Calculus of variation problems
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This presentation gives example of "Calculus of Variations" problems that can be solved analytical. "Calculus of Variations" presentation is prerequisite to this one. For comments please contact me at solo.hermelin@gmail.com. For more presentations on different subjects visit my website at http://www.solohermelin.com.
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Calculus of variation problems
- 1. 1 Calculus of Variations Problems SOLO HERMELIN
- 2. 2 SOLO Calculus of Variations - Problems Table of Content Introduction 1. Brachistochrone Problems Brachistochrone Problem 1 Brachistochrone Problem 2 2. Isoperimetric Problems Dido Maximum Area Problem Maximum volume with given surface area Shape of a liquid drop on a horizontal surface 3. Problem of Minimum Surface of Revolution 4. Geodesics Problems Geodesic in 3 Dimensional Spaces Geodesics in Riemannian Space Geodesics for an Implicit Equation of the Surface, G (x,y,z) = 0 5. Geometrical Optics and Fermat Principle References
- 3. 3 SOLO Introduction The Mathematical Theory of Calculus of Variation was given in another presentation: “Calculus of Variations”, by the same author. Calculus of Variations tackles problems of finding Functions that Extremize a given Cost Functional . The Solutions must satisfy Boundary Conditions and some giving constraints. Solving those problems require finding solution to Differential Equations. Calculus of Variations - Problems In general few problems have analytic solutions and therefore the results can be obtained only numerically. In this presentation we give example f problems that could be solved analytical, in order to facilitate the understanding of the subject. Return to Table of Content
- 4. 4 SOLO A particle slides on a frictionless wire between two fixed points A(0,0) and B (xfc, yfc) in a constant gravity field g. The curve such that the particle takes the least time to go from A to B is called brachistochrone (βραχιστόσ Greek for “shortest“, χρόνοσ greek for “time). The brachistochrone problem was posed by John Bernoulli in 1696, and played an important part in the development of calculus of variations. The problem was solved by Johann Bernoulli, Jacob Bernoulli, Isaac Newton, Gottfried Leibniz and Guillaume de L’Hôpital. Let choose a system of coordinates with the origin at point A (0,0) and the y axis in the constant g direction x y V ( )tγ ( )fcfc yxB ,fcx fcy N ( )0,0A Calculus of Variations - Problems 1. Brachistochrone Problems Brachistochrone Problem 1
- 5. 5 SOLO Since the motion of the particle is in a frictionless fixed gravitational field the total energy is conserved ( ) ygVyVygVV 2 2 1 2 1 2 0 22 0 +=→−= x y V ( )tγ ( )fcfc yxB ,fcx fcy N ( )0,0A Second way to get this relation is: ( ) ygVVdygdVV sd yd ggV sd Vd td sd sd Vd td Vd =−→=→==== 2 0 2 2 1 sinγ where V0 is the velocity of the particle at point A and ( ) ( )22 ydxdsd += td xd xd yd td yd td xd td sd V 222 1 += + == We have xd ygV xd yd xd V xd yd td 2 11 2 0 22 + + = + = The cost function is ∫∫ = + + = cfcf xx xd xd yd yxFxd ygV xd yd J 00 2 0 2 ,, 2 1 Calculus of Variations - Problems 1. Brachistochrone Problems Brachistochrone Problem 1
- 6. 6 HISTORY OF CALCULUS OF VARIATIONS The brachistochrone problem In 1696 proposed the Brachistochrone (“shortest time”) Problem: Given two points A and B in the vertical plane, what is the curve traced by a point acted only by gravity, which starts at A and reaches B in the shortest time. Johann Bernoulli 1667-1748 SOLO
- 7. 7 The brachistochrone problem Jacob Bernoulli (1654-1705) Gottfried Wilhelm von Leibniz (1646-1716) Isaac Newton (1643-1727) The solutions of Leibniz, Johann Bernoulli, Jacob Bernoulli and Newton were published on May 1697 publication of Acta Eruditorum. L’Hôpital solution was published only in 1988. Guillaume François Antoine de L’Hôpital (1661-1704) SOLO HISTORY OF CALCULUS OF VARIATIONS
- 8. 8 SOLO A particle slides on a frictionless wire between two fixed points A(0,0) and B (xfc, yfc) in a constant gravity field g. The curve such that the particle takes the least time to go from A to B is called brachistochrone. x y V ( )tγ ( )fcfc yxB ,fcx fcy N ( )0,0A ∫∫ = + + = cfcf xx xd xd yd yxFxd ygV xd yd J 00 2 0 2 ,, 2 1 We derived the cost function: xd yd y ygV y xd yd yxF = + + = : 2 1 :,, 2 0 2 wher e F doesn’t depend explicitly on the free variable x, therefore if we replace and we use the result obtained for F not depending explicitly on x, we obtain ( ) ( )xtyx ,, → ( ) ( ) const ygVy y ygV y yyFyyyF y == ++ − + + =− α 212 1 ,, 2 0 2 2 2 0 2 const ygVy == ++ α 21 1 2 0 2 or Calculus of Variations - Problems 1. Brachistochrone Problems Brachistochrone Problem 1
- 9. 9 SOLO x y V ( )tγ ( )fcfc yxB ,fcx fcy N ( )0,0A const ygVy == ++ α 21 1 2 0 2 Let define a parameter τ such that τcos 1 1 2 = + xd yd and const ygV ygV xd yd == + = + + α τ 2 cos 21 1 2 02 0 2 From which ( ) ( )ττ αα τ 2cos12cos1 4 1 2 cos 2 22 22 0 +=+==+ r ggg V y Tacking the derivative of this equation with respect to τ we obtain τ τ 2sin2r d yd −= Calculus of Variations - Problems 1. Brachistochrone Problems Brachistochrone Problem 1
- 10. 10 SOLO τ ττ τ ττ 2 22 2 2 cos /1 1 = + = + d yd d xd d xd d xd d yd ( ) ( )ττ ττ τ τττ τ τττ ττ 2sin2 2cos12cos4 cossin16sin cos2sin4cos 0 2 4222 2 2222 22 +±=→ +±=±=→ = → + = → rxx rr d xd r d xd r d xd d xd Let change variables to 2τ = θ – π, to get ( ) ( )θ θθ cos1 2 sin 2 0 0 −=+ −+= r g V y rxx θsinr θcosr θr x y 0x 0V g V 2 2 0 r r A B ),( yx θ We obtain the equation of a cycloid generated by a circle of radius r rolling upon the horizontal line and starting at the point g V y 2 2 0 −= −− g V x 2 , 2 0 0 Calculus of Variations - Problems 1. Brachistochrone Problems Brachistochrone Problem 1
- 11. 11 HISTORY OF CALCULUS OF VARIATIONS The brachistochrone problem ( ) ( ) −−= −+= g V ry rxx 2 cos1 sin 2 0 0 θ θθ Cycloid Equation ∫∫∫∫ = + + === cfcfcf xxxt xd xd yd yxFxd ygV xd yd V sd tdJ 00 2 0 2 00 ,, 2 1 Minimization Problem Solution of the Brachistochrone Problem: SOLO Johann Bernoulli 1667-1748
- 12. 12 SOLO ( ) ( )θ θθ cos1 2 sin 2 0 0 −=+ −+= r g V y rxx θsinr θcosr θr x y 0x 0V g V 2 2 0 r r A B ),( yx θ Calculus of Variations - Problems We have ( ) ( ) td d r td d r td d d xd d xd V r d yd r d xd θθθ θ θ θθ θ θ θ θ =−= + = = −= 2 sin2cos12 sin cos1 22 ( ) rgrgrgygVV = =−=+= 2 sin2 2 sin4cos122 22 0 θθ θ const r g td d == θ ( )0 * 0 θθθ θ θ −=== ∫∫ fAB g r d g r tdt f From those two equations we obtain and the minimum time to travel between A and B will be 1. Brachistochrone Problems Brachistochrone Problem 1 Return to Table of Content
- 13. 13 SOLO Calculus of Variations - Problems Brachistochrone Problem 2 (Bryson Sec 2.7, Problem 6, pg. 81, Sec. 3.11, Example 1, pg. 119, Sec. 4.3, Example 1, pg. 142) A particle slides on a frictionless wire between two points A (0,0) and B (xfc, yfc) in a constant gravity field. The particle has an initial velocity V0 at point A. What is the shape of the wire: • that will produce a minimum time path between the two points with the constraint that y ≤ x tanθ + h, where θ and h are constant. • That provides a maximum xf for given tf and yf. hxy +< θtan x y V ( )tγ ( )fcfc yxB ,fcx fcy N 1. Brachistochrone Problems
- 14. 14 SOLO Calculus of Variations - Problems Brachistochrone Problem 2 Solution to Brachistochrone Problem 2 Since the motion of the particle is in a frictionless fixed gravitational field the total energy is conserved ( ) ygVyVygVV 2 2 1 2 1 2 0 22 0 +=→−= The equations of motions are ( ) ( ) ( ) ( ) ( ) ( ) fcf fcf ytytyyVy xtxtxyVx === === 0sin 0cos 0 0 γ γ We have 2 First Order Differential Equations in x (t) and y (t) and 1 control γ (t) with 4 Boundary Conditions To find the wire path we must define the angle γ (t) such that ftJ γγ minmin = 1. Brachistochrone Problems
- 15. 15 SOLO Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 1) The unconstrained problem: y ≤ x tanθc + h ( ) =− =− = 0 0 :, fcf fcf ff yy xx txψ Let adjoin the constraints to obtain the augmented cost function: ( ) ( ) ( )( ) ( )( )[ ]∫ −+−+−+−+= ft t yxfcffcff dtyyVxyVxxxxtJ 0 sincos21 γλγλνν 1. The Hamiltonian ( )( )γλγλ sincos: yxyVH += ( ) ( ) ( )fcffcffff xxxxttx −+−+=Φ 21:, ννand 2. Euler-Lagrange Equations ( )fxfx T tH Φ=−= λλ ( ) ( ) ( ) 10 νλλλλ ==→= ∂ ∂ −= fxfxxx ttt x H ( ) ( ) 2sincos νλγλγλλ =+−= ∂ ∂ −= fyyxy t yd Vd y H 1. Brachistochrone Problems
