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How do i best work with Polar vectors in Maple?...

Asked by: Mads758 10
vector polar vectorcalculus
June 18 2025
2  3

Hello,

I’m currently a student working with vectors as part of my electrical AC calculations. Up until now, I’ve been using GeoGebra to add and plot vectors, but I’m trying to transition fully to Maple, since I already use it for everything else in my studies.

All of my assignments are either given in — or require answers in — polar form (magnitude and angle), so I would really like to work directly in polar coordinates without converting everything to Cartesian and back.

I’ve already tried to figure this out on my own, but so far, I’ve only been able to make it work using Cartesian notation. I’ve attached a PDF with a typical example of the kind of tasks I work on, in case that helps clarify my needs.

Could you guide me on how to best set up and work with polar vectors in Maple — including how to define, add, and plot them directly in polar notation?

Vector_Help.pdf

Thanks in advance for your help!

Best regards,
Mads Bach Nielsen

- Yes this was written with the help of Chat GPT

ODE error in plots ...

Asked by: KIRAN SAJJAN 40
ode dsolve numeric
June 17 2025
0  13

why this error is getting. but in the published paper for the same parameter it is converging. what is the mistake in this worksheet please rectify it in sachin_base_paper.mw

Laplace Adomian Decomposition method for solving...

by: dharr 8235 Maple
symbolic pde differential-equations
June 17 2025
7  14

Thanks to @salim-barzani, I became interested in the Laplace Adomian Decompositon method to solve partial differential equations. It produces a sum of analytical components that converge to the exact solution. Probably it is not that efficient for numerical solutions, but the possibility of finding an exact solution to nonlinear PDEs by finding a formula for the components and therefore the infinite sum is interesting.

The version here covers many of the common forms of PDEs, but is still a work in progress. The method for finding the Adomian polynomials could be improved by using one of the reccurence formulas in the literature. Extension to more PDE forms, ODEs etc are possible and I will attempt some of these before uploading an improved version to the application centre. In the meantime, feedback about bugs, usability etc. are appreciated.

Laplace Adomian Decomposition method to solve partial differential equations.
 D.A. Harrington, June 2025, v 1.16.
 Procedure LAD is in the startup code region.

Arguments:

1. 

Partial differential equation in one "time" variable and several other variables (called spatial variables here). Specified as an equation, or an expression implicitly equal to zero.
Comprising: (1) one time derivative term of order m (denoted L), (2) linear terms that are derivatives in the spatial variables (R), (3) nonlinear terms with derivatives in the spatial variables (N). Together, (1)-(3) is a multivariate polynomial in the dependent variable, its derivatives, and any parameters. (4) inhomogenous terms in the time and spatial variables that do not contain the dependent variable or its derivatives (G). The general form is L = R+N-G.
Any symbolic parameters are assumed to be real. Any numeric coefficients or parameters should be of type algebraic. Floating point values will be converted to rationals with convert(value, rational).
2. 

Function to solve for, e.g., u(x, y, t).
3. 

m initial condition(s) (at time zero). For m > 1, these must be in a list, and are the values of the function and its successive derivatives with respect to time, evaluated at time zero.
4. 

name of time variable.
5. 

number of iterations.
6. 

(optional) order for a fractional derivative, specified as fracorder = name or fracorder = numeric*value (floating point values are converted to type rational). The permissible values alpha are related to the value of m: m-1 < alpha and alpha <= m, i.e., an mth order time derivative is specified in the pde and m initial conditions are provided, as well as the value of alpha in the correct range. A symbolic alpha is assumed to be in this range.

infolevel[LAD] may be set to different values to print out additional information as follows. Greater numbers include the information from smaller values.
1. 

The nonlinear expansion variables (those in the Adomian polynomials).
2. 

The different parts of the the pde (L, R, N, G) in jet notation.
3. 

Progress is indicated, as the time at each iteration.
4. 

Values of the components of the solution as they are produced (one per iteration).
> 

restart

It may be useful to load the Physics package if derivatives contain abs. LAD converts these to a form without abs but with the conjugate of the dependent variable. The Physics package affects the internal simplifications and may affect the form of the output or the running time.

> 

PDEtools:-declare(U(x, t), quiet)

Example from A-M. Wazwaz, Applied Mathematics and Computation 111 (2000) 53. Sec. 4.

> 

pde := diff(U(x, t), t)+U(x, t)^2*(diff(U(x, t), x)); inx := 3*x

diff(U(x, t), t)+U(x, t)^2*(diff(U(x, t), x))

3*x

> 

infolevel[LAD] := 2
> 

approx := LAD(pde, U(x, t), inx, t, 10)

LAD: L = U[t]; R = 0; N = -U^2*U[x]; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: [U, U[x]]

175692092397588*t^10*x^11-5650915252554*t^9*x^10+184670433090*t^8*x^9-6155681103*t^7*x^8+210450636*t^6*x^7-7440174*t^5*x^6+275562*t^4*x^5-10935*t^3*x^4+486*t^2*x^3-27*t*x^2+3*x

By extrapolating to an infinite number of terms, we can find an exact solution. (Wazwaz calculates the coefficient of x^4*t^3 incorrectly and deduces an incorrect exact solution.) Multiplying by t gives a series in y = x*t for which guessgf can use the coefficients to find the exact solution.

> 

algsubs(x*t = y, expand(approx*t)); [seq(coeff(%, y, i), i = 0 .. degree(%, y))]; gfun:-guessgf(%, y)[1]; exact := (eval(%, y = x*t))/t

175692092397588*y^11-5650915252554*y^10+184670433090*y^9-6155681103*y^8+210450636*y^7-7440174*y^6+275562*y^5-10935*y^4+486*y^3-27*y^2+3*y

[0, 3, -27, 486, -10935, 275562, -7440174, 210450636, -6155681103, 184670433090, -5650915252554, 175692092397588]

6*y/(1+(1+36*y)^(1/2))

6*x/(1+(36*t*x+1)^(1/2))

Check

> 

pdetest(U(x, t) = exact, [pde, U(x, 0) = inx])

[0, 0]

An example with inhomogeneous terms (but no nonlinear part). Ex. 3 from R. Shah et al, Entropy 21 (2019) 335.

> 

pde := diff(U(x, t), t)+diff(U(x, t), `$`(x, 3)) = -sin(Pi*x)*sin(t)-Pi^3*cos(Pi*x)*cos(t); inx := sin(Pi*x); exact := sin(Pi*x)*cos(t); pdetest(U(x, t) = exact, [pde, U(x, 0) = inx])

diff(U(x, t), t)+diff(diff(diff(U(x, t), x), x), x) = -sin(Pi*x)*sin(t)-Pi^3*cos(Pi*x)*cos(t)

sin(Pi*x)

sin(Pi*x)*cos(t)

[0, 0]

> 

approx := LAD(pde, U(x, t), inx, t, 10)

LAD: L = U[t]; R = -U[x,x,x]; N = 0; G = -sin(Pi*x)*sin(t)-Pi^3*cos(Pi*x)*cos(t).

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

sin(Pi*x)-cos(Pi*x)*Pi^3*sin(t)+sin(Pi*x)*(-1+cos(t))-Pi^3*(Pi^3*(-1+cos(t))*sin(Pi*x)-cos(Pi*x)*sin(t))+(Pi^3*(-sin(t)+t)*cos(Pi*x)+sin(Pi*x)*(-1+cos(t)))*Pi^6-(1/2)*Pi^9*(Pi^3*(t^2+2*cos(t)-2)*sin(Pi*x)+2*cos(Pi*x)*(-sin(t)+t))-(1/6)*Pi^12*(Pi^3*(t^3+6*sin(t)-6*t)*cos(Pi*x)-3*sin(Pi*x)*(t^2+2*cos(t)-2))+(1/24)*(Pi^3*(t^4-12*t^2-24*cos(t)+24)*sin(Pi*x)+4*cos(Pi*x)*(t^3+6*sin(t)-6*t))*Pi^15+(1/120)*Pi^18*(Pi^3*(t^5-20*t^3-120*sin(t)+120*t)*cos(Pi*x)-5*sin(Pi*x)*(t^4-12*t^2-24*cos(t)+24))-(1/720)*Pi^21*(Pi^3*(t^6-30*t^4+360*t^2+720*cos(t)-720)*sin(Pi*x)+6*cos(Pi*x)*(t^5-20*t^3-120*sin(t)+120*t))-(1/5040)*(Pi^3*(t^7-42*t^5+840*t^3+5040*sin(t)-5040*t)*cos(Pi*x)-7*sin(Pi*x)*(t^6-30*t^4+360*t^2+720*cos(t)-720))*Pi^24+(1/40320)*Pi^27*(Pi^3*(t^8-56*t^6+1680*t^4-20160*t^2-40320*cos(t)+40320)*sin(Pi*x)+8*cos(Pi*x)*(t^7-42*t^5+840*t^3+5040*sin(t)-5040*t))+(1/362880)*Pi^30*(Pi^3*(t^9-72*t^7+3024*t^5-60480*t^3-362880*sin(t)+362880*t)*cos(Pi*x)-9*sin(Pi*x)*(t^8-56*t^6+1680*t^4-20160*t^2-40320*cos(t)+40320))

Expand the exact solution as a series in t to the same order and verify that it is the same as above.

> 

series(exact-approx, t, 11, oterm = false)

0

An example with a complex solution

> 

pde := I*(diff(U(x, t), t))+diff(U(x, t), `$`(x, 2))+2*abs(U(x, t))^2*U(x, t); inx := sech(x); exact := sech(x)*exp(I*t); `assuming`([simplify(pdetest(U(x, t) = exact, [pde, U(x, 0) = inx]))], [real])

I*(diff(U(x, t), t))+diff(diff(U(x, t), x), x)+2*abs(U(x, t))^2*U(x, t)

sech(x)

sech(x)*exp(I*t)

[0, 0]

> 

approx := LAD(pde, U(x, t), inx, t, 10)

LAD: L = I*U[t]; R = -U[x,x]; N = -2*U^2*C; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: [C, U]

sech(x)+I*sech(x)*t-(1/2)*sech(x)*t^2-((1/6)*I)*sech(x)*t^3+(1/24)*sech(x)*t^4+((1/120)*I)*sech(x)*t^5-(1/720)*sech(x)*t^6-((1/5040)*I)*sech(x)*t^7+(1/40320)*sech(x)*t^8+((1/362880)*I)*sech(x)*t^9-(1/3628800)*sech(x)*t^10

The series for exp(I*t) is apparent.

> 

collect(approx, sech); series(approx-exact, t, 11, oterm = false)

(1+I*t-(1/2)*t^2-((1/6)*I)*t^3+(1/24)*t^4+((1/120)*I)*t^5-(1/720)*t^6-((1/5040)*I)*t^7+(1/40320)*t^8+((1/362880)*I)*t^9-(1/3628800)*t^10)*sech(x)

0

An example with fractional order differentiation wrt t of order alpha. For m-1 < alpha and alpha <= m, the derivative must be entered with order m and there must be a list of m initial conditions:f(x, 0), (D[2](f))(x, 0), (D[2, 2](f))(x, 0), () .. ()

 Ex. 1 from R. Shah et al, Entropy 21 (2019) 335

> 

pde := diff(U(x, t), t) = -2*(diff(U(x, t), x))-(diff(U(x, t), `$`(x, 3))); inx := sin(x)

diff(U(x, t), t) = -2*(diff(U(x, t), x))-(diff(diff(diff(U(x, t), x), x), x))

sin(x)

Symbolic order is accepted.

> 

approx := LAD(pde, U(x, t), inx, t, 6, fracorder = alpha)

LAD: L = U[t]; R = -2*U[x]-U[x,x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

sin(x)-cos(x)*t^alpha/GAMMA(1+alpha)-sin(x)*t^(2*alpha)/GAMMA(1+2*alpha)+cos(x)*t^(3*alpha)/GAMMA(1+3*alpha)+sin(x)*t^(4*alpha)/GAMMA(1+4*alpha)-cos(x)*t^(5*alpha)/GAMMA(1+5*alpha)-sin(x)*t^(6*alpha)/GAMMA(1+6*alpha)

Giving a numerical value for the order will generally be more efficient. For alpha = 1/2, we can find an exact solution.

