# Physics-driven Learning of the Steady Navier-Stokes Equations using Deep
Convolutional Neural Networks

Hao Ma<EMAIL_ADDRESS>Yuxuan Zhang<EMAIL_ADDRESS>Nils Thuerey
<EMAIL_ADDRESS>Xiangyu Hu<EMAIL_ADDRESS>Oskar J. Haidn
<EMAIL_ADDRESS>Department of Aerospace and Geodesy, Technical University
of Munich, 85748 Garching, Germany  Beijing Aerospace Propulsion Institute,
100076, Beijing, China  Technical University of Munich, 85748, Garching,
Germany  Department of Mechanical Engineering, Technical University of Munich,
85748 Garching, Germany

###### Abstract

Recently, physics-driven deep learning methods have shown particular promise
for the prediction of physical fields, especially to reduce the dependency on
large amounts of pre-computed training data. In this work, we target the
physics-driven learning of complex flow fields with high resolutions. We
propose the use of _Convolutional neural networks_ (CNN) based U-net
architectures to efficiently represent and reconstruct the input and output
fields, respectively. By introducing Navier-Stokes equations and boundary
conditions into loss functions, the physics-driven CNN is designed to predict
corresponding steady flow fields directly. In particular, this prevents many
of the difficulties associated with approaches employing fully connected
neural networks. Several numerical experiments are conducted to investigate
the behavior of the CNN approach, and the results indicate that a first-order
accuracy has been achieved. Specifically for the case of a flow around a
cylinder, different flow regimes can be learned and the adhered “twin-
vortices” are predicted correctly. The numerical results also show that the
training for multiple cases is accelerated significantly, especially for the
difficult cases at low Reynolds numbers, and when limited reference solutions
are used as supplementary learning targets.

###### keywords:

Deep learning , Physics-driven method , Convolutional neural networks ,
Navier–Stokes equations

††journal: X. X. X

##  1 Introduction

In some practical fluid mechanics problems such as real-time or frequent query
analysis, a large number of solutions for different initial/boundary condition
combinations are to be considered [1, 2, 3]. For the traditional discrete
analysis, numerical simulations have to be conducted repeatedly and the
computational cost quickly becomes overly expensive [4]. In contrast to
classical computational methods, machine learning approaches, and especially
the field of deep learning that employs _Neural Networks_ (NN), have
demonstrated their capabilities to predict flow fields rapidly and accurately
[5, 6, 7].

The previous research on flow field prediction using NN is mainly focused on
data-driven methods. Besides the indirect way using closure model [8, 9], the
field solution can also be directly obtained from the network trained with a
large number of samples [10, 11]. However, for complex flows in practical
engineering problems, the training samples very often require extraction, pre-
processing and may be hard to obtain [12]. Some data-driven learning work
utilizes _Computational Fluid Dynamics_ (CFD) approach to generate the data
sets [13, 14, 15], and it does not really solve the demand of avoiding the big
computational cost of discrete methods.

In order to remedy the above-mentioned shortcomings, physics-driven methods
are a relatively new development. By providing physics information, NN are
able to directly obtain the field solution with much less or even no training
data. Based on _Multi-layer Perceptron_ (MLP) [16], Raissi et al. designed a
_Physics Informed Neural Networks_(PINN). Due to the constraint of loss
function employing _Partial Differential Equations_ (PDEs), the outputs
gradually approach the ones obeying the physics laws [17, 18, 19]. Also, Sun
et al. used MLP to predict fluid flows, in which a specific prior ansatz was
devised to force the network satisfying the geometric boundary of a flow [20].
However, due to the full connectivity between the neurons, MLP suffer from
extensive memory requirements and statistical inefficiencies [21]. Therefore,
it is difficult to handle well the multi-dimensional learning space with high-
resolution physics fields containing much more details. Taking the highest
resolution solution in Ref. [22] as an example, the MLP with one single
temperature channel has over 1 million weights. Considering more complex fluid
dynamics problems requiring multiple feature channels, the weight count would
increase even further. One avenue for alleviating this problem is to employ
the reduced-order modeling to compress and reconstruct the flow fields apart
from training the network [10, 23]. However, this operation not only is
complicated but also may introduce additional errors from the projection onto
reduced space [24]. In addition, MLP architecture by itself does not take into
account the spatial structure of data. The data points in the learning domain
irrespective of their distance are treated in a same way [25]. However, the
physics laws represented with PDEs are based on the localities of data points,
which suggests that the capability of NN to reflect this spatial relationship
can be very important, especially for the physics-driven methods which are
constrained only by PDEs.

On the other hand, _Convolutional Neural Networks_(CNN) represent a
specialized and well-established type of NN to tackle the aforementioned
challenges [26]. In previous research using data-driven methods, CNNs have
presented a good performance to predict high-fidelity physics solutions. E.g.,
without an extra reduced-order modeling step, the CNN can directly compress
and reconstruct high-fidelity flow fields with a series of convolutional
calculations [7]. In physics-driven methods, CNNs succeeded in solving simple
physics problems which obey a single PDE, such as Laplace equation [22] and
Darcy’s law [27], achieving high computational efficiency in capturing multi-
scale features of the physics fields. Meanwhile, CNNs also show the capacity
to learn spatial connections between the adjacent data points [28], or the
long-term control of fluids with physical losses [29].

In this paper, based on the idea of physics constraints and a specific CNN
architecture, we propose a _Physics-driven Convolutional Neural Networks_ (PD-
CNN) method. With this method, _Navier-Stokes_ (N-S) equations and boundary
conditions are introduced as a loss function that is discretized on the
computational mesh in a controlled manner. The geometry of the object and
other flow conditions are additionally embedded in the input layer. To our
knowledge, this is the first attempt that using complex, discrete PDE
formulations to drive CNNs for predicting physics fields.

##  2 Methodology

In this section, the CNN architecture used to compress input and to
reconstruct the output are described. Then the physics-driven learning
framework for N-S equations is introduced, while an accelerating approach
employing reference targets is presented at last.

###  2.1 U-net architecture of CNN

The U-net is a widely-used architecture of CNN which is first designed for
biomedical image segmentation [30] and has previously been used for flow field
reconstruction with data-driven learning [7]. In this paper, we modify it to
suit physics-driven learning approaches.