- 16. 16 SOLO Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 2) ( )( ) x y yxyVH λ λ γγλγλγ =→=+−= tan0cossin: Since H is not an explicit function of time we have: ( ) ( )ftHconsttH ** == Using the Boundary Conditions: ( ) ( ) 1* −=Φ−= ftf ttH Therefore instead of solving the differential equation, let use:( )tyλ ( )( ) ( ) ( ) ( ) ( ) 1 sin sincoscotsincossincos:* −==+= +=+= γ λ γγγλγγ λ λ λγλγλ y y y x yyx yVanyVyVyVH From which ( )yV y γ λ sin −= Note: If for the end point B (xfc, yfc) only xfc, is defined and yfc is free, then ( ) ( ) ( ) ( ) 0sin0 sin =→=−= f f f fy t yV t t c γ γ λ End Note 1. Brachistochrone Problems
- 17. 17 SOLO Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 3) Also ( )( ) ( ) ( ) ( ) ( ) 1 cos sintancossincossincos:* −==+= +=+= γ λ γγγλγ λ λ γλγλγλ x x x y xyx yVyVyVyVH Therefore ( ) 0 0 cos 1 cos γλγ V const yV x ==−= or ( ) x yV λ γcos −= Note: ( ) 0 * cos 01 γ λx VH =−= 2 0 π γ =If V0 = V (0)=0, to keep we must have End Note If we take the time derivative of the last equation we obtain: ( ) ( ) ( ) td d yV yd yVd td yd yd yVd x γ λ γ γ sin sin == or using ( ) ygVyV 2 2 0 += ( ) ( ) ( ) ( ) constgyV yV g yV yd yVd td d xxx ===== ωλλλ γ 1. Brachistochrone Problems
- 18. 18 SOLO Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 4) Let find now the shape of the wire Let use and( ) ( ) 0cos 0 == txyV td xd γ ( ) x yV λ γcos −= ( ) x x yVg d xd td d d xd td xd λ γ γλ γ γ γ 2 cos cos −==== or ( ) −+−−= + −=−= →−= ∫∫ 002 2 2 2 2 2 2sin2sin22 4 1 2 2cos11 cos 1 cos 00 γγγγ λ γ γ λ γγ λ λ γ γ γ γ γ γ x xx x g d g d g x gd xd From the equations: and , we obtain:( ) ygVyV 2 2 0 += ( ) x yV λ γcos −= ( ) g V gg V g yygV xxx 2 2cos1 4 1 2 cos 2 1cos 2 2 0 2 2 0 2 2 2 2 2 0 −+=− =→=+ γ λλ γ λ γ 1. Brachistochrone Problems
- 19. 19 SOLO Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 5) The wire shape is given by: ( ) ( ) g V g y g x x x 2 2cos1 4 1 2sin2sin22 4 1 2 0 2 002 −+= −+−−= γ λ γγγγ λ For V0 ≠ 0 from which( ) 0 00 cos 0 cos VyV x γγ λ −= = −= 0 0 cosγλω γ V g g td d x −=== ( ) ( ) g V g V y g V x 2 2cos1 cos4 2sin2sin22 cos4 2 0 0 2 2 0 00 0 2 2 0 −+= −+−−= γ γ γγγγ γ The wire shape is given by: 1. Brachistochrone Problems
- 20. 20 SOLO Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 6) By using unknown g V r 0 0 2 2 0 cos4 : γ γ = πγθ += 2: ( ) unknownrx 0000 sin: θθθ −−= we obtain: ( ) ( ) EquationCycloid g V ry rxx −−= −+= 2 cos1 sin 2 0 0 θ θθ To find r and x0 we must solve the following 4 equations with additional 2 unknowns θ0 and θf ( ) ( ) ( ) ( ) ( )ff fff r g V y rxx g V g V rr g V rx c c θ θθ θθ θ θθ cos1 2 sin 2 sin4 cos12 cos1 2 sin0 2 0 0 02 2 0 0 2 0 0 2 0 000 −=+ −+= = − =→−= −+= For V0 ≠ 0 1. Brachistochrone Problems
- 21. 21 SOLO Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 7) For V0 = 0 we found 0,0 2 000 ==→= xθ π γ ( ) ( ) −= −= θ θθ cos1 sin ry rx where 2 4 1 xg r λ = and it can be calculated using the following 2 equations with 2 unknowns r and θf ( ) ( ) −= −= ff fff ry rx c c θ θθ cos1 sin θf is given by the following transcedental equation ( ) ( )f ff f f c c y x θ θθ cos1 sin − − = and ( )f fc y r θcos1− = Now we can compute ( ) cf f x ygrg θ λ cos1 2 1 2 1 − == ( ) const y g y g r g g td d td d cc f f f f x = = − ===== 22 sin cos1 2 1 2 1 2 1 θθ λω θγand ∫∫ === f c g y dtdt f f f AB θ θ θ θ ω0 * 2 2 sin 2 2 1 1. Brachistochrone Problems
- 22. 22 SOLO 1. Brachistochrone Problems Calculus of Variations - Problems Solution to Brachistochrone Problem 2 (continue – 8) If yfc is free, we found that ( ) ( ) ( ) ( ) ( ) ( ) πππγθπγγ γ λ 2,2,00sin0 sin =+=→=→=→=−= ffff f f fy ttt yV t t c Since for V0 = 0 we have θ0 = 0, we must have θf = π (γf = 0) Using this result we obtain: ( ) π πθθ c c f fff x rrrx =→=−= sin ( ) π θ c c f ff x rry 22cos1 ==−= cf x xgrg π λ 2 1 2 1 == const x g r g g td d td d cf x ====== π λω θγ 2 1 2 1 2 1 ∫∫ === π π θ ω0 * 2 1 g x dtdt cf AB For V0 = 0 Return to Table of Content
- 23. 23 HISTORY OF CALCULUS OF VARIATIONSSOLO “When the Tyrian princess Dido landed on the Mediterranean sea she was welcomed by a local chieftain. He offered her all the land that she could enclose between the shoreline and a rope of knotted cowhide. While the legend does not tell us, we may assume that Princess Dido arrived at the correct solution by stretching the rope into the shape of a circular arc and thereby maximized the area of the land upon which she was to found Carthage. This story of the founding of Carthage is apocryphal. Nonetheless it is probably the first account of a problem of the kind that inspired an entire mathematical discipline, the calculus of variations and its extensions such as the theory of optimal control.” (George Leitmann “The Calculus of Variations and Optimal Control – An Introduction” Plenum Press, 1981) 1. Dido Maximum Area Problem 1. Isoperimetric Problems
- 24. 24 Given a rope of length P connected to each end of straight line of length 2 a < P find the shape of the rope necessary to enclose the maximum area between the rope and the straight line. ( ) ( ) ( ) ∫∫∫∫∫ −−− =+= +=+== a a a a a a dxdxdx xd yd ydxdsdP θθ sectan11 2 2 22 subject o the isoperimetric constraint: where: θtan= xd yd SOLO Rope of length P ( )xθ x y a+a− y dx Calculus of Variations - Problems ( ) ∫∫ − == a a xy dxyAdJ maxmaxmaxThe problem can be formulated as: Solution 1. Dido Maximum Area Problem Isoperimetric Problems
- 25. 25 SOLO Calculus of Variations - Problems Therefore we have the following differential equations as constraint ( ) ( ) 0&0tan ==−= ayay xd yd θ ( ) ( ) Papap xd pd ==−= &0 cos 1 θ 1. Define the Hamiltonian: θ λθλ cos 1 tan: pyyH ++= and the end constraint function ( ) ( )Ppypyx py −+=Φ νν:,, 2. Euler-Lagrange and Boundary Conditions: ( ) ( ) ( )xax y a y H xd d yyyy y −+=→= ∂ Φ∂ =−= ∂ ∂ −= νλνλ λ 1 ( ) ( ) pypy p x p a p H xd d νλνλ λ =→= ∂ Φ∂ == ∂ ∂ −= 0 p y p y py xaH ν ν λ λ θ θ θ λ θ λ θ −+ −=−=→+= ∂ ∂ = * 22 sin cos sin cos 1 0 3. The extremal of the Hamiltonian is given for * * ** * ** cos cos 1 tansin cos 1 tan θλ θ θθλ θ θ λ λ λ pp p y p yyyH += +−+= ++= Rope of length P ( )xθ x y a+a− y dx Solution (continue – 1) 1. Dido Maximum Area Problem Isoperimetric Problems
- 26. 26 SOLO Calculus of Variations - Problems Since H is not an explicit function of x and ( ) constpyxH x H xd Hd py =→= ∂ ∂ = λλθ ,,,,,0 Therefore ( ) ( ) ** 0 ** 0 ** 0 coscos cos0cos0 ff fpp axHaxH θθθθ θλθλ ±=→=→ +===+=−= From p y xa ν ν θ −+ −=* sin we have ( ) ( ) p y p y ax a ax ν ν θ ν ν θ ± −==±= + −=−= ** sin 2 sin This equation is defined only when ± → -, and in this case p y p y a ν ν ν ν = + − 2 ( ) ( ) 0 ** θπθπθ −==−=−= axax and ay −=ν p x ν θ =→ * sin We have and 2 * 1cos −= p x ν θ 2 * 1 tan − == p p x x xd yd ν ν θ Rope of length P ( )xθ x y a+a− y dx Solution (continue – 2) 1. Dido Maximum Area Problem Isoperimetric Problems
- 27. 27 SOLO Calculus of Variations - Problems We obtained 2 * 1 tan − == p p x x xd yd ν ν θ Let integrate this equation ( ) 22 2 2 0 11 12 1 −+ −−= − − −=−− ∫− p p p p x a p p p ax x x d ayy ν ν ν ν ν ν ν or 2222 xay pp −−=−− νν Rope of length P ( )ax +=θ x y a+a− ( )ax −=θ pR ν= 22 ap −ν 0θ and finally 22 2 22 pp xay νν =+ −− This the equation of a circular arc with radius R = vp and center at − 22 ,0 apν Solution (continue – 3) 1. Dido Maximum Area Problem Isoperimetric Problems
- 28. 28 SOLO Calculus of Variations - Problems 1. Dido Maximum Area Problem Isoperimetric Problems Rope of length P ( )ax +=θ x y a+a− ( )ax −=θ pR ν= 22 ap −ν 0θ The radius R of the circular arc is found from the perimeter constraint = = − == − = −= − −− ∫∫ p p ax axp p a a p a a ax x xdxd P ν ν ν ν ν θ 11 2* sin2sin 1 cos or = p p P a ν ν 2 sin From the equations * sinθν px = * 0 * 22 coscos11 θνθν ν ν ν ν pp p p p p ax y −= −− −= The maximum Area enclosed by the Rope is ( ) −= 00 2 2sin 2 1 * θθν pA ( ) ( ) ( )∫∫∫∫ −−−− − + =−== 0 0 0 0 0 0 ** 0 2* * 2*** 0 ** sincos 2 2cos1 coscoscos θ θ θ θπ θ θπ θθνθ θ νθθνθνθν dddxdyA ppppp a a ( ) ( ) ( ) ( ) −= −+= − + = − 00 2 000 2 0 2 2sin 2 1 2sin2sin 2 1 sincos 2 2sin 2 1 0 θθνθθθνθθ θθ ν θ θπ ppp The maximum Area Solution (continue – 4) Return to Table of Content
- 29. 29 SOLO Calculus of Variations - Problems Given an area of canvas, A, to built a tent, find the shape of the canvas necessary to cover a circular floor area of radius a ( π a2 < A ) with maximum volume inside the tent. x y z r dr Solution ( ) ( )∫∫ == aa drrrzdrrrzV 00 max22maxmax ππ where z (r) is the height of the canvas surface at radius r. Define ( ) ( ) ( ) rdrd rd zd zdrd rd rzd θ θ cos 1 1tan 2 22 = +=+→= and ( ) ( ) ∫∫∫ = +=+= aa rd r rd rd zd zdrdrA 00 2 22 cos 2122 θ πππ Isoperimetric Problems 2. Maximum volume with given surface area
