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The Calculus of Variations: an Introduction

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The Calculus of Variations: an Introduction The Calculus of Variations: An Introduction By Kolo Sunday Goshi Some Greek Mythology Queen Dido of Tyre – Fled Tyre after the death of her husband – Arrived at what is present day Libya Iarbas’ (King of Libya) offer – “Tell them, that this their Queen of theirs may have as much land as she can cover with the hide of an ox.” What does this have to do with the Calculus of Variations? What is the Calculus of Variations “Calculus of variations seeks to find the path, curve, surface, etc., for which a given function has a stationary value (which, in physical problems, is usually a minimum or maximum).” (MathWorld Website) Variational calculus had its beginnings in 1696 with John Bernoulli Applicable in Physics Calculus of Variations Understanding of a Functional Euler-Lagrange Equation – Fundamental to the Calculus of Variations Proving the Shortest Distance Between Two Points – In Euclidean Space The Brachistochrone Problem – In an Inverse Square Field Some Other Applications Conclusion of Queen Dido’s Story What is a Functional? The quantity z is called a functional of f(x) in the interval [a,b] if it depends on all the values of f(x) in [a,b]. Notation b z f x a – Example 1 1 x22cos x dx 0 0 Functionals The functionals dealt with in the calculus of variations are of the form b f x F x, y ( x ), y ( x ) dx a The goal is to find a y(x) that minimizes Г, or maximizes it. Used in deriving the Euler-Lagrange equation Deriving the Euler-Lagrange Equation I set forth the following equation: y x y x g x Where yα(x) is all the possibilities of y(x) that extremize a functional, y(x) is the answer, α is a constant, and g(x) is a random function. y1 y(b) y0 = y y(a) y2 a b Deriving the Euler-Lagrange Equation Recalling b It can now be said that: b y F x,, y y dx f x F x, y ( xa ), y ( x ) dx a At the extremum yα = y0 d = y and 0 d 0 The derivative of the b functional with respect d F x,, y y dx to α must be evaluated da and equated to zero Deriving the Euler-Lagrange Equation The mathematics involved y x y x g x db Fyy F dx a d y y – Recalling So, we can say b db F Fd b F b F dg g g dx gdxF x,, y y dx dx ad aa a d y y y y dx Deriving the Euler-Lagrange Equation dbb F F dg gdx dx aa d y y dx Integrate the first part by parts and get b dF g dx a dx y So db F d F g dx a d y dx y Since we stated earlier that the derivative of Г with respect to α equals zero at α=0, the extremum, we can equate the integral to zero Deriving the Euler-Lagrange Equation So b F d F 0 g dx a y00 dx y We have said that y0 = y, y being the extremizing y1 function, therefore y0 = y y2 Since g(x) is an arbitrary b F d F function, the quantity in the 0 g dx a brackets must equal zero y dx y The Euler-Lagrange Equation We now have the Euler-Lagrange Equation F d F 0 y dx y When F F y , y , where x is not included, the modified equation is F F y C y The Shortest Distance Between Two Points on a Euclidean Plane What function describes the shortest distance between two points? – Most would say it is a straight line Logically, this is true Mathematically, can it be proven? The Euler-Lagrange equation can be used to prove this Proving The Shortest Distance Between Two Points Define the distance to be s, so s ds b ds dy dx a Therefore s dx22 dy Proving The Shortest Distance Between Two Points Factoring a dx2 inside the square root and taking its square root we obtain 2 b dy s1 dx a dx dy Now we can let y dx b so 222 ss1 dx y dx dy a Proving The Shortest Distance Between Two Points b Since 1 y2 dx a b And we have said fthat x F x, y ( x ), y ( x ) dx a 2 we see that Fy1 Fy F therefore 0 y y 1 y2 Proving The Shortest Distance Between Two Points Recalling the Euler-Lagrange equation F d F 0 Knowing that y dx y A substitution can be made dy 0 dx 2 1Fyy F 0 2 Therefore the term in brackets y y 1 y must be a constant, since its derivative is 0. Proving The Shortest Distance Between Two Points More math to reach the solution y C 1 y2 y2 C 21 y 2 y21 C 2 C 