Fysikens matematiska metoder week: 13-23 Semester: Spring 2015 (1FA121) Lokal: Bergsbrunnagatan 15, Sal 2 Tid: 08:00-13:00 General remark: Material allowed to be taken with you to the exam: Physics Handbook, Mathematics Handbook. For each problem, you will need to show your work in order to earn full credits. A correct answer without the necessary steps that lead to it will only earn partial credits. Problem 1: (10 pt) Consider the equation of a 1D vibrating string, placed between x = 0 and x = L. The equation governing its motion is 2 2 h(t, x) = c2 h(t, x), t2 x2 where c is a real constant, and h(t, x) is the displacement of the string. 1. What is the dimension (unit) of the constant c and what is its physical meaning? (0.5pt) 2. Use the separation of variables and write down the most general solution, without taking into account any boundary conditions for the moment. (1pt) 3. Now first assume that the string is held fixed at the two ends, as in fig.1. write down the corresponding boundary condition and write down the most general solution. The solutions that you get for this problem are called the characteristic modes of the string. (3 pt) 4. Now suppose that the string is subject to an external force, so the wave equation reads 2 2 h(t, x) = c2 h(t, x) f(x) sin νt, t2 x2 { x x [0, L/2] f(x) = L x x (L/2, L] with the same boundary condition that you had for part 3. Now solve this inhomogeneous problem, you are welcome to write down the most general solution, but one specific solution will earn you the full credit. (4.5 pt) 5. Explain what happens if I let ν approach cnπ/l for some integer n > 0, just in words. (1 pt) 0 h(x) x L Figure 1: For problem 1 part 3. Solution 1: c has the dimension of m/s and it is the wave velocity. 2. The most general solution is the superposition of the following e iλt e λ c x. 1
3. The boundary condition is h(t, 0) = h(t, L) = 0. The solution is h(t, x) = Re n>0 where a n are some arbitrary complex coefficients. 4. Expanding f(x) and h(t, x) Plugging this into the equation A specific solution is given by h(t, x) = sin νt n 0 a n exp ( i nπct ) nπx sin L L f(x) = 4( 1) n L (2n + 1)πx π 2 sin, (2n + 1) 2 L n 0 h(t, x) = h k (t) sin kπx L. k>0 ḧ 2n+1 (t) = c 2 (2n + 1)2 π 2 ḧ 2n (t) = c 2 (2n)2 π 2 h 2n (t). a 2n+1 sin L 2 L 2 h 2n+1 (t) + 4( 1)n L π 2 sin νt (2n + 1) 2 (2n + 1)πx, a 2n+1 = c2 (2n + 1) 2 π 2 L L 2 v 2 4( 1)n L π 2 (2n + 1) 2 v 2. Problem 2: (10 pts) Consider the 1D heat conducting problem. A heat conductor is placed between x = 0 and x = L, the temperature profile is given by the function T (t, x), which satisfies the equation c t 2 T (t, x) = k T (t, x), x2 where c is the heat capacity of the material and k is the heat conductivity. 1. Now suppose the end x = 0 is insulated, but the end at x = L, there is a constant in flow of heat of intensity q. Write down the boundary condition in this case (you will need the Fourier s law of heat conducting). (2 pt) 2. Now we turn to the boundary condition x T (t, x) x=0 = x T (t, x) x=l = 0, which corresponds to the case that both ends x = 0 and x = L are insulated. Solve the equation under this boundary condition, you will need to write down the most general solution. (3 pt) 3. Now suppose further that the temperature profile at t = 0 is given by T (0, x) = cos πx L. 2
With the boundary condition at x = 0 and x = L as the previous part, solve this problem. (5 pt) Solution 1: 2. 3. Setting t = 0 x T (t, x) x=0 = 0, T (t, x) = n=0 x T (t, x) x=l = q k. a n cos nπx L exp ( n2 π 2 k 2 c 2 L 2 t), a n R. T (0, x) = n=0 a n cos nπx L πx = cos L, so a 1 = 1, and all others are zero. So T (t, x) = cos πx L exp ( π2 k 2 c 2 L 2 t). Problem 3: (10 pt) Consider the problem of a vibrating membrane described by the equation 2 t 2 h(t, r) = c2 2 h(t, r), r R 2, r 2 L 2 where c is a real constant, and h(t, r) is the displacement of the membrane. 1. Write down the Laplace operator 2 in polar coordinates: r = (x, y), x = r cos θ, y = r sin θ. Now write h(t, r) = T (t)r(r)s(θ) and apply the method of separation of variable, write down the equations that are satisfied by T (t), R(r) and S(θ) respectively. (2 pt) 2. Write down the most general solution to the above equations you get. (3 pt) 3. Now suppose that the boundary condition is r h(t, r, θ) r=l = 0, and also assume that the solution is θ-independent. Use this to narrow down the previous general solution. You will need the formula d J m (x) xdx x m = J m+1(x) x m+1, and your answer should be written in terms of the zeros x (i) m d dx (xm J m (x)) = x m J m 1 (x) (1) of the Bessel functions J m (x). (5 pt) Solution 1: The equations satisfied by T, R, S are 2 = 2 r 2 + 1 r r + 1 2 r 2 θ 2. S (θ) = m 2 S, R (r) + r 1 R r 2 m 2 R = k 2 R, T = k 2 c 2 T. 3
