tomleslie

As I said in my original response...

Answered: tomleslie 13579

March 15 2023

0 0

@delvin

Pretty unlikely that you will be able to solve analytically
Given some additional information, you may be able to solve numerically

HAve you even read...

Answered: tomleslie 13579

March 02 2023

0 0

@MANUTTM

the error message which you are receiving?

Error, (in Optimization:-NLPSolve) could not store 3233180.16916754-HFloat(infinity)*I in a floating-point rtable

I warned you before that generating complex numbers in an optization problem was a very bad idea - so what do think is going to happen when you generate a complex number with an INFINITE imaginary part?

'D'...

Answered: tomleslie 13579

March 02 2023

0 0

is a particularly unfortunate example because, by default, Maple will interpret this as the differential operator, rather than the simple literal 'D'. In consequence it will appear upright rather than in italic, unless you force italic, as in the attached

>	restart: with(Units): plot( x, x=0..1, labels=[ typeset(DUnit(kg/m^3)), typeset(AUnit(m) ) ], labelfont=[times, bolditalic, 14] );

Automatically loading the Units[Simple] subpackage

>

Download labs2.mw

Again the appearance of the figure on this site is misleading - the way it appears in a worksheet is shown below

As it...

Answered: tomleslie 13579

March 02 2023

0 0

appears in a Maple worksheet

Doesn't make...

Answered: tomleslie 13579

February 28 2023

0 0

@lemelinm

muh of a difference - see the attached

Download simpleInt2.mw

This is not...

Answered: tomleslie 13579

February 23 2023

0 0

@KIRAN SAJJAN

a code writing service. You are expected to do most of the work!

See the attached two files which produce very similar results using Maple's buil-in numeric solvers, and the HPM method

>

a2 := (1-p2)*(1-p1+p1*rs1/rf)+p2*rs2/rf

>

a3 := 1+3*((p1*sigma1+p2*sigma2)/`σf`-p1-p2)/(2+(p1*sigma1+p2*sigma2)/((p1+p2)*`σf`)-((p1*sigma1+p2*sigma2)/`σf`-p1-p2))

>

a4 := (1-p2)*(1-p1+p1*rs1*cps1/(rf*cpf))+p2*rs2*cps2/(rf*cpf)

>

a5 := (ks1+2*kf-2*p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2))/((ks1+2*kf+p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))+2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2)))

>

  OdeSys:= diff(U(Y), Y, Y)/(a1*a2) + Theta(Y) + N*Theta(Y)*Theta(Y) - a3*M*M*U(Y)/a2 - kp*kp*U(Y)/(a1*a2),
           a5*diff(Theta(Y), Y, Y)/a4 + Pr*Ec*(diff(U(Y), Y)^2 + U(Y)^2*kp*kp)/(a1*a2);

  MVals:= [0.5, 1.0, 1.5];
  Cond := U(0) = lambda*D(U)(0), Theta(0) = A + g*D(Theta)(0),
          U(1) = 0, Theta(1) = B;

  for j to numelems(MVals) do
      Ans[j]:= dsolve( eval([OdeSys, Cond], M = MVals[j]),
                       numeric,
                       output = listprocedure
                    );
      Theta_b[j]:= evalf( Int(eval(U(Y), Ans[j])(Y)*eval(Theta(Y), Ans[j])(Y), Y = 0 .. 1)
                          /
                          Int(eval(U(Y), Ans[j])(Y), Y = 0 .. 1)
                        );
  end do;

.7732859212*(diff(diff(U(Y), Y), Y))+Theta(Y)+.5*Theta(Y)^2-.7903642147*M^2*U(Y)-0.6959573287e-1*U(Y), 1.281183414*(diff(diff(Theta(Y), Y), Y))+2.783829316*(diff(U(Y), Y))^2+.2505446384*U(Y)^2

(1)

>

  with(plots):
  cols := [red, blue, black, green, cyan];
  display( [ seq( odeplot( Ans[j],
                           [Y, diff(U(Y), Y)],
                           Y = 0 .. 1,
                           color = cols[j]
                         ),
                  j = 1 .. numelems(MVals)
                )
           ],
           title = typeset(diff(U(Y), Y), "vs ", Y),
           legend = [typeset(M = 0, 5), typeset(M = 1), typeset(M = 1.5)]
         );
  display( [ seq( odeplot( Ans[j],
                           [Y, Theta(Y)],
                           Y = 0 .. 1,
                           color = cols[j]
                         ),
                  j = 1 .. numelems(MVals)
                )
           ],
           title = typeset(Theta(y), "vs ", Y),
           legend = [typeset(M = 0, 5), typeset(M = 1), typeset(M = 1.5)]
         );

