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Unformatted text preview: CAAM 336 · DIFFERENTIAL EQUATIONS Problem Set 5 · Solutions Posted Thursday 23 September 2010. Due Wednesday 29 September 2010, 5pm. All of the problems on this set use the inner product ( u,v ) = Z 1 u ( x ) v ( x ) dx. 1. [30 points: 6 points each for (a) and (b); 9 points each for (c) and (d)] Consider the linear operator L b : C 2 b [0 , 1] → C [0 , 1] defined by L b u = d 2 u dx 2 , where C 2 b [0 , 1] = n u ∈ C 2 [0 , 1] : du dx (0) = u (1) = 0 o . (a) Is L b symmetric? (b) What is the null space of L b ? That is, find all u ∈ C 2 b [0 , 1] such that L b u ( x ) = 0 for all x ∈ [0 , 1]. (c) Show that ( L b u,u ) ≥ 0 for all nonzero u ∈ C 2 b [0 , 1] and explain why this implies that λ ≥ 0 for all eigenvalues λ . (d) Find the eigenvalues and eigenfunctions of L b . Solution. (a) Yes, L b is symmetric . To prove this, we shall show that for any u,v ∈ C 2 b [0 , 1], we have ( L b u,v ) = ( u,L b v ). He we shall use primes to denote derivation with respect to x . Integrating by parts twice, we have ( L b u,v ) = Z 1 u 00 ( x ) v ( x ) dx = [ u ( x ) v ( x )] 1 + Z 1 u ( x ) v ( x ) dx = [ u ( x ) v ( x )] 1 + [ u ( x ) v ( x )] 1 Z 1 u ( x ) v 00 ( x ) dx. Since u,v ∈ C 2 b [0 , 1] we have u (0) = 0 and v (1) = 0, and hence the first term in square brackets must be zero. Again using u,v ∈ C 2 b [0 , 1] we have v (0) = 0 and u (1) = 0, and hence the second term in square brackets is also zero. It follows that ( L b u,v ) = Z 1 u ( x )( v 00 ( x )) dx = ( u,L b v ) . (b) The general solution to the differential equation d 2 u dx 2 = 0 has the form u ( x ) = A + Bx for constants A and B that are determined by the boundary conditions. In order for u to be in C 2 b [0 , 1], we must have u (0) = 0; since u ( x ) = B , we have B = 0. Now u ∈ C 2 b [0 , 1] also requires u (1) = 0, and since u (1) = A , we conclude that A = 0 too, meaning that u ( x ) = A + Bx = 0 for all x ∈ [0 , 1]. Thus, the only element of the null space is the zero function, that is, the null space is trivial, N ( L b ) = { } . (c) [ GRADERS : please count ( L b u,u ) ≥ 0 for 5 points, and count λ ≥ 0 for 4 points.] We wish to show that ( L b u,u ) ≥ 0 for all ∈ C 2 b [0 , 1]. Using the first integration by parts from part (a), we have ( L b u,u ) = [ u ( x ) u ( x )] 1 + Z 1 u ( x ) u ( x ) dx = Z 1 ( u ( x )) 2 dx. Thus, ( L b u,u ) is the integral of a nonnegative function, so it is nonnegative. This statement implies that all eigenvalues are nonnegative, since λ ( u,u ) = ( λu,u ) = ( L b u,u ) ≥ , and we know that ( u,u ) > 0 for all nonzero u due to the positivity of the inner product. (Though not asked for in the problem statement, one can show that u ( x ) 6 = 0 for some x ∈ [0 , 1]: otherwise u would be a constant, and the only constant that satisfies the boundary conditions is u ( x ) = 0, which is prohibited from consideration in the problem statement. Thus, we can actually say ( L b u,u ) > 0 for all nonzero u .) (d) Suppose that...
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This note was uploaded on 01/21/2012 for the course CAAM 330 taught by Professor Tompson during the Fall '09 term at UVA.
 Fall '09
 Tompson

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