Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

sstf.mw

In Graph has curves that are overlapping each other and the values of  S[t] = -1, -2, 1, 2 and also the y axis to range from 0 to 1.

I am trying to separate the real and imaginary parts of a complex expression in Maple to get Eq. (1.5) as in the attached image, but the Re and Im functions do not seem to return the expected results. Instead, Maple leaves the expression unchanged. PD_OD.mw

Is it possible to set a worksheet or maple up to only express trigonometric functions as sines? This is very helpful for certain exotic coordinate systems. Meaning cos(x)=sqrt(1-sin(x)^2)  and NOT equivalently  adding Pi/2 to the argument cos(x)=sin(x+Pi/2).
It is completely useless and futile to do algsubs, as Maple will later just revert back to cosines, so you want to do it in the headers so output is always expressed as sines in the document by default.

Basically you want to set something in the header of a worksheeet to always use expsincos on output, and then in addition cos(x)=k sqrt(1-sin(x)^2)   k\e{1,-1} on the output.

The modified Liouville equation

How to solve this pde for a general solution ?

The general solution in this form exist.

restart;

with(PDEtools): declare(u(x,t)); U:=diff_table(u(x,t));
PDE1:=U[t,t]=a^2*U[x,x]+b*exp(beta*U[]);
Sol11:=u(x,t)=1/beta*ln(2*(B^2-a^2*A^2)/(b*beta*(A*x+B*t+C)^2));
Sol12:=S->u(x,t)=1/beta*ln(8*a^2*C/(b*beta))
-2/beta*ln(S*(x+A)^2-S*a^2*(t+B)^2+S*C);
Test11:=pdetest(Sol11,PDE1);
Test12:=pdetest(Sol12(1),PDE1);
Test13:=pdetest(Sol12(-1),PDE1);

u(x, t)*`will now be displayed as`*u

 

table( [(  ) = u(x, t) ] )

 

diff(diff(u(x, t), t), t) = a^2*(diff(diff(u(x, t), x), x))+b*exp(beta*u(x, t))

 

u(x, t) = ln(2*(-A^2*a^2+B^2)/(b*beta*(A*x+B*t+C)^2))/beta

 

proc (S) options operator, arrow; u(x, t) = ln(8*a^2*C/(b*beta))/beta-2*ln(S*(x+A)^2-S*a^2*(t+B)^2+S*C)/beta end proc

 

0

 

0

 

0

(1)

The Soll11 can be plotted with a Explore plot in this form of soll11 with th eparameters , but suppose i try to get the general solution in Maple ?

infolevel[pdsolve] := 3

pdsolve(PDE1, generalsolution)

ans := pdsolve(PDE1);

What solvin gstrategy to follow ? : the pde is a non-linear wave eqation  with a exponentiel sourceterm
It seems that the pde can reduced to a ode? :

 

with(PDEtools):
declare(u(x,t));

# Stap 1: Definieer de PDE
PDE := diff(u(x,t), t,t) = a^2 * diff(u(x,t), x,x) + b * exp(beta * u(x,t));

# Stap 2: Definieer de transformatie naar karakteristieke variabelen
# Nieuw: x en t uitgedrukt in ξ en η
tr := {
    x = (xi + eta)/2,
    t = (eta - xi)/(2*a)
};

# Pas de transformatie toe op de PDE
simplified_PDE := dchange(tr, PDE, [xi, eta], params = [a, b, beta], simplify);

# Stap 3: Definieer de algemene oplossing
solution := u(x,t) = (1/beta) * ln(
    (-8*a^2/(b*beta)) *
    diff(_F1(x - a*t), x) * diff(_F2(x + a*t), x) /
    (_F1(x - a*t) + _F2(x + a*t))^2
);

# Stap 4: Controleer de oplossing (optioneel)
pdetest(solution, PDE);  # Moet 0 teruggeven als correct

u(x, t)*`will now be displayed as`*u

 

diff(diff(u(x, t), t), t) = a^2*(diff(diff(u(x, t), x), x))+b*exp(beta*u(x, t))

 

{t = (1/2)*(eta-xi)/a, x = (1/2)*xi+(1/2)*eta}

 

a^2*(diff(diff(u(xi, eta), xi), xi)-2*(diff(diff(u(xi, eta), eta), xi))+diff(diff(u(xi, eta), eta), eta)) = a^2*(diff(diff(u(xi, eta), xi), xi))+2*a^2*(diff(diff(u(xi, eta), eta), xi))+a^2*(diff(diff(u(xi, eta), eta), eta))+b*exp(beta*u(xi, eta))

 

u(x, t) = ln(-8*a^2*(D(_F1))(-a*t+x)*(D(_F2))(a*t+x)/(b*beta*(_F1(-a*t+x)+_F2(a*t+x))^2))/beta

 

0

(2)

missing some steps here : solution u  without  the pde reduced ?
there is a ode ?