>

  interface(rtablesize = 100):
  interface(displayprecision = 5):
  Matrix( [ [0, M = 0.5, M = 0.5, M = 1, M = 1, M = 1.5, M = 1.5],
            [Y, Cf, Nu, Cf, Nu, Cf, Nu],
            seq( [ j,
                   seq( [ eval(diff(U(Y), Y)/a1, Ans[k])(j),
                          eval(-diff(Theta(Y), Y)/(Theta_b[k]*a5), Ans[k])(j)
                        ][],
                        k = 1 .. numelems(MVals)
                      )
                 ],
                 j = 0 .. 1, 0.1
               )
          ]
        );
  interface(rtablesize = 10):
  interface(displayprecision = -1):

(2)

>

Download more3.mw

>

restart:

>

(1)

>

(2)

>

a2 := (1-p2)*(1-p1+p1*rs1/rf)+p2*rs2/rf

>

a3 := 1+3*((p1*sigma1+p2*sigma2)/`σf`-p1-p2)/(2+(p1*sigma1+p2*sigma2)/((p1+p2)*`σf`)-((p1*sigma1+p2*sigma2)/`σf`-p1-p2))

>

a4 := (1-p2)*(1-p1+p1*rs1*cps1/(rf*cpf))+p2*rs2*cps2/(rf*cpf)

>

a5 := (ks1+2*kf-2*p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2))/((ks1+2*kf+p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))+2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2)))

>

Z := 4:

>	U(Y):= add((p^(i))*u[i](Y), i = 0 .. Z):

>	Theta(Y):= add((p^(i))*theta[i](Y), i = 0 .. Z):

>

    odeSys:= { (1-p)*(diff(U(Y),Y,Y))/(a1*a2)
              +
              p*((diff(U(Y),Y,Y))/(a1*a2)+Theta(Y)+N*(Theta(Y)*Theta(Y))-a3*(M*M)*U(Y)/a2-(kp*kp)*U(Y)/(a1*a2)),

              (1-p)*a5*(diff(Theta(Y),Y,Y))/a4
              +
              p*(a5*(diff(Theta(Y),Y,Y))/a4+Pr*Ec*((diff(U(Y),Y))*(diff(U(Y),Y))+(U(Y)*U(Y))*(kp*kp))/(a1*a2))
            }:

   MVals:=[0.5,1.0,1.5]:

   for j from 1 by 1 to numelems(MVals) do

       cond:= u[0](0) = lambda*(D(u[0]))(0), theta[0](0) = A+g*(D(theta[0]))(0),
              u[0](1) = 0, theta[0](1) = B:
       ans[j]:= {}:

       for k from 0 by 1 to Z do
           ans[j]:= evalf( `union`
                           ( ans[j],
                             dsolve
                             ( { eval
                                 ( coeff~(eval(odeSys, M=MVals[j]), p, k),
                                   ans[j]
                                 )[],
                                 cond
                               }
                             )
                           )
                        ):
           cond:= u[k+1](0) =lambda*(D(u[k+1]))(0), theta[k+1](0) = 0, u[k+1](1) = 0, theta[k+1](1) = 0:
       od:

       Theta_b[j]:= int( eval(U(Y), [ans[j][], p=1])*eval(Theta(Y), [ans[j][], p=1]), Y=0..1)
                    /
                    int( eval(U(Y), [ans[j][], p=1]), Y=0..1):
   od:

>

  with(plots):
  cols := [red, blue, black, green, cyan]:

  display
  ( [ seq
      ( plot
        ( diff(eval( U(Y), [ans[j][], p=1]),Y),
          Y=0..1,
          color=cols[j]
        ),
        j=1..numelems(MVals)
      )
    ],
    title = "diff(U(Y), Y) vs Y)",
    legend = [typeset(M = 0.5), typeset(M = 1), typeset(M = 1.5)]
  );
  display
  ( [ seq
      ( plot
        ( eval( Theta(Y), [ans[j][], p=1]),
          Y=0..1,
          color=cols[j]
        ),
        j=1..numelems(MVals)
      )
    ],
    title = "Theta(Y) vs Y)",
    legend = [typeset(M = 0.5), typeset(M = 1), typeset(M = 1.5)]
  );

