i+v+σuu0 (1)
u
t
u
xx
2
|
u(x,t)
x∈R,t>0
i
ν,σ∈R
u(x,t)sech(x-2κνt)exp[i(κx-Ωt)](2)
2
Ω-ν
2
κ
σ
Ω-ν
2
κ
ν
κ,Ω∈R
Ω2
κνσ1
t0,2.5,3.5
x∈[-50,50]
u
Re(u)
Im(u)
u
In[]:=
Clear["Global`*"](*defineparameters*)v=1;k=1;σ=1;Ω=2;(*definesolitonsolution*)u[x_,t_]=Sqrt[2]*Sqrt[Abs[(Ω-k^2*v)/σ]]*Sech[Sqrt[Abs[(Ω-k^2*v)/v]]*(x-2*k*v*t)]*Exp[I(k*x-Ω*t)]
Out[]=
2
(-2t+x)
For t0
For
t0
In[]:=
fig1=Plot[Evaluate[Im[u[x,0]]],{x,-50,50},PlotRange->All,PlotLabel->"Imu"]
Out[]=
In[]:=
fig2=Plot[Evaluate[Re[u[x,0]]],{x,-50,50},PlotRange->All,PlotLabel->"Re[u]"]
Out[]=
In[]:=
fig3=Plot[Evaluate[Abs[u[x,0]]],{x,-50,50},PlotRange->All,PlotStyle->{Thick,Dashed,Blue},PlotLabel->"Abs[u]"]
Out[]=
For t2.5
For
t2.5
In[]:=
fig4=Plot[Evaluate[Im[u[x,2.5]]],{x,-50,50},PlotRange->All,PlotLabel->"Im[u]"]
Out[]=
In[]:=
fig5=Plot[Evaluate[Re[u[x,2.5]]],{x,-50,50},PlotRange->All,PlotLabel->"Re[u]"]
Out[]=
In[]:=
fig6=Plot[Evaluate[Abs[u[x,2.5]]],{x,-50,50},PlotRange->All,PlotStyle->{Thick,Dashed,Red},PlotLabel->"Abs[u]"]
Out[]=
For t3.5
For
t3.5
In[]:=
fig7=Plot[Evaluate[Im[u[x,3.5]]],{x,-50,50},PlotRange->All,PlotLabel->"Im[u]"]
Out[]=
In[]:=
fig8=Plot[Evaluate[Re[u[x,3.5]]],{x,-50,50},PlotRange->All,PlotLabel->"Re[u]"]
Out[]=
In[]:=
fig9=Plot[Evaluate[Abs[u[x,3.5]]],{x,-50,50},PlotRange->All,PlotStyle->{Thick,Dashed,Purple},PlotLabel->"Abs[u]"]
Numerical Solution
Numerical Solution
Problem 1
Problem 1
3-D visualization
3-D visualization
Problem 2
Problem 2
CITE THIS NOTEBOOK
CITE THIS NOTEBOOK
Nonlinear Schrödinger equation: solution and visualization
by Athanasios Paraskevopoulos
Wolfram Community, STAFF PICKS, April 24 2023
https://community.wolfram.com/groups/-/m/t/2906358
by Athanasios Paraskevopoulos
Wolfram Community, STAFF PICKS, April 24 2023
https://community.wolfram.com/groups/-/m/t/2906358