sand15

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What are the ranges for epsilon, u0 and v0 (those you want to investigate, not those you coded) ?

@KIRAN SAJJAN 

Non convergence appears when yu < yl, which depends on the values of x and t.

This can be shown in a very simple way.
Thus, for instance when tfixed = 1 dsolve does not converge over the range xfixed = 0..2 , but only over the subrange xfixed = 0.366825212..1.633174787.

Read the attached file carefully and try to figure out what you can do with it XT_non_convergence_issue.mw

@KIRAN SAJJAN 

You asked for a plot in the range x=0..600.
In your last wotksheet you took x=5 and claim "it's converging"... yes indeed, for this value of x.

Now take x=6 and tell me if "it's converging".

@KIRAN SAJJAN 

you understood what I said when I told you to try and fix yourself the "Newton non converging" issue

Upload your worksheet by using the big green up-arrow in the menubar

@Math-dashti 

Have you given up on this question, or is it still of interest to you?

Is it you who deleted your last question, which I spent the time to answer?

If that's the case, let me tell you that you are being particularly impolite and scornful. 

For the record your question was


To which I answered

@Math-dashti 

look at it

@Math-dashti 

restart; with(plots); f := proc (u, v) options operator, arrow; v*(2*chi*u^2+p) end proc; g := proc (u, v) options operator, arrow; -beta[2]*u^5-2*chi*u*v^2+u*p*k^2-beta[1]*u^3+u*w end proc; equilibria := solve({f(u, v) = 0, g(u, v) = 0}, {u, v}, explicit)

# Find the parameters in f(u, và and g(u, v)

indE := convert( indets({f(u, v) = 0, g(u, v) = 0}, name) minus {u, v}, list)

[chi, k, p, w, beta[1], beta[2]]

(1)

# Define a function 'Equilibria" whch maps 'indE' to 'equilibria'

Equilibria := unapply({equilibria}, indE):

# Example of use: a change of name

Equilibria(X, k, p, alpha[1], alpha[2], alpha[3])

{{u = 0, v = 0}, {u = -(1/2)*(-(2*alpha[2]+2*(4*k^2*p*alpha[3]+4*alpha[1]*alpha[3]+alpha[2]^2)^(1/2))/alpha[3])^(1/2), v = 0}, {u = (1/2)*(-(2*alpha[2]+2*(4*k^2*p*alpha[3]+4*alpha[1]*alpha[3]+alpha[2]^2)^(1/2))/alpha[3])^(1/2), v = 0}, {u = -(1/2)*2^(1/2)*((-alpha[2]+(4*k^2*p*alpha[3]+4*alpha[1]*alpha[3]+alpha[2]^2)^(1/2))/alpha[3])^(1/2), v = 0}, {u = (1/2)*2^(1/2)*((-alpha[2]+(4*k^2*p*alpha[3]+4*alpha[1]*alpha[3]+alpha[2]^2)^(1/2))/alpha[3])^(1/2), v = 0}, {u = -(1/2)*(-2*X*p)^(1/2)/X, v = -(1/4)*2^(1/2)*(X*(4*X^2*k^2*p+4*X^2*alpha[1]+2*X*p*alpha[2]-p^2*alpha[3]))^(1/2)/X^2}, {u = -(1/2)*(-2*X*p)^(1/2)/X, v = (1/4)*2^(1/2)*(X*(4*X^2*k^2*p+4*X^2*alpha[1]+2*X*p*alpha[2]-p^2*alpha[3]))^(1/2)/X^2}, {u = (1/2)*(-2*X*p)^(1/2)/X, v = -(1/4)*2^(1/2)*(X*(4*X^2*k^2*p+4*X^2*alpha[1]+2*X*p*alpha[2]-p^2*alpha[3]))^(1/2)/X^2}, {u = (1/2)*(-2*X*p)^(1/2)/X, v = (1/4)*2^(1/2)*(X*(4*X^2*k^2*p+4*X^2*alpha[1]+2*X*p*alpha[2]-p^2*alpha[3]))^(1/2)/X^2}}

(2)

# Example of use: some parameters are numeric

Equilibria(chi, k, p, 1, 3/5, -7)

