Physics+ Pendulum in Lagrangian Mechanics: UNIZOR.COM - Physics+ 4 All -...

Notes to a video lecture on UNIZOR.COM

Mathematical Pendulum

Plain Pendulum

We will illustrate the application of Lagrangian mechanics by analyzing the movement of a mathematical pendulum - a problem we have already discussed in the Physics 4 Teens - Mechanics - Pendulum, Spring - Pendulum using the Newtonian approach.

We recommend reviewing the lecture mentioned above and refresh the Newtonian method of deriving the main equation of motion of the pendulum:
α"(t) = −(g/l)·sin(α(t))
which was obtained from properly determining the force F that moves a pendulum as a vector sum of the gravity force directed vertically down and the tension of an unstretchable thread that keeps an object at the free end of a thread on a constant distance from the fixed end of a thread.

The two forces involved in formation of a resulting force F, gravity P=m·g and tension of a thread T, had to be combined using the rules for addition of vectors, which required some thinking.

Let's apply the Lagrangian mechanics to this problem using an angle of a thread with a vertical α as the one and only parameter that determines a position of an object.
This way of identification of a position is more convenient than Cartesian coordinates originated at the fixed end of a thread because both of them can be easily derived from α
x = l·sin(α)
y = −l·cos(α)

The Lagrangian is the difference between kinetic and potential energies.
L(α(t),α'(t)) =
= E_kin(α'(t)) − E_pot(α(t))

Kinetic energy depends on mass M and linear speed of an object along its circular trajectory v=l·α'(t)
E_kin = ½M·v² =
= ½M·l²·[α'(t)]²

Potential energy depends on a mass of an object M, its height over the ground h(t) and an acceleration of free fall g.
E_pot(t) = M·g·h(t)
If the origin of our coordinates, the fixed end of a thread, is at height H over the ground,
h = H − l·cos(α)
and, therefore,
E_pot(t) = M·g·[H−l·cos(α(t))]

Now we can construct the Euler-Lagrange equation
(∂/∂α)L(α(t),α'(t)) =
= (d/dt)(∂/∂α')L(α(t),α'(t))

Calculate left and right sides separately.
(∂/∂α)L(α(t),α'(t)) =
= (∂/∂α)[E_kin(α'(t)) − E_pot(α(t))] =
= (∂/∂α)[−E_pot(α(t))] =
= (∂/∂α)[−M·g·[H−l·cos(α(t))]] =
= −M·g·l·sin(α(t))

(d/dt)(∂/∂α')L(α(t),α'(t)) =
=(d/dt)(∂/∂α')[E_kin(α'(t))−E_pot(α(t))]=
= (d/dt)(∂/∂α')E_kin(α'(t)) =
= (d/dt)(∂/∂α')½M·l²·[α'(t)]² =
= (d/dt)M·l²·α'(t) =
= M·l²·α"(t)

The Euler-Lagrange equation is
−M·g·l·sin(α(t)) = M·l²·α"(t)
or
α"(t) = −(g/l)·sin(α(t))
which is exactly as applying Newtonian mechanics.
If you don't think the Lagrangian mechanics is simpler than Newtonian for those who are familiar with Calculus of partial derivatives, consider the next example.


Spring Pendulum

A weightless spring replaces an unstretchable thread of the previous problem.
The spring and an object on its end are in a weightless frictionless tube that maintains a straight form, so an object has two degrees of freedom - radial inside a tube stretching and squeezing a spring and pseudo-circular as it moves together with a tube in a pendulum like motion.
The problem of specifying the motion of an object is much more complex here because the spring tension is changing not only with an angle α(t) but also because of the movement of an object within a tube.

However, using the Langrangian mechanics, this problem can be analyzed with much less efforts and the corresponding differential equation can be constructed relatively easy.

As before, let's calculate the kinetic and potential energies of an object.

