Maple 2019 Questions and Posts

These are Posts and Questions associated with the product, Maple 2019

I need to create a slider plot for A10, A11, and A12 by varying the parameters theta, Pu, and a.
I have a syntax ready — could you suggest modifications to make it work correctly and generate the plot?

Additionally, is it possible to compute the values of A13 and A14 by substituting the obtained A10, A11, and A12 values for each combination of theta, Pu, and a from the slider plot?

Sheet attached: Slider_Q.mw

I want both Pc and r(Pc) to be greater than or equal to zero. The only constraint is that all parameters for which I’ve provided data must remain positive. Can we identify the key parameters that significantly affect Pc? Also, what condition ensures that Pc ≥ 0? Ideally, I’d like Pc to be less than or equal to Pu. Could you suggest what changes in the numerical values I should make to ensure Pc becomes a positive value?

Attaching file: Q2.mw

Can we include a graph that shows how Pc changes with respect to variations in the most sensitive parameters?

The maple worksheet shows an incorrect evaluation of the integral in (1) which is a standard integral representation of a Bessel function.  Equations (2)-(5) along with the graph show the incorrectness of the evaluation.  What is going on?

Bessel.mw

Loop is not executing all values just considerinng 1st value of tau0 =0.3.

Also how to plot a 2D graph from the loop result

Attaching sheet with questions in the sheet:
for_loop_question_rec.mw

Hello Everyone;

I need to find the bifurcation point and further bifarcation diagram for the given model. But facing error. Can anybody help to do this? Can you refer some library for bifurcation analysis of ODE's? Code is attched. Thanks in Advance. 

123.mw

 

 

 

 

restart

C_m := 1.0; g_K := 36.0; I_inj := 0; g_L := .3; E_Na := 50.0; E_K := -77.0; E_L := -54.4

alpha_m := (.1*(V-25.0))/(1-exp(-(V-25.0)*(1/10))); beta_m := 4*exp(-V/(18.0)); alpha_h := 0.7e-1*exp(-V/(20.0)); beta_h := 1/(1+exp(-(V-30)*(1/10))); alpha_n := (0.1e-1*(V-10.0))/(1-exp(-(V-10.0)/(10.0))); beta_n := .125*exp(-V/(80.0)); I_Na := g_Na*m^3*h*(V-E_Na); I_K := g_K*n^4*(V-E_K); I_L := g_L*(V-E_L)

.125*exp(-0.1250000000e-1*V)

(1.1)

eq1 := (I_inj-I_Na-I_K-I_L)/C_m; m := alpha_m/(alpha_m+beta_m); n := alpha_n/(alpha_n+beta_n); h := alpha_h/(alpha_h+beta_h)

-16.32000000-1.000000000*g_Na*m^3*h*(V-50.0)-36.00000000*n^4*(V+77.0)-.3000000000*V

 

.1*(V-25.0)/((1-exp(-(1/10)*V+2.500000000))*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V)))

 

0.1e-1*(V-10.0)/((1-exp(-.1000000000*V+1.000000000))*(0.1e-1*(V-10.0)/(1-exp(-.1000000000*V+1.000000000))+.125*exp(-0.1250000000e-1*V)))

 

0.7e-1*exp(-0.5000000000e-1*V)/(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3)))

(1.2)

bif_eq1 := eq1 = 0;

-16.32000000-0.7000000000e-4*g_Na*(V-25.0)^3*exp(-0.5000000000e-1*V)*(V-50.0)/((1-exp(-(1/10)*V+2.500000000))^3*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V))^3*(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3))))-0.3600000000e-6*(V-10.0)^4*(V+77.0)/((1-exp(-.1000000000*V+1.000000000))^4*(0.1e-1*(V-10.0)/(1-exp(-.1000000000*V+1.000000000))+.125*exp(-0.1250000000e-1*V))^4)-.3000000000*V = 0

bif_eq2 := diff( eq1, V) = 0;

