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Ramakrishnan

Ramakrishnan Vaidyanathan

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11 years, 154 days

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Contact Ramakrishnan Vaidyanathan
With twenty years of Industrial experience and twenty years of teaching experience, I am now as retired Professor, using Maple to teach mathematics subject for students studying X to XII standards. Published XII Mathematics books.

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These are replies submitted by Ramakrishnan

@EugeneKalentev  I have found out t...

Ramakrishnan 399
December 07 2015
0 0

@EugeneKalentev 

I have found out the answer  thanks to Professor Robert Lopez.

Since l__e was redefined in the line where q__0 was defined, % is taken. The doubt was clarified by Professor Robert Lopez.

My sincere thanks go on record for him for the immediate responses from him whenever i send a mail.

Thanks to you also for the attempt. Since I have not used dimensions in the command mode, problem was not from units and dimensions. Cheers.

Ramakrishnan V

Embedded components...

Ramakrishnan 399
December 07 2015

@Lenin Araujo Castillo 

Thanks.I will try maplesim. Hope to use this later. I am learning but difficult without a book it seems. Some basic learning from class room is essential. Thanks for your response. Regards.

Ramakrishnan V

Product line...

Ramakrishnan 399
December 06 2015
0 0

@Markiyan Hirnyk 

Maple 2015.2 is the one i use and Windows10 is in my PC. I shall fill up the product(s) data in my questionnaire hereafter. Thanks for letting me know my mistakes,however small it may look from one end. Regards.

Ramakrishnan V

one dimensional heat equation solution...

Ramakrishnan 399
December 06 2015
0 0

@Markiyan Hirnyk 

Hope the attached equation and solution for the heat transfer problems would help understand the method.There may be unprofessional use of commands and manual mode since i am only learning 

 

# Determine the temperature distribution in a fin of circular cross section shown in Fig. Base of the fin is maintained at 100 0C and convection occurs at the tip:

 


restart

h := .2:    W/m2K; k := 2:W/cmK:   D__1 := 1:cm; D__2 := 1:  cm;

``phi__f := 20:0C;
A__1 := 22*D__1^2/(7*4) = 11/14``

A__2 := 22*D__2^2/(7*4) = 11/14``

l__1 := 4 cm:p := (22/7)*D__1 = 22/7 cm
l__2 := 4
 = 4cm:p := (22/7)*D__2 = 22/7NULL cm

Discretisation:

Number of linear one dimensional elements = 2;
Total Number of nodes = 3;
Global stiffness matrix is 3 x 3
For element 1
Stiffness matrix for conduction:
                                            Vector[row](2, {(1) = 1, (2) = 2})
K__1a := k*A__1/l__1.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[11/28, -11/28], [-11/28, 11/28]])  Vector(2, {(1) = 1, (2) = 2})

 

Stiffness matrix for convection:
                                            Vector[row](2, {(1) = 1, (2) = 2})
K__1b := (1/6)*h*p*l__1.(Matrix(2, 2, {(1, 1) = 2, (1, 2) = 1, (2, 1) = 1, (2, 2) = 2})) = Matrix([[.838095238000000, .419047619000000], [.419047619000000, .838095238000000]])  Vector(2, {(1) = 1, (2) = 2})

                                          Vector[row](2, {(1) = 1, (2) = 2})
K__1 := K__1a+K__1b = Matrix([[1.23095238085714, 0.261904761428571e-1], [0.261904761428571e-1, 1.23095238085714]]) Vector(2, {(1) = 1, (2) = 2})
``

For element 2
Stiffness matrix for conduction:
                                            Vector[row](2, {(1) = 2, (2) = 3})
K__2a := k*A__1/l__1.(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1})) = Matrix([[11/28, -11/28], [-11/28, 11/28]])  Vector(2, {(1) = 2, (2) = 3})

 

Stiffness matrix for convection:
                                            Vector[row](2, {(1) = 2, (2) = 3})
K__2b := (1/6)*h*p*l__1.(Matrix(2, 2, {(1, 1) = 2, (1, 2) = 1, (2, 1) = 1, (2, 2) = 2})) = Matrix([[.838095238000000, .419047619000000], [.419047619000000, .838095238000000]])  Vector(2, {(1) = 2, (2) = 3})

Stiffness matrix for convection:

 

