Maple 2022 Questions and Posts

These are Posts and Questions associated with the product, Maple 2022

I am trying to separate the real and imaginary parts of a complex expression in Maple to get Eq. (1.5) as in the attached image, but the Re and Im functions do not seem to return the expected results. Instead, Maple leaves the expression unchanged. PD_OD.mw

Hello,
I have another vector problem, and I honestly have no idea how to solve it. (I am not even sure if there is a solution). I tried to simplify it as much as possible (also the attached maple file)unknown_vector.mw

We are looking for the vectors v2, v3 and v4 
Given are vector v1 (with its unit vector e1 and magnitude m1), the unit vector of v2 (e2) and the unit vector of v3 (e3)
From vector v4 we don't have any information.
The following should be true:
v2 = v1 + v3 + v4, with the condition that v4 shall be as small as possible, with the goal to find the combination of v1 and v3 that comes as close as possible to v2. Of course, depending on the unit vectors, v4 can also be zero. But we assume that the given unit vectors can not be changed. 
I would be really grateful for any help. Thanks in advance! 
Roman

restart; with(PDEtools); declare(F(x, t), G(x, t), H(x, t))

F(x, t)*`will now be displayed as`*F

 

G(x, t)*`will now be displayed as`*G

 

H(x, t)*`will now be displayed as`*H

(1)

q := 1-(diff(diff(log(F(x, t)), x), t)); r := G/F; s := H/F

1-(diff(diff(F(x, t), t), x))/F(x, t)+(diff(F(x, t), x))*(diff(F(x, t), t))/F(x, t)^2

 

G/F

 

H/F

(2)

r1s1 := r*s; r1s1der := diff(r1s1(x, t), x)

qt := diff(q(x, t), t)

eq1B := F(x, t)^3*(qt+r1s1der) = 0; eq12B := simplify(expand(eq1B))

-F(x, t)^3*(diff((diff(diff(F(x, t), t), x))(x, t), t))/(F(x, t))(x, t)+F(x, t)^3*(diff(diff(F(x, t), t), x))(x, t)*(diff((F(x, t))(x, t), t))/(F(x, t))(x, t)^2+F(x, t)^3*(diff((diff(F(x, t), x))(x, t), t))*(diff(F(x, t), t))(x, t)/(F(x, t))(x, t)^2-2*F(x, t)^3*(diff(F(x, t), x))(x, t)*(diff(F(x, t), t))(x, t)*(diff((F(x, t))(x, t), t))/(F(x, t))(x, t)^3+F(x, t)^3*(diff(F(x, t), x))(x, t)*(diff((diff(F(x, t), t))(x, t), t))/(F(x, t))(x, t)^2+F(x, t)*(diff(G(x, t), x))*H(x, t)-2*G(x, t)*H(x, t)*(diff(F(x, t), x))+F(x, t)*G(x, t)*(diff(H(x, t), x)) = 0

(3)

D_x_x_G_F := (diff(G(x, t), x, x))*F(x, t)-2*(diff(G(x, t), x))*(diff(F(x, t), x))+G(x, t)*(diff(F(x, t), x, x)); D_t_t_F_F := F(x, t)*(diff(F(x, t), `$`(t, 2)))-2*(diff(F(x, t), t))^2

(diff(diff(G(x, t), x), x))*F(x, t)-2*(diff(G(x, t), x))*(diff(F(x, t), x))+G(x, t)*(diff(diff(F(x, t), x), x))

 

F(x, t)*(diff(diff(F(x, t), t), t))-2*(diff(F(x, t), t))^2

(4)

NULL

rxt := diff(diff(r(x, t), x), t)

eq2B := -2*q*r+rxt = 0

eq22B := simplify(expand(eq2B))

