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Has this been eliminated? I've been at 31 or 32 for 30 days for awhile, but no badge. 10 badge was automatic.

Formal Proof of Smooth Solutions for Modified Navier-Stokes Equations

1. Introduction

We address the existence and smoothness of solutions to the modified Navier-Stokes equations that incorporate frequency resonances and geometric constraints. Our goal is to prove that these modifications prevent singularities, leading to smooth solutions.

2. Mathematical Formulation

2.1 Modified Navier-Stokes Equations

Consider the Navier-Stokes equations with a frequency resonance term R(u,f)\mathbf{R}(\mathbf{u}, \mathbf{f})R(u,f) and geometric constraints:

∂u∂t+(u⋅∇)u=−∇pρ+ν∇2u+R(u,f)\frac{\partial \mathbf{u}}{\partial t} + (\mathbf{u} \cdot \nabla) \mathbf{u} = -\frac{\nabla p}{\rho} + \nu \nabla^2 \mathbf{u} + \mathbf{R}(\mathbf{u}, \mathbf{f})∂t∂u​+(u⋅∇)u=−ρ∇p​+ν∇2u+R(u,f)

where:

• u=u(t,x)\mathbf{u} = \mathbf{u}(t, \mathbf{x})u=u(t,x) is the velocity field.

• p=p(t,x)p = p(t, \mathbf{x})p=p(t,x) is the pressure field.

• ν\nuν is the kinematic viscosity.

• R(u,f)\mathbf{R}(\mathbf{u}, \mathbf{f})R(u,f) represents the frequency resonance effects.

• f\mathbf{f}f denotes external forces.

2.2 Boundary Conditions

The boundary conditions are:

u⋅n=0 on Γ\mathbf{u} \cdot \mathbf{n} = 0 \text{ on } \Gammau⋅n=0 on Γ

where Γ\GammaΓ represents the boundary of the domain Ω\OmegaΩ, and n\mathbf{n}n is the unit normal vector on Γ\GammaΓ.

3. Existence and Smoothness of Solutions

3.1 Initial Conditions

Assume initial conditions are smooth:

u(0)∈C∞(Ω)\mathbf{u}(0) \in C^{\infty}(\Omega)u(0)∈C∞(Ω) f∈L2(Ω)\mathbf{f} \in L^2(\Omega)f∈L2(Ω)

3.2 Energy Estimates

Define the total kinetic energy:

E(t)=12∫Ω∣u(t)∣2 dΩE(t) = \frac{1}{2} \int_{\Omega} \mathbf{u}(t)^2 \, d\OmegaE(t)=21​∫Ω​∣u(t)∣2dΩ

Differentiate E(t)E(t)E(t) with respect to time:

dE(t)dt=∫Ωu⋅∂u∂t dΩ\frac{dE(t)}{dt} = \int_{\Omega} \mathbf{u} \cdot \frac{\partial \mathbf{u}}{\partial t} \, d\OmegadtdE(t)​=∫Ω​u⋅∂t∂u​dΩ

Substitute the modified Navier-Stokes equation:

dE(t)dt=∫Ωu⋅[−∇pρ+ν∇2u+R] dΩ\frac{dE(t)}{dt} = \int_{\Omega} \mathbf{u} \cdot \left[ -\frac{\nabla p}{\rho} + \nu \nabla^2 \mathbf{u} + \mathbf{R} \right] \, d\OmegadtdE(t)​=∫Ω​u⋅[−ρ∇p​+ν∇2u+R]dΩ

Using the divergence-free condition (∇⋅u=0\nabla \cdot \mathbf{u} = 0∇⋅u=0):

∫Ωu⋅∇pρ dΩ=0\int_{\Omega} \mathbf{u} \cdot \frac{\nabla p}{\rho} \, d\Omega = 0∫Ω​u⋅ρ∇p​dΩ=0

Thus:

dE(t)dt=−ν∫Ω∣∇u∣2 dΩ+∫Ωu⋅R dΩ\frac{dE(t)}{dt} = -\nu \int_{\Omega} \nabla \mathbf{u}^2 \, d\Omega + \int_{\Omega} \mathbf{u} \cdot \mathbf{R} \, d\OmegadtdE(t)​=−ν∫Ω​∣∇u∣2dΩ+∫Ω​u⋅RdΩ

Assuming R\mathbf{R}R is bounded by a constant CCC:

∫Ωu⋅R dΩ≤C∫Ω∣u∣ dΩ\int_{\Omega} \mathbf{u} \cdot \mathbf{R} \, d\Omega \leq C \int_{\Omega} \mathbf{u} \, d\Omega∫Ω​u⋅RdΩ≤C∫Ω​∣u∣dΩ

Applying the Poincaré inequality:

∫Ω∣u∣2 dΩ≤Const⋅∫Ω∣∇u∣2 dΩ\int_{\Omega} \mathbf{u}^2 \, d\Omega \leq \text{Const} \cdot \int_{\Omega} \nabla \mathbf{u}^2 \, d\Omega∫Ω​∣u∣2dΩ≤Const⋅∫Ω​∣∇u∣2dΩ

Therefore:

dE(t)dt≤−ν∫Ω∣∇u∣2 dΩ+C∫Ω∣u∣ dΩ\frac{dE(t)}{dt} \leq -\nu \int_{\Omega} \nabla \mathbf{u}^2 \, d\Omega + C \int_{\Omega} \mathbf{u} \, d\OmegadtdE(t)​≤−ν∫Ω​∣∇u∣2dΩ+C∫Ω​∣u∣dΩ

Integrate this inequality:

E(t)≤E(0)−ν∫0t∫Ω∣∇u∣2 dΩ ds+CtE(t) \leq E(0) - \nu \int_{0}^{t} \int_{\Omega} \nabla \mathbf{u}^2 \, d\Omega \, ds + C tE(t)≤E(0)−ν∫0t​∫Ω​∣∇u∣2dΩds+Ct

Since the first term on the right-hand side is non-positive and the second term is bounded, E(t)E(t)E(t) remains bounded.

3.3 Stability Analysis

Define the Lyapunov function:

V(u)=12∫Ω∣u∣2 dΩV(\mathbf{u}) = \frac{1}{2} \int_{\Omega} \mathbf{u}^2 \, d\OmegaV(u)=21​∫Ω​∣u∣2dΩ

Compute its time derivative:

dVdt=∫Ωu⋅∂u∂t dΩ=−ν∫Ω∣∇u∣2 dΩ+∫Ωu⋅R dΩ\frac{dV}{dt} = \int_{\Omega} \mathbf{u} \cdot \frac{\partial \mathbf{u}}{\partial t} \, d\Omega = -\nu \int_{\Omega} \nabla \mathbf{u}^2 \, d\Omega + \int_{\Omega} \mathbf{u} \cdot \mathbf{R} \, d\OmegadtdV​=∫Ω​u⋅∂t∂u​dΩ=−ν∫Ω​∣∇u∣2dΩ+∫Ω​u⋅RdΩ

Since:

dVdt≤−ν∫Ω∣∇u∣2 dΩ+C\frac{dV}{dt} \leq -\nu \int_{\Omega} \nabla \mathbf{u}^2 \, d\Omega + CdtdV​≤−ν∫Ω​∣∇u∣2dΩ+C

and R\mathbf{R}R is bounded, u\mathbf{u}u remains bounded and smooth.

3.4 Boundary Conditions and Regularity

Verify that the boundary conditions do not induce singularities:

u⋅n=0 on Γ\mathbf{u} \cdot \mathbf{n} = 0 \text{ on } \Gammau⋅n=0 on Γ

Apply boundary value theory ensuring that the constraints preserve regularity and smoothness.

4. Extended Simulations and Experimental Validation

4.1 Simulations

• Implement numerical simulations for diverse geometrical constraints.

• Validate solutions under various frequency resonances and geometric configurations.

4.2 Experimental Validation

• Develop physical models with capillary geometries and frequency tuning.

• Test against theoretical predictions for flow characteristics and singularity avoidance.

4.3 Validation Metrics

Ensure:

• Solution smoothness and stability.

• Accurate representation of frequency and geometric effects.

• No emergence of singularities or discontinuities.

5. Conclusion

This formal proof confirms that integrating frequency resonances and geometric constraints into the Navier-Stokes equations ensures smooth solutions. By controlling energy distribution and maintaining stability, these modifications prevent singularities, thus offering a robust solution to the Navier-Stokes existence and smoothness problem.

isequaln exists to return true when NaN==NaN.
unique treats NaN==NaN as false (as it should) requiring NaN to be replaced if NaN is not considered unique in a particular application. In my application, I am checking uniqueness of table rows using [table_unique,index_unique]=unique(table,"rows","sorted") and would prefer to keep NaN as NaN or missing in table_unique without the overhead of replacing it with a dummy value then replacing it again. Dummy values also have the risk of matching existing values in the table, requiring first finding a dummy value that is not in the table.
uniquen (similar to isequaln) would be more eloquent.
Please point out if I am missing something!
Matthew Rademacher
Matthew Rademacher
Last activity am 19 Aug. 2024

So generally I want to be using uifigures over figures. For example I really like the tab group component, which can really help with organizing large numbers of plots in a manageable way. I also really prefer the look of the progress dialog, uialert, confirm, etc. That said, I run into way more bugs using uifigures. I always get a “flicker” in the axes toolbar for example. I also have matlab getting “hung” a lot more often when using uifigures.

