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Course: M 612, Fall 2009
School: Texas A&M
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Assignment M612, 1, due Friday Jan. 30 NOTE. In Problems 1 and 2, we will work fairly hard to establish that the shortest distance between any two points is a straight line. This is the classical first problem in the calculus of variations, and even though the result is trivial, an enormous amount of theory becomes apparent. 1. [10 pts] We know from first-semester calculus that the length of the graph of a C 1...

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Assignment M612, 1, due Friday Jan. 30 NOTE. In Problems 1 and 2, we will work fairly hard to establish that the shortest distance between any two points is a straight line. This is the classical first problem in the calculus of variations, and even though the result is trivial, an enormous amount of theory becomes apparent. 1. [10 pts] We know from first-semester calculus that the length of the graph of a C 1 function y(t) connecting the point (0, a) to the point (T, b) is given by the arclength formula L[y(t)] = 0 T 1 + y (t)2 dt. Compute L [y(t)], and use your result to show that the shortest such curve must be a straight line. 2. [10 pts] (This problem refers to Problem 1.) As observed in class, the derivative L [y(t)] is a linear mapping on the space A := { C 2 [0, T ] : (0) = (T ) = 0}. We can compute the derivative of L [y(t)] (and so the second derivative of L[y(t)]) as the bilinear form L [y(t)]((t), (t)) := lim L [y(t) + (t)] - L [y(t)] ((t)). 0 For the arclength functional from Problem 1, compute L [y(t)], where y(t) is taken as the minimizer found in Problem 1, and show that L [y(t)] is a positive definite bilinear form: that is, show that for any A, not identically 0, L [y(t)]((t), (t)) > 0. This is the variational second-order derivative condition for a minimizer, and ensures that y(t) is in fact a local minimizer. 3. [10 pts] This problem fills in the final step of our derivation from class of the Hamilton Jacobi equation. Suppose L C 2 (Rn Rn ; R), that and q(t) satisfies the EulerLagrange equations d Dq L(q(t), q (t), t) - Dq L(q(t), q (t), t) = 0 dt for t [0, T ]. Let q(0) = be fixed, and fix , but let T vary, where q(T ) = . In this context, it's useful to write q = q(t; T ). Set A(T ) = 0 T L(q(t), q (t), t)dt. Show that A = -H(p(T ), q(T )). T 1 Notes. After cleverly integrating by parts, notice that q(0; T ) = 0 T q(T ; T ) 0 = q(T ; T ) = qt (T ; T ) + qT (T ; T ), T q(0; T ) and then use the fact that q(t) satisfies the EulerLagrange equations to clean things up. 4. [10 pts] The Schrodinger equation for a single free quantum object is iut = - 2m u in Rn R+ , where = h/2, and h is Planck's constant. (Planck's constant is the constant of proportionality between the energy of a photon E and the frequency of the photon ; that is, E = h. Of course, this makes no sense: when I write "a photon" I mus...

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boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -0.693147180559945286 eta= 0.100000000000000006 nu = 0.299999999999999989 alpha = 0.490907905324351579 beta= 0.509092094675648421 kappa0= -1 M2 = 18 NJ= 3
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boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -2.30258509299404590 eta= 0.299999999999999989 nu = 0.299999999999999989 alpha = 0.472782819701179080 beta= 0.527217180298820920 kappa0= -1 M2 = 18 NJ= 30
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boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 0.693147180559945286 eta= 0.299999999999999989 nu = 0.299999999999999989 alpha = 0.472782819701179080 beta= 0.527217180298820920 kappa0= -1 M2 = 18 NJ= 30
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boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 2.30258509299404546 eta= 0.100000000000000006 nu = 0.299999999999999989 alpha = 0.490907905324351579 beta= 0.509092094675648421 kappa0= -1 M2 = 18 NJ= 30
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