UNIVERSITY of LIMERICK OLLSCOIL LUIMNIGH College of Informatics and Electronics END OF SEMESTER ASSESSMENT PAPER MODULE CODE: MS425 SEMESTER: Autumn 25/6 MODULE TITLE: Applied Analysis DURATION OF EXAMINATION: 2 1/2 hours LECTURER: Dr. J. Kinsella PERCENTAGE OF TOTAL MARKS: 1% EXTERNAL EXAMINER: Prof. J. King INSTRUCTIONS TO CANDIDATES: Answer four questions correctly for full marks. See the Appendix at the end of the paper for some useful results.
1 Suppose a shopkeeper knows that a fraction k(t) of goods remain at a time t after their purchase. At what rate should goods be purchased to keep stock constant? (a) Let u(t) be the rate (goods per unit time) at which goods are to be bought. Let A be the initial stock or opening balance. In each time interval [τ, τ + t] the shop will buy u(τ) t quantity of goods. Show that: 8% Stock at time t = Ak(t) + t k(t τ)u(τ)dτ so the problem to be solved is to find a function u(t) (given k(t)) such that: A = Ak(t) + t k(t τ)u(τ)dτ (1) (b) Now solve (1) for u(t) given k(t) = e t/t for all t >. Hint: use the 6% fact that the Laplace Transform of a convolution is just the product of the transforms: t a(x)b(t x)dx = a(s)b(s). (c) Now re-solve (1) using the differentiation method i.e. differentiate twice w.r.t t and eliminate the term involving the integral to find an expression for u (t). 1% (d) Interpret this result. 1% 2 (a) Given a Fredholm integral equation of the second kind u = f + λku, with symmetric kernel (k(x, t) = k(t, x)) show that the solution u(x) can be expressed as: 8% u(x) = f(x) + λ f k φ k (x) λ k λ (2) where the functions φ k (x) are the eigenfunctions of K corresponding to the eigenvalues λ k (provided that λ λ k ) so that we have φ k (x) = λ k (Kφ k )(x). 1
(b) Solve the following Fredholm integral equation of the second kind : with as follows. u = f + λku u(x) = e x + λ k(x, t) = π 2 { sin x cos t k(x, t)u(t)dt x t sin t cos x t x π/2 (i) First show by differentiating the eigenvalue equation u(x) = λ(ku)(x) twice wrt x that: 6% u (x) = (1 + λ)u(x). (ii) Next show that the definition of the kernel implies that u() = u(π/2) =. 2% (iii) Solve this boundary value problem and show that the eigenfunctions and eigenvalues are given by: 4% λ n = 4n 2 1 φ n (x) = 2 π sin 2nx (iv) Finally, using (2) above, express u(x) in terms of the eigenfunctions φ n (x). 4% (v) Suppose we fix λ = 3, can we still find a solution? Explain. 1% 2
3 The solution to a Sturm-Liouville problem (see Appendix A) can be expressed as u(x) = b a g(x, y)f(y)dy. (3) where g(x, y) is the Green s Function for the problem (see Appendix B). (a) show that the Green s Function has the following properties: (i) Ag(x, y) = for all x y. 1% (ii) g(x, y) satisfies the boundary conditions : 2% (1) B 1 g(a, y) = (2) B 2 g(b, y) = (iii) g(x, y) is continuous on [a, b] including x = y (w.r.t. both x and y separately). 2% (iv) g (x, y) is not continuous at x = y: 4% (b) Solve the o.d.e. g (x, y) x=y = 1 p(y). x 2 u + 2xu 6u = f(x) with u() = 1 and u(1) = 1 on the interval [, 1] using the Green s Function method. (i) First set v(x) = u(x) + Ax + B and choose A and B so that v(x) satisfies the homogeneous boundary conditions v() = v(1) =. 4% You should find that v(x) satisfies: x 2 v + 2xv 6v = f(x) 8x + 6. (ii) Now solve for the Green s Function g(x, y) that satisfies: 4% x 2 g + 2xg 6g = δ(x y) by noting that for x y the problem is just x 2 g + 2xg 6g = (Use the substitution g = x n.) (iii) Next use the four conditions listed in part (a) together with the boundary conditions to find an expression for g(x, y). 5% (iv) If f(x) = e x, find an expression for the solution u(x) do not evaluate the integrals. 3% 3
