Euler-Lagrange Equation
Based on theoretical knowledge of functionals, their extremums and calculus of variations (method of finding these extremums using directional derivatives), let's see what is the result of application of this method in many cases occurring in Physics.
The spectrum of functions, that these functionals are defined on, in many cases is reduced to smooth (sufficiently differentiable) real functions defined on some segments [a,b] with fixed values at the ends of this segment:
f(a)=A; f(b)=B.
For examples, to find the best in some sense trajectory we have to know the starting and ending points of this trajectory and search for the best one among functions with fixed values at the beginning (start) point and at the ending (finish) point of movement.
Many problems in Physics are related to finding extremums of specific type of functionals of the above mentioned functions:
Φ[f(x)] = ∫[a,b] F[x,f(x),f '(x)]dx
where F[...] is some known smooth real function of three arguments - real variable x, real value of function f(x) and real value of derivative f '(x).
This function F[...] is derived from known laws of Physics.
The function f(x), the argument to a functional Φ[f(x)], is the function from a class of smooth functions defined on some segment with fixed values at the ends described above.
Our task is to find function f0(x) where functional Φ[f(x)] has an extremum.
The plan is:
1. Assuming f0(x) is a point where Φ[f(x)] reaches its extremum, increment this function by some Δ(x) to f1(x)=f0(x)+Δ(x), keeping in mind that f1(x) should belong to the same class as f0(x), that is it should be smooth, defined on the same segment [a,b] and have the same values A and B at the ends of this segment, which means that Δ(x) should be smooth, defined on the same [a,b] and be equal to zero at the ends of this segment.
2. Consider all functions of type g(x,t)=f0(x)+t·Δ(x).
This set of functions is parameterized by parameter t and g(x,0)=f0(x).
This set of functions can be considered as filling some neighborhood of function f0(x) on a "line" from f0(x) in a direction defined by increment Δ(x).
3. Since f0(x) is a point where functional Φ[f(x)] has an extremum, changing the value of t closer and closer to zero would result in moving Φ[g(x,t)] closer and closer to Φ[f0(x)], which is an extremum.
Therefore, a derivative of Φ[g(x,t)] by t at t=0 (that is, when g(x,t)=g(x,0)=f0(x)) should be equal to zero.
That gives an equation where f0(x) and Δ(x) participate.
4. Since increment Δ(x) can be chosen relatively freely (as long as it's a smooth function with values of zero on both ends of [a,b]), the equation obtained at step 3 above should be true for any Δ(x), which gives additional condition that might lead to identification of f0(x).
Let's follow the plan step by step.
1. f1(x)=f0(x)+Δ(x)
Δ(a)=Δ(b)=0
2. g(x,t)=f0(x)+t·Δ(x)
Φ[g(x,t)] =
= ∫[a,b] F[x,g(x,t),g'(x,t)]dx
(apostrophe is a derivative by variable x)
where g(x,t)=f0(x)+t·Δ(x).
Now the functional Φ[g(x,t)] can be considered as a function of one real argument t that characterizes how close function g(x,t) is to an assumed point of extremum f0(x).
3. The derivative of functional Φ[g(x,t)] considered as a function of one real argument t by parameter t must be equal to zero at t=0 and g(x,0)=f0(x):
(d/dt)Φ[g(x,t)]|t=0 = 0
For many practical problems in Physics functional Φ[g(x,t)] is an integral by x presented in the beginning, while differentiation above is by t.
These two variables (x and t) and corresponding operations (integration by x and differentiation by t) are independent. If, instead of integration by x we had a sum by some index i, we would not hesitate to replace a derivative of a sum with a sum of derivatives.
Integration is just a sum of infinitesimal parts and the same rule of interchanging the order of operations can be applied.
Therefore,
(d/dt)Φ[g(x,t)] =
= (d/dt)∫[a,b] F[x,g(x,t),g'(x,t)]dx =
= ∫[a,b] (d/dt)F[x,g(x,t),g'(x,t)]dx
Since argument of differentiation t is contained inside a function F[...] of multiple arguments, the derivative by t should be taken using partial derivative by each argument (∂F/∂x, ∂F/∂g, ∂F/∂g') multiplied by an inner derivative of this argument by t (correspondingly, dx/dt, dg/dt, dg'/dt): (d/dt)F[x,g(x,t),g'(x,t)] =
= (∂/∂x)F[x,g(x,t),g'(x,t)]·(dx/dt) +
+ (∂/∂g)F[x,g(x,t),g'(x,t)]·(dg/dt) +
+ (∂/∂g')F[x,g(x,t),g'(x,t)]·(dg'/dt)
Since the first argument of function F[...] is just x and does not depend on t, its derivative by t is zero:
dx/dt = 0
Since g(x,t)=f0(x)+t·Δ(x), its derivative by t is
dg(x,t)/dt = Δ(x)
Since g'(x,t)=f '0(x)+t·Δ'(x),
dg'(x,t)/dt = Δ'(x)
Now the derivative by t looks like
(d/dt)F[x,g(x,t),g'(x,t)] =
= (∂/∂g)F[x,g(x,t),g'(x,t)]·Δ(x) +
+ (∂/∂g')F[x,g(x,t),g'(x,t)]·Δ'(x)
Therefore, the expression for a derivative of our functional Φ[g(x,t)] by parameter t is:
(d/dt)Φ[g(x,t)] =
= ∫[a,b] (d/dt)F[x,g(x,t),g'(x,t)]dx =
=∫[a,b] (∂/∂g)F[x,g(x,t),g'(x,t)]·Δ(x)dx+
+∫[a,b] (∂/∂g')F[x,g(x,t),g'(x,t)]·Δ'(x)dx
As mentioned above, the condition
(d/dt)Φ[g(x,t)]|t=0 = 0
is necessary for f0(x)=g(x,0) to be a function-argument where our functional reaches its extremum.