2 yD2 yM Proving The Shortest Distance Between Two Points Since yM We see that the derivative or slope of the minimum path between two points is a constant, M in this case. The solution therefore is: y Mx B The Brachistochrone Problem Brachistochrone – Derived from two Greek words brachistos meaning shortest chronos meaning time The problem – Find the curve that will allow a particle to fall under the action of gravity in minimum time. Led to the field of variational calculus First posed by John Bernoulli in 1696 – Solved by him and others The Brachistochrone Problem The Problem restated – Find the curve that will allow a particle to fall under the action of gravity in minimum time. The Solution – A cycloid – Represented by the parametric equations D x 2 sin 2 2 D y 1 cos 2 2 Cycloid.nb The Brachistochrone Problem In an Inverse Square Force Field The Problem – Find the curve that will y allow a particle to fall 1 under the action of an inverse square force field 2 r0 k 2 Fr ˆ defined by k/r in r 2 minimum time. x – Mathematically, the force is defined as k F r r 2 The Brachistochrone Problem In an Inverse Square Force Field Since the minimum time is being considered, an 2 ds expression for time must be t 1 v determined An expression for the 1 k velocity v must found and mv2 E this can be done using the 2 r fact that KE + PE = E The Brachistochrone Problem In an Inverse Square Force Field r The initial position 0 is 1 kk known, so the total energy E mv2 is given to be –k/r0, so 2 rr0 2k 1 1 An expression can be found v for velocity and the desired m r r0 expression for time is found m2 ds t 1 2k 11 rr0 The Brachistochrone Problem In an Inverse Square Force Field Determine an expression for ds rdΘ r ds dr r + dr ds22 dr22 r d The Brachistochrone Problem In an Inverse Square Force Field We continue using a polar coordinate system 2 222 dr ds d r d An expression can be determined for ds to put 22 into the time expression ds r r d ds22 dr22 r d The Brachistochrone Problem In an Inverse Square Force Field Here is the term for 22 m 2 rr() r r time t t 0 1 2k r0 r The function F is the rr() r22 r term in the integral F 0 rr0 The Brachistochrone Problem In an Inverse Square Force Field Using the modified Euler-Lagrange equation F F r C r 22 rr00() r r2 rr rC22 r00 r r r() r r The Brachistochrone Problem In an Inverse Square Force Field More math involved in finding an integral to be solved 22 r() r r2 r rD22 r00 r r r() r r rr2 r5 D G 22 22 r0 r() r r r0 r() r r The Brachistochrone Problem In an Inverse Square Force Field Reaching the integral dr r52 r G() r r r 0 d G() r0 r G() r r 0 dr d r52 r G() r r Solving the integral for r(Θ) 0 finds the equation for the path that minimizes the time. The Brachistochrone Problem In an Inverse Square Force Field Challenging Integral to Solve – Brachistochrone.nb Where to then? – Use numerical methods to solve the integral – Consider using elliptical coordinates Why Solve this? – Might apply to a cable stretched out into space to transport supplies Some Other Applications The Catenary Problem – Derived from Greek for “chain” – A chain or cable supported at its end to hang freely in a uniform gravitational field – Turns out to be a hyperbolic cosine curve Derivation of Snell’s Law nn1sini 2 sin 2 Conclusion of Queen Dido’s Story Her problem was to find the figure bounded by a line which has the maximum area for a given perimeter Cut ox hide into infinitesimally small strips – Used to enclose an area – Shape unknown – City of Carthage Isoperimetric Problem – Find a closed plane curve of a given perimeter which encloses the greatest area – Solution turns out to be a semicircle or circle References Atherton, G., Dido: Queen of Hearts, Atherton Company, New York, 1929. Boas, M. L., Mathematical Methods in the Physical Sciences, Second Edition, Courier Companies, Inc., United States of America, 1983. Lanczos, C, The Variational Principles of Mechanics, Dover Publications, Inc., New York, 1970. Ward, D., Notes on Calculus of Variations Weinstock, R., Calculus of Variations, Dover Publications, Inc., New York, 1974..
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