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Breather solution of the sine-Gordon equation

  1. Jul 12, 2016 #1
    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. jcsd
  3. Jul 12, 2016 #2
    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
    $$\partial_\Theta u = \frac{4}{\Theta^2+1}\\
    \partial_{\Theta\Theta} u = \frac{-8\Theta}{(\Theta^2+1)^2}$$
     
  4. Jul 12, 2016 #3

    TSny

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    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)##.
     
  5. Jul 12, 2016 #4

    TSny

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    OK
    I get something different for ##u_{tt}## and ##u_{xx}##.
    OK
     
  6. Jul 12, 2016 #5
    This could be true since I have used the "wrong" derivatives in the problem.

    I have converted it to:
    $$u_t=A_t B\partial_\Theta u\\
    u_{tt} = \partial_t(A_t B \partial_\Theta u) = \partial_t(A_t B)\partial_\Theta u + A_tB\partial_{\Theta}(\partial_t u)\\
    =A_{tt}B\partial_\Theta u + A_t B \partial_{\Theta}(A_t B\partial_\Theta u)= A_{tt}B\partial_\Theta u + (A_t B)^2 \partial_{\Theta\Theta} u$$
     
  7. Jul 12, 2016 #6

    TSny

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    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##)?
     
  8. Jul 12, 2016 #7
    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.
     
  9. Jul 12, 2016 #8

    TSny

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

    Sounds good.
     
  10. Jul 12, 2016 #9
    Yes I saw where the mistake came from, I am going to revise everything a bit tomorrow when I am a bit more fresh.
     
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