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Hyperbolic substitution in KdV equation resolution

  1. May 20, 2015 #1
    Hello,

    I am trying to understand the resolution of the following KdV equation. I try to demonstrate it by myself.

    upload_2015-5-20_7-38-44.png

    The solitary wave solution is :
    upload_2015-5-20_7-39-26.png


    At first, I created new variable as follows so I could transform the PDE in to an ODE.
    A = A(p)
    p = g(x,t)
    g(x,t) = x - ct

    I succeeded to transform the PDE to ODE by the chain rule. My problem is when I arrive at that integral :

    upload_2015-5-20_7-42-0.png

    I read a lot of article and I have found that that integral needs to be solve by hyperbolic trigonometric substitution :

    upload_2015-5-20_7-43-57.png

    I have found that this is the substitution, but I have found anywhere why it needs that specific one. It might be because I just don't see it as the last time I hade to integrate by trigonometric substitution is a few years ago.

    Is there someone on PF that knows why it needs that specific substitution?

    Thank you!!!

    (I apologize for my bad english as the language I use everyday is french)
     

    Attached Files:

  2. jcsd
  3. May 24, 2015 #2

    hunt_mat

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    You need to use that the solution and it's derivatives tend to zero as [itex]w\rightarrow\pm\infty[/itex]
     
  4. May 24, 2015 #3
    Yes I succeeded to get the final answer. I want to know where the substitution comes from. I read in many articles that you have to use it. I am trying to see where it comes from. A sech2 substitution is not obvious when looking at that integral.
     
  5. May 24, 2015 #4

    hunt_mat

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    It's basically about knowing your identities backwards and being able to spot which is the correct one to use. you know that [itex]\cosh^{2}w-\sinh^{2}w=1[/itex] right. Then dividing through by [itex]\cosh^{2}w[/itex] to show that [itex]1-\tanh^{2}w=\textrm{sech}^{2}w[/itex] then [itex]1-\textrm{sech}^{2}w=\tanh^{2}w[/itex]

    Now you just have to convince yourself that the substitution will work.
     
  6. May 24, 2015 #5
    Yes I know that the substitution works............... I succeded to demonstrate the solution. What I want to know is where the substitution comes from. How by looking at the integral do you come at the conclusion that you have to substitute A by a sech2(w). Have you ever seen a table that propose that type of substitution for a integral of that type?

    Right now, it seems akward to me to just say at one moment let's substitute by sech2(w), I am looking for an explanation why we have to do that.

    I really appreciate your help. Thank you.

    (My language is not english, so I am sorry if how I write seems horrible.)
     
  7. May 24, 2015 #6

    hunt_mat

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    Like I said before, it's about knowing your trig and hyperbolic identities and spotting what's the right one to do.

    There must be one somewhere, it seems relatively elementary to me. Try a generic book on calculus or a book on integrals.

    It's one that I would spot just by looking and thinking for a bit.
     
  8. May 24, 2015 #7
    Ok I will look at a book on integrals. As you, I come up to the conclusion it is only about spotting an identity. If I find one, I will inform you.

    Thanks for your help.
     
  9. May 24, 2015 #8

    hunt_mat

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    I know about this stuff as my PhD thesis was in equations like this and more general.
     
  10. May 25, 2015 #9

    epenguin

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    Your English is not horrible and it is clear what you mean which matters more.

    The reason you can't find about where this comes from is that it is a secret.

    Some guy was playing around differentiating functions of sech and tanh and, not very different, sin and cos and found he could naturally eliminate the original functions from the answer. Then he said to himself "This is a wonderful answer, I must find a question for it". Which he did in physics or geometry, and said "look look I have this brilliant solution to the question you were asking!" (His name was probably J. Bernoulli but as that is hard to pronounce he wrote under the pseudonym J. Bernoulli.)

    In 20 or 40 years' time these integration "methods", which consist in knowing the answer, will have been dropped from all the syllabuses as people can get it from whatever devices they have then like you did, and can always check a solution is true, as you can.
     
    Last edited: May 26, 2015
  11. May 25, 2015 #10

    epenguin

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    ...having said which if we want to play this game there is some semblance of logic in:

    first getting rid of distracting constants by defining a variable y = A/3c (you take care of the constants accounting) so the key thing you are integrating has the form dy/[ y √(1 - y)] , by the way we must have y < 1 for this to be real, I think that will be all right physically;

    then saying the square root is nasty so let us change variable again to x, where x2 = (1 - y).
    You end up with a constant × ∫dx/(1 - x2). Factorise and express as partial fractions and get solution

    ½ ln [(1 + x)/(1 - x)]

    which actually is tanh-1 x if | x | < 1 i.e. -1 < x < 1 but is coth-1 if x < 1 (which I think will be all right as it corresponds to a positive time?) so a bit complicated but seeming to give the right kind of functions.
     
    Last edited: May 26, 2015
  12. May 28, 2015 #11
    Thank you. I arrived at that with regular substitution. However, I have integration limits problem as I have to integrate from 0 to A. When I evaluate at 0, the arctanh gives infinite...... which doesnt help me.
     
  13. May 29, 2015 #12
    upload_2015-5-29_23-13-43.png

    There must be something wrong with the integration limits......
     
  14. May 30, 2015 #13

    pasmith

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    The lower limit in the integral should be [itex]A(0)[/itex]. Now [itex]A = 0[/itex] is a fixed point of the ODE [tex]
    A'' = cA - \tfrac12A^2,[/tex] so either you start there and don't go anywhere ([itex]A(p) \equiv 0[/itex]), or you approach it only in the limits [itex]p \to \pm\infty[/itex].
     
    Last edited: May 30, 2015
  15. May 30, 2015 #14

    epenguin

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    Can someone please :oldsmile: plot us out solutions with time, or link us to some pics? KdV is about solitons right? Not everyone who comes here might know why this is interesting.
     
  16. May 30, 2015 #15
    I will as soon as we are all gonna be fixed about that integral. I use SCILAB to modelize it using different initial conditions.
     
  17. May 30, 2015 #16
    upload_2015-5-30_7-47-50.png

    I am sure what to do next about the right term. It should be equal to zero.

    The integral is equal to p which was integrated from 0 to p.
     

    Attached Files:

  18. May 30, 2015 #17
    I think I got something...

    A(p=0) is gonna be equal to the amplitude of the solution so 3c => arctanh(0)=0
     
  19. May 30, 2015 #18

    pasmith

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    Graph of sech^2 on wolfram alpha. As time increases, the graph is translated to the right.

    In full generality, the solution is [tex]
    A(x,t) = 3 \alpha^2 \mathrm{sech}^2(\tfrac12(\alpha x - \alpha^3 t) + C)
    [/tex] for some constant [itex]C[/itex]. If you choose to have the peak coincide with [itex]x - \alpha^2 t = 0[/itex] then [itex]C = 0[/itex].

    The shape of the wave depends on the wave speed [itex]c = \alpha^2[/itex]; as [itex]c[/itex] increases the peak becomes taller and narrower.
     
  20. Jun 1, 2015 #19
    For p, is it correct to integrate from 0 to p? at least it works that way....
     
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