  K__2c := h*A__2.(Matrix(2, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = 0, (2, 2) = 1})) = Matrix([[0., 0.], [0., .157142857100000]])  Vector(2, {(1) = 2, (2) = 3})

NULL
                                          Vector[row](2, {(1) = 1, (2) = 2})
K__2 := K__2a+K__2b+K__2c = Matrix([[1.23095238085714, 0.261904761428571e-1], [0.261904761428571e-1, 1.38809523795714]]) Vector(2, {(1) = 1, (2) = 2})

NULL                                                                       Vector[row](3, {(1) = 1, (2) = 2, (3) = 3})
K__g := Matrix(3, 3, {(1, 1) = K__1(1), (1, 2) = K__1(3), (1, 3) = 0, (2, 1) = K__1(2), (2, 2) = K__1(4)+K__2(1), (2, 3) = K__2(3), (3, 1) = 0, (3, 2) = K__2(2), (3, 3) = K__2(4)}) = Matrix([[K__1(1), K__1(3), 0], [K__1(2), K__1(4)+K__2(1), K__2(3)], [0, K__2(2), K__2(4)]])``  Vector(3, {(1) = 1, (2) = 2, (3) = 3}) 

Load vector:
for element 1
     
F__1 := (1/2)*h*p*l__1*`φ__f`.(Vector(2, {(1) = 1, (2) = 1})) = Vector[column]([[25.1428571400000], [25.1428571400000]]) Vector(2, {(1) = 1, (2) = 2})

``

for element 2

F__2 := (1/2)*h*p*l__2*`φ__f`.(Vector(2, {(1) = 1, (2) = 1}))+h*A__2*`φ__f`.(Vector(2, {(1) = 0, (2) = -1})) = Vector[column]([[25.1428571400000], [21.9999999980000]]) Vector(2, {(1) = 2, (2) = 3})

F__g := eval(Vector(3, {(1) = F__1(1), (2) = F__1(2)+F__2(1), (3) = F__2(2)})) = Vector[column]([[F__1(1)], [47.6666666657143], [F__2(2)]])`` Vector(3, {(1) = 1, (2) = 2, (3) = 3}) 
Tip insulation does not contribute to any load vector since heat transferred is zero.
Assembling matrices give the system equation

K__g, (Vector(3, {(1) = `φ__1`, (2) = `φ__2`, (3) = `φ__3`})) = F__g = Matrix(3, 3, {(1, 1) = K__1(1), (1, 2) = K__1(3), (1, 3) = 0, (2, 1) = K__1(2), (2, 2) = K__1(4)+K__2(1), (2, 3) = K__2(3), (3, 1) = 0, (3, 2) = K__2(2), (3, 3) = K__2(4)}), (Vector(3, {(1) = `#msub(mi("φ",fontstyle = "normal"),mi("1"))`, (2) = `#msub(mi("φ",fontstyle = "normal"),mi("2"))`, (3) = `#msub(mi("φ",fontstyle = "normal"),mi("3"))`})) = (Vector(3, {(1) = F__1(1), (2) = HFloat(47.66666666571429), (3) = F__2(2)}))

 

# Base of the fin is at 1000C;  phi__1 := 100:

Multiply the first column by `φ__1` to get the modified column.

We know phi__1 := 100: 

   = Substituting and eliminating the first row and column and transferring the modified first column to RHS

``

 

F__g[2] := -phi__1*K__g[2, 1]+F__g[2]

HFloat(47.66666666571429)

 (1)

(Matrix(2, 2, {(1, 1) = K__g[2, 2], (1, 2) = K__g[2, 3], (2, 1) = K__g[3, 2], (2, 2) = K__g[3, 3]})).(Vector(2, {(1) = `φ__2`, (2) = `φ__3`})) = (Vector(2, {(1) = F__g[2], (2) = F__g[3]})) = (Vector(2, {(1) = 2.4619*`#msub(mi("φ",fontstyle = "normal"),mi("2"))`+0.262e-1*`#msub(mi("φ",fontstyle = "normal"),mi("3"))`, (2) = 0.262e-1*`#msub(mi("φ",fontstyle = "normal"),mi("2"))`+1.3881*`#msub(mi("φ",fontstyle = "normal"),mi("3"))`})) = (Vector(2, {(1) = F__g[2], (2) = F__g[3]}))``

  Solving for `φ__1`,`φ__2` 

NULL

solve({0.262e-1*phi__2+1.3881*phi__3 = 22.000, 2.4619*phi__2+0.262e-1*phi__3 = 45.0476}) = {phi__2 = 18.13287427, phi__3 = 15.50674929}NULL

 

0.262e-1*16.98+1.3881*20.05 = 28.276281

``

NULL

  # Determine the temperatures at the nodal interfaces for the two layered wall shown in Fig. The left face is supplied with heat flux of `q"` := 5 W/cm2;

  # and right face is maintained at phi__3 := 20 0C;

``

restart

 

``

````

``

`q"__1` := 5:  W/m2K; k__1 := .2:W/mK;  = W/mKA := 1:cm2;
We shall consider two one dimensional linear elements
For element 1
l__12 := 4: cm;

                                                       Vector[row](2, {(1) = 1, (2) = 2})
[K] = K__1 := k__1*A*(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1}))/l__12 = Matrix([[0.500000000000000e-1, -0.500000000000000e-1], [-0.500000000000000e-1, 0.500000000000000e-1]]) 
Vector(2, {(1) = 1, (2) = 2})

[F]1= F__1 := `q"`*A.(Vector(2, {(1) = 1, (2) = 0})) = `q"`.(Vector(2, {(1) = 1, (2) = 0})) Vector(2, {(1) = 1, (2) = 2})NULL

or element 2
l__23 := 2: cm;

                                                      Vector[row](2, {(1) = 2, (2) = 3})
[K] = K__2 := k__2*A*(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1}))/l__23 = Matrix([[0.300000000000000e-1, -0.300000000000000e-1], [-0.300000000000000e-1, 0.300000000000000e-1]]) Vector(2, {(1) = 2, (2) = 3})

 

[F]1= F__2 := `q"`*A.(Vector(2, {(1) = 0, (2) = 0})) = `q"`.(Vector(2, {(1) = 0, (2) = 0})) Vector(2, {(1) = 2, (2) = 3})

Assembling the matrices, we get the global system equation,

Global Stiffness matrix

           Vector[row](3, {(1) = 1, (2) = 2, (3) = 3})
K__g := Matrix(3, 3, {(1, 1) = K__1(1), (1, 2) = K__1(3), (1, 3) = 0, (2, 1) = K__1(2), (2, 2) = K__1(4)+K__2(1), (2, 3) = K__2(3), (3, 1) = 0, (3, 2) = K__2(2), (3, 3) = K__2(4)})
 = Matrix([[K__1(1), K__1(3), 0], [K__1(2), K__1(4)+K__2(1), K__2(3)], [0, K__2(2), K__2(4)]])``

Global load matrix
 = F__g := Vector(3, {(1) = F__1(1), (2) = F__1(2)+F__2(1), (3) = F__2(2)}) = Vector[column]([[F__1(1)], [F__1(2)+F__2(1)], [F__2(2)]])``

K__g.(Vector(3, {(1) = `φ__1`, (2) = `φ__2`, (3) = `φ__3`})) = F__g = (Vector(3, {(1) = 5.0000-0.500e-1*`#msub(mi("φ",fontstyle = "normal"),mi("2"))`, (2) = -5.0000+0.800e-1*`#msub(mi("φ",fontstyle = "normal"),mi("2"))`-0.300e-1*`#msub(mi("φ",fontstyle = "normal"),mi("3"))`, (3) = -0.300e-1*`#msub(mi("φ",fontstyle = "normal"),mi("2"))`+0.300e-1*`#msub(mi("φ",fontstyle = "normal"),mi("3"))`})) = (Vector(3, {(1) = F__1(1), (2) = F__1(2)+F__2(1), (3) = F__2(2)}))``

We know phi__3 := 20;Substituting and solving for `φ__1`,`φ__2` 

``

solve({-0.500e-1*u__1+0.800e-1*u__2 = .6, 0.500e-1*u__1-0.500e-1*u__2 = 5})

{u__1 = 286.6666667, u__2 = 186.6666667}

 (2)

`φ__1` = 286.67; `φ__2` = 186.670C

NULL

``


Download 1Dheattransfersolutions.mw1Dheattransfersolutions.mw

maple.

variational methods...

Ramakrishnan 399
December 06 2015
0 0

@Markiyan Hirnyk 

I am sorry if I am not confusing. I know a reduction method in which the second order differential equations are converted to lower order and solvd. Galerkin and Ritz are the contributers. I attach with a doc detailing this methods with complete answers. Hope this will be useful. Ramakrishnna V

