Maple 2022 Questions and Posts

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

Dear Maple Community,

I come to you with a question about the reduced involutive form (rif) package. Namely, I decided to try the classic example from the "LONG GUIDE TO THE STANDARD FORM PACKAGE", which dates back to 1993. Here is the link to the complete documentation:

https://wayback.cecm.sfu.ca/~wittkopf/files/standard_manual.txt

So, the example is the following:

2.1 SIMPLE EXAMPLES

EXAMPLE A

Consider the system of nonlinear PDEs:       

y Zxxx - x Zxyy  =  Zyy - y Zy

                        2     2    2
2 y x Zxxx Zxyy + x Zxxx + x y Zxyy  =  0

                  2    2
y Zxyy - x W + 2 x  y Z  =  0

                 2    2
Zyy - y Zy  + 2 x  y W  =  x W

where the dependent variables W and Z are functions of the
independent variables x and y, and Zxxx denotes the third partial
derivative of Z with respect to x etc.

After making computations back in 1993 with Maple V, they obtain the following involutive form:

In our original notation the (considerably) simplified system is:
                                            2
  Zxxx = 0, Zxy = 0, Zyy = y Zy, W = 2 x y Z

So, I tried the same system of PDEs in the modern Maple and the modern rifsimp() command. You can find the result below:

demo_question.mw

The problem is that nowadays [Maple 2022.1] , I get only the trivial solution: z = 0 and w = 0.

Could someone clarify, please, where the truth is and what am I doing wrong?

Thanks a lot in advance for any help and clarification!

Best regards,

Dr. Denys D.
 

restart:

with(DETools):

PDE1 := y*diff(z(x,y), x$3) - x*diff(z(x,y),x,y$2) = diff(z(x,y),y$2) - y*diff(z(x,y), y);

y*(diff(diff(diff(z(x, y), x), x), x))-x*(diff(diff(diff(z(x, y), x), y), y)) = diff(diff(z(x, y), y), y)-y*(diff(z(x, y), y))

(1)

PDE2 := 2*x*y*diff(z(x,y),x$3)*diff(z(x,y),x,y$2) + x*(diff(z(x,y),x$3))^2 + x*y^2*(diff(z(x,y),x,y$2))^2 = 0;

2*x*y*(diff(diff(diff(z(x, y), x), x), x))*(diff(diff(diff(z(x, y), x), y), y))+x*(diff(diff(diff(z(x, y), x), x), x))^2+x*y^2*(diff(diff(diff(z(x, y), x), y), y))^2 = 0

(2)

PDE3 := y*diff(z(x,y),x,y$2) - x*w(x,y) + 2*x^2*y*z(x,y)^2 = 0;

y*(diff(diff(diff(z(x, y), x), y), y))-x*w(x, y)+2*x^2*y*z(x, y)^2 = 0

(3)

PDE4 := diff(z(x,y), y$2) - y*diff(z(x,y),y) + 2*x^2*y*w(x,y)^2 = x*w(x,y);

diff(diff(z(x, y), y), y)-y*(diff(z(x, y), y))+2*x^2*y*w(x, y)^2 = x*w(x, y)

(4)

sys := [PDE1, PDE2, PDE3, PDE4]:

rif := rifsimp(sys, [[w], [z]], indep = [x,y]);

table( [( Case ) = [[z(x, y)*(8*z(x, y)^2*y^2*x^2-1) = 0, diff(z(x, y), x), "false split"]], ( Solved ) = [w(x, y) = 0, z(x, y) = 0] ] )

(5)
 

 

I am working with an expression in Maple that involves complex terms and an integral. After applying the simplify command, some terms remain unsimplified, even though they seem reducible (see (7)). Additionally, an integral in my expression remains unevaluated (see (9)).
 

restart;

kernelopts(version);

