MaplePrimes Questions

I am looking at the d'Almbert operator for the Minkowski metric in cartesian coordinates. When using the dAlmbertian command and converting to partial derivatives it gets the coefficients wrong, the negatives are not in the correct place based on the metric. But when I manually write the operator with the metric and derivative operator it returns the correct answer. Any thoughts on this? 

See attached, thank you. 

restart;

with(Physics):

Setup(coordinatesystems=cartesian,signature=`-+++`,metric=Minkowski)

 

`Systems of spacetime coordinates are:`*{X = (t, x, y, z)}

 

_______________________________________________________

 

`The Minkowski metric in coordinates `*[t, x, y, z]

 

`Signature: `(`- + + +`)

 

_______________________________________________________

 

Physics:-g_[mu, nu] = Matrix(%id = 36893488152289591652)

 

[coordinatesystems = {X}, metric = {(1, 1) = -1, (2, 2) = 1, (3, 3) = 1, (4, 4) = 1}, signature = `- + + +`]

(1)

g_[lineelement]

-Physics:-d_(t)^2+Physics:-d_(x)^2+Physics:-d_(y)^2+Physics:-d_(z)^2

(2)

CompactDisplay(phi(t,x,y,z))

phi(t, x, y, z)*`will now be displayed as`*phi

(3)

dAlembertian(phi(X))

Physics:-dAlembertian(phi(X), [X])

(4)

convert(%,diff)

-(diff(diff(phi(X), t), t))-(diff(diff(phi(X), x), x))-(diff(diff(phi(X), y), y))+diff(diff(phi(X), z), z)

(5)

g_[~mu,~nu]*d_[nu](d_[mu](phi(X)))

Physics:-g_[`~mu`, `~nu`]*Physics:-d_[mu](Physics:-d_[nu](phi(X), [X]), [X])

(6)

SumOverRepeatedIndices((6))

-(diff(diff(phi(X), t), t))+diff(diff(phi(X), x), x)+diff(diff(phi(X), y), y)+diff(diff(phi(X), z), z)

(7)
 

 

dAlmbert.mw

in this pde I have to find R, and already in the paper finded but  in my openion is not correct even when i tested is not give me solution, in some paper like that i finded R variable by finding some condition in pde but i want a coding for arrange that condition for example in pde we have 4 function of t like C(t),B(t),N(t),and E(t) i have to change them for finding R  like exchange C(t)=C(t), B(t)=C(t),N(t)=-C(t),E(t)=E(t) also we can equal them to one too, this is one of the condition how many condition we can have for substitute each on like that and tested for finding R, one of this condition will make help to find that R  like  this paper

but my equation is different is this equation  in below 

i tested the author outcome and not make my pde to zero, also dipersion relation without changing any any function of t is so different and i think he just use the penlevy test variable 6/beta*diff(phi,x) as a tranformation but i think is mistake, so where is problem and how we can find all the condition 

pde-condition.mw

in this function contain another function which contain integral but when i replace is just take some of the function and other function is not acceptable i want ask  there is any way for plotting the function when i substitute all other function, is work for csc,sec,cosh, and for other not work and  infinity sign appear likesome other  like t,sin,cos,tan,tanh,sinh, so on 

 t-plot.mw

I would like to experiment with error estimation in the symbolic solution of ordinary differential equations. I've written a simple example in the attached file. I would now like to plot both the left side of the ODE and the solution y(x) together in the same coordinate system. I can't do this, and I'm asking for help. How can the cumbersome numerical terms in the solution y(x) be converted to floating-point numbers?

