Calculus of Variations Summer Term 2016

Transcription

1 Calculus of Variations Summer Term 2016 Lecture 14 Universität des Saarlandes 28. Juni 2016 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

2 Purpose of Lesson Purpose of Lesson: To consider aerospace example c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

3 Example 14.1 Lauching a rocket Launch a rocket (with one stage) to deliver its payload into Low-Earth Orbit (LEO) at some height h above the Earth s surface. Assumptions: ignore drag, and curvature and rotation of Earth LEO so assume gravitational force at ground and orbit are approximately the same thrust will generate acceleration a, which is predefined by rocket parameters we thrust for some time T, then follow a ballistic trajectory until (hopefully) we reach height h, at zero vertical velocity, and with horizontal velocity matching the required orbital injection speed. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

4 Example 14.1 Launching a rocket-2 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

5 Example 14.1 Launching a rocket-3 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

6 Example 14.1 Launching a rocket-4 Notation: x = horizontal position y = vertical position u = horizontal velocity v = vertical velocity Initial conditions x(0) = y(0) = u(0) = v(0) = 0. Thrust stops at time T, and then at some later time S, we reach the peak of the trajectory where y(s) = h u(s) = u 0, orbital velocity v(s) = 0 We don t actually care about the final position x(s). c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

7 Example 14.1 Launching a rocket-5 Control: thrust profile is pre-determined. The only thing we can control (in this problem) is the angle of thrust. Thrust a(t) is constant for our example. Measure the angle of thrust θ(t) relative to horizontal. want to minimize fuel but this is equivalent to minimizing time, e.g., need to get to height h J = t 0 T adt = a 1dt 0 need to get to horizontal velocity u 0 to enter orbit. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

8 Constrained equations Example 14.1 Launching a rocket-6 Thrust component: t T Ballistic component: T < t S ẋ = u ẋ = u ẏ = v ẏ = v u = a cos θ u = 0 v = a sin θ g v = g Initial point Initial point: fixed x(0) = y(0) = u(0) = v(0) = 0. x(t ), y(t ), u(t ), v(t ) Final point: free Final point: x(s) free y(s) = h, v(s) = 0, u(s) = u 0 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

9 1st consider ballistic component Example 14.1 Launching a rocket-7 For t [T, S] we have no control, and ẋ = u ẏ = v u = 0 v = g we can calculate the top of the resulting parabola as and x(t ) and x(s) are free. u(s) = u(t ) v(s) = 0 y(s) = y(t ) + (v(t )) 2 /2g c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

10 Coordinate transform Example 14.1 Launching a rocket-8 So we can change variables: make the final point t = T, and take variables u, v as before, and z = y + v 2 2g. We can differentiate this and combine with previous results to get the new system DEs u = acosθ v = a sin θ g ż = ẏ + v v/g = v (1 + v/g) = av g sin θ c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

11 Optimization functional Example 14.1 Launching a rocket-9 Time minimization problem T = T 0 1dt. Including Lagrange multipliers for the 3 system constraints we aim to minimize T 0 (1 + λ u ( u a cos θ) + λ v ( v a sin θ + g) + λ z (ż avg sin θ )) dt subject to u(0) = v(0) = z(0) = 0, θ(0) = free, u(t ) = u 0, v(t ) = free, z(t ) = h, θ(t ) = free. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

12 Euler-Lagrange equations Example 14.1 Launching a rocket-10 u : v : z : θ : h u d h dt u = 0 λ u = 0 h v d h dt v = 0 a λ v = λ z g sin θ h z d h dt ż = 0 λ z = 0 h θ d h dt θ = 0 (λ equations give back systems DEs). aλ u sin θ λ v a cos θ λ z av g cos θ = 0 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

13 Solving the E-L equations Example 14.1 Launching a rocket-11 Take the v equation, and noting that v = a sin θ g λ v = λ z a g sin θ = λ z g ( v + g), λ v = λ z g (v + gt + c) = λ zv g λ zt + b. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

14 Solving the E-L equations Example 14.1 Launching a rocket-12 Substitute λ v = λ zv g λ zt + b into the θ E-L equation (dropping the common factor a) and we get ( λz v λ u sin θ + g λ u sin θ λ v cos θ λ z v g cos θ = 0 ) + λ v zt b cos θ λ z g cos θ = 0 λ u sin θ + (λ z t b) cos θ = 0 tan θ = λ zt b λ u c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

15 Solution Optimal control - aerospace example Example 14.1 Launching a rocket-13 Remember that λ u and λ v and b are all constants, so the equation tan θ = λ zt b λ u angle of thrust now specified ( θ = tan 1 λ ) zt b λ u but we need to determine constants c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

16 End-point conditions Example 14.1 Launching a rocket-14 Final end-points conditions T = free z(t ) = h u(t ) = u 0, orbital velocity v(t ) = free θ(t ) = free λ u = free λ v = free λ z = free c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

17 Natural boundary conditions Example 14.1 Launching a rocket-15 The free-end point boundary condition for J[q, q] = F(t, q, q)dt is In our problem [ m k=1 ( F δq k + δt F q k m k=1 )] F q k q k t=t = 0. F λ k = 0, F θ = 0, F u = λ u, F v = λ v, F ż = λ z c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

18 Natural boundary conditions Example 14.1 Launching a rocket-16 Consider δq k for each coordinate: for fixed coordinates u and z, we have δq k = 0 it is free for θ, λ u, λ v, λ z, but in each case the corresponding = 0, so we can ignore these. F q k only case where it matters is δv, which we can vary, and for which F v = λ v. Also δt is free, so we get two end-point conditions at t = T λ v (T ) = 0 H(T ) := [F uλ u vλ v żλ z ] t=t = 0 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

