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Hello everyone,
Over the past few weeks, our community members have shared some incredible insights and resources. Here are some highlights worth checking out:

Interesting Questions

Johnathan is seeking help with implementing a complex equation into MATLAB's curve fitting toolbox. If you have experience with curve fitting or MATLAB, your input could be invaluable!

Popular Discussions

Athanasios continues his exploration of the Duffing Equation, delving into its chaotic behavior. It's a fascinating read for anyone interested in nonlinear dynamics or chaos theory.
John shares his playful exploration with MATLAB to find a generative equation for a sequence involving Fibonacci numbers. It's an intriguing challenge for those who love mathematical puzzles.

From File Exchange

Ayesha provides a graphical analysis of linearised models in epidemiology, offering a detailed look at the dynamics of these systems. This resource is perfect for those interested in mathematical modeling.
Gareth brings some humor to MATLAB with a toolbox designed to share jokes. It's a fun way to lighten the mood during conferences or meetups.

From the Blogs

Ned Gulley interviews Tim Marston, the 2023 MATLAB Mini Hack contest winner. Tim's creativity and skills are truly inspiring, and his story is a must-read for aspiring programmers.
Sivylla discusses the integration of AI with embedded systems, highlighting the benefits of using MATLAB and Simulink. It's an insightful read for anyone interested in the future of AI technology.
Thank you to all our contributors for sharing your knowledge and creativity. We encourage everyone to engage with these posts and continue fostering a vibrant and supportive community.
Happy exploring!
David
David
Last activity am 16 Sep. 2024

Explore the newest online training courses, available as of 2024b: one new Onramp, eight new short courses, and one new learning path. Yes, that’s 10 new offerings. We’ve been busy.
As a reminder, Onramps are free to all. Short courses and learning paths require a subscription to the Online Training Suite (OTS).
  1. Multibody Simulation Onramp
  2. Analyzing Results in Simulink
  3. Battery Pack Modeling
  4. Introduction to Motor Control
  5. Signal Processing Techniques for Streaming Signals
  6. Core Signal Processing Techniques in MATLAB (learning path – includes the four short courses listed below)
Create a struct arrays where each struct has field names "a," "b," and "c," which store different types of data. What efficient methods do you have to assign values from individual variables "a," "b," and "c" to each struct element? Here are five methods I've provided, listed in order of decreasing efficiency. What do you think?
Create an array of 10,000 structures, each containing each of the elements corresponding to the a,b,c variables.
num = 10000;
a = (1:num)';
b = string(a);
c = rand(3,3,num);
Here are the methods;
%% method1
t1 =tic;
s = struct("a",[], ...
"b",[], ...
"c",[]);
s1 = repmat(s,num,1);
for i = 1:num
s1(i).a = a(i);
s1(i).b = b(i);
s1(i).c = c(:,:,i);
end
t1 = toc(t1);
%% method2
t2 =tic;
for i = num:-1:1
s2(i).a = a(i);
s2(i).b = b(i);
s2(i).c = c(:,:,i);
end
t2 = toc(t2);
%% method3
t3 =tic;
for i = 1:num
s3(i).a = a(i);
s3(i).b = b(i);
s3(i).c = c(:,:,i);
end
t3 = toc(t3);
%% method4
t4 =tic;
ct = permute(c,[3,2,1]);
t = table(a,b,ct);
s4 = table2struct(t);
t4 = toc(t4);
%% method5
t5 =tic;
s5 = struct("a",num2cell(a),...
"b",num2cell(b),...
"c",squeeze(mat2cell(c,3,3,ones(num,1))));
t5 = toc(t5);
%% plot
bar([t1,t2,t3,t4,t5])
xtickformat('method %g')
ylabel("time(second)")
yline(mean([t1,t2,t3,t4,t5]))
Local large language models (LLMs), such as llama, phi3, and mistral, are now available in the Large Language Models (LLMs) with MATLAB repository through Ollama™!
Read about it here:
Mike Croucher
Mike Croucher
Last activity am 15 Sep. 2024