As I said, this is a recreational experiment ;-) .DGL_test.mw

restart

ode := x^2 = sum(x*(diff(y(x), [`$`(x, i)]))/factorial(i), i = 0 .. 3)

x^2 = x*y(x)+x*(diff(y(x), x))+(1/2)*x*(diff(diff(y(x), x), x))+(1/6)*x*(diff(diff(diff(y(x), x), x), x))

(1)

``

``

ics := y(0) = 0, (D(y))(0) = 0, ((D@@2)(y))(0) = 0

y(0) = 0, (D(y))(0) = 0, ((D@@2)(y))(0) = 0

(2)

dsolve({ics, ode})

y(x) = x-1+((1/12)*(2-2^(1/2))*(1+2^(1/2))^(2/3)+(1/12)*2^(1/2)*(1+2^(1/2))^(1/3)+1/3)*exp(((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)-1)*x)+((1/12)*(2^(1/2)-2)*(1+2^(1/2))^(2/3)-(1/12)*2^(1/2)*(1+2^(1/2))^(1/3)+2/3)*exp(-(1/2)*((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)+2)*x)*cos((1/2)*3^(1/2)*(1+2^(1/2))^(1/3)*(2^(1/2)*(1+2^(1/2))^(1/3)-(1+2^(1/2))^(1/3)+1)*x)-3^(1/2)*(-1+(2^(1/2)-1)*(1+2^(1/2))^(1/3))*2^(1/2)*exp(-(1/2)*((1+2^(1/2))^(2/3)*2^(1/2)-(1+2^(1/2))^(2/3)-(1+2^(1/2))^(1/3)+2)*x)*sin((1/2)*3^(1/2)*(1+2^(1/2))^(1/3)*(2^(1/2)*(1+2^(1/2))^(1/3)-(1+2^(1/2))^(1/3)+1)*x)/((1+2^(1/2))^(2/3)*(12*2^(1/2)-12))

(3)

NULL``

Download DGL_test.mw

I have a lot of maple code for analysing electronic design uncertainties and would like to place a clip from the schematic before outputing the analysis results.  I have been working for hours on this with no success but I'm sure it must be simple.

I see that ImageTools,Preview kind of works but it outputs a very small, blurry image.  Is there just a simple command that I can read in my .png image and output it to the display?

Hello,

I have full access on .../.../version.txt but Maple tells me I have no write access...

Sorry for the previous post, thank you.

If I try with the install command I get the same error.

Kind regards to all

Jean-Michel

PS: have a look on the two inserted screenshots

I'm not sure of the Maple function I have to use to represent the modified Bessel function of the third kind with index 1.

The Bessel function of the third kind is also named the Hankel function.

I mistakenly started to use HankelH1 or HankelH2 but quickly realized they are both complex-valued functions while the paper I'm working on uses a real-valued modified Bessel function of the third kind: finally, does this latter exist in Maple 2015?

Thanks in advance

undesired_textplot_format.mw

When using textplot() AND display(), how do I ensure that the expression I want in the figure is not modified by display()? See my attached spreadsheet where I demostrate this undesired behaviour. 

NULL

Evaluation with `ϕ__0`subscripted

-t(0)+t(`ϕ`(t)) = Int(1/sqrt(2*C*cos(`ϕ`)-2*C*cos(`ϕ__0`)), `ϕ` = 0 .. `ϕ`(t), continuous)

-t(0)+t(varphi(t)) = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi(t), continuous)

(1)

t(0) = 0, `ϕ`(t) = `ϕ__0`, t(`ϕ__0`) = (1/4)*T

t(0) = 0, varphi(t) = varphi__0, t(varphi__0) = (1/4)*T

(2)

subs(t(0) = 0, varphi(t) = varphi__0, t(varphi__0) = (1/4)*T, -t(0)+t(varphi(t)) = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi(t), continuous))

(1/4)*T = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi__0, continuous)

(3)

isolate(`assuming`([simplify(value((1/4)*T = Int(1/(2*C*cos(varphi)-2*C*cos(varphi__0))^(1/2), varphi = 0 .. varphi__0, continuous)))], [C > 0, 0 < `&varphi;__0` and `&varphi;__0` < Pi]), T)

T = 4*InverseJacobiAM((1/2)*varphi__0, csc((1/2)*varphi__0))*csc((1/2)*varphi__0)/C^(1/2)

(4)

NULL

NULL

Replacing now in the above varphi__0 by varphi[0]