19 Natural boundary conditions Example 14.1 Launching a rocket-17 Given λ v (T ) = 0, and from previous work λ v = λ zv g λ zt + b we get λ z v(t ) g = λ z T + b = λ u tan θ(t ) v(t ) = λ ug λ z tan θ(t ) c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

20 Natural boundary conditions Example 14.1 Launching a rocket-18 H(T ) = [F uλ u vλ v żλ z ] t=t = 0. Substituting F and taking into account that λ v (T ) = 0 we get aλ u cos θ(t ) + a λ zv(t ) g Combining the latter with v(t ) = λug λ z sin θ(t ) = 1 tan θ(t ) we arrive at aλ u cos θ(t ) + aλ u tan θ(t ) sin θ(t ) = 1 λ u = cos θ(t ) a c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

21 Acceleration profile Example 14.1 Launching a rocket-19 The next step depend on the acceleration profile a(t), but lets take a simple case a = const. First we can solve the DEs, with respect to θ, using the chain rule dx dt = dx dθ dθ dt = cos 2 θ λ z dx λ u dθ e.g. from the system DE u = a cos θ u = cos 2 θ λ z du λ u dθ du dθ = λ u λ z cos 2 θ u = aλ u λ z cos θ c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

22 Acceleration profile Example 14.1 Launching a rocket-20 dx dx dθ = dt = dθ dt dx dt cos 2 θ λz λ u The complete set of system DEs becomes du dθ = aλ u λ z cos θ dv dθ = aλ u sin θ λ z cos 2 θ + gλ u λ z cos 2 θ dz dθ = aλ u sin θ gλ z cos 2 θ v(θ) These can just be integrated with respect to θ. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

23 Acceleration profile Example 14.1 Launching a rocket-21 The system DEs can be directly integrated (with respect to θ) including initial conditions u(0) = v(0) = z(0) = 0 to get u(θ) = aλ ( ) u sec θ0 + tan θ 0 log λ z sec θ + tan θ v(θ) = aλ u λ z z(θ) = a2 λ 2 u gλ 2 z + aλ2 u 2λ 2 z (sec θ 0 sec θ) gλ u λ z (tan θ 0 tan θ) sec θ 1 (sec θ 0 sec θ) a2 λ 2 u 2gλ 2 z [ tan θ 0 sec θ 0 tan θ sec θ + log ) θ = tan 1 ( λ zt b λ u ( tan 2 θ 0 tan 2 θ ) ( )] sec θ0 + tan θ 0 sec θ + tan θ c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

24 Calculating the constants Example 14.1 Launching a rocket-22 There are five constants to calculate: θ 0 the initial angle of thrust θ 1 the final angle of thrust λ u λ z b and also we need to calculate T. Solving for end-point conditions is non-trivial, but a method that works well follows. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

25 Calculating the constants Example 14.1 Launching a rocket-23 Take the equation for v at time T, and substitute λ z v(t ) = gλ u tan θ 1 to get gλ u λ z v(θ 1 ) = aλ u λ z tan θ 1 = aλ u λ z sec θ 1 = sec θ 0 g a tan θ 0 which gives us a way to calculate θ 1 from θ 0. (sec θ 0 sec θ 1 ) gλ u λ z (tan θ 0 tan θ 1 ) (sec θ 0 sec θ 1 ) gλ u λ z (tan θ 0 tan θ 1 ) Once we know θ 1 we can calculate λ u using aλ u = cos θ 1, and b from tan θ = ( (λ z t b)/λ u ) at t = 0. Then we can calculate λ z from u(θ 1 ) = u 0, the orbital injection velocity. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

26 Calculating the constants Example 14.1 Launching a rocket-24 So the only remaining question is how to calculate θ 0. We do so numerically, by take a range of θ 0 calculate all of the above use this to calculate z(t ) = z 1 as a function of θ 0 look for the point where z 1 (θ 0 ) = h the orbit height. That gives us the θ 0, from which we can derive everything else. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

27 Restricting choice of θ 0 Example 14.1 Launching a rocket-25 Calculating the range of θ 0 to search The maximum (reasonable) value for θ 0 is π/2. The minimum value of θ 0 will be determined by the minimum possible value of θ 1, i.e., θ 1 = 0 sec θ 1 = sec θ 0 g a tan θ 0 sec 0 = 1 = sec θ 0 g a tan θ 0 1 = 1 + tan2 (θ 0 /2) 1 tan 2 (θ 0 /2) g a 2 tan (θ 0 /2) 1 tan 2 (θ 0 /2) 1 tan 2 (θ 0 /2) = 1 + tan 2 (θ 0 /2) 2g a tan (θ 0/2) c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

28 Restricting choice of θ 0 Example 14.1 Launching a rocket-26 1 tan 2 (θ 0 /2) = 1 + tan 2 (θ 0 /2) 2g a tan (θ 0/2) 2 tan 2 (θ 0 /2) 2g a tan (θ 0/2) = 0 ( tan (θ 0 /2) tan (θ 0 /2) g ) = 0. a Now θ 0 can t be zero, so the last step implies that the minimum value of θ 0 is ( θ 0 = 2 tan 1 g ). a Note the existence of a minimum critical h below which we can t find a trajectory of this type. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

29 Parameters Parameters of previous example consistent with a LEO. h = 500km u 0 = 8000m/s g = 9.8m/s 2 a = 3g Derived constants θ 0 = 0.234π θ 1 = π λ u = λ z = e 0.5 b = T = seconds S = seconds c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

30 Trajectory Optimal control - aerospace example c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

31 Generalizations Optimal control - aerospace example More realistic assumptions non-zero drag (depends on velocity and height) Thrust is constant, but rocket mass changes, so that acceleration isn t constant multiple stages centripetal forces c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31