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
The vector of gravity force is directed down along Y-axis.
Therefore,
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 Ua of our object at the beginning of its motion is
Ua = 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 Ka 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
Ea = 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
Ux = m·g·f(x)
New kinetic energy equals to
Kx =m·v²(x)/2.
The decrease in potential energy Ua−Ux should be compensated by an increase in kinetic energy Kx.
From the Energy Conservation Law the total energy should remain the same Ea = Ex
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) = |
|
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] |
|
dx |
So, our task is to minimize a functional
| Φ[f(x)]=∫[a,b] |
|
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] |
|
dx |
From this follows that an expression under an integral is
| F[x,f,h]= |
|
(∂/∂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·(−½)·(A−f)−3/2·(−1) =
| = |
|
Right side of an equation is
(d/dx)(∂/∂h)F [x,f,h] =
| = (d/dx)(∂/∂h) |
|
= |
| = (d/dx) |
|
= |
| = |
|
− |
| − |
|
+ |
| + |
|
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 = ±√ |
|
This differential equation can be transformed to integrate separately by x and y:
| −√ |
|
·dy = dx |
Without getting into the details of integration, we can just write the result of the integration by y of the left size.
| − ∫√ |
|
·dy = |
| = − C·arctan√ |
|
− |
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√ |
|
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.
No comments:
Post a Comment