1homeworksol - First Homework, due Monday September 14,...

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First Homework, due Monday September 14, 2009 Please solve problems 1, 4, 5, 8 on page 85-87 of L.C.Evans’ book. Solution to problem 1 on page 85 of L.C.Evans: Set u ( x,t ) = e - ct v ( x,t ) yields the equation - ce - ct v ( x,t ) + e - ct v t ( x,t ) + e - ct b · Dv ( x,t ) + ce - ct v ( x,t ) = 0 with the initial condition v ( x, 0) = g ( x ). Thus v t + b · Dv = 0 which yields the solution v ( x,t ) = g ( x - bt ) , and thus u ( x,t ) = e - ct g ( x - bt ) . Solution to problem 4 on page 86 of L.C.Evans: a) As for the harmonic case consider the expression φ x ( r ) = 1 | ∂B r ( x ) | Z ∂B r ( x ) v ( y ) dS ( y ) and differentiate with respect to r . You get φ 0 x ( r ) = 1 | ∂B r ( x ) | Z ∂B r ( x ) ∂v ∂n ( y ) dS ( y ) = 1 | ∂B r ( x ) | Z B r ( x ) Δ v ( y ) dS ( y ) , by Gauss’ theorem. Since Δ v 0 this shows that φ 0 x ( r ) 0 and hence v ( x ) = lim r 0 φ x ( r ) φ x ( r ) = 1 | ∂B r ( x ) | Z ∂B r ( x ) v ( y ) dS ( y ) . Further since Z B r ( x ) v ( y ) dy = Z r 0 Z ∂B r ( x ) v ( y ) dS ( y ) dr v ( x ) Z r 0 | ∂B r ( x ) | dr = v ( x ) | B r ( x ) | the claim follows. b) By restricting ourselves to a connected component of
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1homeworksol - First Homework, due Monday September 14,...

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