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Unformatted text preview: CAAM 336 Â· DIFFERENTIAL EQUATIONS Problem Set 3 Â· Solutions Posted Wednesday 8 September 2010. Corrected 14 September. Due Wednesday 15 September 2010, 5pm. 1. [20 points: 5 points per part] Determine whether each of the following functions ( Â· , Â· ) determines an inner product on the vector space V . If not, show all the properties of the inner product that are violated. (a) V = C 1 [0 , 1], ( u,v ) = Z 1 u ( x ) v ( x ) dx (b) V = C [0 , 1]: ( u,v ) = Z 1  u ( x )  v ( x )  dx (c) V = C [0 , 1]: ( u,v ) = Z 1 u ( x ) v ( x ) e x dx (d) V = C 1 [0 , 1]: ( u,v ) = Z 1 u ( x ) v ( x ) dx Solution. For each example, we shall check all three properties required for an inner product. (a) This function is not an inner product : there exist nonzero u for which ( u,u ) = 0. The function ( u,v ) is symmetric, as ( u,v ) = Z 1 u ( x ) v ( x ) dx = Z 1 v ( x ) u ( x ) dx = ( v,u ) . Similarly, the function is also linear: ( Î±u + Î²v,w ) = Z 1 ( Î±u + Î²v ) ( x ) w ( x ) dx = Z 1 ( Î±u ( x ) + Î²v ( x ) ) w ( x ) dx = Î± Z 1 u ( x ) w ( x ) dx + Î² Z 1 v ( x ) w ( x ) dx = Î± ( u,w ) + Î² ( v,w ) . Finally, we note that ( u,u ) = Z 1 u ( x ) u ( x ) dx = Z 1 ( u ( x )) 2 dx â‰¥ . However, it is possible that ( u,u ) = 0 even when u 6 = 0. For example, if u ( x ) = 1, then u âˆˆ C 1 [0 , 1], but u ( x ) = 0 for all x , and hence ( u,u ) = 0. (b) This function is not an inner product : it is not linear. The function ( u,v ) is symmetric, as ( u,v ) = Z 1  u ( x )  v ( x )  dx = Z 1  v ( x )  u ( x )  dx = ( v,u ) . However, it is not linear: ( Î±u,w ) = Z 1  Î±u ( x )  w ( x )  dx = Z 1  Î±  u ( x )  w ( x )  dx =  Î±  ( u,w ) . If u,w 6 = 0 and Î± < 0, then we have ( Î±u,w ) 6 = Î± ( u,w ). The function ( u,u ) is nonnegative and positive when u 6 = 0: ( u,u ) = Z 1  u ( x )  2 dx is the integral of a nonnegative function, and hence is nonnegative; ( u,u ) = 0 if and only if u ( x ) = 0 for all x , i.e., u = 0. (c) This function is an inner product. The function ( u,v ) is symmetric, as ( u,v ) = Z 1 u ( x ) v ( x ) e x dx = Z 1 v ( x ) u ( x ) e x dx = ( v,u ) . Similarly, the function is also linear: ( Î±u + Î²v,w ) = Z 1 ( Î±u ( x ) + Î²v ( x )) w ( x ) e x dx = Î± Z 1 u ( x ) w ( x ) e x dx + Î² Z 1 v ( x ) w ( x ) e x dx = Î± ( u,w ) + Î² ( v,w ) . Lastly, we check ( u,u ) = Z 1 u ( x ) 2 e x dx. The function e x is positive valued on [0 , 1], so we have that ( u,u ) is the integrand of a nonnegative function, and hence is also nonnegative. If ( u,u ) = 0, then u ( x ) 2 e x = 0 for all x âˆˆ [0 , 1], which means that u ( x ) = 0 for all such x , i.e., u = 0. (d) This function is not an inner product : it is not symmetric and ( u,u ) need not be positive....
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 Fall '09
 Tompson
 dx, inner product, Dirichlet boundary condition, Nvec

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