Hot off the heels of my High Performance Computing experience in the Czech republic, I've just booked my flights to Atlanta for this year's supercomputing conference at SC24.
Will any of you be there?
syms u v
atan2alt(v,u)
ans = 
function Z = atan2alt(V,U)
% extension of atan2(V,U) into the complex plane
Z = -1i*log((U+1i*V)./sqrt(U.^2+V.^2));
% check for purely real input. if so, zero out the imaginary part.
realInputs = (imag(U) == 0) & (imag(V) == 0);
Z(realInputs) = real(Z(realInputs));
end
As I am editing this post, I see the expected symbolic display in the nice form as have grown to love. However, when I save the post, it does not display. (In fact, it shows up here in the discussions post.) This seems to be a new problem, as I have not seen that failure mode in the past.
You can see the problem in this Answer forum response of mine, where it did fail.
David
David
Last activity am 12 Sep. 2024

In case you haven't come across it yet, @Gareth created a Jokes toolbox to get MATLAB to tell you a joke.
Chen Lin
Chen Lin
Last activity am 12 Sep. 2024

Dear MATLAB contest enthusiasts,
In the 2023 MATLAB Mini Hack Contest, Tim Marston captivated everyone with his incredible animations, showcasing both creativity and skill, ultimately earning him the 1st prize.
We had the pleasure of interviewing Tim to delve into his inspiring story. You can read the full interview on MathWorks Blogs: Community Q&A – Tim Marston.
Last question: Are you ready for this year’s Mini Hack contest?
As far as I know, starting from MATLAB R2024b, the documentation is defaulted to be accessed online. However, the problem is that every time I open the official online documentation through my browser, it defaults or forcibly redirects to the documentation hosted site for my current geographic location, often with multiple pop-up reminders, which is very annoying!
Suggestion: Could there be an option to set preferences linked to my personal account so that the documentation defaults to my chosen language preference without having to deal with “forced reminders” or “forced redirection” based on my geographic location? I prefer reading the English documentation, but the website automatically redirects me to the Chinese documentation due to my geolocation, which is quite frustrating!
----------------2024.12.13 update-----------------
Although the above issue was resolved by technical support, subsequent redirects are still causing severe delays...
In the past two years, MATHWORKS has updated the image viewer and audio viewer, giving them a more modern interface with features like play, pause, fast forward, and some interactive tools that are more commonly found in typical third-party players. However, the video player has not seen any updates. For instance, the Video Viewer or vision.VideoPlayer could benefit from a more modern player interface. Perhaps I haven't found a suitable built-in player yet. It would be great if there were support for custom image processing and audio processing algorithms that could be played in a more modern interface in real time.
Additionally, I found it quite challenging to develop a modern video player from scratch in App Designer.(If there's a video component for that that would be great)
-----------------------------------------------------------------------------------------------------------------
BTW,the following picture shows the built-in function uihtml function showing a more modern playback interface with controls for play, pause and so on. But can not add real-time image processing algorithms within it.
goc3
goc3
Last activity am 3 Dez. 2024 um 13:54

I was browsing the MathWorks website and decided to check the Cody leaderboard. To my surprise, William has now solved 5,000 problems. At the moment, there are 5,227 problems on Cody, so William has solved over 95%. The next competitor is over 500 problems behind. His score is also clearly the highest, approaching 60,000.
Please take a moment to congratulate @William.
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.

