Looking through the code, the only thing this really does different to the current implementation is that it solves the primary problem on a mesh that is padded beyond the current mesh correct? Is there any reason we don't just use the analytic response for this in general?

Looking through the code, the only thing this really does different to the current implementation is that it solves the primary problem on a mesh that is padded beyond the current mesh correct? Is there any reason we don't just use the analytic response for this in general?

I wasn't quite sure what boundary conditions were being imposed originally because the underlying math wasn't documented. I only knew that they weren't correct. One thing this PR is working towards is better organization of the NSEM module and setting things up to add the fictitious sources approach.

I think I see more now looking at it. Currently the Primary field solution uses a Dirichlet Boundary condition on the top and bottom nodes with values taken from the analytic solution.

$$e_{sol}(z_{bottom}) = e_{ana}(z_{bottom})$$ $$e_{sol}(z_{top}) = e_{ana}(z_{top})$$

It would appear that in this proposal that you're using a Neumann boundary on $e$ at the top,

$$h_{top} = 1 \rightarrow \frac{\partial e}{\partial z}|_{top} = - i \omega \mu_0,$$

and at the bottom you're using a Robin boundary condition (or at least something close to one),

$$\frac{\partial e}{\partial z} - i k e_{bot} = 0$$

If so, then it is indeed different, and we can likely simplify up this implementation a decent amount using some of discretize's pre-built operators and be a little more rigorous with the derivation. It would be good to explicitly state these boundary conditions more clearly in the docstring though.

It would also appear there's some errors in the bottom boundary condition in the implementation as well, Some quick comparison's on my end:

As you can see, there is a large phase error at the bottom boundary in this proposal.

Went through my own derivation to double check the boundary conditions, see my notes below, but I do get to the same result as your code, but I'm just not convinced it is a better boundary condition than just using the analytic as dirichlet conditions, here are the real and imaginary parts for a simple layered model:

Derivation

First the PDE you are solving, with boundary conditions clearly stated.

$$ \begin{align} \frac{\partial e}{\partial z} + i\omega b = 0\\ \frac{\partial h}{\partial z} + \sigma e = 0 \end{align} $$

s.t.

$$ \begin{align} \frac{\partial e}{\partial z} = -i \omega \mu \hspace{10pt}\textrm{for } z = z_{top}\\ \frac{\partial e}{\partial z} - i k e = 0 \hspace{10pt}\textrm{for } z = z_{bot} \end{align} $$

and in a weak form,

$$ \begin{align} < u, \partial_z e > + i \omega < u, b > = 0\\ < v, \partial_z \mu^{-1} b > + < v, \sigma e > = 0 \end{align} $$

Discretizing this with an E-B formulation with $e$ on faces (nodes in 1D) and $b$ on edges (cell centers in 1D),

$$ \begin{align} u^T M_{e} G e + i \omega u^T M_{e} b = 0\\ -v^T G^T M_{e, \mu^{-1}} b + (v \mu^{-1} b)|^{top - bot} + v^T M_{f, \sigma} e = 0 \end{align} $$

Using the two boundary conditions to solve for $b_{top}$ and $b_{bot}$ from Faraday's Law:

$$ \begin{align} i\omega b_{top} = -\partial_z e |^{top} = i\omega \mu\\ b_{top} = \mu \end{align} $$

$$ \begin{align} i \omega b_{bot} = -\partial_z e |^{bot} = -ike\\ b_{bot} = -\frac{k}{\omega} e \end{align} $$

Which gives

$$ -v^T G^T M_{e,\mu^{-1}} b + v_{top} + v_{bot}\frac{k}{\mu\omega} e + v^T M_{f, \sigma} e = 0 $$

multiply by $i\omega$

$$ v^T G^T M_{e,\mu^{-1}} (-i \omega b) + i \omega (v_{top} + v_{bot}\frac{k}{\mu\omega} e) + i\omega v^T M_{f, \sigma} e = 0 $$

Substitute in discretized Faraday's Law,

$$ v^T G^T M_{e,\mu^{-1}} G e + i \omega (v_{top} + v_{bot}\frac{k}{\mu\omega} e) + i\omega v^T M_{f, \sigma} e = 0 $$

Re-arrange

$$ v^T G^T M_{e,\mu^{-1}} G e + i(v \frac{k}{\mu} e)|^{bot} + i\omega v^T M_{f, \sigma} e = -i \omega (v_{top}) $$

So then system of equations is then:

$$ (G^T M_{e,\mu^{-1}} G +i\omega M_{f, \sigma} + i P_{c, bot} \frac{k_{bot}}{\mu_{bot}} ) e = -i \omega p_{c, top} $$

Where $P_{c, bot}$ is a matrix of zeros, with a single 1 along the diagonal at the bottom node index, and $p_{c, top}$ is a vector of 0's with a single 1 at the top node index.