The object's kinetic energy can be calculated as a sum of its radial movement's kinetic energy inside the tube and kinetic energy of its pseudo-circular movement perpendicularly to the tube.
The reason is simple. The object's linear velocity vector can be represented as a sum of two perpendicular to each other vectors, one is inside the tube and another perpendicular to it.
v = v_|| + v_⊥
Since kinetic energy depends on a square of the linear speed, according to Pythagorean Theorem
v² = v_||² + v_⊥²
from which follows that
E_kin = ½M·v² =
= ½M·v_||² + ½M·v_⊥²

Distance of an object from the fixed point of oscillation l is variable and depends on time: l=l(t).
Therefore, v_||(t)=l'(t).
Perpendicular to l component of an object speed is v_⊥(t)=l(t)·α'(t).
Therefore, kinetic energy of an object is
E_kin = ½M·[l'(t)²+l(t)²·α'(t)²]

The object's potential energy can be calculated as a sum of its potential energy due to gravity and potential energy of a stretched or a squeezed spring.
If the fixed point of oscillation is at the height H above the ground, an object is at height h(t)=H−l(t)·cos(α(t)) above the ground, and potential energy of an object related to its position in the gravitational field is
E_grav = M·g·[H−l(t)·cos(α(t))]

Potential energy of a spring depends on the degree of its stretching or squeezing.
Assume, the length of a spring in a neutral state is l₀. Then the length of stretching or squeezing at time t is l(t)−l₀.
Therefore, potential energy of an object related to a spring is
E_spr = ½k·[l(t)−l₀]²
where k is a coefficient of elasticity of a spring.

Total potential energy of an object is
E_pot = E_grav + E_spr =
= M·g·[H−l(t)·cos(α(t))] +
+ ½k·[l(t)−l₀]²

Construct Lagrangian
L(l,l',α,α') = E_kin − E_pot =
= ½M·[l'(t)²+l(t)²·α'(t)²] −
− M·g·[H−l(t)·cos(α(t))] −
− ½k·[l(t)−l₀]²

All which remains is to write the Euler-Lagrange equation for this Lagrangian.

The problem is, we are familiar only with the Euler-Lagrange equation for a system with one degree of freedom, like x(t). Now we have two degrees of freedom - l(t) and α(t).

Fortunately, the Euler-Lagrange equation can be specified for each degree of freedom independently, which will be proven in the next lecture.

Therefore, we can write two independent Euler-Lagrange equations (skipping (t) for brevity):
(∂/∂l)L(l,l',α,α') =
= (d/dt)(∂/∂l')L(l,l',α,α')
and
(∂/∂α)L(l,l',α,α') =
= (d/dt)(∂/∂α')L(l,l',α,α')

The first equation is
M·l·α'²+M·g·cos(α)−k·(l−l₀) =
= M·l"

The second equation is
−M·g·l·sin(α) = M·l²·α"
or
−g·sin(α) = l·α"
or
−(g/l)·sin(α) = α"
which looks exactly the same as in the case above with unstretchable thread instead of a spring.

Physics+ Lagrangian: UNIZOR.COM - Physics+ 4 All - Lagrangian

Notes to a video lecture on UNIZOR.COM

Lagrangian

First of all, let's stipulate that the Laws of Newton are based on experiment, they are not derived from some more fundamental theories.

Lagrangian mechanics presents a different approach to analyze the motion than Newtonian mechanics.
In many cases it presents a simpler, more universal way to describe the motion of a mechanical system than Newtonian one.

Let's start with an example where both methodologies lead to the same result.

Spring Oscillation

Consider an ideal spring with one end fixed and a point-mass attached to another end.
The oscillations will occur along the length of a spring that coincides with X-axis.
Position of a point-mass on the spring's end will be described by it X-coordinate x(t) as a function of time t with initial position at time t=0 being an origin of X-coordinate, that is x(0)=0.


According to the Hooke's Law, the force F of a spring applied to an object attached to its end is proportional to a length x by which this spring is stretched or squeezed from its neutral position and directed always towards a neutral point x=0 of no stretching or squeezing.
F = −k·x
where k is a coefficient of elasticity that characterizes physical properties of a spring.

According to the Newton's Second Law, the acceleration a of an object is proportional to a force F applied to it
F = m·a
where m is the object's mass being a coefficient of proportionality.

