Maple 2022 Questions and Posts

How to check solutions to a first-order system of ...

Asked by: zenterix 405

February 14 2025

0 2

The document below is long because I go through the context of my question step-by-step.

The context is solving a first-order system of differential equations where we have complex eigenvalues.

But essentially, my question has to do with the end of this reasoning. I find four solutions x1, x2, x3, and x4, and I would like to check whether they are indeed solutions by subbing them into the original system x'=Ax.

When we make such a substitution, we get an equation of the form yl=yr, where yl and yr are 4x1 vectors. If the equality is true then we can say we have a solution.

My question is about how to check for this equality. I amusing Maple's "Equals" but this doesn't seem to work for all four solutions.

I explain the entire reasoning that leads to the final vectors that I compare

The following matrix has complex eigenvalues , each with multiplicity 2.

Matrix(%id = 36893488151974184588)

(1)

Consider the eigenvalue . The associated eigenvectors are

Vector[column](%id = 36893488151974164828)

(2)

We see this gives us only one linearly independent eigenvector.

We can find another one by forming a rank 2 generalized eigenvector from one of the eigenvectors above.

The pair of such eigenvectors is a chain of generalized eigenvectors of length 2 and they satisfy

(A-I*`λ__1`)*w__2 = w__1*(A-I*`λ__1`)*w__1 and w__1*(A-I*`λ__1`)*w__1 = 0

These equations come from trying the solution and , where is an eigenvector for .

The equations above lead to

which we can solve for the generalized eigenvector as follows

Vector[column](%id = 36893488151974138212)

(3)

One such generalized eigenvector is

$w__2 := subs({c[2] = I, c[4] = I}, `w__2,gen`)$

Vector[column](%id = 36893488151982985812)

(4)

which gives us

Vector[column](%id = 36893488151982979180)

(5)

which is indeed an eigenvector for .

Thus, we have the two complex solutions

Vector[column](%id = 36893488151982971708)

(6)

Vector[column](%id = 36893488151982963516)

(7)

But we want real solutions.

Vector[column](%id = 36893488151974145788)

(8)

Vector[column](%id = 36893488151982991948)

(9)

Vector[column](%id = 36893488151982977980)

(10)

Vector[column](%id = 36893488151982967252)

(11)

Let's check that each of the vectors above (the real and imaginary parts of our complex solutions) are real solutions.

I would like to substitute, say, into the first order system of differential equations .

At first I tried simply

$subs({x = x__1}, diff(x, t) = A.x)$

0 = Matrix(%id = 36893488151974184588).Vector[column](%id = 36893488151982948468)

(12)

This doesn't work because the command evaluates to zero before we do the desired substitution.

Then I tried

$subs({x = x__1}, 'diff(x, t)' = A.x)$

diff(Vector[column](%id = 36893488151982948468), t) = Matrix(%id = 36893488151974184588).Vector[column](%id = 36893488151982948468)

(13)

$result := eval(subs({x = x__1}, 'diff(x, t)' = A.x))$

Vector[column](%id = 36893488151991655348) = Vector[column](%id = 36893488151982967612)

(14)

By inspection, it seems the two sides are equal and so indeed satisfies the system.

But how do I get maple to tell me this?

(15)

(16)

Okay, but does this use of always work?

$sols := [x__1, x__2, x__3, x__4]; for i to 4 do result := eval(subs({x = sols[i]}, 'diff(x, t)' = A.x)); print(simplify(result))*print(Equal(rhs(result), lhs(result))) end do$

(17)

Apparently not.

What is going on here?

It sure looks like the righthand and lefthand sides of each expression are the same.

Download evs_equal.mw

Reduced Involutive Form...

Asked by: dds 20

February 13 2025

2 8

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:

I 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 := ydiff(z(x,y), x$3) - xdiff(z(x,y),x,y$2) = diff(z(x,y),y$2) - y*diff(z(x,y), y);

(1)

>	PDE2 := 2xydiff(z(x,y),x$3)diff(z(x,y),x,y$2) + x(diff(z(x,y),x$3))^2 + xy^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 := ydiff(z(x,y),x,y$2) - xw(x,y) + 2x^2y*z(x,y)^2 = 0;

(3)

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

(4)

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

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

(5)

Why Doesn’t simplify Fully Simplify My Expression ...

Asked by: FZ 30

February 10 2025

1 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);

(1)

>

>	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));

(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 := psi2phi1conjugate(lambda1) + psi1lambda1phi2;

(4)

>	Bnumexp := subs(rel,Bnum):

>	Den := -phi1psi2 - phi2psi1;

(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 - (4I/beta(t))ratiosr

(7)

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

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

(8)

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

(9)

Download simplify.mw

Why does this limit work using "assume" but not "a...

Asked by: zenterix 405

February 05 2025

2 1

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?

Download assuming.mw

How to simplify an expression by substituting a ra...

Asked by: zenterix 405

February 05 2025

1 3

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?

Download simplify_side.mw

How to get Maple to write vector times scalar with...

Asked by: zenterix 405

February 05 2025

1 2

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?

Download expr.mw

How can I obtain the Hamiltonian?...

Asked by: FZ 30

February 05 2025

0 9

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

>

(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)

>

(-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)

>

(4)

>

(-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)

>

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)

Download Hamiltonian.mw

How to integrate it?...

Asked by: Ahmed111 140

February 03 2025

0 2

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

Download integration.mw

How to Separate Real and Imaginary Parts?...

Asked by: FZ 30

February 02 2025

1 5

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

Unknown vector with given unit vectors ...

Asked by: romanrieme 55

January 31 2025

1 4

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

How to simplify Eqs. (3), (5) and (6) and write in...

Asked by: FZ 30

January 31 2025

0 5

>

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

(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

(2)

>

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

>

(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

(4)

>

((-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)

>

((-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)

>

Download BE.mw

`How to calculate u = `((∂)/(∂ t...

Asked by: FZ 30

January 25 2025

1 10

`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))

(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);

(2)

>

NULL

Download integ.mw

Vector calculus with list/function as input...

Asked by: romanrieme 55

January 23 2025

1 2

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?...

Asked by: Rebecca Awerigiya 5

January 20 2025

1 2

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.

(1)

(2)

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

(3)

(4)

(5)

(6)

[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)

(8)

(9)

(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)]]