Consider a cantilever beam subjected to uniformly distributed load q__0as shown in Fig. Calculate the displacement and compare with exact solution.

  restart
The governing differential equation is  I*E*(diff(v(x), x, x, x, x))-q__0 = 0; Boundary conditions are v(0) = 0; (diff(v(x), x))*0 = 0; (diff(v(x), x, x))(L) = 0; (diff(v(x), x, x, x))(L) = 0;
Boundary conditions v(0) = 0; (diff(v(x), x))*0 = 0These conditions enforce zero displacement and slope at fixed end;
Boundary conditions (diff(v(x), x, x))(L) = 0; (diff(v(x), x, x, x))(L) = 0These conditions enforce zero bending moment and shear force at free end;
Let us find the approximate solution using single continuous trial function.
Step 1: Assumption of trial solution (guess solution)
"v(x) = conjugate(v)(x) = `c__0` + `c__1`*x + `c__2`*x^(2) + `c__3`*x^(3) + `c__4`*x^(4)"
The constants c__0, c__1 are determined using boundary conditions v(0) = 0; (diff(v(x), x))*0 = 0; c__0 = c__1 and c__1 = 0
The constants c__2, c__3 are determined in terms of c__4 using boundary conditions (diff(v(x), x, x))(L) = 0; (diff(v(x), x, x, x))(L) = 0;
c__2 = -6*L^2*c__4-3*L*c__3; c__3 = -4*c__4*L
Substituting the values for c__0, c__1, c__2, c__3, c__4 in the trial solution, we get
v(x) = c__4*(6*L^2*x^2-4*L*x^3+x^4) [Note that finding the trial function satisfying all the boundary conditions is thus cumbursome]
Step 2: Find the domain residual.
Substituting the trial solution in the governing equation, we get
    diff(c__4*(6*L^2*x^2-4*L*x^3+x^4), x, x, x, x) = 24*c__4

 

The governing equation is
R__d = I*E*(diff(v(x), x, x, x, x))-q__0 and I*E*(diff(v(x), x, x, x, x))-q__0 = 24*EI*c__4-q__0;
Step 3: Minimise the residual.24*EI*c__4-q__0 = 0"(->)"[[c__4 = (1/24)*q__0/EI]]NULL 
The final solution is v(x) = 0.4167e-1*q__0*(6*L^2*x^2-4*L*x^3+x^4)/EI

``

The following differential equation is available for a physical phenomenon.
  diff(y(x), x, x)+50 = 0, 0 <= x and x <= 10
Trial function is y = a__1*x*(10-x)
Boundary conditions are y(0) = 0; y(10) = 0
 Find the solution byGalerkin method

``

  unassign('a__1')

Step 1: Assumption of trial solution (guess solution)

y(x) = ((D@@2)(y))(x) and ((D@@2)(y))(x) = a__1*x*(10-x) and a__1*x*(10-x) = -a__1*x^2+10*a__1*x

diff(-a__1*x^2+10*a__1*x, x, x) = -2*a__1NULL

Step 2: Find the residue by substituting in the LHS of the governing equation
The governing equation is
diff(y(x), x, x)+50 = 0"->"R := -2*a__1+50 = -2*a__1+50:
The condition for Galerkin's method
Weighing function = Trial function (solution guessed)

w := y(x) = y(x) 

   

w*R = y(x)*(-2*a__1+50) 

int((-2*a__1+50)*(-a__1*x^2+10*a__1*x), x) = (-2*a__1+50)*(-(1/3)*x^3*a__1+5*a__1*x^2)int((-2*a__1+50)*(-a__1*x^2+10*a__1*x), x = 0 .. 10) = 0 = (500/3)*(-2*a__1+50)*a__1 = 0"(->)"[[a__1 = 0], [a__1 = 25]]

Answer considered is a__1 := 25 = 25``

   Hence the final solution is y(x) = a__1*x(10-x) = y(x) = 25*x(10-x)    ````

restartNULL

``

NULL


Download Approximate_solutions.mwApproximate_solutions.mw

evalf, assign - working fine - thanks...

Ramakrishnan 399
December 03 2015
0 1

@Kitonum 

Your suggestion is very useful and working when i call back eta later. Thanks. Ramakrishnan V

midrich...

Ramakrishnan 399
November 26 2015
0 0

