- June 2, 2020

Please read the instructions below before you start the exam. April/May 2020 MA3PD2 2019/0 A 800 UNIVERSITY OF READING PARTIAL DIFFERENTIAL EQUATIONS II (MA3PD2) Two hours Answer ALL questions in section A and at least ONE question from section B. (If more than one question from section B is attempted then marks from the BETTER section B question will be used. If the exam mark calculated in this way is less than 40%, then marks from the other section B question which has been attempted will be added to the exam mark until 40% is reached). vc2 Page 2 SECTION A 1. Let (r, ✓) denote polar coordinates, and suppose that u(r, ✓) satisfies the Neumann Laplace problem⇢ urr + r 1ur + r 2u✓✓ = 0 in D = {0 < r < 1, 0 ✓ < 2⇡}, @u/@r = F (✓) on C = {r = 1, 0 ✓ < 2⇡}, in which the boundary forcing function F (✓) satisfies R 2⇡ 0 F (✓) d✓ = 0. The solution u must be bounded in D and 2⇡-periodic in ✓. (a) Use separation of variables to show that the most general solution u(r, ✓) which is bounded and satisfies Laplace’s equation in D, and is 2⇡-periodic in ✓, can be written as u(r, ✓) = 1 2 a0 + 1X n=1 rn[an cos(n✓) + bn sin(n✓)] for constants an, bn. [27 marks] (b) Now apply the boundary condition on C, determine the constants an, and hence show that the solution can be written as u(r, ✓) = 1 2 a0 + 1 ⇡ Z 2⇡ 0 F (✓0) 1X n=1 rn n cos[n(✓ ✓0)] d✓0. (1) Explain why the constant a0 is indeterminate. [10 marks] (c) Use the identity P1 n=1 z n/n = ln(1 z) (which is valid for |z| < 1) to explicitly evaluate the summation in (1), and hence show that u(r, ✓) = 1 2 a0 1 2⇡ Z 2⇡ 0 F (✓0) ln[1 2r cos(✓ ✓0) + r2] d✓0, for r < 1. [10 marks] (d) Write down the solution u(r, ✓) in the case F (✓) = cos(4✓) + sin ✓. [13 marks] MA3PD2 2019/0 A 800 vc2 Page 3 SECTION B 2. (a) Let D = {0 < x < 1, 0 < y < ⇡}, and consider the Dirichlet Green’s function problem⇢ r2GD = (x x0) (y y0) in D, with (x0, y0) 2 D, GD = 0 on C, the boundary of D, By expanding GD as a Fourier sine series in y, derive the representation GD(x, y|x0, y0) = 1X n=1 sin(ny) sin(ny0) n⇡ sinhn (cosh[n(1 x x0)] cosh[n(1 |x x0|)]) for distinct points (x, y), (x0, y0) in D. [27 marks] [The identity cosh(A+B) cosh(A B) = 2 sinhA sinhB may prove useful. You may also use the symmetry property GD(x, y|x0, y0) = GD(x0, y0|x, y) without proof, if required..] (b) Now consider the general Neumann Helmholtz Green’s function problem⇢ r2GN + cGN = (x x0) (y y0) in D, with (x0, y0) 2 D, @GN/@n = 0 on C, (2) in which C is the boundary of D, D is a bounded domain, and c < 0. Show that the solution to (2) satisfies the symmetry property GN(x, y|x0, y0) = GN(x0, y0|x, y) for distinct (x, y), (x0, y0) in D. [13 marks] [Green’s first identity may be assumed in the formRR D r2 +r ·r dxdy = RC @ @n dc.] MA3PD2 2019/0 A 800 vc2 Turn over Page 4 3. Let D = {x > 0, 0 < y < ⇡}, and C be its boundary. Suppose that u = u(x, y) satisfies the Neumann Helmholtz problem8>><>>: r2u+ k2u = 0 in D, @u/@n = F on C1 = {0 < x < 1, y = 0}, @u/@n = 0 on remainder of boundary C2 = C\C1, u outgoing and bounded as x!1 in D. It is given that the corresponding Green’s function GN(x, y|x0, y0) satisfies8r2GN + k2GN = (x x0) (y y0) in D, with (x0, y0) 2 D, @GN/@n = 0 on C, GN outgoing and bounded as x!1 in D, and can be written as GN(x, y|x0, y0) = 1X n=0 ✏n cos(ny) cos(ny0) 2↵n⇡ n e ↵n|x x0| + e ↵n(x+x0) o , (3) for (x, y) 6= (x0, y0), where ✏0 = 1, ✏n = 2 for n 1, and ↵n = ⇢ p n2 k2 if n > k ipk2 n2 ⌘ i n if n < k. (4) (a) Write DL = {0 < x < L, 0 < y < ⇡}, for L > 1. Use a suitable Green’s identity to show that u(x0, y0) = Z 1 0 F (x)GN(x, 0|x0, y0) dx+ Z ⇡ 0 ✓ u @GN @x GN @u @x ◆ x=L dy (5) for (x0, y0) 2 DL. [Green’s first identity may be assumed in the formRR D r2 +r ·r dxdy = RC @ @n dc.] [12 marks] MA3PD2 2019/0 A 800 vc2 Page 5 (b) Starting from (3), explain why GN(x, y|x0, y0) ⇠ NX n=0 ✏n cos(ny) cos(ny0) i n⇡ ei nx cos( nx0) (6) as x!1, where n is defined in (4). How is the integer N determined? Explain why a suitable expansion of u(x, y) as x!1 is u(x, y) ⇠ NX n=0 ✏nB + n e i nx cos(ny), (7) for some constants B+n . [11 marks] (c) Finally, use the expansions (6) and (7) to show that the second integral on the right-hand side of (5) vanishes in the limit as L!1. [17 marks] [End of Question Paper] MA3PD2 2019/0 A 800 vc2