i cant find the error the program is saying i have 

> L := -1.576674; MU := 0; DE13 := {D(y)*t = -x(t)*(1-2*x(t)*x(t))+MU*y(t)*(x(t)*x(t)-3*y(t)*y(t)-L), (D(x))(t) = y(t)*(1-y(t)*y(t))+MU*y(t)*(x(t)*x(t)-3*y(t)*y(t)-L)}; DEplot(DE13, [x(t), y(t)], t = -20 .. 20, [[x(0) = 0.1e-1, y(0) = .99], [x(0) = 0.5e-1, y(0) = .95], [x(0) = .1, y(0) = .9], [x(0) = 0.4e-1, y(0) = .96]], stepsize = 0.1e-2, scene = [x(t), y(t)], title = "phaseplane plot", linecolor = black, thickness = 1, number = 1000);
 
Error, (in DEtools/DEplot/CheckInitial) too few initial conditions: [x(0) = 0.1e-1, y(0) = .99]

it might be hard to read but if someone could help me it would be very appreciated 

Hi

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

wz

Download heating.mw

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, i'm wondering if theres anything i can do printf and print and have the outputs on the same line. I want the string wihtout any quotations so i think i can only do printf to get this, but then i'm putting an entry out of an array after this. This is a small example out of my code, i would like the output to be all on one line:

The synchronising vector is (S[j-1]) The synchronising word is (X[j-1]) The length of the synchronising word is ...

if (sync=true) then 
printf("The Synchronising vector is"); print(S[j-1]);
printf("The sychronising word is");
print(X[j-1]);
printf("The length of the synchronising word is");
print(length(X[j-1]));
else printf("There is no synchronising word"); 
end if;

 At the moment its outputing each print on a new line.

Any help would be much appreciated, thanks.

hi all.
i have a system of ODE's including 9 set of coupled OED's . 

for example

a*b v a^-1 = b

i guess Disj or Conj are Max or Min respectively

however i do not know how to max(a,b) where a and b are permutation group

reference from L group in 

http://books.google.com.hk/books/about/Lattice_Ordered_Groups.html?id=_qAFX38-pKkC&redir_esc=y

if can not calculate this, do not know how to determine whether equal in a*b v a^-1 = b

I came to know that Chrome no longer supports MathML. Will this be an issue with Maple T.A.9/9.5? Should I require my students to switch to Firefox instread? 

One annoying feature in using document mode that I found is this. As I hit the return key to execute one command and automatically jump to the command below it (a feature that I like), but after few returns, I find myself at the bottom edge of the window, with the cursor pointing at the last command in the window.

Now when hitting return, the current command is execuated, and the next command below it appears, and it is also at the bottom edge of the window.  The cursor is from now on, always sitting at the bottom edge of the window, and I am not able to fully see the rest of the command and what is below it any more.

What would be better, is to try to keep the window centered, so that as I hit a return, the actual cursor remains in the middle of the window, and the window itself scrolls up. This way the command I am about to evaluate, and anything below it, remain in full view.

What I end up doing now, is when I get to the bottom of the window, is to manually scroll the window down using the mouse, so that now the cursor is up higher, and I can see more the commands which I am about to evaluate, then I repeate this process.

Is there  a way to configure Maple to try to keep the window centered? it is really very annoying to have to keep adjusting the window every few returns, and when one has 100 commands to evaluate, one by one (since I want to see the result before I go to the next command), this process gets tiring.

(does Maple do any usability studies and evaluation on its user interface by any independent outside group?)

Hi!

I have the following function:

and I want to calculate it for a certain set of r, theta and t, but when I use subs, the theta variable is also subsituted int the integral's parameters and I get the folllowing result:

I really don't understand why it does that given that there is no issue for the r variable...

Thanks in advance for your help!

P-S: please forgive my poor English : I'm a French student...

Florent

Dear all

Please I need your help to simplify by the coefficient a in this Matrix

I have The matrix A defined by  A:=Matrix(2,2, [[a,a],[3*a,4*a]]);

I want with maple transform A to  A:=a*Matrix(2,2, [[1,1],[3,4]]);

Thanks for your Help.

Dear all;

Please see only the last lines of this procedure,

1)I ask if this procedure give an output epsilon(x0,h). Really, I need your help. Thanks.

2) Can we plot epsilon(x0,h), versus h

> heun := proc (x0, h)

local x, y, i, N, k, f, ode, k1, k2, x1, x2, y1, y2;

f :=(x,y)-> 1/(1+cos(y)) ;

ode := diff(y(x), x) = f(x, y);

N := round((1/2)*x0/h);

y2 := Array(0 .. N);

x2 := Array(0 .. N);

y1 := Array(0 .. 2*N);

x1 := Array(0 .. 2*N);

x2[0] := 0; y2[0] := (1/4)*Pi;

for i from 0 to N-1 do

x1[2*i+2] := (2*i+2)*h;  k1[1] := f(x1[2*i], y1[2*i]); 

k1[2] := f(x1[2*i]+h, y1[2*i]+h*k1[1]);

y1[2*i+2] := y1[2*i]+(1/2)*h*(k1[1]+k1[2]);

x2[i+1] := (2*i+2)*h;

k2[1] := f(x2[i], y2[i]);

k2[2] := f(x2[i]+2*h, y2[i]+2*h*k2[1]);

y2[i+1] := y2[i]+h*(k2[1]+k2[2])

