Module Code: MATH265001 Module Title: Calculus of Variations School of Mathematics Semester Two 201819 Calculator instructions: • You are allowed to use a calculator which has had an approval sticker issued by the School of Mathematics. Dictionaries: • You are not allowed to use your own dictionary in this exam. A basic English dictionary is available to use: raise your hand and ask an invigilator, if you need it. Exam information: • There are 4 pages to this exam. • There will be 2 hours to complete this exam. • Answer all questions. • All questions are worth equal marks. Page 1 of 4 Turn the page over Module Code: MATH265001 1. (a) Consider the problem of finding the extremal of the integral I = ∫ x2 x1 F (x, y(x), y′(x)) dx, with boundary conditions y(x1) = y1, y(x2) = y2. Derive the Euler-Lagrange equation that must be satisfied by the extremal y(x). [You may assume without proof, but should state, the Fundamental Lemma of the Calculus of Variations.] (b) Find the extremal of the integral I = ∫ 2 0 ( y′2 − 2yy′ − 2y′) dx, with y(0) = 1−α and y(2) = 1. Calculate the value of I and minimise the answer with respect to α. (c) Suppose now that the extremal in part (a) has no boundary condition specified at x = x1. Derive the natural boundary condition that must be satisfied there. (d) Now consider the problem in part (b) but without a boundary condition applied at x = 0. By applying the natural boundary condition at x = 0, derive the extremal. Comment on the relation between this answer and that derived in part (b). 2. Let A(y) be the area of the surface of revolution formed by rotating the curve y(x) about the x-axis, where y > 0 and the curve has fixed end-points y(x1) = a and y(x2) = b, with x2 > x1. (a) Assuming that a solution exists, show that the curve that minimises A is of the form y(x) = y0 cosh ( x− x0 y0 ) , and give a brief interpretation of the two constants x0 and y0. Sketch the curve y(x) for x1 < x < x2. Hint: you may wish to make use of the Beltrami identity, which states that if the integrand F (x, y, y′) is independent of x, then a first integral of the Euler-Lagrange equation is y′ ∂F ∂y′ − F = C, where C is a constant. (b) Suppose now that x1 = −1, x2 = 1 and a = b. What is the value of x0? Show graphically that, depending on the value of a, there may be zero, one or two solutions for y0. (c) The surfaces of revolution in part (b) can be realised experimentally by soap films extending between circular rings of radius a at x = ±1. What would happen experimentally in the case when there is no solution for y0? (d) Briefly discuss how the form of the solution in part (a) might differ if there were the additional constraint that the curve y(x) should have a specified length. Page 2 of 4 Turn the page over Module Code: MATH265001 3. (a) Prove that the shortest distance between two points on a plane is given by the straight line joining the points. (b) The surface g(x, y, z) = 0 may be expressed parametrically as x = x(u, v), y = y(u, v), z = z(u, v). The square of the differential of arc length ds can then be written as ds2 = dx2 + dy2 + dz2 = P (u, v)du2 + 2Q(u, v)du dv +R(u, v)dv2. Derive the expressions for P , Q and R. If Q = 0, what does this say about the (u, v) coordinate system? In the special case when both Q = 0 and P and R are functions only of u, show that the geodesic (i.e. the shortest distance on the surface) is given by v = C ∫ √ P√ R2 − C2R du, where C is a constant of integration. Derive the corresponding expression for when both Q = 0 and P and R are functions only of v. (c) A surface is given in parametric form by: x(u, v) = u, y(u, v) = v, z(u, v) = 2av3/2 3 , where v ≥ 0 and where a is a positive constant. Using the appropriate result from part (b), and evaluating the integral, derive the general expression for a geodesic on the surface. Show that the geodesic with end points at the origin and at (x, y, z) = (1, 1, 2a/3) is given by u(v) = (1 + a2v)3/2 − 1 (1 + a2)3/2 − 1 . Comment briefly on the limiting form of this solution as a→ 0. Page 3 of 4 Turn the page over Module Code: MATH265001 4. (a) Consider the Sturm-Liouville system composed of the ordinary differential equation d dx ( p(x) dy dx ) + q(x)y + λr(x)y = 0, (4.1) with boundary conditions y(x1) = y1, y(x2) = y2. (4.2) Define Λ by Λ = ∫ x2 x1 (p(x)y′2 − q(x)y2) dx∫ x2 x1 r(x)y2 dx . Show that the problem of determining functions that make Λ stationary is equiva- lent to the problem of determining the eigenfunctions of (4.1), subject to the same boundary conditions (4.2). (b) For the remainder of the question, consider the case when p(x) = 1, q(x) = 0 and r(x) = 1. Suppose first that the boundary conditions are y(0) = y(1) = 0. By using the result in part (a) and adopting the trial function y = cx(1−x), obtain an estimate for the lowest eigenvalue λ. By solving (4.1) directly, with these values of p, q and r, derive also the exact value of the eigenvalue. Explain why your estimate is greater than the true eigenvalue. (c) Now suppose that the boundary conditions are changed to y(0) = y′(1) = 0. Explain why the result of part (a) still holds even with the given condition on y′ at x = 1. (d) By using the result in part (a) and adopting a suitable quadratic trial function, obtain an estimate of the smallest eigenvalue λ to the problem in part (c). What is the relative error of your estimate compared to the lowest eigenvalue of the problem with this boundary condition (which you will need to find)? 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