By Alexander G Ramm
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Extra resources for Mathematical And Analytical Techniques With Applications To Engineering
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As and let Then This implies and we claim This claim we prove later. Thus Let us One has Here the estimate and the relation as were used. To prove the first estimate, one uses the formula: and the polar representation of A* yields, where V is an isometry, One has where the spectral representation for T was used. Let us prove the second relation: where P is the orthoprojector onto N(A), and is the resolution of the identity of the self-adjoint operator B (see [KA]). If then Finally, let us prove the claim used above.
4. If Proof. 5. 1) is a closed, nonlinear, injective map. If K is a compactum, then the inverse operator is continuous on A(K). Proof. Since A is injective, Then closed, and A, one has is well-defined on A(K ). 5 is proved. Claim: Let us assume that and Then is monotone, continuous, D(A) = H, A(u) = f, Proof. 1). 6. 1) is unique and Proof. 3. 1) is unique if A is monotone and continuous, because in this case the set of solutions N is convex and closed. Its closedness is obvious, if A is continuous.
13 case (3) is a non-overdetermined problem, and uniqueness of its solution is an open problem. Recently Ramm ([R198]) proved that the spectral data and all determine q (x) uniquely provided that all the eigenvalues are simple. Here in is the kernel of the resolution of the identity of the selfadjoint Neumann operator in The above inverse problem is not overdetermined, because depends on three variables in and q (x) is also a function of three variables in It is an open problem to find out if this result remains valid without the assumption about the simplicity of all the eigenvalues.