kh2n

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14 years, 327 days
Glasgow, United Kingdom

MaplePrimes Activity


These are replies submitted by kh2n

@Axel Vogt 

Its quite hard for me to stop bothering you. Now I am in a situation to find out wether we have complex solutions are not? 

I have attached a paper in which one page 194 there is some explanation about zeros and poles, which maybe related to that "straight line" and about complex solutions. 

 

For the particular case where I doubt about the complex sols is in the following sheet

KVasymptData.mws

Gravitational_Stabil.pdf

 

@Axel Vogt 

Its quite hard for me to stop bothering you. Now I am in a situation to find out wether we have complex solutions are not? 

I have attached a paper in which one page 194 there is some explanation about zeros and poles, which maybe related to that "straight line" and about complex solutions. 

 

For the particular case where I doubt about the complex sols is in the following sheet

KVasymptData.mws

Gravitational_Stabil.pdf

 

@Axel Vogt 

At last i have some good results. plz see the attached file.

Thx for your help.

plots.PDF

@Axel Vogt 

At last i have some good results. plz see the attached file.

Thx for your help.

plots.PDF

@Axel Vogt 

it worked well. Thx

oldDigits:=Digits: 
Digits:=70:
for K from 1 by 0.1 to 10 do
tmp:= eval(G, l=K);
sol:=fsolve(tmp, R3=-1.9332, maxsols=1);
eval(tmp, R3=sol);
evalf[5](abs(%));
lprint('l'=K, evalf[20]('R3'=sol));
end do:
K:='K': Digits:=oldDigits:
 
l = 1, R3 = -1.1125176896515278095
l = 1.1, R3 = -1.1498313611138531951
l = 1.2, R3 = -1.1905214693893404126
l = 1.3, R3 = -1.2345293318398085679

@Axel Vogt 

it worked well. Thx

oldDigits:=Digits: 
Digits:=70:
for K from 1 by 0.1 to 10 do
tmp:= eval(G, l=K);
sol:=fsolve(tmp, R3=-1.9332, maxsols=1);
eval(tmp, R3=sol);
evalf[5](abs(%));
lprint('l'=K, evalf[20]('R3'=sol));
end do:
K:='K': Digits:=oldDigits:
 
l = 1, R3 = -1.1125176896515278095
l = 1.1, R3 = -1.1498313611138531951
l = 1.2, R3 = -1.1905214693893404126
l = 1.3, R3 = -1.2345293318398085679

@Axel Vogt 

It is working very good but in a situation sometime still I need to have high Digits,

that okay but i want that the output should be upto 20 decimal places.

i.e.,

Digits=60:

The output should is

l = 65, R3 = -.715676069242813981481606224909048513255686162604960503675637

 use Digits whatever you want but the output decimal places should also be our choice to choose.

any idea to get the following

i.e.,

Digits=60:

The output should be

l = 65, R3 = -.715676069242813

 

cheers

khan

 

@Axel Vogt 

It is working very good but in a situation sometime still I need to have high Digits,

that okay but i want that the output should be upto 20 decimal places.

i.e.,

Digits=60:

The output should is

l = 65, R3 = -.715676069242813981481606224909048513255686162604960503675637

 use Digits whatever you want but the output decimal places should also be our choice to choose.

any idea to get the following

i.e.,

Digits=60:

The output should be

l = 65, R3 = -.715676069242813

 

cheers

khan

 

@Axel Vogt 

 

from your sheet

lambda1:=65;
eval(F, lambda=lambda1): evalf(%):
r1:=fsolve(%, r=-2*lambda1);

lambda1 := 65
r1 := -0.6145715399135967845940454761730148356205
but if we increase the Digits then the correct value is
                        lambda1 := 65
r1 := -.7150108137031811713404211889874700864929
I am trying to get the data using the following
for K from 0 by 0.1 to 50 do lprint(K, simplify(evalf(fsolve(subs(S0=S01,parvals,l=K, A0),
R3=-1..-.69))));od;
for l=0 .. 50 its gives the values for R3 easly but for l=50 .. 100, the computer take 
to0 much time and the mechine is burning (very hot).
As there any way to get the appropriate data for R3 when l=0 .. 150. 
I commonly use the above
lprint command but it is very tedious work to adjust the range of the dependent variable.
KVasympt.mws
 

