2
$\begingroup$

After doing some googling, I failed to find any explicit solution for the Biharmonic Nonlinear Schrodinger Equation, which states: $$ i\psi (x,t) _t - \Delta ^2 \psi (x,t) + |\psi (x,t) | ^{2 \sigma} \psi (x,t) = 0 $$ $$ \psi(x,0) = \psi_0 (x) \in H^2, x \in\ \mathbb{R}^d $$

Since my main purpose is to run a sanity check to my simulation, other more "soft" checks will be welcomed. I work with $ d = 1 $ and various integer values of $\sigma$.

$\endgroup$

3 Answers 3

2
$\begingroup$

Well, the obvious thing to test first is whether for $d=1$, $\sigma=4$ your solution conserves the $L^2$ norm under the scaling transformation $\psi(x,t)\mapsto L^{-1/2}\psi(x/L,t/L^4)$. For other values of $\sigma$ you could test for the variance identity derived in section 2.1 of Singular solutions of the biharmonic Nonlinear Schrodinger equation (2009).

$\endgroup$
2
$\begingroup$

If you are concerned about testing the consistency of your code, I recommend solving an inhomogeneous problem with known solution. To this end, choose a function $\psi$. For this function, apply the operator on the left and you get another function. Replace the zero on the right by this function.

If you run this program with different discretization parameters $\Delta t$ and $\Delta x$, you should observe the theoretical convergence orders of your method when you compare the computed solution with your previously chosen function $\psi$.

$\endgroup$
1
$\begingroup$

So, as my professor pointed out, there is a solitonic solution for the BNLS. It is not numerically stable and not Explicit, but it still serves as a good check. A solitonic solution for the BNLS is the solution of the following radial ODE:

$ - \Delta _r ^2 R(r) - R(r) + |R(r)|^{2\sigma} R(r) = 0 $

$ R'(0) = R'''(0) = 0, $ $R(\infty) = 0 $

In the case of $d = 1 $ , $\sigma = 4$ , things get even simpler and you get the following ODE:

$ -R^{(4)} -R + R^{9} = 0 $

This solution has an asymptotic sanity check for $r >> 1$, which states : $R(r) \sim ce^{-r/ \sqrt{2} } cos(r/ \sqrt{2} ) $

Again, not a perfect sanity check but it should work. for more details, look at section 4 in this paper: http://www.math.tau.ac.il/~fibich/Manuscripts/dispersion.pdf

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .