dharr

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Name: Dr. David Harrington

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Member for: 20 years, 355 days

Company/Institution: University of Victoria

Title: Professor or university staff

Location: Victoria, British Columbia, Canada

Website: http://web.uvic.ca/~dharr

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Contact Dr. David Harrington

I am a retired professor of chemistry at the University of Victoria, BC, Canada. My research areas are electrochemistry and surface science. I have been a user of Maple since about 1990.

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These are answers submitted by dharr

add...

Answered: dharr 8270

January 14 2025

0 0

In general if you don't want a formula, use add, not Sum or sum. You have sums 5 deep, so it will take some time. I get the answer in 2 minutes on my machine using add.

add.mw

Applying simplify to the output of EulerLagrange solves the apparent singularity problem - odeplot can now plot the solutions. However x is not changing with time so something is wrong with the physics.

Rolling_circle.mw

ImageTools...

Answered: dharr 8270

January 13 2025

2 0

As far as I know, that can't be done directly with plot structures, as you surmised. You can convert to images and then manipulate as required with the ImageTools package (but maybe you figured this out already, because you did produce such an image).

Main plot

Inset plot

Download plots.mw

An answer...

Answered: dharr 8270

January 13 2025

2 5

Maple can find a solution to your system if given a hint. This may not be the set of solutions you are looking for. In the paper it looks like u is replaced by f,g and phi (eq 6), but you have u and f and g. So it is hard to understand the relationship of your worksheet to the paper; perhaps yoou can explain further about what you have done to get to this point. Or parhaps you like this solution.

Download PDEs_system_solution.mw

Pieces of piecewise...

Answered: dharr 8270

January 08 2025

1 1

I played aound with the record form. It is nice to look at (similar to Display(dec,R) on the default output), but hard to extract the pieces. The pieces from the piecewise output can be extracted as follows.

$eq_5382 := {(-x-y+1)*p+m*x*y = 0, y*(m*x-m-n+1)+(-x+1)*n-x = 0, (-p+t)*k+p*y-t = 0, (k-x-y)*t-k*p+y = 0, (-x-y+1)*t+(-k+y)*n+s*x = 0, (-x-y+1)*p+m*y^2+x-y = 0, m*x^2+(-m-n+1)*x+(-y+1)*n+t*y-1 = 0, -k*n+s*x = 0, 0 < k, 0 < m, 0 < s, 0 < x, 0 < y, 0 < n+(t-1)*p, 0 < (m*y-1)*n+(1-p)*(m*x-m+1), 0 < (m*x-m-t+1)*p+m*y*(t-n), 1 < y+x, k < 1, m < 1, s < t, t < 1}$

Triangularize with output as piecewise. You need to look at this to see when conditions are Anded. (Could be automated.)

sys := eq_5382; SuggestVariableOrder(sys); R := PolynomialRing(%); dec_5382 := RealTriangularize(sys, R, output = piecewise)

The sequence of pieces may be extracted by using op. They are condition,solution,condition, solution, ..., otherwise as in the ?piecewise help page.
The conditions (inequalities), if more than one, are Anded together, and can be separated out by another op

nops(dec_5382); sol := op(dec_5382); sol1conds := op(sol[1]); sol1eqns := sol[2][][]; sol2conds := sol[3]; sol2eqns := sol[4][][]; solotherwise := sol[5]

s*x-k*n = 0, (p-t)*k+(y+x)*t-y = 0, (x+y-1)*n+(-x*y+y)*m+x-y = 0, (x+y-1)*p-m*y^2-x+y = 0, m*y-1 = 0, t*y^2+x^2+(y-1)*x-y^2 = 0, x^3+(3*y-2)*x^2+(2*y^2-3*y+1)*x-y^3+y^2 = 0, 0 < k, 0 < s, 0 < x, 0 < -2*x^2*y^2-2*x*y^3+2*y^4+x^2*y+3*x*y^2-3*y^3-x*y+y^2, 0 < x^2*y^2+2*x*y^3+y^4-x^2*y-4*x*y^2-3*y^3+2*x*y+3*y^2-y, 0 < -x^2*y-x*y^2+y^3+x*y-y^2, 0 < x+y-1, 0 < 1-k, 0 < t-s, 0 < 1-t

Download Output_of_RegularChains.mw

self-contained package...

