MaplePrimes - Maple 2018 Posts and Questions

How to compute flow rate, impedance, and shear stress?

KIRAN SAJJAN — Wed, 11 Mar 2026 16:59:18 Z

I am solving a hybrid nanofluid flow problem in a bifurcated artery using Maple. The governing equations for velocity and temperature are solved using dsolve(..., numeric, method=bvp[midrich]).

My Maple code successfully produces velocity profiles for different Hartmann numbers Ha for both the artery parentartery_and_daughter_artery_error.mw.

The velocity profiles are obtained correctly using odeplot.

However, I want to compute additional physical quantities and generate plots similar to the velocity profiles.

Specifically I want to plot:

Flow rate Q versus axial distance z
Impedance (flow resistance) λ versus z
Wall shear stress τ versus z

for different values of Hartmann number Ha.

The formulas I am using are

Flow rate:

Q=2π(R∫01ηw(η) dη+R2∫01w(η) dη)Q = 2\pi \left( R \int_0^1 \eta w(\eta)\,d\eta + R_2 \int_0^1 w(\eta)\,d\eta \right)Q=2π(R∫01ηw(η)dη+R2∫01w(η)dη)

Wall shear stress:

τ=μ∣dwdr∣\tau = \mu \left|\frac{dw}{dr}\right|τ=μdrdw

Impedance:

λ=∣dp/dz∣Q\lambda = \frac{|dp/dz|}{Q}λ=Q∣dp/dz∣
Please help me to solve this question.

Paths on integer lattices

Kitonum — Thu, 05 Mar 2026 08:36:24 Z

This post was inspired by the following discussion thread https://mapleprimes.com/questions/242266-Count-The-Number-Of-Paths , which considered the problem of finding all Hamiltonian paths on an integer lattice in R^2 that connect two distinct vertices. The AllPaths procedure solves a more general problem: it finds all self-disjoint paths connecting two distinct vertices not only in the plane R^2 but also in space R^3 . Of course, it also finds all Hamiltonian paths or allows one to determine their absence. The procedure does not use commands from GraphTheory package (only direct manipulation of sets and lists).
Required parameters of the procedure: S is a set or list of lattice vertices specified by their coordinates, Start and Finish are the initial and final vertices. Optional parameter R (defaults it's NULL if all paths are searched) and any symbol (if only Hamiltonian paths are searched). S can be either a rectangular integer lattice or a union of several such lattices.

Code of the procedure:

restart;
AllPaths:=proc(S::{set(list),listlist},Start::list,Finish::list,R::symbol:=NULL)
local N:=nops(S), S1:=convert(S, set), L, n, m, k, i, j, s, p, q, P, a, b, c;

L:={[Start]};
for n from 2 to N do

if R=NULL then

P:='P';
m:=nops(L);
for k from 1 to m do
if nops(Start)=2 then
i,j:=L[k][-1][];
s:={[i-1,j],[i,j+1],[i+1,j],[i,j-1]} else
a,b,c:=L[k][-1][];
s:={[a-1,b,c],[a,b+1,c],[a+1,b,c],[a,b-1,c],[a,b,c-1],[a,b,c+1]} fi; 
s:=`intersect`(s,S1) minus convert(L[k],set);
if s={} and L[k][-1]=Finish then P[k]:=L[k] else
if s={} and L[k][-1]<>Finish then P[k]:=NULL else
P[k]:=`if`(L[k][-1]=Finish,L[k],seq([L[k][],s[i]],i=1..nops(s)))  fi; fi;
od;
L:=convert(P,set) else

P:='P';
m:=nops(L);
for k from 1 to m do
if nops(Start)=2 then
i,j:=L[k][-1][];
s:={[i-1,j],[i,j+1],[i+1,j],[i,j-1]} else
a,b,c:=L[k][-1][];
s:={[a-1,b,c],[a,b+1,c],[a+1,b,c],[a,b-1,c],[a,b,c-1],[a,b,c+1]} fi;
s:=`intersect`(s,S1) minus convert(L[k],set);
if n<N then 
if s={} or L[k][-1]=Finish then P[k]:=NULL else
P[k]:=seq([L[k][],s[i]],i=1..nops(s)); fi else 
if L[k][-1]=Finish and s<>{} then P[k]:=NULL else P[k]:=seq([L[k][],s[i]],i=1..nops(s));
fi; fi;  
od;
L:=convert(P,set)

fi; od;

L;
end proc:

Examples of use.