@Preben Alsholm 

Thanks

What is midpoint method? Maple suggests ...

Ramakrishnan 399
November 24 2015
0 0


> 

NULL

NULL

sol1 := dsolve([diff(diff(y(x), x), x) = (diff(y(x), x))^2-2*x^2*y(x), y(-1) = 0, y(1) = 1], numeric, method = bvp);

  

NULL

``

``

sol1 := dsolve([diff(diff(y(x), x), x) = y(x)*(diff(y(x), x))^2+diff(y(x), x)-2*x*y(x), y(-1) = 0, y(1) = 1], numeric, method = bvp); 1; sol1(0.); 1; plots[odeplot](sol1, color = red)

  

ol1 := dsolve([diff(diff(y(x), x), x) = y(x)*(diff(y(x), x))^2+diff(y(x), x)-2*y(x)/x, y(0) = 1, y(1) = .5], numeric, method = bvp); sol1(0.); plots[odeplot](sol1, color = red)

  

``


Download possiblesolnto_yr_problem.mwpossiblesolnto_yr_problem.mw

@tomleslie 

solvingODE-equation clarity...

Ramakrishnan 399
November 24 2015
0 0

@tomleslie 

Dear maple user, I am sorry for not giving a staight answer to the question raised. The reason is i could not follow the f(x) in the equation and unclear Boundary Conditions. I am also not supposed to give a numerical answer to a question where the answer may be in functional form.

All I have given is a similar problem for which answer is available and can be used to understand more about alternative answers. Thanks for your comment and I confess that I have no full answer and  i know clearly that they two are not exactly same.

Probably I shall try to put the equation provided by tomleslie and get an answer with my boundary conditions.

Cheers.

My complete document for inline result e...

Ramakrishnan 399
November 23 2015
0 0

heattransfer1Dfea_answer.mw

Example 1: Find the normal temperatures in a composite wall shown in Fig. The wall is maitained at 1000 C at the left face and convection mode of heat transfer occurs between the right face and surrounding fluid. Thermal conduction  

 

restart:

K__1 := 0.6e-1:W/cm 0C;

Convection heat transfer coefficient between wall and fluid, h := .1: W/m2 0C; Field variable is temperature phi; Temperature at node 1, phi__1 := 100: 0C;

Temperature of the fluid, phi__f := 25:  0C:       

Area of cross section, A := 1:``  cm2 ; Length of element 1, l__1 := 2:cm:  Length of element 2, l__2 := 4:cm;

``

Solution:``

 

Finite Element Equation for elements

For element 1 (nodes 1-2), the finite element equation is

For element 1 (nodes 2-3), the finite element equation is

                           Vector[row](2, {(1) = 1, (2) = 2})

(Vector(2, {(1) = F__1, (2) = F__2})) = K__1*A*(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1}))/l__1.(Vector(2, {(1) = u__1, (2) = u__2}))     Vector(2, {(1) = 1, (2) = 2}) 

``

                        Vector[row](2, {(1) = 2, (2) = 3})

(Vector(2, {(1) = F__2, (2) = F__3})) = K__2*A*(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1}))*(Vector(2, {(1) = u__2, (2) = u__3}))/l__2    Vector(2, {(1) = 2, (2) = 3})

``

 

Stiffness Matrices for elements

Stiffness matrix for element 1

Stiffness matrix for element 2

K__1*A/l__1 = 0.3000000000e-1

M1 := K__1*A*(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1}))/l__1 = Matrix([[0.300000000000000e-1, -0.300000000000000e-1], [-0.300000000000000e-1, 0.300000000000000e-1]])

``

K__2*A/l__2 = 0.5000000000e-1

for conduction only