> 

approx := collect(LAD(pde, U(x, t), inx, t, 20, fracorder = 1/2), [sin, cos])

LAD: L = U[t]; R = -2*U[x]-U[x,x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

(1-t+(1/2)*t^2-(1/6)*t^3+(1/24)*t^4-(1/120)*t^5+(1/720)*t^6-(1/5040)*t^7+(1/40320)*t^8-(1/362880)*t^9+(1/3628800)*t^10)*sin(x)+(-2*t^(1/2)/Pi^(1/2)+(4/3)*t^(3/2)/Pi^(1/2)-(8/15)*t^(5/2)/Pi^(1/2)+(16/105)*t^(7/2)/Pi^(1/2)-(32/945)*t^(9/2)/Pi^(1/2)+(64/10395)*t^(11/2)/Pi^(1/2)-(128/135135)*t^(13/2)/Pi^(1/2)+(256/2027025)*t^(15/2)/Pi^(1/2)-(512/34459425)*t^(17/2)/Pi^(1/2)+(1024/654729075)*t^(19/2)/Pi^(1/2))*cos(x)

The first series is exp(-t), the second series is not so obvious, After some scaling, the second series may be passed to guessgf

> 

ser2 := `assuming`([simplify(expand(sqrt(Pi)*op(2, approx)/(sqrt(t)*cos(x))))], [t > 0]); [seq(coeff(ser2, t, i), i = 0 .. 9)]; ser2ex := gfun:-guessgf(%, t)[1]

-2+(4/3)*t-(8/15)*t^2+(16/105)*t^3-(32/945)*t^4+(64/10395)*t^5-(128/135135)*t^6+(256/2027025)*t^7-(512/34459425)*t^8+(1024/654729075)*t^9

-exp(-t)*Pi^(1/2)*erfi(t^(1/2))/t^(1/2)

So the full solution is

> 

exact := exp(-t)*sin(x)+ser2ex*sqrt(t)*cos(x)/sqrt(Pi)

exp(-t)*sin(x)-exp(-t)*erfi(t^(1/2))*cos(x)

Check. fracdiff's default method returns an integral, method = laplace fails, but method = series works with cancellation of all terms but the "left over" last half-integral-order one.

> 

Ord := 20; fracdiff(exact, t, 1/2, method = series, methodoptions = [order = Ord])-series(eval(rhs(pde), U(x, t) = exact), t, Ord)

20

-(1048576/319830986772877770815625)*sin(x)*t^(39/2)/Pi^(1/2)+O(t^(39/2))-O(t^20)

A second-order example (linear)

> 

pde := diff(U(x, t), t, t) = -c^2*(diff(U(x, t), `$`(x, 2))); inx := [exp(I*x/c), exp(I*x/c)]; exact := exp(t+I*x/c)

diff(diff(U(x, t), t), t) = -c^2*(diff(diff(U(x, t), x), x))

[exp(I*x/c), exp(I*x/c)]

exp(t+I*x/c)

> 

pdetest(U(x, t) = exact, [pde, U(x, 0) = inx[1], (D[2](U))(x, 0) = inx[2]])

[0, 0, 0]

> 

approx := LAD(pde, U(x, t), inx, t, 6)

LAD: L = U[t,t]; R = -c^2*U[x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

exp(I*x/c)*(1+t)+(1/6)*exp(I*x/c)*t^2*(3+t)+(1/120)*exp(I*x/c)*t^4*(t+5)+(1/5040)*exp(I*x/c)*t^6*(7+t)+(1/362880)*exp(I*x/c)*t^8*(9+t)+(1/39916800)*exp(I*x/c)*t^10*(t+11)+(1/6227020800)*exp(I*x/c)*t^12*(13+t)

> 

series(exact-approx, t, 13, oterm = false)

0

Fractional order example

> 

approx := collect(LAD(pde, U(x, t), inx, t, 10, fracorder = 3/2), [exp, t])

LAD: L = U[t,t]; R = -c^2*U[x,x]; N = 0; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: []