- 30. 30 SOLO Calculus of Variations - Problems x y z r dr Solution (continue – 1) Summarize ∫= a drrzJ 0 2π ( ) ( ) ( ) ( ) 0&0tan &00 cos 2 == === azdefinednotz rd zd AaAA r rd Ad θ θ π where 1. Define Hamiltonian θλ θ πλπ tan cos 22: zA r rzH ++= 2. Euler-Lagrange Equations and Boundary Conditions const A H rd d A A == ∂ ∂ −= λ λ 0 ( ) ( )( ) 2 0sin002 rdefinednotzcer z H rd d zz z πλλπ λ −=→=−= ∂ ∂ −= 22 22 2 1cos 222 sin cos 1 cos sin 20 −=→==−=→++= ∂ ∂ = AAAA z zA rr r r r r H λ θ λλπ π λπ λ θ θ λ θ θ πλ θ 3. The extremal of the Hamiltonian is given for Isoperimetric Problems 2. Maximum volume with given surface area
- 31. 31 SOLO Calculus of Variations - Problems x y z r dr Solution (continue – 2) The following relations are developed 2 2 1 2 cos sin tan − === A A r r rd zd λ λ θ θ θ from which ∫∫ − −= − = − =− a A A A A A a A A r r r d rd r r tzz 0 2 2 2 0 2 0 1 2 12 2 1 2 2 2 1 2 λ λ λ λ λ λ λ that can be written as ( ) ( )222 0 22 AA rzz λλ =++− Since 222 yxr += we obtain ( ) ( )2222 0 22 AA yxzz λλ =+++− Isoperimetric Problems 2. Maximum volume with given surface area
- 32. 32 SOLO Calculus of Variations - Problems Solution (continue – 3) we obtained ( ) ( )2222 0 22 AA yxzz λλ =+++− This is the equation of a spherical surface with radius 2 λ A and center on the vertical axis passing trough the center of the circular floor and located at .Azz λ20 −= We want to find the two unknowns z0 and λA (1) by using the given A, the area of the canvas ( ) ( ) ( ) −−= −−= −−= − == = = ∫∫ 22 2 2 0 2 2 0 2 2 2 0 2222 2 118 2 18 2 1 2 22 cos 2 a a r r r d rd r A AAA A A ar r A A a A A A a λλλπ λ πλ λ πλ λ λ λπ θ π This is a transcedental equation in 2 λ A . Isoperimetric Problems 2. Maximum volume with given surface area
- 33. 33 SOLO Calculus of Variations - Problems Isoperimetric Problems 2. Maximum volume with given surface area Solution (continue – 4) (2) by using the condition z (r=a) = 0 ( ) ( ) ( ) 22 0 222 0 2222 azaz AAAA −±=→=++− λλλλ We can see that for 2 λ A = a 2 2 aA π= 020 =+− Az λ the spherical surface has the radius a and its center on the center floor. For A > 2 π a2 we have ( ) 22 0 22 az AA −=− λλ the center of the spherical canvas surface is above the floor For A< 2 π a2 we have ( ) 22 0 22 az AA −−=− λλ the center of the spherical canvas surface is below the floor Return to Table of Content
- 34. 34 SOLO Calculus of Variations - Problems A liquid drop on a horizontal surface assume an axially symmetric shape that minimize the sum of the liquid’s potential energy in the earth’s gravitational field plus the liquid’s surface energy. Solution: Because of the axially symmetry let consider a) a disc, at a height z ,that touches the external surface and has a radius r(z) , and width dz that has the potential energy ( )( )zdzrzgmdzgUd vol 2 πρ−=−= from which we obtain ( ) rd rd zd rzgzdzrzgU a vol ∫∫ −=−= 0 22 ρπρπ b) the change in potential energy is given by the change in the axially symmetric surface due to surface tension ( ) ( ) ( )( ) rd rd zd rrzdrdrsdrUd surface 2 22 1222 +=+== πσπσπσ from which we obtain ∫ += a surface rd rd zd rU 0 2 12 σπ x y r ds σσ dr σ σ σ dz a )(zrz Isoperimetric Problems 3. Shape of a liquid drop on a horizontal surface
- 35. 35 SOLO Calculus of Variations - Problems Solution (continue – 1): x y r ds σσ dr σ σ σ dz a )(zrz The sum of the liquid’s potential energy in the earth’s gravitational field plus the liquid’s surface energy that is minimized is given by: ∫∫ ++−=+= aa surfacevol rd rd zd rrd rd zd rzgUUU 0 2 0 2 12 σπρπ The volume of the liquid drop is fixed rd rd zd rV a ∫ = 0 2 π Let define u rd zd == θtan Summarizing ∫∫ ++−= aa rdurrdurzgJ 0 2 0 2 12 σπρπ a is not defined and we have the constraints ( ) ( ) ( ) ( ) VaVVur rd Vd azdefinednotzu rd zd === == &00 0&0 2 π Isoperimetric Problems 3. Shape of a liquid drop on a horizontal surface
- 36. 36 SOLO Calculus of Variations - Problems Solution (continue – 2): x y r ds σσ dr σ σ σ dz a )(zrz1. The Hamiltonian is given by uruururgzH Vz 222 12 πλλσπρπ ++++−= 2. Euler-Lagrange Equations and Boundary Conditions const V H rd d V V == ∂ ∂ −= λ λ 0 ( ) ( )( )definednotzceugr z H rd d z z 0sin002 == ∂ ∂ −= λρπ λ 2 2 2 1 20 r u u rrgz u H Vz πλλσπρπ ++ + +−= ∂ ∂ = 3. The extremal of the Hamiltonian is given for Let take the derivative of the last equation relative to r: ( ) rrgu rd ud u u u r u u rgzrgu Vπλρπσπσπρπρπ 2 11 1 2 1 220 2 22 2 22 2 ++ + − + + + +−−= ( ) σ λ σ ρ V z g u u rrd ud u −=− + + + 232 1 1 1 1 or Isoperimetric Problems 3. Shape of a liquid drop on a horizontal surface
- 37. 37 SOLO Calculus of Variations - Problems Isoperimetric Problems 3. Shape of a liquid drop on a horizontal surface Solution (continue – 3): x y r ds σσ dr σ σ σ dz a )(zrz ( ) σ λ σ ρ V z g u u rrd ud u −=− + + + 232 1 1 1 1 For u << 1 and using we obtainu rd zd = σ λ σ ρ V z g rd zd rrd zd −=−+ 1 2 2 which has the solution ( ) + += r g KBr g IA g rz V σ ρ σ ρ ρ λ 00 where I0 and K0 are Bessel functions of order zero with imaginary argument, and A, B and λV to be defined. Return to Table of Content
- 38. 38 SOLO Problem of Minimum Surface of Revolution Given two points A (a,ya) and B (b, yb) a≠b in the plane. Find the curve that joints these two points with a continuous derivative, in such a way that the surface generated by the rotation of this curve about the x axis has the smallest possible area. x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Minimum Surface of Revolution The surface generated by the rotation of y (x) curve about the x – axis can be calculated using ( ) ( ) xd xd yd yydxdysdydS 2 22 1222 +=+== πππ Therefore ( )∫ +== b a xd xd yd xySJ 2 12: π 2 1,, += xd yd y xd yd yxF We can see that F is not an explicit function of x. Calculus of Variations - Problems
- 39. 39 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Minimum Surface of Revolution Solution: Calculus of Variations - Problems 1. Euler-Lagrange Equation 0=− yy F xd d F 2 1,, += xd yd y xd yd yxF 2 12 yFy += π 2 1 2 y y yFy + = π ( ) ( ) + + + = + − + + = 3 22 2 3 2 2 2 2 11 2 11 2 y yy y y y yyy y yyy F xd d y ππ with Substituting in the Euler-Lagrange Equation gives ( ) ( ) ( ) 0 1 2 1 1 2 11 1 2 11 12 2 32 2 322322 2 2 = + + −+ = + − + = + − + −+=− y y xd d y or y yyy y yy yy yy y y yF xd d F yy π π ππ
- 40. 40 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Minimum Surface of Revolution Solution (continue – 1): Calculus of Variations - Problems ( ) 0 1 2 1 1 2 2 32 2 = + + −+ =− y y xd d y or y yyy F xd d F yy π π E-L. Equation 01 2 =−+ yyy (1) nonlinear second order differential equation 0 1 2 = + y y xd d )2( Second Way to solve Euler-Lagrange Equation Since F is not an explicit function of x, we can write: ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) −=−−+=− yyF xd d yyFyyyF xd d yyyFyyyFyyyFyyyFyyyF xd d yyyyyyy ,,,,,,,, In our case ( ) ( ) C y y yyyyyFyyyF y ππ 2 1 12,, 2 2 2 = + −+=− 2 1 yCy +=or We obtained the same result We obtain two equivalent equations constC y y == + 2 1
- 41. 41 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Solution (continue – 2): Calculus of Variations - Problems Third Way to obtain Euler-Lagrange Equation Start from the expression of d S and change the free variable from x to y ( ) ( ) yd yd xd yydxdysdydS 2 22 1222 +=+== πππ Define ( ) yd xd xxyxxyF =←+= :12:,, ~ 2 π Euler-Lagrange Equation ( ) ( ) 0,, ~ ,, ~ =− xxyF yd d xxyF xx Since is not a function of x( ) 2 12:,, ~ xyxxyF += π ( ) 0,, ~ =xxyFx Therefore ( ) ( ) 1 2 1 1 2 1 2 ,, ~ 0,, ~ 2 2 2 + = + = + ==→= xd yd y x y x xy xxyFconstxxyF yd d xx πππ We recovered equation 2 1 yCy +=
- 42. 42 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Solution (continue – 3): Calculus of Variations - Problems Solving the Euler-Lagrange Equation 1 2 − = C y yRewrite the E-L Equation as Separating variables, we obtain C xd C y C yd = − 1 2 Integration of this equation, gives − +=− 1ln 2 1 C y C y CCx from which 1exp 2 1 − += − C y C y C Cx take the square 1exp211212122exp 1 222 1 − − =− − +=− +− = − C Cx C y C y C y C y C y C y C y C Cx From this equation we can compute 2 expexp 11 − −+ − = C Cx C Cx C y
- 43. 43 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Solution (continue – 4): Calculus of Variations - Problems Solving the Euler-Lagrange Equation (continue – 1) ( ) − = C Cx Cxy 1 coshWe obtained The solution is a curve called a catenary (catena = chain in Latin) and the Surface of Revolution which is generated is called a catenoid of revolution. The two parameters C and C1 are defined by the two End Conditions: − = − = C Cb Cy C Ca Cy b a 1 1 cosh cosh Since we have only two parameters C and C1 and three independent variables ya, yb, b-a , that define the end conditions, a solution not always exists.
