First variation. (onevariable problem) January 21, 2015


 Melanie Richard
 2 years ago
 Views:
Transcription
1 First vrition (onevrible problem) Jnury 21, 2015 Contents 1 Sttionrity of n integrl functionl Euler eqution (Optimlity conditions) First integrls: Three specil cses Vritionl problem s limit of finitedimensionl problem Sttionrity of boundry terms Vrition of boundry conditions Broken extreml nd the WeierstrssErdmn condition Functionl dependent on higher derivtives 13 4 Nonfixed intervl Trnsverslity condition Extreml broken t n unknown point Severl minimizers Euler equtions nd first integrls Vritionl boundry conditions Geometric optics Geometric optics problem Snell s lw of refrction Brchistochrone Miniml surfce of revolution
2 Since, however, the rules for isoperimetric curves or, in modern terms, extreml problems were not sufficiently generl, the fmous Euler undertook the tsk of reducing ll such investigtions to generl method which he gve in the work Essy on new method of determining the mxim nd minim of indefinite integrl formuls ; n originl work in which the profound science of the clculus shines through. Even so, while the method is ingenious nd rich, one must dmit tht it is not s simple s one might hope in work of pure nlysis. In Essy on new method of determining the mxim nd minim of indefinite integrl formuls, by Lgrnge, Sttionrity of n integrl functionl The technique ws developed by Euler, who lso introduced the nme Clculus of vritions in The method is bsed on n nlysis of infinitesiml vritions of minimizing curve. The min scheme of the vritionl method is s follows: ssuming tht the optiml curve u(x) exists mong smooth (twicedifferentible curves), we compre the optiml curve with closeby trjectories u(x) + δu(x), where vrition δu(x) is smll in some sense. Using the smllness of δu, we simplify the comprison, deriving necessry conditions for the optiml trjectory u(x). The method is quite nlogous to the computing of differentil d f(x) = f (x) dx function f(x), minimum point is solution of the eqution f (x) = 0 becuse of the rbitrriness of differentil dx. Generlly speking, vritionl methods yield to only necessry conditions of optimlity becuse it is ssumed tht the compred trjectories re close to ech other; on the other hnd, these methods re pplicble to gret vriety of extreml problems clled vritionl problems. Similrly to the clculus problems, the proof of optimlity of trjectory should be complemented by inequlities tht distinguish minimum from mximum of sddle points nd n existence theorem tht either gurntees tht the minimizer is indeed twicedifferentible function or suggests n lterntive pproch (relxting or regulriztion techniques) 1.1 Euler eqution (Optimlity conditions) Consider the problem clled the simplest problem of the clculus of vritions min u I(u), I(u) = F (x, u, u )dx, u() = u, u(b) = u b, (1) Here integrnt F is clled Lgrngin, it is ssumed to be twice differentible function of its three rguments, I(u) is clled the cost functionl. It is not known priori whether the minimizer u 0 (x) is smooth, but let us ssume tht it is twice differentible function of x. 2
3 Exmple 1.1 Consider the re of the surfce of revolution round the xes OX tht is supported by two prllel coxil circles of rdii R nd R b, the distnce between the centers of circles is b. According to the clculus, the re J of the surfce is A(r) = π r(x) 1 + r (x) 2 dx, where r(x) is the vrible distnce from the OXxes. The problem of miniml re of such surfce I = min r(x) A(u), r() = R, r(b) = R b is vritionl problem, r(x) is n unknown function. To derive necessry condition of optimlity of minimizer u 0 we use the ides of clculus, computing n nlog of the derivtive of I with respect to u (clled the functionl derivtive) nd setting it to zero. We suppose tht function u 0 = u 0 (x) is minimizer nd replce u 0 with test function u 0 + δu. The test function u 0 + δu stisfies the sme boundry conditions s u 0. If indeed u 0 is minimizer, then the increment of the cost is δi(u 0 ) = I(u 0 + δu) I(u 0 ) is nonnegtive: δi(u 0 ) = 1 0 (F (x, u 0 + δu, (u 0 + δu) ) F (x, u 0, u 0))dx 0. (2) If δu is not specified, the eqution (2) is not too informtive. However, prticulr form of the vrition δu simplifies (2) nd llows for finding n eqution for the minimizer. Clculus of vritions suggests set of tests tht differ by vrious form of vritions δu. Euler Lgrnge Equtions The sttionry vritionl condition (the Euler Lgrnge eqution) is derived ssuming tht the vrition δu is infinitesimlly smll nd loclized: { ρ(x) if x [x0, x δu = 0 + ε], (3) 0 if x is outside of [x 0, x 0 + ε]. Here ρ(x) is continuous function tht vnishes t points x 0 nd x 0 + ε nd is constrined s follows: ρ(x) < ε, ρ (x) < ε x, ρ(x 0 ) = ρ(x 0 + ε) = 0 (4) The integrnd t the perturbed trjectory cn be expnded into Tylor series, F (x, u 0 + δu, (u 0 + δu) ) = F (x, u 0, u 0) + (x, u 0, u 0) δu + (x, u 0, u 0) δu + o(δu, δu ) 3
4 Here, δu is derivtive of the vrition δu, δu = (δu), o(δu, δu ) denotes higher order terms which norms re smller thn δu nd δu when ε 0. Substituting this expression into (2) nd collecting liner (with respect to ε) terms, we rewrite (2) s ( ) δi(u 0 ) = (δu) + (δu) dx + o(ε) 0. (5) where The F is clculted t the exmined trjectory u 0. To simplify nottions, we omit index ( 0 ) below. The vritions δu nd (δu) re mutully dependent nd (δu) cn be expressed in terms of δu. Integrtion by prts of the underlined term in (5) gives nd we obtin (δu) dx = 0 δi(u 0 ) = ( d dx ) δu dx + S F (x, u, u )δu dx + δu where S F denotes the functionl derivtive, δu x=b x= x=b x= + o(ε), (6) S F (x, u, u ) = d dx +. (7) The nonintegrl term in the righthnd side of (6) is zero, becuse the boundry vlues of u re prescribed s u() = u nd u(b) = u b ; therefore their vritions δu x= nd δu x=b equl zero, δu x= = 0, δu x=b = 0 Due to the rbitrriness of δu, we rrive t the following Theorem 1.1 (Sttionrity condition) Any differentible nd bounded minimizer u 0 of the vritionl problem (1) is solution to the boundry vlue problem S F (x, u, u ) = d dx = 0 x (, b); u() = u, u(b) = u b, (8) clled the Euler Lgrnge eqution. The Euler Lgrnge eqution is lso clled the sttionry condition of optimlity becuse it expresses sttionrity of the vrition. Using the chin rule, the lefthnd side of eqution (8) cn be rewritten to the form of explicit secondrnk differentil eqution. S F (x, u, u ) = 2 F 2 u + 2 F u + 2 F x In this derivtion, it is indirectly ssumed tht the extreml u(t) is twice differentible function of x. (9) 4
