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Unformatted text preview: Partial Differential Equations Math 442 C13/C14 Fall 2009 Homework 3 Solutions 1. Here we will prove that solutions to the heat equation satisfy (some of) the invariance principles mentioned in class, or in the book in 2.4. That is, if u ( x,t ) is a solution to u t = ku xx for x R ,t > 0, then so are (a) u ( x y,t ) for any fixed y , (b) u x ,u t , (c) v ( x,t ) = integraltext u ( x y,t ) g ( y ) dy where g has finite support, (d) v ( x,t ) = u ( ax,at ) for any a > 0. Solution: (a) Let v ( x,t ) = u ( x y,t ). Then v t ( x,t ) = u t ( x y,t ) 1 = u t ( x y,t ) , v x ( x,t ) = u x ( x y,t ) 1 = u x ( x y,t ) , 2 v x 2 ( x,t ) = x v x ( x,t ) = x u x ( x y,t ) = 2 u x 2 ( x y,t ) . Then v t ( x,t ) k 2 v x 2 ( x,t ) = u t ( x y,t ) k 2 u x 2 ( x y,t ) = 0 , since u solves the heat equation. (b) We compute for u x , the other is similar. Denoting v = u x gives v t = ( u x ) t = u xt , v xx = ( u x ) xx = u xxx . Then v t kv xx = u xt u xxx = u tx u xxx = ( u t u xx ) x = 0 x = 0 . (c) Since g has compact support, we can exchange derivatives and integration (see e.g. Theorem A.3.2 from Strauss), and thus we have v t = t integraldisplay u ( x y,t ) g ( y ) dy = integraldisplay u t ( x y,t ) g ( y ) dy and 2 v x 2 = 2 x 2 integraldisplay u ( x y,t ) g ( y ) dy = integraldisplay 2 u x 2 ( x y,t ) g ( y ) dy. But then v t kv xx = integraldisplay u t ( x y,t ) g ( y ) dy integraldisplay 2 u x 2 ( x y,t ) g ( y ) dy = integraldisplay parenleftbigg u t ( x y,t ) 2 u x 2 ( x y,t ) parenrightbigg g ( y ) dy = integraldisplay dy = 0 . 1 (d) We have v t ( x,t ) = a u t ( ax,at ) , v x ( x,t ) = a u x ( ax,at ) , 2 v x 2 ( x,t ) = a 2 u x 2 ( ax,at ) ....
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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
 s
 Equations, Variance

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