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The GP equation is

$$i\frac{\partial u}{\partial z}+\nabla^2u+|u|^2u+\int e^{-[(x-x')^2+(y-y')^2]}|u(x',y')|^2 dx'dy'u(x,y)=0$$

with Neumann boundary condition. The initial condition is given by a Gaussian function, and $x, y\in [-10,10], z>0$.

Question: How do we deal with partial differential equations with a convolution integral?

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    $\begingroup$ What does GP stands for? $\endgroup$ – nicoguaro♦ Jul 19 '19 at 15:21
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    $\begingroup$ One simple way is to do a fixed-point iteration where you calculate the integral first with an initial guess in place of $u$ (say $u_0$) and then solve for $u_k$ from $i \partial_z u_k + \nabla^2 u_k + |u_k|^2 u_k + f(u_{k-1}) = 0$ using standard PDE technique iterating over $k$. $\endgroup$ – knl Jul 19 '19 at 17:03
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    $\begingroup$ How should we read this? E.g. is $u$ a function of $x,y,z$, and we integrate over $x', y'$, i.e. $$i \frac{\partial u}{\partial z}(x,y,z)+\nabla^{2} u(x,y,z)+|u(x,y,z)|^{2} u(x,y,z)+\int e^{-\left[\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}\right]}\left|u\left(x^{\prime}, y^{\prime}, z\right)\right|^{2} u(x, y,z)dx'dy'=0?$$ Or something else? $\endgroup$ – doetoe Jul 22 '19 at 20:48
  • $\begingroup$ @doetoe, it's right $\endgroup$ – yun shi Jul 25 '19 at 7:27

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