Showing the sine-Gordon equation is satisfied.

[varth]

Homework Statement

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).

Homework 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)}) ]

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!

 

[szynkasz]

$$\sin\alpha=\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}}\Rightarrow \sin(\arctan 4y)=\frac{4y}{\sqrt{1+16y^2}}$$

 

[varth]

Thanks for the reply! But I believe the expression sin(4arctan(...)), not sin(arctan(4...). I believe this complicates things?

 

[szynkasz]

Sorry, I misread it.
$$\sin\alpha=\frac{\tan\alpha}{\sqrt{1+\tan^2\alpha}},\,\cos\alpha=\frac{1}{\sqrt{1+\tan^2\alpha}}\\<br /> \sin 4\alpha=4\sin\alpha\cos^3\alpha-4\sin^3\alpha\cos\alpha\\<br /> \sin(4\arctan y)=\frac{4y-4y^3}{(1+y^2)^2}$$

 

[varth]

I believe that has done the trick, many thanks!