5 Exmple 1.2 Compute the Euler eqution for the problem I = min u(x) 1 We compute = u, The minimizer u 0 (x) is 0 F (x, u, u )dx u(0) = 1, u(1) =, F = 1 2 (u ) u2 = u nd the Euler eqution becomes u u = 0 in (0, 1), u(0) = 1, u(1) =. u 0 (x) = cosh(x) + cosh(1) sinh(1) sinh(x). Remrk 1.1 The sttionrity test lone does not llow to conclude whether u is true minimizer or even to conclude tht solution to (8) exists. For exmple, the function u tht mximizes I(u) stisfies the sme Euler Lgrnge eqution. The tests tht distinguish miniml trjectory from other sttionry trjectories re discussed in Chpter??. Remrk 1.2 The reltion (6) request tht differentil eqution (8) is stisfied in ech infinitesiml intervl but not in in ech point. The eqution my be not defined in isolted points, if F or its prtil derivtives re not defined in these points. The minimizer cn chnge its vlues t severl points, or even t set of zero mesure without effecting the objective functionl. Such solutions re clled wek solutions?? of differentil eqution. The definition of the wek solution nturlly rises from the vritionl formultion tht does not check the behvior of the minimizer in every p[oint but only t intervls of nonzero mesure. In mbiguous cses, one should specify in wht sense (Riemnn, Lebesgue) the integrl is defined nd define of vrition ccordingly. 1.2 First integrls: Three specil cses In severl cses, the secondorder Euler eqution (8) cn be integrted t lest once. These re the cses when Lgrngin F (x, u, u ) does not depend on one of the rguments. Assume tht F = F (x, u), nd the mini Lgrngin is independent of u miztion problem is I(u) = min u(x) 1 0 F (x, u)dx (10) In this cse, Euler eqution (8) becomes n lgebric reltion for u = 0 (11) The vrition does not involve integrtion by prts, nd the minimizer does not need to be continuous. 5
6 Monimizer u(x) is determined in ech point independently of neighboring points. The boundry conditions in (8) re stisfied by possible jumps of the extreml u(x) in the end points; these conditions do not ffect the objective functionl t ll. Exmple 1.3 Consider the problem I(u) = min u(x) 1 0 (u sin x) 2 dx, u(0) = 1; u(1) = 0. The miniml vlue J(u 0 ) = 0 corresponds to the discontinuous minimizer sin x if 0 x 1 u 0 (x) = 1 if x = 0 0 if x = 1 Remrk 1.3 Formlly, the discontinuous minimizer contrdicts the ssumption posed when the Euler eqution ws derived. To be consistent, we need to repet the derivtion of the necessry condition for the problem (10) without ny ssumption on the continuity of the minimizer. This derivtion is quite obvious. Lgrngin is independent of u If Lgrngin does not depend on u, F = F (x, u ), Euler eqution (8) becomes S F (x, u, u ) = d dx = 0 x (, b); u() = u, u(b) = u b, (12) it cn be integrted once: = constnt (13) The differentil eqution (13) for u is the first integrl of eqution (8); it defines quntity tht stys constnt everywhere long n optiml trjectory. It remins to integrte the first order eqution (13) nd determine the constnts of integrtion from the boundry conditions. Exmple 1.4 Consider the problem I(u) = min u(x) 1 (compre with exmple 1.3) The first integrl is Integrting, we find the minimizer, 0 (u cos x) 2 dx, u(0) = 1; u(1) = 0. = u (x) cos x = C u(x) = sin x + C 1 x + C 2. 6
7 The constnts C 1 nd C 2 re found from nd the boundry conditions: C 2 = 1, C 1 = 1 sin 1, minimizer u 0 nd the cost of the problem become, respectively u 0 (x) = sin x (sin 1 + 1)x + 1 I(u 0 ) = (sin 1 + 1) 2. Notice tht the Lgrngin in the exmple (1.3) is the squre of difference between the minimizer u nd function sin x, nd the Lgrngin in the exmple (1.4) is the squre of difference of their derivtives. In the problem (1.3), the minimizer coincides with sin x, nd jumps to stisfy the prescribed boundry vlues. The minimizer u in the exmple (1.4) cnnot jump. Indeed, continuous pproximtion of derivtive u of discontinuous functionis unfunded in the proximity of the point of discontinuity, such behvior increses the objective functionl, nd therefore it is nonoptiml. Lgrngin is independent of x If F = F (u, u ), eqution (8) hs the first integrl: W (u, u ) = constnt (14) where W (u, u ) = u F Indeed, compute the xderivtive of W (u, u ) which must be equl to zero by virtue of (14): d dx W (u, u ) = [ ( 2 u + F u u + 2 F u 2 )] u u = 0 where the expression in squre brckets is the derivtive of the first term of W (u, u ). Cncelling the equl terms, we bring this eqution to the form ( u 2 F 2 u + 2 F u ) = 0 (15) The expression in prenthesis coincides with the lefthndside term S(x, u, u ) of the Euler eqution in the form (9), simplified for the considered cse (F is independent of x, F = F (u, u )). W is constnt t ny solution u(x) of Euler eqution. Insted of solving the Euler eqution, we my solve the firstorder eqution W = C. Exmple 1.5 Consider the Lgrngin for hrmonic oscilltor: F = 1 2 [ (u ) 2 ω 2 u 2] 7
8 The Euler eqution is The first integrl is u + ω 2 u = 0 W = 1 2 ( ω 2 u 2 + (u ) 2) = C 2 = constnt Let us check the constncy of the first integrl. The solution u of the Euler eqution is equl u = A cos(ωx) + B sin(ωx) where A nd B re constnts. Substituting the solution into the expression for the first integrl, we compute W = (u ) 2 + ω 2 u 2 = [ Aω sin(cx) + Bω cos(ωx)] 2 +ω 2 [A cos(ωx) + B sin(ωx)] 2 = ω 2 (A 2 + B 2 ) We hve shown tht W is constnt t the optiml trjectory. Notice tht W is the totl energy (sum of the potentil nd kinetic energy) of the oscilltor. 1.3 Vritionl problem s limit of finitedimensionl problem Here, we derive Euler eqution for finitedimensionl problem tht pproximte the simplest vritionl problem min I(u), I(u) = u(x) F (x, u, u )dx Consider clss of piecewise constnt discontinuous functions U N : ū(x) U N, if ū(x) = u i x [ + in ] (b ) A function ū in U N is defined by n Ndimensionl vector {u 1,... u N }. Rewriting the vritionl problem for this clss of minimizers, we replce the derivtive u (x) with finite difference Diff (u i ) Diff (u i ) = 1 (u i u i 1 ), = b N ; (16) when N, this opertor tends to the derivtive. The vritionl problem is replced by finitedimensionl optimiztion problem: min u 1,...,u N 1 I N I N = N F i (u i, z i ), z i = Diff (z i ) = 1 (z i z i 1 ) (17) i=1 Compute the sttionry conditions for the minimum of I N (u) I N i = 0, i = 1...., N. 8
