## 220 Reputation

17 years, 31 days
Glasgow, United Kingdom

## complex solution??...

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

## complex solution??...

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

## Re: afew plots...

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

plots.PDF

## Re: afew plots...

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

plots.PDF

## Re: decimal places in Iprint output...

it worked well. Thx

`oldDigits:=Digits: Digits:=70:for K from 1 by 0.1 to 10 dotmp:= 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 = -1.1125176896515278095l = 1.1, R3 = -1.1498313611138531951l = 1.2, R3 = -1.1905214693893404126l = 1.3, R3 = -1.2345293318398085679`

## Re: decimal places in Iprint output...

it worked well. Thx

`oldDigits:=Digits: Digits:=70:for K from 1 by 0.1 to 10 dotmp:= 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 = -1.1125176896515278095l = 1.1, R3 = -1.1498313611138531951l = 1.2, R3 = -1.1905214693893404126l = 1.3, R3 = -1.2345293318398085679`

## Re: Iprint...

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

## Re: Iprint...

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

## Re: numerical noise...

`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`
` `

## Re: numerical noise...

`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`
` `

## Re: understanding R3=0...

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

## Re: understanding R3=0...

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

## Re:...

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

## Re:...

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

## shortwave disturbance...

# 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)

 1 2 3 4 5 6 7 Last Page 3 of 11
﻿