Introduction Johann Bernoulli’s . . . Euler’s Solution
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Title Page The Brachistochrone Curve
JJ II Paige MacDonald J I May 16, 2014 Page1 of 21
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Quit Introduction Johann Bernoulli’s . . . Euler’s Solution
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Abstract Title Page This papers explores the brachistochrone problem, which is to find the path of quickest descent of a particle moving between two fixed points in a vertical plane. A brief history of the problem is given, followed by different approaches to solving the problem. JJ II
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Quit 1. Introduction
The brachistochrone problem seeks to find the curve between two points, A and B, in a vertical plane and not in the same vertical line, along which a particle will slide in Introduction the shortest amount of time under the force of gravity and neglecting friction. Johann Bernoulli’s . . . Johann Bernoulli posed the problem in 1696. Solutions were found by Gottfried Wil- Euler’s Solution helm von Leibniz, Isaac Newton, Guillaume de l’Hopital, Jacob Bernoulli, and Johann Bernoulli himself. All of their answers agreed, although each used different methods of Home Page derivation. In 1744, Euler published a work generalizing the work done by the Bernoulli brothers and came up with what is now known as the Euler-Lagrange differential equation (which Title Page Lagrange later independently derived) in order to minimize the value of a definite integral over a family of functions, which led to the calculus of variations. Lagrange provided an analytic method to solve the brachistochrone problem and JJ II other problems of its type, and introduced partial derivatives to the equation.
J I 2. Johann Bernoulli’s Solution
A creative solution like Johann Bernoulli’s uses an indirect approach, but results in Page1 of 21 mathematics that are much easier to understand, so this will be a good starting point. Consider a ray of light that travels from A to P with velocity v1 and then from P to Go Back B with velocity v2 as in Figure1. Using the equation distance time = velocity Full Screen and solving for time leads to the following (in terms of the notation in Figure1): √ a2 + x2 pb2 + (c − x)2 Close T = + v1 v2 Quit Introduction A Johann Bernoulli’s . . . v1 Euler’s Solution a α1 Home Page
x P c − x Title Page
α 2 JJ II v2 b J I
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B c Go Back
Figure 1: A ray of light travels from A to P with velocity v1 and then from P to B with velocity v2. Full Screen
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Quit Taking the derivative of each side gives: √ ! dT d a2 + x2 pb2 + (c − x)2 = + Introduction dx dx v1 v2 Johann Bernoulli’s . . . 1 2 2 − 1 1 2 2 − 1 (a + x ) 2 (2x) (b + (c − x) ) 2 (2)(c − x)(−1) Euler’s Solution = 2 + 2 v1 v2 x c − x = √ − Home Page 2 2 p 2 2 v1 a + x v2 b + (c − x)
In order to minimize T , set dT/dx = 0. So, Title Page x c − x 0 = √ − 2 2 p 2 2 v1 a + x v2 b + (c − x) (1) JJ II x c − x √ = 2 2 p 2 2 v1 a + x v2 b + (c − x) J I Referring back to Figure1, it can be seen that x sin α1 = √ (2) Page3 of 21 a2 + x2 and c − x Go Back sin α2 = (3) pb2 + (c − x)2
Substituting equations (2) and (3) into equation (1) results in the following: Full Screen sin α sin α 1 = 2 (4) v1 v2 Close
Quit Introduction Johann Bernoulli’s . . . Euler’s Solution v Home Page 1 α1
