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### oldex1sol

Course: MATH 425, Spring 2011
School: UPenn
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Word Count: 988

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425 Math Midterm 1 Dr. DeTurck February 8, 2007 1. Solve ux + yuy + u = 0, u(0, y ) = y . In what domain in the plane is your solution valid? As usual, we construct the graph of the solution by propagating the initial data o the line in the xy -plane where the data are given, namely x = 0. To get the part of the solution emanating from the point (0, b), we need to solve the system of ODEs: dx = 1, ds together...

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425 Math Midterm 1 Dr. DeTurck February 8, 2007 1. Solve ux + yuy + u = 0, u(0, y ) = y . In what domain in the plane is your solution valid? As usual, we construct the graph of the solution by propagating the initial data o the line in the xy -plane where the data are given, namely x = 0. To get the part of the solution emanating from the point (0, b), we need to solve the system of ODEs: dx = 1, ds together with initial data x(0) = 0, y (0) = b, u(0) = b. dy = y, ds du +u=0 ds There is no coupling in this system, and each of the three ODEs can be solved by separation of variables to yield: x(s; b) = s, y (x; b) = bes , u(s; b) = bes . So we have u as a function of b and s. We use the solutions for x and y to calculate b and s as functions of x and y : y y s = x, b = s = x. e e Now substitute this into the equation for u(s; b) to get u(x, y ) = y x e = ye2x . ex Since any point (x, y ) can be expressed in terms of (b, s) via x = s, y = bes , our solution is valid for all x and y . 2. Let u(x, t) be the temperature in a rod of length L that satises the partial dierential equation: ut = kuxx for (x, t) (0, L) (0, ), where k is a positive constant, together with the initial condition u(x, 0) = (x) for x [0, L], where satises (0) = (L) = 0 and (x) > 0 for x (0, L). (a) If u also satises the Neumann boundary conditions ux (0, t) = 0, ux (L, t) = 0, show that the average temperature in the rod at time t, which is given by A(t) = is a constant (independent of t). 1 L L u(x, t) dx 0 (b) On the other hand, if u satises the Dirichlet boundary conditions u(0, t) = 0, u(L, t) = 0, show that it must be the case the u(x, t) 0 for all (x, t) satisfying 0 < x < L and t > 0. (c) Still under the assumption that u satises the Dirichlet boundary conditions, show that A(t) is a non-increasing function of t. (Hint for (a) and (c): Use an argument similar to an energy argument). (a) To show that A(t) is a constant, we calculate its derivative with respect to t, and use that u satises the PDE ut = kuxx : 1 dA = dt L L ut (x, t) dx = 0 k L T uxx (x, t) dx = 0 k (ux (L, t) ux (0, t)). L But then the Neumann boundary conditions imply that both terms in the last dierence are zero. So A (t) = 0, hence A is constant. (b) Because u(0, t) = u(L, t) = 0, and u(x, 0) = (x) > 0 for x (0, L), its clear that the minimum value of u on the standard U-shaped set in the maximum principle is zero. By the minimum (or principle the maximum principle applied to u), the minimum of u for all (x, t) with t > 0 and x (0, L) must be bigger than or equal to zero. (c) From part (a), we already know that dA k = (ux (L, t) ux , (0, t)). dt L And from part (b) we know that u(x, t) 0 for x (0, L) (in fact, if we use the strong version of the maximum principle, we know that u(x, t) > 0 for x (0, L)). So ux (L, t) cannot be positive (or else u would be increasing to zero as x L from the left, contradicting u 0) and ux (0, t) cannot be negative (for a similar reason). Therefore dA/dt 0, and so A is non-increasing. 