D.R. Kaprekar was a self taught recreational mathematician, perhaps known mostly for some numbers that bear his name.
Today, I'll focus on Kaprekar's constant (as opposed to Kaprekar numbers.)
The idea is a simple one, embodied in these 5 steps.
1. Take any 4 digit integer, reduce to its decimal digits.
2. Sort the digits in decreasing order.
3. Flip the sequence of those digits, then recompose the two sets of sorted digits into 4 digit numbers. If there were any 0 digits, they will become leading zeros on the smaller number. In this case, a leading zero is acceptable to consider a number as a 4 digit integer.
4. Subtract the two numbers, smaller from the larger. The result will always have no more than 4 decimal digits. If it is less than 1000, then presume there are leading zero digits.
5. If necessary, repeat the above operation, until the result converges to a stable result, or until you see a cycle.
Since this process is deterministic, and must always result in a new 4 digit integer, it must either terminate at either an absorbing state, or in a cycle.
For example, consider the number 6174.
7641 - 1467
ans = 6174
We get 6174 directly back. That seems rather surprising to me. But even more interesting is you will find all 4 digit numbers (excluding the pure rep-digit nmbers) will always terminate at 6174, after at most a few steps. For example, if we start with 1234
4321 - 1234
ans = 3087
8730 - 0378
ans = 8352
8532 - 2358
ans = 6174
and we see that after 3 iterations of this process, we end at 6174. Similarly, if we start with 9998, it too maps to 6174 after 5 iterations.
9998 ==> 999 ==> 8991 ==> 8082 ==> 8532 ==> 6174
Why should that happen? That is, why should 6174 always drop out in the end? Clearly, since this is a deterministic proces which always produces another 4 digit integer (Assuming we treat integers with a leading zero as 4 digit integers), we must either end in some cycle, or we must end at some absorbing state. But for all (non-pure rep-digit) starting points to end at the same place, it seems just a bit surprising.
I always like to start a problem by working on a simpler problem, and see if it gives me some intuition about the process. I'll do the same thing here, but with a pair of two digit numbers. There are 100 possible two digit numbers, since we must treat all one digit numbers as having a "tens" digit of 0.
N = (0:99)';
Next, form the Kaprekar mapping for 2 digit numbers. This is easier than you may think, since we can do it in a very few lines of code on all possible inputs.
Ndig = dec2base(N,10,2) - '0';
Nmap = sort(Ndig,2,'descend')*[10;1] - sort(Ndig,2,'ascend')*[10;1];
I'll turn it into a graph, so we can visualize what happens. It also gives me an excuse to employ a very pretty set of tools in MATLAB.
G2 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G2)
Do you see what happens? All of the rep-digit numbers, like 11, 44, 55, etc., all map directly to 0, and they stay there, since 0 also maps into 0. We can see that in the star on the lower right.
G2cycles = cyclebasis(G2)
G2cycles = 2x1 cell array
{1x1 cell} {1x5 cell}
G2cycles{1}
ans = 1x1 cell array
{'00'}
All other numbers eventually end up in the cycle:
G2cycles{2}
ans = 1x5 cell array
{'09'} {'45'} {'27'} {'63'} {'81'}
That is
81 ==> 63 ==> 27 ==> 45 ==> 09 ==> and back to 81
looping forever.
Another way of trying to visualize what happens with 2 digit numbers is to use symbolics. Thus, if we assume any 2 digit number can be written as 10*T+U, where I'll assume T>=U, since we always sort the digits first
syms T U
(10*T + U) - (10*U+T)
ans = 
So after one iteration for 2 digit numbers, the result maps ALWAYS to a new 2 digit number that is divisible by 9. And there are only 10 such 2 digit numbers that are divisible by 9. So the 2-digit case must resolve itself rather quickly.
What happens when we move to 3 digit numbers? Note that for any 3 digit number abc (without loss of generality, assume a >= b >= c) it almost looks like it reduces to the 2 digit probem, aince we have abc - cba. The middle digit will always cancel itself in the subtraction operation. Does that mean we should expect a cycle at the end, as happens with 2 digit numbers? A simple modification to our previous code will tell us the answer.
N = (0:999)';
Ndig = dec2base(N,10,3) - '0';
Nmap = sort(Ndig,2,'descend')*[100;10;1] - sort(Ndig,2,'ascend')*[100;10;1];
G3 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G3)
This one is more difficult to visualize, since there are 1000 nodes in the graph. However, we can clearly see two disjoint groups.
We can use cyclebasis to tell us the complete story again.
G3cycles = cyclebasis(G3)
G3cycles = 2x1 cell array
{1x1 cell} {1x1 cell}
G3cycles{:}
ans = 1x1 cell array
{'000'}
ans = 1x1 cell array
{'495'}
And we see that all 3 digit numbers must either terminate at 000, or 495. For example, if we start with 181, we would see:
811 - 118
ans = 693
963 - 369
ans = 594
954 - 459
ans = 495
It will terminate there, forever trapped at 495. And cyclebasis tells us there are no other cycles besides the boring one at 000.
What is the maximum length of any such path to get to 495?
D3 = distances(G3,496) % Remember, MATLAB uses an index origin of 1
D3 = 1x1000
Inf 6 5 4 3 1 2 3 4 5 6 6 5 4 3 1 2 3 4 5 5 5 5 4 3 1 2 3 4 5
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
D3(isinf(D3)) = -inf; % some nodes can never reach 495, so they have an infinite distance
plot(D3)
The maximum number of steps to get to 495 is 6 steps.
find(D3 == 6) - 1
ans = 1x54
1 10 11 100 101 110 112 121 122 211 212 221 223 232 233 322 323 332 334 343 344 433 434 443 445 454 455 544 545 554
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
So the 3 digit number 100 required 6 iterations to eventually reach 495.
shortestpath(G3,101,496) - 1
ans = 1x7
100 99 891 792 693 594 495
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
I think I've rather exhausted the 3 digit case. It is time now to move to the 4 digit problem, but we've already done all the hard work. The same scheme will apply to compute a graph. And the graph theory tools do all the hard work for us.
N = (0:9999)';
Ndig = dec2base(N,10,4) - '0';
Nmap = sort(Ndig,2,'descend')*[1000;100;10;1] - sort(Ndig,2,'ascend')*[1000;100;10;1];
G4 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G4)
cyclebasis(G4)
ans = 2x1 cell array
{1x1 cell} {1x1 cell}
ans{:}
ans = 1x1 cell array
{'0000'}
ans = 1x1 cell array
{'6174'}
And here we see the behavior, with one stable final point, 6174 as the only non-zero ending state. There are no circular cycles as we had for the 2-digit case.
How many iterations were necessary at most before termination?
D4 = distances(G4,6175);
D4(isinf(D4)) = -inf;
plot(D4)
The plot tells the story here. The maximum number of iterations before termination is 7 for the 4 digit case.
find(D4 == 7,1,'last') - 1
ans = 9985
shortestpath(G4,9986,6175) - 1
ans = 1x8
9985 4086 8172 7443 3996 6264 4176 6174
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Can you go further? Are there 5 or 6 digit Kaprekar constants? Sadly, I have read that for more than 4 digits, things break down a bit, there is no 5 digit (or higher) Kaprekar constant.
We can verify that fact, at least for 5 digit numbers.
N = (0:99999)';
Ndig = dec2base(N,10,5) - '0';
Nmap = sort(Ndig,2,'descend')*[10000;1000;100;10;1] - sort(Ndig,2,'ascend')*[10000;1000;100;10;1];
G5 = graph(N+1,Nmap+1,[],cellstr(dec2base(N,10,2)));
plot(G5)
cyclebasis(G5)
ans = 4x1 cell array
{1x1 cell} {1x2 cell} {1x4 cell} {1x4 cell}
ans{:}
ans = 1x1 cell array
{'00000'}
ans = 1x2 cell array
{'53955'} {'59994'}
ans = 1x4 cell array
{'61974'} {'63954'} {'75933'} {'82962'}
ans = 1x4 cell array
{'62964'} {'71973'} {'83952'} {'74943'}
The result here are 4 disjoint cycles. Of course the rep-digit cycle must always be on its own, but the other three cycles are also fully disjoint, and are of respective length 2, 4, and 4.
J.K043006
J.K043006
Last activity am 29 Aug. 2024

