# What does the central thm of calculus of variation says?

1. Aug 3, 2006

### quasar987

Based on the proof given in the action article of wiki (http://en.wikipedia.org/wiki/Action_(physics)#Euler-Lagrange_equations_for_the_action_integral), it would seems that the statement of the "central thm of calculus of variation"(http://en.wikipedia.org/wiki/Euler-Lagrange_equations#Statement) is in fact an $\Leftrightarrow$ one. I.e. not just "IF action is extremized, THEN L satisfies E-L", but rather "Action is extremized, iff L satisfies E-L"

someone can confirm?

2. Aug 4, 2006

### HallsofIvy

Staff Emeritus
Neither of those statements have anything to do with calculus of variations because "action" is a physics term, not a mathematics term! Please learn the difference between the two.

In any case iff is not correct. It is quite possible that E-L is satisfied while the integral does not have an extremum, in exactly the same way that a function of a single variable may have derivative 0 where it does not have an extremum (f(x)= x3 for example).

3. Aug 4, 2006

### quasar987

sorry, replace the word "extremize" by "criticalize" (i.e. is a critical point)

4. Aug 4, 2006

### quasar987

Aaah, I understand everything! A correct statement would be

Given a functional L(f(x),f'(x),x) with continuous first partial derivatives, a function f "criticalizes" the cost functional (i.e. $\delta J/\delta f(x)=0$) iff it satisfies the E-L equations.

5. Aug 4, 2006

### HallsofIvy

Staff Emeritus
What is the definition of "criticalize"? The definition of "critical point" in calculus is "the derivative either is 0 or does not exist at that point". If you are using an equivalent definition for functionals, then what you are saying is trivial.

6. Aug 5, 2006

### quasar987

See post #4. By criticalize, I mean $\delta J/\delta f(x)=0$.

How trivial is it? Do you mean it in a "Of course it's true, stop wasting my time" way, or in a "Well it's almost the definition" way?

7. Aug 5, 2006

### HallsofIvy

Staff Emeritus
Okay, I'll concede. I was afraid you were using something like "the E-L equations" are true (the direct analog of "df/dx= 0 or does not exist" as a definition of critical point) which I am sure you would agree would make the statement trivial. If you are using $\delta J/\delta f(x)$= 0 then it is not trivial. It does, of course, follow from the derivation of the E-L equation, which is far from trivial!