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Unformatted text preview: Name: Math 362, Spring 2006, Professor Callahan Final Exam, Tue., May 9
As was usually the case with scientific problems, what seemed clear and simple in an abstract or a summary became more complicated the more precise an explanation I required. --Stanislaw Lem, "Return From the Stars" Note 1: This test is open note, open homework, open solutions and open book. Feel free to quote from the lecture and books and to take reasonable shortcuts. (Show your work, but don't reinvent the wheel.) Note 2: Show your work. Clarity counts. If I can't follow your reasoning I can't give credit. Problem 1: Given the vector field F = (zex , zx2 , x2 + sin y), calculate the following: Answer: We have F = P, Q, R = (zex , zx2 , x2 + sin y), so div F = Q R P + + = (zex ), + (zx2 ), + (x2 + sin y) = zex . x y z x y z 1 2 grad(div F) = (zex ) (zex ) (zex ) x y z = (zex , 0, ez ). 3 ^ i curl F = x ^ y ^ k z ^ i = ze x x ^ zx y 2 ^ k z P Q R x2 + sin y ^ =^ i(cos y - x2 ) + ^ x - 2x) + k(2xz - 0) (e = (cos y - x2 , ex - 2x, 2xz). ^ i curl(curl F) = x ^ y 2 x ^ k z cos y - x e - 2x 2xz = ^ - 0) +^ - 2z) + ^ x - 2 + sin y) = (0, -2z, ex - 2 + sin y). i(0 (0 k(e 2 F = = 2 2 2 + 2+ 2 x2 y z
2 x2 zex , zx2 , x2 + sin y (zex ),
2 x2 + 2 y 2 + 2 z 2 + 2 y 2 + 2 z 2 (zx2 ), 2 x2 + 2 y 2 + 2 z 2 (x2 + sin y) = (zex , 2z, 2 - sin y). Note that 2 F means 2 acting on each component of F separately. Hint: Your results should satisfy curl(curl F) = grad(div F) - 2 F. (Do not use this hint to calculate any of the above; it's just a check for errors.) 1 Problem 2: Consider a flat square loop of side L parallel to the x-y plane so that 0 x L and 0 y L. The loop is spanned by a film of bubble fluid and is moving in the positive z-direction with velocity V0 . The film is flat while the loop is moving. At time t = 0 the loop reaches the x-y plane and is abruptly stopped. Thus the initial position of the membrane is z t=0 = 0 and the initial vertical velocity is z/t t=0 = V0 . Find the subsequent motion z(x, y, t) of the membrane. (Note: The on p. 110 should read mn , as it depends upon m and n.) Answer: The loop is flat, so the equilibrium solution is zeq = 0. For the homogenized problem, the penultimate equation on page 110 says that for a square of side L, zh (x, y, t) =
m=1 n=1 sin ny mx sin L L Amn cos(mn t) + Bmn sin(mn t) , with mn given by mn = c m2 2 L2 + n2 2 L2 m2 + n2 c = . L Now we satisfy the initial conditions. At t = 0 the position of the membrane is z(x, y, 0) = 0, so 0=
m=1 n=1 sin mx ny sin Amn . L L We could use Fourier series and orthogonality, but we can just see that we must have Amn = 0. The vertical velocity of the membrane is given by z mx ny = sin sin t L L m=1 n=1 At t = 0, and remembering that Amn = 0, we get -mn Amn sin(mn t) + mn Bmn cos(mn t) . V0 =
m=1 n=1 sin ny mx sin mn Bmn . L L y x We can multiply both sides by sin mL sin n L , integrate and use orthogonality, or we can just quote the equations at the bottom of page 69, with mn Bmn instead of Amn : mn Bmn = so that Bmn = Now,
L 4 L2 L 0 L 0 L V0 sin ny mx sin L L
L dx dy, 4V0 L2 mn sin
0 mx L dx
0 sin ny L dy . sin
0 mx L dx = 0, 2L m , m odd . m even The same holds for the y-integral. Thus Bmn = Our solution is therefore z(x, y, t) = 16V0 2 mx ny 1 sin sin sin(mn t). mnmn L L 16V0 , m and n both odd mn 2 mn 0, m or n even. m odd n odd 2 Problem 3: Recall the paraboloidal cordinates from Homework #6, Problem 5.3, p. 16. To solve Laplace's o equation 2 u = 0 you separated variables with u(s, t, ) = S(s)T (t)() to get d2 = -m2 , d2 1 d dS m2 s + - - 2 s ds ds s S = 0, 1 d dT m2 t + - 2 t dt dt t T = 0. The first gives = eim for integer m. Consider the lens-shaped region between the two paraboloids given by s = s0 and t = t0 . The lower surface t = t0 is held at temperature 0. The whole upper surface s = s0 , all the way from t = 0 at the top to t = t0 at the edge, is held at a constant temperature U0 . Using solutions to the above equations that vanish at t = t0 , find the equilibrium temperature everywhere. z s=s0 t=t0 Answer: If > 0 then the T equation is Bessel's equation, and the S equation is the modified Bessel equation. If < 0 then they are the other way around. We want solutions that vanish at t = t0 , so T must oscillate. Thus we must have > 0, so we let = k 2 . Our solutions must be good for t = 0 and for s = 0, so we cannot have any K's or Y 's. Thus our typical solutions have T (t) = Jm (kt), S(s) = Im (ks).
To have T vanish at t = t0 we must have kt0 = zmn , so that k = zmn /t0 . The general solution is an arbitrary combination of our typical solutions, so u(s, t, ) =
m=- n=1 Amn Im zmn s t0 Jm zmn t t0 eim . We have the boundary condition u(s0 , t, ) = U0 , so at s = s0 we get U0 =
m=- n=1 Amn Im zmn s0 t0 Jm zmn t t0 eim . We quote formula (36) from page 102, only with Amn Im Amn Im zmn s0 t0 = 1 t2 0 Jm+1 (zmn )
2 0 zmn s0 t0
2 0 t0 instead of Amn : zmn t t0 e-im t dt d. U0 Jm There is no -dependence to the boundary conditions, so we are not surprised to find that the -integral vanishes for m = 0. For m = 0 the integral gives us 2. We are left with A0n = 2U0 z0n s0 J1 (z0n ) t0
t0 2 t2 I0 0 tJ0
0 z0n t t0 dt. For this integral we quote the result of Problem 28.1 on page 105. For m = 0 this becomes
0 z0n t t0 dt = t2 0 J1 (z0n ). z0n Thus we have A0n = z0n I0 Now the solution is 2U0 . z0n s0 J1 (z0n ) t0 J0 z0n t t0 . J1 (z0n ) z0n s I 2U0 0 t0 u(s, t, ) = z0n s0 z n=1 0n I0 t0 3 ...
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- Spring '06