Answered: dharr 8270

January 07 2025

3 1

I agree that the geometry and geom3d packages can be frustrating in the sense that they work differently from the rest of Maple. For some reason, these packages make extensive use of attributes, which are hidden behind the scenes; attributes are rarely used in the rest of Maple. So c1 is actually just a symbol - it doesn't have a value, but its attached attributes can be seen using attributes(c1).

From a practical point of view this means you need to do all your geometry calculations with the commands in the package (or decide to calculate outside the package). In particular, use the draw command to plot objects created in the package, e.g. draw([c1,A,B,C],printtext=true);

It is possible to assign the result of draw to a variable, and then it can be combined with other regular plots with display.

As for Equation vs equation, that is frustrating though clearly outlined on the circle help page.

ThueSolve...

Answered: dharr 8270

January 05 2025

2 0

Well, as is often the case, Maple needs help to go the extra mile. Actually, ThueSolve sends this case to isolve, but without the inequalities.

Download Thue.mw

No solution by elimination of x,y,z...

Answered: dharr 8270

December 31 2024

1 4

My take on it is that the elimination process does not lead to any valid solutions.

restart;

>	expr :=[a, b, c]^~2 =~ 4[yz/((x + y)(x + z)), zx/((y + z)(y + x)), xy/((z + x)*(z + y))];

The following assumes denominators are not zero

>	expr2:=map(w->numer(normal(lhs(w)-rhs(w))),expr);

>	denoms1:={map(denom@rhs,expr)[]};

Interesting result - probably an error message or null result would be better

>	eliminate(expr2,{x,y,z});

Let's do it "by hand" - eliminate x between pairs, leaving 2 eqns in y and z. Denominators in expressions for x must be nonzero

>	e12:=eliminate(expr2[1..2],x); e23:=eliminate(expr2[2..3],x); expr3 := e12[2] union e23[2];

$[{x = -b^2*y*(y+z)/(b^2*y+b^2*z-4*z)}, {4*y*z*(a^2*b^2*y^2-a^2*b^2*z^2-b^4*y^2-2*b^4*y*z-b^4*z^2+4*a^2*z^2+8*b^2*y*z+8*b^2*z^2-16*z^2)}]$

$[{x = -c^2*z*(y+z)/(c^2*y+c^2*z-4*y)}, {b^2*c^2*y^3+b^2*c^2*y^2*z-b^2*c^2*y*z^2-b^2*c^2*z^3-4*b^2*y^3-4*b^2*y^2*z+4*c^2*y*z^2+4*c^2*z^3}]$

${4*y*z*(a^2*b^2*y^2-a^2*b^2*z^2-b^4*y^2-2*b^4*y*z-b^4*z^2+4*a^2*z^2+8*b^2*y*z+8*b^2*z^2-16*z^2), b^2*c^2*y^3+b^2*c^2*y^2*z-b^2*c^2*y*z^2-b^2*c^2*z^3-4*b^2*y^3-4*b^2*y^2*z+4*c^2*y*z^2+4*c^2*z^3}$

>	denoms2:={denom(rhs(e12[1][])), denom(rhs(e23[1][]))};

${b^2*y+b^2*z-4*z, c^2*y+c^2*z-4*y}$

Now eliminate y between the remaining 2 equations - denom has no restrictions on z

>	e13:=eliminate(expr3,y);

$[{y = -z*c^2*(-a^2*b^4*c^2+a^4*b^2+2*a^2*b^4+5*a^2*b^2*c^2+b^6+3*b^4*c^2-4*a^4-12*a^2*b^2-4*a^2*c^2-16*b^4-12*b^2*c^2+16*a^2+48*b^2)/(-a^2*b^4*c^4+a^4*b^2*c^2+2*a^2*b^4*c^2+3*a^2*b^2*c^4+b^6*c^2+3*b^4*c^4-16*b^4*c^2-4*b^2*c^4-16*a^4-16*a^2*b^2-32*a^2*c^2-16*c^4+128*a^2+64*b^2+128*c^2-256)}, {c*z*(b-2)*(b+2)*(a-2)*(a+2)*(-a*b*c+a^2+b^2+c^2-4)*(a*b*c+a^2+b^2+c^2-4)}]$

>	denoms3:=denom(rhs(e13[1][]));