end do;

return firstresulat = evalf([seq([x2[i], y2[i]], i = 0 .. N)]);

return secondresulat =evalf([seq([x1[2*i], y1[2*i]], i = 0 .. N)]);

epsilon:=(x0,h)->add((firsttest[i][2]-secondtest[i][2])^2 , i=1..round(x0/(2*h))+1)

end proc;

how maple calculate exp(x) with e.g. 100000 decimal numbers

a divsion of the series x^k/k! with e.g. 1/25000!/25001 lasts longer than the exp(1.xx) calculation

is there a faster way to calculate exp(x) than with the x^k/k! series

thanks

I've been poking around with convolutions on Maple, and some weird behavior came up---if I let it compute the convolution of a piecewise function, then take the convolution of that, it comes out differently than if I enter a function from scratch as the middle step---file attached (PiecewiseProblem.mw).  I'm not really a Maple pro, so am I'm doing something crazy here?

Thanks!

Dear collegues

I wrote the following code

restart:
Digits := 15;
a[k]:=0;
b[k]:=7.47;
a[mu]:=39.11;
b[mu]:=533.9;
mu[bf]:=9.93/10000;
k[bf]:=0.597;
ro[p]:=3880 ;
ro[bf]:= 998.2;
c[p]:= 773;
c[bf]:= 4182;
#mu[bf]:=1;
Gr[phi]:=0; Gr[T]:=0;
#dp:=0.1;
Ree:=1;
Pr:=1;
Nbt:=cc*NBTT+(1-cc^2)*6;

#######################
slip:=0.1;         ####
NBTT:=2;           ####
lambda:=0.1;       ####
phi_avg:=0.02;    ####
#######################


eq1:=diff( (1+a[mu]*phi(eta)+b[mu]*phi(eta)^2)*diff(u(eta),eta),eta)+dp/mu[bf]+Gr[T]*T(eta)-Gr[phi]*phi(eta);
eq2:=diff((1+a[k]*phi(eta)+b[k]*phi(eta)^2)*diff(T(eta),eta),eta)+lambda*T(eta)/k[bf];
eq3:=diff(phi(eta),eta)+1/Nbt*diff(T(eta),eta);
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10;
global Q1,Q2;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({subs(dp=pp2,eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=fi0}, numeric,output=listprocedure,continuation=cc);
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
INT0:=evalf(Int(F0(eta),eta=0..1));
INT10:=evalf(Int(F0(eta)*F1(eta),eta=0..1));
a[1]:=evalf(Int(F0(eta),eta=0..1))-Ree*Pr;;
a[2]:=INT10/INT0-phi_avg;
Q1(_passed):=a[1];
Q2(_passed):=a[2];
if type(procname,indexed) then a[op(procname)] else a[1],a[2] end if
end proc;
Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;
Optimization:-LSSolve([Q1,Q2],initialpoint=[0.3,0.0007]);




se:=%[2];
res2 := dsolve({subs(dp=se[1],eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=se[2]}, numeric,output=listprocedure,continuation=cc);
G0,G1,G2:=op(subs(res2,[u(eta),phi(eta),T(eta)])):
TTb:=evalf(Int(G0(eta)*G2(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1))/evalf(Int(G0(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1));
with(plots):
odeplot(res2,[[eta,phi(eta)/phi_avg]],0..1);
odeplot(res2,[[eta,T(eta)/TTb]],0..1);
odeplot(res2,[[eta,u(eta)/(Ree*Pr)]],0..1);

res2(1);
Nuu:=(1/TTb);
1/((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2));
(1/TTb)*(((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2)));
>

I want to run the code for the value of NBTT in the range of 0.2 to 10. this code gave the results in the range of 4-10 easily. So, I used the continuation which improve the range of the results between 2-10. However, I coudnt gave the results when 0.2<NBTT<2. Would you please help me in this situation.

Also, It is to be said that the values of phi should be positive. in some ranges, I can see that phi(1) is negative. Can I place a condition in which the values phi restricted to be positive.

Thanks for your attentions in advance

Amir

Dear all;

Than you for help.

how  many steps are required to achieve a error of 1.e-3 in the numerical value of y(1).

Here The 3 -step procedure  Range Kutta Method.

## Exact  solution  

### We will modifty N ( number of steps to get error =10^(-3). )

## Procedure Range Kutta

> RK3 := proc (f, a, b, y0, N)

local x, y, n, h, k, vectRK3;

y := Array(0 .. N);

x := Array(0 .. N);

h := evalf(b-a)/N;

x[0] := a; y[0] := 1;

for n from 0 to N-1 do

x[n+1] := a+(n+1)*h;

k[1] := f(x[n], y[n]);

k[2] := f(x[n]+(1/2)*h, y[n]+(1/2)*h*k[1]);

k[3] := f(x[n]+h, y[n]+h*(-k[1]+2*k[2]));

y[n+1] := y[n]+(1/6)*h*(k[1]+4*k[2]+k[3])

end do;

[seq([x[n], y[n]], n = 0 .. N)]; y[1];

end proc;

## Now  we compute the error between y(1) and exact  solution for different value of  N

### I have a problem in this part

 errorRk3 := array(1 .. 29);
 for N from  2 to 30 do

errorrRk3[N] := abs(eval(rhs(res), x = 1)-RK3((x,y)->-y,0,1,N));

if errorrRk3[N] =10^{-3} end ;
end  do ;