Stationary points

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5.2 Stationary points

For a function we speak of a stationary point when the function doesn’t change under a small change, i.e., if we take x → x + δx, and thus f → f + δf, the change in f, δf = 0, to first order in δx. This leads to the obvious relation {df\over dx} = 0.

As for functions, we are extremely interested in stationary points of functionals:

A functional has a stationary point for any function y such that a small change y(x) → y(x) + ϵ(x) leads to no change in I[y] to first order in ϵ(x).

A key difference with the stationary points of functions is that smallness of ϵ(x) only implies that as a function it is everywhere (mathematician would say “uniformly”) close to zero, but can still vary arbitrarily.

An important class of functionals is given by

I[y] ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}dx\kern 1.66702pt F(y(x),y'(x),x)\quad ,

(5.1)

where a, b are fixed, and y is specified at the boundaries, i.e., the values of y(a) and y(b) are specified as boundary conditions. Thus under a (small) change y(x) → y(x) + ϵ(x), the preservation of the boundary conditions implies

ϵ(a) = ϵ(b) = 0.

(5.2)

Now substitute and expand to first order in ϵ(x),

\eqalignno{ δI[y] & = I[y + ϵ] − I[y] & & \cr & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}dx\kern 1.66702pt F(y(x) + ϵ(x),y'(x) + ϵ'(x),x) −{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}dx\kern 1.66702pt F(y(x),y'(x),x) & & \cr & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}dx\kern 1.66702pt \left ( {∂F\over ∂y(x)}ϵ(x) + {∂F\over ∂y'(x)}ϵ'(x)\right ) & & \cr & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}dx\kern 1.66702pt \left ( {∂F\over ∂y(x)}ϵ(x) − {d\over dx} {∂F\over ∂y'(x)}ϵ(x)\right ) & & \cr & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{a}^{b}dx\kern 1.66702pt ϵ(x)\left [ {∂F\over ∂y(x)} − {d\over dx} {∂F\over ∂y'(x)}\right ] & & }

where we have integrated by parts to obtain the penultimate line, using the boundary conditions on ϵ.

Since ϵ(x) is allowed to vary arbitrarily, this is only zero if the quantity multiplying ϵ is zero at every point x. For example, you can choose a set of ϵ’s that are all peaked around a particular value of x. We thus see that the term proportional to ϵ vanishes at each point x, and we get the Euler-Lagrange equation

{∂F\over ∂y(x)} − {d\over dx} {∂F\over ∂y'(x)} = 0.

(5.3)

Remarks

we often don’t write the dependence on x, but it is implicitly assumed that you understand that this is a “functional equation”.
We use the functional derivative notation ${δI[y]\over δy}$
for the term in the functional I[y + δy] proportional to δy,
$I[y + δy] = I[y] +\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dx\kern 1.66702pt δy(x)\kern 1.66702pt {δI[y]\over δy} (x) + \mathop{\mathop{…}}\quad .$
If the functional is not an integral, e.g., $I[y] = y(0)\quad ,$
we can turn it into an integral by adding a delta function. In the case above,
$I[y] =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits y(x)δ(x)\kern 1.66702pt dx.$
The notations {δI[y]\over δy} (x) and {δI[y]\over δy(x)} are used interchangeably.

For a general functional, the equation

{δI[y]\over δy} = 0

is called the Euler-Lagrange equation.

Solutions to this equation define stationary points of the functional.

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