I'm writing a simple Maple program to test the Generalized Finite Element Method: main_screened_Poisso.mw

When trying to define the Neumann boundary conditions, I have to define a directional derivative dudn=dudx*n. However, I can't seem to define a unit vector normal to Gamma, which is defined by a LineSegments objects.

Other than that, the row reduction is very slow, even though I'm using floating point arithmatic and not exact arithmatic, I believe.

How can I solve these problems? Thanks in advance!

Hello, 
I have tried to make a densityplot with a logarithmic scale on one axis using : 

So the plot is displayed with a log scale but the grid used to generate the plot is still linear, with a regular spacing.
I don't know how to make it logarithmic too. 
Maybe it's not very clear. For example I would like  a spacing of 0.1 from 0.1 to 1 then a spacing of 1 from 1 to 10.
How can I do that ? 
I have not find any clue in the help ...

EDIT : The option "axis[1]=[mode=log]" only change how the plot is displayed.
I would like to change the grid, that is to say how the plot is calculated.

Hi!

I still have a problem and im looking forward to any suggestions

this is the previous code that i have wrote

P := array([[8, 4], [8, 3], [8, 2], [7, 1], [6, 0], [5, 0], [4, 0], [2, 1], [1, 1], [1, 4]]);

> for j from 2 to 5 do k[j] := j+1;

x[j] := add(P[j, 1], j = j-1 .. j+2);

X[j] := add(P[j, 1]^2, j = j-1 .. j+2);

y[j] := add(P[j, 2], j = j-1 .. j+2);

Y[j] := add(P[j, 2]^2, j = j-1 .. j+2);

xy[j] := add(P[j, 1]*P[j, 2], j = j-1 .. j+2);

cx[j] := evalf(x[j]/k[j]);

cy[j] := evalf(y[j]/k[j]);

c11[j] := evalf(X[j]/k[j]-cx[j]^2);

c22[j] := evalf(Y[j]/k[j]-cy[j]^2);

c12[j] := evalf(xy[j]/k[j]-cx[j]*cy[j]);

C[j] := evalf(Matrix(2, 2, [[c11[j], c12[j]], [c12[j], c22[j]]]));

E[j] := simplify(fnormal(LinearAlgebra[Eigenvalues](C[j])));

if E[j][1] > E[j][2] then a[j] := E[j][2]/(E[j][1]+E[j][2]) else b[j] := E[j][1]/(E[j][1]+E[j][2])

 end if;

a[j];b[j]

 end do;

now, my question is how to put the output from the above looping in a  matrix form.

in my matrix, i need to call a[j],b[j], E[j][1], E[j][2] and the coordinate points. so my matrix dimension 4 x 5

I guess its a simple task but i tried hard and didnt get it worked.

thank you.

> sol := pdsolve({ICS, sys1, sys2, sys3, sys4, sys5, sys6, sys7}, numeric, method = rkf45, parameters = [a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p], output = listprocedure);

Error, (in pdsolve) invalid input: `pdsolve/numeric` expects its 2nd argument, IBCs, to be of type {set, list}, but received method = rkf45

restart;

x11:=[1.05657970467127, .369307407127487, .400969917393968, .368036162749865, .280389875142339, .280523489139136, .283220960827744, .373941285224253, .378034013792196, .384412762008662, .358678988563716, .350625923673556, .852039817522304, .362240519978640, 1.03197080591829, .343650441408896, .982510654490390, .404544012440991, .422063867224247, 1.20938803285209, .455708586000668, 1.22503869712995, .388259397947667, .472188904769827, 1.31108028794286, 1.19746589728366, .572669348193002];

y11:= [.813920951682113, 10.3546712426210, 2.54581301217449, 10.2617298458172, 3.82022939508992, 3.81119683373741, 3.90918914917183, 10.5831132713329, 10.8700088489538, 11.0218056177585, 10.5857571473115, 9.89034057997145, .271497107157453, 9.77706473740146, 2.23955104698355, 4.16872072216206, .806710906391666, 11.9148193656260, 12.0521411908477, 2.52812993540440, 12.6348841508094, 2.72197067934160, 5.10891266728297, 13.3609183272238, 3.03572692234234, 1.07326033849793, 15.4268962507711];