A linear acceleration a(t), as a function of time, is a derivative of a linear speed v(t) by time t
a(t) = dv(t)/dt = v'(t)
A linear speed v(t) is, in turn, a derivative of a position of an object x(t) by time
v(t) = dx(t)/dt = x'(t)
Therefore, an acceleration is a second derivative of a position by time
a(t) = d²x(t)/dt² = x"(t)

Equating the value of force by Hooke's Law to that of Newton's Second Law, we get a differential equation that defines a motion of the object
−k·x(t) = m·a(t)
or, equivalently,
−k·x(t) = m·x"(t)
Solution to this differential equation of the second order is a trajectory of our object.

Let's approach the same problem from another side.

An object attached to a spring's end that has mass m and linear speed v has kinetic energy that is equal to
E_kin = ½m·v²
Since speed v(t), as a function of time t is just a derivative of a position x(t) by time, we can express kinetic energy in terms of position, as in the case of potential energy above
E_kin = ½m·[x'(t)]²
NOTICE:
E_kin depends explicitly only on speed x'(t) and
d/dt[∂(E_kin)/∂x'] =
= d/dt[m·x'] = mx"(t) = F

A stretched or a squeezed spring has potential energy equal to the amount of work needed to stretch or squeeze it against the force of its elasticity (you can refer to a lecture Physics 4 Teens - Energy - Potential Energy - Spring on UNIZOR.COM).
Thus, a potential energy of a spring squeezed or stretched by the length x(t), as a function of time t, equals to
E_pot = ½k·[x(t)]²
where k is the same coefficient of elasticity as above that characterizes the physical properties of a spring.
NOTICE:
E_pot depends explicitly only on position x(t) and
∂(−E_pot)/∂x = −k·x(t) = F

Based on two NOTICEs above, it is IMPORTANT to see that
∂(−E_pot)/∂x =
= d/dt[∂(E_kin)/∂x'] = F

Since E_pot depends explicitly only on position x(t), not on speed x'(t), and E_kin depends explicitly only on speed x'(t), not on position x(t),
∂(E_kin−E_pot)/∂x =
= ∂(−E_pot)/∂x =
= d/dt[∂(E_kin)/∂x'] =
= d/dt[∂(E_kin−E_pot)/∂x']

At this point it's essential to recall the Euler-Lagrange equation (you can refer to a lecture Physics+ 4 All - Variations - Euler-Lagrange on UNIZOR.COM) - a differential equation of the second order that defines a function f₀(x) that minimizes or maximizes a functional
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 Euler-Lagrange differential equation looks like this:
(∂/∂f)F [x,f₀(x),f '₀(x)] =
= (d/dx)(∂/∂f ')F [x,f₀(x),f '₀(x)]

Let's change more abstract symbols x and f(x) to those applicable to our task.
The argument will be time t instead of abstract x. The function will be a position x(t) instead of abstract f₀(x).
Now the Euler-Lagrange equation looks like
(∂/∂x)F [t,x(t),x'(t)] =
= (d/dt)(∂/∂x')F [t,x(t),x'(t)]

Compare this to the equation above that equates partial derivative from E_kin−E_pot by x with its partial derivative by x'.

Obviously, L=E_kin−E_pot satisfies the Euler-Lagrange equation
(∂/∂x)L[t,x(t),x'(t)] =
= (d/dt)(∂/∂x')L[t,x(t),x'(t)]
which in this case is exactly the same as the equation obtained from the Newton's Second Law
−k·x(t) = m·x"(t)
Expression
L[x(t),x'(t)]=E_kin(x')−E_pot(x)
is called Lagrangian.

Consider an object moving along some trajectory x(t) from the moment of time t₁ to the moment of time t₂.
At any moment it has certain kinetic and potential energy, so we can constract a Lagrangian
L(t) = E_kin(x'(t)) − E_pot(x(t))

Consider an integral of this Lagrangian by time
S = ∫_[t1,t2]L(t)·dt
This integral is call action.
The trajectory that minimizes or maximizes this action is a solution to an Euler-Lagrange equation
(∂/∂x)L[x(t),x'(t)] =
= (d/dt)(∂/∂x')L[t,x(t),x'(t)]
which has the same solution as Newtonian F=m·a.

