MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 1 Exmples 5 Qu.1 Show tht the extreml funtion of the funtionl I[y] = 1 0 [(y ) + yy + y ] dx, where y(0) = 0 nd y(1) = 1, is y(x) = sinh x sinh 1. Qu. A urve y = y(x) whih strts t the point A( l, ) nd ends t the point B(l, ) is rotted bout the x-xis. Show tht the surfe of revolution whih is generted hs minimum re if y(x) = osh(x/) where the onstnt stisfies = osh(l/). Qu.3 Estblish the first integrls of the Euler-Lgrnge eqution, nmely ) y = onstnt, when F = F(x, y ), nd b) F y y = onstnt, when F = F(y, y ). Qu.4 A hevy, uniform hin of length l hngs with its ends level t distne (< l) prt. Find its shpe y(x) ssuming tht its potentil energy E[y] is minimised, where E[y] = y 1 + (y ) dx. Qu.5 Wht losed urve of length πr in the xy-plne enloses the mximum re? (For urve desribed prmetrilly by r = (x(s), y(s)), the re A is given by x dy or b x dy ds ds, where < s < b for one iruit of the losed urve. With the sme desription, the length of the urve is b (x ) + (y ) ds, where the dsh mens differentition with respet to s. )
MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions Qu.6 The brhistohrone (from the Greek for shortest nd time ) problem desribes the time it tkes prtile to fll under grvity down surfe desribed by y = y(x); it gives rise to the funtionl T[y] for the totl time of trnsit, where T[y] = 1 g x x 1 1 + (y ) y 1 y dx, in whih g is the elertion due to grvity nd y 1 is the initil height of the prtile. Show tht y 1 y = k (1 + os ψ) x l = k (ψ + sin ψ) where k, l re onstnts. (Hint: write y = tn ψ, nd use 1 + tn θ = se θ.)
MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 3 Solutions 5 Qu.1 In the usul nottion, F (x, y, y ) = y + yy + y. y = y + y, = y + y. y The Euler-Lgrnge eqution is ( ) d dx y y = 0, d dx (y + y) (y + y) = 0 y y = 0, seond order o.d.e. with onstnt oeffiients. The generl solution my be written y (x) = A osh x + B sinh x. Applying the boundry onditions, y (0) = 0 A = 0 nd y (1) = 1 B = 1/ sinh(1). y (x) = sinh x sinh 1. Qu. The surfe re is given by the integrl Surfe re = l l πy 1 + y dy. F = πy 1 + y nd x is bsent from F, F = F (y, y ). From the leture notes (see lso ques. 3(b) below) we know tht the Euler-Lgrnge eqution hs the first integrl F y y = onstnt. Now, nd so or y = πyy 1 + y πy ( ) 1 + y y πyy = π, sy. 1 + y y = 1 + y (y ) y = 1 dx dy = y.
MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 4 Integrting, or Now, y (l) = y ( l) k = 0, so ( y ) x = osh 1 + k ( ) x k y = osh. ( x y = osh, ) nd y (±) = = osh whih impliitly determines the onstnt. ( ) l, Qu.3 () The Euler-Lgrnge eqution is ( ) d dx y So, if F = F (x, y ) then nd Integrting gives y = 0. y = 0 ( ) d = 0. dx y y = onstnt. (b) If F = F (x, y, y ) then differentiting the left-hnd-side of the theorem we wish to verify using the hin-rule [relling tht y = y (x), y = y (x)]gives ( d F y ) dx y = x + y y + = x + y = x = 0 [ y d dx [ y ( )] y y y + d y dx y ( )] in the se where x is bsent from F, F = F (y, y ). In this se, integrting gives y F y y = onstnt.
MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 5 Qu.4 The potentil energy to be mde sttionry is E (y) = y 1 + (y ) dx. The length L of the hin is onstnt nd equl to l nd so L (y) = 1 + (y ) dx = l. (.1) Define new funtionl by I (y, λ) = E λl = = (y λ) 1 + (y ) dx, H (y, y, λ)dx, sy. H (y, y, λ) is independent of x nd so first integrl of the Euler-Lgrnge eqution for I is H y H y = onstnt, [ ] (y λ) 1 + (y ) y (y λ) 1 + (y y ) = A, sy, (y λ) 1 + (y ) y (y λ) y = A, 1 + (y ) Solving for y, y λ = A 1 + (y ) (y λ) A y = ±. A dx dy = ± A. (y λ) A Integrting, ( ) y λ x = ±A osh 1 + B, A ( ) x B y = λ + A osh. A Now hoose xes so tht y () = y () = 0 then B = 0. Also hoose xes so tht y (±) = 0, then ( λ = A osh.
MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 6 ( x ( y = A osh A osh. The onstrint given by equ.(.1) gives ( x osh dx = l, ( A sinh = l. Thus, A is determined impliitly in terms of l by ( A sinh = l. Qu.5 Let the urve C be given by the prmetri equtions x = x (s),y = y (s) where s is prmeter, s b. The urve is losed x () = x (b), y () = y (b). The re A (x, y) enlosed by C my be written s A (x, y) = b xy ds, where prime denotes differentition with respet to s. The length onstrint my be written b L (x, y) = (x ) + (y ) ds = πr. Define the ugmented funtionl I (x, y, λ) by b [ ] I (x, y, λ) = xy λ (x ) + (y ) ds = where b H (s, x, y, x, y ) = xy λ (x ) + (y ) H (s, x, y, x, y ) ds Here, s is the independent vrible nd x, y re the dependent vribles. The Euler- Lgrnge equtions re ( ) d H ds x ( ) d H ds y H x = 0 H y = 0 nd H x = y, H λx =, x (x ) + (y ) H y = 0 H λy = x. y (x ) + (y )
MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 7 d ds d x ds λx (x ) + (y ) λy (x ) + (y ) y = 0, = 0. These equtions my both be immeditely integrted to obtin λx + B y = 0, (x ) + (y ) x λy (x ) + (y ) = A where A nd B re rbitrry onstnts of integrtion. Eliminting λ, x A = λy (x ) + (y ), λx y B =, (x ) + (y ) nd this my be integrted to give (x x + (y B) y = 0 (x + (y B) = C. This is the eqution of irle, entre (A, B) nd rdius C. So C = πr. Note: We hve only shown tht irle mkes A (x, y) sttionry. Sine we hve not onsidered suffiient onditions we nnot show tht irle gives the mximum re. Qu. 6 The brhistohrone (from the Greek for shortest nd time ) problem my be stted s follows: Two points A, B re onneted by smooth wire. A bed slides under grvity from A between the two points. Whih urve enbles the bed to reh B in the minimum time? The problem ws first proposed by John Bernoulli in 1696 nd, s ws ommon prtie t the time, he issued hllenge to other mthemtiins to see who ould solve it first. Newton, Liebnitz nd L Hopitl responded to the hllenge... nd it gve rise to the subjet we now ll the Clulus of Vritions. Using the mehnis of prtile, the funtionl for the brhistohrone my be written s T (y) = 1 g x x 1 1 + (y ) y y 1 dx.
MATH3403: Green s Funtions, Integrl Equtions nd the Clulus of Vritions 8 nd F (y, y ) = 1 1 + (y ) g y y 1 F y y = onstnt. The first integrl beomes 1 1 + (y ) y 1 1 1 ( 1 + (y ) ) 1/ y = onstnt g y y 1 g (y y 1 ) 1/ 1 + (y ) y y 1 1 (y ) (y y 1 ) 1/ ( 1 + (y ) ) = 1 1/ k (y y 1 ) 1/ ( 1 + (y ) ) 1/ = k 1/, 1/, sy Let y = tnψ then ( (y y 1 ) 1 + (y ) ) = k (y y 1 )se ψ = k y y 1 = k os ψ = 1 k (1 + os ψ). Differentiting so i.e Integrting with respet to x, i.e Thus, y = k sin ψ dψ dx = tn ψ k os ψ dψ dx = 1 k (os ψ + 1) dψ dx = 1 ( ) 1 k sin ψ + ψ = x + l 1 k (sin ψ + ψ) = x l y y 1 = 1 k (1 + os ψ) x l = 1 k (ψ + sin ψ)