Solution to the non-linear PDE ( Sine-Gordon equation ) by substitution

Question

I am trying to solve the following simple equation:

$$\frac{df}{dx}=\sin{f}$$ This is the "kink" solution to the Sine-Gordon equation. To solve this, I do the following substitution:

$$f=\tan^{-1}g$$

Then we can use $\frac{d}{dz}\tan^{-1}z=\frac{1}{1+z^2}$ to find:

$$\frac{1}{1+g^2}\frac{dg}{dx}=\sin(\tan^{-1}g)=\frac{g}{\sqrt{1+g^2}}$$

This gives:

$$\frac{dg}{dx}=g\sqrt{1+g^2}$$ which I need to solve.

The problem now is that the solution I find is:

$$f=\tan^{-1}{e^{x-a}}$$ which suggests that $\frac{dg}{dx}=g$ and therefore, what I have wrote above is not correct. What did I do wrong?

Answer 1

The trick is to use a slightly modified substitution:

$$f=2\tan^{-1}g$$ Then, the RHS becomes $$\sin{(2\tan^{-1}g)}=2\sin{(\tan^{-1}g)}\cos{(\tan^{-1}g)}$$ solving the problem

Answer 2

You could make it faster switching variables $$\frac{df}{dx}=\sin(f)\implies \frac{dx}{df}=\frac 1{\sin(f)}$$ $$x+C=\log \left(\tan \left(\frac{f}{2}\right)\right)\implies\tan \left(\frac{f}{2}\right)=C e^x\implies f=2\tan ^{-1}(C e^x)$$