Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Method of characteristics and shock waves

  1. Mar 8, 2013 #1

    Leb

    User Avatar

    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. jcsd
  3. Mar 11, 2013 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    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.
     
  4. Mar 11, 2013 #3

    Leb

    User Avatar

    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
  5. Mar 28, 2013 #4

    hunt_mat

    User Avatar
    Homework Helper

    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?
     
  6. May 5, 2013 #5

    Leb

    User Avatar

    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
     
  7. May 6, 2013 #6

    hunt_mat

    User Avatar
    Homework Helper

    We need the actual equation to be able to give you some advice.
     
  8. May 6, 2013 #7

    Leb

    User Avatar

    Sorry, I have not realized that did not fit.

    It's [itex]u_{t}+uu_{x}=0[/itex] and u(x,0) = -1 for x<0 and u(x,0) = 1
     
  9. May 6, 2013 #8

    hunt_mat

    User Avatar
    Homework Helper

    This states that u is constant on the characteristics.
     
  10. May 6, 2013 #9

    Leb

    User Avatar

    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...
     
  11. May 6, 2013 #10

    hunt_mat

    User Avatar
    Homework Helper

    You have to compute the characteristics, write dt/ds=1, dx/ds=u. with t(0)=r and x(0)=0.
     
  12. May 6, 2013 #11

    Leb

    User Avatar

    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 ?
     
  13. May 6, 2013 #12

    hunt_mat

    User Avatar
    Homework Helper

    You're mapping the characteristics.
     
  14. May 7, 2013 #13

    Leb

    User Avatar

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

    I'm having trouble with this question:
    shock1.jpg

    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 ?
     
  15. May 7, 2013 #14

    hunt_mat

    User Avatar
    Homework Helper

    Do you know how to compute characteristics? If I have a hyperbolic PDE, I can write down the characteristics as
    [tex]\frac{dt}{ds}=1\quad\frac{dx}{dt}=u,\quad\frac{du}{ds}=0[/tex]
    Then consider what [itex]u[/itex] is in those separate regions.
     
  16. May 7, 2013 #15

    hunt_mat

    User Avatar
    Homework Helper

    We'll take this one step at a time. Solving PDEs of this worm is handle turning, I'll guide you through the process.
     
  17. May 7, 2013 #16

    Leb

    User Avatar

    Solving your given Monge equations:
    [itex]t = s + t_{0} ; x = ut + x_{0} ; u=F({\sigma}) [/itex]

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

    Then [itex]u(x,t) = F(x-us)[/itex] s=t
     
  18. May 7, 2013 #17

    hunt_mat

    User Avatar
    Homework Helper

    No, you could not have chosen x(0)=0, why can't you do that?
     
  19. May 7, 2013 #18

    Leb

    User Avatar

    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.
    I appreciate your help!
     
  20. May 7, 2013 #19

    hunt_mat

    User Avatar
    Homework Helper

    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?
     
  21. May 7, 2013 #20

    Leb

    User Avatar

    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 ?
     
  22. May 7, 2013 #21

    hunt_mat

    User Avatar
    Homework Helper

    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 [itex]\sigma[/itex]).
     
  23. May 7, 2013 #22

    Leb

    User Avatar

    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...
     
  24. May 7, 2013 #23

    hunt_mat

    User Avatar
    Homework Helper

    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 [itex]\sigma[/itex]), so you have to solve the equations:
    [tex]\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 )=...[/tex]
    Where ... are the two cases mentioned in the problem. Let's examine the case for ...=0, what does this make the equation solution?
     
  25. May 7, 2013 #24

    Leb

    User Avatar

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

    [itex]u(0,\sigma) = \sigma[/itex] and that will let us solve for x i.e. [itex]x = \sigma t + \sigma [/itex] for the given limits (with now [itex] 0 \leq \sigma < 1[/itex] ). I then rearrange for [itex] \sigma = x / 1+t [/itex] now look at limits [itex]0 \leq x/1+t < 1[/itex] 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 [itex]\sigma = 1 [/itex] is not included in the limits, there has to be a fan of chars. The form of u will be [itex] u = \alpha [/itex]. Hence the "fan" will be [itex]x = \alpha t + 1 (\text{here 1 is the sigma which was not included in the limits})[/itex]. 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. [itex]1<s(t)<1+t[/itex] 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
  26. May 7, 2013 #25

    hunt_mat

    User Avatar
    Homework Helper

    I was hoping you would just solve the problem. If [itex]u_{0}(0,x)=0[/itex], then [itex]t(s)=s[/itex] and [itex]du/ds=0[/itex] which then implies [itex]u(\sigma ,s)=0[/itex], then [itex]dx/ds=0[/itex], which then implies [itex]x=\sigma[/itex]. So we have solved the problem when [itex]u_{0}(0,x)=0[/itex]. The next case to tackle is when [itex]u_{0}(0,x)=x[/itex]. How do you think we should proceed?
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook