Chain Rule Confusion (Euler-Lagrange Equation)

  • Thread starter Thread starter laser1
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the application of the chain rule in the context of the Euler-Lagrange equation, specifically addressing the differentiation of a functional F with respect to its variables. Participants clarify that the correct expression for the total derivative dF/dx is given by the formula dF/dx = (∂F/∂y) y' + (∂F/∂y') y'', emphasizing the necessity of using the chain rule. The confusion arises from the lecturer's use of dF/dy instead of ∂F/∂y, which is deemed incorrect in this context. Overall, the conversation reinforces the importance of proper notation in calculus.

PREREQUISITES
  • Understanding of the Euler-Lagrange equation
  • Familiarity with the chain rule in calculus
  • Knowledge of total and partial derivatives
  • Basic concepts of functional analysis
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equation in detail
  • Learn about the differences between total and partial derivatives
  • Explore advanced applications of the chain rule in calculus
  • Review notation conventions in calculus and their implications
USEFUL FOR

Students of calculus, mathematicians, and physicists who are working with the Euler-Lagrange equation and need clarity on differentiation techniques and notation.

laser1
Messages
170
Reaction score
23
Homework Statement
No explicit x dependence for E-L eq.
Relevant Equations
Euler-Lagrange.
1744095877659.png

The above image is from my lecturer's notes.

My concern is when it seems like my lecturer has split up the dF/dx term into dF/dy y' + dF/dy' y''. Why is it this as opposed to ##\frac{\partial F}{\partial y}## etc.? Or would this not matter, because y is an independent variable, and hence, the partial F wrt y = total F wrt y? (although I guess y has to be a function of x...)

Thank you!
 
Last edited:
Physics news on Phys.org
Not ##dF/dx = (\partial F/\partial y) y’ - (\partial F/\partial y’) y’’##, it is ##(\partial F/\partial y) y’ +(\partial F/\partial y’) y’’##. This is just the chain rule.
 
Orodruin said:
Not ##dF/dx = (\partial F/\partial y) y’ - (\partial F/\partial y’) y’’##, it is ##(\partial F/\partial y) y’ +(\partial F/\partial y’) y’’##. This is just the chain rule.
sorry, edited!
 
I don’t see the issue then. The total d/dx has to be decomposed using the chain rule.

It makes no sense to talk about a total derivative wrt y.
 
Orodruin said:
I don’t see the issue then. The total d/dx has to be decomposed using the chain rule.

It makes no sense to talk about a total derivative wrt y.
my lecturer uses the notation ##dF/dy## in the second last term. I am confused why it is not ##\partial F/\partial y## instead.
 
Yeah ok, that’s just sloppy.
 
  • Like
Likes   Reactions: laser1

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
21
Views
2K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
860
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K