9 Only two terms, F i nd F i+1, in the bove sum depend on u i : the first depends on u i directly nd lso through the opertor z i = Diff (u i ), nd the second only through z i+1 = Diff (u i+1 ): df i = i + i 1 du i i z i, df i+1 = i+1 1 du i z i+1. df k = 0, k i, k i + 1 du i Collecting the terms, we write the sttionry condition with respect to u i : I N = i + 1 ( i i i z ) i+1 = 0 z or, reclling the definition (16) of Diff opertor, the form I N = ( ) i i+1 Diff = 0. i i z The initil nd the finl point u 0 nd u N enter the difference scheme only once, therefore the optimlity conditions re different. They re, respectively, N+1 Diff (u N+1 ) = 0; o Diff (u 0 ) = 0. Formlly pssing to the limit N, Diff d dx, z u replcing the index ( i ) with continuous vrible x, vector of vlues {u k } of the piecewise constnt function with the continuous function u(x), difference opertor Diff with the derivtive d dx ; then nd N F i (u i, Diff u i ) i=1 i i Diff ( i+1 z ) F (x, u, u )dx. d d x The conditions for the end points become the nturl vritionl conditions: (0) = 0, (T ) = 0, So fr, we followed the forml scheme of necessry conditions, thereby tcitly ssuming tht ll derivtives of the Lgrngin exist, the increment of the functionl is correctly represented by the first term of its power expnsion, nd the limit of the sequence of finitedimensionl problems exist nd does not depend on the prtition {x 1,... x N } if only x k x k 1 0 for ll k. We lso indirectly ssume tht the Euler eqution hs t lest one solution consistent with boundry conditions. 9
10 If ll the mde ssumptions re correct, we obtin curve tht might be minimizer becuse it cnnot be disproved by the sttionry test. In other terms, we find tht is there is no other closeby clssicl curve tht correspond to smller vlue of the functionl. Remrk 1.4 This sttement bout the optimlity seems to be rther wek but this is exctly wht the clculus of vrition cn give us. On the other hnd, the vritionl conditions re universl nd, being ppropritely used nd supplemented by other conditions, led to very detiled description of the extreml s we show lter in the course. Remrk 1.5 In the bove procedure, we ssume tht the limits of the components of the vector {u k } represent vlues of smooth function in the closeby points x 1,..., x N. At the other hnd, u k re solutions of optimiztion problems with the coefficients tht slowly vry with the number k. We need to nswer the question whether the solution of minimiztion problem tends to is differentible function of x; tht is whether the limit u k u k 1 lim k x k x k 1 exists nd this is not lwys the cse. We ddress this question lter in Chpter?? 2 Sttionrity of boundry terms 2.1 Vrition of boundry conditions Vritionl conditions nd nturl conditions The vlue of minimizer my not be specified on one or both ends of the intervl [, b]. In this cse, these vlues re clculted by the minimiztion of the gol functionl together with the minimizer. Consider vritionl problem where the boundry vlue t the right end b of the intervl is not defined nd the functionl directly depends on this vlue, min I(u), I(u) = u(x):u()=u F (x, u, u )dx + f(u(b)) (18) The Euler eqution for the problem remin the sme, S(x, u, u ) = 0, but this time it must be supplemented by vritionl boundry condition tht comes from the requirement of the sttionrity of the minimizer with respect to vrition of the boundry term. This term is ( + f ) δu(b) x=b The first term comes from the integrtion by prt in the derivtion of Euler eqution, see (6), nd the second is the vrition of the lst term in the objective 10
11 functionl (18): δf(u) = f δu. Becuse the sign of the vrition δu(b) is rbitrry, the sttionrity condition hs the form ( + f ) = 0. (19) This equlity provides the missing boundry condition t the endpoint x = b for the second order Euler eqution. Similr condition cn be derived for the point x = if the vlue t this point is not prescribed. Exmple 2.1 Minimize the functionl x=b 1 1 I(u) = min u 0 2 (u ) 2 dx + Au(1), u(0) = 0 Here, we wnt to minimize the endpoint vlue nd we do not wnt the trjectory be too steep. The Euler eqution u = 0 must be integrted with boundry conditions u(0) = 0 nd (see (19)) u (1) + A = 0 The extreml is stright line, u = Ax. The cost of the problem is I = 1 2 A2. If f = 0, the condition (19) becomes = 0 (20) x=b nd it is clled the nturl boundry condition. Exmple 2.2 Consider the Lgrngin F = (x)(u ) 2 + φ(x, u) where (x) 0 The nturl boundry condition is u x=b = Broken extreml nd the WeierstrssErdmn condition The clssicl derivtion of the Euler eqution requires the existence of ll second prtils of F, nd the solution u of the secondorder differentil eqution is required to be twicedifferentible. In some problems, F is only piecewise twice differentible; in this cse, the extreml consists of severl curves solutions of the Euler eqution tht re computed t the intervls of smoothness. We consider the question: How to join these pieces together? The first continuity condition is continuity of the (differentible) minimizer u(x) [u] + = 0 long the optiml trjectory u(x) (21) Here [z] + = z + z denotes the jump of the vrible z. The extreml u is differentible, the first derivtive u exists t ll points of the trjectory. This derivtive does not need to be continuous. Insted, Euler eqution requests the differentibility of to ensure the existence of the term d in the Euler eqution. dx 11
12 Integrting the sttionrity condition (8), we obtin sttionrity in the integrl form x x ( d S F (x, u, u (x, u, u ) )dx = dx (x, u, ) u ) dx = 0 or If (x, u, u ) x = (x, u, u ) x 0 dx + (x, u, u ) (22) x= is bounded t the optiml trjectory, the righthnd side is continuous function of x, nd so is the lefthnd side. This requirement of continuity of n optiml trjectory is clled the WeierstrssErdmn condition on broken extreml. Theorem 2.1 At ny point of the optiml trjectory, the WeierstrssErdmn condition must be stisfied: [ ] + = 0 long the optiml trjectory u(x). (23) Exmple 2.3 (Broken extreml) Consider the Lgrngin F = 1 2 c(x)(u ) { c1 if x [, x 2 u2, c(x) = ), c 2 if x (x, b] where x is point in (, b). The Euler eqution is held everywhere in (, b) except of the point x, d dx [c 1u ] u = 0 if x [, x ) d dx [c 2u ] u = 0 if x (x, b], At x = x, the continuity conditions hold, u(x 0) = u(x + 0), c 1 u (x 0) = c 2 u (x + 0). The derivtive u (x) itself is discontinuous; its jump is determined by the jump in the coefficients: u (x + 0) u (x 0) = c 1 c 2 These conditions together with the Euler eqution nd boundry conditions determine the optiml trjectory. 12
