Midterm 2 Solution

Midterm 2 Solution - Math 362, Spring 2006, Professor...

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Unformatted text preview: Math 362, Spring 2006, Professor Callahan Test #2, ThuFri, Apr. 1314. Subtly, very subtly, Nowhere neglect the Use of Intelligence. --Sun Tzu, "The Art of War" Note 1: This test is open note, open homework, open solutions and open book. Feel free to quote from the lecture and take reasonable shortcuts. (Show your work, but you don't have to start from first principles.) Note 2: Show your work. Clarity counts. If I can't follow your reasoning I can't give credit. Problem 1: Find the Fourier cosine series for the function f (x) = x on the interval 0 x . Evaluate it at x = /4 to find the value of 1- 1 1 1 1 1 1 1 - 2 + 2 + 2 - 2 - 2 + 2 + . 2 3 5 7 9 11 13 15 Answer: We use equation (10) from page 26, with L = . First, A0 = Then for n > 0 we have An = = 2 1 f (x) dx = 0 1 x dx = 0 x2 2 = 0 . 2 f (x) cos(nx) dx = 0 0 2 x cos(nx) dx = 0 u=x du = dx dv = cos(nx) dx 1 v = n sin(nx) 2 x sin(nx) n 2 cos(nx) n2 - 2 n sin(nx) dx 0 =0 = = Thus f (x) = x = At x = /4 we get 1 1 1 4 3 5 cos + 2 cos + 2 cos + 2 cos = - 4 2 4 3 4 5 4 7 1 1 1 1 1 1 1 1 4 1 - 2 - 2 + 2 + 2 - = - 2 2 3 2 7 2 9 2 2 5 1 1 4 1 = - 1 - 2 - 2 + 2 + . 2 3 5 7 2 Thus 1 1 2 2 1 . 1 - 2 - 2 + 2 + = 3 5 7 16 7 4 + 0 = 2 cos(n) - 1 n2 -4/n2 , n odd 0, n even. 4 - 2 1 cos(nx). n2 n odd 1 Problem 2: A uniform rod of length L (with 0 x L) starts out with initial temperature profile u(x, t) = x2 , for some given constant . The ends are insulating, so that the boundary conditions are u x = 0, x=0 u x = 0. x=L There is no external heating (so Q(x, t) = 0). Find the temperature u(x, t) for all times t > 0. Answer: Step 1 is to find any equilibrium. In this case, ueq (x) = 0 works just fine. (Actually, any constant temperature would be an equilibrium.) For step 2, the general solution to the homogenized problem is found in Section 8, and is given in the box on page 25: uh (x, t) = n=0 An cos nx -n2 2 t/L2 e . L Step 3 is trivial (u = ueq + uh = uh ). For step 4 we evaluate the temperature at t = 0: u(x, 0) = x2 = n=0 An cos nx . L This is just a Fourier cosine series, and the cofficients are given by equation (10) on page 26. First, e A0 = For n > 0 we have An = 2 L 2 L L 0 1 L L 0 x2 dx = x3 3L L = 0 1 2 L . 3 x2 cos nx L dx = L 0 u = x2 du = 2x dx 4 L L n L dv = cos v= x sin 0 nx dx L L nx n sin L = L n x2 sin nx L =0 - u=x nx = L du = dx dv = sin v=- nx dx L L nx n cos L 4 = L 2 L n 2 nx x cos L 2 L 0 - 4 L L n 2 0 L cos nx L dx =0 = Thus 4L 4L cos(n) = 2 2 (-1)n . 2 2 n n 4L2 nx -n2 2 t/L2 L2 e . (-1)n cos + 2 2 3 n L n=1 u(x, t) = 2 Problem 3: We have a cube of side L (with 0 x L, 0 y L and 0 x L) of uniform material. Given the three-dimensional heat equation 2u 2u 2u 1 u + 2 + 2, = t x2 y z use separation of variables to find the general solution uh (x, y, z, t) to the homogenized problem with Dirichlet boundary conditions (so that uh = 0 on all six faces of the cube). You do not need to solve for the arbitrary cofficients, as there are no initial conditions for this problem. Just do "Step 2". e Answer: We separate variables with u(x, y, z, t) = X(x)Y (y)Z(z)T (t) to get dY 2 dZ 2 dT d2 X 1 + XZT 2 + XY T 2 . XY Z = Y ZT dt dx2 dy dz Dividing by XY ZT , we get 1 dT 1 d2 X 1 d2 Y 1 d2 Z = + . + 2 2 T dt X dx Y dy Z dz 2 ++ Everything is now completely separated. For the X equation we want the boundary conditions X(0) = X(L) = 0, so m2 2 mx d2 X , =- 2 , = X X(x) = sin dx2 L L for some positive integer m. Similarly, for Y and Z we get Y (y) = sin ny , L =- n L 2 , Z(z) = sin pz , L =- p L 2 . For the T equation we have dT = ( + + )T dt Thus the general solution is T (t) = e-(m 2 +n2 +p2 ) 2 t/L2 . uh (x, y, z, t) = m=1 n=1 p=1 Amnp sin ny pz -(m2 +n2 +p2 )2 t/L2 mx sin sin e . L L L 3 ...
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