# Obtaining the Lagrangian Formulation for a Linearized System

I have a system of one NLS equation and one Poisson equation:

$$i \dfrac{\partial u}{\partial Z} + \dfrac{1}{2}\nabla^2 u + 2 \ \Phi \ u = 0 \\ \nu \nabla^2 \, \Phi - 2 \, q \, \Phi + 2 \mid u \mid^2 \: = \: 0\\$$

Where u is an envelope function, q and $\nu$ are scalars, and $\Phi$ is a direction function (the angle that the solution takes as it travels through the space).

The "Lagrangian formulation" for this system is:

$$L \ = \ i \big(u^* u_Z - u u_Z^* \big) \ - \ \mid \nabla u \mid^2 \ + \ 4 \, \Phi \mid u \mid^2 - \ \nu \mid \nabla \Phi \mid^2 - 2 \, q \, \Phi^2$$

I understand that the Lagrangian is $$L \left( u, \dot{u} \right) \, = T \left( \dot{u} \right) - V\left( u \right)$$

But I don't see how that, applied to my system, gives the result. For instance, how is a partial derivative suddenly turned into a difference of products involving complex conjugates?

• It may have something to do with the treatments in this article: en.wikipedia.org/wiki/Heisenberg_picture, that would give $i$ times a commutator looking expression. What kind of problem does this relate to, is it quantum mechanical? Commented Aug 10, 2018 at 15:12
• Is there a term like $q |\Phi|^2$ missing in the Lagrangian? My guess is, that the Lagrangian density needs to be a function mapping $(u,\Phi)$ into the reals, therefore one can not use $i u \cdot u_Y$ and a term like $\text{Im}(u \cdot u_Z)$ is used instead. Commented Aug 10, 2018 at 15:21

Interpretation of the Lagrangian.

I will use $t$ instead of $Z$ and denote $u_Z$ by $\dot u$, since the equations look more familar in this case.

We define $\mathcal L(u,\nabla u, \dot u, \Phi, \nabla \Phi, \dot \Phi)$ via $$\mathcal L = \ i \big(u^* \dot u - u \dot u^* \big) - | \nabla u |^2 + 4\Phi | u |^2 - \nu | \nabla \Phi |^2 + 2q |\Phi|^2.$$

Hamiltons principle from classical field theory now implies that we look for stationary points of $$S(u,\Phi) = \int_{\Omega_T} \mathcal L(u,\nabla u, \dot u, \Phi, \nabla \Phi, \dot \Phi) \mathrm{dvol}.$$

($\Omega_T$ denotes the domain of $u$ and $\Phi$, for example $[0,T] \times \Omega$.)

The Euler-Lagrange-Equations are given as

$$\frac{\partial \mathcal L}{\partial u} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot u} - \sum_i \frac{\mathrm d}{\mathrm d X_i}\frac{\partial \mathcal L}{\partial \mathrm D_i u} = 0$$

and

$$\frac{\partial \mathcal L}{\partial \Phi} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot \Phi} - \sum_i \frac{\mathrm d}{\mathrm d X_i}\frac{\partial \mathcal L}{\partial \mathrm D_i \Phi} = 0.$$

The principle idea behind these equations is partial integration, basically the same as here: energy minimization.

Here $\mathrm D_i u = \frac{\partial u}{\partial X_i}$ and the derivative $\frac{\partial \mathcal L}{\partial \mathrm D_i u}$ denotes the derivative of $\mathcal L$ with respect to the components of $\nabla u$ in $\mathcal L$. (To make this mathematical sound, the concept of jets is helpful, but I don't want to go into this. Basically, we consider $u$, $\dot u$ and $\nabla u$ to be independent when computing the derivatives.)

The rest is 'just' a computational task...

The influence of $i (u^* u_Z - u u_Z^*)$?

In my notation, this term is denoted by $i (u^* \dot u - u \dot u^*)$. Let me split real and imaginary parts, $u = v + iw$, then this term becomes \begin{align} i \left( (v-iw)(\dot v + i \dot w) - (v+iw)(\dot v - i \dot w)\right) &= 2 i ~ \text{Im}( (v-iw)(\dot v + i\dot w) ) \\ &= -2 ( v \dot w - w \dot u )\\ &= 2 ( w \dot v - v \dot w ). \end{align}

This term enters the Euler-Lagrange equations trough \begin{align} \frac{\partial \mathcal L}{\partial u} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot u} &= \frac{\partial \mathcal L}{\partial v} - \frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot v} + i \frac{\partial \mathcal L}{\partial w} - i\frac{\mathrm d}{\mathrm dt} \frac{\partial \mathcal L}{\partial \dot w}\\ &= -2 \dot w - \frac{\mathrm d}{\mathrm d t}(2w) + i (2 \dot v - i \frac{\mathrm d}{\mathrm d t}( -2 w ) )\\ &= 4i \frac{\partial u}{\partial t}. \end{align}

The rest...

The complete computation is a bit to long for me to write down here, but not complicated anymore. Therefore, I just demonstrate one step.

For example a term like $\mathcal L_1 = |\nabla \Phi|^2$ leads to the term $$-\sum_i \frac{\mathrm d}{\mathrm d X_i}\frac{\partial \mathcal L_1}{\partial \mathrm D_i \Phi} = -\text{DIV}( 2 \nabla \Phi ) = -2 \Delta \Phi$$ in the Euler-Lagrange equations.

At the end, the first equation is divided by $4$ and the second equation by $2$.

• Thank you for your work! I see how, if you choose the Lagrangian to start, you can get the terms in the original system: using your example, running $$\nu \mid \nabla \Phi \mid^2$$ through the Euler-Lagrange equations will indeed return $$-2 \nu \, \nabla^2 \Phi$$ What I am trying to figure out is how you can derive the Lagrangian, term-by-term, from the original system. If someone gives me the system, and then asks me to calculate the Lagrangian, how do I do that? For one thing, it looks like I am somehow combining $$2 \Phi u$$ and $$2 \mid u \mid^2$$ to get $$4 \Phi \mid u \mid^2$$. Commented Aug 12, 2018 at 18:22
• Somehow the problem is similar to finding a primitive function! There you want to find $F$ sucht that $F' = f$, to find a Lagrangian you want to find $\mathcal L$ such that the Euler-Lagrange equations are correct. Hence I assume that educated guesses or physical insigth are the best approach. Commented Aug 12, 2018 at 18:42
• I don't know enough about the physics of your system. It looks like some quantum theory and electro-mechanics? Why are you interested in the Lagrangian? (It might be easier to prove existence, without explicitly computing it. There are some integrability Lemmas also for the infinite dimensional case.) Commented Aug 12, 2018 at 18:52
• Good morning! I'm modeling 1) the propagation of a light pulse 2) through a nematic liquid crystal that is 3) subjected to an external electric field. So yes, the NLS equation certainly has some quantum flavor to it, and there are electro-mechanics as well. Sala, F., et al, 'Bending reorientational solitons with modulated alignment,' (Journal of the Optical Society of America B/ Vol. 34, No. 12 / December 2017) gives a thorough overview -- if you'd like to see a bit of what is happening in the field. Commented Aug 13, 2018 at 7:24