As Figure 1 shows, including the input and output layers, the U-net
architecture consists of 17 layers and corresponding convolutional blocks. The
input layer consists of four channels. The first two, $u_{0}$ and $v_{0}$, are
inflow velocities in both $x$ and $y$ directions, which are uniform non-
dimensional values in the whole learning domain. The geometry channel $G$
describes the shape of the object in the flow fields. When there is an object
in the flow, all values inside it be marked as 1 and the other as 0. In this
way, the geometry is embedded into the network and it is also used for
evaluating physics loss as shown in later discussion. To study the capability
of the proposed method in characterizing the different patterns of flow, the
Reynolds number is introduced as the fourth channel with the definition

$Re=\frac{\rho vL}{\mu}$ (1)

where $\rho$ is density, $v$ inflow velocity, $L$ characteristic length, and
$\mu$ the dynamic viscosity. Since the $\rho$, $v$, $L$ are all unit values in
this paper, $Re$ is only depends on $\mu$. The output layer consists of three
channels $u$, $v$, and $p$, which are velocities in both x and y directions
and pressure respectively. These outputs are also non-dimensional values.

Figure 1: Schematic of U-net architecture. Each blue box corresponds to a
multi-channel feature map generated by a convolutional layer. The resulting
size are denoted in the box. Black corner arrows denote the down-sampling or
up-sampling operation through convolutional layers. $cX,kX,sX$ shortly denotes
channel factor $c=X$, convolutional kernel size $k=(X,X)$, stride $s=X$
respectively. And the channel number is the product of $c$ and a basic
multiplier 64. The activation functions _Rectified Linear Unit_ (ReLU) is used
to introduce the non-linearity. Orange arrows denote “skip-connections” via
concatenation.

From inputs towards outputs, the network consists of two processes: encoding
and decoding. In the encoding process, the input fields are progressively
down-sampled by convolutional calculations with corresponding kernels. In this
way, the matrices with a size of $128\times 128$ are reduced to a 512
component vector. The decoding part works in an opposite manner, using an
inverse convolutional process mirroring the behavior of the encoding part.
Along with the increase of spatial resolution, the flow fields are
reconstructed by up-sampling operations and convolutions.

The core target of this work is to generate a flow field constrained by given
physics laws. For steady problems, the PDEs, i.e. the mathematical expressions
of underlying physics laws, represent the spatial relationships of adjacent
positions. Similarly, the convolutional kernels extract the spacial feature of
the receptive field consisting of a group of adjacent pixels. In the U-net
architecture we used for our CNN, the encoding part is responsible for
recognizing the geometry and inflow conditions of the flow field, in order to
extract the necessary features representing the physics of the inputs using
convolution operation layer by layer. These features are the basis for the
subsequent decoding part. Here, the layers of the decoding process at
different depths store the physical feature maps and the spacial relationship
are recovered by the inverse convolutional calculation. Eventually, the
decoding part is able to reconstruct the proper flow field under the constrain
of PDEs. In addition, there are the concatenations of the feature channels
between encoding and decoding as the orange arrows denoted in Figure 1.
Duplicating the feature channels from the encoding blocks to the corresponding
decoding ones, the “skip connections” effectively double the number of feature
channels in each decoding layer and enable the network to consider the
information from the encoding layers, which extracted from the geometry and
inflow conditions.

The architecture used in this paper is symmetrical, which means the encoding
and decoding processes have the same depth, meanwhile, the amounts and
dimensions of corresponding blocks are the same. However, the depths of the
two processes are both adjustable. A coarse input compressed by fewer encoding
layers is also able to generate a high-resolution solution reconstructed by
more decoding layers. More details of the U-net architecture and convolutional
block, including activation function, pooling, and dropout, can be found in
Ref. [7] and [31].

###  2.2 Physics-driven learning

The PDEs controlling the behavior of Newtonian fluid are Navier-Stokes (N-S)
Equations. In our study, the steady and incompressible form of N-S Equations
is chosen as follows:

$\nabla\cdot\textbf{\emph{U}}=0,$ (2)
$\textbf{\emph{U}}\cdot\nabla\textbf{\emph{U}}+\nabla\emph{P}-\mu\nabla^{2}\textbf{\emph{U}}=0,$
(3)

where $\textbf{\emph{U}}\equiv\textbf{\emph{U}}(u,v)$, P, $\mu$ are velocity,
pressure and viscosity respectively. Equation (2) is the continuity equation,
which imposes the incompressibilities of the fluid. Equation (3) is the
momentum conservation equation, in which the first term represents the
momentum convection, $\nabla\emph{P}$ the pressure gradient and
$\mu\nabla\textbf{\emph{U}}$ the viscous dissipation.

Figure 2: Physics-driven learning for solving the N-S equations. The U-net CNN
generates the solution. The backpropagation computes the gradient of the loss
function and updates the weights of the CNN to satisfy the discretized N-S
equations and boundary conditions.

As shown in Figure 2, once the preliminary flow field is obtained from the
U-net CNN generator, we apply the physical constraint to this field. The
learning domain is separated as inner domain and boundaries, which are
represented by $\Omega$ and $\Gamma$ respectively. In the inner domain
$\Omega$, the left hand sides of N-S Equations are employed as loss function
and 3 residuals can be obtained as

$\left\\{\begin{array}[]{ccc}\frac{\partial u}{\partial x}+\frac{\partial
v}{\partial y}=e_{1}\\\ u\frac{\partial u}{\partial x}+v\frac{\partial
u}{\partial y}+\frac{\partial p}{\partial
x}-\mu\left(\frac{\partial^{2}u}{\partial x^{2}}+\frac{\partial^{2}u}{\partial
y^{2}}\right)=e_{2}\\\ u\frac{\partial v}{\partial x}+v\frac{\partial
v}{\partial y}+\frac{\partial p}{\partial
y}-\mu\left(\frac{\partial^{2}v}{\partial x^{2}}+\frac{\partial^{2}v}{\partial
y^{2}}\right)=e_{3}\\\ \end{array}\right.$ (4)

So the residuals of the N-S equations in $\Omega$ can be represented as
$E_{\rm{\Omega}}=[e_{1},e_{2},e_{3}]^{\rm T}$.