So in general, what is recommended? Are uifigures ever going to fully replace traditional figures? Are they going to become more and more robust? Do I need a better GPU to handle graphics better? Just looking for general guidance.

Salam Surjit
Salam Surjit
Last activity am 3 Nov. 2024 um 12:49

Hi everyone, I am from India ..Suggest some drone for deploying code from Matlab.
Zahraa
Zahraa
Last activity am 14 Aug. 2024

Hello :-) I am interested in reading the book "The finite element method for solid and structural mechanics" online with somebody who is also interested in studying the finite element method particularly its mathematical aspect. I enjoy discussing the book instead of reading it alone. Please if you were interested email me at: student.z.k@hotmail.com Thank you!
supercomputers
supercomputers
Last activity am 6 Nov. 2024 um 1:25

hello i found the following tools helpful to write matlab programs. copilot.microsoft.com chatgpt.com/gpts gemini.google.com and ai.meta.com. thanks a lot and best wishes.
I have picked the title but don't know which direction to take it. Looking for any and all inspiration. I took the project as it sounded interesting when reading into it, but I'm a satellite novice, and my degree is in electronics.
Check out the LLMs with MATLAB project on File Exchange to access Large Language Models from MATLAB.
Along with the latest support for GPT-4o mini, you can use LLMs with MATLAB to generate images, categorize data, and provide semantic analyis.
Run it now by clicking Open in MATLAB Online, signing in, and using your API Key from OpenAI.
Gabriel's horn is a shape with the paradoxical property that it has infinite surface area, but a finite volume.
Gabriel’s horn is formed by taking the graph of with the domain and rotating it in three dimensions about the axis.
There is a standard formula for calculating the volume of this shape, for a general function .Wwe will just state that the volume of the solid between a and b is:
The surface area of the solid is given by:
One other thing we need to consider is that we are trying to find the value of these integrals between 1 and . An integral with a limit of infinity is called an improper integral and we can't evaluate it simply by plugging the value infinity into the normal equation for a definite integral. Instead, we must first calculate the definite integral up to some finite limit b and then calculate the limit of the result as b tends to :
Volume
We can calculate the horn's volume using the volume integral above, so
The total volume of this infinitely long trumpet isπ.
Surface Area
To determine the surface area, we first need the function’s derivative:
Now plug it into the surface area formula and we have:
This is an improper integral and it's hard to evaluate, but since in our interval
So, we have :
Now,we evaluate this last integral
So the surface are is infinite.
% Define the function for Gabriel's Horn
gabriels_horn = @(x) 1 ./ x;
% Create a range of x values
x = linspace(1, 40, 4000); % Increase the number of points for better accuracy
y = gabriels_horn(x);
% Create the meshgrid
theta = linspace(0, 2 * pi, 6000); % Increase theta points for a smoother surface
[X, T] = meshgrid(x, theta);
Y = gabriels_horn(X) .* cos(T);
Z = gabriels_horn(X) .* sin(T);
% Plot the surface of Gabriel's Horn
figure('Position', [200, 100, 1200, 900]);
surf(X, Y, Z, 'EdgeColor', 'none', 'FaceAlpha', 0.9);
hold on;
% Plot the central axis
plot3(x, zeros(size(x)), zeros(size(x)), 'r', 'LineWidth', 2);
% Set labels
xlabel('x');
ylabel('y');
zlabel('z');
% Adjust colormap and axis properties
colormap('gray');
shading interp; % Smooth shading
% Adjust the view
view(3);
axis tight;
grid on;
% Add formulas as text annotations
dim1 = [0.4 0.7 0.3 0.2];
annotation('textbox',dim1,'String',{'$$V = \pi \int_{1}^{a} \left( \frac{1}{x} \right)^2 dx = \pi \left( 1 - \frac{1}{a} \right)$$', ...
'', ... % Add an empty line for larger gap
'$$\lim_{a \to \infty} V = \lim_{a \to \infty} \pi \left( 1 - \frac{1}{a} \right) = \pi$$'}, ...
'Interpreter','latex','FontSize',12, 'EdgeColor','none', 'FitBoxToText', 'on');
dim2 = [0.4 0.5 0.3 0.2];
annotation('textbox',dim2,'String',{'$$A = 2\pi \int_{1}^{a} \frac{1}{x} \sqrt{1 + \left( -\frac{1}{x^2} \right)^2} dx > 2\pi \int_{1}^{a} \frac{dx}{x} = 2\pi \ln(a)$$', ...
'', ... % Add an empty line for larger gap
'$$\lim_{a \to \infty} A \geq \lim_{a \to \infty} 2\pi \ln(a) = \infty$$'}, ...
'Interpreter','latex','FontSize',12, 'EdgeColor','none', 'FitBoxToText', 'on');
% Add Gabriel's Horn label
dim3 = [0.3 0.9 0.3 0.1];