`Maple 2022.0, X86 64 WINDOWS, Mar 8 2022, Build ID 1599809`

(1)

with(plots)

interface(showassumed=0):

assume(x::real);assume(t::real);assume(lambda1::complex);assume(b::real);

alias(psi1 = psi1(x,t), psi2 = psi2(x,t), phi1 = phi1(x,t), phi2 = phi2(x,t), beta = beta(t), alpha =alpha(t));

psi1, psi2, phi1, phi2, beta, alpha

(2)

rel := {psi1 = exp((-I*lambda1)*x - (1/(4*I*lambda1))*int((alpha + b*beta),t)), psi2 = exp((I*lambda1)*x + (1/(4*I*lambda1))*int((alpha + b*beta),t)), phi1= exp((-I*conjugate(lambda1))*x - (1/(4*I*conjugate(lambda1)))*int((alpha + b*beta),t)), phi2 = exp((I*conjugate(lambda1))*x + (1/(4*I*conjugate(lambda1)))*int((alpha + b*beta),t))}

{phi1 = exp(-I*conjugate(lambda1)*x+((1/4)*I)*(int(b*beta+alpha, t))/conjugate(lambda1)), phi2 = exp(I*conjugate(lambda1)*x-((1/4)*I)*(int(b*beta+alpha, t))/conjugate(lambda1)), psi1 = exp(-I*lambda1*x+((1/4)*I)*(int(b*beta+alpha, t))/lambda1), psi2 = exp(I*lambda1*x-((1/4)*I)*(int(b*beta+alpha, t))/lambda1)}

(3)

Bnum := psi2*phi1*conjugate(lambda1) + psi1*lambda1*phi2;

psi2*phi1*conjugate(lambda1)+psi1*lambda1*phi2

(4)

Bnumexp := subs(rel,Bnum):

Den := -phi1*psi2 - phi2*psi1;

-phi1*psi2-phi2*psi1

(5)

expDen := subs(rel, Den)

-exp(-I*conjugate(lambda1)*x+((1/4)*I)*(int(b*beta+alpha, t))/conjugate(lambda1))*exp(I*lambda1*x-((1/4)*I)*(int(b*beta+alpha, t))/lambda1)-exp(I*conjugate(lambda1)*x-((1/4)*I)*(int(b*beta+alpha, t))/conjugate(lambda1))*exp(-I*lambda1*x+((1/4)*I)*(int(b*beta+alpha, t))/lambda1)

(6)

sr := Bnumexp/expDen: ratiosr := simplify(diff(sr,t), complex):

B := b - (4*I/beta(t))*ratiosr

b+2*(b*beta+alpha)*exp(((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*beta+alpha, t)))/conjugate(lambda1))*exp(-((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*beta+alpha, t)))/conjugate(lambda1))*exp(-((1/4)*I)*(4*lambda1^2*x-(int(b*beta+alpha, t)))/lambda1)*(-conjugate(lambda1)+lambda1)^2*exp(((1/4)*I)*(4*lambda1^2*x-(int(b*beta+alpha, t)))/lambda1)/(beta(t)*(exp(((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*beta+alpha, t)))/conjugate(lambda1))*exp(-((1/4)*I)*(4*lambda1^2*x-(int(b*beta+alpha, t)))/lambda1)+exp(-((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*beta+alpha, t)))/conjugate(lambda1))*exp(((1/4)*I)*(4*lambda1^2*x-(int(b*beta+alpha, t)))/lambda1))^2*lambda1*conjugate(lambda1))

(7)

p := {alpha(t) = t^2, beta = exp(-t)}

{beta = exp(-t), alpha(t) = t^2}

(8)