In order to obtain differentiable formulation of the physics in the loss, we
construct suitable convolutional filters to compute the N-S equations via
finite differences. Similar approaches have been proposed previously for
simple PDEs [22], and the partial differential operators for two different
dimensions are constructed separately [32, 33]. The construction via
convolutions has the advantage that the backpropagation of a deep learning
framework can be used, and the finite difference kernels yield well controlled
accuracies for the derivative calculations. Choosing the x direction as an
example, the weights of the filters are represented as

$\rm{W}_{\frac{\partial}{\partial x}}={\left[\begin{array}[]{ccc}0&-0.5&0\\\
0&0&0\\\ 0&0.5&0\end{array}\right]},\rm{W}_{\frac{\partial^{2}}{\partial
x^{2}}}={\left[\begin{array}[]{ccc}0&1&0\\\ 0&-2&0\\\
0&1&0\end{array}\right]}.$ (5)

After the rotating and moving operation through the matrix obtained from the
last layer, the first- or second-order partial derivatives of local quantities
are calculated. Choosing $u$ as an example, this procedure can be written as:

$g_{i,j}=\sum_{m=0}^{2}\sum_{n=0}^{2}u_{i+m-1,j+n-1}\cdot f_{m,n},$ (6)

where $f_{m,n}$ is the convolutional filter and the $g_{i,j}$ is the central
difference of $u$ in each data point.

On the boundary $\Gamma$, including inflow and outflow side and walls, the
Dirichlet and Neumann boundary conditions are considered as shown in 2. The
residuals of $u$, $v$ and $p$ are represented as
$E_{\Gamma}=[e_{u},e_{v},e_{p}]^{\rm T}$. Combining both as
$E=[E_{\Gamma},E_{\Omega}]^{\rm T}$, the whole residual of the physics-driven
method is obtained.

To reduce the residuals, the CNN is trained in an iterative manner using a
stochastic gradient descent variant (we employ Adam [34]). After the CNN
generator, the preliminary flow fields are introduced in the loss function,
and then the residuals $E$ are obtained. Once the backpropagation is applied,
the weights and bias of CNN are adapted to minimize these physics residuals.
Eventually, the high-resolution flow fields which obey N-S Equations and
corresponding boundary conditions can be obtained.

###  2.3 Acceleration with reference targets

In the process of the physics-driven learning, the weights of CNN are adapted
only to minimize the residuals of PDEs, the solution itself is not
constrained, which means there is no target being offered for reference. In
order to accelerate the convergence and eventually improve the training
performance, besides the physical laws, we provide additional reference
targets for constraining the network.

Similar to data-driven methods, there is a reference loss term comparing the
difference between output and target, which is defined as

$\mathcal{L}_{\rm{ref}}=\sum_{i=1}^{\mathcal{I}}\sum_{n=1}^{\mathcal{N}}\left|\mathcal{X}_{\rm{out}}-\mathcal{X}_{\rm{tar}}\right|.$
(7)

The subscript “ref” here denotes reference targets. Generally,
$\mathcal{X}_{\rm{out}}$ and $\mathcal{X}_{\rm{tar}}$ are output quantities
and corresponding targets, respectively, $\mathcal{I}=targets\ amount$,
meanwhile $\mathcal{N}=batch\ size$ denotes the amount of training data in one
batch operation. The total loss considering both reference targets and physics
laws can be represented as

$\mathcal{L}=\mathcal{L}_{\rm{ref}}+{\rm R}*\mathcal{L}_{\rm{phy}},$ (8)

where $\mathcal{L}_{\rm{phy}}$ is the physics loss term considering the N-S
equations and boundary conditions. R is a constant hyperparameter which is
tuned to adapt the scales. With this weighted loss function, the different
loss terms can be easily scaled to an equivalent magnitude.

In the physics-driven training, a certain amount of randomly picked Reynolds
numbers are input as one batch in each iterative step. In contrast, in the
accelerating approach with reference targets, we also use some constant
Reynolds numbers besides the variable ones. So, the new batch includes two
groups as shown in Figure 3.

Figure 3: Acceleration with reference targets. One batch consists of the
physics and reference groups. The outputs of physics group are introduced to
physics loss function, while the outputs of reference group are introduced to
the reference loss function.

The cases in the random group vary in every iterative step and are trained
with physics loss $\mathcal{L}_{\rm{phy}}$. While the ones in constant group
are fixed in the whole iterative process and are trained with reference loss
$\mathcal{L}_{\rm{ref}}$.

This accelerating approach using reference targets is merely an enhancement of
the original physics-driven method, which means the PD-CNN without references
is sufficient to predict the final solution of the flow field. In order to
clearly represent this property, section 3 only presents the physics-driven-
alone results while the enhancement of this reference acceleration will be
discussed in section 4.

##  3 Results

###  3.1 Overview

In this section, several numerical experiments are conducted to estimate the
capability of the proposed PD-CNN framework to predict steady laminar flow
fields. The experiments follow two distinct patterns, single case and multiple
cases.

Single case means the PD-CNN only predicts the solution of one specific flow
field. Given one specific combination of boundary conditions and fluid
properties, the network is trained and then fixed to generate the unique
corresponding solution. This pattern is similar to a traditional CFD
simulation or a normal PINN training, and it is used to estimate the capacity
of PD-CNN as an alternative to solve a specific problem.

Multiple cases mean the PD-CNN is used to obtain the solutions of multiple
cases with a unique network. The input inflow conditions are varied in a
moderate range in the training stage. After that, given any of parameter
combinations inside the span, the fixed network is able to generate the
corresponding flow field. We use this pattern to estimate the capability of
the PD-CNN learning different physics and whether these physics can coexist
within one unique network. Compared with CFD simulation, this capability
allows the trained network to directly generate the solutions under different
conditions with tiny computational cost, which makes the real-time or many-
query analysis feasible.

In the study of single case, we choose several different 2D cases, such as
Couette flow, Poiseuille flow, and flow around a cylinder. While for multiple
cases, there are only cases of flow around cylinder. All of them are steady-
state flows with low Reynolds numbers. For the Couette and Poiseuille flow,
analytical solutions are used as references. For flow around cylinder cases,
the numerical results calculated by the mature _Finite Volume Method_ (FVM)
are used as references [35]. The mesh for simulations is generated by Gmesh
[36] and an unstructured grid. Then the numerical solutions are interpolated
into a Cartesian grid to compare with the PD-CNN results.

