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Derivation of an integral identity from the kdv equation.

  1. May 22, 2014 #1
    Hi everybody! First post!(atleast in years and years).
    The stationary KdV equation given by
    $$ 6u(x)u_{x} - u_{xxx} = 0 $$.
    It has a solution given by
    $$ \bar{u}(x)=-2\sech^{2}(x) + \frac{2}{3} $$
    This solution obeys the indentity
    $$ \int_{0}^{z}\left(\bar{u}(y) - \frac{2}{3}\right)\int_{0}^{y}\left(\bar{u}(x) - \frac{2}{3}\right)dxdy = \bar{u}(z) + \frac{4}{3} $$

    Is it possible to derive this kind of identity with out using the explicit form of the solution ##\bar{u}(x) ##? That is only by using the fact that ##\bar{u}(x) ## is a solution to (1). I tried differentiating (3) thrice on the left side but that generates a lot of different terms for which it is hard to use (1) to simplify. I really need some hints for this problem.

    Thank you in advance :)
     
  2. jcsd
  3. May 22, 2014 #2

    pasmith

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    Are you sure? The left hand side is, using the express solution, [tex]
    4 \int_0^z \mathrm{sech}^2 (y) \int_0^y \mathrm{sech}^2 (x) \,dx\,dy
    = 4 \int_0^z \mathrm{sech}^2 (y) \tanh (y)\,dy = 2\tanh^2 (z) = 2(1 + \mathrm{sech}^2 (z))[/tex] but the right hand side is
    [tex] 2(1 - \mathrm{sech}^2(z)).[/tex]
     
  4. May 22, 2014 #3
    ## \tanh(x)^{2} = 1- \sech(x)^{2} ##
     
  5. May 22, 2014 #4

    pasmith

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    Oops :blushing:

    On further analysis, I suspect that the integral identity is a consequence of [itex]\bar u - \frac23[/itex] being a multiple of [itex]\mathrm{sech}^2(x)[/itex], not of [itex]\bar u[/itex] being a solution of the ODE.

    If you look for solutions of the ODE which satisfy [itex]u'' \to 0[/itex], [itex]u' \to 0[/itex] and [itex]u \to u_0[/itex] as [itex]|x| \to \infty[/itex] then integrating once yields [tex]
    3u^2 = u'' + 3u_0^2.[/tex] Multiplying by [itex]u'[/itex] and integrating once more yields [tex]
    \frac12 u'^2 = u^3 - 3u_0^2u + 2u_0^3 = (u - u_0)^2(u + 2u_0).[/tex] A further messy integration yields [tex] u(x) = u_0 - 3u_0 \mathrm{sech}^2 \left(\sqrt{\frac{3u_0}{2}} x\right).[/tex]
     
    Last edited: May 22, 2014
  6. May 22, 2014 #5

    pasmith

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    I have worked out how to get the identity. We start from [tex]
    \frac12 u'^2 = (u - u_0)^2(u + 2u_0)[/tex] and make two assumptions: firstly that [itex]-2u_0 \leq u(x) < u_0[/itex] and secondly that [itex]u(x)[/itex] attains its minimum at [itex]x = 0[/itex]. Thus [itex]u'(x) > 0[/itex] in [itex]x > 0[/itex] and [itex]u(0) = -2u_0[/itex].

    Starting from [tex]
    \frac12 u'^2 = (u - u_0)^2(u + 2u_0) [/tex] we take the square root and divide by [itex]u + 2u_0[/itex] to obtain [tex]
    \frac{u'}{\sqrt 2 \sqrt{u + 2u_0}} = u_0 - u.[/tex] Integrating once then yields [tex]
    \left[\sqrt{2}\sqrt{u(x) + 2u_0}\right]_0^y = -\int_0^y u(x) - u_0\,dx.[/tex] Since by assumption [itex]u(0) + 2u_0 = 0[/itex] we have [tex]
    \sqrt{2}\sqrt{u(y) + 2u_0} = -\int_0^y u(x) - u_0\,dx.[/tex] Multiplying by [itex]u_0 - u(y)[/itex] then gives [tex]
    \sqrt{2}\sqrt{u(y) + 2u_0}(u_0 - u(y)) = u'(y) = (u(y)- u_0) \int_0^y u(x) - u_0\,dx.[/tex] A further integration then yields
    [tex]
    u(z) - u(0) = \int_0^z (u(y)- u_0) \int_0^y u(x) - u_0\,dx\,dy
    [/tex] and since [itex]u(0) = -2u_0[/itex] we obtain finally [tex]
    \int_0^z (u(y)- u_0) \int_0^y u(x) - u_0\,dx\,dy = u(z) + 2u_0[/tex] as required.
     
  7. May 22, 2014 #6
    Thank you very much! I've just this minute completed my own similar solution from the simplicfications you did earlier.
    Start from
    $$ \frac{u'^{2}}{(u-u_{0})^{2}} = 2(u + 2u_{0}) $$
    Then use the identity $$\partial_{x}\left( \frac{u'}{u-u_{0}}\right) = \frac{u''}{u-u_{0}} - \frac{u'^{2}}{(u-u_{0})^{2}} $$ and
    $$u'' = 3(u^{2}-u_{0}^{2}) $$
    to find $$\partial_{x}\left( \frac{u'}{u-u_{0}}\right) = u-u_{0}$$
    Integrate twice and we get
    $$ u = \int_{0}^{z}(u(y) - u_{0})\left( \int_{0}^{y}(u(x) - u_{0} )dx+ u'(0)\right) dy + u(0) $$

    So all in all the conclusion must be: yes, you can derive the identity without using the explicit form of ## \bar{u}## as long as we specify some boundary conditions. This was quite a nice problem :)
     
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