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Showing the sine-Gordon equation is satisfied.

  1. Dec 21, 2012 #1
    1. The problem statement, all variables and given/known data
    I am currently trying to show a the sine-Gordon equation is satisfied by a 'soliton-antisoliton' solution. Basically, I need to differentiate twice w.r.t. t and x separately and plug in everything as usual but my expression is getting extremely complicated and I don't know how to deal with the sin(4arctan(...)) term (this will become clear in the relevant equations).


    2. Relevant equations
    sine-Gordon equation: (∂^2 θ)/(∂t^2) - (∂^2 θ)/(∂x^2) + sin(θ)=0
    where the solution is θ(x,t)=4arctan[ (sinh{(ut)/SQRT(1-u^2)}) / (u cosh{x/SQRT(1-u^2)}) ]

    3. The attempt at a solution
    My attempt is rather messy so it might be a better idea if I don't write my working. However, what I have tried is differentiating directly but the expression didn't simplify and I didn't know what to do with sin(4arctan(...)). I was then told to rewrite sinh and cosh as exp functions but that still leaves a very complicated expression. I even tried doing it on Matlab (with cosh and sinh, not exp) but the LHS of the sine-Gordon equation doesn't simplify to 0 on Matlab, possibly due to the sine and arctan combining.. causing complication? Could anyone give me some tips?

    Many thanks. I hope I have posted this in the correct thread!
     
  2. jcsd
  3. Dec 21, 2012 #2
    [tex]\sin\alpha=\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}}\Rightarrow \sin(\arctan 4y)=\frac{4y}{\sqrt{1+16y^2}}[/tex]
     
  4. Dec 21, 2012 #3
    Thanks for the reply! But I believe the expression sin(4arctan(...)), not sin(arctan(4...). I believe this complicates things?
     
  5. Dec 22, 2012 #4
    Sorry, I misread it.
    [tex]
    \sin\alpha=\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}},\,\cos\alpha=\frac{1}{\sqrt{1+\tan^2\alpha}}\\
    \sin 4\alpha=4\sin\alpha\cos^3\alpha-4\sin^3\alpha\cos\alpha\\
    \sin(4\arctan y)=\frac{4y-4y^3}{(1+y^2)^2}
    [/tex]
     
  6. Dec 22, 2012 #5
    I believe that has done the trick, many thanks!!
     
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