Therefore, the trajectory obtained using a Lagrangian approach is the same the one from Newtonian mechanics. BUT IN SOME CASES IT MIGHT BE MUCH MORE CONVENIENT.

The equivalence of a differential equation obtained from the Newton's Second Law and the Euler-Lagrange equation is not just a coincidence peculiar for springs.

In general, kinetic energy always depends on mass and speed
E_kin = ½m·v²
In general, derivative of E_kin by speed v is a momentum of motion p
p = m·v = ∂/∂v(E_kin) =
= ∂/∂v(½mv²)
In general, derivative of momentum p by time t is the force F
dp/dt = d(m·v)/dt = m·a = F
In general, potential energy is, actually, an amount of work W.
Since dW=F·dx, its derivative by x is the force, and a derivative of potential energy by coordinates gives the force as well.

So, the Newton's Second Law and Euler-Lagrange equation are equivalent. Why do we need both?
Practical mechanical problems are rarely as simple as we are taught at high school.
It appears that the more complicated problems with more than one object involved are easier to solve using Lagrangian L=E_kin−E_pot than to deal with complicated forces and their interaction constructing the equations of F=m·a type.

The next lectures will be dedicated to a few important physical problems and their solutions using Euler-Lagrange equation.

Physics+ Brachistochrone: UNIZOR.COM - Physics+ 4 All - Variations

Notes to a video lecture on UNIZOR.COM

Brachistochrone

The approach to choose a path along which any system progresses (light propagates, planet moves around the Sun etc.) based on minimizing some numeric function defined for each path appears to be very valuable in Physics, and it helps to solve certain tasks faster and more efficiently than using only the classic Newton's Laws.

Before generalizing this idea, let's consider a specific problem suggested by Johann Bernoulli in 1696.
It's called the Brachistochrone problem (from Greek 'brachistos' + 'chronus' = 'short' + 'time') and is formulated as follows.

Consider two points A and B in the uniform gravitational field (like near the surface of the Earth) with force of gravity directed vertically down. These points are positioned on different heights and not on the same vertical.

A small object should slide from the top point A(a,A) to the lower point B(b,B) along some frictionless supporting track.

We use a standard Cartesian reference frame with Y-coordinates increasing upwards, and X-coordinates increasing from left to right on a picture above.
The vector of gravity force is directed down along Y-axis.
Therefore, Y-coordinate A is greater then B, and X-coordinate a is less than b.

The supporting track can go straight from A to B or take some curved form.
The straight brown line of descend on a picture above is shorter, but the curved blue or purple lines, while longer, allow for an object to gain speed faster and the resulting time of descend might still be shorter than for a straight line.

The problem is to determine the shape of a supporting track to minimize the time of sliding.

Mathematically speaking, we have to consider all smooth functions f(x) on a segment [a,b] that satisfy the conditions:
f(a) = A and f(b) = B
Then, out of all these functions, we have to find such that represents the curve of fastest descend from A(a,A) to B(b,B).

This simply formulated problem is far from having a simple solution.
Best mathematicians of 17th century worked on it and solved using different methodologies.

Let's solve it using the apparatus developed for finding a minimum of a functional - the Euler-Lagrange equation. This methodology was discussed in the previous lectures of this course.

We have to express the time T of moving from point A to point B as a functional of a trajectory represented by function f(x):
T = Φ[f(x)]
and find a function y=f₀(x) that minimizes this functional.

Hopefully, our functional will look like
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)
and we will be able to apply Euler-Lagrange equation to find y=f₀(x) as its solution.



The picture above illustrates a trajectory of a movement of an object in a uniform gravitational field along a supporting curved track described by a function y=f(x).
The object's weight (the force of gravity vector) is P=m·g, where m is its mass and g is an acceleration of a free falling in the gravitational field.

