Hi

I need a temperature distribution inside a barrier during a heating process.
I will be appreciated for any help.

wz

Download heating.mw

Hi

I use Maple perhaps only one time per year and thus i'm a beginner with the product (even if i use it since 1996). I'm trying to make the following script work in Maple 17:

http://www.maplesoft.com/applications/view.aspx?SID=4268&ref=Feed&L=F

but i fail.... Some help will be appreciate

restart:

Eq1:=1/r*diff(r*diff(w(r),r),r)-(cos(z1))^2*(w(r)+1)-DP;

dsolve({Eq1});

bcs:=D(w)(0)=0,w(r2)=A1;

dsolve({Eq1,bcs},w(r)):

where z1, Dp, r2, A1 are constants.

Hi, can I get some help with this?

The question is:

Consider the following IVP for a mass of m = 2 kg attached to a spring with a spring constant k = 9 N/m. The spring mass system is in a medium with damping constant b.

2y" + by' + 9y = 0

y(0) = 0, D(y)(0) = -3 

It then asks find three values b1, b2, b3 where b1 is underdamped, b2 is critical, b3 is over. 

I set b1 as 1, b2 as sqt 72, b3 as 9. 

Then it asks to find the quasi period. 

I can't get my quasi period right. My answer is 2pi/ sqrt (4.5).

Any help?  

Hi.

Please, I need a code in maple for adaptative setp size control for runge Kutta.

Thank you.

Hi.

Please help to use dsolve.

Suppose I have a Matrix A of size two. Y=[u,v].

Y'(t)=A*y(t).

How can use dsolve this problem.

Thanks

Is there any simple way to print the sum of a series with exceptions?

For example, the following works (sort of), but its not consistent with mathematical notation, where one would expect to see something like a Sigma^' printed out, and the condition, alongw ith ther limits appear under the Sigma sign:

[Sum(1/(k^2-p),k=0..infinity),And(k<>p)]

Hi MaplePrime-ers!

I've been using the Maple(17) toolbox in Matlab(2012b) to quickly enumerate systems of equations by: (i) solving them symbolically, (ii) using unapply to make them functions, (iii) then supplying the points (driver equations) to get the system solution.  Speed is a must, because there may be 3 million+ systems to solve.  Symbolics is also very important because I am evaluating topology, so the structure of the equations may change, and therefore a functional approach will not work.

I have had success (seen in the first code snippet).  I would like similiar behaviour in the second code snippet, but sometimes I get 'solutions may be lost' as an error message,  or 'Error, (in unapply) variables must be unique and of type name'

The system of equations include:  Linear equations, 5th order polynomials, absolute functions, and pieceiwse functions.

Here is code with a topology that solves:

#Interconnection Equations
eq2[1] := FD_T + EM2_T = 0;
eq2[2] := ICE_T + GEN_T = 0;
eq2[3] := EM2_A + GEN_A + BAT_A = 0;
eq2[4] := -FD_W + EM2_W = 0;
eq2[5] := -ICE_W + GEN_W = 0;
eq2[6] := -EM2_V + GEN_V = 0;
eq2[7] := -EM2_V + BAT_V = 0;

#ICE
eq_c[1] := ICE_mdot_g=((671.5) + (-21.94)*ICE_T + (0.1942)*ICE_W + (0.5113)*ICE_T^2 + (-0.01271)*ICE_T*ICE_W + ( -0.0008761)*ICE_W^2 + (-0.006071)*ICE_T^3 + (9.867e-07)*ICE_T^2*ICE_W + (5.616e-05)*ICE_T*ICE_W^2 + (1.588e-06)*ICE_W^3 + (3.61e-05)*ICE_T^4 + (8.98e-07)*ICE_T^3*ICE_W + (-2.814e-07)*ICE_T^2*ICE_W^2 + (-8.121e-08)*ICE_T*ICE_W^3 + ( -8.494e-08 )*ICE_T^5 + (-2.444e-09)*ICE_T^4*ICE_W + (-9.311e-10)*ICE_T^3*ICE_W^2 + ( 5.835e-10)*ICE_T^2*ICE_W^3 ) *1/3600/1000 * ICE_T * ICE_W;

