samiyare

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14 years, 76 days

 

Amir

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These are questions asked by samiyare

Following my previous question

http://www.mapleprimes.com/questions/200627-Lssolve-Midpoint

I wrote the following code

 

restart:
Phiavg:=0.06;
lambda:=0.05;
Ha:=0;
NBT:=0.5;
Nr:=500;
#N[bt]:=cc*NBT+(1-cc)*4; ## cc between 0 and 1
N[bt]:=cc*NBT+(1-cc^2)*0.75;


                              0.06
                              0.05
                               0
                              0.5
                              500
                                           2
                    0.5 cc + 0.75 - 0.75 cc
eq1:=diff(u(eta),eta,eta)+1/(mu(eta)/mu1[w])*(sigma-Nr*(phi(eta)-phi1[w])-(1-phi(eta))*T(eta)-Ha^2*u(eta))+((1/mu(eta)*(mu_phi*diff(phi(eta),eta)))*diff(u(eta),eta));
eq2:=diff(T(eta),eta)-1/(k(eta)/k1[w]);
eq3:=diff(phi(eta),eta)-phi(eta)/(N[bt]*(1-gama1*T(eta))^2)*diff(T(eta),eta);
 /  d   /  d         \\      1                                 
 |----- |----- u(eta)|| + ------- (mu1[w] (sigma - 500 phi(eta)
 \ deta \ deta       //   mu(eta)                              

    + 500 phi1[w] - (1 - phi(eta)) T(eta)))

             /  d           \ /  d         \
      mu_phi |----- phi(eta)| |----- u(eta)|
             \ deta         / \ deta       /
    + --------------------------------------
                     mu(eta)                
                    /  d         \   k1[w]
                    |----- T(eta)| - ------
                    \ deta       /   k(eta)
                                       /  d         \            
                              phi(eta) |----- T(eta)|            
/  d           \                       \ deta       /            
|----- phi(eta)| - ----------------------------------------------
\ deta         /   /                       2\                   2
                   \0.5 cc + 0.75 - 0.75 cc / (1 - gama1 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)):
gama1:=0.00:
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:=1:
phi1[w]:=phi0:
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);
/  d   /  d         \\   //                                    
|----- |----- u(eta)|| + \\0.0009930000000 + 0.03883623000 phi0
\ deta \ deta       //                                         

                      2\                                 
   + 0.5301627000 phi0 / (sigma - 500 phi(eta) + 500 phi0

                           \//               
   - (1 - phi(eta)) T(eta))/ \0.0009930000000

                                                   2\   
   + 0.03883623000 phi(eta) + 0.5301627000 phi(eta) / +

  /                                       /  d           \ /  d  
  |(0.03883623000 + 1.060325400 phi(eta)) |----- phi(eta)| |-----
  \                                       \ deta         / \ deta

         \\//                                        
   u(eta)|| \0.0009930000000 + 0.03883623000 phi(eta)
         //                                          

