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Paras31

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0 years, 174 days
Hellenic Open University
Mathematician

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Teacher of Mathematics with a proven track record of working in education management. Proficient in Ease of Adaptation, Course Design, and Instructional Technology. Strong education professional with a Bachelor's degree with an emphasis in Mathematics from the University of Aegean. Currently, he is pursuing a Master's degree in Applied Mathematics at the Hellenic Open University, with a specific focus on Ordinary and Partial Differential equations. His enthusiasm lies in the application of mathematical models to real-world contexts, such as epidemiology and population growth. Aside from his passion for teaching, Athanasios enjoys football, basketball, and spending time with his dogs.

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These are Posts that have been published by Paras31

Exploring Nonlinear Dynamics: Analysis o...

Paras31 130 Maple 2024
ode chaos differential-equations
August 11 2024
1 8


 

An attractor is called strange if it has a fractal structure, that is if it has a non-integer Hausdorff dimension. This is often the case when the dynamics on it are chaotic, but strange nonchaotic attractors also exist.  If a strange attractor is chaotic, exhibiting sensitive dependence on initial conditions, then any two arbitrarily close alternative initial points on the  attractor, after any of various numbers of iterations, will lead to  points that are arbitrarily far apart (subject to the confines of the attractor), and after any of various other numbers of iterations will  lead to points that are arbitrarily close together. Thus a dynamic system with a chaotic attractor is locally unstable yet globally stable: once some sequences have entered the attractor, nearby points diverge  from one another but never depart from the attractor.


The term strange attractor was coined by David Ruelle and Floris Takens to describe the attractor resulting from a series of bifurcations of a system describing fluid flow. Strange attractors are often differentiable in a few directions, but some are like a Cantor dust, and therefore not differentiable. Strange attractors may also be found  in the presence of noise, where they may be shown to support invariant  random probability measures of Sinai–Ruelle–Bowen type.


Examples of strange attractors include the  Rössler attractor, and Lorenz attractor.

 

 

THOMAS``with(plots); b := .20; sys := diff(x(t), t) = sin(y(t))-b*x(t), diff(y(t), t) = sin(z(t))-b*y(t), diff(z(t), t) = sin(x(t))-b*z(t); sol := dsolve({sys, x(0) = 1.1, y(0) = 1.1, z(0) = -0.1e-1}, {x(t), y(t), z(t)}, numeric); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 600, axes = boxed, numpoints = 50000, labels = [x, y, z], title = "Thomas Attractor")

  

 

 

Dabras``

with(plots); a := 3.00; b := 2.7; c := 1.7; d := 2.00; e := 9.00; sys := diff(x(t), t) = y(t)-a*x(t)+b*y(t)*z(t), diff(y(t), t) = c*y(t)-x(t)*z(t)+z(t), diff(z(t), t) = d*x(t)*y(t)-e*z(t); sol := dsolve({sys, x(0) = 1.1, y(0) = 2.1, z(0) = -2.00}, {x(t), y(t), z(t)}, numeric); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 100, axes = boxed, numpoints = 35000, labels = [x, y, z], title = "Dabras Attractor")

  

Halvorsen

NULLwith(plots); a := 1.89; sys := diff(x(t), t) = -a*x(t)-4*y(t)-4*z(t)-y(t)^2, diff(y(t), t) = -a*y(t)-4*z(t)-4*x(t)-z(t)^2, diff(z(t), t) = -a*z(t)-4*x(t)-4*y(t)-x(t)^2; sol := dsolve({sys, x(0) = -1.48, y(0) = -1.51, z(0) = 2.04}, {x(t), y(t), z(t)}, numeric, maxfun = 300000); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 600, axes = boxed, numpoints = 35000, labels = [x, y, z], title = "Halvorsen Attractor")

  

Chen

 

 

with(plots); alpha := 5.00; beta := -10.00; delta := -.38; sys := diff(x(t), t) = alpha*x(t)-y(t)*z(t), diff(y(t), t) = beta*y(t)+x(t)*z(t), diff(z(t), t) = delta*z(t)+(1/3)*x(t)*y(t); sol := dsolve({sys, x(0) = -7.00, y(0) = -5.00, z(0) = -10.00}, {x(t), y(t), z(t)}, numeric); odeplot(sol, [x(t), y(t), z(t)], t = 0 .. 100, axes = boxed, numpoints = 35000, labels = [x, y, z], title = "Chen Attractor")

  

References

1. 

https://www.dynamicmath.xyz/strange-attractors/
2. 

https://en.wikipedia.org/wiki/Attractor#Strange_attractor

``


 

Download Attractors.mw

Overview of Maple Coding Expert (ChatGp...