Substituting t=0 and changing g(x,0) to f0(x), we have an equation
0 = (d/dt)Φ[f0(x)] =
=∫[a,b] (∂/∂f)F[x,f0(x),f '0(x)]·Δ(x)dx+
+∫[a,b] (∂/∂f ')F[x,f0(x),f '0(x)]·Δ'(x)dx
To shorten the formulas, let's replace
(∂/∂f)F[x,f0(x),f '0(x)]
with
F∂f [x,f0(x),f '0(x)]
and
(∂/∂f ')F[x,f0(x),f '0(x)]
with
F∂f ' [x,f0(x),f '0(x)]
With this substitution the equation above looks like
0 = (d/dt)Φ[f0(x)] =
=∫[a,b] F∂f [x,f0(x),f '0(x)]·Δ(x)dx+
+∫[a,b] F∂f ' [x,f0(x),f '0(x)]·Δ'(x)dx
If only the first integral in the above expression participated in the equation, the requirement that this integral should be equal to zero regardless of Δ(x) would cause the smooth function under an integral to be zero everywhere on [a,b].
Existence of the second integral complicates the picture, but we can change the second integral to contain only Δ(x) instead of Δ'(x) by using the formula of integrating by parts for two functions u(x) and v(x):
∫[a,b]u·dv = u·v|[a,b] − ∫[a,b]v·du
Let's apply this formula for the second integral in the equation above, substituting
u(x) = F∂f '[x,f0(x),f '0(x)]
v(x) = Δ(x)
Then
du(x) = u'(x)·dx =
= (d/dx)F∂f '[x,f0(x),f '0(x)]·dx
dv(x) = v'(x)·dx = Δ'(x)·dx
Using these substitutions we transform the second integral in the above equation as follows
∫[a,b] F∂f ' [x,f0(x),f '0(x)]·Δ'(x)dx =
= ∫[a,b] F∂f ' [x,f0(x),f '0(x)]·dΔ(x) =
= F∂f ' [x,f0(x),f '0(x)]·Δ(x)|[a,b] −
− ∫[a,b] Δ(x)·dF∂f ' [x,f0(x),f '0(x)] =
The first component of the above expression equals to zero:
F∂f ' [x,f0(x),f '0(x)]·Δ(x)|[a,b] = 0
because Δ(a)=Δ(b)=0
So, the final expression for a variation of our functional contains two integrals. both with Δ(x) as a factor:
0 = (d/dt)Φ[f0(x)] =
=∫[a,b] F∂f [x,f0(x),f '0(x)]·Δ(x)dx−
−∫[a,b] Δ(x)·dF∂f ' [x,f0(x),f '0(x)] =
= ∫[a,b] h(x)·Δ(x)·dx
where
h(x) = F∂f [x,f0(x),f '0(x)] −
− (d/dx)F∂f ' [x,f0(x),f '0(x)]
Since the last integral must be equal to zero for any Δ(x), function h(x) must be equal to zero for any x∈[a,b].
Therefore, the necessary condition for function-argument f0(x) to be a point where functional Φ[f(x)] reaches its extremum is that f0(x) is a solution to a differential equation h(x)=0 or, in terms of original functional Φ[f(x)],
F∂f [x,f0(x),f '0(x)] −
− (d/dx)F∂f ' [x,f0(x),f '0(x)] = 0
CONCLUSION
Consider a class Ω of all sufficiently differentiable real functions f(x) on segment [a,b] that take fixed values on the ends of this segment:
f(a)=A and f(b)=B.
Given functional
Φ[f(x)] = ∫[a,b] F[x,f(x),f '(x)]dx
defined for all f(x)∈Ω
and where F[...] is some known sufficiently differentiable real function of three arguments
real variable x∈[a,b],
real value of function f(x)∈Ω
and real value of its derivative f '(x).
If function f0(x) from the same class Ω is where the above functional reaches its extremum (minimum or maximum) than this function should be a solution to Euler-Lagrange differential equation
(∂/∂f)F [x,f0(x),f '0(x)] −
− (d/dx)(∂/∂f ')F [x,f0(x),f '0(x)] = 0
