Maple Questions and Posts

Why this unterminated loop error?...

Asked by: RohanKarthik 0

14 minutes ago

0 0

rfin := proc(m::integer) 
local c, i, flg := 0;
for i from 0 to m do
	local b := i;
	do 
		c := b mod 3;
		if (c <> 2) then next else flg := 1; 
		end if;
		b := 1/3*b - 1/3*c;
	until b = 0;
	if evalb(flg = 0) then print(i); 
	end if; 
end do;
end proc;

Hello everyone. I've written a procedure that outputs all numbers <= a given user I/P whose ternary representation has no 2.

However, I get an "Error, unterminated loop" message. Can someone please out the mistake(s)?

Vectors in Spherical Coordinates using Tensor...

by: ecterrab 7951 Product: Maple

9 hours ago

4 0

Vectors in Spherical coordinates using tensor notation

Edgardo S. Cheb-Terrab¹ and Pascal Szriftgiser²

(2) Laboratoire PhLAM, UMR CNRS 8523, Université de Lille, F-59655, France

(1) Maplesoft

The following is a topic that appears frequently in formulations: given a 3D vector in spherical (or any curvilinear) coordinates, how do you represent and relate, in simple terms, the vector and the corresponding vectorial operations Gradient, Divergence, Curl and Laplacian using tensor notation?

The core of the answer is in the relation between the - say physical - vector components and the more abstract tensor covariant and contravariant components. Focusing the case of a transformation from Cartesian to spherical coordinates, the presentation below starts establishing that relationship between 3D vector and tensor components in Sec.I. In Sec.II, we verify the transformation formulas for covariant and contravariant components on the computer using TransformCoordinates. In Sec.III, those tensor transformation formulas are used to derive the vectorial form of the Gradient in spherical coordinates. In Sec.IV, we switch to using full tensor notation, a curvilinear metric and covariant derivatives to derive the 3D vector analysis traditional formulas in spherical coordinates for the Divergence, Curl, Gradient and Laplacian. On the way, some useful technics, like changing variables in 3D vectorial expressions, differential operators, using Jacobians, and shortcut notations are shown.

The computation below is reproducible in Maple 2020 using the Maplesoft Physics Updates v.640 or newer.

Start setting the spacetime to be 3-dimensional, Euclidean, and use Cartesian coordinates

>

$`Default differentiation variables for d_, D_ and dAlembertian are:`*{X = (x, y, z)}$

Physics:-g_[mu, nu] = Matrix(%id = 18446744078372175734)

(1)

I. The line element in spherical coordinates and the scale-factors

In vector calculus, at the root of everything there is the line element , which in Cartesian coordinates has the simple form

>

(1.1)

To compute the line element in spherical coordinates, the starting point is the transformation

>

(1.2)

>

(1.3)

Since in are just symbols with no relationship to start transforming these differentials using the chain rule, computing the Jacobian of the transformation (1.2). In this Jacobian J, the first line on the right-hand side is , ,

>

So in matrix notation,

>

Vector[column](%id = 18446744078719224758) = Vector[column](%id = 18446744078719224998)

(1.4)

To complete the computation of in spherical coordinates we can now use ChangeBasis , provided that next we substitute (1.4) in the result, expressing the abstract objects in terms of .

In two steps:

>

(1.5)

The line element

>

(1.6)

This result is important: it gives us the so-called scale factors, the key that connect 3D vectors with the related covariant and contravariant tensors in curvilinear coordinates. The scale factors are computed from (1.6) by taking the scalar product with each of the unit vectors , then taking the coefficients of the differentials (just substitute them by the number 1)

>

(1.7)

The scale factors are relevant because the components of the 3D vector and the corresponding tensor are not the same in curvilinear coordinates. For instance, representing the differential of the coordinates as the tensor , we see that corresponding vector, the line element in spherical coordinates , is not constructed by directly equating its components to the components of , so

The vector is constructed multiplying these contravariant components by the scaling factors, as

This rule applies in general. The vectorial components of a 3D vector in an orthogonal system (curvilinear or not) are always expressed in terms of the contravariant components the same way we did in the line above with the line element, using the scale-factors , so that

$`#mover(mi("A"),mo("→"))` = Sum(h[j]*A^j*`#mover(mi("\`e__j\`"),mo("&circ;"))`, j = 1 .. 3)$

where on the right-hand side we see the contravariant components and the scale-factors . Because the system is orthogonal, each vector component satisfies

The scale-factors do not constitute a tensor, so on the right-hand side we do not sum over j. Also, from

it follows that,

A__j = A__j/h__j

where on the right-hand side we now have the covariant tensor components .