Paras31 130 Maple
ai chatgpt
August 07 2024
1 2

 Introduction
Maple Coding Expert is a GPT-based AI tool designed to assist with various mathematical tasks using Maple software. It offers step-by-step guidance and detailed explanations for a range of functions, making it a valuable resource for students, educators, and professionals.

 Core Features and Functions

1.Graph Creation:

   - Function Plotting: Users can plot a wide range of mathematical functions. For instance, to plot the function y = x2, the user would input the command `plot(x^2, x = -10..10);` in Maple. The expert helps in setting up the plotting parameters to visualize the function effectively.
   - Advanced Graphing: Beyond simple functions, the expert can guide users through plotting more complex functions and customizing plots with labels, legends, and different styles.

2. Equation Definition and Manipulation:

   - Defining Equations: The tool assists in defining equations for various calculus operations. For example, to differentiate a function, the command might be `diff(f(x), x);`. This helps in accurately modeling the equations necessary for solving real-world problems.
   - Solving Integrals: For integral calculus, users can get assistance in setting up both definite and indefinite integrals. Commands like `int(f(x), x);` are used to perform integration in Maple.

3. Calculus Problem Solving:
   - Differentiation and Integration: The expert provides guidance on solving derivatives and integrals, which are fundamental operations in calculus. It supports both symbolic and numerical methods, allowing users to choose the best approach for their problem.
   - Differential Equations: Users can solve ordinary and partial differential equations using commands like `dsolve({equations}, {variables});`. The expert offers advice on choosing solution methods and interpreting results.

I recently tried using the Maple Coding Expert for solving some calculus problems. It worked well overall and provided detailed solutions, though sometimes it approached the problems in a more complicated way than expected. Despite this, the accuracy and depth of the explanations were impressive and very helpful for understanding the underlying concepts.

 

Maple Coding Expert stands out as a comprehensive tool for anyone involved with Maple software for mathematical computing. It enhances learning, supports professional tasks, and aids in solving complex mathematical problems with ease.

For more information, you can explore the Maple Coding Expert on [GPTs Hunter](https://www.gptshunter.com/gpt-store/MzExMzI2MzYyMzJlNTAxMjM2) and [YesChat.ai](https://www.yeschat.ai).

 

Standard and extended Lotka-Volterra mod...

Paras31 130 Maple 2024
ode numeric differential-equations
August 05 2024
1 2


 

This project discusses predator-prey system, particularly the Lotka-Volterra equations,which model the interaction between two sprecies: prey and predators. Let's solve the Lotka-Volterra equations numerically and visualize the results.

NULL

NULL

alpha := 1.0; beta := .1; g := 1.5; delta := 0.75e-1; ode1 := diff(x(t), t) = alpha*x(t)-beta*x(t)*y(t); ode2 := diff(y(t), t) = delta*x(t)*y(t)-g*y(t); eq1 := -beta*x*y+alpha*x = 0; eq2 := delta*x*y-g*y = 0; equilibria := solve({eq1, eq2}, {x, y}); print("Equilibrium Points: ", equilibria); initial_conditions := x(0) = 40, y(0) = 9; sol := dsolve({ode1, ode2, initial_conditions}, {x(t), y(t)}, numeric); eq_points := [seq([rhs(eq[1]), rhs(eq[2])], `in`(eq, equilibria))]

[[0., 0.], [20., 10.]]

plots[odeplot](sol, [[t, x(t)], [t, y(t)]], t = 0 .. 100, legend = ["Rabbits", "Wolves"], title = "Prey-Predator Dynamics", labels = ["Time", "Population"])

NULL

NULL

NULL

sol_plot := plots:-odeplot(sol, [[x(t), y(t)]], 0 .. 100, color = "blue"); equilibrium_plot := plots:-pointplot(eq_points, color = "red", symbol = solidcircle, symbolsize = 15); plots:-display([sol_plot, equilibrium_plot], title = "Phase Portrait with Equilibrium Points", labels = ["Rabbits", "Wolves"])

Now, we need to handle a modified version of the Lotka-Volterra equations. These modified equations incorporate logistic growth fot the prey population.