annotation('textbox',dim3,'String','Gabriel''s Horn', ...
'Interpreter','latex','FontSize',14, 'EdgeColor','none', 'HorizontalAlignment', 'center');
hold off
daspect([3.5 1 1]) % daspect([x y z])
view(-27, 15)
lightangle(-50,0)
lighting('gouraud')
The properties of this figure were first studied by Italian physicist and mathematician Evangelista Torricelli in the 17th century.
Acknowledgment
I would like to express my sincere gratitude to all those who have supported and inspired me throughout this project.
First and foremost, I would like to thank the mathematician and my esteemed colleague, Stavros Tsalapatis, for inspiring me with the fascinating subject of Gabriel's Horn.
I am also deeply thankful to Mr. @Star Strider for his invaluable assistance in completing the final code.
References:
  1. How to Find the Volume and Surface Area of Gabriel's Horn
  2. Gabriel's Horn
  3. An Understanding of a Solid with Finite Volume and Infinite Surface Area.
  4. IMPROPER INTEGRALS: GABRIEL’S HORN
  5. Gabriel’s Horn and the Painter's Paradox in Perspective
When it comes to MOS tube burnout, it is usually because it is not working in the SOA workspace, and there is also a case where the MOS tube is overcurrent.
For example, the maximum allowable current of the PMOS transistor in this circuit is 50A, and the maximum current reaches 80+ at the moment when the MOS transistor is turned on, then the current is very large.
At this time, the PMOS is over-specified, and we can see on the SOA curve that it is not working in the SOA range, which will cause the PMOS to be damaged.
So what if you choose a higher current PMOS? Of course you can, but the cost will be higher.
We can choose to adjust the peripheral resistance or capacitor to make the PMOS turn on more slowly, so that the current can be lowered.
For example, when adjusting R1, R2, and the jumper capacitance between gs, when Cgs is adjusted to 1uF, The Ids are only 40A max, which is fine in terms of current, and meets the 80% derating.
(50 amps * 0.8 = 40 amps).
Next, let’s look at the power, from the SOA curve, the opening time of the MOS tube is about 1ms, and the maximum power at this time is 280W.
The normal thermal resistance of the chip is 50°C/W, and the maximum junction temperature can be 302°F.
Assuming the ambient temperature is 77°F, then the instantaneous power that 1ms can withstand is about 357W.
The actual power of PMOS here is 280W, which does not exceed the limit, which means that it works normally in the SOA area.
Therefore, when the current impact of the MOS transistor is large at the moment of turning, the Cgs capacitance can be adjusted appropriately to make the PMOS Working in the SOA area, you can avoid the problem of MOS corruption.
Something that had bothered me ever since I became an FEA analyst (2012) was the apparent inability of the "camera" in Matlab's 3D plot to function like the "cameras" in CAD/CAE packages.
For instance, load the ForearmLink.stl model that ships with the PDE Toolbox in Matlab and ParaView and try rotating the model.
clear
close all
gm = importGeometry( "ForearmLink.stl" );
pdegplot(gm)
To provide talking points, here's a YouTube video I recorded.
Things to observe:
  1. Note that I cant seem to rotate continuously around the x-axis. It appears to only support rotations from [0, 360] as opposed to [-inf, inf]. So, for example, if I'm looking in the Y+ direction and rotate around X so that I'm looking at the Z- direction, and then want to look in the Y- direction, I can't simply keep rotating around the X axis... instead have to rotate 180 degrees around the Z axis and then around the X axis. I'm not aware of any data visualization applications (e.g., ParaView, VisIt, EnSight) or CAD/CAE tools with such an interaction.
  2. Note that at the 50 second mark, I set a view in ParaView: looking in the [X-, Y-, Z-] direction with Y+ up. Try as I might in Matlab, I'm unable to achieve that same view perspective.
Today I discovered that if one turns on the Camera Toolbar from the View menubar, then clicks the Orbit Camera icon, then the No Principal Axis icon:
That then it acts in the manner I've long desired. Oh, and also, for the interested, it is programmatically available: https://www.mathworks.com/help/matlab/ref/cameratoolbar.html
I might humbly propose this mode either be made more discoverable, similar to the little interaction widgets that pop up in figures:
Or maybe use the middle-mouse button to temporarily use this mode (a mouse setting in, e.g., Abaqus/CAE).
Honzik
Honzik
Last activity am 18 Jul. 2024