- 44. 44 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Solution (continue – 5): Calculus of Variations - Problems Solving the Euler-Lagrange Equation (continue – 2) To see when a solution exists , without loss of generality choose aybBCCa /&/&0 1 === λ − = − = C Cb Cy C Ca Cy b a 1 1 cosh cosh We have λcosh ay C = ( ) λ λλ cosh coshcosh − = B y y a b Since ( ) ( ) tt tt t ∀≥ −+ = 2 expexp cosh 1 cosh ≤ λ λ therefore 0 cosh sinhcosh cosh 1 cosh max 2 = − = →≤ λ λλλ λ λ λλ λ λ d d The minimum is obtained for a positive λ0 that is a solution of . 0sinhcosh 000 =− λλλ ( ) λ λ λ λ λ λλ λ λλ coshcoshcosh cosh cosh coshcosh −≥−= − ≥ − = BB BB y y a bwe have We obtained
- 45. 45 SOLO Problem of Minimum Surface of Revolution Solution (continue – 6): Calculus of Variations - Problems Solving the Euler-Lagrange Equation (continue – 3) We can see that if we don't have a solution to the minimum problem. 0 0 cosh λ λ −< aa b y b y y λ λ λ λ coshcosh −=−≥ aa b y b B y y A solution to the minimum surface problem exists if: 0sinhcosh 000 =− λλλ where λ > λ0 and λ0 is given by Let define ( ) a a b y y b y −= 0 0 0 cosh : λ λ λ ( ) existscatenaryayyIf bb 0λ≥ ( ) existstdoesncatenaryayyIf bb '0λ< ( )ayaA , x y ( )bybB , ( ) ( )22 ydxdsd += y x y ( )bybB , ( )ayaA , Catenary exists Catenary does not exists
- 46. 46 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Legendre's Condition for a Weak Local Minimum Let compute ( ) ( ) 0 1 2 11 1 2 1 2 3 2 3 2 2 22 ≥ + = + − + = +∂ ∂ = ∂ ∂ = y y y y y y y y y y F y F yyy π ππ We can see that the Legendre's Necessary Condition for a minimum is satisfied.
- 47. 47 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Weierstrass' Necessary Condition for a Strong Local Minimum Let compute the Weierstrass' Excess Function E ( ) ( ) ( ) ( ) ( ) ( )YyyY y y y yYyy Yy y y yYyYyyyFyYyyFYyFE y −++ + = + −++ −+= + −−+−+=−−−= 22 22 22 2 2 22 11 1 2 1 1 12 1 112,,, π π π By adding to both sides of the inequality , we obtain22 Yy 01 22 >++ yY ( ) ( ) 22222222 111 YyyYYyyY >++=+++ By taking the positive square root of this inequality, we obtain YyyY >++ 22 11 Therefore ( ) 011 1 2 22 2 >−++ + = YyyY y y E π We can see that the Weierstrass' Necessary Condition for a strong minimum is satisfied.
- 48. 48 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Conjugate Points and Jacobi’s Equation To have an extremal for x є [a,b] we must show that there are no conjugate points to A (a, ya) in this interval. This can be shown by finding the non-trivial solution (u(x)≠0) of the Jacobi’s Equation: 02 2 = −+ −++ uP xd Qd xd ud QQ xd Rd xd ud R T T where yyyyyy FRFQFP === ::: 2 12 yFy += πWe found we can see that 0:,0: ==== yyyy FQFP ( ) 0 1 2 3 2 ≥ + == y y FR yy π We also found constC y y == + 2 1 Using in the previous equation we obtain 0 2 1 2 2 3 2 ≥= + == y C y C FR yy ππ − += − == C CxCC C Cx CCyC xd ud 1 2 1222 2cosh1 2 ~ cosh ~~ 2 2 3 2 2 ~2 0 yC xd ud const xd ud y C xd ud R xd ud R xd d xd ud xd Rd xd ud R =→==→= =+ π Substitute those results in the Jacobi’s Equation
- 49. 49 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Conjugate Points and Jacobi’s Equation − += C CxCC xd ud 1 2 2cosh1 2 ~ ( ) ( ) ∫ − +=− x a xd C CxCC auxu 1 2 2cosh1 2 ~ ( ) ( ) − − − +−= − ++= = C Ca C CxC ax CC C CxC x CC auxu x ax 11 2 1 2 0 2sinh2sinh 22 ~ 2sinh 22 ~ ( ) bxa C Cax C ax Cax CC xu ≤<> −+ − +−= 0 2 coshsinh 2 ~ 1 2 Therefore there are non-trivial solution (u(x)≠0) for a < x ≤ a and there are no Conjugate Points in this interval.
- 50. 50 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Canonical Equations ( ) 2 12:,, yyyyxF += π Define 2 1 2 : y yy Fp y + == π from which ( ) 222 22 2 222 2 222222 4 4 1 4 41 py y y py p yyypy − =+→ − =→=+ π π π π Substituting this in equation gives( )yyxF ,, ( ) ( ) 222 22 4 4 ,,:,, ~ py y yyxFpyxF yFp − == = π π 1. The Hamiltonian is ( ) ( ) 222 222 2 222 22 4 44 4 ,,:,, py py p py y ypyyxFpyxH −−= − + − −=+−= π ππ π 2. The canonical equations are ( ) ( ) 222 2 222 4 4,, 4 ,, py y y pyxH xd pd py p p pyxH xd yd − = ∂ ∂ −= − = ∂ ∂ = π π π
- 51. 51 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Canonical Equations The canonical equations are ( ) ( ) 222 2 222 4 4,, 4 ,, py y y pyxH xd pd py p p pyxH xd yd − = ∂ ∂ −= − = ∂ ∂ = π π π If we divide the first equation by the second, we obtain y p pd yd 2 4 π = or pdpydy =2 4 π Integrating this equation gives 22222 4.4 Cconstpy ππ ==− The reason that we choose the constant as const .= 4 π2 C2 is to recover previous results , as we shall see. If we substitute this expression in the Hamiltonian expression we obtain: ( ) CpypyxH ππ 24,, 222 −=−−= The fact that the Hamiltonian is constant on the extremal trajectory is a consequence of the fact that H is not an explicit function of x. The canonical equations become y Cxd pd p Cxd yd π π 2 2 1 = =
- 52. 52 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Canonical Equations The canonical equations are y Cxd pd p Cxd yd π π 2 2 1 = = Let Differentiate first equation with respect to x and use the second, to obtain y Cxd pd Cxd yd 22 2 1 2 1 == π The general solution of this Ordinary Differential Equation is: ( ) − = C Cx Cxy 1 cosh We recovered the “catenary” equation.
- 53. 53 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Hamilton-Jacoby Equation ( ) ( ) ( ) 22 2 2 1 2 1 12,,,,, y y y y yyyyyxFyyyxFyxS yx + = + −+=−= π π ( ) ( ) p y yy yyxFyxS yy = + == 2 1 2 ,,, π Using in the first equation, we obtainC y y = + 2 1 ( ) C y y yxSx π π 2 1 2 , 2 = + = from which we obtain 1 2 2 − = C y y Substituting this equation in the Sy (x,y) equation, we obtain: ( ) 12 12 1 2 , 2 2 2 − = − = + = C y C C y C y y y yy yxSy π π π ( ) ( ) 022,,, =−=+ CCSyxHyxS yx ππWe have Hamilton-Jacoby Equation We find that the Hamiltonian is ( ) CpypyxH ππ 24,, 222 −=−−=
- 54. 54 SOLO Problem of Minimum Surface of Revolution x y ( )bybB , ( )ayaA , ( ) ( )22 ydxdsd += y Calculus of Variations - Problems Hamilton-Jacoby Equation − +=+= yd C y xdCydSxdSSd yx 12 2 π Integrating this, to obtain: 0 22 0 2 1ln 2 1 2 1 212 S C y C yC C y yxCSyd C y xdCS + − +−− +=+ − += ∫∫ ππ Using the expressions for Sx (x,y) and Sy (x,y), we obtain Return to Table of Content
- 55. 55 SOLO 4. Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces Suppose we have a surface specified by two parameters u, v and the vector .( )vur , The shortest path lying on the surface and connecting to points of the surface is called a Geodesic. A B ( )vur , vdrv udru rd The Shortest Path on a Surface The arc length differential is td td vd v r v r td vd td ud v r u r td ud u r u r td td vd v r td ud u r td vd v r td ud u r td td rd td rd td td rd ds 2/122 2/12/1 2 ∂ ∂ ⋅ ∂ ∂ + ∂ ∂ ⋅ ∂ ∂ + ∂ ∂ ⋅ ∂ ∂ = ∂ ∂ + ∂ ∂ ⋅ ∂ ∂ + ∂ ∂ = ⋅== The length of the path between the two points A and B is: ∫ + + == B A r r td td vd G td vd td ud F td ud ESJ 2/122 2: where ( ) ( ) ( ) ∂ ∂ ⋅ ∂ ∂ = ∂ ∂ ⋅ ∂ ∂ = ∂ ∂ ⋅ ∂ ∂ = v r v r vuG v r u r vuF u r u r vuE :,,:,,:,
- 56. 56 SOLO Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces A B ( )vur , vdrv udru rd The Shortest Path on a Surface We have ( ) ( ) ( ) ( ) ( )( ) ( ) ( )[ ] dt ds vvuGvuvuFuvuEvuvutF =++= 2/122 ','',2',',',,, where td vd v td ud u == :',:' 1. Euler-Lagrange Equations ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )[ ] 0 ','',2', ',', ','',2',2 ','',2', ',',,,',',,, 2/1222/122 22 ' = ++ + − ++ ++ = − vvuGvuvuFuvuE vvuFuvuE td d vvuGvuvuFuvuE vvuGvuvuFuvuE vuvutF td d vuvutF uuu uu ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )[ ] ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )[ ] 0 ','',2', ',', ','',2',2 ','',2', ',',,,',',,, 2/1222/122 22 ' = ++ + − ++ ++ = − vvuGvuvuFuvuE vvuGuvuF td d vvuGvuvuFuvuE vvuGvuvuFuvuE vuvutF td d vuvutF vvv vv
