MaplePrimes Questions

I am studying linear operators that have vanishing Nijenhuis Torsion, that is a (1-1) tensor L whose corresponding (1-2) tensor N given by a tensor equation of L (see attached) is identically zero.

I am new to maple, i have used it to plot vector fields and solve systems of equations in the past but i am unfamiliar with the DifferentialGeometry and Physics packages.

Attached is my best effort at solving this problem directly for a simple 2d case without the use of any packages and i am wondering if it is possible to do it all in one line without having to define tensor components one by one.

nijenhuis_torsion.mw

I just realized that voting for a deleted answer is possible. This does not make sense. Does an upvote also change the reputation?

You can try it in the answer to myself below, but please do not upvote permanetly for this question.
 

How can I use Maple to solve a system of nonlinear equations symbolically and display the steps in the solution?

Here are two systems over the reals:

sys__1:=And(r*(387*r+52)+2<r*(226*q+121*s)+9*q*(q*(2*q-5)-3*s+2)+6*s,4*q^3+r*(27*r+4)+s^2=q*(q+18*r),q>=0,r>=0):
sys__2:=And((392-1739*q)*r+4*(2-9*q)**2+2151*r**2<75*r*s,4*q**3+r*(27*r+4)+s**2=q*(q+18*r),q>=0,r>=0):

The following results indicate that both and are satisfiable 

QuantifierElimination:-QuantifierEliminate(exists([s,q,r],sys__1));
                              true
QuantifierElimination:-QuantifierEliminate(exists([s,q,r],sys__1));
                              true

but RealDomain:-solve simply returns an empty list (that is, no solution exists) in both cases

RealDomain:-solve(sys__1,[q,s,r]); # ⟹ sys1 cannot be satisfied
                               []
RealDomain:-solve(sys__2,[q,s,r]); # ⟹ sys2 cannot be satisfied
                               []

As discussed in the previous problem, in contrast to using QuantifierElimination:-QuantifierEliminate, the use of RealDomain:-solve is unsafe. Nevertheless, the above output suggests that even the much-more-sophisticated QuantifierElimination:-QuantifierEliminate is still not always reliable (since the correct returnedvalue appears to be in lieu of ). So, what is the right command to handle polynomial systems over real domains in Maple? 

Can i get this ode in a "standardform"  ?

verg:= (-delta*eta^2 + alpha*eta)*diff(diff(U(xi), xi), xi) - U(xi)*(2*eta*gamma*theta*(delta*eta - alpha)*U(xi)^2 + eta^2*delta*k^2 + (-alpha*k^2 - 2*delta*k)*eta + 2*k*alpha + delta) = 0;

Since the puzzle task "A circle is to be disturbed ..." makes no fun, here is a Maple task:
The term to be simplified step by step:
(2+10/(3*sqrt(3)))^(1/3)+(2-10/(3*sqrt(3)))^(1/3)

A circle with radius r is given. The circle should be divided into three
parts of equal area such that the circumference of each part is equal to the circumference of the circle.
(Note: There are several solutions.)

I have learned that the SPECTRA.mla library can solve SDP problems. I have tried to download and use it, but I am still missing the FGb module

> with(SPECTRA);
> M := Matrix(6, 6, [[10, 1, 0, m[1], -m[3], m[2]], [1, -2*m[1]+27, m[3], -27/2, -m[4], -m[5]], [0, m[3], -2*m[2], m[4], m[5], 0], [m[1], -27/2, m[4], 10, 0, m[6]], [-m[3], -m[4], m[5], 0, -2*m[6], 0], [m[2], -m[5], 0, m[6], 0, 1]]);
> SolveLMI(M);
=> Error, (in SPECTRA:-SolveLMI) `FGb` does not evaluate to a module

 

How can I resolve this issue on a Windows environment, beause I don't see install file for Windows, only MacOS and Linux:

We are working to obtain a fully symbolic dynamic model for a robot. 

Using the CPU (I9-12900K) and 128 GB of DDR5 RAM (5600 MHZ) did not compute a 7x7 Inverse of a symbolic matrix

Is is possible to exploit the CUDA functions to compute it on the GPU? I have a NVidia RTX A6000 (48 GB of DDR6 GPU memory)

I tried this: 

CUDA:-Enable(true)

CUDA: -MatrixInverse(D_Q):

But it does not use the GPU to compute this. 

Maybe I'm doing smth wrong. 

Thank you, 

Calin

Hi

Dear friends, I am a relatively new user and I have a problem in entering and calculating the 2F1 hypergeometric function. My question is how to enter this function in Maple in equations so that Maple recognizes it? Because hypergeom ([1], [2], [3]) Maple itself is a 3-element function, while 2F1 hypergeometric ([1], [2], [3], [4]) is a four-element!