((-F*F(x, t)*G(x, t)+2*G*F(x, t)^2)*(diff(diff(F(x, t), t), x))+(diff(diff(G(x, t), t), x))*F*F(x, t)^2+((2*F*G(x, t)-2*G*F(x, t))*(diff(F(x, t), x))-F*(diff(G(x, t), x))*F(x, t))*(diff(F(x, t), t))-(diff(G(x, t), t))*(diff(F(x, t), x))*F*F(x, t)-2*G*F(x, t)^3)/(F*F(x, t)^3) = 0

(5)

sxt := diff(diff(s(x, t), x), t)

eq3B := -2*q*s+sxt = 0

eq32B := simplify(expand(eq3B))

((-F*F(x, t)*H(x, t)+2*H*F(x, t)^2)*(diff(diff(F(x, t), t), x))+(diff(diff(H(x, t), t), x))*F*F(x, t)^2+((2*F*H(x, t)-2*H*F(x, t))*(diff(F(x, t), x))-F*(diff(H(x, t), x))*F(x, t))*(diff(F(x, t), t))-(diff(H(x, t), t))*(diff(F(x, t), x))*F*F(x, t)-2*H*F(x, t)^3)/(F*F(x, t)^3) = 0

(6)

"#`# How to simplify Eqs. (3), (5) and (6) and write in terms of following bilineat operators` by using (4)"?""

NULL

NULL

Download BE.mw

`How to calculate u = `((∂)/(∂ t) - (∂)/(∂ x))^(-1)(z)? 

restart;

with(PDEtools):

alias(z=z(x,y,t), u=u(x,y,t))

z, u

(1)

z := 32*delta2^3*(exp((2*a*y*delta2^3 + 2*(x + y + t)^3*(B1 + 2*B2)*delta2^2 + (2*a*delta1^2*y + t)*delta2 + 2*delta1^2*(x + y + t)^3*(B1 + 2*B2))/(delta1^2 + delta2^2)) - exp((-2*a*y*delta2^3 + 4*(B1 + B2/2)*(x + y + t)^3*delta2^2 + (-2*a*delta1^2*y - t)*delta2 + 4*delta1^2*(B1 + B2/2)*(x + y + t)^3)/(delta1^2 + delta2^2)))*a/((delta1^2 + delta2^2)^2*(exp((2*(delta1^2 + delta2^2)*(x + y + t)^3*B2 + 2*delta2*(a*(delta1^2 + delta2^2)*y + t/2))/(delta1^2 + delta2^2)) + exp((2*(delta1^2 + delta2^2)*(x + y + t)^3*B1 - 2*delta2*(a*(delta1^2 + delta2^2)*y + t/2))/(delta1^2 + delta2^2)))^3);

32*delta2^3*(exp((2*a*y*delta2^3+2*(x+y+t)^3*(B1+2*B2)*delta2^2+(2*a*delta1^2*y+t)*delta2+2*delta1^2*(x+y+t)^3*(B1+2*B2))/(delta1^2+delta2^2))-exp((-2*a*y*delta2^3+4*(B1+(1/2)*B2)*(x+y+t)^3*delta2^2+(-2*a*delta1^2*y-t)*delta2+4*delta1^2*(B1+(1/2)*B2)*(x+y+t)^3)/(delta1^2+delta2^2)))*a/((delta1^2+delta2^2)^2*(exp((2*(delta1^2+delta2^2)*(x+y+t)^3*B2+2*delta2*(a*(delta1^2+delta2^2)*y+(1/2)*t))/(delta1^2+delta2^2))+exp((2*(delta1^2+delta2^2)*(x+y+t)^3*B1-2*delta2*(a*(delta1^2+delta2^2)*y+(1/2)*t))/(delta1^2+delta2^2)))^3)

(2)

NULL

NULL

Download integ.mw

Hello

I have a question regarding the input of a list that I would like to use for further calculations. 
Attached, you’ll find the maple file with some vector calculations and a 3D visualisation. Three numbers are used as input (F_D3, F_D4, F_S2). I would like to use a function over time for these inputs instead of plain numbers. The function consists of an Excel list with an input number for each time step. How do I do this? Can I also use PlotVector in visu2 to plot a vector at a certain time step? 