 

@Axel Vogt 

 

from your sheet

lambda1:=65;
eval(F, lambda=lambda1): evalf(%):
r1:=fsolve(%, r=-2*lambda1);

lambda1 := 65
r1 := -0.6145715399135967845940454761730148356205
but if we increase the Digits then the correct value is
                        lambda1 := 65
r1 := -.7150108137031811713404211889874700864929
I am trying to get the data using the following
for K from 0 by 0.1 to 50 do lprint(K, simplify(evalf(fsolve(subs(S0=S01,parvals,l=K, A0),
R3=-1..-.69))));od;
for l=0 .. 50 its gives the values for R3 easly but for l=50 .. 100, the computer take 
to0 much time and the mechine is burning (very hot).
As there any way to get the appropriate data for R3 when l=0 .. 150. 
I commonly use the above
lprint command but it is very tedious work to adjust the range of the dependent variable.
KVasympt.mws
 

 

@Axel Vogt 

If we compare the following two plots its seems that

 

1. sigma first becomes  to be postive  for medium l (my guess is l=7) when R3<=-0.6

2. so for some R3<=-0.6 (not for R3<<<=-0.6 ), sigma<0 for very small and for very large l.

3. so, first sigma>0 for medium l, then the rest follows, this means that in  your 3d plot that extra

straight line which falsing try to represents a  line equation, to my understanding if you agree this

line shows

(a). singlurities

(b). maybe the equation has multi-solutions for sigma and maple tries to caputre both.

(c). numerical noise (see (b)).


I have attached your sheet and try to compare the resluts. By the way in ur 3d plot the range of sigma

is alway ploted to be sigma=0..600, can we imposed a range sigma=-200..300.

 

KVasympt.mws

 

@Axel Vogt 

If we compare the following two plots its seems that

 

1. sigma first becomes  to be postive  for medium l (my guess is l=7) when R3<=-0.6

2. so for some R3<=-0.6 (not for R3<<<=-0.6 ), sigma<0 for very small and for very large l.

3. so, first sigma>0 for medium l, then the rest follows, this means that in  your 3d plot that extra

straight line which falsing try to represents a  line equation, to my understanding if you agree this

line shows

(a). singlurities

(b). maybe the equation has multi-solutions for sigma and maple tries to caputre both.

(c). numerical noise (see (b)).


I have attached your sheet and try to compare the resluts. By the way in ur 3d plot the range of sigma

is alway ploted to be sigma=0..600, can we imposed a range sigma=-200..300.

 

KVasympt.mws

 

@Axel Vogt 

 

I am not sure about that linear relationship between R3 and l.   

I have tried to make it more clear in the following pdf.

 

dispersion_eqn.mws

Eignfunengery_vapou.mws

Eignfunengery_liqui.mws

 

Eigenfunpressure.mws

 

new.PDF

 

@Axel Vogt 

 

I am not sure about that linear relationship between R3 and l.   

I have tried to make it more clear in the following pdf.

 

dispersion_eqn.mws

Eignfunengery_vapou.mws

Eignfunengery_liqui.mws

 

Eigenfunpressure.mws

 

new.PDF

 

results_S0=0.125.mws

# This plot shows that there exist a values of R3 for which sigma=0 when l approches to infinity. Now I need your help to find that R3.

 Well my calculated guess is that value of R3 for which sigma=0 is -0.6. But i need to confrom.

your above animation shows that that phase change front S0=0.0275 is stable for shortwave number (l approches to infinity) 

 

 

 

 

 

 

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