunt_mat

I did my MPhil thesis is shock waves and I have a lot of experience with the method of characteristics, what exactly do you need help with?

5. May 5, 2013

### Leb

A very simple problem that is probably the key problem for me is the fact that x and t are easily compared and interchanged. I do not understand how one can compare time and distance. For example saying u = 1 for x<-t. Should I somehow think in 3D ?

This is a very simple problem with the t and x comparison.
http://s12.postimg.org/xv0wzdcn1/Chars.jpg

6. May 6, 2013

### hunt_mat

We need the actual equation to be able to give you some advice.

7. May 6, 2013

### Leb

Sorry, I have not realized that did not fit.

It's $u_{t}+uu_{x}=0$ and u(x,0) = -1 for x<0 and u(x,0) = 1

8. May 6, 2013

### hunt_mat

This states that u is constant on the characteristics.

9. May 6, 2013

### Leb

I think I get that (and I thought that was the whole point of characteristics, that u is the same on a characteristic). What I do not get is how we compare time t and space x...

10. May 6, 2013

### hunt_mat

You have to compute the characteristics, write dt/ds=1, dx/ds=u. with t(0)=r and x(0)=0.

11. May 6, 2013

### Leb

Yes, I know how to algebraically get the expressions in the picture of the solutions I have attached. What I do not understand is why are we allowed to compare x and t ?

12. May 6, 2013

### hunt_mat

You're mapping the characteristics.

13. May 7, 2013

### Leb

Not sure what you mean there, but I will pretend to understand :)

I'm having trouble with this question:

The "shock location is obvious" for me it is only obvious if I draw the picture, maybe I should look at the limits ?

But the real problem to me is understanding how were the limits for s(t) (underlined red) established ?

14. May 7, 2013

### hunt_mat

Do you know how to compute characteristics? If I have a hyperbolic PDE, I can write down the characteristics as
$$\frac{dt}{ds}=1\quad\frac{dx}{dt}=u,\quad\frac{du}{ds}=0$$
Then consider what $u$ is in those separate regions.

15. May 7, 2013

### hunt_mat

We'll take this one step at a time. Solving PDEs of this worm is handle turning, I'll guide you through the process.

16. May 7, 2013

### Leb

$t = s + t_{0} ; x = ut + x_{0} ; u=F({\sigma})$

Then choose $t_{0} = 0$ (Could have equally done that to $x_{0}$ I assume because it does not matter where we start, might as well chose 0)

Then $u(x,t) = F(x-us)$ s=t

17. May 7, 2013

### hunt_mat

No, you could not have chosen x(0)=0, why can't you do that?

18. May 7, 2013

### Leb

I really do not know, my notes said:
"Note that we may choose t_{0} = 0 without loss of generality since the Monge equations are invariant under the change of variable. In other examples it may be more appropriate to choose x_{0} = 0 instead, using the same argument."

I thought that the constants only matter for which characteristic we are looking at ?

P.S.

19. May 7, 2013

### hunt_mat

The characteristic is defined by the initial value of the x in this case. Use int inition conditions u(0,x)=..., what does that tell you for the initial conditions for the characteristic equations?

20. May 7, 2013

### Leb

Not really sure how to answer that, but whatever u(t=0,x) is equal to, will stay the same along the characteristic (note, that I am answering using the definition, not understanding). Is that what you are asking about ?

21. May 7, 2013

### hunt_mat

when s=0, this will be the starting point of your characteristic curve, that is when t=0, so this should imply that t=... at s=0 and also we assign a value to x at x=0 (the solution calls this $\sigma$).

22. May 7, 2013

### Leb

Thank you for your help. However, I think I am just to daft to understand these. I am not sure what I do not understand. I bet it has to do with understanding the surfaces. The Monge equations, as far as I understand, are about the tangent vector ? Sort of like streamlines in fluid mechanics ?

I'm learning bit by bit how to solve it mechanically (i.e. without thinking), I wonder, though, how far this will take me...

23. May 7, 2013

### hunt_mat

To see which variable has the zero initial condition, look at the initial condition for the PDE u(0,x). The characteristics should come from that initial curve, so t=0 when s=0, because s=0 is when you are at the initial condition. However x is not-zero there so set it to some (as yet) unknown value (the solution calls it $\sigma$), so you have to solve the equations:
$$\frac{dt}{ds}=1\quad t(0)=0,\quad\frac{dx}{ds}=u\quad x(0)=\sigma ,\quad\frac{du}{dx}=0,\quad u(0,\sigma )=...$$
Where ... are the two cases mentioned in the problem. Let's examine the case for ...=0, what does this make the equation solution?

24. May 7, 2013

### Leb

I assume you are talking about the most recent problem, so

$u(0,\sigma) = \sigma$ and that will let us solve for x i.e. $x = \sigma t + \sigma$ for the given limits (with now $0 \leq \sigma < 1$ ). I then rearrange for $\sigma = x / 1+t$ now look at limits $0 \leq x/1+t < 1$ and then consider the two cases when t > -1 and t < -1 (the latter just inverts the order of inequalities) (But I think we assume t > 0, because in the solution it does not consider this case). Then, since $\sigma = 1$ is not included in the limits, there has to be a fan of chars. The form of u will be $u = \alpha$. Hence the "fan" will be $x = \alpha t + 1 (\text{here 1 is the sigma which was not included in the limits})$. Now the way I think one should get alpha (apart from drawing and looking for the jump) is to look at the case when t = 0 (not sure if this is general, but the discont. in this case occurs at t = 0 ). In this case alpha ranges from 0 (since that's the u which is after sigma > 1 ) to 1 (since at t=0 we have u=1 for the top expression of u). So we get the expression for u of the fan of chars. Now it also makes sense, why the shock location is at 1.

I now think that the reason for the limits on s(t) i.e. $1<s(t)<1+t$ is because we will always insert the shock, where the fan is (but not on the end points, because I think that has something to do with mass conservation (i.e. equal areas rule?))?

Does that sound OK ?

P.S.

I think I found out when setting x_{0} to zero works. It's when considering boundary condition problem instead of an IVP...

Last edited: May 7, 2013
25. May 7, 2013

### hunt_mat

I was hoping you would just solve the problem. If $u_{0}(0,x)=0$, then $t(s)=s$ and $du/ds=0$ which then implies $u(\sigma ,s)=0$, then $dx/ds=0$, which then implies $x=\sigma$. So we have solved the problem when $u_{0}(0,x)=0$. The next case to tackle is when $u_{0}(0,x)=x$. How do you think we should proceed?