MaplePrimes Questions

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 = 0.4 days-1 and  c=0.1 days-1. 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.

I have used the following code to solve this problem but I face difficulties when I am trying to define and plot the effective reproduction number.

Any suggestions? Thank you in advance!
with(Physics, diff);
with(LinearAlgebra);

with(plots);
with(DEtools);

b := 0.4;
c := 0.1;
n := 10^6;
p := 0.5;
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);

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 of Population"], title = "SIR Model with vaccination")


#define the effective reproduction number
Reff := t -> b*1/c*S(t)*1/n;
Reff(100);

I have been trying to solve the following diffusion equations

  diff(u(z, t), t) = D*diff(u(z, t), z, z) + A*z*exp(lambda*z)

my boundary canditions are

  u(0,t)=0,D(u)(H,t)=0,u(initinity,t)=0

I have followed the examples on tte link below, but I am unable to get the solution when I write the boundary conditions. It there anything I am doing wrongly? I would be most grateful if an email can be sent to me with with the solution that I can download fram an attachment

Thanks

https://www*mapleprimes*com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

SI_MODEL.mw
I have tried the following code to simulate the following SI Model. But when I run it, it gives me several mistakes and most importantly does not calculate the Jacobian Matrix and the eigenvalues. 

Any help would be deeply appreciated


Many of the most interesting dynamics in nature have to do with interactions between organisms. These interactions are often subtle, indirect, and difficult to detect. Interactions in which one organism consumes all or part of another. This includes predator-prey, herbivore-plant, and parasite-host interactions. These linkages are the prime movers of energy through food chains. They are an important factor in the ecology of populations, determining the mortality of prey and the birth of new predators.

Mathematical models and logic suggest that a coupled system of predator and prey should cycle: predators increase when prey is abundant, prey are driven to low numbers by predation, the predators decline, and the prey recover, ad infinitum. One such model that simulates predator-prey interactions is the Lotka - Volterra Model.

 


We will discuss a behavior of 2 D - subsystems: SI - model.


SI Model (without predator)

 

We will study a mathematical model that appears in eco - eco-epidemiology.


The SI model describes the interactions between S-prey with density x and I-prey with density y under the assumption z≡0


Break the population into compartments.

• 

Prey (S)

• 

Infected prey (I)


SI Equations :

(D(x))(t) = r*x(t)-b*x(t)^2-c*x(t)*y(t)-beta*x(t)*y[t]/(a+x(t))

(D(y))(t) = -mu*y(t)+beta*x(t)*y(t)/(a+x(t))

– 

 r stands for the intrinsic growth rate of S-prey.

– 

b defines the intra-class competition in S-prey.

– 

c characterizes the inter-class competition between S-prey and I-prey.

– 

μ stands for the mortality of the I-prey

– 

β considered as the bifurcation parameter.

• 

Note: The dynamics of the system depend on parameter β

 

Modeling with Differential Equations

 

• 

We will fix the parameters in the study to have the following values r = 1, b = 1, c = 1/100, μ = 4/10, a=1/2.75.

• 

The initial values will be x(0)=0.2, y(0)=0.05


As mentioned above the dynamics of the system depend on parameter β. So we will consider two cases when β=0.75 and β=1


We will solve the system for the above values of parameters and initial conditions and for a time interval [0,200] , then we will create its plots

 

Case 1

  restart

beta := .75; r := 1; b := 1; c := 1/100; mu := 4*(1/10); a := 1/2.75; f := r*x(t)-b*x(t)^2-c*x(t)*y(t)-(3/4)*x(t)*y(t)/(a+x(t)); g := -mu*y(t)+(3/4)*x(t)*y(t)/(a+x(t)); deq1 := diff(x(t), t) = f; deq2 := diff(y(t), t) = g; equilibrio := solve({f = 0, g = 0}, {x(t), y(t)}); jacobian := simplify(Matrix(2, 2, [[diff(f, x(t)), diff(f, y(t))], [diff(g, x(t)), diff(g, y(t))]])); A := simplify(eval(jacobian, equilibrio[1])); B := simplify(eval(jacobian, equilibrio[2])); Q := simplify(eval(jacobian, equilibrio[3])); eigen1 := Eigenvalues(A); eigen2 := Eigenvalues(B); eigen3 := Eigenvalues(Q); soln := dsolve({deq1, deq2, x(0) = .2, y(0) = 0.5e-1}, numeric, output = listprocedure); plots:-display(plots:-odeplot(soln, [t, x(t)], 0 .. 200, color = blue, legend = ["x(t)"]), plots:-odeplot(soln, [t, y(t)], 0 .. 200, color = red, legend = ["y(t)"]), labels = ["t", "Population"], title = "Population Dynamics")

