Assignment 7 solutions

# 2 sinkx 1 eky eky 2 1 k 2 e u uk x y ek u

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Unformatted text preview: on the circle 1 3.4, Problem 17. Verify that uk (x, y ) = sin(�kx) sinh(�ky) solves Laplace’s equation for k = 1, 2, . . .. Its sinh(�k) boundary values on the unit square are u0 = sin(�kx) along y = 1 and ub = 0 on the three lower edges. z −z Recall that sinh (x) = e −e . 2 sin(�kx) 1 (e�ky − e−�ky ) 2 1 �k 2 (e u = uk (x, y ) = − e−�k ) . �u e�ky − e−�ky = � k cos(�kx) · �k �x e − e−�k 2 �u e�ky − e−�ky = −� 2 k 2 sin(�kx) · �k � x2 e − e−�k So �u � ke�ky + � ke−�ky = sin(�kx) · �y e�k − e−�k 2 2 2 �ky �u � k (e − e−�ky ) = sin(�kx) · � y2 e�k − e−�k �2u �2u + 2 = 0. � x2 �y 3.6, Problem 6. Solve Poisson’s equation −uxx − uyy = 1 with u = 0 on the standard un...
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## This note was uploaded on 04/12/2011 for the course MATH 18.085 taught by Professor Staff during the Fall '10 term at MIT.

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