# Breather solution of the sine-Gordon equation

1. Jul 12, 2016

### IanBerkman

In a homework problem, the following statement is made:

Sine-Gordon equation for the field u(x, t) in dimensionless units reads:
$$u_{tt}-u_{xx}+\sin u = 0$$
Search for the breather solution in the form of variable separation:
$$u(x,t)=4\arctan (\Theta(x,t)); \Theta(x,t)=A(t)B(x)$$

a. Substitute the above equation into sine-Gordon equation. Transform derivatives of
arbitrariry function of $\Theta$ with the aid of $\tilde{A}=A_t/A; \tilde{B} = B_x/B$ in the following way:
$$\partial_t F(\Theta) = \tilde{A} \partial_\Theta F(\Theta), \partial_x F(\Theta) = \tilde{B} \partial_\Theta F(\Theta)$$

My question about this problem is if the definition given for $\partial_t F(\Theta)$ and $\partial_x F(\Theta)$ is true. Because chain rule gives:
$$\partial_t F(\Theta) = \partial_t(\Theta)\partial_\Theta F(\Theta) = BA_t\partial_\Theta F(\Theta) \neq A_t/A\partial_\Theta F(\Theta)$$

Furthermore, when it is substituted into the sine-Gordon equation, I obtain:
$$(A^2B^2+1)\tilde{A}_t-2\tilde{A}^2AB-A^3B^3=(A^2B^2+1)\tilde{B}_t-2AB\tilde{B}^2-AB$$
The LHS should become a function of A only, and RHS of B, but it seems like this is not possible.

Last edited: Jul 12, 2016
2. Jul 12, 2016

### IanBerkman

These are the quantities I have used:
$$\sin u = \frac{-4(\Theta^3-\Theta)}{(\Theta^2+1)^2}\\ u_{tt} = \tilde{A}_tB\partial_\Theta u +\tilde{A}^2\partial_{\Theta\Theta}u\\ u_{xx} = A\tilde{B}_x\partial_\Theta u +\tilde{B}^2\partial_{\Theta\Theta}u$$
where
$$\partial_\Theta u = \frac{4}{\Theta^2+1}\\ \partial_{\Theta\Theta} u = \frac{-8\Theta}{(\Theta^2+1)^2}$$

3. Jul 12, 2016

### TSny

I agree that what is written above is not correct. To see what they should have written, you have shown
Simplify the last expression by writing it in terms of $\tilde{A}$, $\Theta$, and $F(\Theta)$.

4. Jul 12, 2016

### TSny

OK
I get something different for $u_{tt}$ and $u_{xx}$.
OK

5. Jul 12, 2016

### IanBerkman

This could be true since I have used the "wrong" derivatives in the problem.

I have converted it to:
$$u_t=A_t B\partial_\Theta u\\ u_{tt} = \partial_t(A_t B \partial_\Theta u) = \partial_t(A_t B)\partial_\Theta u + A_tB\partial_{\Theta}(\partial_t u)\\ =A_{tt}B\partial_\Theta u + A_t B \partial_{\Theta}(A_t B\partial_\Theta u)= A_{tt}B\partial_\Theta u + (A_t B)^2 \partial_{\Theta\Theta} u$$

6. Jul 12, 2016

### TSny

I think you will be better off expressing these in terms of $\tilde{A}$.

For example, take $u_t=A_t B\partial_\Theta u$. Can you express this in terms of $\tilde{A}$ (without any explicit appearance of $B$)?

7. Jul 12, 2016

### IanBerkman

That would indeed be $u_t = \tilde{A}\Theta \partial_\Theta u$ and $u_{tt} = \tilde{A}_t\Theta u_\Theta + 2\tilde{A}^2\Theta u_\Theta + \tilde{A}^2\Theta^2u_{\Theta\Theta}$ if I calculated correctly (it is a bit late now and I am a bit sleepy).

This looks a lot more appropriate and I will proceed tomorrow to work out the sine-Gordon equation.

8. Jul 12, 2016

### TSny

OK, except I'm not getting the factor of 2 in the middle term of the expression for $u_{tt}$.

Sounds good.

9. Jul 12, 2016

### IanBerkman

Yes I saw where the mistake came from, I am going to revise everything a bit tomorrow when I am a bit more fresh.