 

 

restart

alpha := 1.0; beta := .1; g := 1.5; delta := 0.75e-1; k := 100; ode1 := diff(x(t), t) = alpha*x(t)*(1-x(t)/k)-beta*x(t)*y(t); ode2 := diff(y(t), t) = delta*x(t)*y(t)-g*y(t); eq1 := alpha*x*(1-x/k)-beta*x*y = 0; eq2 := delta*x*y-g*y = 0; equilibria := solve({eq1, eq2}, {x, y}); print("Equilibrium Points: ", equilibria); initial_conditions := x(0) = 40, y(0) = 9; sol := dsolve({ode1, ode2, initial_conditions}, {x(t), y(t)}, numeric); eq_points := [seq([rhs(eq[1]), rhs(eq[2])], `in`(eq, equilibria))]

[[0., 0.], [100., 0.], [20., 8.]]

plots[odeplot](sol, [[t, x(t)], [t, y(t)]], t = 0 .. 100, legend = ["Rabbits", "Wolves"], title = "Prey-Predator Dynamics", labels = ["Time", "Population"])

NULL

plots:-odeplot(sol, [[x(t), y(t)]], 0 .. 50, color = "blue"); equilibrium_plot := plots:-pointplot(eq_points, color = "red", symbol = solidcircle, symbolsize = 15); plots:-display([plots:-odeplot(sol, [[x(t), y(t)]], 0 .. 50, color = "blue"), equilibrium_plot], title = "Phase Portrait with Equilibrium Points", labels = ["Rabbits", "Wolves"])

NULL


 

Download predator_prey2.mw

Exploring the SIR Model with Vaccination...

Paras31 130 Maple 2024
ode differential-equations epidemiology
May 07 2024
0 3

In our recent project, we're diving deep into understanding the SIR model—a fundamental framework in epidemiology that helps us analyze how diseases spread through populations. The SIR model categorizes individuals into three groups: Susceptible (S), Infected (I), and Recovered (R). By tracking how people move through these categories, we can predict disease dynamics and evaluate interventions.

Key Points of the SIR Model:

  • Susceptible (S): Individuals who can catch the disease.
  • Infected (I): Those currently infected and capable of spreading the disease.
  • Recovered (R): Individuals who have recovered and developed immunity.

Vaccination Impact: One of the critical interventions in disease control is vaccination, which moves individuals directly from the susceptible to the recovered group. This simple action reduces the number of people at risk, thereby lowering the overall spread of the disease.

We're experimenting with a simple model to understand how different vaccination rates can significantly alter the dynamics of an outbreak. By simulating scenarios with varying vaccination coverage, students can observe how herd immunity plays a crucial role in controlling diseases. Our goal is to make these abstract concepts clear and relatable through practical modeling exercises.


 

In this exercise, we are going back to the simple SIR model, without births or deaths, to look at the effect of vaccination. The aim of this activity is to represent vaccination in a very simple way - we are assuming it already happened before we run our model! By changing the initial conditions, we can prepare the population so that it has received a certain coverage of vaccination.
​
We are starting with the transmission and recovery parameters  b = .4/daysand c = .1/days . To incorporate immunity from vaccination in the model, we assume that a proportion p of the total population starts in the recovered compartment, representing the vaccine coverage and assuming the vaccine is perfectly effective. Again, we assume the epidemic starts with a single infected case introduced into the population.​
We are going to model this scenario for a duration of 2 years, assuming that the vaccine coverage is 50%, and plot the prevalence in each compartment over time.