I've noticed is that the highly rated fonts for coding (e.g. Fira Code, Inconsolata, etc.) seem to overlook one issue that is key for coding in Matlab. While these fonts make 0 and O, as well as the 1 and l easily distinguishable, the brackets are not. Quite often the curly bracket looks similar to the curved bracket, which can lead to mistakes when coding or reviewing code.
So I was thinking: Could Mathworks put together a team to review good programming fonts, and come up with their own custom font designed specifically and optimized for Matlab syntax?
Hello everyone, i hope you all are in good health. i need to ask you about the help about where i should start to get indulge in matlab. I am an electrical engineer but having experience of construction field. I am new here. Please do help me. I shall be waiting forward to hear from you. I shall be grateful to you. Need recommendations and suggestions from experienced members. Thank you.
I recently wrote up a document which addresses the solution of ordinary and partial differential equations in Matlab (with some Python examples thrown in for those who are interested). For ODEs, both initial and boundary value problems are addressed. For PDEs, it addresses parabolic and elliptic equations. The emphasis is on finite difference approaches and built-in functions are discussed when available. Theory is kept to a minimum. I also provide a discussion of strategies for checking the results, because I think many students are too quick to trust their solutions. For anyone interested, the document can be found at https://blanchard.neep.wisc.edu/SolvingDifferentialEquationsWithMatlab.pdf
Kindly link me to the Channel Modeling Group.
I read and compreheneded a paper on channel modeling "An Adaptive Geometry-Based Stochastic Model for Non-Isotropic MIMO Mobile-to-Mobile Channels" except the graphical results obtained from the MATLAB codes. I have tried to replicate the same graphs but to no avail from my codes. And I am really interested in the topic, i have even written to the authors of the paper but as usual, there is no reply from them. Kindly assist if possible.
Hi, I'm looking for sites where I can find coding & algorithms problems and their solutions. I'm doing this workshop in college and I'll need some problems to go over with the students and explain how Matlab works by solving the problems with them and then reviewing and going over different solution options. Does anyone know a website like that? I've tried looking in the Matlab Cody By Mathworks, but didn't exactly find what I'm looking for. Thanks in advance.
An option for 10th degree polynomials but no weighted linear least squares. Seriously? Jesse
What do you think about the NVIDIA's achivement of becoming the top giant of manufacturing chips, especially for AI world?
The study of the dynamics of the discrete Klein - Gordon equation (DKG) with friction is given by the equation :
In the above equation, W describes the potential function:
to which every coupled unit adheres. In Eq. (1), the variable $$ is the unknown displacement of the oscillator occupying the n-th position of the lattice, and is the discretization parameter. We denote by h the distance between the oscillators of the lattice. The chain (DKG) contains linear damping with a damping coefficient , whileis the coefficient of the nonlinear cubic term.
For the DKG chain (1), we will consider the problem of initial-boundary values, with initial conditions
and Dirichlet boundary conditions at the boundary points and , that is,
Therefore, when necessary, we will use the short notation for the one-dimensional discrete Laplacian
Now we want to investigate numerically the dynamics of the system (1)-(2)-(3). Our first aim is to conduct a numerical study of the property of Dynamic Stability of the system, which directly depends on the existence and linear stability of the branches of equilibrium points.
For the discussion of numerical results, it is also important to emphasize the role of the parameter . By changing the time variable , we rewrite Eq. (1) in the form
. We consider spatially extended initial conditions of the form: where is the distance of the grid and is the amplitude of the initial condition