#BAT
eq_c[2] := BAT = 271;

#EM2
EM2_ReqPow_eq := (-148.3) + (4.267)*abs(EM2_W) + (12.77)*abs(EM2_T) + (-0.0364)*abs(EM2_W)^2 + ( 1.16)*abs(EM2_W)*abs(EM2_T) + (-0.258)*abs(EM2_T)^2 + ( 0.0001181)*abs(EM2_W)^3 + (-0.0005994)*abs(EM2_W)^2*abs(EM2_T) + ( 0.0001171)*abs(EM2_W)*abs(EM2_T)^2 + (0.001739 )*abs(EM2_T)^3 + (-1.245e-07 )*abs(EM2_W)^4 + ( 1.2e-06)*abs(EM2_W)^3*abs(EM2_T) + ( -1.584e-06)*abs(EM2_W)^2*abs(EM2_T)^2 + ( 4.383e-07)*abs(EM2_W)*abs(EM2_T)^3 + (-2.947e-06)*abs(EM2_T)^4;
eq_c[3] := EM2_P = piecewise( EM2_T = 0, 0, EM2_W = 0, 0, EM2_W*EM2_T < 0,-1 * EM2_ReqPow_eq, EM2_ReqPow_eq);
eq_c[4] := EM2_A = EM2_P/EM2_V;

#GEN
GEN_ReqPow_eq:= (-5.28e-12) + ( 3.849e-14)*abs(GEN_W) + (-71.9)*abs(GEN_T) + (-1.168e-16)*abs(GEN_W)^2 +(1.296)*abs(GEN_W)*abs(GEN_T) + (2.489)*abs(GEN_T)^2 + (1.451e-19)*abs(GEN_W)^3 + (0.0001326)*abs(GEN_W)^2*abs(GEN_T) + (-0.008141)*abs(GEN_W)*abs(GEN_T)^2 + (-0.004539)*abs(GEN_T)^3 +(-6.325e-23)*abs(GEN_W)^4 + (-2.091e-07)*abs(GEN_W)^3*abs(GEN_T) + ( 3.455e-06)*abs(GEN_W)^2*abs(GEN_T)^2 + ( 2.499e-05)*abs(GEN_W)*abs(GEN_T)^3 + (-5.321e-05)*abs(GEN_T)^4;
eq_c[5] := GEN_P = piecewise( GEN_T = 0, 0, GEN_W = 0, 0, GEN_W*GEN_T < 0,-1 * GEN_ReqPow_eq, GEN_ReqPow_eq);
eq_c[6] := GEN_A = GEN_P/GEN_V;

#ASSUMPTIONS
assume(BAT_V::nonnegative);
assume(FD_W::nonnegative);

#FINAL EQUATIONS

sys_eqs2 := convert(eq2,set) union {eq_c[1],eq_c[2],eq_c[3],eq_c[4],eq_c[5],eq_c[6]};

#Selecting which variables to solve for:

drivers2:= { ICE_T,ICE_W,FD_T,FD_W};
symvarnames2:=select(type,indets(convert(sys_eqs2,list)),name);
notdrivers2:=symvarnames2 minus drivers2;

#Symbolic solve

sol2:=solve(sys_eqs2,notdrivers2) assuming real:
symb_sol2:=unapply(sol2,convert(drivers2,list)):


#Enumerate (there will generally be about 40, not 6)

count := 0;
for i1 from 1 to 40 do
     for i2 from 1 to 40 do
          for i3 from 1 to 40 do
               for i4 from 1 to 40 do
                    count := count + 1;
                    solsol2(count) := symb_sol2(i1,i2,i3,i4);
               od;
          od;
     od;
od;
count;

