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MIT18_102s09_lec22

# MIT18_102s09_lec22 - on L 2(0 2 π so we know that it has a...

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MIT OpenCourseWare http://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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132 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 22. Thursday April 30: Dirchlet problem continued I did not finish the proof last time:- Proof. Notice the form of the solution in case V 0 in (21.25). In general, we can choose a constant c such that V + c 0 . Then the equation d 2 w d 2 w (22.1) dx 2 + V w = Tw k ⇐⇒ dx 2 + ( V + c ) w = ( T + c ) w. Thus, if w satisfies this eigen-equation then it also satisfies (22.2) w = ( T + c ) A (Id + A ( V + c ) A ) 1 Aw ⇐⇒ Sw = ( T + c ) 1 w, S = A (Id + A ( V + c ) A ) 1 A. Now, we have shown that S is a compact self-adjoint operator on L 2
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Unformatted text preview: on L 2 (0 , 2 π ) so we know that it has a complete set oF eigenFunctions, e k , with eigenvalues τ k = 0 . ±rom the discussion above we then know that each e k is actually continuous – since it is Aw with w ∈ L 2 (0 , 2 π ) and hence also twice continuously diﬀerentiable. So indeed, these e k satisFy the eigenvalue problem (with Dirichlet boundary conditions) with eigenvalues (22.3) T k = τ k − 1 + c → ∞ as k → ∞ . The solvability part also Follows much the same way....
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MIT18_102s09_lec22 - on L 2(0 2 π so we know that it has a...

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