13 3 Functionl dependent on higher derivtives Consider more generl type vritionl problem with the Lgrngin tht depends on the minimizer nd its first nd second derivtive, J = F (x, u, u, u )dx The Euler eqution is derived similrly to the simplest cse: The vrition of the gol functionl is δj = ( δu + δu + δu ) dx Integrting by prts the second term nd twice the third term, we obtin ( δj = d ) dx + d2 dx 2 δu dx [ + δu + δu d ] x=b δu dx The sttionrity condition becomes the fourthorder differentil eqution x= (24) d 2 dx 2 d dx + = 0 (25) supplemented by two boundry conditions on ech end, δu [ = 0, δu d ] dx = 0 t x = nd x = b (26) or by the correspondent min conditions posed on the minimizer u nd its derivtive u t the end points. Exmple 3.1 The equilibrium of n elstic bending bem correspond to the solution of the vritionl problem min u(x) L 0 ( 1 2 (E(x)u ) 2 q(x)u)dx (27) where u(x) is the deflection of the point x of the bem, E(x) is the elstic stiffness of the mteril tht cn vry with x, q(x) is the lod tht bends the bem. Any of the following kinemtic boundry conditions cn be considered t ech end of the bem. (1) A clmped end: u() = 0, u () = 0 (2) simply supported end u() = 0. (3) free end (no kinemtic conditions). 13
14 Let us find eqution for equilibrium nd the missing boundry conditions in the second nd third cse. The Euler eqution (25) becomes The equtions (26) become (Eu ) q = 0 (, b) δu (Eu ) = 0, δu ((Eu ) ) = 0 In the cse (2) (simply supported end), the complementry vritionl boundry condition is Eu = 0, it expresses vnishing of the bending momentum t the simply supported end. In the cse (3), the vritionl conditions re Eu = 0 nd (Eu ) = 0; the lst expresses vnishing of the bending force t the free end (the bending momentum vnishes here s well). Generliztion The Lgrngin F (x, u, u,..., u (n)) dependent on first k derivtives of udependent on higher derivtives of u is considered similrly. The sttionry condition is the 2korder differentil eqution d dx dk ( 1)k dx k (k) = 0 supplemented t ech end x = nd x = b of the trjectory by k boundry conditions [ ] δu (k 1) (k) x=,b = 0 [ d ] δu (k 2) (k 1) dx (k) x=,b = 0... [ d dx d(k 1) ( 1)k dx (k 1) ] δu (k) x=,b = 0 If u is vector minimizer, u cn be replced by vector but the structure of the necessry conditions sty the sme. 4 Nonfixed intervl 4.1 Trnsverslity condition Free boundry Consider now the cse when the intervl [, b] is not fixed, but the end point is to be chosen so tht it minimizes the functionl. Let us 14
15 compute the difference between the two functionls over two different intervls = δi = +δx (F (x, u + δu, u + δu ) F (x, u, u )) dx + The second integrl is estimted s +δx b F (x, u + δu, u + δu )dx +δx b F (x, u, u )dx F (x, u + δu, u + δu )dx F (x, u + δu, u + δu )dx = F (x, u, u ) x=b δx + o( δu, δx ) nd the first integrl is computed s before with integrtion by prts: S F (x, u, u )δu dx + δu(b) = 0 x=b 1. Suppose tht no boundry conditions re imposed t the minimizer t the point x = b. Becuse of rbitrriness of δx nd δu, we rrive t the conditions: S F (x, u, u ) = 0 x (, b), = 0, x=b nd F (x, u, u ) x=b = 0. (28) Euler eqution for the extreml stisfies n extr boundry condition (28), but hs lso n dditionl degree of freedom: unknown coordinte b. Exmple 4.1 Consider the problem s ( ) 1 min u(x),s 2 u 2 u + x dx u(0) = 0. 0 The Euler eqution u + 1 = 0 nd the condition t u(0) = 0 corresponds to the extreml u = 1 2 x2 + Ax, u = x + A where A is prmeter. The condition = u = 0 t the unknown right end x = s gives s = A. The trnsverslity condition F = 0 or ( u + x) x=a=s = 1 2 s2 s 2 + s = s (1 12 ) s = 0 We find s = 2, u = 1 2 x2 + 2x. 2. Next, consider the problem in which the boundry dt t x = b is prescribed, u = β, but the vlue of b is not known. In the perturbed trjectory, the 15
16 boundry condition is u(b + δx) = β. The vlue of u(b + δx) is n extrpoltion of u(x) s follows u(b + δx) = u(b) + u (b)δx + o( δu, δx ) Therefore, the vlue (u + δu) x=b depends on δx, u(b) = β u (b)δx or δu(b) = u (b)δx. Combining the depending on δx terms, we obtin the condition ( F (b)) x=b u δx Becuse δx is rbitrry, the boundry conditions re: u = β nd ( F (x, u, u ) u ) x=b = 0. (29) Remrk 4.1 Notice tht the lefthnd side expression in (29) t the unknown end is identicl to the expression for the first integrl (14) of the problem in the cse when F (u, u ) is independent of x. This integrl is constnt long n optiml trjectory which mens tht the problem with Lgrngin F (u, u ) does not stisfy (29) t n isolted point. 3. Finlly, consider the problem when the rjectory ends t curve. If the boundry vlue depends on b, u(b) = φ(b) (30) then the vritions δu nd δx re bounded: δu = φ δx. The two sttionrity conditions t the end point ( δu = 0 nd F (x, u, u ) u ) δx = 0 together (30) gives the conditions ( F (u φ ) ) x=b = 0 nd u(b) = φ(b). (31) The next exmple dels with constrint t the unknown length of the intervl nd the boundry dt. Exmple 4.2 Find the shortest pth between the origin nd curve φ(x). The pth length is given by I = min y(x),s s y 2 dx, u(0) = 0 At the end point x the pth meets the curve, therefore y(s) = φ(s) or δy = φ (s)δs (32) 16
17 The Euler eqution y = y = C 1 + y 2 shows tht y = constnt, therefore the pth is stright line, y = Ax s expected. At the point s, the vrition is ( u ) y F δx + y y δy = 1 δx + y δu 1 + y y 2 The sttionrity gives the reltion δx + y δu = 0. Compring it with the constrint (32), we conclude tht y (s)φ (s) = 1, or tht the shortest pth is stright line orthogonl to the curve φ(x), s it is expected. 4.2 Extreml broken t n unknown point Combining the techniques, we my ddress the problem of en extreml broken in n unknown point. The position of this point is determined from the minimiztion requirement. Assume tht Lgrngin hs the form { F (x, u, u F (x, u, u ) = ) if x (, ξ) F + (x, u, u ) if x (ξ, b) where ξ is n unknown point in the intervl (, b) of the integrtion. The Euler eqution is { SF (u) if x (, ξ) S F (u) = S F+ (u) if x (ξ, b) The sttionrity conditions t the unknown point ξ consist of sttionrity of the trjectory + = (33) nd sttionrity of the position of the trnsit point F + (u) u + + = F (u) u. (34) or F + (u) F (u) = (u + u ). (35) They re derived by the sme procedure s the conditions t the end point. The vrition δx of the trnsit point δx = δx + = δx increses the first prt of the trjectory nd decreses the second prt, or vise vers, which explins the structure of the sttionrity conditions. In prticulr, if the Lgrngin is independent of x, the condition (34) expresses the constncy of the first integrl (14) t the point ξ. 17