Title Page α2 v2 α2
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α3 v3 α3 J I v Page4 of 21 4 α4
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Quit Equation (4) is known as Snell’s law of refraction and demonstrates Fermat’s prin- ciple of least time, which says that light travels from one point to another along the path requiring the least time. If the layers are divided as in Figure2, the velocity Introduction of light in each layer is constant, but decreases from one layer to the layer below it. Applying Snell’s law gives Johann Bernoulli’s . . . Euler’s Solution sin α sin α sin α sin α 1 = 2 = 3 = 4 v1 v2 v3 v4 As the layers are divided into smaller and smaller sections, the path approaches a Home Page smooth curve and the velocity decreases continuously so that sin α = c (5) Title Page v where c is a constant. Returning to the Brachistochrone problem, assume that the particle moves along a JJ II curve from A to B in the least possible time as in Figure3. Due to the assumption that there is no friction, the total energy of the particle at point A is that same as the total energy at B, which is the kinetic energy plus the potential energy. Since the particle J I starts from rest,
Ki + Ui = Kf + Uf Page5 of 21 1 0 + mgy = mv2 + 0 2 1 Go Back mv2 = mgy. 2 Solving for velocity results in v = p2gy. (6) Full Screen From Figure3, 1 1 1 Close sin α = cos β = = p = p . (7) sec β 1 + tan2 β 1 + (y0)2
Quit x A Introduction Johann Bernoulli’s . . . Euler’s Solution
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β JJ II α B J I y
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Substituting equations (6) and (7) into equation (5) results in: Go Back
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Quit 1 √ = c 2gyp1 + (y0)2 p 1 p2gy 1 + (y0)2 = Introduction c Johann Bernoulli’s . . . √ p 1 y 1 + (y0)2 = √ Euler’s Solution c 2g 1 y[1 + (y0)2] = 2c2g Home Page Noting that the right hand side of the equation will be replaced by some other arbitrary constant, the differential equation of the brachistochrone is Title Page
y[1 + (y0)2] = C (8) JJ II Using Leibniz notation, the differential equation of the brachistochrone becomes:
" 2# dy J I y 1 + = C dx dy 2 C Page7 of 21 1 + = dx y 2 dy C Go Back = − 1 dx y dy 2 C − y = Full Screen dx y dy C − y 1/2 = Close dx y
Quit Separating variables and solving for dx results in
y 1/2 dx = dy (9) Introduction C − y Johann Bernoulli’s . . . Introducing a new variable and setting Euler’s Solution
y 1/2 = tan φ, (10) Home Page C − y and then solving for y gives Title Page y = tan2 φ C − y y = (C − y) tan2 φ JJ II y = C tan2 φ − y tan2 φ y(1 + tan2 φ) = C tan2 φ J I y sec2 φ = C tan2 φ Page8 of 21 y = C sin2 φ.
Taking the derivative of each side with respect to φ leads to Go Back dy = 2C sin φ cos φdφ. (11)
Next, substitute equations (10) and (11) into equation (9) to obtain Full Screen
dx = (tan φ)(2C sin φ cos φdφ) Close dx = 2C sin2 φdφ
Quit Integrating both sides results in Z Z dx = 2c sin2 φdφ Introduction Z 1 − cos 2φ Johann Bernoulli’s . . . x = 2C dφ 2 Euler’s Solution Z Z x = C 1dφ − C cos 2φdφ Home Page 1 x = Cφ − c sin 2φ + c 2 1 C Title Page x = (2φ − sin 2φ) + c 2 1 Using the initial condition (0, 0) in equation (10) and solving for φ gives φ = 0 when JJ II the particle is at point A. Therefore, c1 is 0 and C x = (2φ − sin 2φ) (12) 2 J I and y = C sin2 φ Page9 of 21 1 − cos 2φ y = C 2 (13) C Go Back y = (1 − cos 2φ). 2 Letting a = C/2 and θ = 2φ, equations (12) and (13) become Full Screen x = a(θ − sin θ) (14) and Close y = a(1 − cos θ). (15)
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a θ J I x 2πa (x, y) Page 10 of 21
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Quit These are the parametric equations of the curve traced out by a point on the circum- ference of a circle of radius a rolling along the x axis (the cycloid) with the condition that the curve passes through B for some value of a. The result is shown in Figure4. Introduction Johann Bernoulli’s . . . 3. Euler’s Solution Euler’s Solution
Euler’s approach to solving this problem is more analytic and involves minimizing a function of a function, called a functional. Problems of this type belong to The Calculus Home Page of Variations, which Euler’s work on this problem has been credited with leading to. This is similar to optimization problems in elementary calculus, except in calculus the Title Page quantity that varies when finding extrema of f(x) is a simple variable, x. In calculus of variations, the quantity that varies is itself a function of the form f(x, y, y0). The functional assigns a numerical value (time, in the case of the brachistochrone problem) JJ II to each function in the family of candidate functions. For the brachistochrone problem, referring to Figure5, suppose two points A and B are in a vertical plane. An infinite number of curves can be drawn connecting the J I points, but which of these curves minimizes the time for a frictionless particle to move from point A to point B under gravity? The problem, then, is to find a function that minimizes the family of functions that are differentiable and whose derivatives and Page 11 of 21 second derivatives are continuous and pass through the points A and B-these are the candidate, or admissable, functions. If the coordinates of points A and B are (x1, y1) Go Back and (x2, y2), as in Figure6, consider the family of functions y = y(x) that satisfy the conditions y(x1) = y1 and y(x2) = y2. The function that will minimize time is of the form Full Screen Z x2 I(y) = f(x, y, y0)dx. (16) x1 Assuming a function y(x) that minimizes the integral in equation (16) exists, consider Close a function η(x) that “disturbs” y(x) slightly. If y(x) minimizes I, then I will inrease
Quit Introduction Johann Bernoulli’s . . . y Euler’s Solution
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Quit as a result. 00 Let η(x) be a function with η (x) continous and η(x1) = η(x2) = 0. If α is a small parameter, then Introduction y¯ = y(x) + αη(x) (17) Johann Bernoulli’s . . . repesents a family of admissable functions of one-parameter with a vertical deviation Euler’s Solution from the minimizing curve y(x), as seen in Figure6. For any choice of the function η(x), the function y(x) belongs to the family and corresponds to α = 0. Substituting y¯(x) = y(x) + αη(x) andy ¯0(x) = y0(x) + αη0(x) into the integral (16) gives a function Home Page of α. Z x2 0 I(α) = f(x, y,¯ y¯ )dx Title Page x1 (18) Z x2 I(α) = f[x, y(x) + αη(x), y0(x) + αη0(x)]dx. x1 JJ II When α = 0, equation (17) givesy ¯(x) = y(x). Since y(x) minimizes the integral (16), I(α) must have a minimum when α = 0, which means that the derivative I0(α) when α = 0 is 0, or I0(0) = 0. To compute the derivative I0(α), differentiate (18) under the J I integral sign. Z x2 ∂ I0(α) = f(x, y,¯ y¯0)dx (19) Page 13 of 21 x1 ∂α Using the chain rule Go Back ∂ ∂f ∂x ∂f ∂y¯ ∂f ∂y¯0 f(x, y,¯ y¯0) = + + (20) ∂α ∂x ∂α ∂y¯ ∂α ∂y¯0 ∂α Full Screen Now, ∂x/∂α = 0, ∂y/∂α¯ = η(x) and ∂y¯0/∂α = η0(x), so equation (20) becomes
