# Deriving a differential equation for momentum from the integral form

The Lugiato-Lefever equation can be written in the form:

$$\frac{\partial \psi}{\partial \tau} = -(1 + i\alpha)\psi - i\frac{\beta}{2}\frac{\partial ^2 \psi}{\partial \theta^2} + i|\psi|^2 \psi + F_0 \exp[i\delta_m \sin\theta]$$

The solutions to this partial differential equation can have soliton solutions. In a paper that I am reading, I found an equation for the "momentum" of these solitons:

$$P = -\frac{i}{2}\int_{-\pi}^{\pi} d\theta \left ( \psi^*\frac{\partial \psi}{\partial \theta} - \psi \frac{\partial \psi^*}{\partial \theta} \right )$$

The authors then proceed to find the time derivative of the above equation and state without showing any calculations that it is given by:

$$\frac{dP}{d \tau} = -2P - i\int_{-\pi}^{\pi} d\theta \left ( \psi^*\frac{\partial F}{\partial \theta} - \psi \frac{\partial F^*}{\partial \theta} \right )$$

where $$F(\theta,\tau) = F_0 \exp[i\delta_m \sin\theta]$$.

I am struggling to derive the third equation from the second. I attempted to take the derivative of the second equation with respect to $$\tau$$, but then I got stuck dealing with too many parameters and I feel like there is an easy way to show this.

• Not an answer to your question but interesting : ncbi.nlm.nih.gov/pmc/articles/PMC8325515 Commented Aug 5, 2023 at 21:28
• It's interesting you mentioned this article. I am a PhD student collaborating with the authors of that paper and I am very familiar with that article. Commented Aug 5, 2023 at 21:39

## 1 Answer

We can use a facsimile of the Ehrenfest theorem by following its proof as far as we can for this nonlinear Schrodinger equation. Rewrite your momentum equation as

$$\langle P \rangle = \int_{-\pi}^\pi \psi^*P\psi \:d\theta$$

by integration by parts, where $$P \equiv -i\frac{\partial}{\partial \theta}$$. I am also assuming based on the angular notation and the integral bounds that $$P$$ is self-adjoint because of periodic boundary conditions at $$\pm \pi$$ (but this crucial context was missing and not stated from the original post!!). Also denote

$$\frac{1}{i}\frac{\partial}{\partial \tau}(\cdot) = \left[(-\alpha+i)-\frac{\beta}{2}\frac{\partial^2}{\partial\theta^2}\right](\cdot) + |\cdot|^2(\cdot) - iF_0\exp(i\delta_m\sin\theta) \equiv H(\cdot) + N(\cdot) - iF$$

Now taking time derivatives, we obtain

$$\frac{d\langle P \rangle}{dt} = i\int_{-\pi}^\pi -(H^*\psi+N\psi-iF)^*(P\psi)+\psi^*(PH\psi+PN\psi-iPF)d\theta$$

$$= i\int_{-\pi}^\pi \psi^*(PH-H^*P)\psi -i \left(\psi^2\psi^*\frac{\partial\psi^*}{\partial\theta}+{\psi^{*}}^2\psi\frac{\partial\psi}{\partial\theta}\right)-\left(\psi^*\frac{\partial F}{\partial\theta}+F^*\frac{\partial\psi}{\partial\theta}\right)\:d\theta$$

$$\require{cancel} = i\int_{-\pi}^\pi \psi^*(2iP)\psi\:d\theta + \cancel{\frac{|\psi|^4}{2}-iF^*\psi\Biggr|_{-\pi}^\pi} + i\int_{-\pi}^\pi\psi\frac{\partial F^*}{\partial\theta}-\psi^*\frac{\partial F}{\partial\theta}\:d\theta$$

$$\implies \boxed{\frac{d\langle P \rangle}{dt} = -2\langle P \rangle -i\int_{-\pi}^\pi\psi^*\frac{\partial F}{\partial\theta}-\psi\frac{\partial F^*}{\partial\theta}\:d\theta}$$

Remark: This implies that on average, energy is being drained away from the system as the momentum decays away if there was no driving force. That is a result of the complex-valued potential with positive imaginary part, not of the nonlinear cubic term which ended up not contributing (or detracting) from energy conservation at all.

• Exactly what I wanted! Thanks! Also, apologies for not specifying the boundary conditions. My colleagues and I look at this equation so often that it is implied between us and I forget that not everyone is familiar with the equation. Commented Aug 6, 2023 at 0:26