I think I see more now looking at it. Currently the Primary field solution uses a Dirichlet Boundary condition on the top and bottom nodes with values taken from the analytic solution.

e s o l ( z b o t t o m ) = e a n a ( z b o t t o m ) e s o l ( z t o p ) = e a n a ( z t o p )

It would appear that in this proposal that you're using a Neumann boundary on e at the top,

h t o p = 1 → ∂ e ∂ z | t o p = − i ω μ 0 ,

and at the bottom you're using a Robin boundary condition (or at least something close to one),

∂ e ∂ z − i k e b o t = 0

If so, then it is indeed different, and we can likely simplify up this implementation a decent amount using some of discretize's pre-built operators and be a little more rigorous with the derivation. It would be good to explicitly state these boundary conditions more clearly in the docstring though.

Ya, basically. I wanted to use the prebuilt operators in discretize for boundary conditions but couldn't quite figure it out.

Went through my own derivation to double check the boundary conditions, see my notes below, but I do get to the same result as your code, but I'm just not convinced it is a better boundary condition than just using the analytic as dirichlet conditions, here are the real and imaginary parts for a simple layered model:

Derivation

First the PDE you are solving, with boundary conditions clearly stated.

∂ e ∂ z + i ω b = 0 ∂ h ∂ z + σ e = 0

s.t.

∂ e ∂ z = i ω μ for  z = z t o p ∂ e ∂ z − i k e = 0 for  z = z b o t

and in a weak form,

< u , ∂ z e > + i ω < u , b >= 0 < v , ∂ z μ − 1 b > + < v , σ e >= 0

Discretizing this with an E-B formulation with e on faces (nodes in 1D) and b on edges (cell centers in 1D),

u T M e G e + i ω u T M e b = 0 − v T G T M e , μ − 1 b + ( v μ − 1 b ) | t o p − b o t + v T M f , σ e = 0

Using the two boundary conditions to solve for b t o p and b b o t from Faraday's Law:

i ω b t o p = − ∂ z e | t o p = i ω μ b t o p = μ

i ω b b o t = − ∂ z e | b o t = − i k e b b o t = − k ω e

Which gives

− v T G T M e , μ − 1 b + v t o p + v b o t k μ ω e + v T M f , σ e = 0

multiply by i ω

v T G T M e , μ − 1 ( − i ω b ) + i ω ( v t o p + v b o t k μ ω e ) + i ω v T M f , σ e = 0

Substitute in discretized Faraday's Law,

v T G T M e , μ − 1 G e + i ω ( v t o p + v b o t k μ ω e ) + i ω v T M f , σ e = 0

Re-arrange

v T G T M e , μ − 1 G e + i ( v k μ e ) | b o t + i ω v T M f , σ e = − i ω ( v t o p )

So then system of equations is then:

( G T M e , μ − 1 G + i ω M f , σ + i P c , b o t k b o t μ b o t ) e = − i ω p c , t o p

Where P c , b o t is a matrix of zeros, with a single 1 along the diagonal at the bottom node index, and p c , t o p is a vector of 0's with a single 1 at the top node index.

My derivation is almost the identical. However, I'm pretty sure you have a sign missing on the boundary conditions at the top. There are two negatives that cancel out.