NULL

Evaluation with `&varphi;`[0] indexed

-t(0)+t(`&varphi;`(t)) = Int(1/sqrt(2*C*cos(`&varphi;`)-2*C*cos(`&varphi;`[0])), `&varphi;` = 0 .. `&varphi;`(t), continuous)

-t(0)+t(varphi(t)) = Int(1/(2*C*cos(varphi)-2*C*cos(varphi[0]))^(1/2), varphi = 0 .. varphi(t), continuous)

(5)

t(0) = 0, `&varphi;`(t) = `&varphi;`[0], t(`&varphi;`[0]) = (1/4)*T

t(0) = 0, varphi(t) = varphi[0], t(varphi[0]) = (1/4)*T

(6)

subs(t(0) = 0, varphi(t) = varphi[0], t(varphi[0]) = (1/4)*T, -t(0)+t(varphi(t)) = Int(1/(2*C*cos(varphi)-2*C*cos(varphi[0]))^(1/2), varphi = 0 .. varphi(t), continuous))

(1/4)*T = Int(1/(2*C*cos(varphi)-2*C*cos(varphi[0]))^(1/2), varphi = 0 .. varphi[0], continuous)

(7)

isolate(`assuming`([simplify(value((1/4)*T = Int(1/(2*C*cos(varphi)-2*C*cos(varphi[0]))^(1/2), varphi = 0 .. varphi[0], continuous)))], [C > 0, 0 < `&varphi;`[0] and `&varphi;`[0] < Pi]), T)

T = 4*2^(1/2)*(-(limit(((-cos(varphi)+cos(varphi[0]))/(-1+cos(varphi[0])))^(1/2)*InverseJacobiAM((1/2)*varphi, csgn(sin((1/2)*varphi[0]))*csc((1/2)*varphi[0]))/(cos(varphi)-cos(varphi[0]))^(1/2), varphi = 0, right))*(cos(varphi[0])-cos(varphi[0][0]))^(1/2)+((-cos(varphi[0])+cos(varphi[0][0]))/(-1+cos(varphi[0][0])))^(1/2)*InverseJacobiAM((1/2)*varphi[0], csgn(sin((1/2)*varphi[0][0]))*csc((1/2)*varphi[0][0])))/(C^(1/2)*(cos(varphi[0])-cos(varphi[0][0]))^(1/2))

(8)

NULL


The double indices in the above limit do not make sense.

In case this is a bug: Is there any workaround for indexed names? I want to use them to have numbers in indices in roman which seems to be not possible with subscripts. 


 

Download Evaluation_of_elliptic_integrals.mw

Is the absence of up/down arrows at the ends of the vertical scroll bar in Maple 2025's worksheet window a bug or by design?

These up/down arrows are helpful for small-increment navigation with the vertical scroll bar.  I hope Maple's designers restore the up/down arrows at the ends of the vertical scroll bar in Maple's worksheet window. These up/down arrows were present in Maple 2024 and earlier.

Any one could find why dsolve can't solve this first order ode when adding IC,  but Student:-ODEs:-ODESteps can?

interface(version);

`Standard Worksheet Interface, Maple 2025.1, Linux, June 12 2025 Build ID 1932578`

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1878 and is the same as the version installed in this computer, created 2025, September 28, 11:35 hours Pacific Time.`

SupportTools:-Version();

`The Customer Support Updates version in the MapleCloud is 29 and is the same as the version installed in this computer, created June 23, 2025, 10:25 hours Eastern Time.`

restart;

ode:=diff(m(t),t) = -k/m(t)^2;
ic:=m(0) = m__0;
DEtools:-odeadvisor(ode);

diff(m(t), t) = -k/m(t)^2

m(0) = m__0

[_quadrature]

sol:=dsolve(ode)

m(t) = (-3*k*t+c__1)^(1/3), m(t) = -(1/2)*(-3*k*t+c__1)^(1/3)-((1/2)*I)*3^(1/2)*(-3*k*t+c__1)^(1/3), m(t) = -(1/2)*(-3*k*t+c__1)^(1/3)+((1/2)*I)*3^(1/2)*(-3*k*t+c__1)^(1/3)

sol:=dsolve([ode,ic])

