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Method of characteristics and shock waves

  1. Mar 8, 2013 #1

    Leb

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    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).
     
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  3. Mar 11, 2013 #2

    Ben Niehoff

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    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

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    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

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    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

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    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

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

    Leb

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    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

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

    Leb

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    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

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    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

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    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

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

    Leb

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    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

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    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

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    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

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    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

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

    Leb

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    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

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    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

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    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 ?
     
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