(1+(1/20922789888000)*t^16+(1/1307674368000)*t^15+(32768/6190283353629375)*t^(29/2)/Pi^(1/2)+(16384/213458046676875)*t^(27/2)/Pi^(1/2)+(1/6227020800)*t^13+(1/479001600)*t^12+(4096/316234143225)*t^(23/2)/Pi^(1/2)+(2048/13749310575)*t^(21/2)/Pi^(1/2)+(1/3628800)*t^10+(1/362880)*t^9+(512/34459425)*t^(17/2)/Pi^(1/2)+(256/2027025)*t^(15/2)/Pi^(1/2)+(1/5040)*t^7+(1/720)*t^6+(64/10395)*t^(11/2)/Pi^(1/2)+(32/945)*t^(9/2)/Pi^(1/2)+(1/24)*t^4+(1/6)*t^3+(8/15)*t^(5/2)/Pi^(1/2)+(4/3)*t^(3/2)/Pi^(1/2)+t)*exp(I*x/c)

There appear to be two separate series in t, with integer and half-integer powers.

> 

t_series := approx/exp(I*x/c); t1, t2 := selectremove(proc (w) options operator, arrow; (degree(w, t))::integer end proc, t_series)

1+(1/20922789888000)*t^16+(1/1307674368000)*t^15+(1/6227020800)*t^13+(1/479001600)*t^12+(1/3628800)*t^10+(1/362880)*t^9+(1/5040)*t^7+(1/720)*t^6+(1/24)*t^4+(1/6)*t^3+t, (32768/6190283353629375)*t^(29/2)/Pi^(1/2)+(16384/213458046676875)*t^(27/2)/Pi^(1/2)+(4096/316234143225)*t^(23/2)/Pi^(1/2)+(2048/13749310575)*t^(21/2)/Pi^(1/2)+(512/34459425)*t^(17/2)/Pi^(1/2)+(256/2027025)*t^(15/2)/Pi^(1/2)+(64/10395)*t^(11/2)/Pi^(1/2)+(32/945)*t^(9/2)/Pi^(1/2)+(8/15)*t^(5/2)/Pi^(1/2)+(4/3)*t^(3/2)/Pi^(1/2)

The integer-powers series is the series for exp(t) with every third term missing: missing powers 2, 5, 8, 11, ...

> 

t1sum := simplify(exp(t)-(sum(t^(3*i+2)/factorial(3*i+2), i = 0 .. infinity))); series(%, t, 17)

(2/3)*exp(t)+(1/3)*(cos((1/2)*3^(1/2)*t)+3^(1/2)*sin((1/2)*3^(1/2)*t))*exp(-(1/2)*t)

series(1+t+(1/6)*t^3+(1/24)*t^4+(1/720)*t^6+(1/5040)*t^7+(1/362880)*t^9+(1/3628800)*t^10+(1/479001600)*t^12+(1/6227020800)*t^13+(1/1307674368000)*t^15+(1/20922789888000)*t^16+O(t^17),t,17)

And for the other one, the following scaled series shows coefficients with powers of two divided by double factorials, every third term missing: missing powers 2,5,8,11,...

> 

expand(sqrt(Pi)*t2/(4*t^(3/2))); all := add(2^i*t^i/doublefactorial(2*(i+1)+1), i = 0 .. 9); missing := add(eval(2^i*t^i/doublefactorial(2*(i+1)+1), i = 3*j+2), j = 0 .. 3)

(8192/6190283353629375)*t^13+(4096/213458046676875)*t^12+(1024/316234143225)*t^10+(512/13749310575)*t^9+(128/34459425)*t^7+(64/2027025)*t^6+(16/10395)*t^4+(8/945)*t^3+(2/15)*t+1/3

1/3+(2/15)*t+(4/105)*t^2+(8/945)*t^3+(16/10395)*t^4+(32/135135)*t^5+(64/2027025)*t^6+(128/34459425)*t^7+(256/654729075)*t^8+(512/13749310575)*t^9

(4/105)*t^2+(32/135135)*t^5+(256/654729075)*t^8+(2048/7905853580625)*t^11

> 

t2sum := 4*t^(3/2)*(sum(2^i*t^i/doublefactorial(2*(i+1)+1), i = 0 .. infinity)-(sum(eval(2^i*t^i/doublefactorial(2*(i+1)+1), i = 3*j+2), j = 0 .. infinity)))/sqrt(Pi)