- 57. 57 SOLO Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces If instead of t we use the variable s, we obtain ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 ,, 2 ,,2, ',',,,',',,, 22 ' = + − + + = − dt ds dt ds ds dv vuF dt ds ds du vuE dt ds sd d dt ds dt ds ds dv vuG dt ds ds dv dt ds ds du vuF dt ds ds du vuE vuvutF td d vuvutF uuu uu A B ( )vur , vdrv udru rd ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 ,, 2 ,,2, ',',,,',',,, 22 ' = + − + + = − dt ds dt ds ds dv vuG dt ds ds du vuF dt ds sd d dt ds dt ds ds dv vuG dt ds ds dv dt ds ds du vuF dt ds ds du vuE vuvutF td d vuvutF vvv vv ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0,,,,2, 2 1 0,,,,2, 2 1 22 22 = +− + + = +− + + ds dv vuG ds du vuF sd d ds dv vuG ds dv ds du vuF ds du vuE ds dv vuF ds du vuE sd d ds dv vuG ds dv ds du vuF ds du vuE vvv uuu or
- 58. 58 SOLO Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces Example: Spherical Surface A spherical surface is defined by or ( ) ( )kjiRr ˆcosˆsinsinˆcossin, θφθφθθφ ++= θ φφ φθ θθ 22 2 sin 0 R rr G rr F R rr E = ∂ ∂ ⋅ ∂ ∂ = = ∂ ∂ ⋅ ∂ ∂ = = ∂ ∂ ⋅ ∂ ∂ = ( ) ( ) ( ) ( ) ( ) ( ) θ θ φ θφθθφφθθ d d d RddRdGddFdEsd 2 222222 sin1sin2 +=+=++= θθθφθφθφ cossin20 2 RGGFFEE ====== Let write the Euler-Lagrange Equations for this particular problem 02 2 1 02 2 1 22 22 = +− + + = +− + + ds dv G ds du F sd d ds dv G ds dv ds du F ds du E ds dv F ds du E sd d ds dv G ds dv ds du F ds du E vvv uuu φθθ dRdvdRdu sin, == 0sin 0cossin 22 2 2 3 2 3 = − =− ds d R sd d sd d R sd d R φ θ θφ θθ φθφθφ cos,sinsin,cossin RzRyRx ===
- 59. 59 SOLO Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces Example: Spherical Surface (continue – 1) 0sin 0cossin 22 2 2 3 2 3 = − =− ds d R sd d sd d R sd d R φ θ θφ θθ From last equation and we obtainθ θ φ θ d d d Rsd 2 2 sin1 += constAR d d d d R ds d d d R ds d R == + == 2 2 2 2222 sin1 sin sinsin θ φ θ θ φ θ θ θ φ θ φ θ 2 222 2 4 2 2 2 sinsin sin1 sin += ⇒= + θ φ θ θ φ θ θ φ θ θ φ θ d d AA d d A d d d d This equation can be rewritten as θ θ θθθ φ 2 2 2 22 sin 1sin sinsin A A A A d d − ± = − ± =
- 60. 60 SOLO Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces Example: Spherical Surface (continue – 2) We obtain θ θ θ φ 2 2 2 sin 1sin A A d d − ± = To solve this equation let write the constant A as the sinus of the constant A = sin α and define a new independent variable w as w = ctg θ. θθ θ 2 2 2 sin 1 1, sin =+−= w d wdTherefore ( ) ( ) α α α α θ θ α θ αθ θ φφ 2222 2 2 2 2 tan1 tan sin11 sin sin sin sin 1sin sin wwdw d d d dw d − = +− =− − ± == −⇒± and ffffffff RzRyRxB RzRyRxA φθφθφ φθφθφ cos,sinsin,cossin: cos,sinsin,cossin: 00000000 === === The Geodesic must pass through the points A and B given by: . tan1 tan 22 constdw w d = − = α α α φThe Geodesic O,D.E. is
- 61. 61 SOLO Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces Example: Spherical Surface (continue – 2) We obtained the O. D. E. The solution is defined by 4 parameters ϕ0, θ0, ϕf. θf that define the initial and final points A and B. This satisfies the point A (ϕ0, θ0 ) constraints. . tan1 tan 22 constdw w d = − = α α α φ ( ) ( ) − +=−+= −− = −− 0 11 00 11 0 tan tan cos tan tan costancostancos θ α θ α φααφφ θctgw ww The O.D.E. can be easily integrated from which we obtain ( ) θ α φφ tan tan cos 0 =− To simplify the expressions, without affect the solution generality, let choose the direction of x and y axes, relative to points A and B such that 0 tan tan cos 0 1 = − θ α ( ) ff θφφα tancostan 0−=and
- 62. 62 SOLO Geodesics Problems Calculus of Variations - Problems 1. Geodesic in 3 Dimensional Spaces Example: Spherical Surface (continue –3) Great Circle ( ) θ α φφ tan tan cos 0 =− Multiply this equation by R sinϕ to obtain αθφφθφφθ tancossinsinsincoscossin 00 RRR =+ Using , in the previous equation, we obtain φθφθφ cos,sinsin,cossin RzRyRx === αφφ tansincos 00 zyx =+ This is the equation of a plane passing through x=y=z=0 (center of the sphere). Thus the geodesic of a spherical surface lies in a plane passing through the center of the sphere, therefore a Great Circle. Return to Table of Content
- 63. 63 SOLO Geodesics Problems Calculus of Variations - Problems 2. Geodesics in Riemannian Space Riemann Spaces Let assume a n dimensional space Vn in which we have a m (m , n) dimensional subspace Vm (Vm submersed in Vn).nm VV ⊂ Let take a point M defined by the vector as function of m independent parameters x1 , x2 ,…,xm : r ( ) ( ) ( ) nkexxxexxxxxxr km kk mkm ,,2,1,,,,,,,,, 212121 === ξξ → 1N M C → 2r → 1r 2V 3V → r Τ dt dr → → 1r → 2r → 1N constx =1 Τ and define the tangent hyper-surface T, defined by the m vectors tangents to C at the point M: mi x r r ii ,,2,1,: = ∂ ∂ = We assume that those vectors are linear independent since the parameters x1 , x2 ,…,xm are independent.
- 64. 64 SOLO Geodesics Problems Calculus of Variations - Problems 2. Geodesics in Riemannian Space Riemann Spaces → 1N M C → 2r → 1r 2V 3V → r Τ dt dr → → 1r → 2r → 1N constx =1 Τ m xxx ,,, 21 m xxx ,,, 21 Change of Coordinates from to ( ) ( )mm xxxrxxxr ,,,,,, 2121 = At each point since the parameters are independent the Jacobian is nonzero: ( ) ( ) 0det ,,, ,,, 1 21 1 2 1 1 1 21 21 ≠ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = ∂ ∂ m mm m m m x x x x x x x x x x x x xxx xxx We have: j j j j i i i i xd x r xd x x x r xd x r rd ∂ ∂ = ∂ ∂ ∂ ∂ = ∂ ∂ = ji j iij i i j i j j j mm i i j m m i i j i i j jj j i i r x x rr x x x r x x x r r x x x x xd x x x x xd x x xdxd x x xd ∂ ∂ = ∂ ∂ = ∂ ∂ ∂ ∂ = ∂ ∂ = = ∂ ∂ ∂ ∂ ⇒ ∂ ∂ ∂ ∂ = ∂ ∂ = ∂ ∂ = &: & δ
- 65. 65 SOLO Geodesics Problems Calculus of Variations - Problems 2. Geodesics in Riemannian Space Riemann Spaces → 1N M C → 2r → 1r 2V 3V → r Τ dt dr → → 1r → 2r → 1N constx =1 Τ mj x r r j j ,...,2,1: = ∂ ∂ = mi x r r i i ,...,2,1: = ∂ ∂ = The vectors are also in T since are given by linear combinations of Metric in T over the base . mi x r r i i ,...,2,1: = ∂ ∂ = Let assume that a Scalar Product is defined in T and define the Metric in T over the base asmi x r r i i ,...,2,1: = ∂ ∂ = ( ) ( ) jiij ji jiij grr x r x r rrg == ∂ ∂ ∂ ∂ == ,,,: Dual Base to . mi x r r i i ,...,2,1: = ∂ ∂ = Since are linear independent; i.e.:miri ,...,2,1= { } ( ) ( ){ } { } 0det,...,10,,0 ,...,100 ≠=⇔=∀=⇔===→ →=∀=⇔= ij i ij i ji i ji i i i i ggmigrrrr mir αααα αα
- 66. 66 SOLO Geodesics Problems Calculus of Variations - Problems 2. Geodesics in Riemannian Space Riemann Spaces → 1N M C → 2r → 1r 2V 3V → r Τ dt dr → → 1r → 2r → 1N constx =1 Τ the mxm matrix is nonsingular and we can define the matrix: { }ijgG = { } { } { }{ } mxmij ij ij ij IgggGg =→== −− ∆ 11 i jkj ik gg δ=or Let define now: mi x r grgr k ik k iki ,...,2,1: == ∂ ∂ == We can see that ( ) ( ) ( ) i jkj ik jk ik jk ik j i ggrrgrrgrr δ==== ,,, mir i ,...,2,1= miri ,...,2,1= Therefore are dual to . Moreover ( ) ( ) ( ) ijj k ikj k ikj k ikji ggrrgrrgrr ==== δ ,,, gik is the Metric of the dual base .mir i ,...,2,1=
- 67. 67 SOLO Geodesics Problems Calculus of Variations - Problems 2. Geodesics in Riemannian Space In a Riemann space the length of a space curve, defined by it’s coordinates xi (t), as a function of a parameter t, is given by ∫ = 1 0 2/1t t dt td xd td xd gs βα βα where we used the tensor index summation rule ∑∑= = ⇔ m m td xd td xd g td xd td xd g 1 1α β βα βα βα βα mi x r r i i ,...,2,1: = ∂ ∂ = ( ) ( ) jiij ji jiij grr x r x r rrg == ∂ ∂ ∂ ∂ == ,,,: Define m td sd td xd td xd g td xd xtF ,,1,:,, 2/1 == = βαβα βα To find the geodesics (the curve of the minimum length) let apply the Euler-Lagrange Equations mi x F x F td d ii ,,10 == ∂ ∂ − ∂ ∂ → 1N M C → 2r → 1r 2V 3V → r Τ dt dr → → 1r → 2r → 1N constx =1 Τ
- 68. 68 SOLO Geodesics Problems Calculus of Variations - Problems 2. Geodesics in Riemannian Space m td sd td xd td xd g td xd xtF ,,1,:,, 2/1 == = βαβα βα 1. To find the geodesics (the curve of the minimum length) let apply the Euler-Lagrange Equations mi x F x F td d ii ,,10 == ∂ ∂ − ∂ ∂ → 1N M C → 2r → 1r 2V 3V → r Τ dt dr → → 1r → 2r → 1N constx =1 Τ ( ) mi td xd g td xd g td sd dtdstd xd td xd x g td xd td xd x g td xd g td xd g dtds dtds td xd g td xd g dt d x F dt d ii ii ii ii i ,,1 /2 1 /2 1 /2 2 2 22 2 2 2 = +− ∂ ∂ + ∂ ∂ ++= + = ∂ ∂ β β α α βα α ββα β αβ β α α β β α α mi td xd td xd x g dtdsx F ii ,,1 /2 1 = ∂ ∂ = ∂ ∂ βαβα We have