>	eqz:={isolate(e13[2][],z)};

From which we conclude y=0

>	eqy:=eval(e13[1],eqz);

So we have a problem at the eliminate x stage, and with the original equations

>	eval(denoms2, eqy union eqz); eval(denoms1, eqy union eqz);

${0, x^2}$

>	solve(expr,{x,y,z},parametric); # correct

Here the c<>0 solution doesn't solve the original equations (zero denoms). And solve is too lazy to do the c=0 case (default option - see ?solve,parametric

>	ans := solve(expr2,{x,y,z},parametric);

$ans := piecewise(c = 0, %SolveTools[Parametric]({-4*y*x, a^2*x^2+a^2*x*y+a^2*x*z+a^2*y*z-4*y*z, b^2*x*y+b^2*x*z+b^2*y^2+b^2*y*z-4*x*z}, {x, y, z}, {a, b}), c <> 0, [{x = 0, y = 0, z = 0}])$

Force it. None of these solve the original equations either.

>	op(2,ans); value(%);

$%SolveTools[Parametric]({-4*y*x, a^2*x^2+a^2*x*y+a^2*x*z+a^2*y*z-4*y*z, b^2*x*y+b^2*x*z+b^2*y^2+b^2*y*z-4*x*z}, {x, y, z}, {a, b})$

piecewise(b = 0, [{x = 0, y = 0, z = z}, {x = 0, y = y, z = 0}], b <> 0, [{x = 0, y = 0, z = z}])

In one step

>	ans := solve(expr2,{x,y,z},parametric=full);

ans := piecewise(c = 0, piecewise(b = 0, [{x = 0, y = 0, z = z}, {x = 0, y = y, z = 0}], b <> 0, [{x = 0, y = 0, z = z}]), c <> 0, [{x = 0, y = 0, z = 0}])

Same result.

>	SolveTools:-PolynomialSystem(expr2,[x,y,z]);

Download eliminate.mw

remove restriction...

Answered: dharr 8270

December 29 2024

1 2

Your question is not clear. Is this what you want to do?

Download removal.mw

Array...

Answered: dharr 8270

December 27 2024

2 2

Put the plots in an Array to achieve this. Attached is an example. (grid is for the fineness of the points plotted, and insequence is, as you found, for animations.)

grid.mw

slightly shorter...

Answered: dharr 8270

December 23 2024

1 0

I don't know a way that doesn't involve a string. But this is slightly shorter:

cat(``,String(x)[1]);

solve...

Answered: dharr 8270

December 21 2024

2 1

I think you just want to solve the 3 equations simultaneously - if there is a solution it lies on the line; if no solution it does not

solve({seq(l[i] = P[i], i = 1 .. 3)}, alpha)

2024-12-21_Q_3D_point_lies_on_3D_line.mw

complex conjugate...

Answered: dharr 8270

December 21 2024

2 0

It means complex conjugate, i.e., every i is replaced by -i.

Hyperbolas...

Answered: dharr 8270

December 19 2024

1 2

No artificial intelligence used.

We consider an ellipse x^2/a^2+y^2/b^2-1=0 and 2 vertices of this ellipse A(a,0) and B(0,b). We imagine a variable equilateral hyperbola passing through the points O, A and B. This curve meets the ellipse at 2 other points A1 and B1. Show that the line A1B1 passes through a fixed point.

https://www.mapleprimes.com/questions/239553-Variable-Equilateral-Hyperbole-Passing

Define ellipse and an arbitrary hyperbola (B^2-4A*C>0) (see Wikipedia)

ellipse := x^2/a^2+y^2/b^2-1; hyperbola := A*x^2+B*x*y+C*y^2+D*x+E*y+F

x^2/a^2+y^2/b^2-1

The hyperbola is equilateral when A=-C

Intersection points between ellipse and hyperbola are at O (assume this means the origin), A, B. Use these to eliminate D,E,F

eqO := eval(hyperbola2, {x = 0, y = 0}); eqA := eval(hyperbola2, {x = a, y = 0}); eqB := eval(hyperbola2, {x = 0, y = b})

A1 and B1 are two points on both the ellipse and the hyperbola. The 4 solutions are points B, A, A1, B1; A1 and B1 are complicated expressions in A.B,a,b