2. The solutions are S(θ) = e imθ, m Z, T = e ±ikct, R(r) = J m (kr). 3. Setting m = 0, the boundary condition requires So the solution reads J 0(kL) = J 1 (kl) = 0, h(t, r, θ) = Re p=1 a p J 0 ( x(p) 1 L k = x(p) 1 L. (p) ix 1 r)e c L t. Problem 4: (10 pt) Consider the Laplace equation on R 3 2 f(x, y, z) = δ(x a)δ(y b)δ(z c), i.e. with a source placed at (a, b, c) R 3. Given that the solution to this problem is f(x, y, z) = 1 1 4π (x a)2 + (y b) 2 + (z c). 2 1. Let now (a, b, c) = (0, 0, 0), and so you can use spherical coordinate to rewrite f(x, y, z) above. Show that away from the point (0, 0, 0), 2 f = 0. (2 pt) 2. Compute the gradient V = f, and compute the flux (2 pt) ds V, where S is a sphere of radius R centred at (0, 0, 0). 3. Consider now the same Laplace equation but on the upper half space z 0 2 f = δ(x)δ(y)δ(z c), S z f(x, y, z) z=0 = 0, c > 0. This equation describes the motion of some fluid in the upper half space, and there is a source placed at x = y = 0 and z = c. Use the method of mirror charge to solve this boundary value problem (pay attention to the kind of boundary condition, you need to decide where to place the mirror source and what is its value). (4 pt) 4. The velocity field of the fluid is given by V = f. Compute V, and what is the value of your V at z = 0? Is it in accordance with the boundary condition placed at y = 0 in part 3? (2 pt) Solution 1: 2 Gf = 1 r 2 rr 2 r 1 4πr = 0. 4
2. 3. Place the same source at (0, 0, c) f = ˆr 4πr 2, ds f = 1. S f = 1 4π 1 x2 + y 2 + (z c) + 1 2 4π 1 x2 + y 2 + (z + c) 2. 4. f = 1 4π 1 ( ) 1 x, y, z c (x 2 + y 2 + (z c) 2 ) 3/2 4π 1 (x 2 + y 2 + (z + c) 2 ) 3/2 ( x, y, z + c ). At the z = 0 plane f z=0 = 1 1 ( ) 2x, 2y, 0. 4π (x 2 + y 2 + c 2 ) 3/2 Problem 5: (10 pt) Consider the 3D Helmholtz problem inside of the sphere of radius a > 0. ( 2 + k 2 )f( r) = 0, r R 3, r a 1. Use the spherical coordinate and separation of variable by writing f = R lm (r)y lm (θ, φ), where Y lm is the spherical harmonics. Write down the equation satisfied by R lm. (2 pt) 2. Write y(r) = R lm (r) r, derive the equation satisfied by y(r). And solve this equation. (4 pts) 3. Now consider the boundary condition f r =a = cos θ, write down the solution to this boundary value problem. You will need to look up the list of spherical harmonics to find out what values of l, m for Y lm to match the boundary condition. And you also need 2 2 J 1/2 (x) = πx sin x, J 1/2(x) = πx cos x plus the formula Eq.1. A specific solution is enough to earn you the full credit. (4 pt) Solution 1: Writing R lm as R R + 2 r R + k 2 R 1 l(l + 1)R = 0. r2 5
2. Writing R(r) = r 1/2 y(r) y + r 1 y + k 2 y 1 r 2 (l + 1/2)2 y = 0. This is solved by y(r) = J l+1/2 (kr), l h(r, θ) = l=0 m= l a lm 1 r J l+1/2 (kr)y lm (θ, φ). The solution J l 1/2 (kr) is excluded since J l 1/2 (0) is divergent. 3. Matching the boundary condition, since cos θ = 2(π/3) 1/2 Y 1,0, which shows that for a specific solution we can pick only a 10 0. So 1 π a 10 a J 3/2 (ka)y 10 (θ, φ) = 2 3 Y 1,0. Hence and J 3/2 (x) is obtained using Eq.1 a 1 h(r, θ) = J 1+1/2 (kr) cos θ, J 3/2 (ka) r J 3/2 (x) = 2 ( x 1/2 cos x + x 3/2 sin x ). π Necessary formulae The general solution to y (x) + p(x)y + q(x)y = f(x) is y = y 2 y1 f + y 1 y2 f, = y 1y 2 y 2y 1, and y 1,2 are the solution to y (x) + p(x)y + q(x)y = 0. For a function f defined on [0, L] with f(0) = f(l) = 0, it can be expanded f(x) = n>0 f n sin nπx L, f n = 2 L L 0 f(x) sin nπx L dx. If f (0) = f (L) = 0, then f(x) = n 0 f n cos nπx L, f n = 2 L L 0 f(x) cos nπx L dx, n > 0, f 1 = 1 L L 0 f(x)dx. 6