###  3.2 Single case

####  3.2.1 Couette flow

Couette flow is the flow of a viscous fluid in the space between two surfaces
moving relatively, which is frequently used in engineering courses to
illustrate shear-driven fluid motion. In our case, the top surface is moving
along the positive direction of x-axis with a constant speed, and the bottom
surface keeps still as shown in Figure 4a. The results on the detecting line
$l:x=0$ and the entire domain are shown in 4b and 4c respectively.

(a) Schematic

(b) learning results on $l$

(c) learning results in the whole domain

Figure 4: Couette Flow.The learning domain is $x\in[-1,1]$ and $y\in[-1,1]$.
The speed of top surface is $u=1$. And the viscosity coefficient $\mu=1$. For
detecting line $l:x=0$, the angle between velocity profile line and horizontal
axis is almost exactly $45^{\circ}$, which is the same as theoretical value.
For the whole domain, $u$ along x-axis is isotropic which match the zero-
gradient condition of $p$. Furthermore, there is no discontinuity on the
boundary.

The Couette flow is a tangential velocity-driven flow. The interference from
$p$ and $v$ are relatively low. The only term that actually affects the flow
is ${\partial^{2}u}/{\partial y^{2}}$, which is the linear term of N-S
Equations. Thus, this case tests the ability of PD-CNN to learn linear
processes. It also means that a Laplace-like equation can be learned by the
PD-CNN.

####  3.2.2 Poiseuille flow

A pressure-driven incompressible flow between two surfaces is calculated in
this section. The boundary conditions are shown in Figure 5a. In the entire
field, a gradient of pressure is imposed, which is the driven force of the
movement of flow. For Poiseuille flow, the flow fields with different $\mu$
are the same, which is presented very well as shown in Figure 5b. This
property indicates the rationality that designing multiple cases by altering
$\mu$ in sub-section 3.3.

(a) Schematic

(b) learning results on $l$

(c) learning results in the whole domain. $\mu=1$.

Figure 5: Poiseuille Flow. The learning domain is $x\in[-1,1]$ and
$y\in[-1,1]$. Both top and bottom wall are stationary walls. For detecting
line $l:y=0$, for all different $\mu$, the angle between velocity profile line
and horizontal axis are almost exactly $135^{\circ}$, which is the same as
theoretical values. For the whole domain, $p$ along y-axis is isotropic and
there is no discontinuity on the boundary.

Although Poiseuille flow is also a linear flow, the flow is induced by the
pressure gradient. The properties of all three variables, $u$, $v$, and $p$,
are well represented. And the obtained flow field is swirl-free and symmetric.
This proves that our treatment of the physics field is isotropic, and the
error is effectively controlled.

####  3.2.3 Flow around cylinder

In this section, the network is trained to solve a flow field around a
cylinder(as shown in Figure 6). Flow around a cylinder is a classic problem in
fluid mechanics and its flow field exhibits different physical phenomena under
different Reynolds numbers. Here, we consider solving the steady
incompressible flows with Reynolds number from 1 to 20. According to Ref.
[37], this range covers the two different regimes which are designated as
creeping laminar state (L1) and laminar flow with steady separation (L2). In
general, when the Reynolds number is smaller than approximately 5, the flow
field will be the creeping laminar flow. Once the Reynolds number is larger
than approximately 5, the flow field will transfer to a steady flow with a
pair of symmetric trapped vortexes behind the cylinder. These two types of
flow regimes are very different.

Figure 6: Schematic of flow around cylinder. The computing domain is
$x\in[-1,1]$ and $y\in[-1,1]$ and diameter of the central cylinder is 0.3.
There are two detecting lines $l_{1}$ and $l_{2}$ for later discussion.

(a) $p$

(b) $u$

(c) $v$

Figure 7: Single case when $Re=1$. Black contour lines are FVM results,
colorful contour flooding is PD-CNN results. For velocities, the learning
results agree reference quite well. While for pressure, especially in the
region behind the cylinder, the learning results have a slight deviation.

(a) $p$

(b) $u$

(c) $v$

Figure 8: Single case when $Re=20$. Black contour lines are FVM results,
colorful contour flooding is PD-CNN results. For all velocities and pressure,
the learning results agree reference quite well.

Figure 7 and Figure 8 show the flow fields predicted by PD-CNN with $Re=1$ and
$Re=20$ respectively. The learning results agree the numerical results
obtained by FVM well.

###  3.3 Multiple cases

For multiple cases training about flow around a cylinder, the CNN is trained
with the variation of Reynolds number while the U-net architecture and inflow
conditions are fixed. The Reynolds numbers, which are represented as the
reciprocal of dynamic viscosity, are randomly picked between 0.8 and 20.2 in
the whole training process. It is similar to the generation of a large
training set in data-driven methods. The random variation of the Reynolds
number provides the PD-CNN inexhaustible training cases and forces the
networks to consider the whole span.

Figure 9 shows the capability of PD-CNN to reconstruct the two flow regimes L1
and L2.

(a) $Re=1$

(b) $Re=5$

(c) $Re=10$

(d) $Re=20$

Figure 9: Stream Traces with different $Re$. The results discriminate two
different flow regimes and represent the evolution of the vortices along with
the growing of $Re$. After the creeping flow regime, (a) and (b), a pair of
symmetric contra-rotating vortices appears in the rear of the cylinder, (c)
and (d).

After the L1 regime, the flow separates on the cylinder surface and the wake
behind started to form the contra-rotating vortices. In the whole range of L2,
the pair of symmetric vortices, adhere stably behind the cylinder. For the
relatively small value of the Reynolds number, the closed wake region which
contains the “twin-vortices” is presented clearly in the rear of the cylinder.
And with the Reynolds number increasing, this pair of vortices grows and
becomes more and more elongated in the flow direction. The results talked
above show a unique network obtained by PD-CNN is able to learn the different
physical properties and predict the distinct flow fields according to input
parameters.

