Stabilization of higher order Schrödinger equations on a finite interval: Part II

Türker Özsarı; Kemal Cem Yılmaz

doi:10.3934/eect.2021037

Article Contents

2022, Volume 11, Issue 4: 1087-1148. Doi: 10.3934/eect.2021037

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Stabilization of higher order Schrödinger equations on a finite interval: Part II

Türker Özsarı^1, , and
Kemal Cem Yılmaz^2,

1.
Department of Mathematics, Bilkent University, Bilkent, Ankara, 06800 Turkey
2.
Department of Mathematics, İzmir Institute of Technology, Urla, İzmir, 35430 Turkey

^*Corresponding author: Türker Özsarı
^*Corresponding author: Türker Özsarı

Received: January 31, 2021

Early access: July 2021

Published: August 2022

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Abstract

Backstepping based controller and observer models were designed for higher order linear and nonlinear Schrödinger equations on a finite interval in [3] where the controller was assumed to be acting from the left endpoint of the medium. In this companion paper, we further the analysis by considering boundary controller(s) acting at the right endpoint of the domain. It turns out that the problem is more challenging in this scenario as the associated boundary value problem for the backstepping kernel becomes overdetermined and lacks a smooth solution. The latter is essential to switch back and forth between the original plant and the so called target system. To overcome this difficulty we rely on the strategy of using an imperfect kernel, namely one of the boundary conditions in kernel PDE model is disregarded. The drawback is that one loses rapid stabilization in comparison with the left endpoint controllability. Nevertheless, the exponential decay of the $ L^2 $-norm with a certain rate still holds. The observer design is associated with new challenges from the point of view of wellposedness and one has to prove smoothing properties for an associated initial boundary value problem with inhomogeneous boundary data. This problem is solved by using Laplace transform in time. However, the Bromwich integral that inverts the transformed solution is associated with certain analyticity issues which are treated through a subtle analysis. Numerical algorithms and simulations verifying the theoretical results are given.

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Mathematics Subject Classification: Primary: 35Q93, 93B52, 93C20, 93D15, 93D20, 93D23; Secondary: 35A01, 35A02, 35Q55, 35Q60.

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Figure 1. Backstepping

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Figure 2. Triangular regions

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Figure 3. Backstepping with an imperfect kernel

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Figure 4. $ H(k, l) : [1, \infty) \times [1, \infty) \to (-1, 1) $

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Figure 5. Integration path for the case $ \alpha^2 + 3\beta\delta > 0 $

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Figure 6. Plot of transformation $ \Im(s) = \omega(\xi) $ when $ \alpha^2 + 3\beta\delta > 0 $

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Figure 7. Integration path for the case $ \alpha^2 + 3\beta\delta = 0 $

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Figure 8. Plot of transformation $ \Im(s) = \omega(\xi) $ when $ \alpha^2 + 3\beta\delta = 0 $

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Figure 9. Integration path for the case $ \alpha^2 + 3\beta\delta < 0 $

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Figure 10. Plot of transformation $ \Im(s) = \omega(\xi) $ when $ \alpha^2 + 3\beta \delta < 0 $

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Figure 11. Up: Plots of $ k(x, y) $ on $ \Delta_{x, y} $. Down: Controller gains for Dirichlet and Neumann boundary conditions. $ L = \pi $, $ \beta = 1 $, $ \alpha = 2 $, $ \delta = 8 $ and $ r = 0.05 $

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Figure 12. $ p(x, y) $ defined on $ \Delta_{x, y} $ for $ L = \pi $, $ \beta = 1 $, $ \alpha = 2 $, $ \delta = 8 $ and $ r = 0.05 $

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Figure 13. Observer gains for $ L = \pi $, $ \beta = 1 $, $ \alpha = 2 $, $ \delta = 8 $ and $ r = 0.05 $

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Figure 14. Numerical results in the presence of controllers. Left: Time evolution of $ |u(x, t)| $. Right: Time evolution of $ \|u(\cdot, t)\|_2 $

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Figure 15. Numerical results. Left: Time evolution of $ |u(x, t)| $. Right: Time evolution of $ \|u(\cdot, t)\|_2 $, $ \|\hat u(\cdot, t)\|_2 $ and $ \|\tilde u(\cdot, t)\|_2 $

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Table 1. Some numerical values for the decay rate $ \lambda $ corresponding to various values of $ r $

$r$	$\lambda = \beta\left(\displaystyle\frac{r}{\beta} - \frac{\\|k_y(\cdot, 0;r)\\|_2^2}{2}\right)$
$0.001$	$0.001981$
$0.01$	$0.018054$
$0.02$	$0.032221$
$0.03$	$0.042507$
$0.04$	$0.048918$
$0.05$	$0.051463$
$0.1$	$0.006407$
$0.11$	$-0.014113$
$0.5$	$-3.729586$
$1$	$-16.379897$

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