B1 := eval(subs(p, B))

b+2*(b*exp(-t)+alpha)*exp(((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*exp(-t)+alpha, t)))/conjugate(lambda1))*exp(-((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*exp(-t)+alpha, t)))/conjugate(lambda1))*exp(-((1/4)*I)*(4*lambda1^2*x-(int(b*exp(-t)+alpha, t)))/lambda1)*(-conjugate(lambda1)+lambda1)^2*exp(((1/4)*I)*(4*lambda1^2*x-(int(b*exp(-t)+alpha, t)))/lambda1)/((exp(-t))(t)*(exp(((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*exp(-t)+alpha, t)))/conjugate(lambda1))*exp(-((1/4)*I)*(4*lambda1^2*x-(int(b*exp(-t)+alpha, t)))/lambda1)+exp(-((1/4)*I)*(4*conjugate(lambda1)^2*x-(int(b*exp(-t)+alpha, t)))/conjugate(lambda1))*exp(((1/4)*I)*(4*lambda1^2*x-(int(b*exp(-t)+alpha, t)))/lambda1))^2*lambda1*conjugate(lambda1))

(9)

NULL


 

Download simplify.mw

Why does the code below work when I use a standalone "assume" statement but not "assuming"?

That is, why don't the first two attempts at calculating the limit use the assumption contained in those statements, ie why don't those statements return infinity and not a signum like the last attempt at the limit?

`ω__b` := proc (alpha) options operator, arrow; `ω__0`*sqrt(1+alpha+sqrt(alpha+alpha^2)) end proc

proc (alpha) options operator, arrow; omega__0*sqrt(1+alpha+sqrt(alpha+alpha^2)) end proc

(1)

`assuming`([limit(`ω__b`(alpha), alpha = infinity)], [`ω__0`::positive]) = signum(omega__0)*infinityNULL

`assuming`([limit(`ω__b`(alpha), alpha = infinity)], [`ω__0` > 0]) = signum(omega__0)*infinityNULL

NULL

assume(`ω__0` > 0)

limit(`ω__b`(alpha), alpha = infinity) = infinityNULL

NULL

Download assuming.mw

I'd like to simplify the simple expression below by dividing numerator and denominator by m_1 to obtain an expression only containing alpha.

Maple doesn't not generate the desired result.

Is there a way to do this?

expr := sqrt(-(-m__1-m__2+sqrt(m__2*(m__1+m__2)))*`ω__0`^2/m__1)

(-(-m__1-m__2+(m__2*(m__1+m__2))^(1/2))*omega__0^2/m__1)^(1/2)

(1)

simplify(expr, {m__2/m__1 = alpha})

(omega__0^2*(alpha*m__1+m__1-(m__1^2*alpha*(alpha+1))^(1/2))/m__1)^(1/2)

(2)

NULL


Download simplify_side.mw

In the code below, expr is a variable that is a vector times a cosine. Maple performs the multiplication of the vector times the scalar and writes it all as a single vector.

For didactical purposes, I wish for Maple to write out the vector times the cosine, without putting the cosine in the vector.

How is this accomplished?

v := `<,>`(`<|>`(a), `<|>`(b))

Matrix(%id = 36893488152047505644)

(1)

expr := v*cos(t*theta+phi)

Matrix(%id = 36893488152047493236)

(2)

expr := 'v'*cos(t*theta+phi)

v*cos(t*theta+phi)

(3)

NULL

Download expr.mw

How can I obtain the Hamiltonian of Eq. (1) in terms of dynamical variables in Maple?

restart

with(PDEtools)

declare(phi(xi), psi(xi))

phi(xi)*`will now be displayed as`*phi

 

psi(xi)*`will now be displayed as`*psi

(1)

r1 := (1/2*(phi(xi)^2-2*c))*(diff(diff(phi(xi), xi), xi))+((-(1/2)*phi(xi)^2+c)*(k*phi(xi)^2-Omega)^2*(phi(xi)^2-c)^2/(phi(xi)^4*(phi(xi)^2-2*c)^2)-(-k*phi(xi)^2+c*k+Omega)*(k*phi(xi)^2-Omega)*(phi(xi)^2-c)/(phi(xi)^2*(phi(xi)^2-2*c))-(1/2)*k^2*phi(xi)^2+Omega*k+(diff(phi(xi), xi))^2+1)*phi(xi) = 0