# Definieer de ODE # vorige stappen ontbreken van de reduktie
ode := (v^2 - a^2) * diff(f(xi), xi, xi) = b * exp(beta * f(xi));

# Algemene oplossing zoeken
sol := dsolve(ode, f(xi));

(-a^2+v^2)*(diff(diff(f(xi), xi), xi)) = b*exp(beta*f(xi))

 

f(xi) = ln((1/2)*c__1*(tan((1/2)*(-c__1*a^2*beta+c__1*beta*v^2)^(1/2)*(c__2+xi)/(a^2-v^2))^2+1)/b)/beta

(3)

 

, ,

Question : how do i arrive on Soll11   in Maple  ?

 

Download liouville_reduced_2-2-2025_mprimes_vraag.mw

I did some example to change them but i didn't get the same result as book did it and i try to figure out by hand i get another result i know all of them are true but polar is for making the simple shap for ploting so How i can get the simple shape and if possible How do step by step like using totur we have for polar or not? 


 

restart

with(PDEtools)

with(LinearAlgebra)

NULL

with(SolveTools)

with(plots); with(plottools)

with(VectorCalculus)

with(Student:-LinearAlgebra)

with(Student:-MultivariateCalculus)

S := x^2+y^2-1

x^2+y^2-1

(1)

convert(S, polar)

polar(abs(x^2+y^2-1), argument(x^2+y^2-1))

(2)

ChangeOfVariables(S, [cartesian[x, y], polar[r, theta]])

r^2*cos(theta)^2+r^2*sin(theta)^2-1

(3)

solve(%, r)

1, -1

(4)

 

 

T := x^4+2*x^2*y^2+y^4+2*x^3+2*x*y^2-y^2

x^4+2*x^2*y^2+y^4+2*x^3+2*x*y^2-y^2

(5)

ChangeOfVariables(T, [cartesian[x, y], polar[r, theta]])

r^4*cos(theta)^4+2*r^4*cos(theta)^2*sin(theta)^2+r^4*sin(theta)^4+2*r^3*cos(theta)^3+2*r^3*cos(theta)*sin(theta)^2-r^2*sin(theta)^2

(6)

simplify(%)

r^2*(-sin(theta)^2+r*(r+2*cos(theta)))

(7)

solve(%, r)

0, 0, -cos(theta)+(cos(theta)^2+sin(theta)^2)^(1/2), -cos(theta)-(cos(theta)^2+sin(theta)^2)^(1/2)

(8)

R := -3*x^2+y^2-4*x-1

-3*x^2+y^2-4*x-1

(9)

ChangeOfVariables(R, [cartesian[x, y], polar[r, theta]])

-3*r^2*cos(theta)^2+r^2*sin(theta)^2-4*r*cos(theta)-1

(10)

simplify(%)

-4*r^2*cos(theta)^2-4*r*cos(theta)+r^2-1

(11)

solve(%, r)

-1/(2*cos(theta)+1), -1/(2*cos(theta)-1)

(12)

NULL


 

Download polar.mw

Hi All,

Maple is changing fast. It is not possible to run some older codes. 

Is it possible those who have a valid Maple license to have the old versions free of charge?

I have Maple 7, 2018, 2021 licenses but still have problem running older codes.

I read the install.html that is installed with Maple2024 and the instructions to increase stack and heap limits are definately outdated.

None of the files they list there for Linux even exists after the Maple2024 installation.

So where exactly can I set the stack and heap size limits on Linux,  as Maple give me stack and heap errors.

Hello,
I have another vector problem, and I honestly have no idea how to solve it. (I am not even sure if there is a solution). I tried to simplify it as much as possible (also the attached maple file)unknown_vector.mw

We are looking for the vectors v2, v3 and v4 
Given are vector v1 (with its unit vector e1 and magnitude m1), the unit vector of v2 (e2) and the unit vector of v3 (e3)
From vector v4 we don't have any information.
The following should be true:
v2 = v1 + v3 + v4, with the condition that v4 shall be as small as possible, with the goal to find the combination of v1 and v3 that comes as close as possible to v2. Of course, depending on the unit vectors, v4 can also be zero. But we assume that the given unit vectors can not be changed. 
I would be really grateful for any help. Thanks in advance! 
Roman

I have two equations on either side of an inequality that contain like terms such as Am and Ce. Could you simplify the expressions by mathematically eliminating these common terms from both sides? For example, if we have an equation like x+y⋅d+hx=g⋅f+hx , it simplifies to x+y⋅d=g⋅f.

Additionally, please solve for Cv and Ce.

Note: All terms are positive except R0er and R0m​.

