Question:hard

The model regularization in a damped least square problem is/are used to

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Regularization stabilizes ill-posed inversions and never turns a well-posed problem into an ill-posed one.
Updated On: Jul 21, 2026
  • introduce additional constraints to the ill-posed problems
  • add stability to the inversion
  • convert well-posed problems into ill-posed problems
  • control the trade-off between the misfit and the model variance
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The Correct Option is A, B, D

Solution and Explanation

Think of regularization through the bias-variance lens used in inverse theory. An unregularized least-squares solution to an ill-posed geophysical problem has very low bias but enormous variance - tiny data noise can produce huge, geologically meaningless swings in the recovered model. Damped least squares deliberately trades a small amount of bias for a large reduction in variance.

It works because it supplies an additional piece of information - the constraint that \(\mathbf{m}\) should stay close to some reference model, or stay small/smooth - that the data alone cannot supply. That is exactly the "additional constraint to an ill-posed problem" described in (A).

That same constraint is what keeps \((\mathbf{G}^T\mathbf{G}+\lambda^2\mathbf{I})\) safely invertible even when \(\mathbf{G}^T\mathbf{G}\) is close to singular - i.e. it adds numerical and physical stability, which is (B).

Regularization is never used to make a nice, well-posed problem worse; a genuinely well-posed problem does not need damping at all. So (C), which describes the exact reverse of what regularization does, is wrong.

Varying \(\lambda\) traces out the classic trade-off (L-curve) between the data misfit \(\|\mathbf{d}-\mathbf{G}\mathbf{m}\|\) and the model norm/variance \(\|\mathbf{m}-\mathbf{m}_0\|\); picking \(\lambda\) is exactly choosing where on that trade-off curve the solution sits, which is (D).

So the valid uses of model regularization are (A), (B) and (D) - options 1, 2 and 4.

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