•	This relationship between the components of a 3D vector and the contravariant and covariant components of a tensor representing the vector is key to translate vector-component to corresponding tensor-component formulas.

II. Transformation of contravariant and covariant tensors

Define here two representations for one and the same tensor: will represent A in Cartesian coordinates, while will represent A in spherical coordinates.

>

${A__c[j], A__s[j], Physics:-Dgamma[a], Physics:-Psigma[a], Physics:-d_[a], Physics:-g_[a, b], Physics:-LeviCivita[a, b, c], Physics:-SpaceTimeVector[a](S), Physics:-SpaceTimeVector[a](X)}$

(2.1)

Transformation rule for a contravariant tensor

We know, by definition, that the transformation rule for the components of a contravariant tensor is $`#mrow(msup(mi("A"),mi("μ",fontstyle = "normal")),mo("⁡"),mfenced(mi("y")),mo("="),mfrac(mrow(mo("&PartialD;"),msup(mi("y"),mi("μ",fontstyle = "normal"))),mrow(mo("&PartialD;"),msup(mi("x"),mi("ν",fontstyle = "normal"))),linethickness = "1"),mo("⁢"),mo("⁢"),msup(mi("A"),mi("ν",fontstyle = "normal")),mfenced(mi("x")))`$ , that is the same as the rule for the differential of the coordinates. Then, the transformation rule from to computed using TransformCoordinates should give the same relation (1.4). The application of the command, however, requires attention, because, as in (1.4), we want the Cartesian (not the spherical) components isolated. That is like performing a reversed transformation. So we will use

where on the left-hand side we get, isolated, the three components of A in Cartesian coordinates, and on the right-hand side we transform the spherical components , from spherical (4^th argument) to Cartesian (3^rd argument), which according to the 5^th bullet of TransformCoordinates will result in a transformation expressed in terms of the old coordinates (here the spherical ). Expand things to make the comparison with (1.4) possible by eye

>

Vector[column](%id = 18446744078703000142) = Vector[column](%id = 18446744078703000622)

(2.2)

We see that the transformation rule for a contravariant vector is, indeed, as the transformation (1.4) for the differential of the coordinates.

Transformation rule for a covariant tensor

For the transformation rule for the components of a covariant tensor , we know, by definition, that it is $`#mrow(msub(mi("A"),mi("μ",fontstyle = "normal")),mo("⁡"),mfenced(mi("y")),mo("="),mfrac(mrow(mo("&PartialD;"),msup(mi("x"),mi("ν",fontstyle = "normal"))),mrow(mo("&PartialD;"),msup(mi("y"),mi("μ",fontstyle = "normal"))),linethickness = "1"),mo("⁢"),mo("⁢"),msub(mi("A"),mi("ν",fontstyle = "normal")),mfenced(mi("x")))`$ , so the same transformation rule for the gradient , where and so on. We can experiment this by directly changing variables in the differential operators , for example

>

Physics:-d_[x] = (proc (u) options operator, arrow; ((-r*cos(theta)^2+r)*cos(phi)*(diff(u, r))+sin(theta)*cos(phi)*cos(theta)*(diff(u, theta))-(diff(u, phi))*sin(phi))/(r*sin(theta)) end proc)

(2.3)

This result, and the equivalent ones replacing x by y or z in the input above can be computed in one go, in matricial and simplified form, using the Jacobian of the transformation computed in . We need to take the transpose of the inverse of J (because now we are transforming the covariant components)

>

Vector[column](%id = 18446744078775738110) = Vector[column](%id = 18446744078775730294)

(2.4)

The corresponding transformation equations relating the tensors and in Cartesian and spherical coordinates is computed with TransformCoordinates as in (2.2), just lowering the indices on the left and right hand sides (i.e., remove the tilde ~)

>

Vector[column](%id = 18446744078372166222) = Vector[column](%id = 18446744078372166702)

(2.5)

We see that the transformation rule for a covariant vector is, indeed, as the transformation rule (2.4) for the gradient.

To the side: once it is understood how to compute these transformation rules, we can have the inverse of (2.5) as follows

>

Vector[column](%id = 18446744078702994846) = Vector[column](%id = 18446744078702995446)

(2.6)

III. Deriving the transformation rule for the Gradient using TransformCoordinates

Start with a 3D vector in Cartesian coordinates

>

(3.1)

>

(3.2)

The gradient of a function f in Cartesian coordinates and spherical coordinates is respectively given by

>

(3.3)

>

(3.4)

What we want now is to depart from (3.3) in Cartesian coordinates and obtain (3.4) in spherical coordinates using the transformation rule for a covariant tensor computed with TransformCoordinates in (2.5). (An equivalent derivation, simpler and with less steps is done in Sec. IV.)

