Maple 2022 Questions and Posts

Approximation of ODEs with cubic splines

by: Kiefer Gerhard 10 Product: Maple 2022

April 06 2025

0 0

Approximation_of_ODEs_with_Cubic_Splines.mw

Show steps for viscous Burger's equation...

Asked by: 130

March 11 2025

0 2

Hi!

I am studying Burger's equation, and I would like to see the steps that Maple takes to solve this. "ShowSteps" doesn't seem to work.

Unfortunately, I am unable to share the worksheet I made.

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How to simplify the arguments of the exponential?...

Asked by: FZ 30

March 07 2025

0 4

How do we simplify the arguments of the exponential in (1)? Further how to express (1) into hyperbolic/trig functions?

>

(1)

>

(2)

>

(3)

>

(4)

>

(5)

>

Download argument.mw

How to perform simplifications to find unknown con...

Asked by: zenterix 400

March 03 2025

2 5

I am taking a course in quantum mechanics and trying to do the calculations in Maple.

In the worksheet below I try to solve a system of equations to find some constants.

It isn't very relevant that the domain is physics.

The issues I have are with manipulations of expressions (specifically, simplifications).

Levine - Ch. 2.4 - Particle in a Rectangular Well

Define

s__1 := sqrt(2*m*(V__0-E))/`&hbar;` = 2^(1/2)*(m*(V__0-E))^(1/2)/`&hbar;`

= -2^(1/2)*(m*(V__0-E))^(1/2)/`&hbar;`

(1)

and are wavefunctions.

=

I would also like to assume that and cannot both be zero simultaneously.

As can be seen above, there are four unknown constants, and .

We can obtain values for these unknowns by imposing boundary conditions.

Some of the boundary conditions involve the first derivatives of the wavefunctions.

=

The boundary conditions are

`ψ__1`(0) = `ψ__2`(0)*`ψ__2`(l) and `ψ__2`(0)*`ψ__2`(l) = `ψ__3`(l)*(D(`ψ__1`))(0) and `ψ__3`(l)*(D(`ψ__1`))(0) = (D(`ψ__2`))(0)*(D(`ψ__2`))(l) and (D(`ψ__2`))(0)*(D(`ψ__2`))(l) = (D(`ψ__3`))(l)

My question is how to find the constants in Maple.

In this problem, and .

Solving the first boundary condition is easy.

$expr1 := solve(`ψ__1`(0) = `ψ__2`(0), {C})$ =

Next, I solve the third boundary condition and sub in the result from solving the first boundary condition. We get in terms of .

$expr3 := eval(solve(`ψ__1,d`(0) = `ψ__2,d`(0), {B}), expr1)$ = ${B = A*(m*(V__0-E))^(1/2)/(m*E)^(1/2)}$

Already here it is not clear to me why Maple does not cancel the 's.

Next, consider the second boundary condition

$expr2 := solve(`ψ__2`(l) = `ψ__3`(l), {G})$ = ${G = (A*cos(s*l)+B*sin(s*l))/exp(s__2*l)}$

We can obtain in terms of just by subbing in previously found relationships, as follows

${G = exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A*(sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2)+cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2))/E^(1/2)}$

(4)

Finally, consider the fourth boundary condition.

$solve(`ψ__2,d`(l) = `ψ__3,d`(l), {G})$

${G = (m*E)^(1/2)*(A*sin(2^(1/2)*(m*E)^(1/2)*l/`&hbar;`)-B*cos(2^(1/2)*(m*E)^(1/2)*l/`&hbar;`))/((m*(V__0-E))^(1/2)*exp(-2^(1/2)*(m*(V__0-E))^(1/2)*l/`&hbar;`))}$

(5)

$expr4 := simplify(subs(expr3, solve(`ψ__2,d`(l) = `ψ__3,d`(l), {G})))$

${G = (sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2)-cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2))*exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A/(V__0-E)^(1/2)}$

(6)

${exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A*(sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2)+cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2))/E^(1/2) = (sin(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*E^(1/2)-cos(2^(1/2)*m^(1/2)*E^(1/2)*l/`&hbar;`)*(V__0-E)^(1/2))*exp(2^(1/2)*m^(1/2)*(V__0-E)^(1/2)*l/`&hbar;`)*A/(V__0-E)^(1/2)}$

(7)

At this point, we have this complicated expression.

The calculations above come from the book "Quantum Chemistry" by Levine. The variable "result" above should, after a few more steps of manipulating the expression, give us

(8)

I've done the calculation by hand. It is not difficult. The exponentials and the cancel on each side and we are left with an equation involving and which easily comes out to the desired equation above.

I would like to be able to do it here in Maple.

