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Euler Lagrange Equation

  1. Jan 19, 2008 #1
    [SOLVED] Euler Lagrange Equation

    Hi there ,
    I am missing a crucial point on the proof of Euler Lagrange equation , here is my question :
    [tex]\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{df}{dy^{'}}\right)=0[/tex] (Euler-Lagrange equation)

    If the function "f" doesn't depend on x explicitly but implicitly and if y satisfies the Euler-Lagrange equation then ;
    [tex]\frac{\partial f}{\partial x}=0[/tex]

    Why is that so ? While ,supposedly , f is dependent to 3 variables : x,y,y' how van that statement be true ?
     
    Last edited: Jan 19, 2008
  2. jcsd
  3. Jan 19, 2008 #2
    If the function [tex]f[/tex] doesn't depend on x explicitly then

    [tex]\frac{\partial f}{\partial x}=0[/tex]

    and this has nothing to do with the

    I don't understand what is the question :confused:
     
  4. Jan 19, 2008 #3
    Still , the function f does depend on x through y and y' . That is why I asked basically .
     
  5. Jan 19, 2008 #4
    That's why we use partial differention. Let

    [tex]f\left(x,y(x),y'(x)\right)=x^2\,e^{y(x)}\,\ln{y'(x)}+\sin\left(x\,y(x)\right)[/tex]

    then [itex]\frac{\partial\,f}{\partial\,x}[/itex] means

    [tex]\frac{\partial\,f}{\partial\,x}=2\,x\,e^{y(x)}\,\ln{y'(x)}+y(x)\,\cos\left(x\,y(x)\right)[/tex]

    i.e. you treat [tex]x,y(x),y'(x)[/tex] as independent variables.
     
  6. Jan 20, 2008 #5
    Thanks for helping out!
     
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