-0.2100000000e-3*g_Na*(V-25.0)^2*exp(-0.5000000000e-1*V)*(V-50.0)/((1-exp(-(1/10)*V+2.500000000))^3*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V))^3*(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3))))+0.2100000000e-4*g_Na*(V-25.0)^3*exp(-0.5000000000e-1*V)*(V-50.0)*exp(-(1/10)*V+2.500000000)/((1-exp(-(1/10)*V+2.500000000))^4*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V))^3*(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3))))+0.2100000000e-3*g_Na*(V-25.0)^3*exp(-0.5000000000e-1*V)*(V-50.0)*(.1/(1-exp(-(1/10)*V+2.500000000))-0.1000000000e-1*(V-25.0)*exp(-(1/10)*V+2.500000000)/(1-exp(-(1/10)*V+2.500000000))^2-.2222222222*exp(-0.5555555556e-1*V))/((1-exp(-(1/10)*V+2.500000000))^3*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V))^4*(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3))))+0.3500000000e-5*g_Na*(V-25.0)^3*exp(-0.5000000000e-1*V)*(V-50.0)/((1-exp(-(1/10)*V+2.500000000))^3*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V))^3*(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3))))+0.7000000000e-4*g_Na*(V-25.0)^3*exp(-0.5000000000e-1*V)*(V-50.0)*(-0.3500000000e-2*exp(-0.5000000000e-1*V)+(1/10)*exp(-(1/10)*V+3)/(1+exp(-(1/10)*V+3))^2)/((1-exp(-(1/10)*V+2.500000000))^3*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V))^3*(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3)))^2)-0.7000000000e-4*g_Na*(V-25.0)^3*exp(-0.5000000000e-1*V)/((1-exp(-(1/10)*V+2.500000000))^3*(.1*(V-25.0)/(1-exp(-(1/10)*V+2.500000000))+4*exp(-0.5555555556e-1*V))^3*(0.7e-1*exp(-0.5000000000e-1*V)+1/(1+exp(-(1/10)*V+3))))-0.1440000000e-5*(V-10.0)^3*(V+77.0)/((1-exp(-.1000000000*V+1.000000000))^4*(0.1e-1*(V-10.0)/(1-exp(-.1000000000*V+1.000000000))+.125*exp(-0.1250000000e-1*V))^4)+0.1440000000e-6*(V-10.0)^4*(V+77.0)*exp(-.1000000000*V+1.000000000)/((1-exp(-.1000000000*V+1.000000000))^5*(0.1e-1*(V-10.0)/(1-exp(-.1000000000*V+1.000000000))+.125*exp(-0.1250000000e-1*V))^4)+0.1440000000e-5*(V-10.0)^4*(V+77.0)*(0.1e-1/(1-exp(-.1000000000*V+1.000000000))-0.1000000000e-2*(V-10.0)*exp(-.1000000000*V+1.000000000)/(1-exp(-.1000000000*V+1.000000000))^2-0.1562500000e-2*exp(-0.1250000000e-1*V))/((1-exp(-.1000000000*V+1.000000000))^4*(0.1e-1*(V-10.0)/(1-exp(-.1000000000*V+1.000000000))+.125*exp(-0.1250000000e-1*V))^5)-0.3600000000e-6*(V-10.0)^4/((1-exp(-.1000000000*V+1.000000000))^4*(0.1e-1*(V-10.0)/(1-exp(-.1000000000*V+1.000000000))+.125*exp(-0.1250000000e-1*V))^4)-.3000000000 = 0

 

 

 

bif_sol := solve({ bif_eq1,bif_eq2}, {V, g_Na});

Warning, solutions may have been lost

 

 

as the solutions, which are then expressed as the points mu, y via

   

[Back to ODE Powertool Table of Contents]

 

 

Hi, I have an homework where I need to find the highest point and the lowest point on an ellipse form by the intersection of two equations wich are 4x-3y+8z=5 and z^2=x^2+y^2 and I have to use the LagrangeMultiplier command. I get how it works but I can't get the correct form. How should I do it ? 