- 69. 69 SOLO Geodesics Problems Calculus of Variations - Problems 2. Geodesics in Riemannian Space → 1N M C → 2r → 1r 2V 3V → r Τ dt dr → → 1r → 2r → 1N constx =1 Τ If we choose 0,1 2 2 ==→= td sd dt ds st mi sd xd sd xd x g x F ii ,,1 2 1 = ∂ ∂ = ∂ ∂ βαβα mi sd xd sd xd x g x g sd xd g sd xd sd xd x g sd xd sd xd x g sd xd g sd xd g x F ds d ii i ii ii i ,,1 2 1 2 1 2 2 2 2 2 2 = ∂ ∂ + ∂ ∂ += ∂ ∂ + ∂ ∂ ++= ∂ ∂ βα α β β αα α βα α ββα β αβ β α α Therefore we obtain: mi sd xd sd xd x g x g x g sd xd g i ii i ,,1 2 1 2 2 = ∂ ∂ − ∂ ∂ + ∂ ∂ + βαβα α β β αα α Multiplying by gri and summing on i, we obtain: mi sd xd sd xd x g x g x gg sd xd i ii ir r ,,1 22 2 = ∂ ∂ − ∂ ∂ + ∂ ∂ + βαβα α β β α We define ∂ ∂ − ∂ ∂ + ∂ ∂ =Γ i ii ir r x g x g x gg βα α β β α βα 2 : Christoffel Symbol of Second Kind mi sd xd sd xd sd xd rr ,,12 2 =Γ+ βα βα Differential Equations of the Geodesics Return to Table of Content
- 70. 70 SOLO Geodesics Problems Calculus of Variations - Problems 3. Geodesics for an Implicit Equation of the Surface Suppose that we want to find the geodesics on a surface defined by an implicit function: ( ) 0,, =zyxG We want to find the extremal of ∫ + + = ft t dt td zd td yd td xd s 0 222 Let adjoin the first equation to the second using the Lagrange multiplier μ (t). ( ) ( )∫ + + + = ft t dtzyxGt td zd td yd td xd s 0 ,, 222 µ Define ( ) ( )zyxGt td zd td yd td xd td zd td yd td xd zyxtF ,,:,,,,,, 222 µ+ + + = dt ds td zd td yd td xd f = + + = 222 :and
- 71. 71 SOLO Geodesics Problems Calculus of Variations - Problems 3. Geodesics for an Implicit Equation of the Surface The Euler-Lagrange Equations are: ( ) ( ) ( ) 0 1 0 1 0 1 = ∂ ∂ − = ∂ ∂ − = ∂ ∂ − z G t td zd ftd d y G t td yd ftd d x G t td xd ftd d µ µ µ ( ) ( )zyxGt td zd td yd td xd f td zd td yd td xd zyxtF ,,,,:,,,,,, µ+ = dt ds td zd td yd td xd f = + + = 222 : μ (t) can be eliminated to obtain ( )t z G td zd ftd d y G td yd ftd d x G td xd ftd d µ= ∂ ∂ = ∂ ∂ = ∂ ∂ 111 Using f = ds/dt we can write the last expression as ( )t z G td zd sd td sd d td sd y G td yd sd td sd d td sd x G td xd sd td sd d td sd µ= ∂ ∂ = ∂ ∂ = ∂ ∂ ( ) td sd t z G sd zd sd d y G sd yd sd d x G sd xd sd d µ = ∂ ∂ = ∂ ∂ = ∂ ∂ or
- 72. 72 SOLO Geodesics Problems Calculus of Variations - Problems 3. Geodesics for an Implicit Equation of the Surface ( ) td sd t z G sd zd sd d y G sd yd sd d x G sd xd sd d µ = ∂ ∂ = ∂ ∂ = ∂ ∂ We found From Differential Geometry we know that for a three dimensional curve ( ) ( ) ( )kszjsyisxr ˆˆˆ ++= we have ( ) ( ) ( ) k sd szd j sd syd i sd sxd sd rd t ˆˆˆ: ++== the unit vector tangent to the curve ( ) ( ) ( ) nk sd szd sd d j sd syd sd d i sd sxd sd d sd rd sd d sd td ˆˆˆˆ κ= + + = = is the unit vector that defines the principal normal to the curve and κ is the magnitude of the curvature. Since is a unit vector nˆ tˆ 0 ˆ ˆˆˆ =⋅=⋅ sd td tnt We also have k z G j y G i x G G ˆˆˆ ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ We can see that the principal normal to any point of the geodesic curve is parallel to the normal to the surface.
- 73. 73 SOLO Geodesics Problems Calculus of Variations - Problems 3. Geodesics for an Implicit Equation of the Surface Example: Geodesics on a Spherical Surface ( ) 0,, 2222 =−++= RzyxzyxGSpherical Surface: kzjyixk z G j y G i x G G ˆ2ˆ2ˆ2ˆˆˆ ++= ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ Let differentiate twice G (x,y,z) = 0 as a function of s 0=++ sd zd z sd yd y sd xd x ( ) 0,, =zyxG sd d ( ) 0,,2 2 =zyxG sd d02 2 2 2 2 2222 =+++ + + sd zd z sd yd y sd xd x sd zd sd yd sd xd 1 222 = + + sd zd sd yd sd xd dt ds td zd td yd td xd f = + + = 222 : dsdt = ( ) td sd t z G sd zd sd d y G sd yd sd d x G sd xd sd d µ = ∂ ∂ = ∂ ∂ = ∂ ∂ ( )tk z sd zd y sd yd x sd xd === 222 2 2 2 2 2 2 ( ) ( ) ( ) ( ) 2 2222 2 1 02121 R tkRtkzyxtk −=→=+=+++
- 74. 74 SOLO Geodesics Problems Calculus of Variations - Problems 3. Geodesics for an Implicit Equation of the Surface Example: Geodesics on a Spherical Surface ( ) 2 2 2 2 2 2 2 2 1 222 R tk z sd zd y sd yd x sd xd −====We found + − + − + − R s C R s Cz R s C R s Cy R s C R s Cx cossin cossin cossin 65 43 21 The solutions of those equation are 0 cos sin 1 65 43 21 = − R s R s CCz CCy CCx 0 0 0 22 2 22 2 22 2 =+ =+ =+ R z sd zd R y sd yd R x sd xd
- 75. 75 SOLO Geodesics Problems Calculus of Variations - Problems 3. Geodesics for an Implicit Equation of the Surface Example: Geodesics on a Spherical Surface 0 cos sin 1 65 43 21 = − R s R s CCz CCy CCx We found The solution exists only if 0det 65 43 21 = CCz CCy CCx 0=++ CzByAx Thus the Geodesic on a Spherical Surface lies in a plane passing through the center of the sphere, therefore a Great Circle. Great Circle Return to Table of Content
- 76. 76 SOLO Geometrical Optics and Fermat Principle The Principle of Fermat (principle of the shortest optical path) asserts that the optical length of an actual ray between any two points is shorter than the optical ray of any other curve that joints these two points and which is in a certain neighborhood of it. An other formulation of the Fermat’s Principle requires only Stationarity (instead of minimal length). ∫ 2 1 P P dsn An other form of the Fermat’s Principle is: Principle of Least Time The path following by a ray in going from one point in space to another is the path that makes the time of transit of the associated wave stationary (usually a minimum). Calculus of Variations - Problems
- 77. 77 SOLO We have: ( ) ( ) ( )∫∫∫∫ = + +=== 2 1 2 1 2 1 ,,,, 1 1,, 1 ,, 1 0 22 00 P P P P P P xdzyzyxF c xd xd zd xd yd zyxn c dszyxn c tdJ Let find the stationarity conditions of the Optical Path using the Calculus of Variations ( ) ( ) ( ) xd xd zd xd yd zdydxdds 22 222 1 + +=++= Define: xd zd z xd yd y == &: ( ) ( ) ( ) 22 22 1,,1,,,,,, zyzyxn xd zd xd yd zyxnzyzyxF ++= + += constS = constdSS =+ sˆ ∫ 2 1 P P dsn 1 P 2 P Paths of Rays Between Two Points Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 78. 78 SOLO Necessary Conditions for Stationarity (Euler-Lagrange Equations) ( ) ( ) ( ) 22 22 1,,1,,,,,, zyzyxn xd zd xd yd zyxnzyzyxF ++= + += 0= ∂ ∂ − ∂ ∂ y F y F dx d ( ) [ ] 2/122 1 ,, zy yzyxn y F ++ = ∂ ∂ [ ] ( ) y zyxn zy y F ∂ ∂ ++= ∂ ∂ ,, 1 2/122 ( ) [ ] [ ] 01 1 ,, 2/122 2/122 = ∂ ∂ ++− ++ y n zy zy yzyxn xd d 0= ∂ ∂ − ∂ ∂ z F z F dx d [ ] [ ] 0 11 2/1222/122 = ∂ ∂ − ++++ y n zy yn xdzy d Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 79. 79 SOLO Necessary Conditions for Stationarity (continue - 1) We have [ ] 0 1 2/122 = ∂ ∂ − ++ y n zy yn sd d y n sd yd n sd d ∂ ∂ = In the same way [ ] 0 1 2/122 = ∂ ∂ − ++ z n zy zn sd d z n sd zd n sd d ∂ ∂ = Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 80. 80 SOLO Necessary Conditions for Stationarity (continue - 2) Using ( ) ( ) ( ) xd xd zd xd yd zdydxdds 22 222 1 + +=++= we obtain 1 222 = + + sd zd sd yd sd xd Differentiate this equation with respect to s and multiply by n sd d 0= + + sd zd sd d n sd zd sd yd sd d n sd yd sd xd sd d n sd xd sd nd sd zd sd nd sd yd sd nd sd xd sd nd = + + 222 sd nd and sd nd sd zd n sd d sd zd sd yd n sd d sd yd sd xd n sd d sd xd = + + add those two equations Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 81. 81 SOLO Necessary Conditions for Stationarity (continue - 3) sd nd sd zd n sd d sd zd sd yd n sd d sd yd sd xd n sd d sd xd = + + Multiply this by and use the fact that to obtain xd sd cd ad cd bd bd ad = xd nd sd zd n sd d xd zd sd yd n sd d xd yd sd xd n sd d = + + Substitute and in this equation to obtain y n sd yd n sd d ∂ ∂ = z n sd zd n sd d ∂ ∂ = xd zd z n xd yd y n xd nd sd xd n sd d ∂ ∂ − ∂ ∂ −= Since n is a function of x, y, z x n xd zd z n xd yd y n xd nd zd z n yd y n xd x n nd ∂ ∂ = ∂ ∂ − ∂ ∂ −→ ∂ ∂ + ∂ ∂ + ∂ ∂ = and the previous equation becomes x n sd xd n sd d ∂ ∂ = Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 82. 82 SOLO Necessary Conditions for Stationarity (continue - 4) We obtained the Euler-Lagrange Equations: x n sd xd n sd d ∂ ∂ = y n sd yd n sd d ∂ ∂ = z n sd zd n sd d ∂ ∂ = k sd zd j sd yd i sd xd sd rd kzjyixr ˆˆˆ ˆˆˆ ++= ++= Define the unit vectors in the x, y, z directionskji ˆ,ˆ,ˆ The Euler-Lagrange Equations can be written as: n sd rd n sd d ∇= This is the Eikonal Equation from Geometrical Optics. Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 83. 