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Kdv solution solitons Bilinear Operator

  1. Dec 22, 2014 #1
    1. The problem statement, all variables and given/known data


    I'm solving for ##f_1## from ##B(f_{1}.1+1.f_{1})## from ## \frac{\partial}{\partial x}(\frac{\partial}{\partial t}+\frac{\partial^{3}}{\partial x^{3}})f_n=-\frac{1}{2}\sum^{n-1}_{m=1}B(f_{n-m}.f_{m}) ##

    where ##B=D_tD_x+D_x^4##, where ##B## is the Bilinear operator.


    2. Relevant equations

    (above)


    3. The attempt at a solution

    I get ##B(f_{1}.1+1.f_{1})=2f_{1xt}## not ##0!##. Whereas the method gets ##\frac{\partial}{\partial x}(\frac{\partial}{\partial t}+\frac{\partial^{3}}{\partial x^{3}})f_1=0##
     
  2. jcsd
  3. Dec 23, 2014 #2

    MathematicalPhysicist

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    How do the functions f_n look like?
     
  4. Dec 23, 2014 #3
    I believe ##f_1## is unknown at this point , and the idea is to use the RHS of the expression on the right of the top line, to find it. So once concluding the RHS=0 we solve ##(\frac{\partial}{\partial t}+\frac{\partial}{dx^{3}})f_1=0##

    The expression I'm talking about, sorry I should have said and cant no longer seem to edit?, came from expressing solving the kdv eq equivalent to solving ##B(f.f)=0##*, where ##f=1+\sum^{\infinty}_{1}\epsilon^nf_n##, and so the expression has came from plugging the latter into * and setting each ##\epsilon^n## coefficient to zero.
     
    Last edited: Dec 23, 2014
  5. Dec 24, 2014 #4

    MathematicalPhysicist

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    Well a general solution to the equation of this form is [tex]f_1 = f(x^3-t)[/tex] but I am still not sure what you are looking for here.
     
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