Besides the gravitational force, a reaction of a supporting curved track vector of force R acts on this object - the force always directed perpendicularly to a tangential line to a curve.

Both the force of gravity P and the reaction force of a curved track R result in the force vector F moving an object along a trajectory and directed along a tangential line to a curved track.

Consider a segment of a trajectory from x to x+dx, where dx is an infinitesimal increment of argument x.
This segment has a length ds and its value satisfies the Pythagorean Theorem
(ds)² = (dx)² + (dy)²
where dy=d(f(x))=f '(x)·dx
so (ds)²=[1+(f '(x))²]·(dx)²
and ds=√1+(f '(x))²·dx

Assume, at point x the linear sliding speed of an object along its trajectory is v(x).
Then the time an object spends passing a segment ds equals to
ds/v(x) = √1+(f '(x))²·dx / v(x)

To find speed of an object v(x), recall the Conservation of Energy Law.

Potential energy of an object depends on its mass and the height over some zero level.
Assume, the zero level of potential energy is at y=0.
Then the initial potential energy U_a of our object at the beginning of its motion is
U_a = m·g·A
where m is an object's mass,
g is an acceleration of free falling and
A is its initial Y-coordinate.

Its kinetic energy K_a at the beginning is zero because its speed along a trajectory is zero at that point.

Then the total initial mechanical energy of an object (potential + kinetic) is
E_a = m·g·A

When our object moved along a curve from its initial position at (a,A) to position (x,f(x)), its potential energy diminished and kinetic energy grew by the same amount because of the Energy Conservation Law.

New potential energy equals to
U_x = m·g·f(x)
New kinetic energy equals to
K_x =m·v²(x)/2.

The decrease in potential energy U_a−U_x should be compensated by an increase in kinetic energy K_x.

From the Energy Conservation Law the total energy should remain the same E_a = E_x
which leads us to an equation
m·g·[A−f(x)] = m·v²(x)/2
Therefore,
v(x) = √2g·[A−f(x)]

Now the time an object spends passing a segment ds equals to

dT(x) =

√1+(f '(x))² √2g·[A−f(x)]

dx

Integrating this by x from a to b gives a total time of moving along a trajectory - the functional we need to minimize
Φ[f(x)] = ∫_[a,b] dT(x)
or

Φ[f(x)]=∫[a,b]

√1+(f '(x))² √2g·[A−f(x)]

dx

At this point we can drop 2g from the denominator, as this does not change the function-argument f₀(x) that minimizes functional Φ[f(x)].
So, our task is to minimize a functional

Φ[f(x)]=∫[a,b]

√1+(f '(x))² √A−f(x)

dx

Recall from the previous lecture that for a given functional
Φ[f(x)] = ∫_[a,b] F[x,f(x),f '(x)]dx
the function f₀(x) that minimizes or maximizes it should satisfy the Euler-Lagrange differential equation
(∂/∂f)F [x,f₀(x),f '₀(x)] −
− (d/dx)(∂/∂f ')F [x,f₀(x),f '₀(x)] = 0

Let's construct this equation for our case.
To shorten formulas, let's temporarily use
h(x) instead of f '(x) and
omit (x) from both f(x) and h(x).

Using this substitution, our functional looks like

Φ[f(x)]=∫[a,b]

√1+h² √A−f

dx

From this follows that an expression under an integral is

F[x,f,h]=

√1+h² √A−f

In terms of these functions, the Euler-Lagrange equation looks like
(∂/∂f)F [x,f,h] =
= (d/dx)(∂/∂h)F [x,f,h]

Let's calculate each term separately.

Left side of an equation is
(∂/∂f)F [x,f,h] =

= (∂/∂f)

√1+h² √A−f

=

= √1+h²·(∂/∂f)(A−f)^−½ =
= √1+h²·(−½)·(A−f)^−3/2·(−1) =

=	√1+h² 2√(A−f)³

Right side of an equation is
(d/dx)(∂/∂h)F [x,f,h] =

= (d/dx)(∂/∂h)

√1+h² √A−f

=

= (d/dx)

h √1+h²·√A−f

=

=

h' √1+h²·√A−f

−

−

h·2h·h' 2√(1+h²)³ ·√A−f

+

+

h·f ' 2√1+h²·√(A−f)³

Note that we've agreed to substitute h(x) for f '(x) in the numerator of the last term.