Student:-ODEs:-ODESteps([ode,ic])

"[[,,"Let's solve"],[,,[(&DifferentialD;)/(&DifferentialD;t) m(t)=-k/((m(t))^2),m(0)=`m__0`]],["&bullet;",,"Highest derivative means the order of the ODE is" 1],[,,(&DifferentialD;)/(&DifferentialD;t) m(t)],["&bullet;",,"Solve for the highest derivative"],[,,(&DifferentialD;)/(&DifferentialD;t) m(t)=-k/((m(t))^2)],["&bullet;",,"Separate variables"],[,,((&DifferentialD;)/(&DifferentialD;t) m(t)) (m(t))^2=-k],["&bullet;",,"Integrate both sides with respect to" t],[,,&int;((&DifferentialD;)/(&DifferentialD;t) m(t)) (m(t))^2 &DifferentialD;t=&int;-k &DifferentialD;t+`c__1`],["&bullet;",,"Evaluate integral"],[,,((m(t))^3)/3=-k t+`c__1`],["&bullet;",,"Solve for" m(t)],[,,{m(t)=(-3 k t+3 `c__1`)^(1/3),m(t)=-((-3 k t+3 `c__1`)^(1/3))/2-(&ImaginaryI; sqrt(3) (-3 k t+3 `c__1`)^(1/3))/2,m(t)=-((-3 k t+3 `c__1`)^(1/3))/2+(&ImaginaryI; sqrt(3) (-3 k t+3 `c__1`)^(1/3))/2}],["&bullet;",,"Simplify"],[,,{m(t)=(-3 k t+3 `c__1`)^(1/3),m(t)=((-3 k t+3 `c__1`)^(1/3) (-1+&ImaginaryI; sqrt(3)))/2,m(t)=-((-3 k t+3 `c__1`)^(1/3) (1+&ImaginaryI; sqrt(3)))/2}],["&bullet;",,"Redefine the integration constant(s)"],[,,{m(t)=(-3 k t+`c__1`)^(1/3),m(t)=((-3 k t+`c__1`)^(1/3) (-1+&ImaginaryI; sqrt(3)))/2,m(t)=-((-3 k t+`c__1`)^(1/3) (1+&ImaginaryI; sqrt(3)))/2}],["&bullet;",,"Use initial condition" m(0)=`m__0`],[,,`m__0`=`c__1`^(1/3)],["&bullet;",,"Solve for" `c__1`],[,,`c__1`=`m__0`^3],["&bullet;",,"Substitute" `c__1`=`m__0`^3 "into general solution and simplify"],[,,m(t)=(`m__0`^3-3 k t)^(1/3)],["&bullet;",,"Use initial condition" m(0)=`m__0`],[,,`m__0`=(`c__1`^(1/3) (-1+&ImaginaryI; sqrt(3)))/2],["&bullet;",,"Solve for" `c__1`],[,,`c__1`=`m__0`^3],["&bullet;",,"Substitute" `c__1`=`m__0`^3 "into general solution and simplify"],[,,m(t)=((`m__0`^3-3 k t)^(1/3) (-1+&ImaginaryI; sqrt(3)))/2],["&bullet;",,"Use initial condition" m(0)=`m__0`],[,,`m__0`=-(`c__1`^(1/3) (1+&ImaginaryI; sqrt(3)))/2],["&bullet;",,"Solve for" `c__1`],[,,`c__1`=`m__0`^3],["&bullet;",,"Substitute" `c__1`=`m__0`^3 "into general solution and simplify"],[,,m(t)=-((`m__0`^3-3 k t)^(1/3) (1+&ImaginaryI; sqrt(3)))/2],["&bullet;",,"Solutions to the IVP"],[,,[m(t)=(`m__0`^3-3 k t)^(1/3),m(t)=((`m__0`^3-3 k t)^(1/3) (-1+&ImaginaryI; sqrt(3)))/2,m(t)=-((`m__0`^3-3 k t)^(1/3) (1+&ImaginaryI; sqrt(3)))/2]]]"

Download why_dsolve_cant_solve_ode_oct_4_2025_maple_2025_1.mw

Given an ode with IC. When solution is explicit, Maple resolves the constant of integration as expected and returns solution with no c__1 in it.

