dharr

URL...

Answered: dharr 8977

Yesterday at 3:37 PM

0 1

The following URL finds my Orbitals package

https://www.maplesoft.com/Applications/Detail.aspx?id=4865

but substituting your ID doesn't find anything, suggesting that the ID is no longer valid or the application is not there.

Code edit region...

Answered: dharr 8977

February 16 2026

2 3

It is much easier to use a code edit region. But you can do it your intended way with EvalString instead of Run. GetVariable returns the variable to Maple. Here are both ways:

python.mw

untrappable...

Answered: dharr 8977

February 16 2026

2 0

As the help page for try says, the "too many levels of recursion" error is one of the few that can't be caught.

nonsense...

Answered: dharr 8977

February 16 2026

2 3

The paper doesn't make much sense.

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For solution 33

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-2^(1/2)*alpha*sech(epsilon*t-x)/beta

The integral is independent of epsilon, as expected

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U := (1/2)*(int(u^2, x = -infinity .. infinity))

2*alpha^2/beta^2

And so its derivative wrt epsilon is zero. The value given in the paper is just a numerical approximation of zero

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dU := diff(U, epsilon); eval(dU, epsilon = -1)

For solution 31

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u := sqrt((-eps^2+1)*lambda^2)*((2*(m*(m+lambda)-mu))/(lambda*(m+(A2-A1*sqrt(mu)-A2*sqrt(mu)*(-eps*t+x))/(A1+A2*(-eps*t+x))))-1)/(sqrt(2)*beta)

(1/2)*((-eps^2+1)*lambda^2)^(1/2)*2^(1/2)*(2*(m*(m+lambda)-mu)/(lambda*(m+(A2-A1*mu^(1/2)-A2*mu^(1/2)*(-eps*t+x))/(A1+A2*(-eps*t+x))))-1)/beta

Complex values for plot if we use eps=-2 because of the factor. For eps = 0.1:

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Integral is infinite

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U := (1/2)*(Int(u^2, x = -infinity .. infinity)); value(%)

signum((4*mu^(3/2)*lambda-2*mu^(1/2)*m*lambda^2-4*mu^(1/2)*m^2*lambda-m^2*lambda^2-4*m^3*lambda-4*m^4-mu*lambda^2+4*m*mu*lambda+8*m^2*mu-4*mu^2)*(eps-1)*(eps+1)/(beta^2*(mu^(1/2)-m)^2))*infinity+(1/2)*piecewise(Im((-A2*mu^(1/2)*eps*t+A2*eps*m*t+A1*mu^(1/2)-A1*m-A2)/(A2*(mu^(1/2)-m))) = 0, -signum((lambda*m+m^2-mu)^2*(eps-1)*(eps+1)/(beta^2*(mu^(1/2)-m)^4))*infinity, 0)

If we integrate under the integral, the result is still infinite.

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(1/2)*piecewise(Im((-A2*mu^(1/2)*eps*t+A2*eps*m*t+A1*mu^(1/2)-A1*m-A2)/(A2*(mu^(1/2)-m))) = 0, undefined, signum(eps*(4*mu^(3/2)*lambda-2*mu^(1/2)*m*lambda^2-4*mu^(1/2)*m^2*lambda-m^2*lambda^2-4*m^3*lambda-4*m^4-mu*lambda^2+4*m*mu*lambda+8*m^2*mu-4*mu^2)/(beta^2*(mu^(1/2)-m)^2))*infinity)

Try some numerical values over the suggested numerical x range

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(1/2)*(Int(2*(-eps^2+1)*(1/(2+(1+2*eps-2*x)/(1-2*eps+2*x))-1)^2, x = -20 .. 20))

Result is still infinity.

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8*(Int((2+(-3-2*x)*eps)/(3-2*eps+2*x)^3, x = -20 .. 20))

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Download S-test.mw

sime...

Answered: dharr 8977

February 15 2026

3 2

It is one of the HTML symbols. If you enter it in 1D as `&sime;` then select the output, right-click and choose comtext panel, one of the options is add to favourites.

assuming positive...

Answered: dharr 8977

February 11 2026

0 5

With inttrans, I can often get an answer by assuming positive, even though this might not be the case. One can then check if it makes sense; not sure about that in this case. I didn't see any intermittent issues.

Download fourier.mw

quadratic...

Answered: dharr 8977

January 27 2026

2 12

[Updated. I've realised C simplifies to zero, so it can be simplified further. There are different cases for zero or positive A, and zero, positive or negative B.

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ineq as output is in the form lhs < rhs, so let's change that to rhs - lhs > 0, and leave off the > 0, i.e., we want to find t that makes q positive, under the assumption that all the named quantities are positive.