Start changing the vector basis in the gradient (3.3)

>

%Nabla(f(X)) = ((diff(f(X), x))*sin(theta)*cos(phi)+(diff(f(X), y))*sin(theta)*sin(phi)+(diff(f(X), z))*cos(theta))*_r+((diff(f(X), x))*cos(phi)*cos(theta)+(diff(f(X), y))*sin(phi)*cos(theta)-(diff(f(X), z))*sin(theta))*_theta+(-(diff(f(X), x))*sin(phi)+cos(phi)*(diff(f(X), y)))*_phi

(3.5)

By eye, we see that in this result the coefficients of are the three lines in the right-hand side of (2.6) after replacing the covariant components by the derivatives of f with respect to the j^th coordinate, here displayed using indexed notation due to using CompactDisplay

>

(3.6)

>

(3.7)

So since (2.5) is the inverse of (2.6), replace A by ∂ f in (2.5), the formula computed using TransformCoordinates, then insert the result in (3.5) to relate the gradient in Cartesian and spherical coordinates. We expect to arrive at the formula for the gradient in spherical coordinates (3.4) .

>

Vector[column](%id = 18446744078744210598) = Vector[column](%id = 18446744078744210478)

(3.8)

>

(3.9)

Simplifying, we arrive at (3.4)

>

(3.10)

>

(3.11)

IV. Deriving the transformation rule for the Divergence, Curl, Gradient and Laplacian, using TransformCoordinates and Covariant derivatives

•	The Divergence

Introducing the vector A in spherical coordinates, its Divergence is given by

>

(4.1)

>

(4.2)

>

%Divergence(%A__s_) = ((diff(A__r(S), r))*r+2*A__r(S))/r+((diff(`A__θ`(S), theta))*sin(theta)+`A__θ`(S)*cos(theta))/(r*sin(theta))+(diff(`A__φ`(S), phi))/(r*sin(theta))

(4.3)

We want to see how this result, (4.3), can be obtained using TransformCoordinates and departing from a tensorial representation of the object, this time the covariant derivative . For that purpose, we first transform the coordinates and the metric introducing nonzero Christoffel symbols

>

$`Default differentiation variables for d_, D_ and dAlembertian are:`*{S = (r, theta, phi)}$

Physics:-g_[a, b] = Matrix(%id = 18446744078775698726)

(4.4)

To the side: despite having nonzero Christoffel symbols, the space still has no curvature, all the components of the Riemann tensor are equal to zero

>

(4.5)

Consider now the divergence of the contravariant tensor, computed in tensor notation

>

(4.6)

>

(4.7)

To the side: the covariant derivative expressed using the D_ operator can be rewritten in terms of the non-covariant d_ and Christoffel symbols as follows

>

(4.8)

Summing over the repeated indices in (4.7), we have

>

%D_[j](%A__s[`~j`]) = diff(A__s[`~1`](S), r)+diff(A__s[`~2`](S), theta)+diff(A__s[`~3`](S), phi)+2*A__s[`~1`](S)/r+cos(theta)*A__s[`~2`](S)/sin(theta)

(4.9)

How is this related to the expression of the $VectorCalculus[Nabla].`#mover(mi("\`A__s\`"),mo("→"))`$ in (4.3)? The answer is in the relationship established at the end of Sec I between the components of the tensor and the components of the vector $`#mover(mi("\`A__s\`"),mo("→"))`$ , namely that the vector components are obtained multiplying the contravariant tensor components by the scale-factors . So, in the above we need to substitute the contravariant by the vector components divided by the scale-factors

>

(4.10)

>

%D_[j](%A__s[`~j`]) = diff(A__r(S), r)+(diff(`A__θ`(S), theta))/r+(diff(`A__φ`(S), phi))/(r*sin(theta))+2*A__r(S)/r+cos(theta)*`A__θ`(S)/(sin(theta)*r)

(4.11)

Comparing with (4.3), we see these two expressions are the same:

>

%Divergence(%A__s_) = diff(A__r(S), r)+(diff(`A__θ`(S), theta))/r+(diff(`A__φ`(S), phi))/(r*sin(theta))+2*A__r(S)/r+cos(theta)*`A__θ`(S)/(sin(theta)*r)

