Transforming PDE to ODE: How Can We Subsitute Correctly?

[thepioneerm]

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

 

[phyzguy]

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.

 

[thepioneerm]

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​

 

[thepioneerm]

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

 

[thepioneerm]

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

I try but failed :(​

 

[phyzguy]

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

 

[thepioneerm]

I will try :(

 

[thepioneerm]

is it right that u is in 3 variables : x and Fi Dash and M ?​

 

[thepioneerm]

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



\partial w / \partial z

= ( \partial w / \partial \varphi ) . ( \partial \varphi / \partial \eta ) . ( \partial \eta / \partial z )

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

= -b \grave{\varphi}​

 

[phyzguy]

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

\frac{\partial{u}}{\partial{x}} = b \phi'(\eta)

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

 

[thepioneerm]

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

\frac{\partial{u}}{\partial{x}} = b \phi'(\eta)

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

is it right that u is in 3 variables : x and Fi Dash and M ?​

 

[phyzguy]

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

\frac{\partial{w}}{\partial{z}}

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

 

[thepioneerm]

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

\frac{\partial{w}}{\partial{z}}

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

ok, but eq (1) say:

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

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

 

[phyzguy]

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

 

[thepioneerm]

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

ok ... I will try :)

Thank you​