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In the form A*t^2+B*t+C, A is explicitly nonnegative (whole expression is squared). Do you know if A can be zero?

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And C happens to be zero, which simplifies things.

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So this is t*(A*t+B). Writing A=alpha and B = beta

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piecewise(And(alpha = 0, 0 < beta), [[0 < t]], And(alpha = 0, beta < 0), [[t < 0]], And(0 < alpha, beta = 0), [[t <> 0]], And(0 < alpha, 0 < beta), [[0 < t], [t < -beta/alpha]], And(0 < alpha, beta < 0), [[t < 0], [-beta/alpha < t]], [])

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Download Q_solve2.mw

Older:

Q_solve.mw

GA method...

Answered: dharr 8977

January 09 2026

3 1

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Eq 3.3 doi: 10.3934/math.2024424

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Eq 3.16

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S := V(xi) = sum((alpha[i]+beta[i]*(diff(G(xi), xi))^i)/G(xi)^i, i = 0 .. 2)

V(xi) = alpha[0]+beta[0]+(alpha[1]+beta[1]*(diff(G(xi), xi)))/G(xi)+(alpha[2]+beta[2]*(diff(G(xi), xi))^2)/G(xi)^2

Eq. 2.8

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Eq. 2.9, n=2,3,4

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Aode2 := (diff(Aode, xi))/(2*(diff(G(xi), xi))); Aode3 := diff(Aode2, xi); diff(Aode3, xi); Aode4 := lhs(%) = eval(rhs(%), Aode2)

Eq 2.10

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G(xi) = kappa*ln(A)*A^xi+(1/4)*sigma/(kappa*ln(A)*A^xi)

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We haven't used Aode yet, so we don't have any sigmas

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Doesn't seem to be a polynomial in dG/G^2 as claimed:

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We get 10, not 8 equations this way.

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${2*a[1]*alpha[1]*beta[1], 2*a[1]*alpha[2]*beta[1], 2*ln(A)^2*rho*sigma*alpha[1]*b[1]-2*ln(A)^2*sigma*a[1]*alpha[1]*beta[2]+2*a[1]*alpha[1]*alpha[2], 6*ln(A)^2*rho*sigma*alpha[2]*b[1]-2*ln(A)^2*sigma*a[1]*alpha[2]*beta[2]+a[1]*alpha[2]^2, 2*ln(A)^2*rho*b[1]*beta[1]+k*theta^2*beta[1]+2*a[1]*alpha[0]*beta[1]+2*a[1]*beta[0]*beta[1]-rho*beta[1]+beta[1], -ln(A)^2*rho*alpha[1]*b[1]+2*ln(A)^2*a[1]*alpha[1]*beta[2]+k*theta^2*alpha[1]+2*a[1]*alpha[0]*alpha[1]+2*a[1]*alpha[1]*beta[0]-rho*alpha[1]+alpha[1], -8*ln(A)^4*rho*sigma*b[1]*beta[2]-ln(A)^2*k*sigma*theta^2*beta[2]-2*ln(A)^2*sigma*a[1]*alpha[0]*beta[2]-2*ln(A)^2*sigma*a[1]*beta[0]*beta[2]-ln(A)^2*sigma*a[1]*beta[1]^2+ln(A)^2*rho*sigma*beta[2]-4*ln(A)^2*rho*alpha[2]*b[1]+2*ln(A)^2*a[1]*alpha[2]*beta[2]-ln(A)^2*sigma*beta[2]+k*theta^2*alpha[2]+2*a[1]*alpha[0]*alpha[2]+a[1]*alpha[1]^2+2*a[1]*alpha[2]*beta[0]-rho*alpha[2]+alpha[2], 6*ln(A)^4*rho*b[1]*beta[2]+ln(A)^2*k*theta^2*beta[2]+2*ln(A)^2*a[1]*alpha[0]*beta[2]+2*ln(A)^2*a[1]*beta[0]*beta[2]+ln(A)^2*a[1]*beta[1]^2-ln(A)^2*rho*beta[2]+k*theta^2*alpha[0]+k*theta^2*beta[0]+ln(A)^2*beta[2]+a[1]*alpha[0]^2+2*a[1]*alpha[0]*beta[0]+a[1]*beta[0]^2-rho*alpha[0]-rho*beta[0]+alpha[0]+beta[0], -6*rho*b[1]*beta[2]+a[1]*beta[2]^2, -2*rho*b[1]*beta[1]+2*a[1]*beta[1]*beta[2]}$