 

NULL

 

 

Case 2

restart``

beta := 1; r := 1; b := 1; c := 1/100; Mu := 4*(1/10); a := 1/2.75; f := r*x(t)-b*x(t)^2-c*x(t)*y(t)-x(t)*y(t)/(a+x(t)); g := -Mu*y(t)+x(t)*y(t)/(a+x(t)); deq1 := diff(x(t), t) = f; deq2 := diff(y(t), t) = g; equilibrio := solve({f = 0, g = 0}, {x(t), y(t)}); jacobian := Matrix(2, 2, [[diff(f, x(t)), diff(f, y(t))], [diff(g, x(t)), diff(g, y(t))]], simplify); J_at_equilibrio := [eval(jacobian, equilibrio)]; eigen1 := Eigenvalues(A); eigen2 := Eigenvalues(B); eigen3 := Eigenvalues(Q); soln := dsolve({deq1, deq2, x(0) = .2, y(0) = 0.5e-1}, numeric); plots:-display(plots:-odeplot(soln, [t, x(t)], 0 .. 200, color = blue, legend = ["x(t)"]), plots:-odeplot(soln, [t, y(t)], 0 .. 200, color = red, legend = ["y(t)"]), labels = ["t", "Population"], title = "Population Dynamics with Beta = 1")

 

NULL


 

Download SI_MODEL.mw

 

 

p eriode:=proc(r::rational)""  local a,b,c,f,i,k,l,p,q,s:  s:=couvert(evalf(abs(r)-trunc(r),50),string):  if  s[1]="." then s:=s[2..length(s)] fi:  a:=numer(simplify(r)): b:=denom(simplify(r)):  if  igcd(b,10)=1 then       q:=0:       p:=1:      while (10^(p) mod b)<>1 do  p:=p+1 od:  else      f:=ifactors(b)[2]:      k:=0:l:=0:      for i  ;
to nops(f) do
        if f[i][1]=2 then k:=f[i][2] fi:
        if f[i][1]=5 then l:=f[i][2] fi: 
     od:
   c:=b/(2^k*5^l):
   q:=max(k,l):
   if c=1 then p:=0 else p:=op(2,periode(1/c)) fi:
fi:
[q,p,[seq(parse(s[k]),k=1+q..q+p)]]
end:
Error, improper op or subscript selector
Error, reserved word `fi` unexpected
I don't understand these errors. Thank you.

How can I make if-statement work in my code?
assuming m,n :: integers
Thank you in advance.

restart;

 

x2 := proc(m,n)
  if ((is(m,even) and is(n,odd)) or (is(m,odd)and is(n,even))) then -8*L*m*n/(m+n)^2/Pi^2/(m-n)^2;
  else  0;
  end if;
end poc:

Error, missing operator or `;`

 


Download test2.mw

The uploaded worksheet is intended to be an example of the Equal Incircles Theorem, but the incircles displayed are not equal.

What is wrong in the worksheet and how can the error(s) be corrected?

Equal_Incircles_theorem.mw

The CauchyRiemann procedure (for older version  of Maple )doesn't work quite right in Maple 2024 .
Also ran the procedure through the AI for so-called code improvement and now it shows what the code stands for 
The output according to the original procedure would look like on the screenshot, but running original procedure does not give this output ? 
I also want to extend the procedure with a plot of the complex function. 
That differentiability of complex functions is not obvious even if the cauchy-riemann equation is satisfied ?

 

restart

"maple.ini in users"

(1)

NULL

CauchyRiemann:=proc(expr::algebraic) # original procedure
  local x, y, u, v, u_x, u_y, v_x, v_y, flag1, flag2;

  u:=evalc(Re(eval(expr, z=x+I*y)));
  v:=evalc(Im(eval(expr, z=x+I*y)));

  u_x:=diff(u,x);
  u_y:=diff(u,y);
  v_x:=diff(v,x);
  v_y:=diff(v,y);

  print('f(z)'=expr);
  printf("\n");
  
  print('u(x,y)'=u);
  print('u[x](x,y)'=u_x);
  print('u[y](x,y)'=u_y);
  printf("\n");

  print('v(x,y)'=v);
  print('v[x](x,y)'=v_x);
  print('v[y](x,y)'=v_y);
  printf("\n");