This works great!  I would like simliar output in my second code snippet, but this time with more inputs to symb_sol.  However, if I try and change the interconnection equations a little, and add a piecewise function, and another driver... (differences in bold)

#Interconnection Equations
eq1[1] := FD_T+EM2_T = 0;
eq1[2] := ICE_T+GBb_T = 0;
eq1[3] := GEN_T+GBa_T = 0;
eq1[4] := EM2_A+GEN_A+BAT_A = 0;
eq1[5] := -FD_W+EM2_W = 0;
eq1[6] := -GEN_W+GBa_W = 0;
eq1[7] := -ICE_W+GBb_W = 0;
eq1[8] := -EM2_V+GEN_V = 0;
eq1[9] := -EM2_V+BAT_V = 0;

#ICE
eq_c[1] := ICE_mdot_g=((671.5) + (-21.94)*ICE_T + (0.1942)*ICE_W + (0.5113)*ICE_T^2 + (-0.01271)*ICE_T*ICE_W + ( -0.0008761)*ICE_W^2 + (-0.006071)*ICE_T^3 + (9.867e-07)*ICE_T^2*ICE_W + (5.616e-05)*ICE_T*ICE_W^2 + (1.588e-06)*ICE_W^3 + (3.61e-05)*ICE_T^4 + (8.98e-07)*ICE_T^3*ICE_W + (-2.814e-07)*ICE_T^2*ICE_W^2 + (-8.121e-08)*ICE_T*ICE_W^3 + ( -8.494e-08 )*ICE_T^5 + (-2.444e-09)*ICE_T^4*ICE_W + (-9.311e-10)*ICE_T^3*ICE_W^2 + ( 5.835e-10)*ICE_T^2*ICE_W^3 ) *1/3600/1000 * ICE_T * ICE_W;

#BAT
eq_c[2] := BAT = 271;

#EM2
EM2_ReqPow_eq := (-148.3) + (4.267)*abs(EM2_W) + (12.77)*abs(EM2_T) + (-0.0364)*abs(EM2_W)^2 + ( 1.16)*abs(EM2_W)*abs(EM2_T) + (-0.258)*abs(EM2_T)^2 + ( 0.0001181)*abs(EM2_W)^3 + (-0.0005994)*abs(EM2_W)^2*abs(EM2_T) + ( 0.0001171)*abs(EM2_W)*abs(EM2_T)^2 + (0.001739 )*abs(EM2_T)^3 + (-1.245e-07 )*abs(EM2_W)^4 + ( 1.2e-06)*abs(EM2_W)^3*abs(EM2_T) + ( -1.584e-06)*abs(EM2_W)^2*abs(EM2_T)^2 + ( 4.383e-07)*abs(EM2_W)*abs(EM2_T)^3 + (-2.947e-06)*abs(EM2_T)^4;
eq_c[3] := EM2_P = piecewise( EM2_T = 0, 0, EM2_W = 0, 0, EM2_W*EM2_T < 0,-1 * EM2_ReqPow_eq, EM2_ReqPow_eq);
eq_c[4] := EM2_A = EM2_P/EM2_V;

#GEN
GEN_ReqPow_eq:= (-5.28e-12) + ( 3.849e-14)*abs(GEN_W) + (-71.9)*abs(GEN_T) + (-1.168e-16)*abs(GEN_W)^2 +(1.296)*abs(GEN_W)*abs(GEN_T) + (2.489)*abs(GEN_T)^2 + (1.451e-19)*abs(GEN_W)^3 + (0.0001326)*abs(GEN_W)^2*abs(GEN_T) + (-0.008141)*abs(GEN_W)*abs(GEN_T)^2 + (-0.004539)*abs(GEN_T)^3 +(-6.325e-23)*abs(GEN_W)^4 + (-2.091e-07)*abs(GEN_W)^3*abs(GEN_T) + ( 3.455e-06)*abs(GEN_W)^2*abs(GEN_T)^2 + ( 2.499e-05)*abs(GEN_W)*abs(GEN_T)^3 + (-5.321e-05)*abs(GEN_T)^4;
eq_c[5] := GEN_P = piecewise( GEN_T = 0, 0, GEN_W = 0, 0, GEN_W*GEN_T < 0,-1 * GEN_ReqPow_eq, GEN_ReqPow_eq);
eq_c[6] := GEN_A = GEN_P/GEN_V;