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$[{rho = rho, alpha[0] = -beta[0], alpha[1] = 0, alpha[2] = 0, beta[0] = beta[0], beta[1] = 0, beta[2] = 0}, {rho = (k*theta^2+1)/(4*ln(A)^2*b[1]+1), alpha[0] = -beta[0], alpha[1] = 0, alpha[2] = -6*ln(A)^2*(k*theta^2+1)*sigma*b[1]/((4*ln(A)^2*b[1]+1)*a[1]), beta[0] = beta[0], beta[1] = 0, beta[2] = 0}, {rho = (k*theta^2+1)/(4*ln(A)^2*b[1]+1), alpha[0] = -(6*k*theta^2*ln(A)^2*b[1]+4*ln(A)^2*a[1]*b[1]*beta[0]+6*ln(A)^2*b[1]+a[1]*beta[0])/((4*ln(A)^2*b[1]+1)*a[1]), alpha[1] = 0, alpha[2] = 0, beta[0] = beta[0], beta[1] = 0, beta[2] = 6*(k*theta^2+1)*b[1]/((4*ln(A)^2*b[1]+1)*a[1])}, {rho = -(k*theta^2+1)/(4*ln(A)^2*b[1]-1), alpha[0] = (2*k*theta^2*ln(A)^2*b[1]-4*ln(A)^2*a[1]*b[1]*beta[0]+2*ln(A)^2*b[1]+a[1]*beta[0])/((4*ln(A)^2*b[1]-1)*a[1]), alpha[1] = 0, alpha[2] = 0, beta[0] = beta[0], beta[1] = 0, beta[2] = -6*(k*theta^2+1)*b[1]/((4*ln(A)^2*b[1]-1)*a[1])}, {rho = -(k*theta^2+1)/(4*ln(A)^2*b[1]-1), alpha[0] = -(4*k*theta^2*ln(A)^2*b[1]+4*ln(A)^2*a[1]*b[1]*beta[0]+4*ln(A)^2*b[1]-a[1]*beta[0])/((4*ln(A)^2*b[1]-1)*a[1]), alpha[1] = 0, alpha[2] = 6*ln(A)^2*(k*theta^2+1)*sigma*b[1]/((4*ln(A)^2*b[1]-1)*a[1]), beta[0] = beta[0], beta[1] = 0, beta[2] = 0}, {rho = (1/6)*a[1]*beta[2]/b[1], alpha[0] = -(1/6)*(6*ln(A)^2*a[1]*b[1]*beta[2]+6*k*theta^2*b[1]+6*a[1]*beta[0]*b[1]-a[1]*beta[2]+6*b[1])/(a[1]*b[1]), alpha[1] = 0, alpha[2] = ln(A)^2*sigma*beta[2], beta[0] = beta[0], beta[1] = 0, beta[2] = beta[2]}, {rho = (1/6)*a[1]*beta[2]/b[1], alpha[0] = -ln(A)^2*beta[2]-beta[0], alpha[1] = 0, alpha[2] = ln(A)^2*sigma*beta[2], beta[0] = beta[0], beta[1] = 0, beta[2] = beta[2]}, {rho = k*theta^2+a[1]*alpha[0]+a[1]*beta[0]+1, alpha[0] = alpha[0], alpha[1] = 0, alpha[2] = 0, beta[0] = beta[0], beta[1] = 0, beta[2] = 0}]$

In Eq 3.18 rho is the same. But beta[2] is on the rhs on theirs, but beta[0] here. The denominators are the same. But overall is not the same

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${rho = (k*theta^2+1)/(4*ln(A)^2*b[1]+1), alpha[0] = -(6*k*theta^2*ln(A)^2*b[1]+4*ln(A)^2*a[1]*b[1]*beta[0]+6*ln(A)^2*b[1]+a[1]*beta[0])/((4*ln(A)^2*b[1]+1)*a[1]), alpha[1] = 0, alpha[2] = 0, beta[0] = beta[0], beta[1] = 0, beta[2] = 6*(k*theta^2+1)*b[1]/((4*ln(A)^2*b[1]+1)*a[1])}$

Eq 3.19. Despite the differences, we get the correct final result.

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eval(S, ans[3]); dsubs(Aode, %); sol := simplify(eval(%, solution))

V(xi) = -(6*k*theta^2*ln(A)^2*b[1]+4*ln(A)^2*a[1]*b[1]*beta[0]+6*ln(A)^2*b[1]+a[1]*beta[0])/((4*ln(A)^2*b[1]+1)*a[1])+beta[0]+6*(k*theta^2+1)*b[1]*(diff(G(xi), xi))^2/((4*ln(A)^2*b[1]+1)*a[1]*G(xi)^2)

V(xi) = -6*ln(A)^2*(k*theta^2+1)*sigma*b[1]/((4*ln(A)^2*b[1]+1)*a[1]*G(xi)^2)

V(xi) = -96*ln(A)^4*(k*theta^2+1)*sigma*b[1]*kappa^2*A^(2*xi)/((4*ln(A)^2*b[1]+1)*a[1]*(4*kappa^2*ln(A)^2*A^(2*xi)+sigma)^2)

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Download f-p-second.mw

typos plus...

