mehdi jafari

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7 years, 84 days

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These are replies submitted by mehdi jafari

@acer yes. Thank you 

@Kitonum yes you are right, maybe it is a bug in maple. I have another question, can this integral be solved symbolically as a function of T ?

@acer i know it can't , but what is wrong with maple? i put numeric option in the middle of the integration and it does give some answer( don't know wrong or true). my first question is why maple is behaving like this. and my second question is how can i get a true output of function T ?
 

@acer thank u, i unassinged the u[10] and the problem solved. thank u

@vv thnx for quick answer. is there a way to know number of repetitition of members in each set? for example [1,1,5]  1 twice and 5 once.
 

@vv thank you for your respone, how you evaluate thin integral like this? could plz explain the surface of integration

@Kitonum tnx for impressive answer,i have one more question, can the volume be computed?

@Kitonum yes you are right it is d3

@tomleslie thnx for the quick answer
the versions are maple 2018.1 windows 10 x64.
i push all the arrows, Pd Dn, Pg Up , ...
but nothing happens and i can not see other pages

i don't think maple has any options related to sound. maybe you want to check your windows setting for this.
how about trying this : open Control panel, on the search box of right top corner, write "sound" . open the sound option and go to sound tab. and change sound scheme to nothing. but it will mute all sounds related to windows events and programs. maybe it solves the problem.

@samen 
 

restart; Digits := 5; with(plots); with(LinearAlgebra)

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a := 0; b := 1; N := 9; h := (b-a)/(N+1); k := 1/1000; phi := .5; K := 10^(-6); mu := 1.67; alpha := K/(phi*mu); lambda := alpha*k/h^2

0.11976e-6

(1)

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for i from 0 while i <= N do u[i, 0] := h*i+1 end do

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for j from 0 while j < N+1 do u[0, j] := .1; u[N+1, j] := .5 end do

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printlevel := 2; for i while i <= N do for j from 0 while j <= N do eq[i, j] := lambda*u[i-1, j]+(2-2*lambda)*u[i, j]+lambda*u[i+1, j] = -lambda*u[i-1, j+1]+(2+2*lambda)*u[i, j+1]-lambda*u[i+1, j+1] end do end do

sys := ([seq])(seq(eq[i, j], j = 0 .. 9), i = 1 .. 9); indets(sys)

fsolve(sys)

{u[0, 10] = u[0, 10], u[1, 1] = 1.1000, u[1, 2] = 1.1000, u[1, 3] = 1.1000, u[1, 4] = 1.1000, u[1, 5] = 1.1000, u[1, 6] = 1.1000, u[1, 7] = 1.1000, u[1, 8] = 1.1000, u[1, 9] = 1.1000, u[1, 10] = 1.1000+0.59880e-7*u[0, 10]+0.98982e-65*u[10, 10], u[2, 1] = 1.2000, u[2, 2] = 1.2000, u[2, 3] = 1.2000, u[2, 4] = 1.2000, u[2, 5] = 1.2000, u[2, 6] = 1.2000, u[2, 7] = 1.2000, u[2, 8] = 1.2000, u[2, 9] = 1.2000, u[2, 10] = 1.2000+0.35856e-14*u[0, 10]+0.16530e-57*u[10, 10], u[3, 1] = 1.3000, u[3, 2] = 1.3000, u[3, 3] = 1.3000, u[3, 4] = 1.3000, u[3, 5] = 1.3000, u[3, 6] = 1.3000, u[3, 7] = 1.3000, u[3, 8] = 1.3000, u[3, 9] = 1.3000, u[3, 10] = 1.3000+0.21470e-21*u[0, 10]+0.27605e-50*u[10, 10], u[4, 1] = 1.4000, u[4, 2] = 1.4000, u[4, 3] = 1.4000, u[4, 4] = 1.4000, u[4, 5] = 1.4000, u[4, 6] = 1.4000, u[4, 7] = 1.4000, u[4, 8] = 1.4000, u[4, 9] = 1.4000, u[4, 10] = 1.4000+0.12856e-28*u[0, 10]+0.46100e-43*u[10, 10], u[5, 1] = 1.5000, u[5, 2] = 1.5000, u[5, 3] = 1.5000, u[5, 4] = 1.5000, u[5, 5] = 1.5000, u[5, 6] = 1.5000, u[5, 7] = 1.5000, u[5, 8] = 1.5000, u[5, 9] = 1.5000, u[5, 10] = 1.5000+0.76980e-36*u[0, 10]+0.76988e-36*u[10, 10], u[6, 1] = 1.6000, u[6, 2] = 1.6000, u[6, 3] = 1.6000, u[6, 4] = 1.6000, u[6, 5] = 1.6000, u[6, 6] = 1.6000, u[6, 7] = 1.6000, u[6, 8] = 1.6000, u[6, 9] = 1.6000, u[6, 10] = 1.6000+0.46096e-43*u[0, 10]+0.12857e-28*u[10, 10], u[7, 1] = 1.7000, u[7, 2] = 1.7000, u[7, 3] = 1.7000, u[7, 4] = 1.7000, u[7, 5] = 1.7000, u[7, 6] = 1.7000, u[7, 7] = 1.7000, u[7, 8] = 1.7000, u[7, 9] = 1.7000, u[7, 10] = 1.7000+0.27602e-50*u[0, 10]+0.21471e-21*u[10, 10], u[8, 1] = 1.8000, u[8, 2] = 1.8000, u[8, 3] = 1.8000, u[8, 4] = 1.8000, u[8, 5] = 1.8000, u[8, 6] = 1.8000, u[8, 7] = 1.8000, u[8, 8] = 1.8000, u[8, 9] = 1.8000, u[8, 10] = 1.8000+0.16528e-57*u[0, 10]+0.35856e-14*u[10, 10], u[9, 1] = 1.9000, u[9, 2] = 1.9000, u[9, 3] = 1.9000, u[9, 4] = 1.9000, u[9, 5] = 1.9000, u[9, 6] = 1.9000, u[9, 7] = 1.9000, u[9, 8] = 1.9000, u[9, 9] = 1.9000, u[9, 10] = 1.9000+0.98970e-65*u[0, 10]+0.59880e-7*u[10, 10], u[10, 10] = u[10, 10]}

(2)

 


 

Download Crank_scheme.mw

as a comment, can the space between a cube with corners (-1,-1,-1,) , (1,1,1) , and this sphere be filled? or can the volumes be subtracted from each other in plotting? 

 

restart

A1:=plots:-implicitplot3d(x^2+y^2+z^2=1,x=-1..1,y=-1..1,z=-1..1,grid= [10$3], style= wireframe, scaling=constrained, axes=boxed, orientation=[25,60]):

A2:=plots:-implicitplot3d(x^2+y^2+z^2=1,x=-1..1,y=-1..1,z=-1..1,grid= [10$3], style= patchnogrid, scaling=constrained, axes=boxed, orientation=[25,60]):

P0:=plots:-display(A1,A2):

A:=plots:-display(plottools:-cuboid([-1,-1,-1], [1,1,1]),scaling=constrained, colorscheme = ["xyzcoloring", proc (x, y, z) options operator, arrow; x+y-z^2 end proc], style = surface,style= wireframe,axes=none):

plots:-display(A,P0)

 

 

 

 

 

 

NULL

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Download comment.mw

@mehran rajabi actually i don't know maple can find maximium value of a complex function or not ! as far as i know , it can't !

as a suggestion, your file name can be dependent to the loop numerator. so it can be done without over writing, but as @Carl Love said, is there need to store the jacobian every loop ?! 

steady state is a state in which variables are independent of time and actucally the "state" of the system is "steady". thus if differential equations are available, the differentiation with respect to the time becomes zero. ( see: https://en.wikipedia.org/wiki/Steady_state). 
then i think you are maximizing your function with respect to the varibale "x" , and your function is complex, is that right?

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