But when asked for implicit solution, also with same IC, it now returns solution with c__1 still there.

Is this by design or a bug? Should not constant of integration be resolved using IC in both cases? If unable to solve for c__1 because solution is implicit, should it then not return solution all?

Does this happen in earlier versions of Maple?

 

restart;

interface(version);

`Standard Worksheet Interface, Maple 2025.1, Linux, June 12 2025 Build ID 1932578`

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1878 and is the same as the version installed in this computer, created 2025, September 28, 11:35 hours Pacific Time.`

SupportTools:-Version();

`The Customer Support Updates version in the MapleCloud is 29 and is the same as the version installed in this computer, created June 23, 2025, 10:25 hours Eastern Time.`

restart;

IC:=D(y)(0)=0;
ode:=diff(y(x),x)^2+x*y(x)=0:
sol:=dsolve([ode,IC]);

 

(D(y))(0) = 0

y(x) = 0, y(x) = -(1/9)*x^3

sol:=dsolve(ode,'implicit');

y(x) = 0, -x^2/((x^3+9*y(x))*(x^2+3*(-x*y(x))^(1/2)))+3*(-x*y(x))^(1/2)/((x^3+9*y(x))*(x^2+3*(-x*y(x))^(1/2)))-c__1 = 0, x^2/((x^3+9*y(x))*(-x^2+3*(-x*y(x))^(1/2)))+3*(-x*y(x))^(1/2)/((x^3+9*y(x))*(-x^2+3*(-x*y(x))^(1/2)))-c__1 = 0

#WHY did not resolve constant of integration here??
sol:=dsolve([ode,IC],'implicit');

x^2/((x^3+9*y(x))*(-x^2+3*(-x*y(x))^(1/2)))+3*(-x*y(x))^(1/2)/((x^3+9*y(x))*(-x^2+3*(-x*y(x))^(1/2)))-c__1 = 0, -x^2/((x^3+9*y(x))*(x^2+3*(-x*y(x))^(1/2)))+3*(-x*y(x))^(1/2)/((x^3+9*y(x))*(x^2+3*(-x*y(x))^(1/2)))-c__1 = 0, y(x) = 0

 


 

Download why_C_still_in_solution_maple_2025_1_oct_4_2025.mw

I looked at help for PDEtools:-Solve and do not see why this would fail. Any idea? Is this by design or could it be a bug?  solve() works on same input, so I expected Solve to also work on same input.

interface(version);

`Standard Worksheet Interface, Maple 2025.1, Linux, June 12 2025 Build ID 1932578`

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1878 and is the same as the version installed in this computer, created 2025, September 28, 11:35 hours Pacific Time.`

SupportTools:-Version();

`The Customer Support Updates version in the MapleCloud is 29 and is the same as the version installed in this computer, created June 23, 2025, 10:25 hours Eastern Time.`

restart;

eq:=D(y)(0)+1 = 0;
solve(eq,D(y)(0));

(D(y))(0)+1 = 0

-1

PDEtools:-Solve(eq,D(y)(0));

Error, (in PDEtools:-Solve) required an indication of the solving variables for the given system

 

 

Download why_PDEtools_Sovle_fail_oct_3_2025.mw

i did seperating but but is seems is not correct what is mistake in here?

issue1.mw

Hello, I have a Maple Flow worksheet that is formatted with a basic style template using headings and sections so that I can auto generate the table of contents. All of the numbered sections are in order, top-to-bottom in the worksheet. However, when Maple Flow auto generates the table of contents, it consistently gets the sections out of order - see a screenshot below - note Section 2 is at the end of the table of contents. The page numbers appear to be correct, but the physical placement of Section 2 is not. 

If anybody has encountered this problem in Maple Flow, I would appreciate some suggestions.

Thanks

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