Nonlinear inversion and Gauss–Newton
Regularize a local linearized step and distinguish local derivative information from global identifiability and convergence.
On this page
The question for today
What you’ll understand
- Derive a damped Gauss–Newton step from a linearized residual model.
- Identify a stationary linearization that misses a valid solution.
- State why local smoothness and initialization assumptions do not imply global uniqueness.
Many forward maps depend nonlinearly on the hidden object, so one fixed singular system no longer describes the entire inversion. Gauss–Newton methods solve successive linearized least-squares problems. Regularization stabilizes those steps, but neither linearization nor damping supplies a global uniqueness theorem. We use a continuously Fréchet differentiable map F on an open subset of a real Hilbert space, with a bounded derivative at each point under consideration. Additional local hypotheses are needed whenever a convergence guarantee is invoked.
Linearization creates a local inverse subproblem
At xₖ, approximate F(xₖ+h) by F(xₖ)+Jₖh, where Jₖ is the derivative. The Gauss–Newton step minimizes the squared norm of Jₖh−rₖ with rₖ=yδ−F(xₖ). In finite dimensions with full column rank it has the usual least-squares solution. If Jₖ is nearly rank deficient or is a compact operator with nonclosed range, that subproblem can itself be unstable. A good forward derivative calculation therefore does not eliminate the need for inverse regularization.
Adding αₖ||h||²/2 to the local squared residual gives the Levenberg–Marquardt-type equation (Jₖ*Jₖ+αₖI)h=Jₖ*rₖ. This penalizes the step, not the absolute object. An iteratively regularized Gauss–Newton method that penalizes xₖ+h relative to a fixed reference has a different right-hand side and different convergence analysis. The distinction matters: a small damped step may indicate strong damping or a weak derivative, rather than an accurate reconstructed object.
Derivative invertibility is a local statement
A smooth map can be locally invertible near a point yet have several preimages globally. The scalar map x↦x² has nonzero derivative at every nonzero x, but positive data have two opposite roots on the unrestricted real domain. Restricting the admissible class to positive x removes that branch ambiguity away from zero. Near zero, however, the derivative degenerates and the square-root inverse loses Lipschitz stability. The domain restriction, derivative bound, and neighborhood must accompany a local identifiability statement.
In a finite-dimensional zero-residual problem, a sufficiently smooth map with a full-column-rank derivative at the solution supports familiar local Gauss–Newton convergence results from a sufficiently close starting point. Those hypotheses do not establish convergence from arbitrary initialization or for a noisy inconsistent residual. In infinite-dimensional inverse problems, bounded invertibility can fail even at the truth. Local regularization theory then uses quantitative conditions on the nonlinear remainder, source behavior, and parameter or stopping rules rather than borrowing a finite-matrix theorem unchanged.
Model agreement and noise determine when to stop trusting a step
A trust-region or line-search test compares the actual residual reduction with that predicted by the linear model. Poor agreement can require a shorter step or stronger damping. Such a safeguard controls local model use but does not reveal an unobserved branch or prove that a local minimum is the desired physical solution. Constraints and multiple initializations may help explore possibilities, but their empirical success should not be presented as a general global recovery theorem.
With noisy data, continuing to reduce the nonlinear residual can fit noise just as in linear inversion. Discrepancy stopping needs a justified data norm and uncertainty level, while parameter updates need to respect the nonlinear method's assumptions. An inaccurately implemented adjoint or derivative can also create an apparently stationary step. Checking the derivative against finite differences and verifying a local normal-equation residual are computational checks; establishing identifiability or a physical error bound requires separate mathematical and modeling evidence.
Work it through
A damped step on one branch
For F(x)=x², target y=4, current x₀=1, and step penalty α=1, compute one damped Gauss–Newton update and compare its residual with the initial one.
The derivative is J₀=2 and residual r₀=4−1=3. The local normal equation is (2²+1)h=2·3, so h=6/5.
The new point is x₁=11/5=2.2. Its forward value is 121/25=4.84, so the new residual is −21/25 and its magnitude 0.84, smaller than the initial magnitude three.
The calculation follows the positive branch near the initial point. The exact unrestricted equation also has solution −2, which this local improvement does not rule out. Neither residual reduction nor α>0 establishes global uniqueness.
The regularized linearized step improves the nonlinear fit while retaining a branch-dependent interpretation.
The living notebook
Explore the model
Interactive lab · function
Compare equal data and different local derivatives
Reset, set coefficient a=1 and position x=1; predict y=1 and slope 2. Compare a fresh Reset, a=1 and x=−1; predict the same y=1 but slope −2. A third fresh Reset, a=1 and x=0, gives slope zero. The exact quadratic displays branch ambiguity and derivative degeneracy, not an automatic Gauss–Newton solver.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forLocal derivative information can differ even when the forward observations agree.
Pause & reason
Check your understanding
Your turn
Try it yourself
On the interval [1,3], show that the inverse of x↦x² is Lipschitz on its image. Explain why this estimate fails uniformly when the admissible interval reaches zero.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Factor x²−z² and bound x+z from below.
Compare with the solution
For x,z in [1,3], |x²−z²|=|x−z|(x+z)≥2|x−z|, giving |x−z|≤|F(x)−F(z)|/2. On intervals including zero, no positive lower bound for x+z holds. Taking z=0 and x→0 would require x≤Cx², impossible for a fixed finite C. The positive branch remains unique on nonnegative reals, but its inverse near zero is only Hölder of order one half rather than uniformly Lipschitz.
Keep the idea
Take it with you
- Each nonlinear iteration contains its own linear inverse problem.
- Step damping, reference penalties, and local convergence assumptions must be distinguished.
- Global ambiguity can remain after a locally successful update.
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