∂f ∂f ∂f Close f(x, y,¯ y¯0) = η(x) + η0(x) ∂α ∂y¯ ∂y¯0
Quit Introduction y Johann Bernoulli’s . . . Euler’s Solution (x1, y1) y¯(x) = y(x) + αη(x) Home Page
αη(x) y(x) Title Page (x2, y2)
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η(x) Page 14 of 21 η(x) x x1 x x2 Go Back
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Quit and equation (19) can be written as
Z x2 0 ∂f ∂f 0 I (α) = η(x) + 0 η (x) dx. (21) Introduction x1 ∂y¯ ∂y¯ Johann Bernoulli’s . . . 0 0 0 Setting α = 0 and recalling that when α = 0, I (α) = 0,y ¯ = y, andy ¯ = y ,equation (21) Euler’s Solution gives Z x2 ∂f ∂f 0 η(x) + 0 η (x) dx = 0 (22) Home Page x1 ∂y ∂y Integrating the the second term by parts to eliminate η0(x) yields Title Page x Z x2 ∂f ∂f 2 Z x2 d ∂f η0(x)dx = η(x) − η(x) dx ∂y0 ∂y0 dx ∂y0 x1 x1 x1 Z x2 d ∂f JJ II = − η(x) 0 dx x1 dx ∂y J I due to the previously stated conditions that η(x1) = η(x2) = 0. Placing the result back into equation (22) gives Page 15 of 21 Z x2 ∂f d ∂f η(x) − η(x) dx = 0 ∂y dx ∂y0 x1 (23) Z x2 ∂f d ∂f Go Back η(x) − 0 dx = 0. x1 ∂y dx ∂y The integral in equation (23) must vanish for every choice of the function η(x), which Full Screen implies that the expression in brackets must also vanish. Therefore, d ∂f ∂f Close − = 0. (24) dx ∂y0 ∂y
Quit Equation (24) is Euler’s equation. So, if y(x) is an admissable function that min- imizes the integral (16), then y(x) satisfies Euler’s equation. However, the converse is not true. That is, if y(x) is an admissable function satisfying Euler’s equation, it Introduction does not necessarily minimize the integral. The conditions for distinguishing admiss- able solutions of Euler’s equation, called stationary functions or stationary curves, are Johann Bernoulli’s . . . complicated and will not be explained here. Rather, the geometry of the problem will Euler’s Solution be used to determine whether a particular stationary function minimizes the integral.
By expanding the first term in (24) according to the chain rule, Home Page ∂ ∂f ∂ ∂f dy ∂ ∂f dy0 ∂f 0 + 0 + 0 0 − = 0, ∂x ∂y ∂y ∂y dx ∂y ∂y dx ∂y Title Page
Euler’s equation becomes
dy dy0 JJ II f 0 + f 0 + f 0 0 − f = 0 y x y y dx y y dx y dy d dy f 0 + f 0 + f 0 0 − f = 0 (25) J I y x y y dx y y dx dx y d2y dy f 0 0 + f 0 + f 0 − f = 0 Page 16 of 21 y y dx2 y y dx y x y where f refers to ∂f and f refers to ∂ f . y ∂y yx ∂x y Go Back An equation like (25) is usually unsolvable, but there are special cases where it is solvable. Namely, if x and y are missing from the function f, if y is missing from the function f, or if x is missing from f, then equation (25) is solvable. For the Full Screen brachistochrone problem, the case of interest is the latter case, where x is missing from the function f. Close
Quit 3.1. Special Case If x is missing from f, equation (25) can be integrated to Introduction 0 ∂f Johann Bernoulli’s . . . y 0 − f = c. (26) ∂y Euler’s Solution This equation is known as Beltrami’s Identity. To show that this is true, begin by 0 rearranging Euler’s equation (24) and multiplying by y to obtain Home Page
0 ∂f 0 d ∂f y = y 0 . (27) ∂y dx ∂y Title Page Now, according to the chain rule, df ∂f ∂f ∂f JJ II = y0 + y00 + . dx ∂y ∂y0 ∂x Rearranging this result yields J I
∂f 0 df ∂f 00 ∂f y = − y − . Page 17 of 21 ∂y dx ∂y0 ∂x
Substituting this into the left hand side of (27) gives Go Back df ∂f ∂f d ∂f − y00 − = y0 dx ∂y0 ∂x dx ∂y0 Full Screen and, rearranging this result leads to