M2__1 := K__2*A*(Matrix(2, 2, {(1, 1) = 1, (1, 2) = -1, (2, 1) = -1, (2, 2) = 1}))/l__2 = Matrix([[0.500000000000000e-1, -0.500000000000000e-1], [-0.500000000000000e-1, 0.500000000000000e-1]])

for convection only

h*A = .1NULL

M2__2 := h*A*(Matrix(2, 2, {(1, 1) = 0, (1, 2) = 0, (2, 1) = 0, (2, 2) = 1})) = Matrix([[0., 0.], [0., .100000000000000]])NULL

 for both conduction and convection combined,

M2 := M2__1+M2__2 = Matrix([[0.500000000000000e-1, -0.500000000000000e-1], [-0.500000000000000e-1, .150000000000000]])NULL

``

NULL

 

Combining the matrices we get global stiffness matrix,

 

Kglobal := Matrix(3, 3, {(1, 1) = M1(1), (1, 2) = M1(3), (1, 3) = 0, (2, 1) = M1(2), (2, 2) = M1(4)+M2(1), (2, 3) = M2(3), (3, 1) = 0, (3, 2) = M2(2), (3, 3) = M2(4)}) = Matrix([[M1(1), M1(3), 0], [M1(2), M1(4)+M2(1), M2(3)], [0, M2(2), M2(4)]])``

``

Nodal Load Matrices for the elements

Load matrix for element1 (nodes 1-2) is

Load matrix for element2 (nodes 2-3) is

                         Nodes

L1 := Vector(2, {(1) = 0, (2) = 0}) = Vector[column]([[0], [0]])   Vector(2, {(1) = 1, (2) = 2})

``

 

h*phi__f*A = 2.5NULL

NULL

                                 Nodes

L2 := h*`&phi;__f`*A*(Vector(2, {(1) = 0, (2) = 1})) = Vector[column]([[0.], [2.50000000000000]])   Vector(2, {(1) = 2, (2) = 3})

Combining the element load matrics, we get global load matrix as

NULL 

F := Vector(3, {(1) = L1(1), (2) = L1(2)+L2(1), (3) = L2(2)}) = Vector[column]([[L1(1)], [L1(2)+L2(1)], [L2(2)]])NULL

``

 

      

Nodal Displacement Matrices for elements

Nodal Displacement matrix is

phi := Vector(3, {(1) = `&phi;__1`, (2) = `&phi;__2`, (3) = `&phi;__3`}) = Vector[column]([[phi__1], [phi__2], [phi__3]])

``

 

  Finite element equation for the system is [Kglobal.[ϕ]=[F];``

  Matrix([[M1(1), M1(3), 0], [M1(2), M1(4)+M2(1), M2(3)], [0, M2(2), M2(4)]])     =      

 

Substituting the nonzweo boundary condition phi__1 = 100 , deleting  the row and column containing `&phi;__1` and transferring the column containing node to rhs, we get 

``

(Matrix(2, 2, {(1, 1) = m[1, 1], (1, 2) = m[1, 2], (2, 1) = m[2, 1], (2, 2) = m[2, 2]})).(Vector(2, {(1) = `&phi;__2`, (2) = `&phi;__3`})) = (Vector(2, {(1) = m[1, 1], (2) = m[2, 1]}))+`&phi;__1`*(Vector(2, {(1) = -M1(2), (2) = 0})) = (Vector(2, {(1) = 0.800e-1*`#msub(mi("&phi;",fontstyle = "normal"),mi("2"))`-0.500e-1*`#msub(mi("&phi;",fontstyle = "normal"),mi("3"))`, (2) = -0.500e-1*`#msub(mi("&phi;",fontstyle = "normal"),mi("2"))`+.1500*`#msub(mi("&phi;",fontstyle = "normal"),mi("3"))`})) = (Vector(2, {(1) = HFloat(3.0), (2) = HFloat(2.5)}))

G := Matrix(2, 3, {(1, 1) = m[1, 1], (1, 2) = m[1, 2], (1, 3) = m[1, 3], (2, 1) = m[2, 1], (2, 2) = m[2, 2], (2, 3) = m[2, 3]}) = Matrix([[0.8e-1, -0.5e-1, 3.0], [-0.5e-1, .15, 2.5]])````

"By Gaussian Elimination,"

 

Matrix([[0.8e-1, -0.5e-1, 3.0], [0., .1187500000, 4.375000000]])

 (1)

`` solve({.1188*phi__3 = 4.375, 0.8e-1*phi__2-0.5e-1*phi__3 = 3}) = {phi__2 = 60.51662458, phi__3 = 36.82659933}``

``

NULL


Download heattransfer1Dfea_answer.mw

@tomleslie 

Thank you for prompt response. I give the full document here. In case you could find time, please do the needful to help me learn more abou the intricacies in using these command. Thanks in advance.

solving ode...

