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NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Nonlinear Schrodinger equation with complexsupersymmetric potentials
D. Nath∗
Department of MathematicsVivekananda College, Kolkata-700063, India.
Supersymmetries & Quantum Symmetries - SQS’2015August 6, 2015
∗in collaboration with P. Roy, ISI, Kolkata, India.
1 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Plan of the talk
1 PT symmetry
2 Supersymmetric potentials
3 Non Linear Schrodinger Equation (NLSE)SolutionsExamples
4 Linear Stability AnalysisStability condition
5 Conclusion
2 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
PT symmetry
• PT symmetric quantum mechanical system are invariantunder the simultaneous action of the P space and T timeinversion operations.
• P:p→ −p, x→ −x, T :i→ −i• These systems possess non-Hermitian Hamiltonians, still
they have some characteristics similar to Hermitianproblems.
• discrete energy spectrum, (partly real or completely real)
• basis states form an orthogonal set
• inner product 〈ψ(x, t)|ψ(x, t)〉PT = 〈ψ(x, t)|PT |ψ(x, t)〉
3 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
PT symmetric
PT symmetric Hamiltonian
• simplest PT symmetric Hamiltonian contains a 1DSchrodinger operator H
• PT H=HPT = p2
2m + V ∗(−x) = H
• such problems have been described by numerical andperturbational techniques and also several PT symmetricpotentials have been described analytically
4 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
PT symmetric potential
Linear Schrodinger Eq. with PT symmetric complex potential
−d2ψ
dx2+ V (x)ψ = Eψ
PT symmetric potentials (P:p→ −p, x→ −x, T :i→ −i)
• V ∗(−x) = V (x), PT symmetric potentials
• V = VR(x) + iVI(x), then VR(−x) = VR(x) &VI(−x) = −VI(x)
• PT symmetric systems can put into the Non LinearSchrodinger Equation(NLSE)
• C. M. Bender and S. Boettcher, Phys. Rev. Lett 80,(1998) 5243.
5 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Supersymmetric potentials
Linear Schrodinger Eq. with supersymmetric potential
−d2ψ
dx2+ V±(x)ψ = Eψ
• V±(x) = w2(x)± w′(x), supersymmetric potentials
• w(x) is the superpotential
• ψ(−)0 (x) = e−
∫w(x) dx is the ground state, with E
(−)0 = 0
• F. Cooper, A. Khare and U. Sukhatme, Supersymmetry inQuantum Mechanics, World Scientific (2001).
6 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Non Linear Schrodinger Equation (NLSE) withcomplex potentials and power law nonlinearity
NLSE
iΨt = −Ψxx + (V (x) + iW (x))Ψ + g|Ψ|2kΨ (1)
• For the optical beam dynamics t is a scaled propagationdistance, g = −1, corresponds to a self-focusingnonlinearity while g = 1 to a defocusing one
• same model as the Gross-Pitaevskii equation inBose-Einstein condensates
• same model as the propagation of laser radiation along thet axis of a medium
• The NLS equations describe physical phenomena in optics,Bose–Einstein condensates, as well as water waves.
• so our model is a prototype NLSE
7 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Potentials
Cubic nonlinearity k = 1
• V (x) + iW (x) = ε δ(x) + i γ δ′(x) Chaos 25, 023112(2015).
• V (x) + iW (x) = x2 − 2 i α x Phys. Rev. A 85, 043840(2012)
• V (x) + iW (x) = w2(x)− iw′(x), w(x) is real,arXiv:1408.2719v3 (2014), (non PT )
• V (x) + iW (x) = sech2x + i sech x tanh x
• k = 2 quintic nonlinearity
8 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Ψ(x, t) = e−iµtψ(x)
−d2ψdx2
+ (V + iW )ψ + g|ψ|2kψ = µψ (2)
• µ is the propagation constant, momentum number
9 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Our present work
Super potential of the form U(x) = UR(x) + i UI(x) andgeneral power law nonlinearity k
• V = V0Re(U2 − U ′) = V0 (U2
R − U ′R − U2I )
• W = W0 Im(U2 − U ′) = W0 (2URUI − U ′I)• V (−x) = V (x), W (−x) = −W (x), i.e.,UR(−x) = −UR(x), UI(−x) = UI(x)
ψ = ψ0 e−
∫(UR+iAUI)dx
(V0 − 1)(U2R − U
′R) + (A2 − V0)U2
I + gψ2k0 e−2k
∫UR(x) dx = µ
(2URUI − U′I)(W0 −A) = 0 (3)
• for linear Schrodinger Eq. g = 0
• for exact SUSY V0 = W0 = A = 1, µ = 0
10 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
• (2URUI − U′I) = 0, generates real potential
• for PT symmetric complex potentials we choose W0 = A
• for NLSE V0 = W0 = A = 1 is not possible
• V0 6= 1 or (and )W0 = A 6= 1
11 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Characteristic of our model
The basic characteristic of our model is the following
• We are trying to use supersymmetric potential of thelinear Schrodeinger Eq. into the nonlinear one.