The contra-rotating vortices can be noticed from the stream trace pictures of
$Re=8$ but not $Re=7$. But it can not be concluded that the critical Reynolds
number at which the pair of symmetric contra-rotating vortices begin to form
is between 7 and 8. Because in the neighborhood of the critical Reynolds
number, the dimension of the contra-rotating vortices is very small and can
not be directly observed in the stream trace pictures. By employing the method
proposed by Taneda [38], the critical Reynolds number can be estimated.

Figure 10: Size of the vortices pair against $Re$. By linear fitting, the
critical Reynolds number at which the vortices pair begin to form in the rear
of a cylinder is obtained.

As shown in Figure 10, the sizes of the twin-vortices are measured against the
Reynolds numbers. It can be observed that as the Reynolds number increases,
the length of twin-vortices grows. Hence, from the linear curve fit the
critical Reynolds number can be deduced. So the pair of the contra-rotating
vortices begin to form in the rear of the cylinder at $Re=6.63$, which is
close with the FVM result, $Re=6.1$ [39].

When $Re<20.2$, reasonably good agreements are observed between the learning
results and the numerical solutions for the size of the closed wake. We also
evaluate the extrapolation capabilities of the network with the Reynolds
numbers between 20.2 and 25. However, as the comparison shows, the predicted
length of the twin-vortices has bigger deviations and the extrapolated flow
fields don’t represent reliable references. It can be concluded that for the
PD-CNN it is hard to accurately recover the neighboring flow fields which are
completely missing at the learning stage. This phenomenon reflects the
fundamentally interpretive characteristic of NN modeling and the model is
merely well approximated in the training span [5]. If the objective field is
relatively complex and the trainable parameters of NN are extensive, the
network deteriorates the extrapolation accuracy badly.

Figure 11 shows the comparison between the learning results and the numerical
solutions.

(a) $p$

(b) $u$

(c) $v$

Figure 11: Comparison between PD-CNN and FVM. The pressure is about
$l_{1}:x=0.5$ and the velocities are about $l_{2}:y=0$. The results of lower
$Re$ have relatively larger deviations.

All the learning results of these four cases are in good agreement with the
FVM results. But the values have an “even-bias”. Compared with the ground
truth, all values are closer to the mean value in the whole domain. For $p$,
the mean value is in the vicinity of 0. For $u$, it is 0.7, and $v$ is 0. That
means the generator has a trend to predict all the flow fields in the whole
parameter space more evenly. In addition, the predictive flow fields which
have lower Reynolds number have bigger deviations. Choosing $p$ as an example,
the flow fields with relatively big Reynolds number have better accuracy,
while the case of $Re=1$ has a bigger deviation. In the generation of the
training cases, the Reynolds numbers are chosen randomly between 0.8 and 20.2,
which means the distribution of training cases is even. However, for the case
with $Re=1$, the flow field is more distinctive and requires more training
effort. So, when designing the training space, more training cases should be
distributed near these exceptional parameter combinations.

##  4 Discussion

###  4.1 Accuracy

The drag and lift coefficient of a cylinder can be expressed as follows [40]

$C_{\rm d}=\frac{2F_{\rm D}}{\rho U^{2}A},\ C_{\rm l}=\frac{2F_{\rm L}}{\rho
U^{2}A},$ (9)

where $A$ represents the frontal area. $F_{\rm D}$ and $F_{\rm L}$ denote the
drag and lift force respectively. When the mean physical residual $E$ of 1k
epochs is lower than $0.4\%$, the training stops and the coefficients are
calculated. These two coefficients obtained by PD-CNN are shown in Table 2.
For $C_{\rm d}$, PD-CNN with the resolution $384\times 384$ predicts achieve a
relative error less than $2.9\%$. The value of $C_{\rm l}$ is very small and
the learning results are in the same magnitude compared with FVM references.
At the same time, it can be clearly seen that with the resolution increasing,
both the drag and lift coefficients gradually approach the numerical
solutions, which proves that the field solutions obtained by PD-CNN are
convergent.

In addition, we define the order of convergence as

$Order=\lvert\frac{{\rm log}({e_{i}}/{e_{j}})}{{\rm
log}({h_{i}}/{h_{j}})}\rvert,$ (10)

where $e$ is the error of drag or lift coefficient under different
resolutions, and $h$ is the resolution size. Based on this definition the PD-
CNN can achieve approximate first-order accuracy.

Table 1: Accuracy of PD-CNN with different resolutions Coefficient | Resolution |  | $Re=1$ |  |  | $Re=20$ |
---|---|---|---|---|---|---|---
Result | Error | Order | Result | Error | Order
$C_{\rm d}$ | FVM | 44.3 | - | - | 5.96 | - | -
$128\times 128$ | 39.7 | -4.6 | - | 5.52 | -0.44 | -
$256\times 256$ | 42.3 | -2.0 | 1.21 | 5.71 | -0.25 | 0.82
$384\times 384$ | 43.0 | -1.3 | 1.08 | 5.89 | -0.07 | 3.09
$C_{\rm l}$ | FVM | 0.0028 | - | - | 0.0097 | - | -
$128\times 128$ | -0.059 | -0.0618 | - | -0.044 | -0.0537 | -
$256\times 256$ | -0.015 | -0.0178 | 1.79 | -0.013 | -0.0227 | 1.24
$384\times 384$ | -0.0059 | -0.0087 | 1.76 | 0.0028 | -0.0069 | 2.94

###  4.2 Acceleration

In the numerical experiments of PD-CNN which are accelerated with reference
targets, there are 18 cases in each batch. The first half batch consists of 9
random $Re$ varied with epoch number, while the other half batch consists of 9
constant $Re$ fixed in the whole training process. According to Equation (8),
a moderate number of constant $Re$ for the reference targets requires being
defined beforehand. As discussed in section 3.3, the cases whose Reynolds
numbers are near to 1 are much more difficult to train. So, the manually
defined 9 constant Reynolds numbers are 1.0, 1.5, 2.0, 3.0, 4.0, 6.0, 8.0,
12.0 and 18.0. In the whole training process, the numerical solutions obtained
by FVM of these 9 constant Reynolds numbers are input as nine targets and the
CNN is trained to minimize Equation (7).

