Analyse 3 NA, FINAL EXAM. * Monday, January 8, 2018, *

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1 Analyse 3 NA, FINAL EXAM * Monday, January 8, 08, * Motivate each answer with a computation or explanation. The maximum amount of points for this exam is 00. No calculators!. (Holomorphic functions) [ points] Given is a function u : R R with u(x, y) = x + 3x + y. (a) Show that u is harmonic. (b) Find a function v : R R such that the complex function f(x + iy) = u(x, y) + iv(x, y) is holomorphic. Is v unique? (c) The function f from (b) is given as a function of x and y. Write it as a function of z = x + iy. (a) [3 points] xu(x, y) + yu(x, y) = 4 + ( 4) = 0 (b) [7 points] v(x, y) = 4xy + 3y + C, C R (c) [ points] f(z) = z + 3z + + ic, C R. (Laurent Series, Residues, Integrals) [6 points] (a) Consider f(z) = z cos(iz) (z sin(z)) sinh(iz). (i) Determine the order of the pole of f at z = 0 and the principal part of the Laurent series of f around z = 0. (ii) Determine the value of the complex line integral f(z) dz. z = (b) Use residue calculus to compute the definite integral 0 3 cos(θ) + sin(θ) dθ. of 6

2 (a) (i) [6 points] f is an even function with a pole of order at z = 0, so its Laurent series is of the form f(z) = a z + a 0 + a z +..., hence, to compute the principal part one only needs to compute a. One way of doing that is a = Res{zf(z), 0} = lim z 0 z f(z) = lim z 0 z ( 6 z3 +...)(iz +...) = 6i. Hence, the principal part is 6iz. (ii) [4 points] f(z) dz = 0 since f is holomorphic in the punctured unit disc and, z = by (i), Res{f(z), 0} = 0, so the claim follows from the residue theorem (b) [6 points] of 6

3 3. (Ordinary differential equations) [5 points] Consider the ODE initial value problem (IVP) 5 y (x) 6 5 y (x) + 5y(x) = e 3x, y(0) = 0, y (0) = 0. (a) Use the Variation of Parameters formula to determine the solution of the IVP. (b) Use Green s function to determine the solution of the IVP. (c) Determine the general solution for the ODE u (t) 5 t u (t) + 5 t u(t) = 0, t > 0. (a) [0 points] y (x) = e 3x cos(4x), y (x) = e 3x sin(4x), W (y, y )(x) = 4e 6x, y p (x) = 5 6 e3x (any other particular solution with multiples of y or y added is also fine at this stage), general solution y(x) = C y (x) + C y (x) e3x, C, C R, solution to IVP has C = 5 6, C = 0 (b) [0 points] The Green s function is of the form { A(z)e 3x cos(4x) + B(z)e 3x sin(4x), 0 x < z <, G(x, z) = C(z)e 3x cos(4x) + D(z)e 3x sin(4x), 0 z < x <. Fitting the initial conditions G(0, z) = 0, d G(0, z) = 0 gives A(z) = B(z) = 0. Enforcing dx continuity at x = z gives C(z) = D(z) sin(4z), while using the jump condition at x = z cos(4z) gives D(z) = 4 e 3z cos(4z) (for the ODE multiplied by 5), so C(z) = 4 e 3z sin(4z). In summary, G(x, z) = { 0, 0 x < z <, 4 e3(x z) sin(4z) cos(4x) + 4 e3(x z) cos(4z) sin(4x), 0 z < x <. The solution of the initial value problem becomes then y(x) = 0 G(x, z)5e 3z dz = 5 6 e3x 5 6 e3x cos(4x) (c) [5 points] This is an Euler ODE which becomes after the substitution t = e x the homogeneous ODE from the previous questions, so y (x) 6y (x) + 5y(x) = 0. Upon using that y (x) = e 3x cos(4x), y (x) = e 3x sin(4x) and back substitution x = ln(t) we finally get u(t) = C t 3 cos(4 ln(t)) + C t 3 sin(4 ln(t)), C, C R. 3 of 6

4 4. (Fourier Analysis) [37 points] (a) (Fourier Series) Consider the -periodic function f(x) = e x, x [, ), f(x + ) = f(x), x R. (i) Compute its Fourier Series. (ii) Using your result from (i), compute the value of the sum ( ) n + n. (b) (Fourier Transform) Consider the function h(x) = exp( x ), x R. n= (i) Show that the Fourier Transform of h is given by ( ) ĥ(k) = γ, k + and compute γ. (ii) Using your result from (i), compute the value of the integral ( k +) dk. (iii) Consider the ODE y (x) y (x) y(x) = h(x), x R. Use Fourier transformation to show that A ŷ(k) = (k + )(k + ik + ) and compute A. Use inverse Fourier transformation and residue calculus to determine the solution y for x > 0. (a) (i) [5 points] and therefore (ii) [6 points] so and finally We have c n = S f (x) = sinh() e x e inx dx = ( ) n sinh() n Z ( ) + in ( ) n + n ( ) + in, + n exp(inx). By continuity of f at x = 0 we have = f(0) = S f (0) = sinh() ( ) + in ( ) n + n n Z sinh() = + ( ) ( ) n + n n N ( ) ( ) n = + n n N ( ) sinh() 4 of 6

5 (b) (i) [5 points] so γ = (ii) [6 points] ĥ(k) = e x e ikx dx = By Plancherel s theorem we have that ( ) dk = k + ĥ(k) dk = ( ), k + e x dx = (iii) [5 points] Fourier transformation of the ODE gives ( k ik )ŷ(k) = ĥ(k), hence, indeed so A = ŷ(k) = and the solution is given by y(x) = ŷ(k) e ikx dk = (k + )(k + ik + ), (k + )(k + ik + ) eikx dk. It remains to compute the integral via residue calculus. To that end, we associate it to the complex line integral along C R, the usual semi-cirlce in the upper half plane, C R (z + )(z + iz + ) eizx dz. Since the integrand has a double pole at z = i and a simple pole at z = i, we have that { } C R (z + )(z + iz + ) eizx dz = i Res (z + )(z + iz + ) eizx, i ( ) e izx ( = i lim = i i ) z i (z + i)(z + i) 36 e x (6x + 5) = 8 e x (6x + 5). Since x > 0 and the integrand decays sufficiently rapidly, we have that y(x) = ( 8 e x (6x + 5)) = 8 e x (6x + 5). 5 of 6

6 5. (Series solutions for ODEs) [0 points] Consider the ODE z y (z) + zy (z) + (z ν )y(z) = 0 Bessel s equation (of order ν) where ν R is a constant. (a) Explain why z = 0 is a regular singular point. (b) Compute the Wronskian of two solutions without solving the problem. (c) Determine the indicial equation at z = 0 and find its roots. (a) [ points] s(z) =, τ(z) = z ν are analytic functions at z = 0 (b) [4 points] By Abel s theorem W (y, y )(z) = C z, C C. (c) [4 points] σ ν = 0 so σ / = ±ν End of Exam 6 of 6