Almost i did 10 method for this ode equation all of them are succes but this one is giving me some confusing and i am looking for  get my answer, the mothod say if we have the auxilary equation if substitute the solution of this auxilary equation in our series solution then substitute in ode equation must be satisfy but it is not satisfy so when he did assumption for the auxilary equation he say it satisfy if we sabstitute this assumption in our series solution!

My question is this how we get thus assumption ? and why finding exact  solution of auxilary equation not satisfy?

restart

with(PDEtools)

with(LinearAlgebra)

with(Physics)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

NULL

Fode := (-delta*eta^2+alpha*eta)*(diff(diff(U(xi), xi), xi))-U(xi)*(2*eta*gamma*theta*(delta*eta-alpha)*U(xi)^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

(-delta*eta^2+alpha*eta)*(diff(diff(U(xi), xi), xi))-U(xi)*(2*gamma*eta*theta*(delta*eta-alpha)*U(xi)^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

(2)

NULL

F := sum(a[i]*G(xi)^i, i = 0 .. 1)

a[0]+a[1]*G(xi)

(3)

``

(4)

D1 := diff(F, xi)

a[1]*(diff(G(xi), xi))

(5)

NULL

S := (diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

(diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

(6)

S1 := diff(G(xi), xi) = sqrt(G(xi)^4+A[2]*G(xi)^2+A[1])

diff(G(xi), xi) = (G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(7)

E1 := subs(S1, D1)

a[1]*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(8)

D2 := diff(E1, xi)

(1/2)*a[1]*(4*G(xi)^3*(diff(G(xi), xi))+2*A[2]*G(xi)*(diff(G(xi), xi)))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(9)

E2 := subs(S1, D2)

(1/2)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(10)

K := U(xi) = F

U(xi) = a[0]+a[1]*G(xi)

(11)

K1 := diff(U(xi), xi) = E1

diff(U(xi), xi) = a[1]*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(12)

K2 := diff(U(xi), xi, xi) = E2

diff(diff(U(xi), xi), xi) = (1/2)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(13)

L := eval(Fode, {K, K1, K2})

(1/2)*(-delta*eta^2+alpha*eta)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)-(a[0]+a[1]*G(xi))*(2*gamma*eta*theta*(delta*eta-alpha)*(a[0]+a[1]*G(xi))^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0

(14)

L1 := normal((1/2)*(-delta*eta^2+alpha*eta)*a[1]*(4*G(xi)^3*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)+2*A[2]*G(xi)*(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2))/(G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)-(a[0]+a[1]*G(xi))*(2*gamma*eta*theta*(delta*eta-alpha)*(a[0]+a[1]*G(xi))^2+eta^2*delta*k^2+(-alpha*k^2-2*delta*k)*eta+2*k*alpha+delta) = 0)

 

collect(L1, {G(xi)})

(-2*delta*eta^2*gamma*theta*a[1]^3+2*alpha*eta*gamma*theta*a[1]^3-2*delta*eta^2*a[1]+2*alpha*eta*a[1])*G(xi)^3+(-6*delta*eta^2*gamma*theta*a[0]*a[1]^2+6*alpha*eta*gamma*theta*a[0]*a[1]^2)*G(xi)^2+(-6*delta*eta^2*gamma*theta*a[0]^2*a[1]+6*alpha*eta*gamma*theta*a[0]^2*a[1]-delta*eta^2*k^2*a[1]+alpha*eta*k^2*a[1]-delta*eta^2*A[2]*a[1]+alpha*eta*A[2]*a[1]+2*delta*eta*k*a[1]-2*alpha*k*a[1]-delta*a[1])*G(xi)-2*gamma*delta*eta^2*theta*a[0]^3+2*gamma*alpha*eta*theta*a[0]^3-delta*eta^2*k^2*a[0]+alpha*eta*k^2*a[0]+2*delta*eta*k*a[0]-2*alpha*k*a[0]-delta*a[0] = 0

(15)

eq0 := -2*delta*eta^2*gamma*theta*a[0]^3+2*alpha*eta*gamma*theta*a[0]^3-delta*eta^2*k^2*a[0]+alpha*eta*k^2*a[0]+2*delta*eta*k*a[0]-2*alpha*k*a[0]-delta*a[0] = 0

eq1 := -6*delta*eta^2*gamma*theta*a[0]^2*a[1]+6*alpha*eta*gamma*theta*a[0]^2*a[1]-delta*eta^2*k^2*a[1]+alpha*eta*k^2*a[1]-delta*eta^2*A[2]*a[1]+alpha*eta*A[2]*a[1]+2*delta*eta*k*a[1]-2*alpha*k*a[1]-delta*a[1] = 0

eq2 := -6*delta*eta^2*gamma*theta*a[0]*a[1]^2+6*alpha*eta*gamma*theta*a[0]*a[1]^2 = 0

eq3 := -2*delta*eta^2*gamma*theta*a[1]^3+2*alpha*eta*gamma*theta*a[1]^3-2*delta*eta^2*a[1]+2*alpha*eta*a[1] = 0