(a) FVM

(b) PD-CNN

(c) PD-CNN with references

Figure 12: Comparison of pressure field of different methods when $Re=1$
(epoch=10k). For this typical case with reference, the training with targets
is able to obtain accurate solution faster.

(a) FVM

(b) PD-CNN

(c) PD-CNN with references

Figure 13: Comparison of pressure field of different methods when $Re=1.2$
(epoch=10k). For this typical case without reference, the training with
targets is also able to obtain accurate solution faster.

In this way, the reference targets restrain the results generated by the
physics-driven method to approximate real solutions faster (as shown in Figure
12 and Figure 13). Since the targets only include a limiting number of cases
and are used through the whole training process, this approach reduces the
expensive data generation cost of the traditional data-driven methods. In
practical engineering applications, the reference targets can be easily picked
from the existing data, e.g., experimental and numerical results.

##  5 Conclusion

In this paper, we proposed a physics-driven method based on a CNN with a U-net
structure. By introducing the discretized Navier-Stokes equations and boundary
conditions as the loss function, the CNN is able to directly predict steady-
state laminar flow fields. Compared with the MLP used in previous research,
the CNN is able to embed the objective geometry and flow structure into the
latent space, and then decode them to reconstruct the corresponding flow
fields with physics consistency. The PD-CNN is able to obtain accurate
solutions for both single case and multiple cases. In the multiple cases
prediction of flow around cylinder, PD-CNN is capable of describing the
transformation of the “twin-vortices” and obtaining the critical Reynolds
number between creeping laminar state and laminar flow with separation. By
constraining with a small number of reference targets, the network training
was accelerated, especially for the difficult cases. Further research will be
carried out for predicting complex flow fields in practical engineering
problems with the present method.

##  6 Acknowledgement

Hao Ma (No. 201703170250) and Yuxuan Zhang (No. 201804980021) are supported by
China Scholarship Council when they conduct the work this paper represents.

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00footnotetext: Support of the research of the first author by IGA, project
PřF 2023 010, and of the third author by the Austrian Science Fund (FWF),
project I 4579-N, entitled “The many facets of orthomodularity”, is gratefully
acknowledged.

# Special filters in bounded lattices

Ivan Chajda, Miroslav Kolařík and Helmut Länger

###### Abstract

M.S. Rao recently investigated some sorts of special filters in distributive
pseudocomplemented lattices. In our paper we extend this study to lattices
which need neither be distributive nor pseudocomplemented. For this sake we
define a certain modification of the notion of a pseudocomplement as the set
of all maximal elements belonging to the annihilator of the corresponding
element. We prove several basic properties of this notion and then define
coherent, closed and median filters as well as $D$-filters. In order to be
able to obtain valuable results we often must add some additional assumptions
on the underlying lattice, e.g. that this lattice is Stonean or $D$-Stonean.
Our results relate properties of lattices and of corresponding filters. We
show how the structure of a lattice influences the form of its filters and
vice versa.

AMS Subject Classification: 06B05, 06B10, 06D15, 06D20

Keywords: Filter, coherent filter, closed filter, median filter, D-filter,
Stonean lattice, $D$-Stonean lattice

##  1 Introduction and basic concepts

M.S. Rao recently studied various types of filters and ideals in distributive
lattices with pseudocomplements, see [5] and [6]. Some of these findings
require the lattice under consideration to be either Stone or weakly Stone, as
indicated in the referenced papers. There is the question whether the
distributivity assumption can be relaxed and, more specifically, if the
pseudocomplementation assumption can be eliminated. This approach, in which we
consider instead of the pseudocomplement a certain subset (in fact an
antichain) of elements behaving similar to the pseudocomplement, has already
been introduced by I. Chajda and H. Länger in [3]. In this paper the authors
show that this construction can be successfully used in order to introduce the
implication and negation connectives in an intuitionistic-like logic based on
any bounded lattice that satisfies the Ascending Chain Condition.

In this paper we will deal with bounded lattices
$\mathbf{L}=(L,\vee,\wedge,0,1)$ satisfying the Ascending Chain Condition
(ACC). We identify singletons with their unique element, i.e. we will write
$a$ instead of $\\{a\\}$. For $a\in L$ and for subsets $A,B$ of $L$ we define

$\displaystyle A\vee B$ $\displaystyle:=\\{x\vee y\mid x\in A\text{ and }y\in
B\\},$ $\displaystyle A\wedge B$ $\displaystyle:=\\{x\wedge y\mid x\in A\text{
and }y\in B\\},$ $\displaystyle A^{0}$
$\displaystyle:=\operatorname{Max}\\{x\in L\mid x\wedge y=0\text{ for all
}y\in A\\},$ $\displaystyle a^{0}$ $\displaystyle:=\operatorname{Max}\\{x\in
L\mid x\wedge a=0\\}.$

Here and in the following $\operatorname{Max}A$ denotes the set of all maximal
elements of $A$ which obviously is an antichain. Observe that $0^{0}=1$ and
$1^{0}=0$ and that $A^{0}\neq\emptyset$. The set $a^{0}$ is in fact a
generalization of the pseudocomplement of $a$ introduced by O. Frink [4] or a
modification of the annihilator since the set

$\\{x\in L\mid x\wedge y=0\text{ for all }y\in A\\}$

is in fact the annihilator of the set $A$ as known in lattice theory. (The
pseudocomplement of $a$ is the greatest element of $\\{x\in L\mid x\wedge
a=0\\}$.) The advantage of our approach is that $a^{0}$ can be defined in any
lattice with $0$. Of course, we must pay for this advantage by the fact that
$a^{0}$ need not be an element of $L$, but may be a subset of $L$, namely an
antichain of $\mathbf{L}$.

It is easy to see that in a bounded completely distributive lattice every
element has a pseudocomplement.

