From PDE to ODE ? + research

  • #1
From PDE to ODE ?! + research

Homework Statement



In the attached research, What are the steps that we work to transform the equation (1) to (8)

Homework Equations



(1) and (8)

The Attempt at a Solution



I know that they used similarity transformations but I do not know how to do it by myself (how to substitute correctly) ...

can you help me to I understand the steps ... then I will do the others In the same way.

Thanks
 

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  • #2


Do you know how to apply the chain rule for partial differentiation? Try going to Wikipedia and searching on "chain rule", and reading the section on several variables. Then I think if you apply this to the research paper you attached you will see how they arrived at the equations. Basically, they chose a new set of variables so that after applying the chain rule and collecting terms, many terms canceled and the original set of PDEs was converted into a set of ODEs.
 
  • #3


phyzguy said:
Do you know how to apply the chain rule for partial differentiation? Try going to Wikipedia and searching on "chain rule", and reading the section on several variables. Then I think if you apply this to the research paper you attached you will see how they arrived at the equations. Basically, they chose a new set of variables so that after applying the chain rule and collecting terms, many terms canceled and the original set of PDEs was converted into a set of ODEs.

Thank you for this hint...I will try it :) and read:

http://en.wikipedia.org/wiki/Chain_rule

https://www.physicsforums.com/library.php?do=view_item&itemid=353
 
  • #4


Q: How do I choose the appropriate similarity transformations to a particular system?!​
 
  • #5


please ... help me ... just the first term in eq(1)

I try but failed :(
 
  • #6


Show us your calculations on the first term in Equation 1. Learn how to use TeX so you can type it in.
 
  • #8


is it right that u is in 3 variables : x and Fi Dash and M ?
 
  • #9


please ... Is this true?! I took the second term in eq(1)



[tex]\partial w[/tex] / [tex]\partial z[/tex]

= ( [tex]\partial w[/tex] / [tex]\partial \varphi[/tex] ) . ( [tex]\partial \varphi[/tex] / [tex]\partial \eta[/tex] ) . ( [tex]\partial \eta[/tex] / [tex]\partial z[/tex] )

= (- [tex]\sqrt{b\nu}[/tex] ) . ( [tex]\grave{\varphi}[/tex] ) .( [tex]\sqrt{b/\nu}[/tex] )

= -b [tex]\grave{\varphi}[/tex]​
 
  • #10


I think this is correct. Now if you look at the first term in equation 1, you see:

[tex]\frac{\partial{u}}{\partial{x}} = b \phi'(\eta)[/tex]

so the two terms cancel and equation 1 is automatically satisfied. Keep going!
 
  • #11


phyzguy said:
I think this is correct. Now if you look at the first term in equation 1, you see:

[tex]\frac{\partial{u}}{\partial{x}} = b \phi'(\eta)[/tex]

so the two terms cancel and equation 1 is automatically satisfied. Keep going!

Thanks ...I appreciate it

but how do I get the equation (8) ?!

and

thepioneerm said:
is it right that u is in 3 variables : x and Fi Dash and M ?
 
Last edited:
  • #12


Yes, you need to consider u to be a function of x, [tex]\phi'[/tex], and M. You need to evaluate all of the partial derviatives, like you did for

[tex]\frac{\partial{w}}{\partial{z}}[/tex]

and plug these into equations 2-5, and you should come out with equations 8-12.
 
  • #13


phyzguy said:
Yes, you need to consider u to be a function of x, [tex]\phi'[/tex], and M. You need to evaluate all of the partial derviatives, like you did for

[tex]\frac{\partial{w}}{\partial{z}}[/tex]

and plug these into equations 2-5, and you should come out with equations 8-12.

ok, but eq (1) say:

[tex]\frac{\partial{u}}{\partial{x}}[/tex] + ( [tex]\partial w[/tex] / [tex]\partial z[/tex]) =0

so, I have evaluate all of the partial derviatives but How can I have eq (8) :( I do not understand !
 
  • #14


Equation 8 comes from equation 2. Since u has a term proportional to [tex]\phi'[/tex], when you evaluate
[tex]\frac{\partial^2{u}}{\partial{z^2}}[/tex] , you will get a term in [tex]\phi'''[/tex] . Do you see?
 
  • #15


phyzguy said:
Equation 8 comes from equation 2. Since u has a term proportional to [tex]\phi'[/tex], when you evaluate
[tex]\frac{\partial^2{u}}{\partial{z^2}}[/tex] , you will get a term in [tex]\phi'''[/tex] . Do you see?

ok ... I will try :)

Thank you
 

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