The Nonlinear Schrödinger Equation

[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.

 

[A. Neumaier]

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

 

[DrClaude]

extra $P_0$

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

 

[roam]

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

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).

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

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.

 

[DrClaude]

I am not sure what you mean. Are you referring to ∂/∂z√P0U(z,τ)\partial / \partial z \sqrt{P_0} U(z, \tau), and ∂2/∂2τ√P0U(z,τ)\partial^2 / \partial^2 \tau \sqrt{P_0} U(z, \tau)? Some more explanation would be very helpful.

I think he meant the product rule.

 

[roam]

I think he meant the product rule.

But how is the product rule applicable in this case?

 

[blue_leaf77]

Do we then need to take the √P0P0\sqrt{P_0} terms out of the derivation and divide both sides by √P0P0\sqrt{P_0}?

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.

Also how do I introduce the "sign β2" expression (sign of the GVD parameter) in there?

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

 

[roam]

Thank you very much for the clarification.