dharr

Ask a Question

Create a Post

Name: Dr. David Harrington

22 Badges

Member for: 20 years, 339 days

Company/Institution: University of Victoria

Title: Professor or university staff

Location: Victoria, British Columbia, Canada

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

Social Networks and Content at Maplesoft.com

Maple Application Center

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.

MaplePrimes Activity

These are answers submitted by dharr

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.

eq2 := 60*g*sigma*(diff(q, y))*(diff(q, x))^3-30*sigma*(diff(q, y))*(diff(q, x))^2*g^2+18*(diff(g, x))*(diff(q, y))*(diff(q, x))^2-15*(diff(q, y))*(diff(q, x))*(diff(g, x))*g+18*g*(diff(q, y))*(diff(q, x))*(diff(q, x, x))-3*(diff(q, y))*g^2*(diff(q, x, x))+6*(diff(g, y))*(diff(q, x))^3+18*g*(diff(q, x))^2*(diff(q, y, x))-9*(diff(g, y))*g*(diff(q, x))^2-3*g^2*(diff(q, x))*(diff(q, y, x)) = 0

eq6 := -3*(diff(f, x))*(diff(f, y, x))-3*(diff(f, y))*(diff(f, x, x))+diff(f, y, x, x, x)+diff(f, y, t) = 0

${f = f__1(x)+c__1+f__3(t), g = f__4(x)+c__2+f__6(t), q = f__7(x)+c__3+f__9(t)}$

So we can replace the subscripted f functions with arbitrary functions of x (or t), e.g.,

>	$sol1 := eval(sol, {f__1(x) = x^2, f__3(t) = sin(t), f__4(x) = exp(x), f__6(t) = t, f__7(x) = cos(x), f__9(t) = t^3})$

${f = x^2+c__1+sin(t), g = exp(x)+c__2+t, q = cos(x)+c__3+t^3}$

Download PDEs_system_solution.mw

Pieces of piecewise...

Answered: dharr 8215

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 8215

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 8215

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 8215

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 8215

December 29 2024

1 2

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

Download removal.mw

Array...

Answered: dharr 8215

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 8215

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 8215

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 8215

December 21 2024

2 0

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

Hyperbolas...

Answered: dharr 8215

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

If the assertion is true, then the line A1-B1 intersects at the same point for different {A,B} pairs.

>	$L1 := Line(eval([A1, B1][], {A = A__1, B = B__1})); L2 := Line(eval([A1, B1][], {A = A__2, B = B__2}))$

And we indeed find a value independent of

[-b^2*a/(a^2+b^2), -a^2*b/(a^2+b^2)]

Construct a diagram for two different hyperbolas

params0 := {a = 1.5, b = 1.}; params := `union`(params0, {A = 1., B = 1.}); params2 := `union`(remove(has, params, B), {B = 3.})

Red points are O,A,B,X; Black and blue points are A1,B1 for the blue and black hyperbolas

Download Ellipse.mw

sort...

Answered: dharr 8215

December 17 2024

1 3

I'm not sure exactly what you mean by labelled sets. You can do something like this

ans := sort([solve({x^2 - x + 1}, x)])

sort will sort in a consistent way from session to session, though what exact the order is is nontrivial.

missing multiplication...

Answered: dharr 8215

December 17 2024

1 0

You are missing the multiplication between the (x+7) and (x-1), so l45 := (x + 7)*(x - 1) = (1 + x)^2;

You can use a space instead of * if you are using 2D input

Octagon...

Answered: dharr 8215

December 17 2024

4 0

Vectors of adjacent sides

Origin and far point C

Midpoints

Area of polygon formula. Accepts a sequence of points in cyclic order.

Area := proc () local n, V, i; n := nargs; V := Array(0 .. n-1, [args]); simplify((1/2)*add(-V[`mod`(i+1, n)][1]*V[i][2]+V[`mod`(i+1, n)][2]*V[i][1], i = 0 .. n-1)) end proc