18 Exmple 4.3 Consider the problem with Lgrngin { F (x, u, u + u ) = 2 + b + u 2 if x (, ξ) u 2 if x (ξ, b) nd boundry conditions The Euler eqution is S F (u) = u() = 0, u(b) = 1 { + u b + u = 0 if x (, ξ) u = 0 if x (ξ, b) The solution to this eqution tht stisfies the boundry conditions is ( ) u + (x) = C 1 sinh b+ + (x ) if x (, ξ) u (x) = C 2 (x b) + 1 if x (ξ, b) ; it depends on three constnts ξ, C 1, nd C 2 (Notice tht the coefficient does not enter the Euler equtions). These constnts re determined from three conditions t the unknown point ξ which express (1) continuity of the extreml (2) WeierstrssErdmn condition (3) trnsverslity condition u + (ξ) = u (ξ), + u +(ξ) = u (ξ), + (u +(ξ)) 2 + b + u(ξ) 2 = (u (ξ)) 2. The trnsverslity condition sttes the equlity of two first integrl. It is simplified to C 2 1b + = C 2 2 From the WeierstrssErdmn condition, we find C 1 + b + cosh q = C 2, where q = b + + (ξ ) The first condition nd the definition of q llows for determintion of ξ: cosh q = +, ξ = + + b + cosh 1 + Finlly, we define constnts C 1 nd C 2 from the continuity C 1 sinh q = 1 + C 2 (ξ b) nd trnsverslity conditions: C 1 = sinh q b + (ξ b), C b+ 2 = sinh q b + (ξ b), 18
19 5 Severl minimizers 5.1 Euler equtions nd first integrls The Euler eqution cn be nturlly generlized to the problem with the vectorvlued minimizer I(u) = min u F (x, u, u )dx, (36) where x is point in the intervl [, b] nd u = (u 1 (x),..., u n (x)) is vector function. We suppose tht F is twice differentible function of its rguments. Let us compute the vrition δi(u) equl to I(u + δu) I(u), ssuming tht the vrition of the extreml nd its derivtive is smll nd loclized. To compute the Lgrngin t the perturbed trjectory u+δu, we use the expnsion F (x, u + δu, u + δu ) = F (x, u, u ) + n i=1 i δu i + n δu i i=1 i We cn perform n independent vritions of ech component of vector u pplying vritions δ i u = (0,..., δu i..., 0). The increment of the objective functionl should be zero for ech of these vrition, otherwise the functionl cn be decresed by one of them. The sttionrity condition for ny of considered vritions coincides with the oneminimizer cse. δ i I(u) = ( δu i + δu i i i ) dx 0 i = 1,..., n. Proceeding s before, we obtin the system of n secondorder differentil equtions, d = 0, i = 1,... n (37) dx i nd the boundry term i n x=b δu i = 0 (38) i=1 i If the vlue of u i () or u i (b) is not prescribed, the nturl boundry conditions x= or i x=b, respectively, must be stisfied. i The vector form of the system (37), S F (u) = d dx x=b = 0, δut = 0 (39) is identicl to the sclr Euler eqution. This system corresponds to n definition of differentition with respect to vector rgument u. x= Exmple 5.1 Consider the problem with the integrnd x= F = 1 2 u u 2 2 u 1 u u2 1 (40) 19
20 The system of sttionrity conditions is computed to be d dx d dx = u 1 + u 2 u 1 = 0 2 = (u 2 u 1 ) = 0. If consists of two differentil equtions of second order for two unknowns u 1 (x) nd u 2 (x). First integrls The first integrls tht re estblished for the specil cses of the sclr Euler eqution, cn lso be derived for the vector eqution. 1. If F is independent of u k, then one of the Euler equtions degenertes into lgebric reltion: = 0 k nd the one of differentil eqution in (37) becomes n lgebric one. The vrible u k (x) cn be discontinuous function of x in n optiml solution. Since the Lgrngin is independent of u k, the discontinuities of u k(x) my occur long the optiml trjectory. 2. If F is independent of u k, the first integrl exists: k = constnt For instnce, the second eqution in Exmple 5.1 cn be integrted nd replced by u 2 u 1 = constnt 3. If F is independent of x, F = F (u, u ) then first integrl exist Here W = u T F = constnt (41) T u = n i=1 u i i For the Exmple 5.1, this first integrl is computed to be ( 1 W = u u 2 (u 2 u 1) 2 u u 2 2 u 1 u ) 2 u2 1 = 1 ( u u 2 2 u 2 1) = constnt These three cses do not exhust ll possible first integrls for vector cse. For exmple, if the functionl depends only on, sy (u 1 + u 2 ), one cn hope to find new invrints by chnging the vribles. We discuss this mtter below in Sections?? nd??. 20
Euler Euler Everywhere Using the EulerLagrange Equation to Solve Calculus of Variation Problems
Euler Euler Everywhere Using the EulerLgrnge Eqution to Solve Clculus of Vrition Problems Jenine Smllwood Principles of Anlysis Professor Flschk My 12, 1998 1 1. Introduction Clculus of vritions is brnch
More information5.2. LINE INTEGRALS 265. Let us quickly review the kind of integrals we have studied so far before we introduce a new one.
5.2. LINE INTEGRALS 265 5.2 Line Integrls 5.2.1 Introduction Let us quickly review the kind of integrls we hve studied so fr before we introduce new one. 1. Definite integrl. Given continuous relvlued
More informationExample A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding
1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde
More informationEQUATIONS OF LINES AND PLANES
EQUATIONS OF LINES AND PLANES MATH 195, SECTION 59 (VIPUL NAIK) Corresponding mteril in the ook: Section 12.5. Wht students should definitely get: Prmetric eqution of line given in pointdirection nd twopoint
More informationUse Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions.
Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd
More informationPolynomial Functions. Polynomial functions in one variable can be written in expanded form as ( )
Polynomil Functions Polynomil functions in one vrible cn be written in expnded form s n n 1 n 2 2 f x = x + x + x + + x + x+ n n 1 n 2 2 1 0 Exmples of polynomils in expnded form re nd 3 8 7 4 = 5 4 +
More informationReview guide for the final exam in Math 233
Review guide for the finl exm in Mth 33 1 Bsic mteril. This review includes the reminder of the mteril for mth 33. The finl exm will be cumultive exm with mny of the problems coming from the mteril covered
More information6 Energy Methods And The Energy of Waves MATH 22C
6 Energy Methods And The Energy of Wves MATH 22C. Conservtion of Energy We discuss the principle of conservtion of energy for ODE s, derive the energy ssocited with the hrmonic oscilltor, nd then use this
More information1 Numerical Solution to Quadratic Equations
cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll
More information4.11 Inner Product Spaces
314 CHAPTER 4 Vector Spces 9. A mtrix of the form 0 0 b c 0 d 0 0 e 0 f g 0 h 0 cnnot be invertible. 10. A mtrix of the form bc d e f ghi such tht e bd = 0 cnnot be invertible. 4.11 Inner Product Spces
More informationAREA OF A SURFACE OF REVOLUTION
AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7.