In the first example from the post above, we find the number of Hamiltonian paths in

L:=CodeTools:-Usage(AllPaths({seq(seq([i,j], i=1..11), j=1..3)}, [2,2], [10,2], 'H')):
nops(L);

In this same example, we find the number of all paths from A to B and the possible lengths of these paths.

L:=CodeTools:-Usage(AllPaths({seq(seq([i,j], i=1..11), j=1..3)}, [2,2], [10,2])):
nops(L);
map(t->nops(t), L);
L1:=select(t->nops(t)=%[-1], L):
nops(L1);

In the following example, we find the number of all paths, as well as the number of Hamiltonian paths, and animate these paths (total 24 one's).

S:={seq(seq([i,j],i=1..5),j=1..3)} union {seq(seq([i,j],i=4..7),j=3..5)}: A:=[1,1]: B:=[7,5]:
P:=plots:-display(plots:-pointplot(S, symbol=solidcircle, color=blue, symbolsize=15, view=[0..7.5,0..6.5], size=[600,500], scaling=constrained), plots:-textplot([[A[],"A"],[B[],"B"]], font=[times,bold,22], align=[left,above])):
L:=AllPaths(S,A,B):
nops(L);
map(t->nops(t), L);
L1:=select(t->nops(t)=%[-1], L):
nops(L1);
plots:-animate((plots:-display)@(plottools:-curve),[L1[round(a)], color=red, thickness=4], a=1..%, frames=180, background=P, size=[700,500], paraminfo=false);

In the final example, we search for Hamiltonian paths in a lattice defined on the surface of a cube. Imagine a cube made of wires, and an ant must crawl along these wires from point A(0,0,0) to point B(2,2,2) , visiting all nodes of this lattice. Is this possible? We see that it is not. The length of the maximum path is 25, and this lattice has 26 vertices. An animation of one of the maximum paths is provided.

S:={seq(seq(seq([i,j,k],i=0..2),j=0..2),k=0..2)} minus {[1,1,1]}: A:=[0,0,0]: B:=[2,2,2]:
P:=plots:-display(plots:-pointplot3d(S, symbol=solidcircle, color=blue, symbolsize=20, scaling=constrained), plots:-textplot3d([[A[],"A"],[B[],"B"]], font=[times,bold,22], align=[left,above]), plottools:-curve([[2,0,1],[2,2,1],[0,2,1],[0,0,1],[2,0,1]], color=black,thickness=0),plottools:-curve([[1,0,2],[1,0,0],[1,2,0],[1,2,2],[1,0,2]], color=black, thickness=0),plottools:-curve([[2,1,0],[0,1,0],[0,1,2],[2,1,2],[2,1,0]], color=black,thickness=0), tickmarks=[3,3,3]):
L:=AllPaths(S,A,B):
nops(L);
map(t->nops(t), L);
L1:=select(t->nops(t)=%[-1], L):
nops(L1);
plots:-animate((plots:-display)@(plottools:-curve),[L1[-1][1..round(a)], color=red, thickness=4], a=1..25, frames=240, background=P, paraminfo=false, axes=box, labels=[x,y,z]);

We can see from this animation that the path does not pass through the vertex (0, 2, 2) .

Edit. A code error that could cause incorrect operation when using the R option has been fixed. Everything now works correctly. If there is no Hamiltonian path passing through all vertices, the procedure returns the empty set { } .