83 SOLO Transversality Conditions for Geometrical Optics and Fermat’s Principle ( ) ( ) ( ) 22 22 1,,1,,,,,, zyzyxn xd zd xd yd zyxnzyzyxF ++= + += For the Geometrical Optics we obtained: Assume that the initial and final boundaries are defined by the surfaces A (x0, y0, z0) and B (xf, yf, zf) respectively. The transversality conditions at the boundaries i=0,f are defined by ( ) ( ) ( )[ ] ( ) ( ) 0,,,,,,,, ,,,,,,,,,,,, =++ −− iziy izy dzzyzyxFdyzyzyxF dxzyzyxFzzyzyxFyzyzyxF ( ) [ ] [ ] [ ] [ ] ( ) sd xd zyxn zy n zy zn z zy yn yzynFzFyzyzyxF zy ,, 1 11 1,,,, 2/122 2/1222/122 2/122 = ++ = ++ − ++ −++=−− ( ) [ ] ( ) ( ) [ ] ( ) sd zd zyxn zy zzyxn z F F sd yd zyxn zy yzyxn y F F z y ,, 1 ,, ,, 1 ,, 2/122 2/122 = ++ = ∂ ∂ = = ++ = ∂ ∂ = For are tangent to the boundary surfaces A (x0, y0, z0) and B (xf, yf, zf).fird i ,0= From Transversality Conditions we can see that the rays are normal (transversal) to the boundary surfaces (see Figure). Transversality Conditions Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 84. SOLO Corner Conditions for Geometrical Optics and Fermat’s Principle ( ) ( ) ( ) 22 22 1,,1,,,,,, zyzyxn xd zd xd yd zyxnzyzyxF ++= + += For the Geometrical Optics we obtained: Let examine the following two cases: 1. The optical path passes between two regions with different refractive indexes n1 to n2 (see Figure) In region (1) we have: In region (2) we have: ( ) ( ) 22 11 1,,,,,, zyzyxnzyzyxF ++= ( ) ( ) 22 22 1,,,,,, zyzyxnzyzyxF ++= ( ) ( ) ( )[ ]{ ( ) ( ) ( )[ ]} ( ) ( )[ ] ( ) ( )[ ] 0,,,,,,,, ,,,,,,,, ,,,,,,,,,,,, ,,,,,,,,,,,, 222111 222111 22222222222 11111111111 =−+ −+ −−− −− dzzyzyxFzyzyxF dyzyzyxFzyzyxF dxzyzyxFzzyzyxFyzyzyxF zyzyxFzzyzyxFyzyzyxF zz yy zy zy The Weierstrass-Erdmann necessary condition at the boundary between the two regions is where dx, dy, dz are on the boundary between the two regions. Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 85. SOLO Corner Conditions for Geometrical Optics and Fermat’s Principle (continue – 1) ( ) [ ] [ ] [ ] [ ] ( ) sd xd zyxn zy n zy zn z zy yn yzynFzFyzyzyxF zy ,, 1 11 1,,,, 2/122 2/1222/122 2/122 = ++ = ++ − ++ −++=−− ( ) [ ] ( ) ( ) [ ] ( ) sd zd zyxn zy zzyxn z F F sd yd zyxn zy yzyxn y F F z y ,, 1 ,, ,, 1 ,, 2/122 2/122 = ++ = ∂ ∂ = = ++ = ∂ ∂ = ( ) ( ) 0 21 21 =⋅ − rd sd rd n sd rd n rayray where is on the boundary between the two regions andrd ( ) ( ) sd rd s sd rd s rayray 2 :ˆ, 1 :ˆ 21 == are the unit vectors in the direction of propagation of the rays. Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 86. SOLO Corner Conditions for Geometrical Optics and Fermat’s Principle (continue – 2) ( ) 0ˆˆ 2211 =⋅− rdsnsn 2211 ˆˆ snsn −Therefore is normal to .rd Since can be in any direction on the boundary between the two regions (see Figure ) is parallel to the unit vector normal to the boundary surface, and we have rd 2211 ˆˆ snsn − 21 ˆ −n ( ) 0ˆˆˆ 221121 =−×− snsnn This the Snell’s Law of Geometrical Optics Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 87. SOLO Corner Conditions for Geometrical Optics and Fermat’s Principle (continue – 3) 2. The optical path is reflected at the boundary. ( ) ( ) ( ) 0ˆˆ 21 21 =⋅−=⋅ − rdssrd sd rd sd rd rayray n1 = n2 , we obtain i.e. is normal to , i.e. to the boundary where the reflection occurs. Also we can write 21 ˆˆ ss − rd ( ) 0ˆˆˆ 2121 =−×− ssn ( ) ( ) ( ) 0ˆˆ 21 221121 =⋅−=⋅ − rdsnsnrd sd rd n sd rd n rayray In this case, if we substitute in the equation Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 88. 88 SOLO Hilbert’s Invariant Integral David Hilbert (1862 – 1943) Geometrical Optics and Fermat’s Principle ( ) ( ) ( ) 22 22 1,,1,,,,,, zyzyxn xd zd xd yd zyxnzyzyxF ++= + += For the Geometrical Optics we obtained: The Hilbert’s Invariant Integral is ( ) ( )( ) ( ) ( )[ ] ( ) ( )( ){ ( ) ( ) ( ) ( )[ ] ( ) ( )( )} xdzyxzzyxyzyxFzyxZzyxz zyxzzyxyzyxFzyxYzyxyzyxzzyxyzyxF z zyxP zyxP yC ffff ,,,,,,,,,,,, ,,,,,,,,,,,,,,,,,,,, ,, ,, 0000 −− ∫ −− This is known as Hilbert’s Invariant Integral because it is invariant on the path C as long as this curve remains in the field of the unique extremal solution. ( ) ( ) ( ) ( )zyx x z zyxzzyx x y zyxy ,,,,,,,,, ∂ ∂ = ∂ ∂ = is the field slope and ( ) ( ) CC x z zyxZ x y zyxY ∂ ∂ = ∂ ∂ = :,,,:,, is the path C slope at the point (x,y,z) of C we have on path C ( ) ( ) dx x z dxzyxZzddx x y dxzyxYyd C C C C ∂ ∂ == ∂ ∂ == ,,,,, Calculus of Variations - Problems
- 89. 89 SOLO Hilbert’s Invariant Integral (continue – 1) David Hilbert (1862 – 1943) ( ) ( ) ( )[ ]{ ( ) ( ) ( ) ( ) }zdzyzyxFydzyzyxFxdzyzyxFzzyzyxFyzyzyxF zy zyxP zyxP zyC ffff ,,,,,,,,,,,,,,,,,,,, ,, ,, 0000 −−−−∫ The Hilbert’s Invariant Integral is We can write ( ) [ ] [ ] [ ] [ ] ( ) sd xd zyxn zy n zy zn z zy yn yzynFzFyzyzyxF zy ,, 111 1,,,, 2/1222/1222/122 2/122 = ++ = ++ − ++ −++=−− ( ) [ ] ( ) ( ) [ ] ( ) sd zd zyxn zy zzyxn z F F sd yd zyxn zy yzyxn y F F z y ,, 1 ,, ,, 1 ,, 2/122 2/122 = ++ = ∂ ∂ = = ++ = ∂ ∂ = Now we can write the Hilbert’s Invariant Integral as ( ) ( ) ( ) ( ) ∫∫ ⋅=⋅ ffffffff zyxP zyxP zyxP zyxP ray rdsnrd sd rd n ,, ,, ,, ,, 1000010000 ˆ This is the Lagrange’s Invariant Integral from Geometrical Optics. Joseph-Louis Lagrange (1736-1813) Integration Path through a Ray Bundle Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 90. 90 SOLO Second Order Conditions: Legendre’s Condition for a Weak Minimum Adrien-Marie Legendre 1752-1833 From ( ) [ ] ( ) sd yd zyxn zy yzyxn y F Fy ,, 1 ,, 2/122 = ++ = ∂ ∂ = we obtain ( ) [ ] [ ] [ ] ( ) [ ] 2/322 2 2/322 2 2/1222/1222 2 1 1 111 ,, zy zn zy yn zy n zy yzyxn yy F ++ + = ++ − ++ = ++∂ ∂ = ∂ ∂ From we obtain ( ) [ ] [ ] 2/3222/122 2 11 ,, zy zyn zy zzyxn yzy F ++ −= ++∂ ∂ = ∂∂ ∂ ( ) [ ] ( ) sd zd zyxn zy zzyxn z F Fz ,, 1 ,, 2/122 = ++ = ∂ ∂ = ( ) [ ] ( ) [ ] 2/322 2 2/1222 2 1 1 1 ,, zy yn zy zzyxn zz F ++ + = ++∂ ∂ = ∂ ∂ From those equations we obtain ( ) [ ] ( ) ( ) +− −+ ++ = ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂ ∂ = 2 2 2/322 2 22 2 2 2 '' 1'' 1 1 ,, yyx zyz zy zyxn z F yz F zy F y F F XX Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 91. 91 SOLO Second Order Conditions: Legendre’s Condition for a Weak Minimum (continue – 1) ( ) [ ] ( ) ( ) +− −+ ++ = ∂ ∂ ∂∂ ∂ ∂∂ ∂ ∂ ∂ = 2 2 2/322 2 22 2 2 2 '' 1'' 1 1 ,, yyx zyz zy zyxn z F yz F zy F y F F XX Let use Sylvester’s Theorem to check if/when is positive definite (i.e. check that the determinants of all principal minors are positive) ''XXF James Joseph Sylvester 1814-1897 [ ] ( ) ( ) +− −+ ++ = 2 2 2/322 '' 1'' 1 det 1 det yyx zyz zy n F XX [ ] ( )( )[ ] [ ] 0 1 11 1 2/122 2222 2/322 > ++ =−++ ++ = zy n zyyz zy n ( ) 01 2 >+ z According to Sylvester’s Theorem is positive definite''XXF ( )0'' >XXF According to Legendre’s Condition ,if the Jacobi’s Condition is satisfied (no conjugate points between P1 and P2), every extremal is a weak minimum. ( )0'' >XXF Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 92. SOLO The Weierstrass Necessary Condition for a Strong Minimum (Maximum) ( ) ( ) ( ) 22 22 1,,1,,,,,, zyzyxn xd zd xd yd zyxnzyzyxF ++= + += For the Geometrical Optics we obtained: Weierstrass E Function is defined as ( ) ( ) ( ) ( ) ( ) ( ) ( )zyzyxFzZzyzyxFyYzyzyxFZYzyzyxFZYzyzyxE zy ,,,,,,,,,,,,,,,,,,:,,,,,, −−−−−= [ ] [ ] ( ) [ ] ( ) [ ] [ ] [ ] ( ) ( )[ ] [ ] [ ] ( ) [ ] [ ] [ ] ( )InequalitySchwarz zyZY zZyY ZYn zZyY zy n ZYn zzZyyYzy zy n ZYn zy zn zZ zy yn yYzynZYn 0 11 1 11 1 1 1 1 1 1 1 '' 1 11 2/1222/122 2/122 2/122 2/122 22 2/122 2/122 2/1222/122 2/1222/122 ≥ ++++ ++ −++= ++ ++ −++= −+−+++ ++ −++= ++ −− ++ −−++−++= According to Weierstrass Condition if the Jacobi Condition (no conjugate points between and ) is satisfied every extremal is a strong minimum. ( )( )0',',',',',',,, ≥ZYXzyxzyxE Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 93. 93 SOLO Hamilton’s Canonical Equations Define ( ) [ ] ( ) ( ) [ ] ( ) sd zd zyxn zy zzyxn z F p sd yd zyxn zy yzyxn y F p z y ,, 1 ,, : ,, 1 ,, : 2/122 2/122 = ++ = ∂ ∂ = = ++ = ∂ ∂ = ( )( ) ( )2222222 1 zynzypp zy +=+++ Adding the square of twose two equations gives ( ) ( ) 2 222 2 22 1 = +− =++ xd sd ppn n zy zy from which Substituting in ( ) ( ) ( ) 22 22 1,,1,,,,,, zyzyxn xd zd xd yd zyxnzyzyxF ++= + += gives ( ) ( )222 2 ,,,, zy zy ppn n ppzyxF +− = Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 94. 