The element $a$ is called dense if $a^{0}=0$ and sharp if $a^{00}=a$. Let $D$
($S$) denote the set of all dense (sharp) elements of $L$. Clearly the
following holds:

$\displaystyle a\in D\text{ if and only if }a^{00}=1,$ $\displaystyle 1\in D,$
$\displaystyle 0,1\in S,$ $\displaystyle D\cap S=1.$

Moreover, we define the following binary relations and unary operators on
$2^{L}$:

$\displaystyle A\leq B$ $\displaystyle\text{ if }x\leq y\text{ for all }x\in
A\text{ and all }y\in B,$ $\displaystyle A\leq_{1}B$ $\displaystyle\text{ if
for every }x\in A\text{ there exists some }y\in B\text{ with }x\leq y,$
$\displaystyle A\leq_{2}B$ $\displaystyle\text{ if for every }y\in B\text{
there exists some }x\in A\text{ with }x\leq y,$ $\displaystyle A=_{1}B$
$\displaystyle\text{ if both }A\leq_{1}B\text{ and }B\leq_{1}A,$
$\displaystyle\overline{A}$ $\displaystyle:=\\{x\in L\mid 1\in x^{00}\vee
y^{00}\text{ for each }y\in A\\},$ $\displaystyle A^{D}$
$\displaystyle:=\\{x\in L\mid x\vee y\in D\text{ for all }y\in A\\}.$

The set $A$ is called closed if $\overline{\overline{A}}=A$. Observe that
$\overline{0}=D$ and $\overline{1}=L$.

######  Lemma 1.1.

Let $(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC. Then
$\overline{D}=L$ and $\overline{L}=D$.

###### Proof.

If $a\in L$ then $1\in a^{00}\vee x^{00}$ for all $x\in D$ and hence
$a\in\overline{D}$. This shows $L\subseteq\overline{D}$, i.e.
$\overline{D}=L$. Now let $a\in\overline{L}$. Then $1\in a^{00}\vee x^{00}$
for all $x\in L$, especially $1\in a^{00}\vee 0^{00}=a^{00}$. Since $a^{00}$
is an antichain we have $a^{00}=1$ whence $a\in D$. This shows
$\overline{L}\subseteq D$. Conversely, if $a\in D$ then $a^{00}=1$ and hence
$1\in a^{00}\vee x^{00}$ for all $x\in L$, i.e. $a\in\overline{L}$. This shows
$D\subseteq\overline{L}$. Altogether, we obtain $\overline{L}=D$. ∎

######  Remark 1.2.

The pair $(A\mapsto\overline{A},A\mapsto\overline{A})$ is the Galois-
correspondence between $(2^{L},\subseteq)$ and $(2^{L},\subseteq)$ induced by
the binary relation $\\{(x,y)\in L^{2}\mid 1\in x^{00}\vee y^{00}\\}$ on $L$.
Hence $A\mapsto\overline{\overline{A}}$ is a closure operator on
$(2^{L},\subseteq)$ and therefore the set of all closed subsets of $L$ forms a
closure system of $L$, i.e. it is closed under arbitrary intersections. This
means that the closed subsets of $L$ form a complete lattice with respect to
inclusion with smallest element $\overline{L}=D$ and greatest element
$\overline{\emptyset}=L$ and we have the following properties:

$\displaystyle A$ $\displaystyle\subseteq B\text{ implies
}\overline{B}\subseteq\overline{A},$ $\displaystyle A$
$\displaystyle\subseteq\overline{\overline{A}},$
$\displaystyle\overline{\overline{\overline{A}}}$
$\displaystyle=\overline{A},$ $\displaystyle A$
$\displaystyle\subseteq\overline{B}\text{ if and only if
}B\subseteq\overline{A}.$

Analogously, the pair $(A\mapsto A^{D},A\mapsto A^{D})$ is the Galois-
correspondence between $(2^{L},\subseteq)$ and $(2^{L},\subseteq)$ induced by
the binary relation $\\{(x,y)\in L^{2}\mid x\vee y\in D\\}$ on $L$.

Recall that a non-empty subset $F$ of $L$ is called a filter of $\mathbf{L}$
if $x\wedge y,x\vee z\in F$ for all $x,y\in F$ and all $z\in L$. Observe that
for every $x\in L$ the set $F_{x}:=\\{y\in L\mid x\leq y\\}$ is a filter of
$\mathbf{L}$, the so-called principal filter generated by $x$. A filter $F$ of
$\mathbf{L}$ is called a $D$-filter if $D\subseteq F$. It is elementary that
every filter of a finite lattice is principal.

##  2 Filters and $D$-filters

It is evident that if a given bounded lattice $\mathbf{L}=(L,\vee,\wedge,0,1)$
is pseudocomplemented then for each element $x\in L$ we have $x^{0}=x^{*}$. We
start this section with emphasizing certain differences between the properties
of our concept $x^{0}$ and pseudocomplements.

It is well-known that for distributive pseudocomplemented lattices
$(L,\vee,\wedge,0)$ with bottom element $0$ and $a,b\in L$ the following
holds:

  1. (i)

$a\leq b$ implies $b^{*}\leq a^{*}$,

  2. (ii)

$a\leq a^{**}$ and $a^{***}=a^{*}$,

  3. (iii)

$(a\vee b)^{*}=a^{*}\wedge b^{*}$,

  4. (iv)

$(a\wedge b)^{**}=a^{**}\wedge b^{**}$.

The following example shows that not all being valid for pseudocomplements,
also holds for $x^{0}$.

######  Example 2.1.

Consider the non-distributive lattice $\mathbf{L}$ depicted in Fig. 1:

$0$$a$$b$$c$$d$$e$$f$$g$$1$Fig. 1Non-distributive non-pseudocomplemented
lattice

We have

$\begin{array}[]{r|r|r|r|r|r|r|r|r|r}x&0&a&b&c&d&e&f&g&1\\\ \hline\cr
x^{0}&1&fg&eg&g&c&c&a&c&0\\\ \hline\cr x^{00}&0&a&b&c&g&g&fg&g&F_{0}\\\
\hline\cr\overline{x}&F_{1}&F_{d}&F_{d}&F_{d}&abcf1&abcf1&abcdefg1&abcf1&L\end{array}$

$D=F_{1}$ and $S:=\\{0,b,c,g,1\\}$. (Here and in the following we often write
$fg$ instead of $\\{f,g\\}$, and so on.) Hence $\mathbf{L}$ is not
pseudocomplemented. In $\mathbf{L}$ we have:

  1. (i)

$a\leq c$ and $c^{0}=g\nleq\\{f,g\\}=a^{0}$, but $c^{0}\leq_{1}a^{0}$.

