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Course: M 412, Fall 2009
School: Texas A&M
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Word Count: 346

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Assignment M412 7, due Friday November 4 1. [15 pts] (Mean Value Property in three space dimensions.) Suppose is an open subset of R3 and u C 2 () solves the Laplace equation in . Show that if (x0 , y0 , z0 ) , and Sr (x0 , y0 , z0 ) is a sphere centered at (x0 , y0 , z0 ) with radius r, contained entirely in , then u(x0 , y0 , z0 ) = 1 4r2 u(x, y, z)dS. Sr (x0 ,y0 ,z0 ) Hints. Laplace's equation in spherical...

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Assignment M412 7, due Friday November 4 1. [15 pts] (Mean Value Property in three space dimensions.) Suppose is an open subset of R3 and u C 2 () solves the Laplace equation in . Show that if (x0 , y0 , z0 ) , and Sr (x0 , y0 , z0 ) is a sphere centered at (x0 , y0 , z0 ) with radius r, contained entirely in , then u(x0 , y0 , z0 ) = 1 4r2 u(x, y, z)dS. Sr (x0 ,y0 ,z0 ) Hints. Laplace's equation in spherical coordinates (r, , ) takes the form sin (r2 ur )r + (sin u ) + 1 u = 0. sin (See Haberman p. 28 for a description of spherical coordinates.) The differential surface increment in spherical coordinates is dS = r2 sin dd. 2. [10 pts] (Maximum/Minimum principle for the Laplace equation in three space dimensions.) Suppose is a bounded, open, connected subset of R3 and u C 2 () C() solves the Laplace equation on . Show that u can only attain its maximum or minimum on the interior of if u is constant on the entirety of . (You may use without proof the following fact: If u is constant on any sphere in u then it is constant throughout the entirety of .) 3. [5 pts] (Uniqueness of solutions to the Laplace equation in three space dimensions.) Suppose is a bounded, open, connected subset of R3 . Show that solutions u C 2 () C() to the Laplace equation u = 0; u = f; are unique. 4. [5 pts] (Stability of solutions to the Laplace equation in three space dimensions.) Suppose is a bounded, open, connected subset of R3 . Show that solutions u C 2 () C() to the Laplace equation u = 0; u = f; are stable with respect to small changes in the boundary data f . 5. [5 pts] Show that a necessary condition for s...

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Texas A&M - M - 412
M412 Assignment 4, due Friday October 71. [10 pts] Haberman Problem 1.4.1, Parts (f) and (g). 2. [10 pts] Haberman Problem 1.4.5. 3. [10 pts] Haberman Problem 1.4.7, Parts (a) and (c). 4. [10 pts] Haberman Problem 1.4.10. 5. [10 pts] Haberman Proble
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M612 Assignment 7, due Friday(-ish) March 131. [10 pts] Suppose E(t) is a fundamental solution for the operator p(Dt , Dx ) and write down (in terms of E) a distribution solution for the PDE p(Dt , Dx )G = 0 in [0, ) CD(G) = (0, 0, . . . , T , 0, 0)
Texas A&M - M - 612
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M412 Assignment 2, due Friday September 91. [10 pts] Use the method of diagonalization to determine a general solution for the ODE system 3 y1 = - y1 + y 2 4 y2 = - 5y1 + 3y2 . 2. [10 pts] Use the method of characteristics to solve the PDE u t + t2
Texas A&M - M - 412
M412 Assignment 3, due Friday September 161. [10 pts] Use the method of characteristics to solve the PDE ux - uy + 2y =0 u(x, y) = xy on the line x + 2y = 1. 2. [10 pts] For the PDE ut + f (u)x = 0 u(0, x) = g(x), use the method of characteristics t
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M442, Assignment 2, due Friday Sept. 141. [5 pts] For a pendulum under the inuence of linearly modeled air resistance, the angle y(t) solves the equation g y (t) = sin y by , l where b is a coecient of air resistance. Taking l = 1, g = 9.81, and
Texas A&M - M - 442
Assignment 5, due Friday Nov. 161. [10 pts] Find a general solution for the system of differential equations dy1 = - 2y1 + y2 dt dy2 = y1 - 2y2 . dt 2. [10 pts] In class I said that imaginary eigenvalues correspond with oscillatory behavior (orbits
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31 .001309 .043208 .004177 .040258 .001487 .041342 .004515 .044896 .010865 .069105 .001759 .053671 .002594 .046613 .004950 .045492 .007115 .053634 .003186 .046923 .002093 .047043 .005202 .042955 .004489 .039754 .003642 .050224 .003960
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boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -0.693147180559945286 eta= 0.100000000000000006 nu = 0.299999999999999989 alpha = 0.490907905324351579 beta= 0.509092094675648421 kappa0= -1 M2 = 18 NJ= 3
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -2.30258509299404590 eta= 0.299999999999999989 nu = 0.299999999999999989 alpha = 0.472782819701179080 beta= 0.527217180298820920 kappa0= -1 M2 = 18 NJ= 30
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 0.693147180559945286 eta= 0.299999999999999989 nu = 0.299999999999999989 alpha = 0.472782819701179080 beta= 0.527217180298820920 kappa0= -1 M2 = 18 NJ= 30
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 2.30258509299404546 eta= 0.100000000000000006 nu = 0.299999999999999989 alpha = 0.490907905324351579 beta= 0.509092094675648421 kappa0= -1 M2 = 18 NJ= 30
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = -2.30258509299404590 eta= 0.000000000000000000E+00 nu = 0.299999999999999989 alpha = 0.500000000000000000 beta= 0.500000000000000000 kappa0= -1 M2 = 18 NJ
Lehigh - MAG - 025
boy = 0.250000000000000000 Number of quadrature points = 21 gamma = 2.30258509299404546 eta= 0.000000000000000000E+00 nu = 0.299999999999999989 alpha = 0.500000000000000000 beta= 0.500000000000000000 kappa0= -1 M2 = 18 NJ=
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Johns Hopkins - MTS - 251
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Johns Hopkins - MTS - 251
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