• However this can not be done exactly and therefore oneneeds to deform the procedure at two stages
• at the level of potential (V0,W0) and secondly at the levelof solution (A0) (which again causes a deformation of thepotential).
• We call it deformed SUSY complex potential
12 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Find UR(x), UI(x) such that they satisfy theconditions (3)
Case 1. V0 = A2, UR(x) = −f ′(x)f(x)
(V0 − 1)f′2(x) +
g0ψ2k0
k+1 f2k+2(x) = µf2(x) (4)
• this is a first order ordinary nonlinear differential equation
• for different parameter values we get different solutions
13 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Solutions for case 1
• (i) f(x) = α sech1k (βx + γ) with µ = (V0−1)β2
k2, and
g(αψ0)2k = − (k+1)(V0−1)β2
k2, PT
• (ii) f(x) = α sec1k (βx + γ) with µ = − (V0−1)β2
k2, and
g(αψ0)2k = − (k+1)(V0−1)β2
k2, PT
• (iii) f(x) = α cosech1k (βx + γ) with µ = (V0−1)β2
k2, and
g(αψ0)2k = − (k+1)(V0−1)β2
k2, non PT
• (iv) f(x) = α cosec1k (βx + γ) with µ = − (V0−1)β2
k2, and
g(αψ0)2k = − (k+1)(V0−1)β2
k2, non PT
14 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Find UR(x), UI(x) such that they satisfy theconditions
Case 2. V0 6= A2, UR(x) = −f ′(x)f(x)
(V0 − 1)f′′
+ (A20 − V0)U2
I (x) f +g0ψ2k
0k+1 f
2k+21 = µ (5)
• this is a NLSE with real potential we can solve Eq.(5) forsome particular UI(x)’s.
• these particular solutions can be obtained from thesolutions for case 1 by properly choosing the parameters.
15 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Example 1. UR = tanh xk , UI = C0sechmx
PT symmetric potential
• V = V0k2
[1− (k + 1) sech2x− k2 C2
0 sech2mx]
• W = W0 C0k (2 + km) sechmx tanh x
Ψ(x, t) = ψ0 sech1k x e−i{µt+W0 C0 2F1[
12, 1+m
2, 32,− sinh2 x] sinh x}
where
• ψ0 =[(k+1)(V0−1)
gk2
] 12k, µ = V0−1
k2, V0 = W 2
0
• 2F1[a, b, c, x] is a Hypergeometric function.
• For V0 6= W 20 the solution also exists when m = 0, 1
• M. Abramowitz and I.A. stegun, Handbook ofMathematical functions with formulas, graphs andMathematical tables, Dover, 1972.
16 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Example 2. UR = − tan xk , UI = C0secmx
PT symmetric periodic potential
• V = V0k2
[−1 + (k + 1) sec2x− k2 C2
0 sec2mx]
• W = −W0 C0k (2 + km) secmx tan x
Ψ(x, t) = ψ0 sec1k x e−i{µt+W0 C0 2F1[
12, 1+m
2, 32,sin2 x] sin x}
• ψ0 =[(k+1)(1−V0)
gk2
] 12k, µ = 1−V0
k2, V0 = W 2
0
• 2F1[a, b, c, x] is a Hypergeometric function.