Paths1.mw

Reproducing figures fom nanofluid paper

KIRAN SAJJAN — Wed, 18 Feb 2026 18:15:14 Z

Hello everyone,

I am trying to reproduce some results from the paper:

Peristaltic flow of a magnetohydrodynamic nanofluid through a bifurcated channel

I want to generate:

Fig 1 (channel geometry)
Fig 8 (pressure gradient vs axial coordinate ξ)
Fig 11 (streamlines)
Table 1 (resistance values)

I am using Maple 2018.

Below is my Maple code:
p1_ravi_sir.mw

Problem:

The plots do not match the paper

p1_ravi_sir.pdf

Why does my resistance value not change with M

Could someone please help me:

Correct the Maple code
Solve equations (18–22) properly in Maple
Generate the correct plots

Thank you very much.

expand of a Sum

RLessard — Wed, 21 Jan 2026 23:17:23 Z

Hi !

When I use the "expand" command on a summation, I've noticed that sometimes the command works and

sometimes it doesn't. I looked through the previous commands to see which one might be causing this problem.

I found that if I use the "changevar" command from the "student" package, (deprecated package,I know )

the "expand" command works correctly. Strange behavior, isn't it ? I have Maple 2018 on Windows 10.

Regards !

There is a small example in the file.

expand-problem.mw

Inscribed square problem

Kitonum — Mon, 05 Jan 2026 17:07:22 Z

The inscribed square problem, also known as the Toeplitz conjecture, is an unsolved quastion in geometry: Does every plane simple closed curve (Jordan curve) contain all four vertices of some square? This is true if the curve is convex or piecewise smooth and in other special cases. The problem was proposed by Otto Toeplitz in 1911. For detailes see https://en.wikipedia.org/wiki/Inscribed_square_problem

The Inscribed_Square procedure finds numerically one or more solutions for a curve defined by parametric equations of its boundary or by the equation F(x,y)=0. The required parameter of procedure L is the list of equations of the boundary links or the equation F(x,y)=0 . Optional parameters: N and R . By default N='onesolution' (the procedure finds one solution), if N is any symbol (for example N='s'), then more solutions. R is the range for the length of the side of the square (by defalt R=0.1..100 ).

The second procedure Pic visualizes the results obtained.

The codes of the procedures:

restart;
Inscribed_Square:=proc(L::{list(list),`=`},N::symbol:='onesolution',R::range:=0.1..100)
local D, n, c, L1, L2, L3, f, L0, i, j, k, m, A, B, C, P, M, eq1, eq2, eq3, eq4, eq5, eq6, eq7, eq8, eq9, sol, Sol;
uses LinearAlgebra;
if L::list then
L0:=map(p->`if`(type(p,listlist),[[p[1,1]+t*(p[2]-p[1])[1],p[1,2]+t*(p[2]-p[1])[2]],t=0..1],p), L);
c:=0;
n:=nops(L);
for i from 1 to n do
for j from i to n do
for k from j to n do
for m from k to n do
A:=convert(subs(t=t1,L0[i,1]),Vector): 
B:=convert(subs(t=t2,L0[j,1]),Vector):
C:=convert(subs(t=t3,L0[k,1]),Vector): 
D:=convert(subs(t=t4,L0[m,1]),Vector):
M:=<0,-1;1,0>;
eq1:=eval(C[1])=eval((B+M.(B-A))[1]);
eq2:=eval(C[2])=eval((B+M.(B-A))[2]);
eq3:=eval(D[1])=eval((C+M.(C-B))[1]);
eq4:=eval(D[2])=eval((C+M.(C-B))[2]);
eq5:=eval(DotProduct(B-A,B-A, conjugate=false))=d^2;
sol:=fsolve([eq1,eq2,eq3,eq4,eq5],{t1=op([2,2,1],L0[i])..op([2,2,2],L0[i]),t2=op([2,2,1],L0[j])..op([2,2,2],L0[j]),t3=op([2,2,1],L0[k])..op([2,2,2],L0[k]),t4=op([2,2,1],L0[m])..op([2,2,2],L0[m]),d=R});
if type(sol,set(`=`)) then if N='onesolution' then return convert~(eval([A,B,C,D],sol),list) else c:=c+1; Sol[c]:=convert~(eval([A,B,C,D],sol),list) fi;
 fi; 
od: od: od: od:
Sol:=fnormal(convert(Sol,list),7);
print(Sol);
ListTools:-Categorize((X,Y)->`and`(seq(is(convert(X,set)[i]=convert(Y,set)[i]),i=1..4)) , Sol);
return map(t->t[1],[%]);
else
A,B,C,D:=<x1,y1>,<x2,y2>,<x3,y3>,<x4,y4>:
M:=<0,-1;1,0>:
eq1:=eval(C[1])=eval((B+M.(B-A))[1]):
eq2:=eval(C[2])=eval((B+M.(B-A))[2]):
eq3:=eval(D[1])=eval((C+M.(C-B))[1]):
eq4:=eval(D[2])=eval((C+M.(C-B))[2]):
eq5:=eval(LinearAlgebra:-DotProduct((B-A,B-A), conjugate=false))=d^2:
eq6:=eval(L,[x=x1,y=y1]):
eq7:=eval(L,[x=x2,y=y2]):
eq8:=eval(L,[x=x3,y=y3]):
eq9:=eval(L,[x=x4,y=y4]):
sol:=fsolve({eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9},{seq([x||i=-2..2,y||i=-2..2][],i=1..4),d=R});
eval([[x1,y1],[x2,y2],[x3,y3],[x4,y4]], sol):
fi;
end proc:

Pic:=proc(L,Sol,R::range:=-20..20)
local P1, P2, P3, T;
uses plots, plottools;
P1:=`if`(L::list,seq(`if`(type(s,listlist),line(s[],color=blue, thickness=2),plot([s[1][],s[2]],color=blue, thickness=2)),s=L), implicitplot(L, x=R,y=R, color=blue, thickness=2, gridrefine=3));
P2:=polygon(Sol,color=yellow,thickness=0);
P3:=curve([Sol[],Sol[1]],color=red,thickness=3):
T:=textplot([[Sol[1][],"A"],[Sol[2][],"B"],[Sol[3][],"C"],[Sol[4][],"D"]], font=[times,18], align=[left,above]);
display(P1,P2,P3,T, scaling=constrained, size=[800,500], axes=none);
end proc:

Examples of use:

The curve consists of a semicircle, a segment and a semi-ellipse (find 1 solution):

L:=[[[cos(t),sin(t)],t=0..Pi],[[t,0],t=-1..0],[[0.5+0.5*cos(t),0.8*sin(t)],t=Pi..2*Pi]]:
Sol:=Inscribed_Square(L);
Pic(L,Sol);

The procedure finds 6 solutions for a non-convex pentagon:

 L:=[[[0,0],[9,0]],[[9,0],[8,5]],[[8,5],[5,3]],[[5,3],[0,4]],[[0,4],[0,0]]]:
Sol:=Inscribed_Square(L,'s');
plots:-display(Matrix(3,2,[seq(Pic(L,Sol[i]),i=1..6)]),size=[300,200]);

For an implicitly defined curve, only one solution can be found:

L:=abs(x)+2*abs(y)-sin((2*x-y))-cos(x+y)^2=3:
Sol:=Inscribed_Square(L);
Pic(L,Sol);

See more examples in the attached file.

Inscribed_Square.mw

Bug in the solve command

Kitonum — Thu, 01 Jan 2026 16:52:11 Z

The system below obviously has the unique solution t=3*Pi/2 , but Maple doesn't return any solution. I wonder if this bug persists in recent versions of Maple?

solve({cos(t)=0,sin(t)=-1,t>=0,t<=2*Pi}, t);

# NULL