4*t^(3/2)*((1/4)*(Pi^(1/2)*exp(t)-2*t^(1/2)-Pi^(1/2)*erfc(t^(1/2))*exp(t))/t^(3/2)-(4/105)*t^2*hypergeom([1], [3/2, 11/6, 13/6], (1/27)*t^3))/Pi^(1/2)

Putting it together suggests

> 

exact := exp(I*x/c)*simplify(t1sum+t2sum)

(1/3)*exp(I*x/c)*(Pi^(1/2)*exp(-(1/2)*t)*sin((1/2)*3^(1/2)*t)*3^(1/2)+Pi^(1/2)*exp(-(1/2)*t)*cos((1/2)*3^(1/2)*t)+3*Pi^(1/2)*exp(t)*erf(t^(1/2))-(16/35)*t^(7/2)*hypergeom([1], [3/2, 11/6, 13/6], (1/27)*t^3)+2*Pi^(1/2)*exp(t)-6*t^(1/2))/Pi^(1/2)

Check.

> 

Ord := 20; fracdiff(exact, t, 3/2, method = series, methodoptions = [order = Ord+1])-series(eval(rhs(pde), U(x, t) = exact), t, Ord)

20

O(t^(39/2))-(1048576/319830986772877770815625)*exp(I*x/c)*t^(39/2)/Pi^(1/2)-O(t^(41/2))

Numerical calculation of a soliton

> 

pde := I*(diff(U(x, t), t))+diff(U(x, t), `$`(x, 2))+U(x, t)^2*conjugate(U(x, t)); exact := sqrt(2*B)*sech(sqrt(B)*(2*k*t-x-x__0))*exp(I*(k*x+(-k^2+B)*t)); `assuming`([simplify(pdetest(U(x, t) = exact, pde))], [real, B > 0]); inx := eval(exact, t = 0)

I*(diff(U(x, t), t))+diff(diff(U(x, t), x), x)+U(x, t)^2*conjugate(U(x, t))

2^(1/2)*B^(1/2)*sech(B^(1/2)*(2*k*t-x-x__0))*exp(I*(k*x+(-k^2+B)*t))

0

2^(1/2)*B^(1/2)*sech(B^(1/2)*(-x-x__0))*exp(I*k*x)

With numerical parameters it runs faster. Float parameters will be converted to rationals.

> 

params := {B = 2, k = -1/2, x__0 = -2}; inxnum := eval(inx, params); exactnum := eval(exact, params)

2*sech(2^(1/2)*(-x+2))*exp(-((1/2)*I)*x)

2*sech(2^(1/2)*(-t-x+2))*exp(I*(-(1/2)*x+(7/4)*t))

> 

infolevel[LAD] := 3; approx := LAD(pde, U(x, t), inxnum, t, 28)

LAD: L = I*U[t]; R = -U[x,x]; N = -U^2*C; G = 0.

LAD: nonlinear expansion variables (conjugated functions denoted by C) are: [C, U]

LAD: calculating component 1 at time 0.

LAD: calculating component 2 at time .221

[Edited for brevity]

LAD: calculating component 27 at time 90.414

LAD: calculating component 28 at time 103.870

LAD: completed at time 105.309

For larger x and t, the approximation diverges at a point that depends on the number of iterations.

> 

plot3d([abs(exactnum), abs(approx)], t = 0 .. 2, x = -3 .. 3, color = [red, blue], view = [default, default, 0 .. 5], orientation = [-30, 60])

> 

NULL

Download Laplace_Adomian_Decomposition_procedure8b.mw

Where is the "Stop Execution" icon?...

Asked by: WD0HHU 40
interrupt gui duplicate-question
June 17 2025
0  1

Maple 2024 had a very well defined "Stop Execution" symbol on the desktop.  

Maple 2025 doesn't have an obvious symbol somewhere.  Also, the two little "dots" that I once had in the lower left bottom in Maple 2025 have disappeared! There is only a bar at the bottom which seems to do nothing.

there is anyway for making fast the loop?...

Asked by: salim-barzani 1560
performance simplification conjugate
June 17 2025
2  4

i need a way for making my loop working fast, i need something for calculating such term fast without showing result but thus term make a lot of time for coming out ?

fasting.mw

substitute equation to equation...

Asked by: ashy 35
substitution incomplete-question
June 17 2025
0  4

I want to substitute equation to equation, how?