• For V0 6= W 20 the solution also exists when m = 0, 1
17 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Example 3. UR = sn(x,m) dn(x,m)k cn(x,m) , UI(x) = C0
cn(x,m)
PT symmetric periodic potential
• V (x) = ξ−1cn−2(x,m) + ξ0 + ξ1cn
2(x,m)
• W (x) =(2k − 1
)W0C0
sn(x,m)dn(x,m)cn2(x,m)
Ψ(x, t) = ψ0 cn1k (x,m) e
−i{µt+
W0C0√1−m
ln[dn(x,m)cn(x,m)
+√1−m sn(x,m)
cn(x,m)]}
• ψ0 =[m(V0−1)(k+1)
gk2
] 12k, µ = (2m−1)(V0−1)
k2
• C0 ={
(k−1)(V0−1)(1−m)k2(W 2
0−V0)
} 12
, V0 6= W 20
• ξ−1 ={
(1−k)(1−m)k2
− C20
}V0
• ξ0 ={
(1−k)(2m−1)k2
+ (V0−1)(2m−1)k
}V0
• ξ1 = −{m(1−k)k2
+ 2m(V0−1)k
}V0
• sn(x,m), cn(x,m), dn(x,m) are Jacobi Elliptic functions.18 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Linear stability analysis
Perturbed solution
Ψ(x, t) = {ψ(x) + [v(x) + w(x)] eλt + [v∗(x)− w∗(x)] eλ∗t}eiµt
(6)
• |v|, |w| << |ψ|, neglecting 2nd and higher order of v, w.
Eigen value Eq.
i
(h0 ∇2 + h1
∇2 + h2 h3
)(vw
)= λ
(vw
)(7)
h0 = g k2
(ψ2 − ψ∗2
)|ψ|2k−2 + i W (x)
h1 = µ+ V (x) + g|ψ|2k +{|ψ|2 − 1
2
(ψ2 + ψ∗2
)}gk|ψ|2k−2
h2 = µ+ V (x) + g|ψ|2k +{|ψ|2 + 1
2
(ψ2 + ψ∗2
)}gk|ψ|2k−2
h3 = −gk2
(ψ2 − ψ∗2
)|ψ|2k−2 + i W (x) (8)
19 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Stability condition
λ = λR + iλI
• [v(x) + w(x)] eλt = [v(x) + w(x)] [cosλIt+ i sinλIt]eλRt
• [v∗(x)− w∗(x)] eλ∗t =
[v∗(x)− w∗(x)] [cosλIt− i sinλIt]eλRt
• If λR > 0, limt→∞
eλRt →∞, and the corresponding solution
is unstable.
• Therefore the solution Ψ(x, t) is stable if λR in nonpositive.
• To solve Eq.(7) we use Fourier Collocation Method (FCM)
20 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Eigen value spectra & time evolution of Ψ(x, t)stable solution (m = 1)
Stable mode of example 1
−4 −3 −2 −1 0 1 2 3 4 5
x 10−12
−2000
−1500
−1000
−500
0
500
1000
1500
2000
Re[λ]
Im[λ
]
(a) k=0.45, g=−1, C0=0.1, V
0=0.995,W
0=−0.1 HbL k=0.45, g=-1, C0=0.1, V0=0.995, W0=-0.1
-40 -20 0 20 40
X
0
20
40
60
80t
0.000
0.005
0.010
ÈYÈ
• (a) λR = Re(λ) < 5× 10−12 ≈ 0
• (b) stable solution
21 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Eigen value spectra & time evolution of Ψ(x, t)stable solution
Stability condition for Re(λ)
−0.05 0 0.05−2000
−1500
−1000
−500
0
500
1000
1500
2000
Re[λ]
Im[λ]
(a) k=2
−0.4 −0.2 0 0.2 0.4−2000
−1500
−1000
−500
0
500
1000
1500
2000
Re[λ]
(b) k=3
−0.4 −0.2 0 0.2 0.4−2000
−1500
−1000
−500
0
500
1000
1500
2000
Re[λ]
(c) k=4
HdL k=2, g=-1, C0=0.1, V0=0.01, W0=-0.1
-40 -20 0 20 40
X
0
20
40
60
80t
0.0
0.5
1.0
ÈYÈ
HeL k=3, g=-1, C0=0.1, V0=0.01, W0=-0.1
-40 -20 0 20 40
X
0
20
40
60
80t
0.0
0.5
1.0
ÈYÈ
HfL k=4, g=-1, C0=0.1, V0=0.01, W0=-0.1
-40 -20 0 20 40
X
0
20
40
60
80t
0.0
0.5
1.0
ÈYÈ
22 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Stable solution
Stable mode of example 1 for focusing nonlinearity (g = −1)
• g = −1, C0 = 0.1, V0 = 0.81,W0 = −0.9
−40 −20 0 20 400
20
6080
0
0.5
1
t
(d) k=3
x−40 −20 0 20 400
20
6080
0
0.5
1
t
(c) k=2
x
|ψ(x
,t)|
−1.5 −1 −0.5 0 0.5 1 1.5
x 10−11
−2000
−1000
0
1000
2000
Re[λ]