The first thing I would do at this point is to start eliminating some of the clutter. For example, I would like to replace with .

This is a first task (that I have asked about in a separate question).

Then, I would want Maple to do the cancellations for me, but I have already used "simplify".

Finally, I think I would like to collect terms involving and .

Here is what happens when I try to get Maple to solve the equations for me in one go

$solve({`ψ__1`(0) = `ψ__2`(0), `ψ__1,d`(0) = `ψ__2,d`(0), `ψ__2`(l) = `ψ__3`(l), `ψ__2,d`(l) = `ψ__3,d`(l)}, {A, B, C, G})$

(9)

Download solve_constants.mw

How to simplify factors that are square roots into...

Asked by: zenterix 400

March 03 2025

0 2

In a calculation, I obtained the following expression from Maple.

>	expr1:=Asqrt(-m(-V__0 + E))/sqrt(m*E)

A*(-m*(-V__0+E))^(1/2)/(m*E)^(1/2)

(1)

The first question I have is why the

Next, consider the expression

>	expr2:=exp(sqrt(2)sqrt(m)sqrt(-V__0 + E)lI/`&hbar;`)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(2)

>	simplify(expr2,[E-V__0=y])

(3)

>	simplify(expr2,[2*m=y])

exp(I*y^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(4)

Why doesn't the simplification below work?

>	simplify(expr2,[2m(E-V__0)=y])

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(5)

I also tried with other commands

>	eval(expr2,2*m=y)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(6)

>	eval(expr2,E-V__0=y)

(7)

>	eval(expr2,2m(E-V__0)=y)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(8)

>	algsubs(2*m=y,expr2)

exp(I*y^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(9)

>	algsubs(E-V__0=y,expr2)

(10)

>	algsubs(2m(E-V__0)=y,expr2)

exp(I*2^(1/2)*m^(1/2)*(-V__0+E)^(1/2)*l/`&hbar;`)

(11)

In reality, I would like to the entirety of sqrt(2m(E-V__0)/hbar) a variable by the name of s.

Download simplify_roots.mw

Why does this procedure definition work when I nam...

Asked by: zenterix 400

March 03 2025

0 2

Below is a simple demonstration of something that I am very confused about in another worksheet I am working on.

Basically, I have a function of x called psi_I and I want to define another function of x called psi_I,d which is just the derivative of psi_I.

In, the snippet below, this works if I use the variable names psi_1 and psi_1,d, but not if I replace the "1" with a "I" as I want to do.

In addition, I don't know why the output (6) is different from (7) or (8). All three are from the same command in Maple.

Download psi_subscript.mw

eigenvector equation discrepancies...

Asked by: zenterix 400

February 16 2025

1 2

Matrix(%id = 36893488151921005020)

(1)

Matrix(%id = 36893488151921002244)

(2)

[Vector[column](%id = 36893488151920990196), Matrix(%id = 36893488151920990316)]

(3)

Below is an eigenvector equation for for the eigenvalue -3.4376...

Matrix(%id = 36893488151920987180)

(4)

When I solve the system below, I expect to get the vector

Vector[column](%id = 36893488151920981284)

(5)

, just like the result of the Eigenvectors command used above.

Vector[column](%id = 36893488151920975988)

(6)

Instead I get the zero vector.

In addition, when I compute the reduced row echelon form, I expect to have one of the rows be zero since matrix S has rank 1.

(7)

Matrix(%id = 36893488151959135044)

(8)

Therefore, I guess my question is maybe about calculations and not about Maple. That being said, I did the calculations manually, then I asked chatgpt, which agreed with me.

Why are the calculations not coinciding with Maple?

I can do the calculations "manually" with Maple:

(9)

(10)

Now, if I try to solve them simultaneously, it seems I have the same issue as with LinearSolve

(11)

I tried something else that worked, but the questions above still remain.

We start with again

= Matrix(%id = 36893488151921005020)

In my problem I have the variable

`ω__a` := (1/2)*`ω__0`*(sqrt(5)-1) =

and it turns out that is an eigenvalue of .

In particular, for we get the values I showed previously.

For example, here is the eigenvalue we were looking at before

(12)

Okay so the eigenvector equation without subbing in a value for is .

Here is what the matrix on the left looks like

Matrix(%id = 36893488151964015780)

(13)

And here we solve for the eigenvectors

Vector[column](%id = 36893488151964001572)

(14)

Vector[column](%id = 36893488151964010244)

(15)

Which is the expected result.

So, Maple can solve the eigenvector equation in this more general case but not if I specify before asking it to solve.

Download evcalc.mw

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

Asked by: zenterix 400

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 400

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 400

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 400

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)