#GB
FiveSpeedGearbox_R := proc(ig)
local i ,eq;
i[1]:=3.32;
i[2]:=2;
i[3]:=1.36;
i[4]:=1.01;
i[5]:=0.82;
eq:= piecewise(ig=1,i[1],ig=2, i[2],ig=3,i[3],ig=4,i[4],ig=5,i[5],1);
return eq(ig);
end proc;

eq_c[7] := GBb_T = -1/GB_R * GBa_T;
eq_c[8] := GBb_W = GB_R * GBa_W;
eq_c[9] := GB_R = FiveSpeedGearbox_R(ig);

 

#System Equations
sys_eqs := convert(eq1,set) union convert(eq_c,set);

 

 #Solve for variables
symvarnames:=select(type,indets(convert(sys_eqs,list)),name);
drivers:= {ig, ICE_T,ICE_W,FD_T,FD_W};
not_drivers := symvarnames minus drivers;

#Assumptinons

assume(BAT_V::nonnegative);
assume(FD_W::nonnegative);

sol:=(solve(sys_eqs,not_drivers) assuming real);

symb_sol:=unapply(sol,convert(drivers,list)): ---> Error, (in unapply) variables must be unique and of type name

Subsequent parts don't work...

count := 0;
for i1 from 1 to 40 do
     for i2 from 1 to 40 do
          for i3 from 1 to 40 do
               for i4 from 1 to 40 do
                    for i5 from 1 to 40 do
                         count := count + 1;
                         solsol2(count) := symb_sol2(i1,i2,i3,i4,5);
                    od;
               od; 
          od;
     od;
od;
count;

While running the last line sol:, 1 of 2 things will happen, depending on the solver. Maple17 will take a long time (30+ minutes) to solve, then report nothing, or sol will solve, but will report "some solutions have been lost".

Afterwards, evaluating symb_sol(0,0,0,0,0) will return a viable solution (real values for each of the variables).  Whereas evaluating symb_sol(0,X,0,0,0), where X <> 0, will return and empty list [].

Does anyone know how to (i) speed up the symbolic solve time?  (ii) Return ALL of the solutions?

Thanks in advance for reading this.  I've really no idea why this isn't working.  I've also attached two worksheets with the code: noGB.mw   withGB.mw

 Adam

How do I know if I have the latest Maple build id?  What is it now?

Let  us consider the general case of symbolic values C(xC,yC). I make use of the idea suggested by edgar in http://www.mapleprimes.com/questions/97743-How-To-Prove-Morleys-Trisector-Theorem : no assumptions.

restart; with(geometry); point(A, 0, 0);
point(B, 1, 0);
point(C, xC, yC);
point(MA, (xC+1)*(1/2), (1/2)*yC);
point(MC, 1/2, 0);
point(MB, (1/2)*xC, (1/2)*yC);
point(E, (0+1+xC)*(1/3), (0+0+yC)*(1/3));# the center of mass
line(l1, x = 1/4, [x, y]);
The coordinates of the center of the first described circle are found as the solutions of the system of the equations of midperpendiculars.

midpoint(ae, A, E); coordinates(ae);


S1 := solve({x = 1/4, ((xC+1)*(1/3))*(x-(xC+1)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

BTW, Maple can't create the midperpendiculars in this case.

point(O1, op(map(rhs, S1)));
                               O1

Simple details are omitted in the above. The coordinates of the centers of the two next described circles are found similarly.
coordinates(midpoint(mce, MC, E));

S2 := solve({x = 3/4, ((-1/2+xC)*(1/3))*(x-5/12-(1/6)*xC)+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O2, op(map(rhs, S2)));

