Talk:Korteweg–De Vries equation
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[edit]The redirect KortewegdeVriesequation has been listed at redirects for discussion to determine whether its use and function meets the redirect guidelines. Readers of this page are welcome to comment on this redirect at Wikipedia:Redirects for discussion/Log/2024 January 26 § KortewegdeVriesequation until a consensus is reached. Colonies Chris (talk) 11:09, 26 January 2024 (UTC)
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There is a back and forth over the meaning of the operator in the Lax pair description
Elsewhere in this formula, is the operator that multiplies scalars by the scalar field , and is the operator of partial differentiation with respect to . According to this interpretation, and standard conventions for operator products,
However, this is not the correct Lax pair. Instead, what is intended is
where the bracket denotes the commutator of the two operators. Tito Omburo (talk) 16:49, 27 June 2024 (UTC)
 If it’s the commutator you are referring to with the square brackets, then this is not in line with your latest “explanatory note” edit. If ambiguity regarding the use of the partial derivative is what’s concerning you, then I don’t understand why this ambiguity suddenly arises halfway down the article rather than at the first description of the KdV equation Roffaduft (talk) 17:12, 27 June 2024 (UTC)
 Do you disagree that ? Tito Omburo (talk) 17:15, 27 June 2024 (UTC)
 That is not the issue here (although it does look like you’re missing an in there somewhere). It’s about consistent use of symbols and brackets throughout the article. I personally would prefer the use of subscript x rather than the partial symbol, because I think it makes the article look less “cluttered”. But again, addressing a “bracket issue” halfway down the article using “ambiguity” as an argument doesn’t make much sense to me Roffaduft (talk) 17:25, 27 June 2024 (UTC)
 Sorry, there was a typo above, now fixed. Elsewhere, this is not a problem because everything is a scalar. But A is an *operator*, and the correct term is the commutator of the two operators, not the operator , which would take a function f to . Tito Omburo (talk) 17:32, 27 June 2024 (UTC)
 Yes, but that is not what you wrote down in your latest edit as “explanatory note” Roffaduft (talk) 17:34, 27 June 2024 (UTC)
 It's not? Tito Omburo (talk) 17:35, 27 June 2024 (UTC)
 Yes, but that is not what you wrote down in your latest edit as “explanatory note” Roffaduft (talk) 17:34, 27 June 2024 (UTC)
 Sorry, there was a typo above, now fixed. Elsewhere, this is not a problem because everything is a scalar. But A is an *operator*, and the correct term is the commutator of the two operators, not the operator , which would take a function f to . Tito Omburo (talk) 17:32, 27 June 2024 (UTC)
 If those are commutator brackets, then shouldn’t it be . Your “explanatory note” introduces commutator brackets but then explains the basics of taking derivatives. It’s all just a bit convoluted. In fact, you don’t need to use commutator brackets (and the subsequent “explanatory note”) if you just changed the “6” with a “3” in the definition of operator “A”. Roffaduft (talk) 17:45, 27 June 2024 (UTC)
 I feel like changing 6 to 3 would be even more confusing because until this point is synonymous with . Tito Omburo (talk) 17:49, 27 June 2024 (UTC)
 Not only is it not confusing, it is also completely in line with page 32 of the Dunajski (2009) reference; discussing the lax representation of the KdV equation Roffaduft (talk) 17:53, 27 June 2024 (UTC)
 I feel like changing 6 to 3 would be even more confusing because until this point is synonymous with . Tito Omburo (talk) 17:49, 27 June 2024 (UTC)
 That is not the issue here (although it does look like you’re missing an in there somewhere). It’s about consistent use of symbols and brackets throughout the article. I personally would prefer the use of subscript x rather than the partial symbol, because I think it makes the article look less “cluttered”. But again, addressing a “bracket issue” halfway down the article using “ambiguity” as an argument doesn’t make much sense to me Roffaduft (talk) 17:25, 27 June 2024 (UTC)
 Do you disagree that ? Tito Omburo (talk) 17:15, 27 June 2024 (UTC)
Note you made a mistake above . Tito Omburo (talk) 17:56, 27 June 2024 (UTC)
 Wouldn’t that require for “f” to be inside the commutator brackets? Regardless, page 32 of Dunajski shows that introducing the commutator brackets in the operator “A” is completely unnecessary. It would only lead to discussions like the one we’re having now. Roffaduft (talk) 18:01, 27 June 2024 (UTC)
 No, write it out carefully yourself. Also, Dunajski uses subscript notation for the partial derivative of a field, reserving d/dx for the operator. Tito Omburo (talk) 18:04, 27 June 2024 (UTC)
 That is not an answer. It is clear what the assumption was that I made when writing out the equation.
 also, what does it matter? Dunajski clearly shows you don’t need commutator brackets in the definition of operator “A”. He also uses d/dx and subscript notation fairly interchangeably. I don’t see why this is such an issue. Please explain why we can’t just change the “6” to a “3”, in line with the reference material, and be done with it. Roffaduft (talk) 18:13, 27 June 2024 (UTC)
 Because if we were to do this change in departure with the rest of the article. Tito Omburo (talk) 18:15, 27 June 2024 (UTC)
 Ah, I see what your issue is now. I will have a look at it tomorrow and compare some additional references just to be sure. Roffaduft (talk) 18:22, 27 June 2024 (UTC)
 Because if we were to do this change in departure with the rest of the article. Tito Omburo (talk) 18:15, 27 June 2024 (UTC)
 No, write it out carefully yourself. Also, Dunajski uses subscript notation for the partial derivative of a field, reserving d/dx for the operator. Tito Omburo (talk) 18:04, 27 June 2024 (UTC)