For further processing, I would like the p-norm (F_E2) of vector RF_E2 to be visualized/plotted as a function over time. How do I approach this? 

Thank you very much for your help in advance!

input_question.xlsx

Input_question.mw

How do I get my actual solution which involves _Z? I have  tried answers that w ere given on questions related to _Z in a solution but mine is not working. I am completely new to Maple ,kindly help me.

eq1 := Lambda[h]+rho[2]*R[h]-(b[h]+kappa+beta[1])*S[h] = 0

Lambda[h]+rho[2]*R[h]-(b[h]+kappa+beta[1])*S[h] = 0

(1)

eq2 := T[h] = 0

T[h] = 0

(2)

eq3 := R[h]-kappa*S[h]/(rho[2]+b[h]) = 0

R[h]-kappa*S[h]/(rho[2]+b[h]) = 0

(3)

eq4 := Lambda[m]*L-(alpha+pi)*S[m] = 0

Lambda[m]*L-(alpha+pi)*S[m] = 0

(4)

eq5 := c(1-L/K)*S[m]-(d+Lambda[m])*L = 0

c(1-L/K)*S[m]-(d+Lambda[m])*L = 0

(5)

sol := solve({eq1, eq2, eq3, eq4, eq5}, [T[h], S[h], R[h], L, S[m]])

[[T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), L = 0, S[m] = 0], [T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), L = (alpha+pi)*RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)/Lambda[m], S[m] = RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)]]

(6)

DF := sol[1]

[T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), L = 0, S[m] = 0]

(7)

DF2 := sol[2]

[T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), L = (alpha+pi)*RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)/Lambda[m], S[m] = RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)]

(8)

"(=)"

[T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(b[h]^2+(kappa+rho[2]+beta[1])*b[h]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(b[h]^2+(kappa+rho[2]+beta[1])*b[h]+beta[1]*rho[2]), L = (alpha+pi)*RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)/Lambda[m], S[m] = RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)]

(9)

"(=)"

[T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(b[h]^2+(kappa+rho[2]+beta[1])*b[h]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(b[h]^2+(kappa+rho[2]+beta[1])*b[h]+beta[1]*rho[2]), L = (alpha+pi)*RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)/Lambda[m], S[m] = RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)]

(10)

solve({eq4, eq5}, [L, S[m]], explicit)

[[L = 0, S[m] = 0], [L = (alpha+pi)*RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)/Lambda[m], S[m] = RootOf(-c((K*Lambda[m]-_Z*alpha-_Z*pi)/(K*Lambda[m]))*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)]]

(11)

theIndets := indets(indets(sol, atomic))

{K, L, alpha, d, kappa, pi, Lambda[h], Lambda[m], R[h], S[h], S[m], T[h], b[h], beta[1], rho[2]}

(12)

allvalues(sol)

[[T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), L = 0, S[m] = 0], [T[h] = 0, S[h] = Lambda[h]*(rho[2]+b[h])/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), R[h] = kappa*Lambda[h]/(kappa*b[h]+b[h]^2+b[h]*beta[1]+b[h]*rho[2]+beta[1]*rho[2]), L = -K*(RootOf(-c(_Z)*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)-1), S[m] = -K*Lambda[m]*(RootOf(-c(_Z)*Lambda[m]+d*alpha+Lambda[m]*alpha+d*pi+Lambda[m]*pi)-1)/(alpha+pi)]]

(13)

NULL

``

Download maple_code_for_disease_equilibruim.mw

Hello 

I have a 3D vector plot with 3 vectors. It would be nice to have a legend for this, but so far, I have struggled to find a solution.

At the moment, I use a caption for this, but I am not fully happy with this.legend_question.mw 

Is there a simple solution?

Thanks in advance 

This time domain diagram has been drawn, but if you want to draw its outer outline, how should you command it?
saopinjifen1230.mw

 

Hello, I have a question regarding finding two vectors that result in a resulting vector with only the directions of the two needed vectors.