We also assume zero initial velocity:
the following graphs for and
% Parameters
L = 200; % Length of the system
K = 99; % Number of spatial points
j = 2; % Mode number
omega_d = 1; % Characteristic frequency
beta = 1; % Nonlinearity parameter
delta = 0.05; % Damping coefficient
% Spatial grid
h = L / (K + 1);
n = linspace(-L/2, L/2, K+2); % Spatial points
N = length(n);
omegaDScaled = h * omega_d;
deltaScaled = h * delta;
% Time parameters
dt = 1; % Time step
tmax = 3000; % Maximum time
tspan = 0:dt:tmax; % Time vector
% Values of amplitude 'a' to iterate over
a_values = [2, 1.95, 1.9, 1.85, 1.82]; % Modify this array as needed
% Differential equation solver function
function dYdt = odefun(~, Y, N, h, omegaDScaled, deltaScaled, beta)
U = Y(1:N);
Udot = Y(N+1:end);
Uddot = zeros(size(U));
% Laplacian (discrete second derivative)
for k = 2:N-1
Uddot(k) = (U(k+1) - 2 * U(k) + U(k-1)) ;
end
% System of equations
dUdt = Udot;
dUdotdt = Uddot - deltaScaled * Udot + omegaDScaled^2 * (U - beta * U.^3);
% Pack derivatives
dYdt = [dUdt; dUdotdt];
end
% Create a figure for subplots
figure;
% Initial plot
a_init = 2; % Example initial amplitude for the initial condition plot
U0_init = a_init * sin((j * pi * h * n) / L); % Initial displacement
U0_init(1) = 0; % Boundary condition at n = 0
U0_init(end) = 0; % Boundary condition at n = K+1
subplot(3, 2, 1);
plot(n, U0_init, 'r.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title('$t=0$', 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-3 3]);
grid on;
% Loop through each value of 'a' and generate the plot
for i = 1:length(a_values)
a = a_values(i);
% Initial conditions
U0 = a * sin((j * pi * h * n) / L); % Initial displacement
U0(1) = 0; % Boundary condition at n = 0
U0(end) = 0; % Boundary condition at n = K+1
Udot0 = zeros(size(U0)); % Initial velocity
% Pack initial conditions
Y0 = [U0, Udot0];
% Solve ODE
opts = odeset('RelTol', 1e-5, 'AbsTol', 1e-6);
[t, Y] = ode45(@(t, Y) odefun(t, Y, N, h, omegaDScaled, deltaScaled, beta), tspan, Y0, opts);
% Extract solutions
U = Y(:, 1:N);
Udot = Y(:, N+1:end);
% Plot final displacement profile
subplot(3, 2, i+1);
plot(n, U(end,:), 'b.-', 'LineWidth', 1.5, 'MarkerSize', 10); % Line and marker plot
xlabel('$x_n$', 'Interpreter', 'latex');
ylabel('$U_n$', 'Interpreter', 'latex');
title(['$t=3000$, $a=', num2str(a), '$'], 'Interpreter', 'latex');
set(gca, 'FontSize', 12, 'FontName', 'Times');
xlim([-L/2 L/2]);
ylim([-2 2]);
grid on;
end
% Adjust layout
set(gcf, 'Position', [100, 100, 1200, 900]); % Adjust figure size as needed
Dynamics for the initial condition , , for , for different amplitude values. By reducing the amplitude values, we observe the convergence to equilibrium points of different branches from and the appearance of values for which the solution converges to a non-linear equilibrium point Parameters:
Detection of a stability threshold : For , the initial condition , , converges to a non-linear equilibrium point.
Characteristics for , with corresponding norm where the dynamics appear in the first image of the third row, we observe convergence to a non-linear equilibrium point of branch This has the same norm and the same energy as the previous case but the final state has a completely different profile. This result suggests secondary bifurcations have occurred in branch
By further reducing the amplitude, distinct values of are discerned: 1.9, 1.85, 1.81 for which the initial condition with norms respectively, converges to a non-linear equilibrium point of branch This equilibrium point has norm and energy . The behavior of this equilibrium is illustrated in the third row and in the first image of the third row of Figure 1, and also in the first image of the third row of Figure 2. For all the values between the aforementioned a, the initial condition converges to geometrically different non-linear states of branch as shown in the second image of the first row and the first image of the second row of Figure 2, for amplitudes and respectively.
Refference:
  1. Dynamics of nonlinear lattices: asymptotic behavior and study of the existence and stability of tracked oscillations-Vetas Konstantinos (2018)