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A new algorithm for finding Hirota bilinear form (D-Operator) of NPDE?

Hi I need help understanding and implementing an algorithm that has been explained in detail, step by step. While I don’t fully understand the provided code, I can tell that it is likely not very difficult for someone experienced in coding. The algorithm involves substituting terms and working with integrals to derive specific parts of the solution. It seems to handle quadratic and quartic terms and outlines the structure for all dimensions.

Additionally, I believe this method could be used to transform non-linear partial differential equations (NPDEs) into D-operators. There are also many examples available for testing this algorithm, which could be helpful for verifying its correctness. enter image description here

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in order to get better view of topic can get pdf enter link description here
POSTED BY: salim mahmood
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Here is just a start:

Clear[u, x, t, \[CapitalPhi]];
eq21 = D[u[x, t], t] + 5 D[u[x, t] D[u[x, t], {x, 2}], x] + 
  5 u[x, t]^2 D[u[x, t], x] + D[u[x, t], {x, 5}]
eq22 = u -> Function[{x, t}, 6 D[Log[\[CapitalPhi][x, t]], {x, 2}]]
subst = (eq21 /. eq22)/6;
eq23 = Integrate[subst, x] // Together // Numerator
POSTED BY: Gianluca Gorni

I have gone a little further:

Clear[u, x, t, \[CapitalPhi]];
eq21 = D[u[x, t], t] + 5 D[u[x, t] D[u[x, t], {x, 2}], x] + 
  5 u[x, t]^2 D[u[x, t], x] + D[u[x, t], {x, 5}]
eq22 = u -> Function[{x, t}, 6 D[Log[\[CapitalPhi][x, t]], {x, 2}]]
subst = (eq21 /. eq22)/6;
eq23 = Integrate[subst, x] // Together // Numerator
(*2.4*)
dOperator[phi_, psi_, n_Integer, {x_, t_}] :=
 Module[{x1, x2},
  Nest[D[#, x1] - D[#, x2] &, phi[x1, t] psi[x2, t], n] /. 
   x1 | x2 -> x]
(*2.5*)dOperator[phi_, psi_, {n_Integer, m_Integer}, {x_, t_}] := 
  Module[{dt, x1, x2, t1, t2}, 
   dt = Nest[D[#, t1] - D[#, t2] &, phi[x1, t1] psi[x2, t2], m];
   Nest[D[#, x1] - D[#, x2] &, dt, n] /. x1 | x2 -> x /. t1 | t2 -> t];
eq26 = dOperator[\[CapitalPhi], \[CapitalPhi], {1, 1}, {x, t}] + 
  dOperator[\[CapitalPhi], \[CapitalPhi], 6, {x, t}]
eq26 == 2 eq23 // Expand
POSTED BY: Gianluca Gorni

First of all, thank you for your answer regarding the Sawada-Kotera (SK) equation, specifically equation (4.13) in the referenced PDF. Upon reviewing the material, it seems that the author of the paper obtained the result using Mathematica. This conclusion is based on the fact that both the results and the writing format match Mathematica's output term by term.

Now, I need assistance with the following:

  1. Algorithm Development for Hirota Bilinear Form I have already written the algorithm for the (1+1)-dimensional case in Maple, but I still need to extend this to the (1+2) and (1+3) dimensions. Could you guide me on how to implement these cases?

  2. Determining the Quadratic Part in Hirota's Method An important step in Hirota's method is identifying the quadratic part of the equation after substitution. For example, in equation (4.7), the paper mentions that the quadratic part is a lengthy expression. However, when I try to determine this part through substitution, I face difficulties, especially when working towards the bilinear form of the equation. Could you explain the procedure to derive the quadratic part systematically? This step is crucial for my understanding and implementation.

  3. Writing the Hirota Bilinear Form I would like to write the Hirota bilinear form for each of these equations in (1+1), (1+2), and (1+3) dimensions. Specifically, I’m struggling to understand how the quadratic part is determined after substitution. Any detailed explanation or step-by-step guidance for this process would be highly appreciated.

Although I am not familiar with Mathematica coding, I can fully understand its logic and structure. I am currently working on this topic in Maple, so any insights tailored to Maple or generalized concepts would be extremely helpful.

Thank you again for your guidance and support!
POSTED BY: salim mahmood

Thank you so much! I really appreciate that you mentioned each equation—this is incredibly helpful. However, the two equations you referred to are among the easier ones. In front of me, I have more than 20 equations, which I am finding difficult to handle. I hope to find a way to construct a well-organized and systematic approach to Hirota's method that can help me manage these equations effectively.
POSTED BY: salim mahmood
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