 

restart
with(plots)

b := .4; c := .1; n := 10^6; p := .5

deS := diff(S(t), t) = -b*S(t)*I0(t); deI := diff(I0(t), t) = b*S(t)*I0(t)-c*I0(t); deR := diff(R(t), t) = c*I0(t)

diff(R(t), t) = .1*I0(t)

 (1)

F := dsolve([deS, deI, deR, S(0) = 1-p, I0(0) = 1/n, R(0) = p], [S(t), I0(t), R(t)], numeric, method = rkf45, maxfun = 100000)

odeplot(F, [[t, S(t)], [t, I0(t)], [t, R(t)]], t = 0 .. 730, colour = [blue, red, green], legend = ["S(t)", "I0(t)", "R(t)"], labels = ["Time (days)", " Proportion\nof Population "], title = "SIR Model with vaccine coverage 50 %", size = [500, 300])

  

F(100)

[t = 100., S(t) = HFloat(0.46146837378273076), I0(t) = HFloat(0.018483974421123688), R(t) = HFloat(0.5200486517961457)]

 (2)

eval(S(:-t), F(100))

HFloat(0.46146837378273076)

 (3)

Reff := proc (s) options operator, arrow; b*(eval(S(:-t), F(s)))/(c*n) end proc; Reff(100)

HFloat(1.845873495130923e-6)

 (4)

plot(Reff, 0 .. 730, size = [500, 300])

  

Increasing the vaccine coverage to 75%

NULL

restart
with(plots)

b := .4; c := .1; n := 10^6; p := .75

deS := diff(S(t), t) = -b*S(t)*I0(t); deI := diff(I0(t), t) = b*S(t)*I0(t)-c*I0(t); deR := diff(R(t), t) = c*I0(t)

diff(R(t), t) = .1*I0(t)

 (5)

NULL

F1 := dsolve([deS, deI, deR, S(0) = 1-p, I0(0) = 1/n, R(0) = p], [S(t), I0(t), R(t)], numeric, method = rkf45, maxfun = 100000)

odeplot(F1, [[t, S(t)], [t, I0(t)], [t, R(t)]], t = 0 .. 730, colour = [blue, red, green], legend = ["S(t)", "I0(t)", "R(t)"], labels = ["Time (days)", " Proportion\nof Population "], title = "SIR Model with vaccine coverage 75%", size = [500, 300])

  

F(1100)

eval(S(:-t), F1(100))

HFloat(0.249990000844159)

 (6)

Reff := proc (s) options operator, arrow; b*(eval(S(:-t), F1(s)))/(c*n) end proc; Reff(100)

HFloat(9.99960003376636e-7)

 (7)

plot(Reff, 0 .. 730, size = [500, 300])

  

Does everyone in the population need to be vaccinated in order to prevent an epidemic?What do you observe if you model the infection dynamics with different values for p?

No, not everyone in the population needs to be vaccinated in order to prevent an epidemic . In this scenario, if p equals 0.75 or higher, no epidemic occurs - 75 % is the critical vaccination/herd immunity threshold . Remember,, herd immunity describes the phenomenon in which there is sufficient immunity in a population to interrupt transmission . Because of this, not everyone needs to be vaccinated to prevent an outbreak .

What proportion of the population needs to be vaccinated in order to prevent an epidemic if b = .4and c = .2/days? What if b = .6 and "c=0.1 days^(-1)?"

In the context of the SIR model, the critical proportion of the population that needs to be vaccinated in order to prevent an epidemic is often referred to as the "herd immunity threshold" or "critical vaccination coverage."

• 

Scenario 1: b = .4and c = .2/days

``

restart
with(plots)

b := .4; c := .2; n := 10^6; p := .5``

deS := diff(S(t), t) = -b*S(t)*I0(t); deI := diff(I0(t), t) = b*S(t)*I0(t)-c*I0(t); deR := diff(R(t), t) = c*I0(t)

diff(R(t), t) = .2*I0(t)

 (8)

F1 := dsolve([deS, deI, deR, S(0) = 1-p, I0(0) = 1/n, R(0) = p], [S(t), I0(t), R(t)], numeric, method = rkf45, maxfun = 100000)

odeplot(F1, [[t, S(t)], [t, I0(t)], [t, R(t)]], t = 0 .. 730, colour = [blue, red, green], legend = ["S(t)", "I0(t)", "R(t)"], labels = ["Time (days)", " Proportion\nof Population "], title = "SIR Model with vaccine coverage 50 %", size = [500, 300])

  


The required vaccination coverage is around 50% .

