Mr. skystar Pak

## 15 Reputation

8 years, 264 days

## @Kitonum  I m still not reciving a...

I m still not reciving a table

## Moreover...

Moreover the graphs of f, f', theata, theta prime, g , g'  are nor the same as per ur file

## is this a way to find infinity...

Dear thanks alot. Here i m attaching a file which i develop for shooting. But i m not getting all the graphs. ur modifcation or precious suggestion can be very much useful for this

thanks

## How did it guessed for 4...

Dear , thanks alot for corrections. you r always a maple hero really. whenever i contacted u , u guided me very weell. thanks once again

1)How did it guessed for 4

2)can we have any coomand which print the guess value for which this problem converging.

3) How will we plot for these different values of  epsilon := 0 ,0.5, 1.0, 1.5, 2.0

4) i required these graphs not in doted, long lines or short lines format.....not in colors

5) how can we apply shooting technique upon this

## shoot9 package...

thanks dear.

can u plz tell me from where i can get this package.

## Plot...

Dear how shall we plot the remaining two means theata and phi

## ode...

@Preben Alsholm

yes dear it is new but after changing these [] , its still sot working

## ODE...

dear, it is not working for this equation and conditions. Moreover how can we get a single graph for all the different values of epsilon

restart;

eq1 := diff(f(eta), eta, eta, eta, eta)+2*(epsilon/(1+epsilon*theta(eta)))^2*(diff(f(eta), eta, eta))*(diff(theta(eta), eta))^2-epsilon*(diff(theta(eta), eta, eta))*(diff(f(eta), eta, eta))/(1+epsilon*theta(eta))+RE*[(diff(f(eta), eta, eta))*(diff(f(eta), eta))-f(eta)*(diff(f(eta), eta, eta))] = 0;

eq2 := diff(theta(eta), eta, eta)+Pr*RE*f(eta)*(diff(theta(eta), eta)) = 0;

RE := 40; Pr := 1.5; epsilon := .25;

bc := f(1) = 0, (D(f))(-1) = 1, (D(f))(1) = 1, f(-1) = 0, theta(-1) = 1, theta(1) = 0;

res := dsolve({bc, eq1, eq2}, numeric);

plots:-odeplot(res, [eta, f(eta)], -1 .. 1);

plots:-odeplot(res, [eta, theta(eta)], -1 .. 1);

## @Kitonum  dear i need code for this...

dear i need code for this method alsoknown as Keller box method

## oh yes...

Dear its working now. thanks

dear do u have any code about a keller box method(implicit finite difference method) for the coupled differentila equations. kindly help to solve the last obtained result with the help of keller box method

thanks once agani.

## correct...

Oh really dear its same as per required. but i am not getting this out put. I am using maple 14. should i use maple 15. i m also obtaing these two lines in my output.

f(x, eta) will now be displayed as f
h(x, y) will now be displayed as h

u can check below. i just copied and paste here, my all output is like this.

/ d \ / d \
u(x, y) |--- u(x, y)| + v(x, y) |--- u(x, y)| =
\ dx / \ dy /

d / d \
--- |--- u(x, y)|
dy \ dy /
-----------------------
1 + epsilon theta(x, y)

/ d \ / d \
epsilon |--- u(x, y)| |--- theta(x, y)|
\ dy / \ dy /
- --------------------------------------- + theta(x, y)
2
(1 + epsilon theta(x, y))
/ d \ / d \
u(x, y) |--- theta(x, y)| + v(x, y) |--- theta(x, y)| =
\ dx / \ dy /

d / d \
--- |--- theta(x, y)|
dy \ dy /
---------------------
Pr
/ (4/5)
| y x f(x, eta)
< eta = --------------------, psi(x, y) = ----------------,
| (1/5) (1/20) (1/20)
\ x (1 + x) (1 + x)

d d
u(x, y) = --- psi(x, y), v(x, y) = ---- psi(x, y),
dy dx

(1/5) \
x h(x, y)|
theta(x, y) = -------------- >
(1/5) |
(1 + x) /
[
[
[/ d \ / d / d \\
[|--- psi(x, y)| |--- |--- psi(x, y)||
[\ dy / \ dy \ dx //
[
[
[
[
[

/ d \ / d / d \\
- |--- psi(x, y)| |--- |--- psi(x, y)|| =
\ dx / \ dy \ dy //

d / d / d \\
--- |--- |--- psi(x, y)||
dy \ dy \ dy //
--------------------------
(1/5)
epsilon x h(x, y)
1 + ----------------------
(1/5)
(1 + x)

/ d / d \\ (1/5) / d \
epsilon |--- |--- psi(x, y)|| x |--- h(x, y)|
\ dy \ dy // \ dy /
- --------------------------------------------------
2
/ (1/5) \
(1/5) | epsilon x h(x, y)|
(1 + x) |1 + ----------------------|
| (1/5) |
\ (1 + x) /

/
(1/5) |
x h(x, y) / d \ | h(x, y)
+ --------------, |--- psi(x, y)| |---------------------
(1/5) \ dy / | (4/5) (1/5)
(1 + x) \5 x (1 + x)

(1/5) / d \ \
x |--- h(x, y)| (1/5) |
\ dx / x h(x, y)|
+ -------------------- - --------------|
(1/5) (6/5)|
(1 + x) 5 (1 + x) /

/ d \ (1/5) / d \
|--- psi(x, y)| x |--- h(x, y)|
\ dx / \ dy /
- ------------------------------------ =
(1/5)
(1 + x)

(1/5) / d / d \\]
x |--- |--- h(x, y)||]
\ dy \ dy //]
--------------------------]
(1/5) ]
(1 + x) Pr ]
]
]
]
]
[ (1/5) (1/5) / d / d \\ / d \
[epsilon x (1 + x) |--- |--- psi(x, y)|| |--- h(x, y)|
[ \ dy \ dy // \ dy /
[---------------------------------------------------------------
[ 2
[ / (1/5) (1/5) \
[ \(1 + x) + epsilon x h(x, y)/

/ d \ / d / d \\
- |--- psi(x, y)| |--- |--- psi(x, y)||
\ dx / \ dy \ dy //

/ d \ / d / d \\
+ |--- psi(x, y)| |--- |--- psi(x, y)||
\ dy / \ dy \ dx //

(1/5) / d / d / d \\\
(1 + x) |--- |--- |--- psi(x, y)||| (1/5)
\ dy \ dy \ dy /// x h(x, y)
- ---------------------------------------- - -------------- = 0,
(1/5) (1/5) (1/5)
(1 + x) + epsilon x h(x, y) (1 + x)

/ d \ (1/5) / d \
|--- psi(x, y)| x |--- h(x, y)|
\ dy / \ dx /
------------------------------------
(1/5)
(1 + x)

/ d \ (1/5) / d \
|--- psi(x, y)| x |--- h(x, y)|
\ dx / \ dy /
- ------------------------------------
(1/5)
(1 + x)

/ d \ (1/5) / d / d \\ ]
h(x, y) |--- psi(x, y)| x |--- |--- h(x, y)|| ]
\ dy / \ dy \ dy // ]
+ ----------------------- - -------------------------- = 0]
(4/5) (6/5) (1/5) ]
5 x (1 + x) (1 + x) Pr ]
]
[ (1/5) (1/5) / d / d \\ / d \
[epsilon x (1 + x) |--- |--- psi(x, y)|| |--- h(x, y)|
[ \ dy \ dy // \ dy /
[---------------------------------------------------------------
[ 2
[ / (1/5) (1/5) \
[ \(1 + x) + epsilon x h(x, y)/

/ d \ / d / d \\
- |--- psi(x, y)| |--- |--- psi(x, y)||
\ dx / \ dy \ dy //

/ d \ / d / d \\
+ |--- psi(x, y)| |--- |--- psi(x, y)||
\ dy / \ dy \ dx //

(1/5) / d / d / d \\\
(1 + x) |--- |--- |--- psi(x, y)||| (1/5)
\ dy \ dy \ dy /// x h(x, y)
- ---------------------------------------- - -------------- = 0,
(1/5) (1/5) (1/5)
(1 + x) + epsilon x h(x, y) (1 + x)

/ d \ (1/5) / d \
|--- psi(x, y)| x |--- h(x, y)|
\ dy / \ dx /
------------------------------------
(1/5)
(1 + x)

/ d \ (1/5) / d \
|--- psi(x, y)| x |--- h(x, y)|
\ dx / \ dy /
- ------------------------------------
(1/5)
(1 + x)

/ d \ (1/5) / d / d \\ ]
h(x, y) |--- psi(x, y)| x |--- |--- h(x, y)|| ]
\ dy / \ dy \ dy // ]
+ ----------------------- - -------------------------- = 0]
(4/5) (6/5) (1/5) ]
5 x (1 + x) (1 + x) Pr ]
]
f(x, eta) will now be displayed as f
h(x, y) will now be displayed as h
[ (1/5) (1/5) / d / d \\ / d \
[epsilon x (1 + x) |--- |--- psi(x, y)|| |--- h(x, y)|
[ \ dy \ dy // \ dy /
[---------------------------------------------------------------
[ 2
[ / (1/5) (1/5) \
[ \(1 + x) + epsilon x h(x, y)/

