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These are replies submitted by Parham2016

@Preben Alsholm 


There is a physical meaning for the relation between C1 nad B in Mechanics. I do know that 'dsolve' works for differential equations, however, the desire result could be derived by it!



I meant why it was ok and done in your computer but not in my laptop. How did you edit the file? For example: alpha1 was changed to alpha[1]...
Did you write the whole file again?



Hello, yes it's ok now. It's ambiguous yet.. What have happened?



Thanks. But it doesn't work for me.... It drove me crazy!!! It workes correctly last week. I had to get some results by the file today...




Great, God bless you and happy new year.



What a simple solution!




The integral term in F2(r) must be a value necessarily as long as those two Nu are close toghether. Indeed in the loop I want this expression " Error= (Nu_final- Nu_guess)<1e-3" to be satisfied and then the loop is going to be interrupted.

The program is correct in the first iteration for \phi= 0.06 and "initial guess Nu=3.6941" and the "final calculated Nu= 3.6942". I already know the Final Nu for \phi= 0.06!!!! But for other values of \phi I could not get any solution precisely.



By the way, it does not matter whether or not k is zero @ initial guess "For The First Time", the first iteration in a potential loop. As you know the initial guess for Nu is just for calculation of the integral and after that I removed it by this command: Nu:= 'Nu'.

The only important thing is this fact that the final calculated for Nu, in the end of the loop, must be equal to the initial guess.



Hello, thanks for your consideration dear.

Yes, I want to find the Nu if the determinat is equal zero. And the problem is exactly here that the at the initial guess the determinant, k, might be any number, not necessarily zero. That is why I want to correct the Nu by replacing the initial guess with the final calculated Nu. It assumed to be difficult for me.



Thanks. I must change the domain then! That's 3.195986267 if

0.2 <r< 1

@Markiyan Hirnyk 


Thanks again.

@Markiyan Hirnyk 


Thanks dear,

Yes you're right. I solve that system in Matlab (written below) and those solutions are the same. 

[ 1.6759649064951345004378236436225, 0.7729562969870007822477027324705, 1.9940322524732747423780039106827]

By the way, I wrote some indices wrongly in my question!!! Now I corrected it and we can see this is true as well.  





dsolve({-29.71468323*C[1]+65.05900958*C[2]+0.6517031249e-2*C[3] = 1/2, 35.31879251*C[1]-74.07655755*C[2]-.9704532947*C[3] = 0, 204.4345743*C[1]-443.4820627*C[2]+0.8379040182e-1*C[3] = 0}, {C[1], C[2], C[3]})

{C[1] = 109146383352530958107750000000/65124504057059038040022691307, C[2] = 50338395499059263144550000000/65124504057059038040022691307, C[3] = 129860361516102353000000000000/65124504057059038040022691307}



solve({-29.71468323*C[1]+65.05900958*C[2]+0.6517031249e-2*C[3] = 1/2, 35.31879251*C[1]-74.07655755*C[2]-.9704532947*C[3] = 0, 204.4345743*C[1]-443.4820627*C[2]+0.8379040182e-1*C[3] = 0}, {C[1], C[2], C[3]})

{C[1] = 1.675964906, C[2] = .7729562970, C[3] = 1.994032252}










I thank you for your comments and answering to my question.

Unfortunately I didn't write the fourth boundary conditions for you.
The four boundary conditions are in here:

V(r1,z)= V(r2,z)=0
V(r,0)= V(r,b)=0

and this point that the coordinates are

  r1 < r <r2   &    0< z < b

And I want to tell you I don't have the Matlab Code I just olnly the final solution attached in the Maple primes websie.


@Carl Love 


Thanks a lot Mr. Carl.


God bless you



Ok, thanks. But I thought Maple could give the right answer instead of  F_1(x+t)+ F_2(t-x)+ cos(x).


Is the answer correct???


Unfortunately, I could not solve it analytically, so it was not possible to me to compare the numerical and Exact solutions too!!!



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