Variational Methods & Optimal Control

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1 Variational Methods & Optimal Control lecture 12 Matthew Roughan Discipline of Applied Mathematics School of Mathematical Sciences University of Adelaide April 14, 216 Variational Methods & Optimal Control: lecture 12 p.1/??

2 Numerical Solutions The E-L equations may be hard to solve Natural response is to find numerical methods Numerical solution of E-L DE we won t consider these here (see other courses) Euler s finite difference method Ritz (Rayleigh-Ritz) In 2D: Kantorovich s method Variational Methods & Optimal Control: lecture 12 p.2/??

3 Euler s finite difference method We can approximate our function (and hence the integral) onto a finite grid. In this case, the problem reduces to a standard multivariable maximization (or minimization) problem, and we find the solution by setting the derivatives to zero. In the limit as the grid gets finer, this approximates the E-L equations. Variational Methods & Optimal Control: lecture 12 p.3/??

4 Numerical Approximation Numerical approximation of integrals: use an arbitrary set of mesh points a=x < x 1 < x 2 < <x n = b. approximate rectangle rule F{y}= b a y (x i )= y i+1 y i x i+1 x i = y i x i f(x,y,y )dx n 1 i= F( ) is a function of the vector y=(y 1,y 2,...,y n ). f ( x i,y i, y ) i x i = F(y) x i Variational Methods & Optimal Control: lecture 12 p.4/??

5 Finite Difference Method (FDM) Treat this as a maximization of a function of n variables, so that we require for all i=1,2,...,n. F y i = Typically use uniform grid so x i = x=(b a)/n. Variational Methods & Optimal Control: lecture 12 p.5/??

6 Simple Example Find extremals for F{y}= 1 [ 1 2 y ] 2 y2 y dx with y()= and y(1)=. E-L equations y y=1. Variational Methods & Optimal Control: lecture 12 p.6/??

7 Simple Example: direct solution E-L equations y y= 1 Solution to homogeneous equations y y= is given by e λx giving characteristic equation λ 2 1=, so λ=±1. Particular solution y = 1 Final solution is y(x)=ae x + Be x + 1 The boundary conditions y() = y(1) = constrain A + B = 1 and Ae+Be 1 = 1, so Ae+(1 A)e 1 = 1, so A= e 1 1 e e 1 Then the exact solution to the extremal problem is y(x)= e 1 1 e e 1 ex + 1 e e e 1 e x 1 and B= 1 e e e 1. Variational Methods & Optimal Control: lecture 12 p.7/??

8 Simple Example: Euler s FDM Find extremals for F{y}= 1 [ 1 2 y ] 2 y2 y dx Euler s FDM. Take the grid x i = i/n, for i=,1,...,n so end points y = and y n = x=1/n y i = y i+1 y i So y i = y i/ x=n(y i+1 y i ) and y 2 i = n 2( y 2 i 2y i y i+1 + y 2 ) i+1 Variational Methods & Optimal Control: lecture 12 p.8/??

9 Simple Example: Euler s FDM Find extremals for F{y}= 1 [ 1 2 y ] 2 y2 y dx Its FDM approximation is F(y) = = = n 1 i= n 1 i= n 1 i= f (x i,y i,y i) x 1 2 n2( ) y 2 i 2y i y i+1 + y 2 i+1 x+(y 2 i /2 y i ) x 1 2 n( y 2 i 2y i y i+1 + y 2 ) y 2 i+1 + i /2 y i n Variational Methods & Optimal Control: lecture 12 p.9/??

10 Simple Example: end-conditions We know the end conditions y()=y(1)=, which imply that y = y n = Include them into the objective using Lagrange multipliers H(y) = n 1 i= 1 2 n( y 2 i 2y i y i+1 + y 2 ) y 2 i+1 + i /2 y i n + λ y + λ n y n Variational Methods & Optimal Control: lecture 12 p.1/??