restart;
Pr:=0.71: n:=-1:

eta0:=0.0699;

EQ1:=diff(H(x), x ) - x*diff(F(x), x ) ;
 

EQ2:=(1+x^2)*diff(F(x), x$2) + (3*x + x*F(x)-H(x))*diff(F(x), x) + F(x)^2 + G(x)^2 +2*P(x) + x*diff(P(x), x) ;

EQ3:=(1+x^2)*diff(G(x), x$2) + (3*x + x*F(x)-H(x))*diff(G(x), x) ;

EQ4:=(1+x^2)*diff(H(x), x$2) + (3*x + x*F(x)-H(x))*diff(H(x), x) + (1+F(x))*H(x)- diff(P(x), x);

EQ5:=(1+x^2)*diff(theta(x), x$2) + x*(1-2*n)*diff(theta(x), x) + n^2*theta(x) - Pr*( n*F(x)*theta(x) + ( H(x)-x*F(x) )*diff(theta(x), x)  ) ;


EQ:={EQ1=0, EQ2=0,EQ3=0,EQ4=0 ,EQ5=0}:


IC:={ F(0)=0, G(0)=12, H(0)=0, theta(0)= 1, F(eta0)=0, G(eta0)=12, H(eta0)=0, theta(eta0)= 0, P(0)=0};
 

sol:= dsolve(EQ union IC,numeric,output=Array([0,0.0699]));

ques.mw

Just wanted to ask, what the issue here is:

restart;
Int(1/(1 - x*ln(x)), x);
IntegrationTools:-Change(%,u=1-x*ln(x),u);

doesn't give the proper transformation. It gives

Int(1/u,u)

Solving for x and writing the transformation in terms of LambertW gives something else, if I'm not mistaken.

I cannot view 3d graphics with my version of Ubuntu 20.04. I've updated all my computer's graphics card drivers and the problem persists. If I run without hardware acceleration, nothing changes; no visualization and no production possible.
Do you have any ideas for solving this problem? Maple uses OpenGL libraries for 3D production and visualization, and these libraries are installed on my computer. Would installing mesa solve the problem, for example?

Thanks in advance.

The following 2D integrals of 0 are seemingly trivial and one would expect them to evaluate to zero, but Maple evaluates them to undefined

int(0, x=0..infinity, y=0..1) # undefined
int(0, x=0..1, y=0..infinity) # undefined

When the 2D integral is split into two 1D integrals, it does evaluate to zero, as the following examples show

int(0,x=0..infinity) # 0
int(int(0,x=0..infinity), y=0..1) # 0
int(int(0,x=0..1),y=0..infinity) # 0

If infinity is replaced by a variable (say 'c'), the first two integrals are also evaluated to zero.

It may be connected by the following

int(a, x=0..infinity, y=0..1) # a*infinity
int(a, x=0..1, y=0..infinity) # a*infinity
int(a,x=0..infinity) # signum(a)*infinity

So for the 1D integrals the signum is applied to 'a' when the interval is infinite, but not for the 2D integrals. I'm not sure about this difference.

I will have excel sheet with minium 500 coulmns and 1000 rows say

For sample to explain my question I attach a demo excel

All my columns have headers

I am looking to find all 2 way multiplication and add them as columns to my excel sheet and return it as a new excel sheet say

The column names for the new 2 way column should be like the

header name of column you are multiply * the name of the other columsn

Now in sample file if i multiple column with name A with column with name B I get a new column with header A*B the header name should be inserted and

Below that all the elements of that A column multiplied  with that of B should come

I am looking to form columns for all possible2  way multiplication for the excel I will give.

As you can see the demo file

Excel_to_explain.xlsx

Kind help please

The transformed boundary conditions are

Hi
Can someone help me write the program for this equation?
I really need this program.
With respect

 

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