Equating left and right sides of the Euler-Lagrange equation, multiplying both sides by 2√(1+h²)³ ·√(A−f)³  and opening the parenthesis leads to a simpler equation
1 + h² − 2h'·(A−f) = 0

Returning back to original symbols, replace f with f(x) and h with f '(x) getting a second order differential equation for function y=f(x)
1+[f '(x)]²−2f "(x)·[A−f(x)] = 0
or, equivalently,
1+y'²−2y"·(A−y) = 0

The solution y=f₀(x) to this second order differential equation is the function that minimizes the functional Φ[f(x)].

A not so obvious transformation can reduce this second order differential equation to the first order one.

Integration is easy if a subject of an integration is a derivative of some function
∫ s'(x)·dx = s(x) + C
where C is some constant.

The left side of the equation above is not a derivative of any function, but let's see what happens if we multiply it by −y'.
−y'·[1+y'²−2y"·(A−y)] =
= −y'·(1+y'²)+2y'·y"·(A−y)

Notice that we can substitute −y' with (A−y)' and 2y'·y" with (1+y'²)' to get
(A−y)'·(1+y'²) +
+ (A−y)·(1+y'²)'
which is a derivative of a product of (A−y) by (1+y'²).

Therefore, if y=f₀(x) is a solution to Euler-Lagrange equation
1+y'²−2y"·(A−y) = 0,
it's also a solution to the equation
−y'·[1+y'²−2y"·(A−y)] =
[(A−y)·(1+y'²)]' = 0
or,
{[A−f₀(x)]·[1+f₀'(x)²]}' = 0

Integrating this, we get the first order differential equation
[A−f₀(x)]·[1+f₀'(x)²] = C
or, in a shorter notation,
(A−y)·(1+y'²) = C
where C is some constant.

We can solve this differential equation for function y=f₀(x) as follows.

Let's resolve this differential equation for y':

dy/dx = ±√

C−A+y A−y

From physical considerations, our function must decrease with increase of x. Therefore, it's derivative must be negative, and we will choose a minus sign in all transformations below.

This differential equation can be transformed to integrate separately by x and y:

−√

A−y    C−A+y

·dy = dx

We can integrate now both sides separately.

Without getting into the details of integration, we can just write the result of the integration by y of the left size.

− ∫√

A−y    C−A+y

·dy =

= − C·arctan√

C−A+y A−y

−

− √(A−y)·(C−A+y) + D
where C and D are some constants.

Integration of dx produces x+E, where E is yet another constant, but we can combine it into constant D that appears in integration by dy.

Therefore, our function y=f₀(x) that minimizes the object's time of sliding down can be expressed as

x = −C·arctan√

C−A+y A−y −

− √(A−y)·(C−A+y) + D
where C and D are some constants.

It's not really expressed as y being a function of x, just the opposite way, but it should not stop us to explore this curve.

As we see, the function depends on two constants C and D. At the same time we have two initial conditions:
f₀(a) = A and
f₀(b) = B
Substituting x=a, y=A into an equation above will produce on equation for C and D.
Substituting x=b, y=B into this equation will produce another equation for C and D.
These two equations determine the values of C and D needed to specify the full solution to a problem.

As an example, we used points A(1,5) and B(4,1) and calculated approximate values
C=5.8 and D=−10.1.
Here is the graph that, in particular, crosses points A(1,5) and B(4,1).


At the end of this discussion about brachistochrone it's appropriate to mention that the curve we have found as a solution to the Euler-Lagrange equation is a cycloid - a trajectory of a point on a circle that is rolling along a straight line.
We have not discussed this because our main purpose was just to show how to use the Euler-Lagrange equation to find an extremum of a functional.

Physics+ Euler Lagrange Equation: UNIZOR.COM - Physics+ 4 All - Variations

Notes to a video lecture on UNIZOR.COM

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