Kdv solution solitons Bilinear Operator

[binbagsss]

Homework Statement

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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

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(above)

The Attempt at a Solution

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

 

[MathematicalPhysicist]

How do the functions f_n look like?

 

[binbagsss]

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 can't 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.

 

[MathematicalPhysicist]

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