I've been working on some matrix problems recently(Problem 55225)
and this is my code
It turns out that "Undefined function 'corr' for input arguments of type 'double'." However, should't the input argument of "corr" be column vectors with single/double values? What's even going on there?
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.

Following on from my previous post The Non-Chaotic Duffing Equation, now we will study the chaotic behaviour of the Duffing Equation
P.s:Any comments/advice on improving the code is welcome.
The Original Duffing Equation is the following:
Let . This implies that
Then we rewrite it as a System of First-Order Equations
Using the substitution for , the second-order equation can be transformed into the following system of first-order equations:
Exploring the Effect of γ.
% Define parameters
gamma = 0.1;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span
tspan = [0 200];
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
% Define parameters
gamma = 0.318;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span
tspan = [0 800];
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
% Define parameters
gamma = 0.338;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span with more points for better resolution
tspan = linspace(0, 200,2000); % Increase the number of points
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
ax = gca;
chart = ax.Children(1);
datatip(chart,0.5581,-0.1126);
% Define parameters
gamma = 0.35;
alpha = -1;
beta = 1;
delta = 0.1;
omega = 1.4;
% Define the system of equations
odeSystem = @(t, y) [y(2);
-delta*y(2) - alpha*y(1) - beta*y(1)^3 + gamma*cos(omega*t)];
% Initial conditions
y0 = [0; 0]; % x(0) = 0, v(0) = 0
% Time span with more points for better resolution
tspan = linspace(0, 400,3000); % Increase the number of points
% Solve the system
[t, y] = ode45(odeSystem, tspan, y0);
% Plot the results
figure;
plot(t, y(:, 1));
xlabel('Time');
ylabel('x(t)');
title('Solution of the nonlinear system');
grid on;
% Plot the phase portrait
figure;
plot(y(:, 1), y(:, 2));
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait');
grid on;
% Define the tail (e.g., last 10% of the time interval)
tail_start = floor(0.9 * length(t)); % Starting index for the tail
tail_end = length(t); % Ending index for the tail
% Plot the tail of the solution
figure;
plot(y(tail_start:tail_end, 1), y(tail_start:tail_end, 2), 'r', 'LineWidth', 1.5);
xlabel('x(t)');
ylabel('v(t)');
title('Phase Portrait - Tail of the Solution');
grid on;
Salam Surjit
Salam Surjit
Last activity am 3 Nov. 2024