More informationAll pay auctions with certain and uncertain prizes a comment
CENTER FOR RESEARC IN ECONOMICS AND MANAGEMENT CREAM Publiction No. 12015 All py uctions with certin nd uncertin prizes comment Christin Riis All py uctions with certin nd uncertin prizes comment Christin
More informationHarvard College. Math 21a: Multivariable Calculus Formula and Theorem Review
Hrvrd College Mth 21: Multivrible Clculus Formul nd Theorem Review Tommy McWillim, 13 tmcwillim@college.hrvrd.edu December 15, 2009 1 Contents Tble of Contents 4 9 Vectors nd the Geometry of Spce 5 9.1
More informationIntegration by Substitution
Integrtion by Substitution Dr. Philippe B. Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is
More informationVectors 2. 1. Recap of vectors
Vectors 2. Recp of vectors Vectors re directed line segments  they cn be represented in component form or by direction nd mgnitude. We cn use trigonometry nd Pythgors theorem to switch between the forms
More informationGraphs on Logarithmic and Semilogarithmic Paper
0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl
More informationVolumes of solids of revolution
Volumes of solids of revolution We sometimes need to clculte the volume of solid which cn be obtined by rotting curve bout the xxis. There is strightforwrd technique which enbles this to be done, using
More informationCalculus of variations with fractional derivatives and fractional integrals
Anis do CNMAC v.2 ISSN 1984820X Clculus of vritions with frctionl derivtives nd frctionl integrls Ricrdo Almeid, Delfim F. M. Torres Deprtment of Mthemtics, University of Aveiro 3810193 Aveiro, Portugl
More informationg(y(a), y(b)) = o, B a y(a)+b b y(b)=c, Boundary Value Problems Lecture Notes to Accompany
Lecture Notes to Accompny Scientific Computing An Introductory Survey Second Edition by Michel T Heth Boundry Vlue Problems Side conditions prescribing solution or derivtive vlues t specified points required
More informationMathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100
hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More informationThe Velocity Factor of an Insulated TwoWire Transmission Line
The Velocity Fctor of n Insulted TwoWire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the
More information4: RIEMANN SUMS, RIEMANN INTEGRALS, FUNDAMENTAL THEOREM OF CALCULUS
4: RIEMA SUMS, RIEMA ITEGRALS, FUDAMETAL THEOREM OF CALCULUS STEVE HEILMA Contents 1. Review 1 2. Riemnn Sums 2 3. Riemnn Integrl 3 4. Fundmentl Theorem of Clculus 7 5. Appendix: ottion 10 1. Review Theorem
More informationLINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES
LINEAR TRANSFORMATIONS AND THEIR REPRESENTING MATRICES DAVID WEBB CONTENTS Liner trnsformtions 2 The representing mtrix of liner trnsformtion 3 3 An ppliction: reflections in the plne 6 4 The lgebr of
More information200506 Second Term MAT2060B 1. Supplementary Notes 3 Interchange of Differentiation and Integration
Source: http://www.mth.cuhk.edu.hk/~mt26/mt26b/notes/notes3.pdf 256 Second Term MAT26B 1 Supplementry Notes 3 Interchnge of Differentition nd Integrtion The theme of this course is bout vrious limiting
More informationFactoring Polynomials
Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur
Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur Instructionl Objectives
More informationReasoning to Solve Equations and Inequalities
Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing
More informationMODULE 3. 0, y = 0 for all y
Topics: Inner products MOULE 3 The inner product of two vectors: The inner product of two vectors x, y V, denoted by x, y is (in generl) complex vlued function which hs the following four properties: i)
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH3432: Green s Functions, Integrl Equtions nd the Clculus of Vritions Section 3 Integrl Equtions Integrl Opertors nd Liner Integrl Equtions As we sw in Section on opertor nottion, we work with functions
More informationWeek 11  Inductance
Week  Inductnce November 6, 202 Exercise.: Discussion Questions ) A trnsformer consists bsiclly of two coils in close proximity but not in electricl contct. A current in one coil mgneticlly induces n
More informationThe Chain Rule. rf dx. t t lim " (x) dt " (0) dx. df dt = df. dt dt. f (r) = rf v (1) df dx
The Chin Rule The Chin Rule In this section, we generlize the chin rule to functions of more thn one vrible. In prticulr, we will show tht the product in the singlevrible chin rule extends to n inner
More informationOstrowski Type Inequalities and Applications in Numerical Integration. Edited By: Sever S. Dragomir. and. Themistocles M. Rassias
Ostrowski Type Inequlities nd Applictions in Numericl Integrtion Edited By: Sever S Drgomir nd Themistocles M Rssis SS Drgomir) School nd Communictions nd Informtics, Victori University of Technology,
More informationEcon 4721 Money and Banking Problem Set 2 Answer Key
Econ 472 Money nd Bnking Problem Set 2 Answer Key Problem (35 points) Consider n overlpping genertions model in which consumers live for two periods. The number of people born in ech genertion grows in
More informationExperiment 6: Friction
Experiment 6: Friction In previous lbs we studied Newton s lws in n idel setting, tht is, one where friction nd ir resistnce were ignored. However, from our everydy experience with motion, we know tht
More informationGeneralized Inverses: How to Invert a NonInvertible Matrix
Generlized Inverses: How to Invert NonInvertible Mtrix S. Swyer September 7, 2006 rev August 6, 2008. Introduction nd Definition. Let A be generl m n mtrix. Then nturl question is when we cn solve Ax
More informationApplications to Physics and Engineering
Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics
More informationMath 135 Circles and Completing the Square Examples
Mth 135 Circles nd Completing the Squre Exmples A perfect squre is number such tht = b 2 for some rel number b. Some exmples of perfect squres re 4 = 2 2, 16 = 4 2, 169 = 13 2. We wish to hve method for
More informationMath 314, Homework Assignment 1. 1. Prove that two nonvertical lines are perpendicular if and only if the product of their slopes is 1.
Mth 4, Homework Assignment. Prove tht two nonverticl lines re perpendiculr if nd only if the product of their slopes is. Proof. Let l nd l e nonverticl lines in R of slopes m nd m, respectively. Suppose
More informationand thus, they are similar. If k = 3 then the Jordan form of both matrices is
Homework ssignment 11 Section 7. pp. 24925 Exercise 1. Let N 1 nd N 2 be nilpotent mtrices over the field F. Prove tht N 1 nd N 2 re similr if nd only if they hve the sme miniml polynomil. Solution: If
More informationCHAPTER 11 Numerical Differentiation and Integration
CHAPTER 11 Numericl Differentition nd Integrtion Differentition nd integrtion re bsic mthemticl opertions with wide rnge of pplictions in mny res of science. It is therefore importnt to hve good methods
More informationCurve Sketching. 96 Chapter 5 Curve Sketching
96 Chpter 5 Curve Sketching 5 Curve Sketching A B A B A Figure 51 Some locl mximum points (A) nd minimum points (B) If (x, f(x)) is point where f(x) reches locl mximum or minimum, nd if the derivtive of
More information6.2 Volumes of Revolution: The Disk Method
mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine soclled volumes of
More informationMatrix Algebra CHAPTER 1 PREAMBLE 1.1 MATRIX ALGEBRA
CHAPTER 1 Mtrix Algebr PREAMBLE Tody, the importnce of mtrix lgebr is of utmost importnce in the field of physics nd engineering in more thn one wy, wheres before 1925, the mtrices were rrely used by the
More informationPure C4. Revision Notes
Pure C4 Revision Notes Mrch 0 Contents Core 4 Alger Prtil frctions Coordinte Geometry 5 Prmetric equtions 5 Conversion from prmetric to Crtesin form 6 Are under curve given prmetriclly 7 Sequences nd
More informationOperations with Polynomials
38 Chpter P Prerequisites P.4 Opertions with Polynomils Wht you should lern: Write polynomils in stndrd form nd identify the leding coefficients nd degrees of polynomils Add nd subtrct polynomils Multiply
More informationDouble Integrals over General Regions
Double Integrls over Generl egions. Let be the region in the plne bounded b the lines, x, nd x. Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing
More informationMATH 150 HOMEWORK 4 SOLUTIONS
MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive
More informationINTERCHANGING TWO LIMITS. Zoran Kadelburg and Milosav M. Marjanović
THE TEACHING OF MATHEMATICS 2005, Vol. VIII, 1, pp. 15 29 INTERCHANGING TWO LIMITS Zorn Kdelburg nd Milosv M. Mrjnović This pper is dedicted to the memory of our illustrious professor of nlysis Slobodn
More informationPROBLEMS 13  APPLICATIONS OF DERIVATIVES Page 1
PROBLEMS  APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses.