I am attaching the relevant sheet for reference. Q_12.mw

restart; with(PDEtools); declare(F(x, t), G(x, t), H(x, t))

F(x, t)*`will now be displayed as`*F

 

G(x, t)*`will now be displayed as`*G

 

H(x, t)*`will now be displayed as`*H

(1)

q := 1-(diff(diff(log(F(x, t)), x), t)); r := G/F; s := H/F

1-(diff(diff(F(x, t), t), x))/F(x, t)+(diff(F(x, t), x))*(diff(F(x, t), t))/F(x, t)^2

 

G/F

 

H/F

(2)

r1s1 := r*s; r1s1der := diff(r1s1(x, t), x)

qt := diff(q(x, t), t)

eq1B := F(x, t)^3*(qt+r1s1der) = 0; eq12B := simplify(expand(eq1B))

-F(x, t)^3*(diff((diff(diff(F(x, t), t), x))(x, t), t))/(F(x, t))(x, t)+F(x, t)^3*(diff(diff(F(x, t), t), x))(x, t)*(diff((F(x, t))(x, t), t))/(F(x, t))(x, t)^2+F(x, t)^3*(diff((diff(F(x, t), x))(x, t), t))*(diff(F(x, t), t))(x, t)/(F(x, t))(x, t)^2-2*F(x, t)^3*(diff(F(x, t), x))(x, t)*(diff(F(x, t), t))(x, t)*(diff((F(x, t))(x, t), t))/(F(x, t))(x, t)^3+F(x, t)^3*(diff(F(x, t), x))(x, t)*(diff((diff(F(x, t), t))(x, t), t))/(F(x, t))(x, t)^2+F(x, t)*(diff(G(x, t), x))*H(x, t)-2*G(x, t)*H(x, t)*(diff(F(x, t), x))+F(x, t)*G(x, t)*(diff(H(x, t), x)) = 0

(3)

D_x_x_G_F := (diff(G(x, t), x, x))*F(x, t)-2*(diff(G(x, t), x))*(diff(F(x, t), x))+G(x, t)*(diff(F(x, t), x, x)); D_t_t_F_F := F(x, t)*(diff(F(x, t), `$`(t, 2)))-2*(diff(F(x, t), t))^2

(diff(diff(G(x, t), x), x))*F(x, t)-2*(diff(G(x, t), x))*(diff(F(x, t), x))+G(x, t)*(diff(diff(F(x, t), x), x))

 

F(x, t)*(diff(diff(F(x, t), t), t))-2*(diff(F(x, t), t))^2

(4)

NULL

rxt := diff(diff(r(x, t), x), t)

eq2B := -2*q*r+rxt = 0

eq22B := simplify(expand(eq2B))

((-F*F(x, t)*G(x, t)+2*G*F(x, t)^2)*(diff(diff(F(x, t), t), x))+(diff(diff(G(x, t), t), x))*F*F(x, t)^2+((2*F*G(x, t)-2*G*F(x, t))*(diff(F(x, t), x))-F*(diff(G(x, t), x))*F(x, t))*(diff(F(x, t), t))-(diff(G(x, t), t))*(diff(F(x, t), x))*F*F(x, t)-2*G*F(x, t)^3)/(F*F(x, t)^3) = 0

(5)

sxt := diff(diff(s(x, t), x), t)

eq3B := -2*q*s+sxt = 0

eq32B := simplify(expand(eq3B))

((-F*F(x, t)*H(x, t)+2*H*F(x, t)^2)*(diff(diff(F(x, t), t), x))+(diff(diff(H(x, t), t), x))*F*F(x, t)^2+((2*F*H(x, t)-2*H*F(x, t))*(diff(F(x, t), x))-F*(diff(H(x, t), x))*F(x, t))*(diff(F(x, t), t))-(diff(H(x, t), t))*(diff(F(x, t), x))*F*F(x, t)-2*H*F(x, t)^3)/(F*F(x, t)^3) = 0

(6)

"#`# How to simplify Eqs. (3), (5) and (6) and write in terms of following bilineat operators` by using (4)"?""

NULL

NULL

Download BE.mw

Hi,

Experiencing the following problem.  One of our servers was cloned, the GUID was replaced and then rejoined to the domain.  All applications are working exept Maple.  The application launches but then closes right away.  No error messages provided so not sure where else to look for possible fixes to this problem.  The application is runing on Server 2022.

Thank you.

