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Integral equations for biharmonic data completion

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  • A boundary integral based method for the stable reconstruction of missing boundary data is presented for the biharmonic equation. The solution (displacement) is known throughout the boundary of an annular domain whilst the normal derivative and bending moment are specified only on the outer boundary curve. A recent iterative method is applied for the data completion solving mixed problems throughout the iterations. The solution to each mixed problem is represented as a biharmonic single-layer potential. Matching against the given boundary data, a system of boundary integrals is obtained to be solved for densities over the boundary. This system is discretised using the Nyström method. A direct approach is also given representing the solution of the ill-posed problem as a biharmonic single-layer potential and applying the similar techniques as for the mixed problems. Tikhonov regularization is employed for the solution of the corresponding discretised system. Numerical results are presented for several annular domains showing the efficiency of both data completion approaches.

    Mathematics Subject Classification: Primary: 35R25, 31B30, 65R20; Secondary: 74K20.

    Citation:

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  • Figure 1.  Errors as a function of the number of iteration steps when reconstructing $ Nu $ and $ Mu $ on the inner boundary $ \Gamma_1 $ in Ex. 2

    Table 1.  The relative $ L_2 $ error for $ Mu $ on $ \Gamma_2 $ with $ u $ solving $ \rm(4) $$ \rm(5) $, and for $ Mv $ on $ \Gamma_1 $ with $ v $ solving $ \rm(11) $$ \rm(12) $, for the source point $ z_1 = (3, \, 0) $ and the setup of Ex 1

    $ m $ $ e_{2M}(\Gamma_2) $ $ \tilde{e}_{2M}(\Gamma_1) $
    8 $ 0.012728889 $ $ 0.0080868877 $
    16 $ 0.000480837 $ $ 0.0000485082 $
    32 $ 0.000000700 $ $ 0.0000000002 $
    64 $ 0.0 $ $ 0.0 $
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    Table 2.  The errors in the second example calculated on $ \Gamma_1 $ for the "direct" single-layer approach with exact ($ \delta = 0\% $) and noisy data ($ \delta = 3\% $)

    $ \delta = 0\% $ $ \delta = 3\% $
    $ m $ $ \alpha $ $ e_{2M} $ $ e_{2N} $ $ \alpha $ $ e_{2M} $ $ e_{2N} $
    8 1E$ - $05 4.0752E$ - $02 4.7541E$ - $03 1E$ - $02 $ 0.23812 $ 0.02612
    16 1E$ - $07 8.1073E$ - $03 2.5088E$ - $04 1E$ - $02 $ 0.23534 $ 0.02959
    32 1E$ - $10 3.4936E$ - $05 9.9830E$ - $07 1E$ - $03 $ 0.23824 $ 0.03871
    64 1E$ - $10 3.4553E$ - $05 9.8739E$ - $07 1E$ - $02 $ 0.23239 $ 0.02678
     | Show Table
    DownLoad: CSV
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