A The Nonlinear Schrödinger Equation

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1. Apr 2, 2016

roam

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. Apr 2, 2016

A. Neumaier

Resolve the remaining differentiations using the product rule and chain rule.

3. Apr 2, 2016

Staff: Mentor

There is no extra $P_0$. Check your derivation again.

4. Apr 2, 2016

roam

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 $\sqrt{P_0}$?

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.

5. Apr 3, 2016

Staff: Mentor

I think he meant the product rule.

6. Apr 3, 2016

roam

But how is the product rule applicable in this case?

7. Apr 3, 2016

blue_leaf77

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 $|N| \textrm{sign }N$.

8. Apr 3, 2016

roam

Thank you very much for the clarification.