Answered: dharr 8977

January 06 2026

2 8

I don't know about (24) - perhaps the paper has a typo in the ode (missing m?).

For (25) and (26), I corrected typos in magenta and the tests are verified.

I don't know about (30) or (31).

There seem to be typos in (40) and (41) but that doesn't solve the problems.

ode-17.mw

You didn't provide explanation about the Weierstraa cases (what is f3?). Maple has the WeierstrassP and WeierstrassPPrime.

<> construction...

Answered: dharr 8977

December 29 2025

4 3

The <> constructors are useful here:

ga_zero := <IdentityMatrix(2), ZeroMatrix(2); ZeroMatrix(2), -IdentityMatrix(2) >;

or

ga_zero := <<IdentityMatrix(2) | ZeroMatrix(2)>, <ZeroMatrix(2) | -IdentityMatrix(2)>>;

One way...

Answered: dharr 8977

December 29 2025

2 1

This method only works since your envelope is always increasing in size. Then you can use the running minimum and maximum values as your envelope.

numsolve-1229.mw

If we know enough......

Answered: dharr 8977

December 28 2025

2 2

What are we allowed to know here? I don't think you could find dio1 from dio without knowing the form we are seeking. If we do know that form, then we can find the missing numbers using solve as follows:

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3*x^2*y^2*z^2-36*x^2*y^2*z-30*x^2*y*z^2-24*x*y^2*z^2+126*x^2*y^2+432*x^2*y*z+93*x^2*z^2+360*x*y^2*z+312*x*y*z^2+66*y^2*z^2-1692*x^2*y-1476*x^2*z-1440*x*y^2-5040*x*y*z-1104*x*z^2-1080*y^2*z-948*y*z^2+6120*x^2+21096*x*y+18576*x*z+4554*y^2+16056*y*z+3540*z^2-79848*x-69300*y-61272*z+268452

Suppose we are allowed to know that we want an expression of the following form

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Then we require the following to be zero

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So we want the coefficients of each term to be zero. There are 26 coefficients

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${-2*d-2*g-2*a+24, -2*e-2*h-2*b+30, -2*f-2*i-2*c+36, a^2+d^2+g^2-66, b^2+e^2+h^2-93, c^2+f^2+i^2-126, 4*a*b+4*d*e+4*g*h-312, -2*a*b^2-2*d*e^2-2*g*h^2+1104, -2*a^2*b-2*d^2*e-2*g^2*h+948, a^2*b^2+d^2*e^2+g^2*h^2-3540, 4*a*c+4*d*f+4*g*i-360, -2*a*c^2-2*d*f^2-2*g*i^2+1440, -2*a^2*c-2*d^2*f-2*g^2*i+1080, a^2*c^2+d^2*f^2+g^2*i^2-4554, 4*b*c+4*e*f+4*h*i-432, -2*b*c^2-2*e*f^2-2*h*i^2+1692, -2*b^2*c-2*e^2*f-2*h^2*i+1476, b^2*c^2+e^2*f^2+h^2*i^2-6120, -8*a*b*c-8*d*e*f-8*g*h*i+5040, 4*a*b*c^2+4*d*e*f^2+4*g*h*i^2-21096, 4*a*b^2*c+4*d*e^2*f+4*g*h^2*i-18576, 4*a^2*b*c+4*d^2*e*f+4*g^2*h*i-16056, -2*a*b^2*c^2-2*d*e^2*f^2-2*g*h^2*i^2+79848, -2*a^2*b*c^2-2*d^2*e*f^2-2*g^2*h*i^2+69300, -2*a^2*b^2*c-2*d^2*e^2*f-2*g^2*h^2*i+61272, a^2*b^2*c^2+d^2*e^2*f^2+g^2*h^2*i^2-268452}$

Solve these equations for the unknowns {a,b,c...i}. Of course there are multiple solutions corresponding to permutations of the three terms in dioguess

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{a = 1, b = 2, c = 3, d = 7, e = 8, f = 9, g = 4, h = 5, i = 6}, {a = 4, b = 5, c = 6, d = 7, e = 8, f = 9, g = 1, h = 2, i = 3}, {a = 1, b = 2, c = 3, d = 4, e = 5, f = 6, g = 7, h = 8, i = 9}, {a = 7, b = 8, c = 9, d = 4, e = 5, f = 6, g = 1, h = 2, i = 3}, {a = 4, b = 5, c = 6, d = 1, e = 2, f = 3, g = 7, h = 8, i = 9}, {a = 7, b = 8, c = 9, d = 1, e = 2, f = 3, g = 4, h = 5, i = 6}