  if u_x=v_y then
    print('u[x]=v[y]');
    print(u_x=v_y);
    flag1:=true;
  else
    print('u[x]<>v[y]');
    print(u_x<>v_y);
    flag1:=false;
  end if;

  if u_y=-v_x then
    print('u[y]=-v[x]');
    print(u_y=-v_x);
    flag2:=true;
  else
    print('u[y]<>-v[x]');
    print(u_y<>-v_x);
    flag2:=false;
  end if;
  
printf("\n");
if flag1=true and flag2=true then
   print(`Fullfill the Cauchy-Riemann Equations`);
   print(`The derivative is:`='u[x]+I*v[y]');
   print('diff(f(z),z)'=u_x+I*v_y);
else
   print(`Cauchy-Riemann ?`);
end if

end proc:

f(z):=1/(z+2):
CauchyRiemann(f(z))

f(z) = 1/(z+2)

 

 

 

u(x, y) = (x+2)/(y^2+(x+2)^2)

 

u[x](x, y) = 1/(y^2+(x+2)^2)-(x+2)*(2*x+4)/(y^2+(x+2)^2)^2

 

u[y](x, y) = -2*(x+2)*y/(y^2+(x+2)^2)^2

 

 

 

v(x, y) = -y/(y^2+(x+2)^2)

 

v[x](x, y) = y*(2*x+4)/(y^2+(x+2)^2)^2

 

v[y](x, y) = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2

 

 

 

u[x] <> v[y]

 

1/(y^2+(x+2)^2)-(x+2)*(2*x+4)/(y^2+(x+2)^2)^2 <> -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2

 

u[y] <> -v[x]

 

-2*(x+2)*y/(y^2+(x+2)^2)^2 <> -y*(2*x+4)/(y^2+(x+2)^2)^2

 

 

 

`Cauchy-Riemann ?`

(2)

 

Also ran the procedure through the AI for so-called code improvement and now it shows what the code stands for

restart;

# Improved and corrected version of the CauchyRiemann procedure :ASKED AI 
CauchyRiemann := proc(expr::algebraic)
    local x, y, u, v, u_x, u_y, v_x, v_y, CR1, CR2;

    # Assign real and imaginary parts of the function
    u := evalc(Re(eval(expr, z = x + I*y)));
    v := evalc(Im(eval(expr, z = x + I*y)));

    # Calculate partial derivatives
    u_x := diff(u, x);
    u_y := diff(u, y);
    v_x := diff(v, x);
    v_y := diff(v, y);

    # Properly format and print function details
    printf("f(z) = %a\n", expr);
    printf("u(x, y) = %a, u_x = %a, u_y = %a\n", u, u_x, u_y);
    printf("v(x, y) = %a, v_x = %a, v_y = %a\n", v, v_x, v_y);

    # Evaluate and print Cauchy-Riemann equations
    CR1 := u_x = v_y;
    CR2 := u_y = -v_x;
    printf("\nCauchy-Riemann Equations:\n");
    printf("u_x = v_y: %a\n", CR1);
    printf("u_y = -v_x: %a\n", CR2);

    # Check both equations
    if CR1 and CR2 then
        printf("The function is analytic (holomorphic) at this point.\n");
        printf("The derivative f'(z) is %a + I*%a\n", u_x, v_y);
    else
        printf("The function does not satisfy the Cauchy-Riemann equations and is not analytic.\n");
    end if;
end proc;

# Test the procedure with a specific function
f := z -> 1/(z + 2);
CauchyRiemann(f(z));

"maple.ini in users"

 

proc (expr::algebraic) local x, y, u, v, u_x, u_y, v_x, v_y, CR1, CR2; u := evalc(Re(eval(expr, z = x+I*y))); v := evalc(Im(eval(expr, z = x+I*y))); u_x := diff(u, x); u_y := diff(u, y); v_x := diff(v, x); v_y := diff(v, y); printf("f(z) = %a
", expr); printf("u(x, y) = %a, u_x = %a, u_y = %a
", u, u_x, u_y); printf("v(x, y) = %a, v_x = %a, v_y = %a
", v, v_x, v_y); CR1 := u_x = v_y; CR2 := u_y = -v_x; printf("
Cauchy-Riemann Equations:
"); printf("u_x = v_y: %a
", CR1); printf("u_y = -v_x: %a
", CR2); if CR1 and CR2 then printf("The function is analytic (holomorphic) at this point.
"); printf("The derivative f'(z) is %a + I*%a
", u_x, v_y) else printf("The function does not satisfy the Cauchy-Riemann equations and is not analytic.
") end if end proc

