Calculus of Variations Summer Term 2015

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1 Calculus of Variations Summer Term 2015 Lecture 12 Universität des Saarlandes 17. Juni 2015 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

2 Purpose of Lesson Purpose of Lesson: First application of the Ritz method. The Ritz method applied to the catenary gives additional insights. Kantorovich s method generalizes Ritz to 2D functions. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

3 Simple Example-2 Application of the Ritz method: Example 12.1 Find extremals for J[y] = with y(0) = 0 and y(1) = [ 1 2 y ] 2 y 2 y dx The Euler-Lagrange equation y y = 1, but we shall bypass the Euler-Lagrange equation to use Ritz s method. y n (x) = φ 0 (x) + n c i φ i (x) i=1 where we take φ 0 (x) = 0 and φ i (x) = x i (1 x) i. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

4 Simple Example-2 Example 12.1 Simple approximation y 1 = c 1 φ 1 (x) we get J 1 [c 1 ] = J[y 1 ] = 1 0 [ 1 2 c2 1 φ c2 1 φ2 1 c 1φ 1 ] dx. Now φ 1 (x) = x(1 x) so φ 1 = 1 2x, and J 1 [c 1 ] = 1 0 = c2 1 2 [ ] c1 2 2 (1 2x)2 + c2 1 2 x 2 (1 x) 2 c 1 x(1 x) dx 1 0 = 11c c 1 6. [ 1 4x + 5x 2 2x 3 + x 4] 1 dx + c 1 0 [ x + x 2] dx c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

5 Simple Example-2 Example 12.1 We solve for c 1 by setting dj 1 dc 1 = 11c = 0 to get c 1 = 5/11, so the approximate extremal is y 1 (x) = 5 x(1 x). 11 The value of the approximate functional at this point is J 1 [5/11] = 11c c 1 6 = which is an upper bound on the true value of the functional on the extremal. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

6 Simple Example-2: results c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

7 Alternate Approach Example 12.1 (alternate approach) Choose φ 1 (x) = sin (πx) (use the first element of a trigonometric series to approximate y). Then, φ 1 (x) = π cos (πx), and so the functional is J 1 [c 1 ] = J[c 1 φ 1 ] = = [ 1 2 c2 1 φ c2 1 φ2 1 c 1φ 1 ] dx [ ] c1 2π2 2 cos2 (πx) + c2 1 2 sin2 (πx) c 1 sin (πx) dx. 1 1 Observe that cos 2 (πx)dx = sin 2 (πx)dx = 1/2, and 0 1 sin (πx)dx = 0 [ 1 π cos (πx) ] 1 0 = 2/π. 0 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

8 Alternate Approach Example 12.1 (alternate approach) So J 1 [c 1 ] = c2 1 4 [ ] π π c 1. Once again we solve for c 1 by setting dj 1 = c [ ] 1 π dc 1 2 π = 0 4 to get c 1 =, so the approximate extremal is π(π 2 +1) y 1 (x) = 4 π(π 2 sin (πx). + 1) c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

9 Simple Example-2: alternative results c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

10 Ritz and Catenary: Numerical Solutions (continued) Example: the Catenary, again Example 12.2 (the catenary, again) The functional of interest (the potential energy) is J p [y] = mg x 1 x 0 y 1 + y 2 dx. Take symmetric problem with fixed end points y( 1) = a and y(1) = a. We know that the solution looks like ( ) x y(x) = c 1 cosh where c 1 is chosen to match the end points. c 1 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

11 Example: the Catenary, again Example 12.2 (the catenary, again) y(1) = 2 gives c 1 = 0.47 or c 1 = Are they both local minima? c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

12 Ritz and the Catenary Example 12.2 (Ritz and the Catenary) Lets try approximating the curve by a polynomial y(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x Note that symmetry of problem implies y is an even function, and hence the odd terms a 1 = a 3 = = 0. So, to second order we can approximate y(x) a 0 + a 2 x 2. We have fixed y(1) = y 1, so we can simplify to get y(x) a 0 + (y 1 a 0 )x 2. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

13 Ritz and the Catenary Example 12.2 (Ritz and the Catenary) y a 0 + (y 1 a 0 )x 2 y 2(y 1 a 0 )x Taking into account y(1) = 2 we get a 0 + a 2 = 2. We can substitute into the functional J p [y] = mg x 1 x 0 y 1 + y 2 dx and integrate to get a function J p [a 2 ] with respect to a 2. But this function is pretty complicated. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

14 Ritz and the Catenary Example 12.2 (Ritz and the Catenary) From Maple we have the value for J p [a 2 ], (a := a 2 ) O f x d 2 K a C a$x 2 $ 1 C 4$a 2 $x 2 : O int f x, x =K a 2 16 a 2 ln K2 a C 1 C 4 a 2 csgn a csgn a C C 4 a 2 a 2 csgn a (1) K 64 a 3 1 C 4 a 2 csgn a K 32 a ln K2 a C 1 C 4 a 2 csgn a csgn a C ln K2 a C 1 C 4 a 2 csgn a csgn a K 4 1 C 4 a 2 a csgn a C 8 1 C 4 a 2 3 / 2 a csgn a K 16 a 2 ln 2 a C 1 C 4 a 2 csgn a csgn a C 32 a ln 2 a C 1 C 4 a 2 csgn a csgn a K ln 2 a C 1 C 4 a 2 csgn a csgn a O csgn a c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

15 Ritz and the Catenary Example 12.2 (Ritz and the Catenary) Its a pain to find the zeros of dj p, but its easy to plot, and find da them numerically. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

