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# pde6 - 62 BRUCE K DRIVER 6 Elliptic Ordinary Differential...

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62 BRUCE K. DRIVER 6. Elliptic Ordinary Differential Operators Let o R n be a bounded connected open region. A function u C 2 ( ) is said to satisfy Laplace’s equation if 4 u = 0 in . More generally if f C ( ) is given we say u solves the Poisson equation if 4 u = f in . In order to get a unique solution to either of these equations it is necessary to impose “boundary" conditions on u. Example 6.1. For Dirichlet boundary conditions we impose u = g on and for Neumann boundary conditions we impose ∂u ∂ν = g on , where g : R is a given function. Lemma 6.2. Suppose f : C 0 −→ R , ∂ is C 2 and g : R is continuous. Then if there exists a solution to 4 u = f with u = g on such that u C 2 ( ) C 1 ( ) then the solution is unique. De fi nition 6.3. Given an open set R n we say u C 1 ( ) if u C 1 ( ) C ( ) and u extends to a continuous function on . Proof. If e u is another solution then v = e u u solves 4 v = 0 , v = 0 on . By the divergence theorem, 0 = Z 4 v · vdm = Z | v | 2 dm + Z v v · ndσ = Z | v | 2 dm, where the boundary terms are zero since v = 0 on . This identity implies R | u | 2 dx = 0 which then shows v 0 and since is connected we learn v is constant on . Because v is zero on we conclude v 0 , that is u = ˜ u. For the rest of this section we will now restrict to n = 1 . However we will allow for more general operators than in this case. 6.1. Symmetric Elliptic ODE. Let a C 1 ([0 , 1] , (0 , )) and (6.1) Lf = ( af 0 ) 0 = af 00 a 0 f 0 for f C 2 ([0 , 1]) . In the following theorem we will impose Dirichlet boundary conditions on L by restricting the domain of L to D ( L ) := { f C 2 ([0 , 1] , R ) : f (0) = f (1) = 0 } . Theorem 6.4. The linear operator L : D ( L ) C ([0 , 1] , R ) is invertible and L 1 : C ([0 , 1] , R ) D ( L ) C 2 ([0 , 1] , R ) is a bounded operator. Proof. (1) (Uniqueness) If f, g D ( L ) then by integration by parts (6.2) ( Lf, g ) := Z 1 0 ( Lf )( x ) g ( x ) dx = Z 1 0 a ( x ) f 0 ( x ) g 0 ( x ) dx. Therefore if Lf = 0 then 0 = ( Lf, f ) = Z 1 0 a ( x ) f 0 ( x ) 2 dx

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PDE LECTURE NOTES, MATH 237A-B 63 and hence f 0 0 and since f (0) = 0 , f 0 . This shows L is injective. (2) (Existence) Given g C ([0 , 1] , R ) we are looking for f D ( L ) such that Lf = g, i.e. ( af 0 ) 0 = g. Integrating this equation implies a ( x ) f 0 ( x ) = C + Z x 0 g ( y ) dy. Therefore f 0 ( z ) = C a ( z ) Z 1 y z 1 a ( z ) g ( y ) dy which upon integration and using f (0) = 0 gives f ( x ) = Z x 0 C a ( z ) dz Z 1 y z x 1 a ( z ) g ( y ) dz dy. If we let (6.3) α ( x ) := Z x 0 1 a ( z ) dz the last equation may be written as (6.4) f ( x ) = ( x ) Z x 0 ( α ( x ) α ( y )) g ( y ) dy. It is a simple matter to work backwards to show the function f de fi ned in Eq. (6.4) satis fi es Lf = g and f (0) = 0 for any constant C. So it only remains to choose C so that 0 = f (1) = C α (1) Z 1 0 ( α (1) α ( y )) g ( y ) dy.
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pde6 - 62 BRUCE K DRIVER 6 Elliptic Ordinary Differential...

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