(4.12)

•

The Curl

The Curl of the the vector $`#mover(mi("\`A__s\`"),mo("→"))`$ in spherical coordinates is given by

>

%Curl(%A__s_) = ((diff(`A__φ`(S), theta))*sin(theta)+`A__φ`(S)*cos(theta)-(diff(`A__θ`(S), phi)))*_r/(r*sin(theta))+(diff(A__r(S), phi)-(diff(`A__φ`(S), r))*r*sin(theta)-`A__φ`(S)*sin(theta))*_theta/(r*sin(theta))+((diff(`A__θ`(S), r))*r+`A__θ`(S)-(diff(A__r(S), theta)))*_phi/r

(4.13)

One could think that the expression for the Curl in tensor notation is as in a non-curvilinear system

But in a curvilinear system is not a tensor, we need to use the non-Galilean form , where is the determinant of the metric. Moreover, since the expression has one free covariant index (the first one), to compare with the vectorial formula (4.12) this index also needs to be rewritten as a vector component as discussed at the end of Sec. I, using

A__j = A__j/h__j

The formula (4.13) for the vectorial Curl is thus expressed using tensor notation as

>

(4.14)

>

%Curl(%A__s_) = Physics:-LeviCivita[i, j, k]*Physics:-D_[`~j`](A__s[`~k`](S), [S])/%h[i]

(4.15)

followed by replacing the contravariant tensor components by the vector components using (4.10). Proceeding the same way we did with the Divergence, expand this expression. We could use TensorArray , but Library:-TensorComponents places a comma between components making things more readable in this case

>

(4.16)

Replace now the components of the tensor by the components of the 3D vector $`#mover(mi("\`A__s\`"),mo("→"))`$ using (4.10)

>

(4.17)

>

%Curl(%A__s_) = [((diff(`A__φ`(S), theta))*sin(theta)+`A__φ`(S)*cos(theta)-(diff(`A__θ`(S), phi)))/(r*sin(theta)), (diff(A__r(S), phi)-(diff(`A__φ`(S), r))*r*sin(theta)-`A__φ`(S)*sin(theta))/(r*sin(theta)), ((diff(`A__θ`(S), r))*r+`A__θ`(S)-(diff(A__r(S), theta)))/r]

(4.18)

We see these are exactly the components of the Curl (4.13)

>

%Curl(%A__s_) = ((diff(`A__φ`(S), theta))*sin(theta)+`A__φ`(S)*cos(theta)-(diff(`A__θ`(S), phi)))*_r/(r*sin(theta))+(diff(A__r(S), phi)-(diff(`A__φ`(S), r))*r*sin(theta)-`A__φ`(S)*sin(theta))*_theta/(r*sin(theta))+((diff(`A__θ`(S), r))*r+`A__θ`(S)-(diff(A__r(S), theta)))*_phi/r

(4.19)

•	The Gradient

Once the problem is fully understood, it is easy to redo the computations of Sec.III for the Gradient, this time using tensor notation and the covariant derivative. In tensor notation, the components of the Gradient are given by the components of the right-hand side

>

%Nabla(f(S)) = `&dtri;`[j](f(S))/%h[j]

%Nabla(f(S)) = Physics:-d_[j](f(S), [S])/%h[j]

(4.20)

where on the left-hand side we have the vectorial Nabla differential operator and on the right-hand side, since is a scalar, the covariant derivative becomes the standard derivative .

>

(4.21)

The above is the expected result (3.4)

>

(4.22)

•	The Laplacian

Likewise we can compute the Laplacian directly as

>

(4.23)

In this case there are no free indices nor tensor components to be rewritten as vector components, so there is no need for scale-factors. Summing over the repeated indices,

>

%Laplacian(f(S)) = Physics:-dAlembertian(f(S), [S])+2*(diff(f(S), r))/r+cos(theta)*(diff(f(S), theta))/(sin(theta)*r^2)

(4.24)

Evaluating the Vectors:-Laplacian on the left-hand side,

>

((diff(diff(f(S), r), r))*r+2*(diff(f(S), r)))/r+((diff(diff(f(S), theta), theta))*sin(theta)+cos(theta)*(diff(f(S), theta)))/(r^2*sin(theta))+(diff(diff(f(S), phi), phi))/(r^2*sin(theta)^2) = Physics:-dAlembertian(f(S), [S])+2*(diff(f(S), r))/r+cos(theta)*(diff(f(S), theta))/(sin(theta)*r^2)