>

  interface(rtablesize=100):
  interface(displayprecision=5):
  Matrix( [ [ Y, Cf, Nu, Cf, Nu, Cf, Nu],
              seq
              ( [ j,
                  seq( [eval(diff(U(Y),Y)/a1,[ans[k][], Y=j,p=1]),
                        eval(-1/(Theta_b[k])*diff(Theta(Y),Y)/a5,[ans[k][],Y=j,p=1])
                       ][],
                       k=1..numelems(MVals)
                     )

                ],
                j=0..1, 0.1
              )
          ]
      );
  interface(rtablesize=100):
  interface(displayprecision=-1)

(3)

>

Download bhpm3.mw

Ids there no end...

Answered: tomleslie 13579

February 23 2023

0 0

@KIRAN SAJJAN

to what yoiu want?

There are two attached files: one solve the problem using Maple's built-in hunerical solvers, and one using the "HPM method". They produce the following two tables (where I have restricted the display precision (but not the precision used in actual calculation), just to make comparisons easier.

Maple's numeric solvers

HPM method

OK, these are different, but generally they are within about 3%. Since I have absolutely no idea how accurate I would expect a 4-term HPM method to be, such a discrepancy doesn't bother me much

more2.mw

bhpm2.mw

Worksheet...

Answered: tomleslie 13579

February 23 2023

0 0

@KIRAN SAJJAN

with requested graphs and table is attached.

At no time did I enter as input either U(Y)(Y) or Theta(Y)(Y).

In Maple output, this is an artefact of using the option output=listprocedure in the dsolve() command. Maple will give the relevant procedures the names U(Y) and Theta(Y), but these are just literal procedureNames. Applying these literal procedure names to a variable 'Y' is what results in U(Y)(Y) and Theta(Y)(Y)

>

a2 := (1-p2)*(1-p1+p1*rs1/rf)+p2*rs2/rf

>

a3 := 1+3*((p1*sigma1+p2*sigma2)/`σf`-p1-p2)/(2+(p1*sigma1+p2*sigma2)/((p1+p2)*`σf`)-((p1*sigma1+p2*sigma2)/`σf`-p1-p2))

>

a4 := (1-p2)*(1-p1+p1*rs1*cps1/(rf*cpf))+p2*rs2*cps2/(rf*cpf)

>

a5 := (ks1+2*kf-2*p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2))/((ks1+2*kf+p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))+2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2)))

>

interface(rtablesize = 100); Matrix([[0, M = .5, M = .5, M = 1, M = 1, M = 1.5, M = 1.5], [Y, Cf, Nu, Cf, Nu, Cf, Nu], seq([j, seq([(eval((diff(U(Y), Y))/a1, Ans[k]))(j), (eval(-(diff(Theta(Y), Y))/(Theta_b[k]*a5), Ans[k]))(j)][], k = 1 .. numelems(MVals))], j = 0 .. 1, .1)]); interface(rtablesize = 100)

Matrix(%id = 36893488148298344076)

(1)

>

Download more.mw

That depends a lot...

Answered: tomleslie 13579

February 22 2023

0 0

@KIRAN SAJJAN

on whether you can define "stream line", "Nussel number" and "skin friction" in terms of the quantities produced in the worksheet.

I can't.

If you can't - then you are stuffed!

Well...

Answered: tomleslie 13579

February 22 2023

0 0

@KIRAN SAJJAN

your file a.mw produces

Theta_b := 0.418387093869607
Q := 0.0869578833401966

The attached file where I have corrected all the syntax errors (too many to list), fixed the logical errors (too many to list), deeleted all the code which you don't need (about half of it) produces

Theta_b := 0.4131557662
Q := 0.08652473337

Whilst these don't match exactly, the HPM method is an "approximation", so I would say that the agreement is "pretty good". Maybe with more terms in the power series expansion, the agreement would becoome even better, but I haven't got the time/patience to explore this.

