Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A The Nonlinear Schrödinger Equation

  1. Apr 2, 2016 #1
    According to my textbook the nonlinear Schrödinger equation:

    $$\frac{\partial A(z,T)}{\partial z} = -i \frac{\beta_2}{2} \frac{\partial^2A}{\partial T^2} + i \gamma |A|^2 A \ \ (1)$$

    can be cast in the form

    $$\frac{\partial U(z,\tau)}{\partial z} = -i \frac{sign \beta_2}{2} \frac{1}{L_D} \frac{\partial^2 U}{\partial \tau^2} + i \frac{1}{L_{NL}} |U|^2 U \ \ (2)$$

    by normalizing with: ##\tau = \frac{T}{T_0},## and ##A(z,T) = \sqrt{P_0} U(z, \tau).##

    But my textbook does not show the steps involved, and I can't arrive at equation (2) when I try to do this myself.

    So substituting the two parameters into (1) we get

    $$\frac{\partial (\sqrt{P_0} U(z, \tau))}{\partial z} = -i \frac{\beta_2}{2} \frac{\partial^2(\sqrt{P_0} U(z, \tau))}{\partial (\tau T_0)^2} + i \gamma |(\sqrt{P_0} U(z, \tau))|^2 (\sqrt{P_0} U(z, \tau))$$

    We know that the dispersion length is given by ##L_D = \frac{T_0^2}{|\beta_2|}## and the the nonlinear length is ##L_{NL} = \frac{1}{\gamma P_0}.## When substituting these two the expression becomes

    $$\frac{\partial (\sqrt{P_0} U(z, \tau))}{\partial z} = -i \frac{1}{2} \frac{1}{L_D} \frac{\partial^2(\sqrt{P_0} U(z, \tau))}{\partial \tau^2} + i \frac{1}{L_{NL}} \sqrt{P_0} P_0 U^2 U.$$

    So, what can we do about the extra ##\sqrt{P_0}##'s and the extra ##P_0## (peak power)? What is wrong here?

    Any explanation is greatly appreciated.
     
  2. jcsd
  3. Apr 2, 2016 #2

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    Resolve the remaining differentiations using the product rule and chain rule.
     
  4. Apr 2, 2016 #3

    DrClaude

    User Avatar

    Staff: Mentor

    There is no extra ##P_0##. Check your derivation again.
     
  5. Apr 2, 2016 #4
    Sorry I meant that we get:

    $$\frac{\partial (\boxed{\sqrt{P_0}} U(z, \tau))}{\partial z} = -i \frac{1}{2} \frac{1}{L_D} \frac{\partial^2 (\boxed{\sqrt{P_0}} U(z, \tau))}{\partial \tau^2} + i \frac{1}{L_{NL}} \boxed{ \sqrt{P_0}} U^2 U.$$

    with the the P0 terms boxed. Do we then need to take the ##\sqrt{P_0}## terms out of the derivation and divide both sides by ##\sqrt{P_0}##? :confused:

    Also how do I introduce the "sign β2" expression (sign of the GVD parameter) in there? That is either ##\pm 1## (for the focusing/defocusing case).

    I am not sure what you mean. Are you referring to ##\partial / \partial z \sqrt{P_0} U(z, \tau)##, and ##\partial^2 / \partial^2 \tau \sqrt{P_0} U(z, \tau)##? Some more explanation would be very helpful.
     
  6. Apr 3, 2016 #5

    DrClaude

    User Avatar

    Staff: Mentor

    I think he meant the product rule.
     
  7. Apr 3, 2016 #6
    But how is the product rule applicable in this case?
     
  8. Apr 3, 2016 #7

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    Check in your book what ##P_0## is, if it turns out to be a constant then you can take it out from the derivatives.
    In defining ##L_D##, you use ##|\beta_2|## instead of ##\beta_2##. That explains why ##\textrm{sign }\beta_2## appears in the final equation - for any real number ##N##, you can always write it as ##|N| \textrm{sign }N ##.
     
  9. Apr 3, 2016 #8
    Thank you very much for the clarification.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: The Nonlinear Schrödinger Equation
  1. Schrödinger equation (Replies: 11)

Loading...