11 Simple Example: Euler s FDM Taking derivatives, note that y i only appears in two terms of the FDM approximation H(y) = H(y) y i = n 1 i= 1 2 n( y 2 i 2y i y i+1 + y 2 ) y 2 i+1 + i /2 y i n + λ y + λ n y n n(y y 1 )+ y 1 n + λ for i= n(2y i y i+1 y i 1 )+ y i n 1 n for i=1,...,n 1 n(y n y n 1 )+λ n for i=n We need to set the derivatives to all be zero, so we now have n+3 linear equations, including y = y n =, and n+3 variables including the two Lagrange multipliers. We can solve this system numerically using, e.g., matlab. Variational Methods & Optimal Control: lecture 12 p.11/??

12 Simple Example: Euler s FDM Example: n = 4, solve Az=b where A= and b= first n+1 terms of z give y last two terms given the Lagrange multipliers λ and λ n Variational Methods & Optimal Control: lecture 12 p.12/??

13 Simple example: results.2 n = 2.15 x f(x) Variational Methods & Optimal Control: lecture 12 p.13/??

14 Simple example: results.2 n = 4.15 x f(x) Variational Methods & Optimal Control: lecture 12 p.13/??

15 Simple example: results.15 n = 6.1 x f(x) Variational Methods & Optimal Control: lecture 12 p.13/??

16 Simple example: results.2 n = 1.15 x f(x) Variational Methods & Optimal Control: lecture 12 p.13/??

17 Simple example: results.15 n = 2.1 x f(x) Variational Methods & Optimal Control: lecture 12 p.13/??

18 Convergence of Euler s FDM F(y) = n 1 i= f ( x i,y i, y ) i x and y i = y i+1 y i x Only and two terms in the sum involve y i, so F y i = y i f = 1 x f y i + f y i ( x i 1,y i 1, y ) i 1 + f x y i ( x i 1,y i 1, y ) i 1 ( x i,y i, y i x = f y i (x i,y i,y i) f y i x ) 1 f x y i ( ) x i,y i, y i x ( x i,y i, y ) i x ( x i,y i, y i x f y i x ) ( x i 1,y i 1, y i 1 x ) Variational Methods & Optimal Control: lecture 12 p.14/??

19 Convergence of Euler s FDM The condition for a stationary point becomes F y i = f y i (x i,y i,y i) f y i ( ) ( ) x i,y i, y i x f y i x i 1,y i 1, y i 1 x x = In limit n, then x, and so we get f y d dx ( ) f y which are the Euler-Lagrange equations. = i.e., the finite difference solution converges to the solution of the E-L equations Variational Methods & Optimal Control: lecture 12 p.15/??

20 Comments There are lots of ways to improve Euler s FDM use a better method of numerical quadrature (integration) trapezoidal rule Simpson s rule Romberg s method use a non-uniform grid make it finer where there is more variation We can use a different approach that can be even better Variational Methods & Optimal Control: lecture 12 p.16/??

21 Ritz s method In Ritz s method (called Kantorovich s methods where there is more than one independent variable), we approximate our functions (the extremal in particular) using a family of simple functions. Again we can reduce the problem into a standard multivariable maximization problem, but now we seek coefficients for our approximation. Variational Methods & Optimal Control: lecture 12 p.17/??

22 Ritz s method Assume we can approximate y(x) by y(x)=φ (x)+c 1 φ 1 (x)+c 2 φ 2 (x)+ +c n φ n (x) where we choose a convenient set of functions φ j (x) and find the values of c j which produce an extremal. For fixed end-point problem: Choose φ (x) to satisfy the end conditions. Then φ j (x )=φ j (x 1 )= for j= 1,2,...,n The φ can be chosen from standard sets of functions, e.g. power series, trigonometric functions, Bessel functions, etc. (but must be linearly independent) Variational Methods & Optimal Control: lecture 12 p.18/??