16 Ritz and the Catenary Example 12.2 (Ritz and the Catenary) Stationary points local max: a = a local min: a = a c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

17 Ritz and the Catenary Example 12.2 (Ritz and the Catenary) c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

18 Ritz and the Catenary Example 12.2 (Ritz and the Catenary) c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

19 Ritz and the Catenary Ritz and the Catenary Doesn t just give us an approximation to the extremal curves, its also give us some insight into the nature of these extremals. If approximations are near to the actual extrema There are no other extrema so close by The functional is smooth (it can t have jumps either) Then the type of extrema we get for the approximation will be the same for the real extrema, i.e., local max: a local max for c 1 = local min: a local min for c 1 = c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

20 More than one independent variables More than one independent variables 2D Case: We are approximating a surface with series of functions, e.g. z(x, y) z n (x, y) = φ 0 (x, y) + where n c i φ i (x, y) φ 0 (x, y) satisfies the boundary conditions, e.g. φ 0 (x, y) = z 0 (x, y) for (x, y) Ω, the boundary of the region on interest Ω, and the φ i (x, y) satisfy the homogeneous boundary conditions φ i (x, y) = 0 for (x, y) Ω. i=1 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

21 More than one independent variables 2D Case: As before, we approximate the functional by J[z] J[z n ] = J n (c 1,..., c n ). As before we determine the c j by requiring that the partial derivatives are zero, e.g. for all i = 1, 2,..., n. J n c i = 0 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

22 Kantorovich s Method Kantorovich s Method Approximate with z(x, y) z n (x, y) = φ 0 (x, y) + n c i (x)φ i (x, y). i=1 Again the φ i are suitable chosen, but the c i are no longer constants, but rather functions of one independent variable. This allows a larger class of functions to be used. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

23 Kantorovich s Method Kantorovich s Method Note that the integral function J[z n ] = z n (x, y)dxdy = Ω n i=0 c i (x) y 1 (x) y 0 (x) φ i (x, y)dy dx We integrate the inner integral, and get J[z n ] = n i=0 c i (x)φ i (x)dx. Now we just have a function of x, and so we may apply the Euler-Lagrange machinery. The method approx. separates the variables x and y. c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

24 Kantorovich s Method Example 12.3 Find the extremals of J[z(x, y)] = with z = 0 on the boundary. b a b a ( ) zx 2 + zy 2 2z dxdy The Euler-Lagrange equation reduces to the Poisson equation, e.g. F z d dx F z x d dy F z y = 0 2 d dx (2z x) d dy (2z y) = 0 z xx + z yy = 1 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

25 Kantorovich s Method Example 12.3 Approximate z 1 (x, y) = c(x) (b 2 y 2) Note z 1 (x, ±b) = 0 (as required) and ( ) 2 z1 [ = c (x) (b 2 y 2)] 2 x = c (x) 2 [ b 4 2b 2 y 2 + y 4], ( ) 2 z1 = [c(x)2y] 2 y = 4c(x) 2 y 2 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

26 Kantorovich s Method Example 12.3 Hence, we approximate J[z(x, y)] J[z 1 (x, y)] = = = = a a a a a b b b a b a ( ) zx 2 + zy 2 2z dxdy [ ( c (x) 2 b 2 y 2) 2 ( + 4c(x) 2 y 2 2c(x) b 2 y 2)] dy dx [ ( ) c (x) 2 b 4 y 2b 2 y 3 /3 + y 5 /5 + 4c(x) 2 y 3 /3 [ b5 c (x) b3 c(x) 2 8 ] 3 b3 c(x) dx ( )] b +2c(x) b 2 y y 3 /3 b dx a c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

27 Kantorovich s Method Example 12.3 So we can write J[z(x, y)] J[z 1 (x, y)] = J[c(x)] = a a F(x, c, c )dx We can use the simple Euler-Lagrange equation, where F(x, c, c ) = b5 c (x) b3 c(x) b3 c(x) F c = 16 3 b3 c(x) 8 3 b3 F c = b5 c (x) d F dx c = b5 c (x) c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

28 Kantorovich s Method Example 12.3 The Euler-Lagrange equation 16 3 b3 c(x) 8 3 b b5 c (x) = 0 c (x) 5 2b 2 c(x) = 5 4b 2 Solutions ( ) ( ) 5 x 5 x c(x) = k 1 cosh + k 2 sinh b 2 b 2 c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

29 Kantorovich s Method Example 12.3 Note that the function must be zero on the boundary, so z(±a, y) = 0. We look for an even function c(x), and so k 2 = 0. Also c(±a) = 0, so ( ) 5 a c(a) = k 1 cosh b 2 ( ) = k a 1 cosh 2 b 1 k 1 = ( 2 cosh 5 a 2 b ) c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

30 Kantorovich s Method Example 12.3 Solution ( z 1 (x, y) = 1 2 (b2 y 2 cosh ) 1 ( cosh 5 x 2 b 5 a 2 b ) ) If we want a more exact approximation, we could try z 2 (x, y) = (b 2 y 2 )c 1 (x) + (b 2 y 2 ) 2 c 2 (x). c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

31 Lower bounds Remarks Obviously, quality of solution depends on family of functions chosen number of terms used, n Could test convergence by increasing n and seeing the difference in J[y n+1 ] J[y n ], but this is not guaranteed to be a good indication. A better way to assess convergence is to have a lower bound lower bound J[y] upper bound c Daria Apushkinskaya (UdS) Calculus of variations lecture Juni / 31

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