Devin' Derivation

The system:

$$ \frac{\partial e_x}{\partial z} + i\omega b_y = 0 $$ $$ \frac{\partial h_y}{\partial z} + \sigma e_x = 0 $$

Boundary condition at the top:

We set:

$$ h_y^{(top)} = 1 $$

Therefore from Faraday's law:

$$ \frac{\partial e_x^{(top)}}{\partial z} = - i\omega\mu_0 $$

Boundary condition at the bottom:

At the bottom, there is only a downgoing wave of the form:

$$ e_x = E^- \exp (ikz) $$

where

$$ k = \sqrt{-i\omega\mu_0\sigma} = (1 - i)\sqrt{\frac{\omega \mu_0\sigma}{2}} $$

So if $\Delta z$ is negative, the downgoing wave decays. From Faraday's law:

$$ ik E^- \exp (ikz) + i\omega b_y = 0 $$ $$ \implies ik e_x + i \omega b_y = 0 $$ $$ \implies k e_x + \omega b_y = 0 $$

And the boundary condition here is:

$$ \frac{\partial e_x}{\partial z} - i k e_x = 0 $$

Inner products:

Inner-products:

$$ \langle u , \partial_z e_x \rangle + i\omega \langle u , b \rangle = 0 $$ $$ \langle f , \partial_z h_y \rangle + \langle f, \sigma e \rangle = 0 $$

Integrate second equation by parts:

$$ \langle u , \partial_z e_x \rangle + i\omega \langle u , b_y \rangle = 0 $$ $$ -\langle \partial_z f , h_y \rangle + f h_y \bigg |_{bot}^{top} + \langle f, \sigma e_x \rangle = 0 $$

Discrete system:

Get discrete systems (multiply second equation by $i \omega$):

$$ \mathbf{G_n e} = - i\omega \mathbf{b} $$ $$ -i \omega \mathbf{G_n^T M_{\mu} b} + i \omega\mathbf{M_\sigma e} + i\omega h_y \bigg |_{bot}^{top} = 0 $$

Which gives us:

$$ \big [ \mathbf{G_n^T M_{\mu} G_n} + i \omega\mathbf{M_\sigma} \big ] \mathbf{e} + i\omega h_y \bigg |_{bot}^{top} = 0 $$

Or if $\mu = \mu_0$, we can multiply through an obtain:

$$ \big [ \mathbf{G_n^T G_n} + i \omega \mu_0 , diag (\sigma) \big ] \mathbf{e} + i\omega \mu_0 h_y \bigg |_{bot}^{top} = 0 $$

Went through my own derivation to double check the boundary conditions, see my notes below, but I do get to the same result as your code, but I'm just not convinced it is a better boundary condition than just using the analytic as dirichlet conditions, here are the real and imaginary parts for a simple layered model:

My derivation is almost the identical. However, I'm pretty sure you have a sign missing on the boundary conditions at the top. There are two negatives that cancel out.

Ah yep, missed that on the copy over.

Went through my own derivation to double check the boundary conditions, see my notes below, but I do get to the same result as your code, but I'm just not convinced it is a better boundary condition than just using the analytic as dirichlet conditions, here are the real and imaginary parts for a simple layered model:

My derivation is almost the identical. However, I'm pretty sure you have a sign missing on the boundary conditions at the top. There are two negatives that cancel out.

Ah yep, missed that on the copy over.

Can you clarify what is meant by 'analytic as Dirichlet conditions'? Do you just mean setting ex=1 on the top?

I remember there being a reason I liked setting the top boundary condition using the Neumann condition but I can't remember why at the moment.

The current formulation sets Dirichlet conditions on E at the top and bottom boundary. Their values come from the analytic solution of a layered earth, normalized to 1 at the top.

Here is what I have for both Dirichlet and Neumann conditions on the top of the mesh. So we can normal the electric field (Dirichlet) or magnetic field (Neumann) to be 1 at the top of the mesh. By default, we will set the Dirichlet condition since that is the current implementation within SimPEG. By adding some additionally padding cells at the bottom, we do a much better job at enforcing the "downgoing wave" boundary condition.

The plot below is the amplitude of things on a log-scale. As you can see, the absolute error for both implementations is extremely small.

Codecov Report

❌ Patch coverage is 86.91099% with 25 lines in your changes missing coverage. Please review.
✅ Project coverage is 80.39%. Comparing base (5f060ec) to head (74186b9).

@@            Coverage Diff             @@
##             main    #1661      +/-   ##
==========================================
- Coverage   81.52%   80.39%   -1.13%     
==========================================
  Files         420      418       -2     
  Lines       55174    54778     -396     
  Branches     5254     5259       +5     
==========================================
- Hits        44978    44040     -938     
- Misses       8790     9344     +554     
+ Partials     1406     1394      -12     

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