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What does the central thm of calculus of variation says?

  1. Aug 3, 2006 #1

    quasar987

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    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 [itex]\Leftrightarrow[/itex] 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. jcsd
  3. Aug 4, 2006 #2

    HallsofIvy

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    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).
     
  4. Aug 4, 2006 #3

    quasar987

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    sorry, replace the word "extremize" by "criticalize" (i.e. is a critical point)
     
  5. Aug 4, 2006 #4

    quasar987

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    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. [itex]\delta J/\delta f(x)=0[/itex]) iff it satisfies the E-L equations.
     
  6. Aug 4, 2006 #5

    HallsofIvy

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    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.
     
  7. Aug 5, 2006 #6

    quasar987

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

    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?
     
  8. Aug 5, 2006 #7

    HallsofIvy

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    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 [itex]\delta J/\delta f(x)[/itex]= 0 then it is not trivial. It does, of course, follow from the derivation of the E-L equation, which is far from trivial!
     
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