Given:

- Vector F_RC1

- Unit Vector of F_RC1Y (called EV_C1Y)

- Unit Vector of F_RC1_ST (called EV_C1_ST)

Needed

- F_RC1_ST

- F_RC1Y

The picture below and the file should make it clear. 

vector_question.mw

How do I approach this? Best regards,

Hello 

I am working on my vector file and have another question:

1. How do I display any vectors in a matrix notation next to each other or above each other (like: [1;2;3]) 

Find attached the file:question_vector_.mw

I really appreciate any help you can provide.

I have a simple question, but don't find a neat solution for it. I have 4 3d Vectors that result in the resulting Vector V5. I now want the unit vector of V5. How do I do it the most efficient way? 

My attempt is in the file attached. I get the error message: "Error, (in rtable/Power) exponentiation operation not defined for Vectors"

Thanks a lot for your help.3d_vector.mw

Hi!  I am trying to calculate a complex integral.  To double check my work, I directly coded the calculation.  The result is that the two results don't match.  Have I done something wrong?

here is the file

Complex_valued_integrals.mw

Is it possible to configure newer maple versions to use a different file extension ?

I have a lot of  Maple 9 files  with .mw extension. I dont want them opened and saved by a later version as they are valuable and actually work better than later maple versions which throws unneccesary signum errors. The results are verified so the signum errors are bogus in the recent maple versions running the same file.

Particularly, can I configure e.g. Maple 2022 to save to a different file extension say .mww or similar rather than .mw so I can avoid destroying to original 9.5 files?

Hi everyone!  I'm trying to study a certain function.  The output says that, from -5 to 5, the output is undefined.  Am I doing something wrong?

Thanks in advance.

range_not_undefined.mw

Solved a "simple" dynamics problem involving the law of cosines... etc.  Part of the solution involves selecting the positive root from the resulting 2nd order equation in x(t)- below.  

eq1 := (r1 + r2)^2 = r1^2 + x(t)^2 - 2*r1*x(t)*cos(theta(t))  (law of cosines)

eq2:=diff(eq1,t)

eq3 := subs(diff(x(t), t) = v, diff(theta(t), t) = omega, eq2)  (remove xdot and theta dot)

assume(0 < x(t));
assume(0 < t);
assume(0 < theta(t) and theta(t) < 2*Pi);

xx := solve(eq3, v)  (creates an relationship between v, theta and x)

     xx := -r1*x(t~)*omega*sin(theta(t~))/(-cos(theta(t~))*r1 + x(t~))

Not clear what this t~ notation means?

when I solve eq1 above for x(t) there are two roots.  I select the positve root, but receive this warning...

eqx := solve(eq1, x(t))[1]

Warning, solve may be ignoring assumptions on the input variables.
        eqx := cos(theta(t))*r1 + sqrt(cos(theta(t))^2*r1^2 + 2*r1*r2 + r2^2)

How can I correct or address the warning message?

Finally,  to piece together the solution I perform this step

v := subs(x(t) = eqx, xx)

The result has a fairly obvious "simplification", but using "simplify" and "combine" only makes it worse...  Is there a combination of commands to accomplish a simplification.

Here is the entirety of the Maple...

restart;
with(Typesetting);
Settings(typesetdot = true);

eq1 := (r1 + r2)^2 = r1^2 + x(t)^2 - 2*r1*x(t)*cos(theta(t));
eq2 := diff(eq1, t);
eq3 := subs(diff(x(t), t) = v, diff(theta(t), t) = omega, eq2);
assume(0 < x(t));
assume(0 < t);
assume(0 < theta(t) and theta(t) < 2*Pi);
xx := solve(eq3, v);
eqx := solve(eq1, x(t))[1];
v := subs(x(t) = eqx, xx);

examine the result for v... easy to see the "simpificaion"... but not sure how to manipulate Maple to accomplish this...

JD

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