Im[λ
]
(a) k=2
−3 −2 −1 0 1 2 3 4
x 10−12
−2000
−1000
0
1000
2000
Re[λ]
(b) k=3
23 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Stable solution
Stable mode of example 1 for defocusing nonlinearity (g = 1)
• g = 1, C0 = 0.1, V0 = 1.01,W0 = −1.05
−5 0 5 10
x 10−12
−2000
−1000
0
1000
2000
Re[λ]
Im[λ
]
(a) k=2
−5 0 5
x 10−12
−2000
−1000
0
1000
2000
Re[λ]
(b) k=3
−5 0 5
x 10−12
−2000
−1000
0
1000
2000
Re[λ]
(c) k=4
−40 −20 0 20 40 0
20
60
80
0
0.2
0.4
t
(d) k=2
x
|ψ(x
,t)|
−40 −20 0 20 40 0
20
60
80
0
0.5
1
t
(e) k=3
x −40 −20 0 20 40 0
20
60
80
0
0.5
1
t
(f) k=4
x
24 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Eigen value spectra & time evolution of Ψ(x, t)unstable solution
Unstable mode of example 1
• (a) g = −1, C0 = 1, V0 = 0.25,W0 = −0.5
−0.2 −0.1 0 0.1 0.2−5
−4
−3
−2
−1
0
1
2
3
4
5
Re[λ]
Im[λ]
(a) k=2
−1 −0.5 0 0.5 1−5
−4
−3
−2
−1
0
1
2
3
4
5
Re[λ]
(b) k=3
HcL k=2, g=-1, C0=1, V0=0.25, W0=-0.5
-40 -20 0 20 40
X
0
20
40
60
80t
0
1´ 1016
2´ 1016
3´ 1016
4´ 1016
ÈYÈ
HdL k=3, g=-1, C0=1, V0=0.25, W0=-0.5
-40 -20 0 20 40
X
0
20
40
60
80t
0
2
4
ÈYÈ
25 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Eigen value spectra & time evolution of Ψ(x, t)unstable solution (m = 1)
Unstable mode of example 2
• g = 1, C0 = 0.1, V0 = 0.25,W0 = −0.5
−4 −2 0 2 4
x 104
−2000
−1500
−1000
−500
0
500
1000
1500
2000
Re[λ]
Im[λ
]
(a) k=2
−4 −2 0 2 4
x 104
−2000
−1500
−1000
−500
0
500
1000
1500
2000
Re[λ]
(b) k=3
−2 −1 0 1 2
x 104
−2000
−1500
−1000
−500
0
500
1000
1500
2000
Re[λ]
(c) k=4
26 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Eigen value spectra & time evolution of Ψ(x, t)unstable solution
Unstable mode of example 3
• g = −1,m = 0.75, V0 = 0.5,W0 = −0.5
−0.08 −0.06 −0.04 −0.02 0 0.02 0.04 0.06 0.08−1.5
−1
−0.5
0
0.5
1
1.5x 10
4
Im[λ
]
(a) k=2
−4000 −3000 −2000 −1000 0 1000 2000 3000 4000−1.5
−1
−0.5
0
0.5
1
1.5x 10
4 (b) k=3
−5000 0 5000−1
−0.5
0
0.5
1x 10
4
Re[λ]
Im[λ
]
(c) k=4
−6000 −4000 −2000 0 2000 4000 6000−1
−0.5
0
0.5
1x 10
4
Re[λ]
(d) k=4.45
27 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Conclusion
• A method of constructing exact solutions of NLSE withpower law nonlinearity and complex PT symmetricpotentials has been suggested.
• We have analyzed Linear stability analysis of thesesolutions .
• In this case we find that for suitably chosen parametervalues stable solutions can be found for g = 1, k = 2, 3 aswell as for g = −1, k = 2, 3.
• Stable solutions for non integral values of the nonlinearityparameter k has also been found for the example 1 forfocusing (g=-1) as well as de-focusing (g=1) nonlinearity.
• Using these method we can solve NLSE with nonPT symmetric potentials as well.
28 / 29
NonlinearSchrodingerequation withcomplex su-persymmetricpotentials
D. Nath∗
PTsymmetry
Supersymmetricpotentials
Non LinearSchrodingerEquation(NLSE)
Solutions
Examples
LinearStabilityAnalysis
Stabilitycondition
Conclusion
Thank You
29 / 29
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