(1/2)*(phi(xi)^2-2*c)*(diff(diff(phi(xi), xi), xi))+((-(1/2)*phi(xi)^2+c)*(k*phi(xi)^2-Omega)^2*(phi(xi)^2-c)^2/(phi(xi)^4*(phi(xi)^2-2*c)^2)-(-k*phi(xi)^2+c*k+Omega)*(k*phi(xi)^2-Omega)*(phi(xi)^2-c)/(phi(xi)^2*(phi(xi)^2-2*c))-(1/2)*k^2*phi(xi)^2+Omega*k+(diff(phi(xi), xi))^2+1)*phi(xi) = 0

(2)

r2 := simplify(%)

(-4*phi(xi)^3*(-(1/2)*phi(xi)^2+c)^2*(diff(diff(phi(xi), xi), xi))+(-2*phi(xi)^6+4*phi(xi)^4*c)*(diff(phi(xi), xi))^2-2*phi(xi)^6+(-c^2*k^2+2*Omega*c*k-Omega^2+4*c)*phi(xi)^4+Omega^2*c^2)/(-2*phi(xi)^5+4*c*phi(xi)^3) = 0

(3)

pa := {diff(diff(phi(xi), xi), xi) = diff(psi(xi), xi), diff(phi(xi), xi) = psi(xi)}

{diff(diff(phi(xi), xi), xi) = diff(psi(xi), xi), diff(phi(xi), xi) = psi(xi)}

(4)

r3 := subs(pa, r2)

(-4*phi(xi)^3*(-(1/2)*phi(xi)^2+c)^2*(diff(psi(xi), xi))+(-2*phi(xi)^6+4*phi(xi)^4*c)*psi(xi)^2-2*phi(xi)^6+(-c^2*k^2+2*Omega*c*k-Omega^2+4*c)*phi(xi)^4+Omega^2*c^2)/(-2*phi(xi)^5+4*c*phi(xi)^3) = 0

(5)

psixif := isolate(r3, diff(psi(xi), xi))

diff(psi(xi), xi) = -(1/4)*(-(-2*phi(xi)^6+4*phi(xi)^4*c)*psi(xi)^2+2*phi(xi)^6-(-c^2*k^2+2*Omega*c*k-Omega^2+4*c)*phi(xi)^4-Omega^2*c^2)/(phi(xi)^3*(-(1/2)*phi(xi)^2+c)^2)

(6)

NULL

Download Hamiltonian.mw

How to integrate eq (4)? Since 'a', 'b', and 'c' are constant. 

restart

with(DEtools)

declare(z(x), y(x))

declare(z(x), y(x))

(1)

eq1 := (1/2)*(-z(x)^3-2*c*z(x))*(diff(diff(y(x), x), x))-((z(x)^2+2*c)*(diff(y(x), x))+b*z(x)^2+c*k+a)*(diff(z(x), x)) = 0

(1/2)*(-z(x)^3-2*c*z(x))*(diff(diff(y(x), x), x))-((z(x)^2+2*c)*(diff(y(x), x))+b*z(x)^2+c*k+a)*(diff(z(x), x)) = 0

(2)

eq2 := simplify(z(x)*eq1)

-z(x)*(z(x)*((1/2)*z(x)^2+c)*(diff(diff(y(x), x), x))+((z(x)^2+2*c)*(diff(y(x), x))+b*z(x)^2+c*k+a)*(diff(z(x), x))) = 0

(3)

eq3 := eval(int(lhs(eq2), x))

int(-z(x)*(z(x)*((1/2)*z(x)^2+c)*(diff(diff(y(x), x), x))+((z(x)^2+2*c)*(diff(y(x), x))+b*z(x)^2+c*k+a)*(diff(z(x), x))), x)

(4)

NULL

Download integration.mw

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 = `((&PartialD;)/(&PartialD; t) - (&PartialD;)/(&PartialD; 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

 

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