Difference between Essential a Natural BC's

  • Thread starter bugatti79
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  • #1
bugatti79
792
1
Folks,

I am looking at the Ritz method for the following problem

##\displaystyle -\frac{d^2 u}{dx^2}-u+x^2=0## for ##0<x<1##

with boundary conditions ##u(0)=0## and ##\displaystyle \frac{du}{dx} |_{x=1} =1##

The last derivative term, how do I know whether that is a natural or essential BC?

I have googled the following guidelines but I am still confused.

Specification of the primary variable ( u in this case) is an essential BC*
Specification of a secondary variable (like a force F, not present in this example) is a natural boundary condition

IF a boundary condition involves one or more variables in a 'direct' way it is essential otherwise it is natural.
Direct implies excluding derivative of the primary function.**

I find this info conflicting based on * and **
I think the book states it is a natural BC.

Would appreciate some clarification...
THanks
 

Answers and Replies

  • #2
Studiot
5,441
9
IF a boundary condition involves one or more variables in a 'direct' way it is essential otherwise it is natural.
Direct implies excluding derivative of the primary function.**

As I understand the difference:

What is meant is that direct gives an expression that yields a definite value for (in this case) u.

for example u(0) = 0 says that at x=0 the value of u is zero.

This is contrasted by natural expression which does not lead to a definite value of u.

for example


[tex]{\left[ {\frac{{du}}{{dx}}} \right]_{x = 1}} = 1[/tex]

does not yield a definite value for u at x = 1 since a curve of slope 1 can be drawn through any value of u.

However this is really just classification for the sake of it and nothing to worry about.
 
  • #3
bugatti79
792
1
As I understand the difference:

What is meant is that direct gives an expression that yields a definite value for (in this case) u.

for example u(0) = 0 says that at x=0 the value of u is zero.

This is contrasted by natural expression which does not lead to a definite value of u.

for example


[tex]{\left[ {\frac{{du}}{{dx}}} \right]_{x = 1}} = 1[/tex]

does not yield a definite value for u at x = 1 since a curve of slope 1 can be drawn through any value of u.

However this is really just classification for the sake of it and nothing to worry about.

Thank you sir, that explains it nicely. I might be back with other BC type q's :-)
 

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