{{u = 0, v = 0}, {u = -(1/2)*(6/35+(2/7)*(-28*k^2*p-691/25)^(1/2))^(1/2), v = 0}, {u = (1/2)*(6/35+(2/7)*(-28*k^2*p-691/25)^(1/2))^(1/2), v = 0}, {u = -(1/2)*2^(1/2)*(3/35-(1/7)*(-28*k^2*p-691/25)^(1/2))^(1/2), v = 0}, {u = (1/2)*2^(1/2)*(3/35-(1/7)*(-28*k^2*p-691/25)^(1/2))^(1/2), v = 0}, {u = -(1/2)*(-2*chi*p)^(1/2)/chi, v = -(1/4)*2^(1/2)*(chi*(4*chi^2*k^2*p+(6/5)*chi*p+7*p^2+4*chi^2))^(1/2)/chi^2}, {u = -(1/2)*(-2*chi*p)^(1/2)/chi, v = (1/4)*2^(1/2)*(chi*(4*chi^2*k^2*p+(6/5)*chi*p+7*p^2+4*chi^2))^(1/2)/chi^2}, {u = (1/2)*(-2*chi*p)^(1/2)/chi, v = -(1/4)*2^(1/2)*(chi*(4*chi^2*k^2*p+(6/5)*chi*p+7*p^2+4*chi^2))^(1/2)/chi^2}, {u = (1/2)*(-2*chi*p)^(1/2)/chi, v = (1/4)*2^(1/2)*(chi*(4*chi^2*k^2*p+(6/5)*chi*p+7*p^2+4*chi^2))^(1/2)/chi^2}}

(3)
 

 

Download Is_this_you_are_looking_for.mw

@Math-dashti 

I'm not sure I trully understand what you are saying.
Read carefully this worksheet and let me know if it answers your problem simpler_sand15.mw

with Maple 2015, worksheet mode , input display = Maple Notation, output display = 2-D Math Notation

restart

kernelopts(version)

`Maple 2015.2, APPLE UNIVERSAL OSX, Dec 20 2015, Build ID 1097895`

(1)

with(Units):

d := 10*Unit('cm');
c__w := 0.47;
F__luft := 0.040146*Unit('N');

10*Units:-Unit('cm')

 

.47

 

0.40146e-1*Units:-Unit('N')

(2)

F__luft = 1/2*rho__luft*A*c__w*v^2;

0.40146e-1*Units:-Unit('N') = .2350000000*rho__luft*A*v^2

(3)

A := Pi*(d/2)^2

25*Pi*Units:-Unit('cm')^2

(4)

rho__luft := 1.2*Unit('kg/m^3')

1.2*Units:-Unit(('kg')/('m')^3)

(5)

under_radical := 2*F__luft/(rho__luft*A*c__w);

under_radical := simplify(under_radical);
 

0.1812605488e-2*Units:-Unit('N')/(Units:-Unit(('kg')/('m')^3)*Units:-Unit('cm')^2)

 

18.12605488*Units:-Unit(('m')^2/('s')^2)

(6)

v = simplify(surd(under_radical, 2))

v = 4.257470479*Units:-Unit(('m')/('s'))

(7)
 

 

Download No_problem_with_Maple_2015.mw
 

restart

kernelopts(version)

`Maple 2015.2, APPLE UNIVERSAL OSX, Dec 20 2015, Build ID 1097895`

(1)

with(Units):

d := 10*Unit('cm');
c__w := 0.47;
F__luft := 0.040146*Unit('N');

10*Units:-Unit('cm')

 

.47

 

0.40146e-1*Units:-Unit('N')

(2)

F__luft = 1/2*rho__luft*A*c__w*v^2;

0.40146e-1*Units:-Unit('N') = .2350000000*rho__luft*A*v^2

(3)

A := Pi*(d/2)^2

25*Pi*Units:-Unit('cm')^2

(4)

rho__luft := 1.2*Unit('kg/m^3')

1.2*Units:-Unit(('kg')/('m')^3)

(5)

under_radical := 2*F__luft/(rho__luft*A*c__w);

under_radical := simplify(under_radical);
 

0.1812605488e-2*Units:-Unit('N')/(Units:-Unit(('kg')/('m')^3)*Units:-Unit('cm')^2)

 

18.12605488*Units:-Unit(('m')^2/('s')^2)

(6)

v = simplify(surd(under_radical, 2))

v = 4.257470479*Units:-Unit(('m')/('s'))

(7)
 

 

Download No_problem_with_Maple_2015.mw

@Andiguys 

The things you will have to adjust for future data are pink written in the attached file Scalling_issue_fixsd.mw
.

@acer 

When I posted my answer I have no clue at all about what the OP wanted.
I simply guessed he wanted to identify the regions where some TM dominates the others: so I proposed him this notional inequal plot without knowing if it asked or not its question (what puzzed me by the time was the image he joined at the end if its question).
And doing so I didn't pay attention to the phrase "(TM1, TM2, TM3 are all positive)"

So you're likely right that the conditions TMs > 0 should be included in some inequal plot... which I see you did in your answer.

@acer 

Thanks for your reply. 

See you soon on another thread

@Andiguys 

... implies not the same code to apply.

Indeed in your initial question you had L := something = 0, now you have L := M1=M2 (with M2 <> 0).
So you have to replace the line

A := expand(lhs(L)):

by the line

A := expand(M1-M2):

to be in the same conditions than previously.

Here is your file corrected.mw

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