                               O2
coordinates(midpoint(bma, B, MA)); coordinates(midpoint(be, B, E));
  

                

S3 := solve({(xC-1)*(x-(xC+3)*(1/4))+yC*(y-(1/4)*yC) = 0, ((-2+xC)*(1/3))*(x-(4+xC)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O3, op(map(rhs, S3)));

                               O3

Now we find the equation of the circumference which passes through O1, O2, and O3.

eq := a*x+b*y+x^2+y^2+c = 0:
sol := solve({eval(eq, S1), eval(eq, S2), eval(eq, S3)}, {a, b, c});

A long output can be seen in the attached .mw file.

eq1 := eval(eq, sol);

  Now we find (in suspense)  the coordinates of the next center and verify whether it belongs to the sircumference O1O2O3.

coordinates(midpoint(mac, C, MA)); coordinates(midpoint(ec, E, C)); S4 := solve({(xC-1)*(x-(3*xC+1)*(1/4))+yC*(y-3*yC*(1/4)) = 0, ((2*xC-1)*(1/3))*(x-(4*xC+1)*(1/6))+(2*yC*(1/3))*(y-4*yC*(1/6)) = 0}, {x, y});

 point(O4, op(map(rhs, S4)));

                               O4
simplify(eval(eq1, S4));

                             0 = 0

Hope the reader will have a real pleasure to find the two residuary centers and to verify these on his/her own.

geom2.mw

It is well known that the medians of a triangle divide it into 6 triangles.
It is less known that the centers of their circumscribed circles belong to one circumference as drawn below

This remarkable theorem  was proved in the 21st century! Unfortunately, I lost its source.
I can't prove this difficult  theorem by hand. However, I can prove it with Maple.
The aim of this post is to expose these proofs. Everybody knows that it is scarcely possible
to construct a general triangle with help of the geometry package of Maple.
Without loss of generality one may assume that the vertex A is placed at the origin,
the vertex B is placed at (1,0), and the vertex C(xC,yC). We firstly consider the theorem
in the case of concrete values of xC and yC.

restart; with(geometry):with(plots):
point(A, 0, 0);
point(B, 1, 0);
xC := 15*(1/10); yC := sqrt(3); point(C, xC, yC);
triangle(T, [A, B, C]);
median(mA, A, T, MA);
median(mB, B, T, MB);
median(mC, C, T, MC);
line(m1, [A, MA]);
line(m2, [B, MB]);
intersection(E, m1, m2);
triangle(AEMB, [A, E, MB]);
circumcircle(c1, AEMB, 'centername' = C1);
circumcircle(c2, triangle(CEMB, [C, E, MB]), 'centername' = C2);
circumcircle(c3, triangle(CEMA, [C, E, MA]), 'centername' = C3);
circumcircle(c4, triangle(BEMA, [B, E, MA]), 'centername' = C4);
circumcircle(c5, triangle(BEMC, [B, E, MC]), 'centername' = C5);
circumcircle(c6, triangle(AEMC, [A, E, MC]), 'centername' = C6);
circle(CC, [C1, C2, C3]);
IsOnCircle(C4, CC);
                              true

IsOnCircle(C5, CC);
                              true
IsOnCircle(C6, CC);
                              true
display([draw([T(color = black), mA(color = black), mB(color = black), mC(color = black), C1(color = blue), C2(color = blue), C3(color = blue), C4(color = blue), C5(color = blue), C6(color = blue), CC(color = red)], symbol = solidcircle, symbolsize = 15, thickness = 2, scaling = constrained), textplot({[-0.5e-1, 0.5e-1, "A"], [.95, 0.5e-1, "B"], [xC-0.5e-1, yC+0.5e-1, "C"]})], axes = frame, view = [-.1 .. max(1, xC)+.1, 0 .. yC+.1]);

This can be done as a procedure in such a way.

