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Definition/Summary
Also known as the Euler equation. It is the solution to finding an extrema of a functional in the form of
v(y)=\int_{x_{1}}^{x_{2}} F(x,y,y') dx \ .
The solution usually simplifies to a second order differential equation.
Equations
F_{y}-D_{x}F_{y'} \ = \ 0
or
\frac{\partial F}{\partial y}<br /> \ - \ \frac{\mathrm{d} }{\mathrm{d} x} \ \frac{\partial F}{\partial y'} \ = \ 0
Extended explanation
PROOF
Let us find the extrema of the functional
v(y)=\int_{x_{1}}^{x_{2}}F(x,y,y')dx \ .
Such a functional could be arc length, for example. For the variation of v,
\delta v = \frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} \ ,
let Δy be an arbitrary differentiable function such that Δy(x1)=Δy(x2)=0.
Now, to find the extrema, the variation must be zero. i.e.
\delta v=\frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} = 0
or
\frac{\partial }{\partial a}\int_{x_{1}}^{x_{2}} F(x,y+a\Delta y,y'+a\Delta y')dx|_{a=0}=0 \ .
Using the chain rule of multiple variables, this simplifies to
\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 \ .
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: F_{y}=\frac{\partial F}{\partial y})
\int_{x_{1}}^{x_{2}} (\Delta y F_{y}+\Delta y'F_{y'})dx
Using integration by parts on the right side with "u"=Fy' and "dv"=Δy'dx:
\int_{x_{1}}^{x_{2}}\Delta yF_{y} dx \ <br /> + \ [\Delta yF_{y'}]_{x_{1}}^{x_{2}} \ <br /> - \ \int_{x_{1}}^{x_{2}}\Delta y\frac{\mathrm{d} F_{y'}}{\mathrm{d} x} \ dx \ .
However, Δy(x1)=Δy(x2)=0. Thus the middle term is zero, so:
\int_{x_{1}}^{x_{2}}\Delta y(F_{y}-\frac{\partial }{\partial x}F_{y'})dx=0 \ .
Applying the Fundamental Lemma of Calculus of Variations, we find
F_{y}-\frac{\partial}{\partial x}F_{y'}=0 \ .
Or, more compactly,
F_{y}-D_{x}F_{y'} = 0 \ ,
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!
Also known as the Euler equation. It is the solution to finding an extrema of a functional in the form of
v(y)=\int_{x_{1}}^{x_{2}} F(x,y,y') dx \ .
The solution usually simplifies to a second order differential equation.
Equations
F_{y}-D_{x}F_{y'} \ = \ 0
or
\frac{\partial F}{\partial y}<br /> \ - \ \frac{\mathrm{d} }{\mathrm{d} x} \ \frac{\partial F}{\partial y'} \ = \ 0
Extended explanation
PROOF
Let us find the extrema of the functional
v(y)=\int_{x_{1}}^{x_{2}}F(x,y,y')dx \ .
Such a functional could be arc length, for example. For the variation of v,
\delta v = \frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} \ ,
let Δy be an arbitrary differentiable function such that Δy(x1)=Δy(x2)=0.
Now, to find the extrema, the variation must be zero. i.e.
\delta v=\frac{\partial }{\partial a}v(y+a\Delta y)|_{a=0} = 0
or
\frac{\partial }{\partial a}\int_{x_{1}}^{x_{2}} F(x,y+a\Delta y,y'+a\Delta y')dx|_{a=0}=0 \ .
Using the chain rule of multiple variables, this simplifies to
\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 \ .
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: F_{y}=\frac{\partial F}{\partial y})
\int_{x_{1}}^{x_{2}} (\Delta y F_{y}+\Delta y'F_{y'})dx
Using integration by parts on the right side with "u"=Fy' and "dv"=Δy'dx:
\int_{x_{1}}^{x_{2}}\Delta yF_{y} dx \ <br /> + \ [\Delta yF_{y'}]_{x_{1}}^{x_{2}} \ <br /> - \ \int_{x_{1}}^{x_{2}}\Delta y\frac{\mathrm{d} F_{y'}}{\mathrm{d} x} \ dx \ .
However, Δy(x1)=Δy(x2)=0. Thus the middle term is zero, so:
\int_{x_{1}}^{x_{2}}\Delta y(F_{y}-\frac{\partial }{\partial x}F_{y'})dx=0 \ .
Applying the Fundamental Lemma of Calculus of Variations, we find
F_{y}-\frac{\partial}{\partial x}F_{y'}=0 \ .
Or, more compactly,
F_{y}-D_{x}F_{y'} = 0 \ ,
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!