94 SOLO Hamilton’s Canonical Equations (continue – 1) From ( ) [ ] ( ) ( ) [ ] ( ) sd zd zyxn zy zzyxn z F p sd yd zyxn zy yzyxn y F p z y ,, 1 ,, : ,, 1 ,, : 2/122 2/122 = ++ = ∂ ∂ = = ++ = ∂ ∂ = solve for ( ) ( )222 2 ,,,, zy zy ppn n ppzyxF +− =and ( ) ( )222 222 zy z zy y ppn p z ppn p y +− = +− = Define the Hamiltonian ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) sd xd zyxnppzyxn ppn p ppn p ppn n zpypppzyxFppzyxH zy zy z zy y zy zyzyzy ,,,, ,,,,:,,,, 222 222 2 222 2 222 2 −=+−−= +− + +− + +− −= ++−= Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 95. 95 SOLO Hamilton’s Canonical Equations (continue – 2) From We obtain the Hamilton’s Canonical Equations ( ) ( ) ( ) ( ) sd xd zyxnppzyxnppzyxH zyzy ,,,,,,,, 222 −=+−−= ( ) ( )222 222 zy z z zy y y ppn p p H xd zd z ppn p p H xd yd y +− = ∂ ∂ == +− = ∂ ∂ == ( ) ( )222 222 zy z zy y ppn z n n z H xd pd ppn y n n y H xd pd +− ∂ ∂ −= ∂ ∂ −= +− ∂ ∂ −= ∂ ∂ −= Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 96. 96 SOLO Hamilton’s Canonical Equations (continue – 3) From ( ) ( ) ( ) ( ) sd xd zyxnppzyxnppzyxH zyzy ,,,,,,,, 222 −=+−−= ( )222 zy ppn n sd xd +− = By similarity with ( ) sd yd zyxnpy ,,= define ( ) ( ) ( ) ( )222 ,,,,,,,,: zyzyx ppzyxnppzyxH sd xd zyxnp +−=−== Let differentiate px with respect to x ( ) x H xd Hd ppn x n n xd pd zy x ∂ ∂ −=−= +− ∂ ∂ = 222 Let compute ( ) ( ) x n n ppn ppn x n n sd xd xd pd sd pd zy zy xx ∂ ∂ = +− +− ∂ ∂ == 222 222 Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 97. 97 SOLO Hamilton’s Canonical Equations (continue – 4) and ( ) ( ) x n n ppn ppn x n n sd xd xd pd sd pd zy zy xx ∂ ∂ = +− +− ∂ ∂ == 222 222 ( ) ( ) y n n ppn ppn y n n sd xd xd pd sd pd zy zy yy ∂ ∂ = +− +− ∂ ∂ == 222 222 ( ) ( ) z n n ppn ppn z n n sd xd xd pd sd pd zy zy zz ∂ ∂ = +− +− ∂ ∂ == 222 222 n sd pd ∇= xp nsd xd 1 = ( ) ( ) y zy zy y p nn ppn ppn p sd xd xd yd sd yd 1 222 222 = +− +− == ( ) ( ) z zy zy z p nn ppn ppn p sd xd xd zd sd zd 1 222 222 = +− +− == p nsd rd ray 1 = We recover the result from Geometrical Optics Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 98. 98 William Rowan Hamilton (1805-1855) Canonical Equations of Motion 1835 where H is the Hamiltonian defined as: Hamilton-Jacobi Theory SOLO CALCULUS OF VARIATIONS William Rowan Hamilton 1805-1855 Carl Gustav Jacob Jacobi 1804-1851 Hamilton-Jacobi Equation
- 99. 99 SOLO Hamilton-Jacobi Equation (continue – 1) ( ) ( ) ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) sd xd zyxn zy zyxn zy zzyxn z zy yzyxn y zyzyxnFzFyzyzyxFzyxS zyx ,, 1 ,, 1 ,, 1 ,, 1,,,,,,,, 2/1222/1222/122 2/122 = ++ = ++ − ++ − ++=−−= ( ) ( ) [ ] ( ) sd yd zyxnp zy yzyxn y F zyxS yy ,,: 1 ,, ,, 2/122 == ++ = ∂ ∂ = ( ) ( ) [ ] ( ) sd zd zyxnp zy zzyxn z F zyxS zz ,,: 1 ,, ,, 2/122 == ++ = ∂ ∂ = We obtain sn sd rd nS ray ˆ==∇ From this 22 ˆˆ nssnSS =⋅=∇⋅∇ We recovered again the Eikonal Equation Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 100. 100 SOLO Hamilton-Jacobi Equation (continue – 2) ( ) ( ) sdzyxnsd sd zd sd yd sd xd zyxn sd sd zd Ssd sd yd Ssd sd xd SzdSydSxdSSd zyxzyx ,,,, 1 222 = + + = ++=++= ( ) ( ) ( )∫=− sdzyxnzyxSzyxS ,,,,,, 000Along an extremal trajectory we obtain ( ) ( ) sd yd zyxnzyxSy ,,,, = ( ) ( ) sd zd zyxnzyxSz ,,,, = ( ) ( ) 2 222 2 22 1 = +− =++ xd sd ppn n zy zy ( ) ( ) ( ) [ ] ( ) sd xd zyxn zy zyxn FzFyzyzyxFzyxS zyx ,, 1 ,, ,,,,,, 2/122 = ++ =−−= ( ) ( ) [ ] ( )[ ] ( ) zy ppzyxH zyx ppn zy zyxn zyxS ,,,, 2/1222 2/122 1 ,, ,, − +−= ++ = ( ) ( ) 0,,,,,, ≡+ zyx ppzyxHzyxS Hamilton-Jacobi Equation is an identity. Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 101. 101 SOLO If the regularity condition (optical ray not intersecting) doesn’t hold, the optical ray may not be a minimum, as we can see from the Figure, where the optical ray reflected from the planar mirror and reaches the point P2 (P1MP2) is longer than the direct ray from P1 to P2. Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 102. 102 SOLO On other example is given in Figure bellow on the rays from a point source refracted by a lens. The refracted rays form an envelope called caustic. The point P’2 where the refracted ray touches the caustic is called a conjugate point. From the Figure we can see that this point is reached by, at least, two rays with different optical paths. Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 103. 103 SOLO Example of the stationarity of the Fermat’s Principle Suppose that we have a elliptical mirror and a point source locate at one of it’s foci P1. The elliptical mirror has the following properties: 1. The sum of the distances from the two foci to any point R on the ellipse is constant. 2121 PRRPPRRP EE +=+ 2. The normal at any point R on the ellipse bisects the angle P1RP2. 2P 1P Point Source Elliptic Mirror RER Rnˆ ERsˆ Rsˆ According to Snell’s Law, all the rays originated at the focus P1 will be reflected by the elliptical mirror and intersect at the second foci P2. Since the rays travel in the same media and the geometrical paths are equal, the optical paths will be equal also. ( ) ( )2121 PRRPnPRRPn EE +=+ Since all the optical paths reflected by the mirror reach the point P2, we call P2 the conjugate point to P1. Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 104. 104 SOLO Example of the stationarity of the Fermat’s Principle (continue – 1) Now replace the elliptical mirror with a planar one normal to at the point R.Rnˆ 2P 1P Point Source Planar Mirror Elliptic Mirror RER PR Rnˆ ERsˆ PRsˆ Rsˆ For this reason the ray will be reflected at R and reach the point P2, in the same way as for the elliptical mirror. RP1 From the Figure we can see that: ( ) ( ) 222121 PRRRPRPRRRPnPRRPn PPEEPPE +<←+<+ In this case the Fermat’s Principle will give a minimum for the optical path ( )21 PRRPn + Geometrical Optics and Fermat Principle Calculus of Variations - Problems
- 105. 105 SOLO Example of the stationarity of the Fermat’s Principle (continue – 3) Now replace the elliptical mirror with a circular one normal to at the point R (The mirror diameter is smaller than the maximum axis of the ellipse). Rnˆ For this reason the ray will be reflected at R and reach the point P2, in the same way as for the elliptical mirror. RP1 From the Figure we can see that: In this case the Fermat’s Principle will give a maximum for the optical path ( )21 PRRPn + 2P 1P Point Source Elliptic Mirror Circular Mirror R CR ER RnˆCRsˆ ERsˆ PRsˆ Rsˆ ( ) ( ) 222121 PRRRPRPRRPnPRRPn EECCCC +<←+>+ Geometrical Optics and Fermat Principle Calculus of Variations - Problems Return to Table of Content
- 106. 106 SOLO Calculus of Variations - Problems References Bryson & Ho “Applied Optimal Control”, Ginn and Company, 1969, Sec 3.1, pg. 91-95 George Leitmann “The Calculus of Variations and Optimal Control – An Introduction”, Plenum Press, 1981 O. Bolza, “Lectures on the Calculus of Variations”, Dover Publications, New York, 1961, Republication of a work published by Univ. of Chicago 1904, pp.1,27-28,41,48-49,64, 77-78, 153 W.S. Kimball, “Calculus of Variations, by Parallel Displacement”, Butterworths Scientific Publications, 1952, § 13, pp.432-475 L.E. Elsgolc, , “Calculus of Variations”, Pergamon Press, Addison-Wesley, 1962, pp.37-38 I.M. Gelfand, S.V. Fomin, “Calculus of Variations”, Prentice-Hall, 1963, pp.20-21 H. Sagan, “Introduction to Calculus of Variations”, Dover Publication, New York, 1969, pp.62-65 H. Tolle, “Optimization Methods”, Springer Verlag, Berlin Heidelberg New York, 1975, pp.15-18
- 107. 107 SOLO Calculus of Variations - Problems References S. Hermelin, “Calculus of Variations”, http://www.solohermelin.com Math Folder S. Hermelin, “Foundation of Geometrical Optics”, http://www.solohermelin.com Optics Folder Return to Table of Content
- 108. February 23, 2015 108 SOLO Technion Israeli Institute of Technology 1964 – 1968 BSc EE 1968 – 1971 MSc EE Israeli Air Force 1970 – 1974 RAFAEL Israeli Armament Development Authority 1974 – 2013 Stanford University 1983 – 1986 PhD AA
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