z11:= [8.93290500985527, 8.96632856524217, 15.8861149154785, 9.16576669760908, 3.20341865536950, 3.11740291181539, 3.22328961317946, 8.71094047480794, 8.60596466961827, 9.15440788281943, 10.2935566768586, 10.5765776143026, 16.3469510439066, 9.36885507010739, 2.20434678689869, 3.88816077008078, 17.9816287534802, 10.1414228793737, 10.7356141216242, 4.00703203725441, 12.0105837616461, 3.77028605914906, 5.01411979976607, 12.7529165152417, 3.66800269682059, 21.2178824031985, 13.9148746721034];

u11 := [5.19, 5.37, 5.56, 5.46, 5.21, 5.55, 5.56, 5.61, 5.91, 5.93, 5.98, 6.28, 6.24, 6.44, 6.58, 6.75, 6.78, 6.81, 7.59, 7.73, 7.75, 7.69, 7.73, 7.79, 7.91, 7.96, 8.05];

u11 := [seq(close3(t+t3), t3=0..26)];

sys1:=Diff(a1(s,t),s) = a*a1(s,t)+ b*a2(s,t)+ c*a3(s,t)+ d*u(t);

sys2:=Diff(a2(s,t),s) = e*a1(s,t)+ f*a2(s,t)+ g*a3(s,t)+ h*u(t);

sys3:=Diff(a3(s,t),s) = i*a1(s,t)+ j*a2(s,t)+ k*a3(s,t)+ l*u(t);

sys4:=Diff(y(t),t) = m*a1(s,t)+n*a2(s,t)+ o*a3(s,t)+ p*u(t);

sys5:= Diff(a1(s,t),t) = a1(s,t);

sys6:= Diff(a2(s,t),t) = a2(s,t);

sys7:= Diff(a3(s,t),t) = a3(s,t);

sol := pdsolve([sys1, sys2, sys3,sys4,sys5,sys6,sys7]);

t2 := [seq(i, i=1..27)];

xt1 := subs(_C1=1,sol[1]); # a1(t)

xt2 := subs(_C1=1,sol[2]); # a2(t)

xt3 := subs(_C1=1,sol[3]); # a3(t)

ut1 := subs(_C1=1,sol[4]); # u(t)

tim := [seq(n, n=1..27)];

N:=nops(tim):

ICS:=a1(1)=x11[1],a2(1)=y11[1],a3(1)=z11[1],u1(1)=u11[1];

sol:=pdsolve({sys1, sys2, sys3,sys4,sys5,sys6,sys7,ICS}, numeric, method=rkf45, parameters=[ a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p],output=listprocedure);

ans(.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003,.003,.003,.003,.003);

ans:=proc(a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p) sol(parameters=[ a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p]);

add((xt1(tim[i])-x11[i])^2,i=1..N)+add((xt2(tim[i])-y11[i])^2,i=1..N)+add((xt3(tim[i])-z11[i])^2,i=1..N)+add((ut1(tim[i])-u11[i])^2,i=1..N);

end proc;

result1 := Optimization:-Minimize(ans,initialpoint=[.001,.002,.003,.001,.002,.003,.001,.002,.003,.003,.003,.003,.003,.003,.003,.003]);

https://drive.google.com/file/d/0B2D69u2pweEvU3NpWWQwS3U1XzQ/edit?usp=sharing
https://drive.google.com/file/d/0B2D69u2pweEvMnFabkdiX1hpYVk/edit?usp=sharing

a1 := Diff(x1(s,t),s$2) = a*x1(s,t)+b*x2(s,t)+c*x3(s,t)+d*u(t);
a2 := Diff(x1(s,t),t)=x1(s,t);
b1 := Diff(x2(s,t),s$2) = e*x1(s,t)+f*x2(s,t)+g*x3(s,t)+h*u(t);
b2 := Diff(x2(s,t),t)=x2(s,t);
c1 := Diff(x3(s,t),s$2) = i*x1(s,t)+j*x2(s,t)+k*x3(s,t)+l*u(t);
c2 := Diff(x3(s,t),t)=x3(s,t);
sys := [a1, a2, b1, b2, c1, c2];
sol := pdsolve(sys);

length exceed limit

Hello, I have different values to substitute into my theta in a matrix. I tried to use the subs function but the results came out are all dashes. Have you any idea how to fix it please?