                          2\
   + 0.5301627000 phi(eta) /
           /  d         \     0.597 + 4.45959 phi0  
           |----- T(eta)| - ------------------------
           \ deta       /   0.597 + 4.45959 phi(eta)
                                        /  d         \
                            1. phi(eta) |----- T(eta)|
         /  d           \               \ deta       /
         |----- phi(eta)| - --------------------------
         \ deta         /                           2
                             0.5 cc + 0.75 - 0.75 cc  
Q:=proc(pp2,fi0) option remember; local res,F0,F1,F2,a,INT0,INT10,B;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if;
res := dsolve(subs(sigma=pp2,phi0=fi0,{eq1=0,eq2=0,eq3=0,u(1)=-lambda*D(u)(1),u(0)=lambda*D(u)(0),phi(0)=phi0,T(0)=0}), numeric,output=listprocedure,initmesh=10, continuation=cc);
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)]));
INT0:=evalf(Int((abs(F0(eta)),eta=0..1)));
INT10:=evalf(Int(abs(F0(eta))*F1(eta),eta=0..1));
a[1]:=evalf(Int(F0(eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf),eta=0..1));
#a[1]:=evalf(Int((F0(eta),eta=0..1)));
a[2]:=(INT10/INT0-Phiavg)/Phiavg; #relative
[a[1],a[2]]
end proc:
Q1:=proc(pp2,fi0) Q(_passed)[1] end proc;
Q2:=proc(pp2,fi0) Q(_passed)[2] end proc;
proc(pp2, fi0)  ...  end;
proc(pp2, fi0)  ...  end;
#Q(116,0.0041);
#tempe:=Optimization:-LSSolve([Q1,Q2],initialpoint=[130,0.01]);
#tempe:=Optimization:-LSSolve([Q1,Q2],initialpoint=[43.55,0.39]);
tempe:=Optimization:-LSSolve([Q1,Q2],initialpoint=[5.65,0.00036]);
#tempe:=Optimization:-LSSolve([Q1,Q2],initialpoint=[12,0.003]); # khoob ba 1
#tempe:=Optimization:-LSSolve([Q1,Q2],initialpoint=[5,0.01]);
                  HFloat(5.65), HFloat(3.6e-4)
           HFloat(5.650000070103341), HFloat(3.6e-4)
           HFloat(5.65), HFloat(3.600105456508193e-4)
     HFloat(29.63242379055208), HFloat(0.0205927592420527)
    HFloat(12.803902258015825), HFloat(0.006395385884750864)
    HFloat(12.803902403534572), HFloat(0.006395385884750864)
    HFloat(12.803902258015825), HFloat(0.00639539649402585)
   HFloat(12.804004931505949), HFloat(0.0063954867657199386)
    HFloat(12.804107604996073), HFloat(0.006395587646689013)
    HFloat(12.80400483062498), HFloat(0.006498160255844027)
    HFloat(12.803902157134855), HFloat(0.006498059374874952)
   HFloat(-1.0206939292143726), HFloat(-3.32764179807047e-4)
   HFloat(-1.0206939079125088), HFloat(-3.32764179807047e-4)
   HFloat(-1.0206939292143726), HFloat(-3.327536344433438e-4)
    HFloat(18.749500943683863), HFloat(0.01993840615828979)
    HFloat(3.9953780262640484), HFloat(0.00481041471606933)
     HFloat(6.166152606930136), HFloat(0.00703619658484674)
    HFloat(7.3193201827812295), HFloat(0.008218585352824569)
Error, (in Optimization:-LSSolve) complex value encountered
sigma:=tempe[2](1);
                          tempe[2](1)
phi0:=tempe[2](2);
                          tempe[2](2)
with(plots):

res2 := dsolve({eq1=0,eq2=0,eq3=0,u(1)=-lambda*D(u)(1),u(0)=lambda*D(u)(0),phi(0)=phi0,T(0)=0}, numeric,output=listprocedure,continuation=cc);
Error, (in dsolve/numeric/process_input) boundary conditions specified at too many points: {0, 1, 2}, can only solve two-point boundary value problems
G0,G1,G2:=op(subs(res2,[u(eta),phi(eta),T(eta)])):
ruu:=evalf((Int(abs(G0(eta))*(G1(eta)*rhop+(1-G1(eta))*rhobf ),eta=0..zet)))/(Phiavg*rhop+(1-Phiavg)*rhobf);
phb:=evalf((Int(abs(G0(eta))*G1(eta),eta=0..1))) / evalf((Int(abs(G0(eta)),eta=0..1))) ;
TTb:=evalf(Int(abs(G0(eta))*G2(eta)*(G1(eta)*rhop*cp+(1-G1(eta))*rhobf*cbf ),eta=0..1))/evalf(Int(abs(G0(eta))*(G1(eta)*rhop*cp+(1-G1(eta))*rhobf*cbf ),eta=0..1));
Error, invalid input: subs received res2, which is not valid for its 1st argument
                /  /1.                                        \
                | |                                           |
0.0008538922115 | |    |G0(eta)| (2881.8 G1(eta) + 998.2) deta|
                | |                                           |
                \/0.                                          /
                    /1.                       
                   |                          
                   |    |G0(eta)| G1(eta) deta
                   |                          
                  /0.                         
                  ----------------------------
                        /1.                   
                       |                      
                       |                      
                       |    |G0(eta)| deta    
                      /                       
                       0.                     
                                                              /Int(
                              1                               |     
------------------------------------------------------------- |     
  /1.                                                         |     
 |                                                            \     
 |              /             6                       6\            
 |    |G0(eta)| \-1.1752324 10  G1(eta) + 4.1744724 10 / deta       
/                                                                   
 0.                                                                 

                    /             6                       6\ , eta = 0. .. 1.)
  |G0(eta)| G2(eta) \-1.1752324 10  G1(eta) + 4.1744724 10 /                  

  \
  |
  |
  |
  /
#rhouu:=evalf((Int((G1(eta)*rhop+(1-G1(eta))*rhobf)*G0(eta),eta=0..1)));

odeplot(res2,[[eta,u(eta)/ruu],[eta,phi(eta)/phb],[eta,T(eta)/TTb]],0..1);
#odeplot(res2,[[eta,u(eta)],[eta,phi(eta)],[eta,T(eta)]],0..1);
Error, (in plots/odeplot) input is not a valid dsolve/numeric solution
rhou:=evalf((Int(abs(G0(eta))*(G1(eta)*rhop+(1-G1(eta))*rhobf ),eta=0..zet))):