restart; SixPoints := proc (xC, yC) geometry:-point(A, 0, 0); geometry:-point(B, 1, 0); geometry:-point(C, xC, yC); geometry:-triangle(T, [A, B, C]); geometry:-median(mA, A, T, MA); geometry:-median(mB, B, T, MB); geometry:-median(mC, C, T, MC); geometry:-line(m1, [A, MA]); geometry:-line(m2, [B, MB]); geometry:-intersection(E, m1, m2); geometry:-triangle(AEMB, [A, E, MB]); geometry:-circumcircle(c1, AEMB, 'centername' = C1); geometry:-circumcircle(c2, geometry:-triangle(CEMB, [C, E, MB]), 'centername' = C2); geometry:-circumcircle(c3, geometry:-triangle(CEMA, [C, E, MA]), 'centername' = C3); geometry:-circumcircle(c4, geometry:-triangle(BEMA, [B, E, MA]), 'centername' = C4); geometry:-circumcircle(c5, geometry:-triangle(BEMC, [B, E, MC]), 'centername' = C5); geometry:-circumcircle(c6, geometry:-triangle(AEMC, [A, E, MC]), 'centername' = C6); geometry:-circle(CC, [C1, C2, C3]); return geometry:-IsOnCircle(C4, CC), geometry:-IsOnCircle(C5, CC), geometry:-IsOnCircle(C6, CC), geometry:-draw([CC(color = blue), C1(color = red), C2(color = red), C3(color = red), C4(color = red), C5(color = red), C6(color = red), T(color = black), mA(color = black), mB(color = black), mC(color = black), c1(color = green), c4(color = green), c2(color = green), c3(color = green), c5(color = green), c6(color = green)], symbol = solidcircle, symbolsize = 15, thickness = 2) end proc;
SixPoints(1.5, 1.2);

true, true, true, PLOT(...)
 SixPoints(1.5, 1.2)[4];

See geom1.mw

To be continued (The general case will be considered in  part 2http://www.mapleprimes.com/posts/200210-Six-Points-On-Circumference-2 .).

computer a Gröbner basis for <f_[1] = x^2*y - 2*y*z + 1, f_[2] = x*y^2-z^2+ 2*x,  f_[3] = y^2*z - x^2+ 5 > belong to Q[x,y,z], using ≺= <_grlex with x≺y≺z. compare your output to the Gröbner basis the Maple computers with a different order.

I have no idea where to start with this problem.

I know how to create arrays, but I want to assign characteristics to each position in the array; the array being a population with each member having a set of characteistics like height, weight etc. I want to be able to have some of the characteristics able to change with time, and some remain constant (those that change with time will be defined by generic equations)

I have tried looking on maplesoft for answers, and the help within maple 16, but have been unable to make any progress.

Sorry if the question is not specific enough.

Thanks in advance x

Please I need programme to solve this nonlinear couple ode with values of k_1 from 1.5 to 2.5 and eta from 0.0 to 2.0 real_work_3.mw Thanks for your help.

Hi all,

We're programming with Maple objects, and would like to have a Maple object that is appliable (i.e. acts as an operator).  We're really after partial differential operator (PDO) objects, but I give a simpler example below.  The problem is this: the ?ModuleApply proc needs to be non-static (so that each object has its own ?ModuleApply), and it may want to refer to the object instance, via ?thismodule.  Code like this works great when it is typed at the command line or read in from .mpl file, but it barfs as soon as saved to a .mla.  Why is that?  And is there an easy fix / workaround?

Here is an example worksheet, that creates a basic "compute mod p" object

BugThismoduleMLA.mw

The issue seems to be that, for code typed at command line, ?thismodule is resolved when I would expect, i.e. in the instance object; but saving to the .mla seems to cause ?thismodule to be resolved then-and-there, which means it resolves to the prototype object.  For this particular case there is an easy workaround, but we really want to refer to ?thismodule if possible.  Is there a way to do that?

BTW I'm running Maple 17.02 (sorry don't have access to 18 yet).

Thanks,
Ian and Tracy