df ∂f ∂f d ∂f Close − y00 − − y0 = 0. (28) dx ∂y0 ∂x dx ∂y0
Quit Noting that, by the product rule
d ∂f d d ∂f f − y0 = f − y0 dx ∂y0 dx dx ∂y0 Introduction Johann Bernoulli’s . . . d 0 ∂f df 00 ∂f 0 d ∂f f − y = − y − y , Euler’s Solution dx ∂y0 dx ∂y0 dx ∂y0 then, Home Page d ∂f df ∂f d ∂f y0 = − y00 − f − y0 . dx ∂y0 dx ∂y0 dx ∂y0 Plugging this result for the last term on the left hand side of (28) gives Title Page
df ∂f ∂f df ∂f d ∂f − y00 − − − y00 − f − y0 = 0 dx ∂y0 ∂x dx ∂y0 dx ∂y0 JJ II ∂f d 0 ∂f − − f − y 0 = 0 (29) ∂x dx ∂y J I d ∂f ∂f f − y0 = − . dx ∂y0 ∂x Page 18 of 21 Since ∂f/∂x = 0 for this case, integrating (29) yields
∂f Go Back f − y0 = c ∂y0 which is equation (26). Full Screen Now, referring back to equation (16), the integral we wish to minimize is
Z x2 Close I(y) = f(x, y, y0)dx. x1 Quit It is useful to invert the y-axis and place point A at the origin, as in Figure3. The velocity of the particle is√v = ds/dt, so the total time of descent is the integral of ds/v. From equation (7), v = 2gy and length of the curve s = p1 + (y0)2, so the integral Introduction to be minimized becomes Z x2 p1 + (y0)2 Johann Bernoulli’s . . . √ dx Euler’s Solution x 2gy √ 1 Since f = p1 + (y0)2/ 2gy is missing x, referring to Case C and Beltrami’s identity, Euler’s equation becomes Home Page ! ∂ p1 + (y0)2 p1 + (y0)2 √ y0 − √ = c ∂y0 2gy 2gy Title Page 1 1 p1 + (y0)2 √ [1 + (y0)2]−1/22y0 y0 − √ = c 2gy 2 2gy JJ II p (y0)2 1 + (y0)2 p √ − √ = c 2g p 0 2 y y 1 + (y ) J I (y0)2 − (1 + (y0)2) = cp2g √ p 0 2 y 1 + (y ) Page 19 of 21 −1 p √ = c 2g yp1 + (y0)2 Go Back √ 1 yp1 + (y0)2 = − √ c 2g 1 Full Screen y[1 + (y0)2] = 2c2g 2 and, replacing 1/2c g with an arbitrary constant, Close y[1 + (y0)2] = C
Quit which is the differential equation (8) solved in the first section. The resulting curve is the cycloid generated by a circle with radius a rolling under the x axis whose first arch passes through the point (x2, y2). Introduction Johann Bernoulli’s . . . 4. Conclusion Euler’s Solution
The brachistochrone problem’s significance lies in the fact that it led to the calculus of variations, which is a branch of analysis that is applicable to a large number of Home Page problems. It has greatly influenced mechanics, leading to Hamilton’s principle, as well as being used in Einstein’s theory on general relativity. Shrodinger used it to discover Title Page the wave equation, a cornerstone of quantum mechanics.
References JJ II
[1] A special thanks to Dave Arnold for all his help and support. J I [2] Till Tantau. The Tikz and PGF Packages. [3] George Simmons. Differential Equations with Applications and Historical Notes. Page 20 of 21 [4] David Arnold. Writing Scientific Papers in Latex. Go Back [5] Douglas S. Shafer. 2007 The Brachistochrone: Historical Gateway to the Calculus of Variations. Full Screen [6] Nils P. Johnson. 2009 The Brachistochrone Problem [7] J J O’Connor and E F Robertson 2002 History Topic: The Brachistochrone Prob- Close lem http://www.history.mcs.stand.ac.uk/PrintHT/Brachistochrone.html
Quit [8] Unknown author. The Brachistochrone Problem: http://www.math.utk.edu/ ~freire/teaching/m231f08/m231f08brachistochrone.pdf
[9] Frank Porter. Calculus of Variations: http://www.hep.caltech.edu/~fcp/math/ Introduction variationalCalculus/variationalCalculus.pdf Johann Bernoulli’s . . . Euler’s Solution
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