COEFFS := solve({eq0, eq1, eq2, eq3}, {alpha, eta, a[0], a[1]}, explicit)

case1 := COEFFS[4]

{alpha = delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k), eta = eta, a[0] = 0, a[1] = 1/(-gamma*theta)^(1/2)}

(16)

NULL

S

(diff(G(xi), xi))^2 = G(xi)^4+A[2]*G(xi)^2+A[1]

(17)

S1

diff(G(xi), xi) = (G(xi)^4+A[2]*G(xi)^2+A[1])^(1/2)

(18)

S2 := dsolve(S, G(xi))

G(xi) = -(1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2), G(xi) = (1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2), G(xi) = -(1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2), G(xi) = (1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2), G(xi) = JacobiSN((1/2)*(2*(A[2]^2-4*A[1])^(1/2)-2*A[2])^(1/2)*xi+c__1, (-2*(A[2]*(A[2]^2-4*A[1])^(1/2)-A[2]^2+2*A[1])*A[1])^(1/2)/(A[2]*(A[2]^2-4*A[1])^(1/2)-A[2]^2+2*A[1]))*A[1]*2^(1/2)/(A[1]*(-A[2]+(A[2]^2-4*A[1])^(1/2)))^(1/2)

(19)

K

U(xi) = a[0]+a[1]*G(xi)

(20)

K4 := subs(case1, K)

U(xi) = G(xi)/(-gamma*theta)^(1/2)

(21)

NULL

K5 := subs(S2, K4)

U(xi) = -(1/2)*(-2*A[2]-2*(A[2]^2-4*A[1])^(1/2))^(1/2)/(-gamma*theta)^(1/2)

(22)

NULL

F2 := subs(case1, Fode)

(-eta^2*delta+delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)*eta/(eta*k^2+eta*A[2]-2*k))*(diff(diff(U(xi), xi), xi))-U(xi)*(2*gamma*eta*theta*(delta*eta-delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k))*U(xi)^2+eta^2*delta*k^2+(-k^2*delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k)-2*k*delta)*eta+2*k*delta*(eta^2*k^2+eta^2*A[2]-2*eta*k+1)/(eta*k^2+eta*A[2]-2*k)+delta) = 0

(23)

``

(24)

NULL

odetest(K5, F2)

-(1/2)*delta*eta*(A[2]^2-4*A[1])^(1/2)*(-2*(A[2]+(A[2]^2-4*A[1])^(1/2))/gamma)^(1/2)/((eta*k^2+eta*A[2]-2*k)*(-theta)^(1/2))

(25)


and i hope mapleprimes don't delete this question becuase of this pictures also it help for undrestanding

 

there is other picture for different auxilary equation just  add one multiply term for G(xi)^4 in case anyone needed i will upload

Download odetest.mw

how fixed that?

NULL

restart

NULL

A := Vector[row]([4, 0, 5])

Vector[row](%id = 36893490313998292860)

(1)

B := Vector[row]([3, 1, 0])

Vector[row](%id = 36893490313998279372)

(2)

C := Vector[row]([2, 1, 1])

Vector[row](%id = 36893490313998273820)

(3)

C.`&x`(A, B)

9

(4)

NULL

Download we.mw

In many papers, I've noticed that the solution to an ODE (ordinary differential equation) often emerges directly when there's only a single function involved. My question is: is there a way to generate solutions to an ODE by producing specific parameters?

 

restart

with(PDEtools)

with(LinearAlgebra)

with(Physics)

with(SolveTools)

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)

NULL

S := diff(F(xi), xi) = sqrt(P*F(xi)^4+Q*F(xi)^2+R)

diff(F(xi), xi) = (P*F(xi)^4+Q*F(xi)^2+R)^(1/2)

(2)

S1 := dsolve(S, F(xi))

xi-Intat(1/(P*_a^4+Q*_a^2+R)^(1/2), _a = F(xi))+c__1 = 0

(3)

S2 := (diff(F(xi), xi))^2 = P*F(xi)^4+Q*F(xi)^2+R

(diff(F(xi), xi))^2 = P*F(xi)^4+Q*F(xi)^2+R

(4)

S3 := dsolve(S2, F(xi))

F(xi) = -(1/2)*(-2*P*(Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = (1/2)*(-2*P*(Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = -(1/2)*2^(1/2)*(P*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = (1/2)*2^(1/2)*(P*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)/P, F(xi) = JacobiSN((1/2)*(2*(-4*P*R+Q^2)^(1/2)-2*Q)^(1/2)*xi+c__1, (-2*(Q*(-4*P*R+Q^2)^(1/2)+2*R*P-Q^2)*R*P)^(1/2)/(Q*(-4*P*R+Q^2)^(1/2)+2*R*P-Q^2))*R*2^(1/2)/(R*(-Q+(-4*P*R+Q^2)^(1/2)))^(1/2)

(5)
 

NULL

Download get_all_solution_of_ode_by_generation.mw

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