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Unformatted text preview: 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 . 119 LECTURE NOTES FOR 18.102, SPRING 2009 Lecture 21. Tuesday, April 28: Dirichlet problem on an interval I want to do a couple of ‘serious’ applications of what we have done so far. There are many to choose from, and I will mention some more, but let me first consider the Diriclet problem on an interval. I will choose the interval [0 , 2 π ] because we looked at it before. So, what we are interested in is the problem of solving d 2 u ( x ) (21.1) − dx 2 + V ( x ) u ( x ) = f ( x ) on (0 , 2 π ) , u (0) = u (2 π ) = 0 where the last part are the Dirichlet boundary conditions. I will assume that (21.2) V : [0 , 2 π ] −→ R is continuous. Now, it certainly makes sense to try to solve the equation (21.1) for say a given f ∈ C ([0 , 2 π ]) , looking for a solution which is twice continuously differentiable on the interval. It may not exist, depending on V but one thing we can shoot for, which has the virtue of being explicit is the following: Proposition 28. If V ≥ as in (21.2) then for each f ∈ C ([0 , 2 π ]) there exists a unique twice continuously differentiable solution to (21.1) . You will see that it is a bit hard to approach this directly – especially if you remember some ODE theory from 18.03. There are in fact various approaches to this but we want to go through L 2 theory – not surprisingly of course. How to start? Well, we do know how to solve (21.1) if V ≡ since we can use (Riemann) integration. Thus, ignoring the boundary conditions we can find a solution to − d 2 v/dx 2 = f on the interval by integrationg twice: x y (21.3) v ( x ) = − f ( t ) dtdy satifies − d 2 v/dx 2 = f on (0 , 2 π ) . Moroever v really is twice continuously differentiable if f is continuous. So, what has this got to do with operators? Well, we can change the order of integration in (21.3) to write v as x x 2 π (21.4) v ( x ) = − f ( t ) dydt = a ( x,t ) f ( t ) dt, a ( x,t ) = ( t − x ) H ( x − t ) t where the Heaviside function H ( y ) is 1 when y ≥ and when y < . Thus a ( x,t ) is actually continuous on [0 , 2 π ] × [0 , 2 π ] since the t − x factor vanishes at the jump in H ( t − x ) . Thus v is given by applying an integral operator to f. Before thinking more seriously about this, recall that there are also boundary conditions. Clearly, v (0) = since we integrated from there. However, there is no particular reason why 2 π (21.5) v (2 π ) = ( t − 2 π ) f ( t ) dt should vanish. However, we can always add to v any linear function and still satify the differential equation. Since we do not want to spoil the vanishing at x = 0 we can only afford to add cx but if we choose c correctly, namely consider 2 π 1 (21.6) c = − 2 π ( t − 2 π ) f ( t ) dt, then ( v + cx )(2 π ) = 0 . 120 LECTURE NOTES FOR 18.102, SPRING 2009...
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 Topology, Continuous function, Metric space, Hilbert space, Compact space

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