(4.25)

On the right-hand side we see the dAlembertian , in curvilinear coordinates; rewrite it using standard diff derivatives and expand both sides of the equation for comparison

>

diff(diff(f(S), r), r)+(diff(diff(f(S), theta), theta))/r^2+(diff(diff(f(S), phi), phi))/(r^2*sin(theta)^2)+2*(diff(f(S), r))/r+cos(theta)*(diff(f(S), theta))/(sin(theta)*r^2) = diff(diff(f(S), r), r)+(diff(diff(f(S), theta), theta))/r^2+(diff(diff(f(S), phi), phi))/(r^2*sin(theta)^2)+2*(diff(f(S), r))/r+cos(theta)*(diff(f(S), theta))/(sin(theta)*r^2)

(4.26)

This is an identity, the left and right hand sides are equal:

>

(4.27)

>

Download Vectors_and_Spherical_coordinates_in_tensor_notation.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

I can't remove rootOf...

Asked by: fmoraga 10

9 hours ago

2 3

I am trying to simplify an expression, and leave W in terms of other variables. But unfortunately I can't delete the RootOf, and I can't see the expressions. Could you please help me.

optimization problem postive integer solution ...

Asked by: Zeineb 160

10 hours ago

0 5

Dear all

I have an optimization problem. I would like to add a condition to obtain only a positive integer as the solution to the problem

Positive_integer.mw

thanks

New Bug of Maple 2020？...

Asked by: Glowing 90

11 hours ago

1 1

The .mw files have already been associated with Maple 2020, but when clicking on any of them, Maple 2020 won't lunch as expected and nothing actually will happen.
I have tried to reinstall Maple 2020, but the problem just keep existing.

The Windows version where the bug occurs is Windows 10 version 1909.

Problem with solving an equation ...

Asked by: briyola 5

17 hours ago

1 2

Hello everyone I need your help please, I found a problem with solving an equation to explain this in detail ,here is the problem how it is posed :

first of all we have N(q)=(exp(q)/sqrt(q))*sqrt(3/2)*D(3*q/2) . (1)

with: D(3*q/2)=exp(-3*q/2)*integral(t^2)*dt with t varie from 0 to sqrt(3*q/2) this integral is known as Dawson's name (2)

then we have q=(24/(n-3))*(x*k*m)/t with : t=0.6,n=6,k=1.3, so q=18.66666667*x*m (3)

finaly we have m = 1/2*[(exp(q)/(q*N(q)))-1-1/q] (4)

the quetion : is we must find the expression of m (4) as a function of x only .

thank you evryone .

How to generate a random integer with customized p...

Asked by: rstellian 40

18 hours ago

2 5

Hi everyone,

In the RandomTools package, the Generate(integer(range = A..B)) function generates a random integer in the range A..B. All integers in that range have the same probability to be generated, that is, 1/nops([seq(A..B)]). However, I would like to specify the probabilities of each integer. How to do so?

Example: range = 1..5. Instead of P(X=j)=1/5 with j =1,2,3,4,5, let's say the probabilities should be as follows:

P(X=1) = 0.2, P(X=2) = 0.5, P(X=3) = P(X=4) = P(X=5) = 0.1

How to generate a random integer between 1 and 5 with these probabilities?

Thank you in advance.

Error, numeric exception: division by zero...

Asked by: Jameel123 20

19 hours ago

0 3

Error, numeric exception: division by zeroprpblem_maple_2.mw

Download prpblem_maple_2.mw

can someone help me with error ' Error, unable to ...

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22 hours ago

2 3

Looking for tips to solve system of two (complicat...

Asked by: ajfriedlan 40

Yesterday at 12:50 AM

1 5

Hello experts,

I am trying to use solve to find a solution to a system of two equations. The two equations involved are quite complicated, and so sometimes Maple has difficulty with this. In particular, the solve command tries evaluating but never comes up with a solution (I've waited at least an hour, before giving up).

I recently learned about fsolve, which gives approximate numeric solutions (which would be fine for my purposes), but fsolve too struggles with a solution and simply returns my input to me. I tried plotting the system of equations using plots:-implictplot to see if a solution existed, and as expected it does. I was hoping to get some tips on trying to solve a difficult system like this, perhaps given the knowledge that a solution definitely exists. Unfortunately, I need a solution for many variations of the same system, so simply reading off the approximate solution isn't really an option.

In my attatched code, the system with W = 49 is the first one where Maple really begins to struggle, and I believe that solutions for W>49 are also difficult.