More informationDETERMINANTS. ] of order n, we can associate a number (real or complex) called determinant of the matrix A, written as det A, where a ij. = ad bc.
Chpter 4 DETERMINANTS 4 Overview To every squre mtrix A = [ ij ] of order n, we cn ssocite number (rel or complex) clled determinnt of the mtrix A, written s det A, where ij is the (i, j)th element of
More informationReview Problems for the Final of Math 121, Fall 2014
Review Problems for the Finl of Mth, Fll The following is collection of vrious types of smple problems covering sections.,.5, nd.7 6.6 of the text which constitute only prt of the common Mth Finl. Since
More informationPhysics 6010, Fall 2010 Symmetries and Conservation Laws: Energy, Momentum and Angular Momentum Relevant Sections in Text: 2.6, 2.
Physics 6010, Fll 2010 Symmetries nd Conservtion Lws: Energy, Momentum nd Angulr Momentum Relevnt Sections in Text: 2.6, 2.7 Symmetries nd Conservtion Lws By conservtion lw we men quntity constructed from
More information9 CONTINUOUS DISTRIBUTIONS
9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete
More informationA.7.1 Trigonometric interpretation of dot product... 324. A.7.2 Geometric interpretation of dot product... 324
A P P E N D I X A Vectors CONTENTS A.1 Scling vector................................................ 321 A.2 Unit or Direction vectors...................................... 321 A.3 Vector ddition.................................................
More informationORBITAL MANEUVERS USING LOWTHRUST
Proceedings of the 8th WSEAS Interntionl Conference on SIGNAL PROCESSING, ROBOICS nd AUOMAION ORBIAL MANEUVERS USING LOWHRUS VIVIAN MARINS GOMES, ANONIO F. B. A. PRADO, HÉLIO KOII KUGA Ntionl Institute
More informationAnswer, Key Homework 4 David McIntyre Mar 25,
Answer, Key Homework 4 Dvid McIntyre 45123 Mr 25, 2004 1 his printout should hve 18 questions. Multiplechoice questions my continue on the next column or pe find ll choices before mkin your selection.
More informationSection 74 Translation of Axes
62 7 ADDITIONAL TOPICS IN ANALYTIC GEOMETRY Section 74 Trnsltion of Aes Trnsltion of Aes Stndrd Equtions of Trnslted Conics Grphing Equtions of the Form A 2 C 2 D E F 0 Finding Equtions of Conics In the
More informationPhysics 43 Homework Set 9 Chapter 40 Key
Physics 43 Homework Set 9 Chpter 4 Key. The wve function for n electron tht is confined to x nm is. Find the normliztion constnt. b. Wht is the probbility of finding the electron in. nmwide region t x
More informationLectures 8 and 9 1 Rectangular waveguides
1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves
More information4 Approximations. 4.1 Background. D. Levy
D. Levy 4 Approximtions 4.1 Bckground In this chpter we re interested in pproximtion problems. Generlly speking, strting from function f(x) we would like to find different function g(x) tht belongs to
More informationThe Definite Integral
Chpter 4 The Definite Integrl 4. Determining distnce trveled from velocity Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: If we know
More informationOnline Multicommodity Routing with Time Windows
KonrdZuseZentrum für Informtionstechnik Berlin Tkustrße 7 D14195 BerlinDhlem Germny TOBIAS HARKS 1 STEFAN HEINZ MARC E. PFETSCH TJARK VREDEVELD 2 Online Multicommodity Routing with Time Windows 1 Institute
More information9.3. The Scalar Product. Introduction. Prerequisites. Learning Outcomes
The Sclr Product 9.3 Introduction There re two kinds of multipliction involving vectors. The first is known s the sclr product or dot product. This is soclled becuse when the sclr product of two vectors
More informationAAPT UNITED STATES PHYSICS TEAM AIP 2010
2010 F = m Exm 1 AAPT UNITED STATES PHYSICS TEAM AIP 2010 Enti non multiplicnd sunt preter necessittem 2010 F = m Contest 25 QUESTIONS  75 MINUTES INSTRUCTIONS DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD
More informationMATLAB Workshop 13  Linear Systems of Equations
MATLAB: Workshop  Liner Systems of Equtions pge MATLAB Workshop  Liner Systems of Equtions Objectives: Crete script to solve commonly occurring problem in engineering: liner systems of equtions. MATLAB
More informationSlow roll inflation. 1 What is inflation? 2 Equations of motions for a homogeneous scalar field in an FRW metric
Slow roll infltion Pscl udrevnge pscl@vudrevnge.com October 6, 00 Wht is infltion? Infltion is period of ccelerted expnsion of the universe. Historiclly, it ws invented to solve severl problems: Homogeneity:
More informationVector differentiation. Chapters 6, 7
Chpter 2 Vectors Courtesy NASA/JPLCltech Summry (see exmples in Hw 1, 2, 3) Circ 1900 A.D., J. Willird Gis invented useful comintion of mgnitude nd direction clled vectors nd their higherdimensionl counterprts
More informationWarmup for Differential Calculus
Summer Assignment Wrmup for Differentil Clculus Who should complete this pcket? Students who hve completed Functions or Honors Functions nd will be tking Differentil Clculus in the fll of 015. Due Dte:
More informationRadius of the Earth  Radii Used in Geodesy James R. Clynch February 2006
dius of the Erth  dii Used in Geodesy Jmes. Clynch Februry 006 I. Erth dii Uses There is only one rdius of sphere. The erth is pproximtely sphere nd therefore, for some cses, this pproximtion is dequte.