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Computation of left and right eigenvectors...

Asked by: Elisha 50
bifurcation eigenvector
June 17 2025
0  0

Someone should please help me compute the left and right eigenvectors of the system below. The purpose is to compute values for 'a' and 'b' in the bifurcation formula.

Thank you

``

> 

with(VectorCalculus)

 

 (1)
> 

interface(imaginaryunit = I)

I

 (2)
> 

I

I

 (3)
 
> 

diff(S(t), t) := `&Lambda;__p`-(`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p]+µ__C)*S+`&omega;__B`*I__B

Lambda__p-(`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p]+µ__C)*S+omega__B*I__B

 (4)
> 

diff(I__B(t), t) := `#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S/N[p]-`&omega;__B`*I__B-(`&sigma;__B`+µ__C)*I__B

`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&theta;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S/N[p]-omega__B*I__B-(sigma__B+µ__C)*I__B

 (5)
> 

NULL

``

 (6)
> 

diff(S__A(t), t) := `&Lambda;__A`-(µ__A+`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p])*S__A+`&delta;__A`*I__A

Lambda__A-(µ__A+`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`/N[p])*S__A+delta__A*I__A

 (7)
> 

diff(I__A(t), t) := `#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S__A/N[p]-(µ__A+`&delta;__A`)*I__A

`#mrow(mi("&varphi;",fontstyle = "normal"),mo("&sdot;"),msub(mi("&alpha;",fontstyle = "normal"),mi("B")),mo("&sdot;"),msub(mi("I"),mi("B")))`*S__A/N[p]-(µ__A+delta__A)*I__A

 (8)
> 

NULL

``

 (9)

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remove the parentheses (for the dᵢ values)...

Asked by: jalal 435
table typesetting task
June 17 2025
0  6

Hi,

I’m trying to remove the parentheses (for the dᵢ values) to achieve optimal display. Ideas ? Thanks

S4_Droites_Implicite.mw

Problem with dsolve()...

Asked by: Rouben Rostamian 8506
ode dsolve differential-equations
June 16 2025
2  14

A colleage asked me for the solution of an elementary boundary value problem for an ODE that models the steady-state temperature distribution in a nonhomogeneous rod.  I solved it in Maple and passed the solution on to her without checking the result.  She struggled for a day or two, attempting to "debug" her finite element code whose result was not agreeing with Maple's.

Upon closer inspection, it turned out that her code was correct and the solution returned by Maple was not.  Here are the details.  I can't tell whether Maple's dsolve can be improved to provide correct solutions to this and similar problems.

Problem with dsolve()

> 

restart;
> 

kernelopts(version);

`Maple 2025.0, X86 64 LINUX, Mar 24 2025, Build ID 1909157`

A boundary value problem for an ODE

 

Let's solve the following boundary value problem:

> 

de := Diff(a(x)*Diff(u(x),x),x) = -1;

Diff(a(x)*(Diff(u(x), x)), x) = -1

> 

bc := u(-1)=0, u(1)=0;

u(-1) = 0, u(1) = 0

Maple's dsolve() fails to find a solution:

> 

dsolve({de,bc});

> 

 

Solution obtained by hand

 

The solution may be obtained by hand, as follows.

 

Integrate the DE once

a(x)*(diff(u(x), x)) = c-x.

where c is the integration constant.  Therefore

diff(u(x), x) = (c-x)/a(x) and (c-x)/a(x) = c/a(x)-x/a(x).

To determine the value of the constant c, integrate the above over [-1, 1]

and apply the boundary conditions:

c*(int(1/a(x), x = -1 .. 1))+int(x/a(x), x = -1 .. 1) = u(1)-u(-1) and u(1)-u(-1) = 0

whence
c = (int(x/a(x), x = -1 .. 1))/(int(1/a(x), x = -1 .. 1)).

Having thus determined c, we integrate the expression for diff(u(x), x) obtained above, and arrive at the solution
u(x) = int((c-xi)/a(xi), xi = -1 .. x).

Remark: The solution obtained here is valid for any a(x) as long as

the integrals encountered above make sense.  Note that there is
no requirement on the differentiability or even continuity of a(x).

