# Inverse problem in paleomagnetism: Making the most of the measured data

Dmitry Ponomarev

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Dmitry Ponomarev. Inverse problem in paleomagnetism: Making the most of the measured data. Fourth Complex Days, Rudy Valette, Claire Michel, Feb 2024, NICE, France. hal-04807336

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https://hal.science/hal-04807336v1

Submitted on 27 Nov 2024

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Distributed under a Creative Commons Attribution 4.0 International License
Inverse problem in paleomagnetism:
Making the most of the measured data

Dmitry Ponomarev, Factas team,
Centre Inria d'Université Côte d'Azur

February 6, 2024
* Brief introduction

* Asymptotic estimation of the net magnetisation

* Extrapolation of the measurements and denoising

* Recap and outlook
* Brief introduction

* Asymptotic estimation of the net magnetisation

* Extrapolation of the measurements and denoising

* Recap and outlook
Paleomagnetism studies the record of the Earth's magnetic field.

* Certain rocks contain remanent magnetisation from the time they were cooled down.

* Extraction and interpretation of this information brings knowledge of the past behavior of the magnetic field on Earth.

* Studying extraterrestrial rocks may yield information on existence of magnetic dynamo and atmosphere on other planets / satellites.
<figure><img src="image_2.png" /><figcaption>( EAPS paleomagnetism lab, MIT )</figcaption></figure>

<figure><img src="image_3.png" /></figure>

* SQUID magnetometer

* sapphire window

* magnetic rock

<figure><img src="image_4.png" /></figure>
<figure><img src="image_5.png" /></figure>

* $\vec{M}$ is unknown but localised ($Q := \text{supp } \vec{M}$).

* Only one field component ($B_3$) can be measured.

* Measurements of $B_3$ are over a portion of the plane above Q.
<figure><img src="image_6.png" /></figure>

Magnetostatics: $\nabla \cdot \vec{M} = \Delta\Phi$, $\nabla \cdot \vec{B} = -\nabla\Phi$.

↓

↓

↓

$$ B_3(\mathbf{x}, \mathbf{h}) = -\frac{1}{4\pi} \frac{\partial}{\partial \mathbf{h}} \iiint_Q \frac{\nabla \cdot \vec{M}(\mathbf{t}, t_3)}{(|\mathbf{x} - \mathbf{t}|^2 + (\mathbf{h} - t_3)^2)^{1/2}} d^3\mathbf{t} $$
<figure><img src="image_7.png" /></figure>

$$ B_3(\mathbf{x}, \mathbf{h}) = -\frac{1}{4\pi} \frac{\partial}{\partial \mathbf{h}} \iint_Q \frac{\nabla \cdot \vec{\mathcal{M}}(\mathbf{t}, t_3)}{(|\mathbf{x} - \mathbf{t}|^2 + (\mathbf{h} - t_3)^2)^{1/2}} d^3\mathbf{t} $$

* Recovery $\vec{\mathcal{M}}(\vec{x}) \leftarrow B_3(\mathbf{x}, \mathbf{h})$ is a severely ill-posed problem!

* Net magnetisation (moment) $\vec{m} := \iiint_Q \vec{\mathcal{M}}(\vec{x}) d^3x \in \mathbb{R}^3$ can be uniquely determined.
* Brief introduction

* Asymptotic estimation of the net magnetisation

* Extrapolation of the measurements and denoising

* Recap and outlook
<figure><img src="image_8.png" /></figure>

* Measurement area is the disk $D_A$ of the radius $A$.

* For "$A \gg 1$", we can estimate

$$m_{1,2} = 2 \iint_{D_A} x_{1,2} B_3(x, h) \, d^2 x + \mathcal{O}\left(\frac{1}{A}\right),$$

$$m_3 = 2A \iint_{D_A} B_3(x, h) \, d^2 x + \mathcal{O}\left(\frac{1}{A^2}\right).$$
One can obtain estimates of an arbitrary high order, e.g.:

$$m_{1,2} = \frac{2}{105} \iint_{D_A} \left[ 105 - 2016 \left(\frac{x_{1,2}}{A}\right)^4 + 19200 \left(\frac{x_{1,2}}{A}\right)^6 - 22400 \left(\frac{x_{1,2}}{A}\right)^8 \right] x_{1,2} B_3(x, h) \, d^2x + \mathcal{O}\left(\frac{1}{A^4}\right),$$

$$m_3 = \frac{A}{24} \iint_{D_A} \left[ 35 + 1792 \left(\frac{x_j}{A}\right)^6 - 3200 \left(\frac{x_j}{A}\right)^8 \right] B_3(x, h) \, d^2x + \mathcal{O}\left(\frac{1}{A^4}\right).$$
<figure><img src="image_9.png" /></figure>
Numerics: Noiseless $B_3$ field

<figure><img src="image_10.png" /></figure>
Numerics: $B_3$ field with 10% of noise

<figure><img src="image_11.png" /></figure>
Numerics: $B_3$ field with 10% of noise

<figure><img src="image_12.png" /></figure>
Numerics: $B_3$ field, 10% of noise, with post-processing [16]

<figure><img src="image_13.png" /></figure>
* Brief introduction

* Asymptotic estimation of the net magnetisation

* Extrapolation of the measurements and denoising

* Recap and outlook
* The main obstacles for the (net) magnetisation recovery are **insufficient measurement area** and **presence of noise**.