06042020_Predicting_w_AB_Ratio_Maple_Primes.mw

Thanks!

How to count the number of nonlinear terms in a li...

Asked by: emendes 140

Yesterday at 8:14 PM

0 12

Hello

I need to count and separate the nonlinear terms in a list. Example:

w:=[[z, y, x, 1], [x*z, x*y, y, 1], [x*z, z, x*y]];

there are 4 nonlinear terms, x*z, x*y, x*z, and x*y.

The terms can be any combination of the given variables, that is, x, y, and z.

My solution to the problem of counting the nonlinear terms is

aux1:=[seq([seq(nops(w[j,i]),i=1..nops(w[j]))],j=1..nops(w))];

aux2:=[seq(selectremove(x->x>1,aux1[i])[1],i=1..nops(aux1))]

res:=convert([seq(convert(nops(aux2[i]),`+`),i=1..nops(aux2))],`+`);

It works but I wonder whether there is a better solution that includes showing the nonlinear terms themselves.

Many thanks

Ed

Cannot find integration...

Asked by: Jameel123 20

Yesterday at 7:29 PM

0 0

Cannot find integration proplem_in_maple.mw

>

(1)

>

(2)

>

a := 0; b := 1; N := 5; h := (b-a)/N; B[0] := 1; B[1] := x; n := 2; B[2] := x^2+2; alpha := 1/2

(3)

>

(4)

>

(5)

>

y := sum(c[s]*B[s], s = 0 .. N)

(6)

>

(7)

>

k := expand(int(yt*sin(t)*x, t = 0 .. x))

x*c[0]+22*x*c[4]-c[1]*cos(x)*x^2-c[2]*cos(x)*x^3+2*c[2]*sin(x)*x^2-c[3]*cos(x)*x^4+3*c[3]*sin(x)*x^3+5*c[3]*cos(x)*x^2-c[4]*cos(x)*x^5+4*c[4]*sin(x)*x^4+12*c[4]*cos(x)*x^3-24*c[4]*sin(x)*x^2-c[5]*cos(x)*x^6+5*c[5]*sin(x)*x^5+21*c[5]*cos(x)*x^4-63*c[5]*sin(x)*x^3-123*c[5]*cos(x)*x^2-x*cos(x)*c[0]-22*x*cos(x)*c[4]+x*c[1]*sin(x)-5*x*c[3]*sin(x)+123*x*c[5]*sin(x)

(8)

>

(9)

>

f := (8*x^3*(1/3)-2*x^(1/2))*y/GAMMA(1/2)+(1/1260)*x+k4

(10)

>

(11)

>

proc (f, alpha, x) options operator, arrow; Int((x-s)^(alpha-1)*f(s)/GAMMA(alpha), s = 0 .. x) end proc

(12)

>

proc (f, alpha, x) options operator, arrow; Int((x-s)^(alpha-1)*f(s)/GAMMA(alpha), s = 0 .. x) end proc

(13)

>

(14)

>

(15)

>

(16)

>

Download proplem_in_maple.mw

initial condition...

Asked by: r66 15

Yesterday at 1:31 PM

0 1

hi

i have n initial condition like x(0)=a₀,x^'(0)=a₁.....x^(n-1)(0)=a_n-1 and also n equations like S[i]. i d like to write following code in maple

S[i]- a_i-1=0 for i=1,2,..,n

would you please help me how should i do it

thanks a lot

Two problems with solve()...

Asked by: John Blacksad 10

April 06 2020

0 2

Hello again, it's my first time using maple, so I have more problems :(

I need solve three equations, but maple shows an error:

Error, (in solve) a constant is invalid as a variable, gamma

gamma.mw

I have used g instead of gamma and solve works, but I don't understand what happen.

The second problem is when I use g instead of gamma. In some tau values solve() doesn't show the solution. In each solve() always there are one unnecessary equation. Maybe could be that. But I don't know.

g.mw

Thank you in advance!!!

P.S. Sorry for my bad English

Empty result set on `pdsolve` in Maple2019...

Asked by: lcz 115

April 06 2020

0 2

Hello! I'm trying to solve the following:

pde1 := (y+z)*(diff(u(x, y, z), x))+(z+x)*(diff(u(x, y, z), y))+(x+y)*(diff(u(x, y, z), z)) = 0;
{pdsolve(pde1,u(x,y,z))}

Unfortunately, after calling pdsolve , I get an empty result set . Can you help me figure out what's going on? Does it really have no solution?

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