 

proc (z) options operator, arrow; 1/(z+2) end proc

 

f(z) = 1/(z+2)
u(x, y) = (x+2)/(y^2+(x+2)^2), u_x = 1/(y^2+(x+2)^2)-(x+2)/(y^2+(x+2)^2)^2*(2*x+4), u_y = -2*(x+2)/(y^2+(x+2)^2)^2*y
v(x, y) = -y/(y^2+(x+2)^2), v_x = y/(y^2+(x+2)^2)^2*(2*x+4), v_y = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2

Cauchy-Riemann Equations:
u_x = v_y: 1/(y^2+(x+2)^2)-(x+2)/(y^2+(x+2)^2)^2*(2*x+4) = -1/(y^2+(x+2)^2)+2*y^2/(y^2+(x+2)^2)^2
u_y = -v_x: -2*(x+2)/(y^2+(x+2)^2)^2*y = -y/(y^2+(x+2)^2)^2*(2*x+4)
The function does not satisfy the Cauchy-Riemann equations and is not analytic.

 

NULL

Download CAUCHY_RIEMANN_-FORUM_VRAAG.mw

I want to solve or try to solve this equation 

PDE := diff(G(a, H, phi, PI), a)(aH) + diff(G(a, H, phi, PI), H)(k/a^2 - kappa^2/2*PI^2/a^6) + diff(G(a, H, phi, PI), phi)(PI/a^3) = diff(G(a, H, phi, PI), PI)(a^3*diff(V(phi), phi))

with pdsolve(PDE, G)

and maple answer me the next

Error, (in pdsolve/info) first argument does not have a differentiated function with name G

I nw in maple, maybe I´m make a mistake, but I can't find what

How can I  get the result for the integration when m =n or m is not equal to n and How can I add assumption that m, or n can be even or odd?

Thanks in advance for your help.

 

restart;
phi := proc(k,x,L)
  if (type(k,even)) then sqrt(2)*sin(Pi*k*x/L)/sqrt(L)
  else sqrt(2)*cos(Pi*k*x/L)/sqrt(L)
  end if;
end proc:

Int(phi(m,x,L)*_h^2/m2*diff(phi(n,x,L),x,x),x=-L/2..L/2);

Int(-2*cos(Pi*m*x/L)*_h^2*Pi^2*n^2*cos(Pi*n*x/L)/(L^3*m2), x = -(1/2)*L .. (1/2)*L)

(1)
 

 

Download test1.mw

 

I have  4 worksheets with derived equations. So I export the equations and  possibly some procedures (but they can be handled seperately if needed)  from each worksheet as a .mpl file. 

I want to combine the .mpl files  together without using copy/paste. Then I can open that single file in the VS code editor.
There may be other ways to achieve this so I am open to suggestions.

Good evening!

I am Athanasios Paraskevopoulos, a graduate student specializing in applied mathematics. Recently, I've started exploring Maple through its trial version and I'm considering making a purchase. My question for you all is: Am I restricted to buying only the graduate student package, or am I free to choose from any of the Maple packages available? Any guidance or personal experiences with different packages would be greatly appreciated!

Thank you in advance for your help!

I chased down a problem to factoring a square that has sqrt in the coefficients. All numbers are real,
The code is inside a procedure in a package. Iso I could do with something robust.

expand((sqrt(A+B)*x+sqrt(7-K)*y)^2)
     2      2            (1/2)          (1/2)        2      2
  A x  + B x  + 2 (A + B)      x (7 - K)      y - K y  + 7 y 

factor(%) 

 

Using a command in succession can coax maple into different outputs. 

eq:=(a*x+b)/(c*x+d)=1
                                
Getting rid of the denominator is probably the most deciding factor in how maple displays an output.

isolate(eq,b)
                      

isolate(%,x,1)
                      

Using the isolate command, you couldn't arrive at that output without using it twice. 

I was trying to solve a system of polynomial equations, which contains three equations and six variables $a_0,a_1,a_2,b_0,b_1,b_2$. However, as I swap the variable name, Maple solve function gives me a totally different solutions. Only the solutions before swapping the variables are useful for the problem I study. I have already attached the file. Could anyone tell me if the choice of variable name really matters, or if i just misuse this function?

Choice_of_name_infolevel.mw

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