* We can attempt to solve both issues at once achieving **field extrapolation** and **denoising**.

<figure><img src="image_14.png" /></figure>

Spectral extrapolation !
Extrapolation on the measurement plane:

<figure><img src="image_15.png" /></figure>

Note that here we assume that magnetisation $\vec{M}$ is planar.
* Construct special basis functions $(\varphi_n)_{n=1}^{\infty} \subset L^2(Q)$ related to integral operators of the direct problem.

* Expand the measured field

$$B_3^{\text{meas}}(\mathbf{x}) \approx \sum_{n=1}^{N} b_n \varphi_n(\mathbf{x}), \quad \mathbf{x} \in \mathbb{Q},$$

with $N$ chosen according to a noise level in the data.

* Compute the extrapolant:

$$B_3^{\text{ext}}(\mathbf{x}) = \sum_{n=1}^{N} b_n \tilde{\varphi}_n(\mathbf{x}), \quad \mathbf{x} \in \mathbb{R}^2,$$

where $(\tilde{\varphi}_n)_{n=1}$ are “analytic continuations” of $(\varphi_n)_{n=1}$.
* Construct special basis functions $(\varphi_n)_{n=1}^{\infty} \subset L^2(Q)$ related to integral operators of the direct problem.

Solve the eigenvalue problem

$$K_1 \star \chi_Q \varphi_n + K_2 \star \chi_Q \mathcal{R}[\varphi_n] = \lambda_n \varphi_n,$$

where $K_1(x) := \frac{h}{4\pi(|x|^2+h^2)^{3/2}}$, $K_2(x) := -\partial_h \frac{h}{4\pi(|x|^2+h^2)^{3/2}}$, $\chi_Q$ is the characteristic function of $Q$, and $\mathcal{R}$ is the operator

$$\mathcal{R}[f](x) := \iint_Q R(x, t) f(t) \, d^2 t,$$

with $R(x, t)$ *approximately* satisfying

$$\iint_Q K_1(y-t) R(x, t) \, d^2 t = K_2(x-y), \quad x, y \in Q.$$
* Expand the measured field

$$B_3^{\text{meas}}(\mathbf{x}) \approx \sum_{n=1}^{N} b_n \varphi_n(\mathbf{x}), \quad \mathbf{x} \in Q,$$

with $N$ chosen according to a noise level in the data.

Compute the coefficients according to

$$b_n = \langle B_3^{\text{meas}}, \varphi_n + \mathcal{R}^* \mathcal{R}[\varphi_n] \rangle_{L^2(Q)},$$

where $\mathcal{R}^*$ is the operator adjoint to $\mathcal{R}$.
• Compute the extrapolant:

$$B_3^{\text{ext}}(\mathbf{x}) = \sum_{n=1}^{N} b_n \tilde{\varphi}_n(\mathbf{x}), \quad \mathbf{x} \in \mathbb{R}^2,$$

where $(\tilde{\varphi}_n)_{n=1}$ are “analytic continuations” of $(\varphi_n)_{n=1}$.

Recalling

$$K_1 \star \chi_Q \varphi_n + K_2 \star \chi_Q \mathcal{R}[\varphi_n] = \lambda_n \varphi_n,$$

and that $K_1, K_2$ are defined on the whole $\mathbb{R}^2$, we set

$$\tilde{\varphi}_n := \frac{1}{\lambda_n} (K_1 \star \chi_Q \varphi_n + K_2 \star \chi_Q \mathcal{R}[\varphi_n]), \quad n \in \mathbb{N}_+.$$
Numerical test: input field (with 5% of noise)

[ 24 ]

<figure><img src="image_16.png" /></figure>
<figure><img src="image_17.png" /></figure>
Numerical test: extrapolated field

<figure><img src="image_18.png" /></figure>
<figure><img src="image_19.png" /></figure>
<figure><img src="image_20.png" /></figure>
* Brief introduction

* Asymptotic estimation of the net magnetisation

* Extrapolation of the measurements and denoising

* Recap and outlook
* Asymptotic formulas for estimating net magnetisation require field data available over a large area.

* Extrapolation can “unpack” densely made measurements over a portion of the plane onto the whole plane.

* Combining two approaches might be tricky due to sensitivity to errors: high level of accuracy is needed!
The End

<figure><img src="image_21.png" /></figure>

Thank You !