More informationSPECIAL PRODUCTS AND FACTORIZATION
MODULE  Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come
More informationHelicopter Theme and Variations
Helicopter Theme nd Vritions Or, Some Experimentl Designs Employing Pper Helicopters Some possible explntory vribles re: Who drops the helicopter The length of the rotor bldes The height from which the
More informationCOMPONENTS: COMBINED LOADING
LECTURE COMPONENTS: COMBINED LOADING Third Edition A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering 24 Chpter 8.4 by Dr. Ibrhim A. Asskkf SPRING 2003 ENES 220 Mechnics of
More informationSection 54 Trigonometric Functions
5 Trigonometric Functions Section 5 Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form
More informationBabylonian Method of Computing the Square Root: Justifications Based on Fuzzy Techniques and on Computational Complexity
Bbylonin Method of Computing the Squre Root: Justifictions Bsed on Fuzzy Techniques nd on Computtionl Complexity Olg Koshelev Deprtment of Mthemtics Eduction University of Texs t El Pso 500 W. University
More informationMathematics Higher Level
Mthemtics Higher Level Higher Mthemtics Exmintion Section : The Exmintion Mthemtics Higher Level. Structure of the exmintion pper The Higher Mthemtics Exmintion is divided into two ppers s detiled below:
More informationwww.mathsbox.org.uk e.g. f(x) = x domain x 0 (cannot find the square root of negative values)
www.mthsbo.org.uk CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input
More informationLecture 5. Inner Product
Lecture 5 Inner Product Let us strt with the following problem. Given point P R nd line L R, how cn we find the point on the line closest to P? Answer: Drw line segment from P meeting the line in right
More informationBasic Analysis of Autarky and Free Trade Models
Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently
More informationAssuming all values are initially zero, what are the values of A and B after executing this Verilog code inside an always block? C=1; A <= C; B = C;
B26 Appendix B The Bsics of Logic Design Check Yourself ALU n [Arthritic Logic Unit or (rre) Arithmetic Logic Unit] A rndomnumer genertor supplied s stndrd with ll computer systems Stn KellyBootle,
More informationExample 27.1 Draw a Venn diagram to show the relationship between counting numbers, whole numbers, integers, and rational numbers.
2 Rtionl Numbers Integers such s 5 were importnt when solving the eqution x+5 = 0. In similr wy, frctions re importnt for solving equtions like 2x = 1. Wht bout equtions like 2x + 1 = 0? Equtions of this
More informationTreatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3.
The nlysis of vrince (ANOVA) Although the ttest is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the ttest cn be used to compre the mens of only
More informationMath Review 1. , where α (alpha) is a constant between 0 and 1, is one specific functional form for the general production function.
Mth Review Vribles, Constnts nd Functions A vrible is mthemticl bbrevition for concept For emple in economics, the vrible Y usully represents the level of output of firm or the GDP of n economy, while
More informationLecture 3 Gaussian Probability Distribution
Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike
More informationPROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY
MAT 0630 INTERNET RESOURCES, REVIEW OF CONCEPTS AND COMMON MISTAKES PROF. BOYAN KOSTADINOV NEW YORK CITY COLLEGE OF TECHNOLOGY, CUNY Contents 1. ACT Compss Prctice Tests 1 2. Common Mistkes 2 3. Distributive
More informationCURVES ANDRÉ NEVES. that is, the curve α has finite length. v = p q p q. a i.e., the curve of smallest length connecting p to q is a straight line.
CURVES ANDRÉ NEVES 1. Problems (1) (Ex 1 of 1.3 of Do Crmo) Show tht the tngent line to the curve α(t) (3t, 3t 2, 2t 3 ) mkes constnt ngle with the line z x, y. (2) (Ex 6 of 1.3 of Do Crmo) Let α(t) (e
More informationNewton s Three Laws. d dt F = If the mass is constant, this relationship becomes the familiar form of Newton s Second Law: dv dt
Newton s Three Lws For couple centuries before Einstein, Newton s Lws were the bsic principles of Physics. These lws re still vlid nd they re the bsis for much engineering nlysis tody. Forml sttements
More information. At first sight a! b seems an unwieldy formula but use of the following mnemonic will possibly help. a 1 a 2 a 3 a 1 a 2
7 CHAPTER THREE. Cross Product Given two vectors = (,, nd = (,, in R, the cross product of nd written! is defined to e: " = (!,!,! Note! clled cross is VECTOR (unlike which is sclr. Exmple (,, " (4,5,6
More informationGENERALIZED QUATERNIONS SERRETFRENET AND BISHOP FRAMES SERRETFRENET VE BISHOP ÇATILARI
Sy 9, Arlk 0 GENERALIZED QUATERNIONS SERRETFRENET AND BISHOP FRAMES Erhn ATA*, Ysemin KEMER, Ali ATASOY Dumlupnr Uniersity, Fculty of Science nd Arts, Deprtment of Mthemtics, KÜTAHYA, et@dpu.edu.tr ABSTRACT
More informationMechanics Cycle 1 Chapter 5. Chapter 5
Chpter 5 Contct orces: ree Body Digrms nd Idel Ropes Pushes nd Pulls in 1D, nd Newton s Second Lw Neglecting riction ree Body Digrms Tension Along Idel Ropes (i.e., Mssless Ropes) Newton s Third Lw Bodies
More information19. The FermatEuler Prime Number Theorem
19. The FermtEuler Prime Number Theorem Every prime number of the form 4n 1 cn be written s sum of two squres in only one wy (side from the order of the summnds). This fmous theorem ws discovered bout
More informationPHY 222 Lab 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS
PHY 222 Lb 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS Nme: Prtners: INTRODUCTION Before coming to lb, plese red this pcket nd do the prelb on pge 13 of this hndout. From previous experiments,
More informationCypress Creek High School IB Physics SL/AP Physics B 2012 2013 MP2 Test 1 Newton s Laws. Name: SOLUTIONS Date: Period:
Nme: SOLUTIONS Dte: Period: Directions: Solve ny 5 problems. You my ttempt dditionl problems for extr credit. 1. Two blocks re sliding to the right cross horizontl surfce, s the drwing shows. In Cse A
More informationaddition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix.
APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The
More informationROMAI J., 2, 1(2006), Consider three system in fig.1.
ROMAI J.,, 1(6), 19 5 AN INVESTIGATION OF MATHEMATICAL MODELS OF A PIPELINE  PRESSURE SENSOR MECHANICAL SYSTEM Petr A. Velmisov, Yuliy V. Pokldov Ulynovsk Stte Technicl University, Russi velmisov@ulstu.ru
More informationOr more simply put, when adding or subtracting quantities, their uncertainties add.
Propgtion of Uncertint through Mthemticl Opertions Since the untit of interest in n eperiment is rrel otined mesuring tht untit directl, we must understnd how error propgtes when mthemticl opertions re
More informationA5682: Introduction to Cosmology Course Notes. 4. Cosmic Dynamics: The Friedmann Equation. = GM s R 2 s(t).
4. Cosmic Dynmics: The Friedmnn Eqution Reding: Chpter 4 Newtonin Derivtion of the Friedmnn Eqution Consider n isolted sphere of rdius R s nd mss M s, in uniform, isotropic expnsion (Hubble flow). The
More informationITS HISTORY AND APPLICATIONS
NEČAS CENTER FOR MATHEMATICAL MODELING, Volume 1 HISTORY OF MATHEMATICS, Volume 29 PRODUCT INTEGRATION, ITS HISTORY AND APPLICATIONS Antonín Slvík (I+ A(x)dx)=I+ b A(x)dx+ b x2 A(x 2 )A(x 1 )dx 1 dx 2
More informationExponentiation: Theorems, Proofs, Problems Pre/Calculus 11, Veritas Prep.
Exponentition: Theorems, Proofs, Problems Pre/Clculus, Verits Prep. Our Exponentition Theorems Theorem A: n+m = n m Theorem B: ( n ) m = nm Theorem C: (b) n = n b n ( ) n n Theorem D: = b b n Theorem E:
More information1. Find the zeros Find roots. Set function = 0, factor or use quadratic equation if quadratic, graph to find zeros on calculator
AP Clculus Finl Review Sheet When you see the words. This is wht you think of doing. Find the zeros Find roots. Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor.
More information