shakuntala

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These are replies submitted by shakuntala

@mmcdara 

  Yes, your right but this boundary value problem can be changed into initial value problem. Later you can solve it. For                 instance

             f'''+f*f''=0

             t''+f*t'=0

       Boundary conditions

     f(0)=0,f'(0)=1; f(infinity or some limit or 5) = 1; t(0) = 1; t(infinity or some limit or 5)=0

     Simplify above equations as follows

               y = f

               y= f'

               y= f''

                y= f'''

                y= t

                y= t'

    Now write the above equations as

              f'''=-f*f'' and t''= -f*t'  and t'' = - f*t', i.e., F=-f*f'' and T= -f*t'

           f1= -f'' = -y2

              f2  = 0 = 0

           f3=  -f = -y

             t1= - t' = -y5

          t5= -f = -y1

 Initial value conditions

               y1[0] = 0

                y[0]= 1

                y3[0]= Slope1

                y4[0] = 1

                y5[0] = Sllope2

     Differentiate w.r.t slope1 above IVP, we get as

                 u1[0]=0

                 u2[0]=0

                u3[0]=1

                u4[0]=0

                u5[0]=0

        Differentiate w.r.t slope2 above IVP, we get as

                 v1[0]=0

                 v2[0]=0

                v3[0]=0

                v4[0]=0

                v5[0]=1

  

 

          

 

 

@tomleslie 

 

    Good, Can you tell which method you have used to solve this problem?

@tomleslie 

 Well I tell you the method then you may get an idea

                           y'= f(x,y)

                     y=y0, when x=x0, x1=x0+h

                    y1=y0+1/6*(k1+2*k2+2*k3+k4)

                where  k1=h*f(x0,y0)

                            k2=h*f(x0+h/2, y0+k1/2)

                            k3= h*(x0+h/2, y0+k2/2)

                            k4=h*f(x0+h, y0+k3)

    this is called Runge Kutta fourth order method. Now , I want to implement a code for the equations as follows

       Equations:      diff (diff (diff (f (eta), eta), eta), eta)+ f (eta)* diff (diff (f (eta), eta), eta)=0

                              diff (diff (theta (eta), eta), eta)+ f (eta)* diff (theta (eta), eta)=0

     boundary conditions : f(0) = 0, (D(f))(0) = 0, (D(f))(5) = 1

                                         theta(0) = 1, theta(5) = 0

 

 

@Carl Love 

I have not done intentionally. I thought you may get confused. Moreover, this problem is small and easiest to write a code so I sent it repeatedly. Can you help me to write a code using Runge Kutta Fourth order method?

@Carl Love 

Excuse me for making you confused. can you help in this Runge Kutta fourth order method regard?

 

 

  I

Thank you for helping me with plotting the graphs and finally, I got by using the paint... I mean I tried in the paint and got it.

@vv  what can I check for filename ?? I using MAPLE15 if it is needed to search, can you tell how? otherwise what steps can I follow

@Rouben Rostamian  

Thank you, sir.

But I didn't understand that what software you used and how you have plotted this? If it is other than maple15 please can you send me the link of that software (if it is free) so that I can download easily...

@Kitonum 

 

Thank you for sharing your answer. But here, I working on a analytical problem in that after applying all the methods like converting PDE to ODE and similarity methods, at the last step we need compare the co-efficient of the x terms. While comparing we get three equations with respect to the same boundary conditions. So for this I trying to apply numerical method, same answers I got too. But I couldnot understand ? Can you tell why are you removing third equation ?

Or can we solve this problem with those three equations  also ??

@tomleslie Yes, we can do this in another way that is finding the iterative values for each set and taking these values in origin then plotting the graph. But take's time.

@tomleslie since two months I'm trying for this program but I do not get an idea about the program so I asked you to verify it.

I tried for each set of values( phi=0=a,b,c,d) for each execution and moreover, it takes time.

@tomleslie  Actually, I varying a variable called as a nanoparticle (phi) and values are in the form of a,b,c, and d and values of these variables also given in the program please once again verify it, stretching_cynlider2.mw

stretching_cylinder_new.mw

Sir, I change the value of N  please verify once 

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