/ d \ / d / d \\
- |--- psi(x, y)| |--- |--- psi(x, y)||
\ dx / \ dy \ dy //

/ d \ / d / d \\
+ |--- psi(x, y)| |--- |--- psi(x, y)||
\ dy / \ dy \ dx //

(1/5) / d / d / d \\\
(1 + x) |--- |--- |--- psi(x, y)||| (1/5)
\ dy \ dy \ dy /// x h(x, y)
- ---------------------------------------- - -------------- = 0,
(1/5) (1/5) (1/5)
(1 + x) + epsilon x h(x, y) (1 + x)

/ d \ (1/5) / d \
|--- psi(x, y)| x |--- h(x, y)|
\ dy / \ dx /
------------------------------------
(1/5)
(1 + x)

/ d \ (1/5) / d \
|--- psi(x, y)| x |--- h(x, y)|
\ dx / \ dy /
- ------------------------------------
(1/5)
(1 + x)

/ d \ (1/5) / d / d \\ ]
h(x, y) |--- psi(x, y)| x |--- |--- h(x, y)|| ]
\ dy / \ dy \ dy // ]
+ ----------------------- - -------------------------- = 0]
(4/5) (6/5) (1/5) ]
5 x (1 + x) (1 + x) Pr ]
]

## working...

Dear, thanks . alot. its working well but the answer is not correct it in the form of psi derivatives which is wrong. The answer should be in the form of f,h,  and its higher derivaties while the cofficents are almost correct. kindly see once again. actually in phi we have f which is funtion of x and eta again. the  anwer which i obtaing from your given program is :

[epsilon*x^(1/5)*(1+x)^(1/5)*psi[y, y]*h[y]/((1+x)^(1/5)+epsilon*x^(1/5)*h)^2-psi[x]*psi[y, y]+psi[y]*psi[x, y]-(1+x)^(1/5)*psi[y, y, y]/((1+x)^(1/5)+epsilon*x^(1/5)*h)-x^(1/5)*h/(1+x)^(1/5) = 0, psi[y]*x^(1/5)*h[x]/(1+x)^(1/5)-psi[x]*x^(1/5)*h[y]/(1+x)^(1/5)+(1/5)*h*psi[y]/(x^(4/5)*(1+x)^(6/5))-x^(1/5)*h[y, y]/((1+x)^(1/5)*Pr) = 0]

while it should be :

(diff(f(x), x, x, x))/(1+`&epsilon;`*x^(1/5)*h(x, eta)/(1+x)^(1/5))+(16+15*x)*f(x)*(diff(f(x), x, x))/(20*(1+x))-(6+5*x)*(diff(f(x), x))^2/(10*(1+x))+h(x) = `&epsilon;`*(x/(1+x))^(1/5)*(diff(f(x), x, x))*(diff(h(x), x))/(1+`&epsilon;`*x^(1/5)*h(x, eta)/(1+x)^(1/5))^2+x*(diff(f(x), x))*(diff(diff(f(x), x), x))-x*(diff(f(x), x, x))*(diff(f(x), x))

and

(diff(h(x), x, x))/pr+(16+15*x)*f(x)*(diff(h(x), x))/(20*(1+x))-1/(5*(1+x)) = x*((diff(f(x), x))*(diff(h(x), x))-(diff(f(x), x))*(diff(h(x), x)))

## problem...

I want to change the following equations

u*(diff(u, x))+v*(diff(u, y)) = (diff(u, y, y))/(1+`&epsilon;&theta;`)-`&epsilon;`*(diff(u, y))*(diff(theta, y))/(1+`&epsilon;&theta;`)^2+theta

and

u*(diff(theta, x))+v*(diff(theta, y)) = (diff(theta, y, y))/Pr

by using the following

Psi = x^(4/5)*f(x, eta)/(1+x)^(1/20)

eta = y/(x^(1/5)*(1+x)^(1/20))

u = diff(Psi, y)

v = -(diff(Psi, x))

into the follwing form

(diff(f(x), x, x, x))/(1+`&epsilon;`*x^(1/5)*h(x, eta)/(1+x)^(1/5))+(16+15*x)*f(x)*(diff(f(x), x, x))/(20*(1+x))-(6+5*x)*(diff(f(x), x))^2/(10*(1+x))+h(x) = `&epsilon;`*(x/(1+x))^(1/5)*(diff(f(x), x, x))*(diff(h(x), x))/(1+`&epsilon;`*x^(1/5)*h(x, eta)/(1+x)^(1/5))^2+x*(diff(f(x), x))*(diff(diff(f(x), x), x))-x*(diff(f(x), x, x))*(diff(f(x), x))

and

(diff(h(x), x, x))/pr+(16+15*x)*f(x)*(diff(h(x), x))/(20*(1+x))-1/(5*(1+x)) = x*((diff(f(x), x))*(diff(h(x), x))-(diff(f(x), x))*(diff(h(x), x)))

## Problem...

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