It been a while i try to figure out How they find dispersion parameter and phase shift i figure out how find dispersion in some of pde but some of them is not give me even dispresion parameter i don't know they wrong or i am , but for finding phase shift there is three cenarios, when we change pde to bilinear form we have linear term in bilinear form so after substitute in linear term f bilinear form we can get dispersion parameter which is a parameter beside (t) also we can generalized for all of solution by changing the number of parameter as mention in the paper, but for phase shift parameter i don't know how find it i must substitute our solution in linear term or whole  bilinear form of in first pde linear term i try all  but i don't know what is i did mistake the paper say put in pde but i think he mention the the bilinear form i did all part for one soliton is w[1] for 2soliton is w[2] file i just want find parameter a[12] in paper for 2-soliton eq(19)  then i will find for other just i need to find one of them, thanks for any help  in this topic .

 

Here's a problem that I think works well with Maple.
Three runners, A, B, and C, are running on parallel tracks (each running at a constant speed, but not necessarily at the same speed as the others). At the start, the area of ​​triangle ABC is  , and after 5 seconds it is  . What might this area be after another   seconds?

I am trying to show that the eigenvalues of a matrix are described by my proposed formula. I managed to show this numerically, but I would like to show this symbolically. There are two issues here - the orders of the two lists are different, and the forms are different (sums of complex exponentials vs RootOfs). Any suggestions?

restart;

with(LinearAlgebra): with(GraphTheory): with(plots):

L:=9;

9

Generate a matrix and its eigenvalues

C := AdjacencyMatrix(CycleGraph(L, directed)):
Id := IdentityMatrix(L):
A := KroneckerProduct(C, Id) + KroneckerProduct(Id, C) + KroneckerProduct(C, C):
evs := Eigenvalues(A, output = list):
plotevs := complexplot(evs, style = point, color = blue, scaling = constrained):

My guess as to their values

evstheory:=[seq(seq(exp((2*Pi)*I*k/L) + exp((2*Pi)*I*m/L) + exp((2*Pi)*I*(m + k)/L), k = 0 .. L - 1), m = 0 .. L - 1)]:
plotevstheory:=complexplot(evstheory, style = point, color = red, scaling = constrained):

They look to be the same

display(Array([plotevs,plotevstheory]));

 

 

 

 

 

Even showing they are the same numerically is nontrivial because the sorting is not consistent

fnormal(sort(evalf(evs))-sort(evalf(evstheory)));

[0., 0., 0., 0.*I, 0.*I, 0.*I, 0.*I, 0.+0.*I, 0.+0.*I, 0.-1.285575219*I, 0.-1.285575219*I, 0.-.9216049846*I, 0.-.9216049846*I, 0.-1.732050808*I, 0.-1.732050808*I, 0.+1.732050808*I, 0.+1.732050808*I, 0.+.9216049846*I, 0.+.9216049846*I, 0.+1.285575219*I, 0.+1.285575219*I, 0.+0.*I, 0.+0.*I, 0.-0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+.6840402864*I, 0.+.6840402864*I, 0.-.6840402858*I, 0.-.6840402858*I, 0.+0.*I, 0.+0.*I, 0.-3.701666314*I, 0.-3.701666314*I, 0.-.6840402857*I, 0.-.6840402857*I, 0.-.6840402863*I, 0.-.6840402863*I, 0.+.6840402868*I, 0.+.6840402868*I, 0.+.6840402862*I, 0.+.6840402862*I, -0.+3.701666314*I, -0.+3.701666314*I, 0.-4.623271298*I, 0.-.6840402860*I, 0.-.6840402860*I, 0.+.6840402865*I, 0.+.6840402865*I, -0.+4.623271298*I, 0.-4.987241533*I, 0.-4.987241533*I, 0.-.6840402866*I, 0.-.6840402866*I, 0.-1.732050808*I, 0.-1.732050808*I, 0.+1.732050808*I, 0.+1.732050808*I, 0.+.6840402863*I, 0.+.6840402863*I, -0.+4.987241532*I, -0.+4.987241532*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.-4.540765944*I, 0.-0.*I, 0.-0.*I, 0.-0.*I, 0.-0.*I, -0.+4.540765944*I, 0.-1.285575219*I, 0.-1.285575219*I, 0.-1.285575219*I, 0.-1.285575219*I, -0.+2.571150438*I, -0.+2.571150438*I]

This succeeds, so they are the same

fnormal(sort(evalf[20](evs),key=evalf)-sort(evalf[20](evstheory),key=evalf));

[0., 0., 0., 0.*I, 0.*I, 0.*I, 0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.+0.*I, 0.-0.*I, 0.-0.*I, 0.+0.*I, 0.+0.*I]

What about symbolically? [Edit - only part of output shown]

ans1:=simplify(sort(evs,key=evalf)-sort(evstheory,key=evalf));

NULL

Download verification.mw

A regular polygon with n sides of length 1 is drawn. Then, the n midpoints of the edges are used to create another regular polygon inside the first. What is the length of an edge of the new polygon? As n approaches infinity, what does this length approach?

Hopefully this task is suitable for Maple?

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