3. (a) Solve the wave equation with friction: uxx = utt + 2ut for 0 < x < and t > 0 with the initial conditions u(x, 0) = sin x, ut (x, 0) = 0, and the boundary conditions u(0) = u( ) = 0. (Hint: Look for separated solutions) (b) If E (t) = show that t 1 2 u2 + u2 dx, t x 0 lim E (t) = 0. (Hint: To do this, you can calculate E (t) explicitly). (a) As usual, write u(x, t) = X (x)T (t). Substituting into the equation and dividing by X (x)T (t) gives us T (t) + 2T (t) X (x) = = X (x) T (t) where is a constant. Well short-circuit the process (see the solutions to the other set of practice problems) and notice that for = 1, we have X = X , for which X (x) = sin x is a solution having X (0) = X ( ) = 0. Now we turn our attention to T + 2T = T , which has solutions T = c1 et + c2 tet . Our separated solution is thus u(x, t) = sin x(c1 et + c2 tet ) and we have to choose c1 and c2 so that u(x, 0) = sin x and ut (x, 0) = 0. The rst condition certainly gives c1 = 1. We calculate: ut (x, t) = sin x(et + c2 et c2 tet ) which, when t = 0, gives c2 = 1 as well. So the solution of the whole problem is u(x, t) = sin x(1 + t)et . (b) Since ux = cos x(1 + t)et and ut = sin x(tet ), its clear that u2 + u2 has a factor of x t e2t . So, even without calculating the integral exactly, its clear that E (t) = e2t times a quadratic polynomial in t. So the limit of E (t) as t is zero by lHospitals rule. 4. Find as general a solution u(x, y, z ) as you can to the third-order equation uxyz = 0 Integrate rst with respect to x to get uyz is a constant plus a function depending only on y and z (the latter is the constant of integration with respect to x) and continue in this way. Eventually end up with: u(x, y, z ) = F (x, y ) + G(x, z ) + H (y, z ) for three arbitrary functions F , G and H of two variables each.
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UPenn - MATH - 425
Math 425/525, Spring 2011Jerry L. KazdanPeriodic Solutions of ODEsIn class we discussed some aspects of periodic solutions of ordinary differential equations. Fromthe questions I received, my presentation was not so clear. Here Ill give a detailed for
UPenn - MATH - 425
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UPenn - MATH - 425
Math 425Hints and Solutions to Practice Midterm 1Dr. DeTurckFebruary 20101. Suppose f is a function of one variable that has a continuous second derivative. Show thatfor any constants a and b, the functionu(x, y ) = f (ax + by )is a solution of the
UPenn - MATH - 425
Math 425, Spring 2011Jerry L. KazdanThe Wave Equation in R2 and R3To Solveutt = c2u,x Rnwithu(x, 0) = (x),ut (x, 0) = (x)In R3 [Poisson: Kirchoffs formula]1u(x0, t0) =4c2t01(x) dS +t0 4c2t0SZZZZ(x) dS ,Swhere S is the sphere centered
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Exam 1Math 508October 12, 2006Jerry L. Kazdan12:00 1:20Directions This exam has three parts, Part A has 4 problems asking for Examples (20 points, 5points each), Part B asks you to describe some sets (20 points), Part C has 4 traditional problems(6
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Exam 1Math 508October 12, 2006Jerry L. Kazdan12:00 1:20Directions This exam has three parts, Part A has 4 problems asking for Examples (20 points, 5points each), Part B asks you to describe some sets (20 points), Part C has 4 traditional problems(6
UPenn - MATH - 508
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UPenn - MATH - 508
Exam 2Math 508December 8, 2006Jerry L. Kazdan12:00 1:20Directions This exam has two parts, Part A has 3 shorter problems (8 points each, so 24 points),Part B has 5 traditional problems (15 points each, so 75 points).Closed book, no calculators but
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UPenn - MATH - 508
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UPenn - MATH - 508
Exam 1Math 508October 14, 2010Jerry L. Kazdan9:00 10:20Directions This exam has three parts, Part A asks for 4 examples (20 points, 5 points each).Part B has 4 shorter problems (36 points, 9 points each. Part C has 3 traditional problems (45points,
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Exam 2Math 508December 9, 2010Jerry L. Kazdan9:00 10:20Directions This exam has three parts, Part A asks for 3 examples (5 points each, so 15 points).Part B has 4 shorter problems (8 points each so 32 points). Part C has 4 traditional problems (15p
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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UPenn - MATH - 508
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