sol3 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 3 ...

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Unformatted text preview: CAAM 336 · DIFFERENTIAL EQUATIONS Problem Set 3 · 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 ( · , · ) 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: ( αu + βv,w ) = Z 1 ( αu + βv ) ( x ) w ( x ) dx = Z 1 ( αu ( x ) + βv ( x ) ) w ( x ) dx = α Z 1 u ( x ) w ( x ) dx + β Z 1 v ( x ) w ( x ) dx = α ( u,w ) + β ( v,w ) . Finally, we note that ( u,u ) = Z 1 u ( x ) u ( x ) dx = Z 1 ( u ( x )) 2 dx ≥ . However, it is possible that ( u,u ) = 0 even when u 6 = 0. For example, if u ( x ) = 1, then u ∈ 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: ( αu,w ) = Z 1 | αu ( x ) || w ( x ) | dx = Z 1 | α || u ( x ) || w ( x ) | dx = | α | ( u,w ) . If u,w 6 = 0 and α < 0, then we have ( αu,w ) 6 = α ( 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: ( αu + βv,w ) = Z 1 ( αu ( x ) + βv ( x )) w ( x ) e- x dx = α Z 1 u ( x ) w ( x ) e- x dx + β Z 1 v ( x ) w ( x ) e- x dx = α ( u,w ) + β ( 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 ∈ [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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sol3 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 3 ...

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