>

restart:

>

(1)

>

(2)

>

a2 := (1-p2)*(1-p1+p1*rs1/rf)+p2*rs2/rf

>

a3 := 1+3*((p1*sigma1+p2*sigma2)/`σf`-p1-p2)/(2+(p1*sigma1+p2*sigma2)/((p1+p2)*`σf`)-((p1*sigma1+p2*sigma2)/`σf`-p1-p2))

>

a4 := (1-p2)*(1-p1+p1*rs1*cps1/(rf*cpf))+p2*rs2*cps2/(rf*cpf)

>

a5 := (ks1+2*kf-2*p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2))/((ks1+2*kf+p1*(kf-ks1))*(ks2+2*kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))+2*p2*(kf*(ks1+2*kf-2*p1*(kf-ks1))/(ks1+2*kf+p1*(kf-ks1))-ks2)))

>

Z := 4;

(3)

>	U(Y):= add((p^(i))*u[i](Y), i = 0 .. Z) ;

(4)

>	Theta(Y):= add((p^(i))*theta[i](Y), i = 0 .. Z) ;

(5)

>

  odeSys:= {(1-p)*(diff(U(Y), Y, Y))/(a1*a2)+p*((diff(U(Y), Y, Y))/(a1*a2)+Theta(Y)+N*(Theta(Y)*Theta(Y))-a3*(M*M)*U(Y)/a2-(kp*kp)*U(Y)/(a1*a2)), (1-p)*a5*(diff(Theta(Y), Y, Y))/a4+p*(a5*(diff(Theta(Y), Y, Y))/a4+Pr*Ec*((diff(U(Y), Y))*(diff(U(Y), Y))+(U(Y)*U(Y))*(kp*kp))/(a1*a2))}:

  cond:= u[0](0) = lambda*(D(u[0]))(0), theta[0](0) = A+g*(D(theta[0]))(0), u[0](1) = 0, theta[0](1) = B:
  ans:={}:

  for k from 0 by 1 to Z do
      ans:= evalf(`union`
            ( ans,
              dsolve
              ( { eval
                  ( coeff~(odeSys, p, k),
                    ans
                  )[],
                  cond
                }
              )
           )):
       cond:= u[k+1](0) =lambda*(D(u[k+1]))(0), theta[k+1](0) = 0, u[k+1](1) = 0, theta[k+1](1) = 0:
  od:

>	Theta_b:=int( eval(U(Y), [ans[], p=1])*eval(Theta(Y), [ans[], p=1]), Y=0..1) / int( eval(U(Y), [ans[], p=1]), Y=0..1); Q:=int( eval(U(Y), [ans[], p=1]), Y=0..1);

(6)

>

  interface(rtablesize=100):
  Matrix( [ [ M, Cf, Nu],
              seq
              ( [ j,
                  eval(diff(U(Y),Y)/a1,[ans[], Y=j,p=1]),
                  eval(-1/(Theta_b)*diff(Theta(Y),Y)/a5,[ans[],Y=j,p=1])
             ],
                j=0..2, 0.5
              )
          ]
      );
  interface(rtablesize=100):

Matrix(6, 3, {(1, 1) = 1, (1, 2) = Cf, (1, 3) = Nu, (2, 1) = 0, (2, 2) = .2375370263, (2, 3) = 1.257689358, (3, 1) = .5, (3, 2) = -.1672057841, (3, 3) = 1.277606271, (4, 1) = 1.0, (4, 2) = -.2779064137, (4, 3) = 1.351492988, (5, 1) = 1.5, (5, 2) = -.1898708173, (5, 3) = 1.429803746, (6, 1) = 2.0, (6, 2) = 0.3158676368e-1, (6, 3) = 1.453241683})

(7)

>

Download bhpm.mw

In your worksheets...

Answered: tomleslie 13579

February 22 2023

0 0

@KIRAN SAJJAN

a.mw b_hpm.mw

you have Pr=7.2, M=1. In your original question (and my original response), you have Pr=0.3, M := .5. Would you expect to get the same answers when you use different parameters?

I have no idea which set of parameters to use.

It is simple...

Answered: tomleslie 13579

February 21 2023

0 0

@KIRAN SAJJAN

I repeat the same two questiions - what are the answers?

How are these "Streamlines, Isotherms and Microrotations". defined in terms of the variables and parameters of the ODE system listed above?

According to the graphic you supply, the "Streamlines, Isotherms and Microrotations" are parameterised by Re , Pr, Gr^, and Ha. However Re, Gr, and Ha occur nowhere in the definition of the ODE system. What are these parameters?

And...

Answered: tomleslie 13579

February 21 2023

0 0

@KIRAN SAJJAN

until you answer the two simple questions in my original reply, it is impossible for anyone here to help you

Try...

Answered: tomleslie 13579

February 21 2023

0 0

@KIRAN SAJJAN

answering the questions in my original reply

A useful...

Answered: tomleslie 13579

February 21 2023

0 0

first step would be to post a MAple worksheet illustrating the problem, rather than some jumble of images and code

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