• 

Scenario 1: b = .6and c = .1/days

restart
with(plots)

b := .6; c := .1; n := 10^6; p := .83NULL

deS := diff(S(t), t) = -b*S(t)*I0(t); deI := diff(I0(t), t) = b*S(t)*I0(t)-c*I0(t); deR := diff(R(t), t) = c*I0(t)

diff(R(t), t) = .1*I0(t)

 (9)

NULL

F1 := dsolve([deS, deI, deR, S(0) = 1-p, I0(0) = 1/n, R(0) = p], [S(t), I0(t), R(t)], numeric, method = rkf45, maxfun = 100000)

odeplot(F1, [[t, S(t)], [t, I0(t)], [t, R(t)]], t = 0 .. 730, colour = [blue, red, green], legend = ["S(t)", "I0(t)", "R(t)"], labels = ["Time (days)", " Proportion\nof Population "], title = "SIR Model with vaccine coverage 83% ", size = [500, 300])

  

"The required vaccination coverage is around 83 `%` ."


Download SIR_simple_vaccination_example.mw

Directly Integrable Equations...

Paras31 130 Maple
calculus education differential-equations
April 22 2024
0 2


 

When a derivative can be written as a function of the independent variable only for example

y'=f(x)

y''=f(x)

y'''(x)=f(x)

etc.

 

We call that a directly integrable equation.

 

Example 1:

 

Find the general solution for the following directly integrable equation

diff(y, x) = 6*x^2+4*y(1) and 6*x^2+4*y(1) = 0

That means

int(6*x^2+4, x)

y = 2*x^3+c+4*x", where" c is an arbitary solution

``

 

 
equation1 := diff(y(x), x) = 6*x^2+4

diff(y(x), x) = 6*x^2+4

 (1)

NULL

NULL

sol1 := dsolve(equation1, y(x))

y(x) = 2*x^3+c__1+4*x

 (2)

And  if we have the initial condition
y(1) = 0
particular_sol1 := dsolve({equation1, y(1) = 0}, y(x))

y(x) = 2*x^3+4*x-6

 (3)

"(->)"

  

 

 

 

 

"Example 2:"NULL

NULL

" Find the particular solution for the following equation with condition"

 

x^2*(diff(y(x), x)) = -1

y(1)=3

So we will need to get the y' by itself

int(-1/x^2, x)

so,

y = 1/x+c , where c is an arbitary constant

And this is our general solution. Now we plug in the initial condition when x = 1, y = 3.

 

That means c = 2.

 

Thus, the particular solution is

 

y = 1/x+2``

eq := x^2*(diff(y(x), x)) = -1

x^2*(diff(y(x), x)) = -1

 (4)

NULL

NULL

sol := dsolve(eq, y(x))

y(x) = 1/x+c__1

 (5)

NULL

particular_sol := dsolve({eq, y(1) = 3}, y(x))

y(x) = 1/x+2

 (6)

NULL

NULL

plot(1/x+2, x = -20 .. 20, color = "Red", axes = normal, legend = [typeset(1/x+2)])

  

NULL

NULL

NULL

NULL

" Example 3:"

 

" Find the particular solution for the following equation with condition"

 

diff(y, t, t) = cost, (D(y))(0) = 0, y(0) = 1

eq1 := diff(y(t), t, t) = cos(t)

diff(diff(y(t), t), t) = cos(t)

 (7)

particular_sol := dsolve({eq1, y(0) = 1, (D(y))(0) = 0}, y(t))

y(t) = -cos(t)+2

 (8)

"(->)"

  

 

 

 

NULL


 

Download integral.mw

 

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