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Kdv solution solitons Bilinear Operator

  • Thread starter binbagsss
  • Start date
  • #1
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Homework Statement



[/B]
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.


Homework Equations


[/B]
(above)


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##
 

Answers and Replies

  • #2
MathematicalPhysicist
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How do the functions f_n look like?
 
  • #3
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How do the functions f_n look like?
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:
  • #4
MathematicalPhysicist
Gold Member
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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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