AA:=RandomMatrix(3,5);
BB:=map(cos,AA*theta);
CC:=RandomMatrix(3,5);

I tried method 1:

subs[inplace](theta=CC,BB);

Method 2:

DD:=Matrix(3,5);
for m from 1 to 3 do
for n from 1 to 5 do
DD(m..m,n..n):=subs(theta(m..m,n..n)=CC(m..m,n..n),BB(m..m,n..n));
end do:
end do:
DD;

Thank you.

Hello,

I tried to install maplesim on windows 8 but it didn't work.

However, i have installed microsoft visual c++ express before and i have made the precedure as described in maple that is to say 

or

hi, i have a problem with maple. i took the codes from a book, it should be true but even i copied and past exactly the same, again the maple gave me error which i couldnt solve. could you please help me? how can i correct it ? incorrect codes are below, but if you need all codes i can write here. i need a solution immediately :(

>relativefrequencies := proc(text, language)
evalf(frequencies(text, language)/StringTools:-Length(text))
end proc:
VigenereKeyFind := proc(ciphertext, max:=floor(StringTools:-Length(ciphertext)/15),
{language:=en})
uses StringTools;
local freqs;
freqs := map(x -> relativefrequencies(x, language),
map(Implode, partit(Explode(ciphertext),
keylength(ciphertext, max, ’:-language’=language))));
Implode(map(li -> frequencyanalysis(li, ’:-language’=language), freqs))
end proc;

>VigenereKeyFind(c);
Error, (in VigenereKeyFind) ``’`` does not evaluate to a module

a1 := Diff(x1(s,t),s$2) = a*x1(s,t)+b*x2(s,t)+c*x3(s,t)+d*u(t);
b1 := Diff(x2(s,t),s$2) = e*x1(s,t)+f*x2(s,t)+g*x3(s,t)+h*u(t);
c1 := Diff(x3(s,t),s$2) = i*x1(s,t)+j*x2(s,t)+k*x3(s,t)+l*u(t);
sys := [a1, b1, c1];
sol := pdsolve(sys);

sol := pdsolve(sys);
[Length of output exceeds limit of 1000000]

I want to solve the equation x^2 + y^2 = 1, where x, y are rational numbers. How can I tell Maple to do that?

Hi all, 

I am quite new to maple. Thanks for your help in advance. 

Could I define a general function f of a single variable x, f(x), without specifying its explicit form? I know it is simple to define functions of specific forms, e.g., f(x)=2*x (In maple: f:=x->2*x). I try to define a function without knowing its form, e.g., f:=x->f(x), expecting maple return f(y) if I type >>f(y). But this does not work (Maple says:

Error, (in temp) too many levels of recursion). 

Is it possible to define a general function f(x) without specific form in maple? Thank you again. 

Hi,

I get the error in the following code

restart:

gama1:=0.01:

zet:=0;
#phi0:=0.00789:
Phiavg:=0.02;
lambda:=0.01;
Ha:=1;


                               0
                              0.02
                              0.01
                               1
rhocu:=2/(1-zet^2)*int((1-eta)*rho(eta)*c(eta)*u(eta),eta=0..1-zet):

eq1:=diff(u(eta),eta,eta)+1/(mu(eta)/mu1[w])*(1-Ha^2*u(eta))+((1/(eta)+1/mu(eta)*(mu_phi*diff(phi(eta),eta)))*diff(u(eta),eta));
eq2:=diff(T(eta),eta,eta)+1/(k(eta)/k1[w])*(-2/(1-zet^2)*rho(eta)*c(eta)*u(eta)/(p2*10000)+( (a[k1]+2*b[k1]*phi(eta))/(1+a[k1]*phi1[w]+b[k1]*phi1[w]^2)*diff(phi(eta),eta)+k(eta)/k1[w]/(eta)*diff(T(eta),eta) ));
eq3:=diff(phi(eta),eta)+phi(eta)/(N[bt]*(1+gama1*T(eta))^2)*diff(T(eta),eta);
      /  d   /  d         \\   mu1[w] (1 - u(eta))
      |----- |----- u(eta)|| + -------------------
      \ deta \ deta       //         mu(eta)      