Hi everyone, I am from India ..Suggest some drone for deploying code from Matlab.
Studying the attached document Duffing Equation from the University of Colorado, I noticed that there is an analysis of The Non-Chaotic Duffing Equation and all the graphs were created with Matlab. And since the code is not given I took the initiative to try to create the same graphs with the following code.
  • Plotting the Potential Energy and Identifying Extrema
% Define the range of x values
x = linspace(-2, 2, 1000);
% Define the potential function V(x)
V = -x.^2 / 2 + x.^4 / 4;
% Plot the potential function
figure;
plot(x, V, 'LineWidth', 2);
hold on;
% Mark the minima at x = ±1
plot([-1, 1], [-1/4, -1/4], 'ro', 'MarkerSize', 5, 'MarkerFaceColor', 'g');
% Add LaTeX title and labels
title('Duffing Potential Energy: $$V(x) = -\frac{x^2}{2} + \frac{x^4}{4}$$', 'Interpreter', 'latex');
xlabel('$$x$$', 'Interpreter', 'latex');
ylabel('$$V(x)$$','Interpreter', 'latex');
grid on;
hold off;
  • Solving and Plotting the Duffing Equation
% Define the system of ODEs for the non-chaotic Duffing equation
duffing_ode = @(t, X) [X(2);
X(1) - X(1).^3];
% Time span for the simulation
tspan = [0 10];
% Initial conditions [x(0), v(0)]
initial_conditions = [1; 1];
% Solve the ODE using ode45
[t, X] = ode45(duffing_ode, tspan, initial_conditions);
% Extract displacement (x) and velocity (v)
x = X(:, 1);
v = X(:, 2);
% Plot both x(t) and v(t) in the same figure
figure;
plot(t, x, 'b-', 'LineWidth', 2); % Plot x(t) with blue line
hold on;
plot(t, v, 'r--', 'LineWidth', 2); % Plot v(t) with red dashed line
% Add title, labels, and legend
title(' Component curve solutions to $$\ddot{x}-x+x^3=0$$','Interpreter', 'latex');
xlabel('t','Interpreter', 'latex');
ylabel('$$x(t) $$ and $$v(t) $$','Interpreter', 'latex');
legend('$$x(t)$$', ' $$v(t)$$','Interpreter', 'latex');
grid on;
hold off;
% Phase portrait with nullclines, equilibria, and vector field
figure;
hold on;
% Plot phase portrait
plot(x, v,'r', 'LineWidth', 2);
% Plot equilibrium points
plot([0 1 -1], [0 0 0], 'ro', 'MarkerSize', 5, 'MarkerFaceColor', 'g');
% Create a grid of points for the vector field
[x_vals, v_vals] = meshgrid(linspace(-2, 2, 20), linspace(-1, 1, 20));
% Compute the vector field components
dxdt = v_vals;
dvdt = x_vals - x_vals.^3;
% Plot the vector field
quiver(x_vals, v_vals, dxdt, dvdt, 'b');
% Set axis limits to [-1, 1]
xlim([-1.7 1.7]);
ylim([-1 1]);
% Labels and title
title('Phase-Plane solutions to $$\ddot{x}-x+x^3=0$$','Interpreter', 'latex');
xlabel('$$ (x)$$','Interpreter', 'latex');
ylabel('$$v(v)$$','Interpreter', 'latex');
grid on;
hold off;
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