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What is the Euler-Lagrange equation

  1. Jul 23, 2014 #1

    Also known as the Euler equation. It is the solution to finding an extrema of a functional in the form of

    [tex]v(y)=\int_{x_{1}}^{x_{2}} F(x,y,y') dx \ .[/tex]

    The solution usually simplifies to a second order differential equation.


    [tex]F_{y}-D_{x}F_{y'} \ = \ 0[/tex]


    [tex]\frac{\partial F}{\partial y}
    \ - \ \frac{\mathrm{d} }{\mathrm{d} x} \ \frac{\partial F}{\partial y'} \ = \ 0[/tex]

    Extended explanation


    Let us find the extrema of the functional

    [tex]v(y)=\int_{x_{1}}^{x_{2}}F(x,y,y')dx \ .[/tex]

    Such a functional could be arc length, for example. For the variation of v,

    [tex]\delta v = \frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} \ ,[/tex]

    let Δy be an arbitrary differentiable function such that Δy(x1)y(x2)=0.

    Now, to find the extrema, the variation must be zero. i.e.

    [tex]\delta v=\frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} = 0[/tex]


    [tex]\frac{\partial }{\partial a}\int_{x_{1}}^{x_{2}} F(x,y+a\Delta y,y'+a\Delta y')dx|_{a=0}=0 \ .[/tex]

    Using the chain rule of multiple variables, this simplifies to

    [tex]\int_{x_{1}}^{x_{2}} (\frac{\partial F}{\partial y}\frac{\mathrm{d} (y+a\Delta y)}{\mathrm{d} a}+\frac{\partial F}{\partial y'}\frac{\mathrm{d}(y'+a\Delta y' ) }{\mathrm{d} a}) dx \ .[/tex]

    We then split d(y+aΔy) and d(y'+aΔy') into dy+Δyda and dy'+Δy'da respectively. Remember that y and y' is independent of a, and da/da=1. We therefore get (using different notation: [itex]F_{y}=\frac{\partial F}{\partial y}[/itex])

    [tex]\int_{x_{1}}^{x_{2}} (\Delta y F_{y}+\Delta y'F_{y'})dx[/tex]

    Using integration by parts on the right side with "u"=Fy' and "dv"=Δy'dx:

    [tex]\int_{x_{1}}^{x_{2}}\Delta yF_{y} dx \
    + \ [\Delta yF_{y'}]_{x_{1}}^{x_{2}} \
    - \ \int_{x_{1}}^{x_{2}}\Delta y\frac{\mathrm{d} F_{y'}}{\mathrm{d} x} \ dx \ .[/tex]

    However, Δy(x1)=Δy(x2)=0. Thus the middle term is zero, so:

    [tex]\int_{x_{1}}^{x_{2}}\Delta y(F_{y}-\frac{\partial }{\partial x}F_{y'})dx=0 \ .[/tex]

    Applying the Fundamental Lemma of Calculus of Variations, we find

    [tex]F_{y}-\frac{\partial}{\partial x}F_{y'}=0 \ .[/tex]

    Or, more compactly,

    [tex]F_{y}-D_{x}F_{y'} = 0 \ ,[/tex]

    where Dx is the differential operator with respect to x.
    This is a second order differential equation which, when solved, gives the desired extrema of the functional.

    * This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
  2. jcsd
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