23 Ritz s method select{φ j } n j= Approximate y n (x)=φ (x)+c 1 φ 1 (x)+c 2 φ 2 (x)+ +c n φ n (x) Approximate F{y} F{y n }= x1 Integrate to get F{y n }=F n (c 1,c 2,...,c n ) x f(x,y n,y n)dx F n is a known function of n variables, so we can maximize (or minimize) it as usual by F n c i = for all i=1,2,...,n. Variational Methods & Optimal Control: lecture 12 p.19/??

24 Upper bounds Assume the extremal of interest is a minimum, then for the extremal F{y}<F{ŷ} for all ŷ within the neighborhood of y. Assume our approximating function y n is close enough to be in that neighborhood, then F{y} F{y n }=F n (c) so the approximation provides an upper bound on the minimum F{y}. Another way to think about it is that we optimize on a smaller set of possible functions y, so we can t get quite as good a minimum. Variational Methods & Optimal Control: lecture 12 p.2/??

25 Simple Example Find extremals for F{y}= 1 [ 1 2 y ] 2 y2 y dx with y()= and y(1)=. E-L equations y y=1, but we shall bypass the E-L equations to use Ritz s method. n y n (x)=φ (x)+ c i φ i (x) i=1 where we take φ (x)= and φ i (x)=x i (1 x) i. Variational Methods & Optimal Control: lecture 12 p.21/??

26 Simple Example Simple approximation y 1 = c 1 φ 1 (x) we get 1 [ ] 1 F 1 (c 1 )=F{y 1 }= 2 c2 1φ c φ2 1 c 1 φ 1 dx Now φ(x)=x(1 x) so φ 1 = 1 2x, and 1 [ ] c 2 F 1 (c 1 ) = 1 2 (1 2x)2 + c2 1 2 x2 (1 x) 2 c 1 x(1 x) dx = c2 1 2 = c2 1 2 = c [ 1 4x+5x 2 x 4] 1 [ dx+c 1 x+x 2 ] dx [ x 2x 2 + 5x 3 /3 x 5 /5 ] 1 + c c 1 6 [ x 2 /2+x 3 /3 ] 1 Variational Methods & Optimal Control: lecture 12 p.22/??

27 Simple Example We solve for c 1 by setting df 1 dc 1 = 11c = to get c 1 = 5/11, so the approximate extremal is y 1 (x)= 5 11 x(1 x) The value of the approximate functional at this point is F 1 (5/11)= c c 1 6 = which is an upper bound on the true value of the functional on the extremal. Variational Methods & Optimal Control: lecture 12 p.23/??

28 Simple example: results exact y 1 = c 1 x(1 x) Variational Methods & Optimal Control: lecture 12 p.24/??

29 Alternate approach Choose φ 1 (x)=sin(πx) (use the first element of a trigonometric series to approximate y). Then, φ (x)=πcos(πx), and so the functional is 1 [ ] 1 F 1 (c 1 ) = F{c 1 φ 1 }= 2 c2 1φ c φ2 1 c 1 φ 1 dx 1 [ c 2 = 1 π 2 ] 2 cos2 (πx)+ c2 1 2 sin2 (πx) c 1 sin(πx) dx Now 1 cos2 (πx)= 1 sin2 (πx)=1/2, and 1 sin(πx)=[ 1 π cos(πx)]1 = 2/π, so F(c 1 )= c [ π ] 2 π c 1 Variational Methods & Optimal Control: lecture 12 p.25/??

30 Alternate approach Once again we solve for c 1 by setting df 1 1 = c 1 dc 1 2 [ π ] 2 π = to get c 1 = 4 π(π 2 +1), so the approximate extremal is 4 y 1 (x)= π(π 2 + 1) sin(πx) Variational Methods & Optimal Control: lecture 12 p.26/??

31 example: alternative results exact y 1 = c 1 sin(π x) Variational Methods & Optimal Control: lecture 12 p.27/??