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Unformatted text preview: Partial Differential Equations Math 442 C13/C14 Fall 2009 Homework 1 Solutions 1. Determine which of the following operators are linear: (a) Lu = u xx + u xy (b) Lu = uu x (c) Lu = 4 x 2 u y 4 y 2 u yy Solution: We will see that the first and third are linear and the second is not. For example, we compute: (a) L ( u + v ) = ( u + v ) xx + ( u + v ) xy = u xx + v xx + u xy + v xy = ( u xx + u xy ) + ( v xx + v xy ) = Lu + Lv, and L ( u ) = ( u ) xx + ( u ) xy = u xx + u xy = ( u xx + u xy ) = Lu. (b) L ( u ) = ( u )( u ) x = u ( u x ) = 2 u x and this is not equal to Lu if 2 negationslash = . (c) L ( u + v ) = 4 x 2 ( u + v ) y 4 y 2 ( u + v ) yy = 4 x 2 ( u y + v y ) 4 y 2 ( u yy + v yy ) = 4 x 2 u y 4 y 2 u yy + 4 x 2 v y 4 y 2 v yy = Lu + Lv, and L ( u ) = 4 x 2 ( u ) y 4 y 2 ( u ) yy = 4 x 2 u y 4 y 2 u x = (4 x 2 u y 4 y 2 u yy ) = Lu. 2. (Strauss, 1.1.4) Show that the difference of two solutions to Lu = g is a solution to Lu = 0 , when L is any linear operator. Solution: Assume that Lu = g and Lv = g . Define w = u v , then Lw = L ( u v ) = Lu Lv = g g = 0 , where the second equality comes from L being linear....
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This note was uploaded on 01/15/2010 for the course MATH 346 taught by Professor S during the Spring '09 term at University of Victoria.
 Spring '09
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 Equations

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