Technically, a(x) can be any function such that 1/a(x) is integrable.

 

> 

 

Wrong solution returned by dsolve()

 

Let us consider a special choice of the coefficient a(x):

> 

a := x -> 1 + 2*Heaviside(x);

proc (x) options operator, arrow; 1+2*Heaviside(x) end proc

According the the calculations above, we have:

> 

c := int(x/a(x),x=-1..1) / int(1/a(x),x=-1..1) ;

-1/4

and therefore the correct solution is

> 

int((c-xi)/a(xi), xi=-1..x):
sol := u(x) = collect(%, Heaviside);

u(x) = ((1/6)*x+(1/3)*x^2)*Heaviside(x)-(1/4)*x-(1/2)*x^2+1/4

Here is what the solution looks like:

> 

plot(rhs(sol), x=-1..1, color=blue, thickness=3);

Let's attempt to solve the same boundary value problem with Maple's dsolve().

> 

dsolve({de,bc}):
bad_sol := collect(%, Heaviside);

u(x) = ((1/3)*x^2+x/(3*exp(-2)+3)-exp(-2)*x/(3*exp(-2)+3))*Heaviside(x)-(1/2)*x^2-x/(3*exp(-2)+3)+(3*exp(-2)+1)/(6*exp(-2)+6)

That's very different from the correct solution obtained above.   To see the problem with it, let's recall that according to the DE, the expression a(x)*(diff(u(x), x))NULLshould evaluate to  c-x for some constant c,  but what we get through bad_sol is nothing like c-x;  in fact, it's not even continuous:

> 

a(x)*diff(rhs(bad_sol), x):
collect(%, Heaviside, simplify);
plot(%, x = -0.5 .. 0.5);

((4*x-2)*exp(-2)+4*x+2)*Heaviside(x)^2/(3*exp(-2)+3)+(-(4/3)*x-1/3)*Heaviside(x)-x-1/(3*exp(-2)+3)

It appears that dsolve goes astray by attempting to expand the expression diff(a(x)*(diff(u(x), x)), x).   It shouldn't.

 

 

Download dsolve-bug.mw

 

How do i print an index of integrals?...

Asked by: paulmcquad 105
integration education student
June 14 2025
1  3

Hi.

So I get this ->

int(x^2, x = 0 .. a)

(1/3)*a^3

 (1)

NULL


But i would like this ->

Download 5.3-Definite-Integral.mw

 

And maybe something similar for indefinite integrals.

This would really help me with education in calculus so thanks everybody.

Mapleflow worksheets in Maple 2025...

Asked by: Christopher2222 6030
worksheet
June 14 2025
0  2

It would appear being able to run Mapleflow worksheets in Maple is only useful to those who have Mapleflow.

Just wondering without being able to see what variables are in the mapleflow worksheet how would I know what values to pass to it?

What does this sum result mean?...

Asked by: paulmcquad 105
sum symbolic summation
June 14 2025
1  1

Hi.

I have been learning Integration for calculus. I came about on output that i not know what it means.

See for yourself. ->
If anyone can help make this mess.

Thanks in advance.

sum((-2)^j/(j-2)^2, j = 3 .. n)

4*polylog(2, -2)+2^(n+1)*(1+(-2*n^2+4*n-2)*LerchPhi(-2, 2, n))*(-1)^n/(-1+n)^2

 (1)

NULL

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there is anyway to export the table from maple to ...

Asked by: salim-barzani 1560
table matrix latex
June 13 2025
0  1

i want to do export  the table to latex there is command to do?

table.mw

How do I shrink the "useless" white area...

Asked by: rlewis 40
worksheet animation 3d
June 13 2025
1  5

In the screen shot below taken from a 3D animation I want to vastly shrink the area I have stippled with spots (it's actually white on the worksheet).  It's outside the cube defining the 3D animation and inside the square that defines the part of the worksheet used for an image.

How make it error be small in iteration and how a...

Asked by: salim-barzani 1560
June 13 2025
0  4

in my iteration when i got answer the error is so big and the method is powerfull i don't know i did mistake or my exact function i have to change it, becuase error is so big , how i can make error be smaller i did any mistake ?
and how i can plot them together exact and approximate in one graph also 3D and 2D

Num-e1.mw

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