4 A fundamental solution to Laplace s Equation is a distributional solution to u = δ(x; ν) ( u 2 u) where the source term is the delta distribution with pole at x = ν R n. (a) Show that in R 2, g(x, y) = 1 4π ln(x2 + y 2 ) 1 ln r is a fundamental 2π solution when ν =. Use polar coordinates see Appendix C. 17% (b) Use this fundamental solution to construct a Green s Function g(x; ν) for the problem: ( ) g (x, y); (x, y ) = for all x, x and for all y, y >. ) g y ((x, ); (x, y ) = for all x, x and for all y, y >. Hint: use the method of images adding rather than subtracting to ensure that g y = on the x-axis. 6% (c) How can the result of (b) be used to construct a solution to the following problem? 2% u xx + u yy = ρ(x, y) u y (x, ) = 5 (a) Prove the Cauchy Integral Formula: Let f(z) be analytic in a simply connected domain D. Then for any point z in D and any contour (simple closed path) C in D we have (taking C counter-clockwise) 7% f(z) dz = 2πif(z ). (4) z z C (b) Using the Cauchy Integral Formula, integrate z 3 /(z 4 +1) anti-clockwise round each of: 4% + (i) z 1 = 1 (ii) z i = 1 2 4% (c) Use the Residue Theorem (see Appendix D) to evaluate the following real integrals: 5% + 2π 1 + cos θ x 2 5% (i) dθ (ii) 3 sin θ (1 + x 4 ) dx. 6 Use the keyhole contour to perform the following inverse Laplace Transform L 1 1 (s k) s = ekt 1 e ut du k π u(u + k) where k >. See Appendix E for the relevant formula. 25% 4
7 A cylinder of radius 1 is kept at a temperature of 1 It lies inside and parallel to a larger cylinder radius 3 kept at a temperature of. (See the Figure.) A material of thermal conductivity k lies between the cylinders. z-plane 3 3 C O 1 1 2 3 A B 1 Figure 1: Temperature between the cylinders (a) Construct a bilinear map that transforms this problem into one where the boundary consists of two concentric cylinders (circles) of radius 1 and ρ where ρ is to be determined later but is taken to be greater than 1. w-plane ρ C ρ O 1 A B 1 ρ 1 Figure 2: Transformed problem Use the table z w 1 A 2 1 A 2 B 3 ρ B 3 O 1 O 5
You should find that: 8% ω = (2ρ 1)z 3(ρ 1) ( ρ + 2)z + 3(ρ 1) (b) Now fix the value of ρ by requiring that the point C : z = 3 map into the point C : ω = ρ take the value that is greater than one. 4% (c) Solve Laplace s Equation for φ E (u, v) (say) in the ω-plane and write an expression for the solution φ(x, y) in the z-plane. 5% (d) Find the curves of constant temperature in the z-plane between the two cylinders (circles). Hint: the transform can be simplified to 8% ω = ρ ρz 3 3ρ z. 6
Appendix of Results A Sturm-Liouville problems take the form: Au (p(x)u (x)) + q(x)u(x) = f(x), B 1 u(a) α 1 u(a) + α 2 u (a) = B 1 u(b) β 1 u(b) + β 2 u (b) = a < x < b B The Green s Function for a Sturm-Liouville problem is given by: { u1 (x)u 2 (y), x < y p(y)w (y) g(x, y) = u 1 (y)u 2 (x), x > y. p(y)w (y) Here u 1 and u 2 are the solutions to Au = with B 1 u(a) = and B 2 u(b) = repectively. The function W (y) is called the Wronskian and is defined by C Polar Coordinates: W (y) = u 1 (y)u 2(y) u 2 (y)u 1(y). dxdy = rdrdθ φ = φ rr + 1 r φ r + 1 r 2 φ θθ D Theorem 1 (Residue Theorem) Let f(z) be a function that is analytic inside a contour C and on C except at a finite number of singular points a 1,..., a m inside C. Then (taking the integral round C anti-clockwise) E For any t > C f(z)dz = 2πi m j=1 Res f(z). (5) z=a j f(t) L 1 F (s) = 1 a+i F (z)e zt dz. (6) 2πi a i (where all the singularities of F (s) are to the left of x = a). 7