Ramakrishnan 399
November 23 2015
0 0


> 

``

``

sol1 := dsolve([diff(diff(y(x), x), x) = (diff(y(x), x))^2-2*x^2*y(x), y(-1) = 0, y(1) = 1], numeric, method = bvp);

  

``

``


Download possiblesolnto_yr_problem.mwpossiblesolnto_yr_problem.mw

In my opinion, f(x) needs to be mentioned. possible soln by num method is attached.Hope this will be useful.

Ramakrishnan V

rukmini_ramki@hotmail.com

real assumption not needed...

Ramakrishnan 399
November 20 2015
0 0

@Carl Love 

Thank you very much for the clarification that simplify command will do. Useful learning for me. Ramakrishnan V

Answer to 'How to avoid duplication'...

Ramakrishnan 399
November 19 2015
0 0

@Thomas Dean

Sorry for the belated response. Thank you very much for the answer It is useful for me to know the basics in Maple. Ramakrishnan V 

Jumbled order i f numberof terms greater...

Ramakrishnan 399
November 19 2015


 

 

With 25 terms it gives the expression with same terms but jumbled in order:

expr1 := t1+t2+t3+t4+t5+t6+t7+t8+t9+t10+t11+t12+t13+t14+t15+t16+t17+t18+t19+t20+t21+t22+t0-t0+t23;

0

 (1)

 

With 24 terms it gives the expression with same terms  in the same order given by us:

expr3 := t1+t2+t3+t4+t5+t6+t7+t8+t9+t10+t11+t12+t13+t14+t15+t16+t17+t18+t19+t20+t21+t0-t0+t23;

0

 (2)

 =

With 25 terms it gives the expression with same terms but jumbled in order (expr5):With 24 terms, it keeps the order given by us.(expr6)

expr5 := t2+t1+t3+t4+t23+t22+t20+t21+t19+t17+t18+t15+t16+t13+t14+t11+t12+t9+t10+t8+t6+t7+t5+t24-t25;

-t25

 (3)

With 25 terms even if there is no arithmatics (maths) involved) it gives the expression with same terms given by us but jumbled in order:

expr7:=t1+t2+t3+t4+t5+t6+t7+t8+t9+t10+t11+t12+t13+t14+t15+t16+t17+t18+t19+t20+t21+t22+t24+t23+t25;

t2+t1+t3+t4+t23+t22+t20+t21+t19+t17+t18+t15+t16+t13+t14+t11+t12+t9+t10+t8+t6+t7+t5+t24+t25

 (4)

 

   

     

NULL


Download jumbled_order_if_terms_are_more_than_24.mwjumbled_order_if_terms_are_more_than_24.mw

I am new to maple. However i see something in yr post and want to comment. I attach my attemps below. I find that the order is retained up to 24 terms with or without maths (subtraction t0-t0=0 in our case) involved.

Linear ODE of second order...

Ramakrishnan 399
November 14 2015
0 1


``

``

Eq := diff(y(x), x, x) = -(x^2+1)*y(x)+k;

diff(diff(y(x), x), x) = -(x^2+1)*y(x)+k

 (1)

"(->)"

y(x) = WhittakerM(-(1/4)*I, 1/4, I*x^2)*_C2/x^(1/2)+WhittakerW(-(1/4)*I, 1/4, I*x^2)*_C1/x^(1/2)+(2*I)*k*((Int(x^(3/2)*WhittakerW(-(1/4)*I, 1/4, I*x^2)/((1+3*I)*WhittakerW(-(1/4)*I, 1/4, I*x^2)*WhittakerM(1-(1/4)*I, 1/4, I*x^2)+(4*I)*WhittakerW(1-(1/4)*I, 1/4, I*x^2)*WhittakerM(-(1/4)*I, 1/4, I*x^2)), x))*WhittakerM(-(1/4)*I, 1/4, I*x^2)-(Int(x^(3/2)*WhittakerM(-(1/4)*I, 1/4, I*x^2)/((1+3*I)*WhittakerW(-(1/4)*I, 1/4, I*x^2)*WhittakerM(1-(1/4)*I, 1/4, I*x^2)+(4*I)*WhittakerW(1-(1/4)*I, 1/4, I*x^2)*WhittakerM(-(1/4)*I, 1/4, I*x^2)), x))*WhittakerW(-(1/4)*I, 1/4, I*x^2))/x^(1/2)

 (2)

"->"

  

``

``


Download ODESolnTWOSent14-11-2015byRVtMaplePrimeoCommunity.mw

I forgot to attach my doc. Enclosed now the cube plot, an answer to the Linear ODE of second order problem for constant k

 

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