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sol5 - CAAM 336 DIFFERENTIAL EQUATIONS Problem Set 5...

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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 0 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 0 - u 00 ( x ) v ( x ) dx = - [ u 0 ( x ) v ( x )] 1 0 + Z 1 0 u 0 ( x ) v 0 ( x ) dx = - [ u 0 ( x ) v ( x )] 1 0 + [ u ( x ) v 0 ( x )] 1 0 - Z 1 0 u ( x ) v 00 ( x ) dx. Since u, v C 2 b [0 , 1] we have u 0 (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) = 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 0 u ( x )( - v 00 ( x )) dx = ( u, L b v ) . (b) The general solution to the differential equation - d 2 u dx 2 = 0
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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) = 0; since u 0 ( 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 ) = { 0 } . (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 0 ( x ) u ( x )] 1 0 + Z 1 0 u 0 ( x ) u 0 ( x ) dx = Z 1 0 ( u 0 ( 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 non-negative, since λ ( u, u ) = ( λu, u ) = ( L b u, u ) 0 , 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 0 ( 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 L b u = λu . The general solution to the equivalent differential equation - d 2 u dx 2 ( x ) = λu ( x ) has the form u ( x ) = A sin( λx ) + B cos( λx ) . Now we must find values of the constants A , B , and λ that will satisfy the boundary conditions.
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