           /             /  d           \\               
           |      mu_phi |----- phi(eta)||               
           | 1           \ deta         /| /  d         \
         + |--- + -----------------------| |----- u(eta)|
           \eta           mu(eta)        / \ deta       /
                                /      /                        
                                |      |                        
/  d   /  d         \\     1    |      |  rho(eta) c(eta) u(eta)
|----- |----- T(eta)|| + ------ |k1[w] |- ----------------------
\ deta \ deta       //   k(eta) |      |         5000 p2        
                                \      \                        

                                /  d           \
     (a[k1] + 2 b[k1] phi(eta)) |----- phi(eta)|
                                \ deta         /
   + -------------------------------------------
                                          2     
         1 + a[k1] phi1[w] + b[k1] phi1[w]      

            /  d         \\\
     k(eta) |----- T(eta)|||
            \ deta       /||
   + ---------------------||
           k1[w] eta      ||
                          //
                                      /  d         \
                             phi(eta) |----- T(eta)|
          /  d           \            \ deta       /
          |----- phi(eta)| + ------------------------
          \ deta         /                          2
                             N[bt] (1 + 0.01 T(eta))
mu:=unapply(mu1[bf]*(1+a[mu1]*phi(eta)+b[mu1]*phi(eta)^2),eta):
k:=unapply(k1[bf]*(1+a[k1]*phi(eta)+b[k1]*phi(eta)^2),eta):
rhop:=3880:
rhobf:=998.2:
cp:=773:
cbf:=4182:
rho:=unapply(  phi(eta)*rhop+(1-phi(eta))*rhobf ,eta):
c:=unapply(  (phi(eta)*rhop*cp+(1-phi(eta))*rhobf*cbf )/rho(eta) ,eta):
mu_phi:=mu1[bf]*(a[mu1]+2*b[mu1]*phi(eta)):

a[mu1]:=39.11:
b[mu1]:=533.9:
mu1[bf]:=9.93/10000:
a[k1]:=7.47:
b[k1]:=0:
k1[bf]:=0.597:
zet:=0.5:
#phi(0):=1:
#u(0):=0:
phi1[w]:=phi0:
N[bt]:=0.2:
mu1[w]:=mu(0):
k1[w]:=k(0):

eq1:=subs(phi(0)=phi0,eq1):
eq2:=subs(phi(0)=phi0,eq2):
eq3:=subs(phi(0)=phi0,eq3):

#A somewhat speedier version uses the fact that you really need only compute 2 integrals not 3, since one of the integrals can be written as a linear combination of the other 2:
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10,B;
global Q1,Q2;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve(subs(p2=pp2,phi0=fi0,{eq1=0,eq2=0,eq3=0,u(1)=lambda/(phi(1)*rhop/rhobf+(1-phi(1)))*D(u)(1),D(u)(0)=0,phi(1)=phi0,T(1)=0,D(T)(1)=1}), numeric,output=listprocedure):
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
INT0:=evalf(Int((1-eta)*F0(eta),eta=0..1-zet));
INT10:=evalf(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet));
B:=(-cbf*rhobf+cp*rhop)*INT10+ rhobf*cbf*INT0;
a[1]:=2/(1-zet^2)*B-10000*pp2;
a[2]:=INT10/INT0-Phiavg;
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;
#The result agrees very well with the fsolve result.
#Now I did use a better initial point. But if I start with the same as in fsolve I get the same result in just about 2 minutes, i.e. more than 20 times as fast as fsolve:

Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;
Optimization:-LSSolve([Q1,Q2],initialpoint=[6.5,exp(-1/N[bt])]);


proc(pp2, fi0)  ...  end;
proc(pp2, fi0)  ...  end;
proc(pp2, fi0)  ...  end;
              HFloat(6.5), HFloat(0.006737946999)

the error is :

Error, (in Optimization:-LSSolve) system is singular at left endpoint, use midpoint method instead

how can I fix it.

Thanks

Amir

the point given as follows:

x=[3,4,5,6,7]...

y=[7.42494922444550, 3.67768248674133, 2.52235142453921, 1.95610223891559, 1.61770309810016]

I have plotted the point and the output is like this:

plot_output.docx

how to smooth the curve and extend the curve longer 

please help me.....

How to draw an object in 3d which is like a spring

3 layers of circle already enough