Nub:=(1/G2(1))*(((1+a[k1]*abs(G1(0))+b[k1]*abs(G1(0))^2)/(1+a[k1]*Phiavg+b[k1]*Phiavg^2)));
                0.6905123602 (1 + 7.47 |G1(0)|)
                -------------------------------
                             G2(1)             
(rhs(res2(0.0000000000001)[3])-rhs(res2(0)[3]))/0.0000000000001;
Error, invalid input: rhs received res2(0.1e-12)[3], which is not valid for its 1st argument, expr
sigma;
                          tempe[2](1)
NBT;
                              0.5
>

 

the above code has been worked for NBT=0.6 and higher, whereas as NBT decreases, the code doesnt converge easily.

How can I fix this problem?

Thanks for your attention in advance

Amir

Dear Sirs,

I actually rigoruos to know what is the algorithm of BVP[midrich]? how it can obtain the solution of ODE with singularities?

 

Did anyone introduce a reference about the algorithm like this?

Thanks for your attention in advance

Amir

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

Hi

I want to solve two odes with their boundary condition. I wrote the code below:

restart:

eq2:=diff(T(eta),eta,eta)+Nb*diff(T(eta),eta)*diff(phi(eta),eta)+Nt*diff(T(eta),eta)*diff(T(eta),eta);

eq3:=diff(phi(eta),eta,eta)+Nt/Nb*diff(T(eta),eta,eta);

sys_ode:=  eq2=0,eq3=0;
bcs := phi(0)=0,phi(h)=1,T(0)=0,T(h)=1;
sol:=dsolve([sys_ode, ics]);


however, this code doesnt get my desired results (the results are complex!). but when I (with hand) integrate Eq3 twice and substitute boundary conditions and replace in Eq2 the answer is easy and straightforward.

How can I change the following algorithm to get my results?

Thanks for your attention in advance

Amir

Following previous question at

http://www.mapleprimes.com/questions/149581-Improve-Algorithm-Dsolve

and also

http://www.mapleprimes.com/questions/149243-BVP-With-Constraining-Integrals

I wrote the following code

***********************

restart:

gama1:=0:


phi0:=0.00789:


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/(eta-1)+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]/(1-eta)*diff(T(eta),eta) )):
eq3:=diff(phi(eta),eta)-phi(eta)/(N[bt]*(1-gama1*T(eta))^2)*diff(T(eta),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,u(0)=0,eq1):
eq2:=subs(phi(0)=phi0,u(0)=0,eq2):
eq3:=subs(phi(0)=phi0,u(0)=0,eq3):

p:=proc(pp2) global res,F0,F1,F2:
if not type([pp2],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({eq1=0,subs(p2=pp2,eq2)=0,eq3=0,u(0)=0,u(1-zet)=0,phi(0)=phi0,T(0)=0,D(T)(0)=1}, numeric,output=listprocedure):
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
evalf(2/(1-zet^2)*Int((1-eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf)*( F1(eta)*rhop*cp+(1-F1(eta))*rhobf*cbf )/(F1(eta)*rhop+(1-F1(eta))*rhobf)*F0(eta),eta=0..1-zet))-pp2*10000:
end proc:


s1:=Student:-NumericalAnalysis:-Secant(p(pp2),pp2=[6,7],tolerance=1e-6);

                   HFloat(6.600456858832996)

p2:=%:



ruu:=evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet))):
phb:=evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet))) / evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet))) :
TTb:=evalf(2/(1-zet^2)*(Int((1-eta)*F2(eta),eta=0..1-zet))):
rhouu:=evalf(2/(1-zet^2)*(Int((1-eta)*(F1(eta)*rhop+(1-F1(eta))*rhobf)*F0(eta),eta=0..1-zet))):
with(plots):
res(parameters=[R0,R1]):
odeplot(res,[[eta,u(eta)/ruu],[eta,phi(eta)/phb],[eta,T(eta)/TTb]],0..zet);

 

*************************************

as you can see at the second line of the code, the value of phi0:=0.00789. however, I want to modify the code in a way that phi0 is calculated with the following addition constraint

evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta)*F1(eta),eta=0..1-zet))) / evalf(2/(1-zet^2)*(Int((1-eta)*F0(eta),eta=0..1-zet)))-0.02=0

I would be most grateful if you could help me in this problem.

Thanks for your attention in advance

Amir

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