  2. (ii)

$f\not\leq\\{f,g\\}=f^{00}$, but $f\leq_{1}f^{00}$.

  3. (iii)

$(a\vee b)^{0}=c^{0}=g\neq\\{d,g\\}=\\{f,g\\}\wedge\\{e,g\\}=a^{0}\wedge
b^{0}$, but $(a\vee b)^{0}\leq_{1}a^{0}\wedge b^{0}$.

  4. (iv)

$(d\wedge f)^{00}=d^{00}=g\neq\\{d,g\\}=g\wedge\\{f,g\\}=d^{00}\wedge f^{00}$,
but $(d\wedge f)^{00}=_{1}d^{00}\wedge f^{00}$.

On the other hand, some of the properties of pseudocomplements are preserved
also here, see the following result.

######  Theorem 2.2.

Let $(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC, $x,y\in L$
and $A,B\subseteq L$. Then the following holds:

  1. (i)

$a\wedge b=0$ for all $a\in A$ and all $b\in A^{0}$, especially, $x\wedge
x^{0}=0$,

  2. (ii)

If $a\wedge b=0$ for all $a\in A$ and all $b\in B$ then $A\leq_{1}B^{0}$,
especially, $x\wedge y=0$ implies $x\leq_{1}y^{0}$,

  3. (iii)

$A\leq_{1}A^{00}$, especially, $x\leq_{1}x^{00}$,

  4. (iv)

$A\leq_{1}B$ implies $B^{0}\leq_{1}A^{0}$, especially, $x\leq y$ implies
$y^{0}\leq_{1}x^{0}$,

  5. (v)

$A\leq_{1}B^{0}$ if and only if $B\leq_{1}A^{0}$, especially, $x\leq_{1}y^{0}$
if and only if $y\leq_{1}x^{0}$,

  6. (vi)

$a=_{1}B$ implies $a=B$,

  7. (vii)

$A^{0}=_{1}B^{0}$ implies $A^{0}=B^{0}$,

  8. (viii)

$A^{000}=A^{0}$, especially, $x^{000}=x^{0}$,

  9. (ix)

$(A\vee B)^{0}\leq_{1}A^{0}\wedge B^{0}$, especially, $(x\vee
y)^{0}\leq_{1}x^{0}\wedge y^{0}$,

  10. (x)

$(A\wedge B)^{00}=_{1}A^{00}\wedge B^{00}$, especially, $(x\wedge
y)^{00}=_{1}x^{00}\wedge y^{00}$,

  11. (xi)

$A^{00}\vee B^{00}\leq_{1}(A^{0}\wedge B^{0})^{0}$ implies $(A^{00}\vee
B^{00})^{0}=_{1}A^{0}\wedge B^{0}$, especially, $x^{00}\vee
y^{00}\leq_{1}(x^{0}\wedge y^{0})^{0}$ implies $(x^{00}\vee
y^{00})^{0}=_{1}x^{0}\wedge y^{0}$.

###### Proof.

  1. (i)

If $b\in A\wedge A^{0}$ then there exists some $c\in A$ and some $d\in A^{0}$
with $c\wedge d=b$. Since $x\wedge d=0$ for all $x\in A$, we have $b=c\wedge
d=0$.

  2. (ii)

If $A\wedge B=0$ and $b\in A$ then $b\wedge x=0$ for all $x\in B$ and hence
there exists some $c\in B^{0}$ with $b\leq c$.

  3. (iii)

This follows from (i) and (ii).

  4. (iv)

If $A\leq_{1}B$ then

$\\{x\in L\mid x\wedge B=0\\}\subseteq\\{x\in L\mid x\wedge A=0\\}$

and hence $B^{0}\leq_{1}A^{0}$.

  5. (v)

If $A\leq_{1}B^{0}$ then $B\leq_{1}B^{00}\leq_{1}A^{0}$ by (iii) and (iv).
Analogously, $B\leq_{1}A^{0}$ implies $A\leq_{1}B^{0}$.

  6. (vi)

Let $b\in B^{0}$. Because of $B^{0}\leq_{1}a$ we have $b\leq a$, and because
of $a\leq_{1}B^{0}$ there exists some $c\in B^{0}$ with $a\leq c$. Together we
obtain $b\leq a\leq c$ and hence $b\leq c$. Since both $b$ and $c$ belong to
the antichain $B^{0}$ we conclude $b=c$ and hence $b=a$ showing $B^{0}=a$.

  7. (vii)

Assume $b\in A^{0}$. Because of $A^{0}\leq_{1}B^{0}$ there exists some $c\in
B^{0}$ with $b\leq c$, and because of $B^{0}\leq_{1}A^{0}$ there exists some
$d\in A^{0}$ with $c\leq d$. Together we obtain $b\leq c\leq d$ and hence
$b\leq d$. Since both $b$ and $d$ belong to the antichain $A^{0}$ we conclude
$b=d$ and hence $b=c\in B^{0}$. This shows $A^{0}\subseteq B^{0}$.
Interchanging the roles of $A^{0}$ and $B^{0}$ gives $B^{0}\subseteq A^{0}$.
Together we obtain $A^{0}=B^{0}$.

  8. (viii)

Because of (iii) we have $A\leq_{1}A^{00}$ from which we conclude
$A^{000}=(A^{00})^{0}\leq_{1}A^{0}$ by (iv). Again by (iii) we have
$A^{0}\leq_{1}(A^{0})^{00}=A^{000}$. Altogether, $A^{000}=_{1}A^{0}$. Applying
(vii) yields $A^{000}=A^{0}$.

  9. (ix)

We have $A,B\leq_{1}A\vee B$ and hence $(A\vee B)^{0}\leq_{1}A^{0},B^{0}$ by
(iv) which implies $(A\vee B)^{0}\leq_{1}A^{0}\wedge B^{0}$.

  10. (x)

We have $A\wedge B\leq_{1}A^{00}\wedge B^{00}$ according to (iii) and hence
$(A\wedge B)^{00}\leq_{1}(A^{00}\wedge B^{00})^{00}$ according to (iv). Using
(ii), (i) and (iv) we see that any of the following statements implies the
next one:

$\displaystyle(A\wedge B)^{0}\wedge A^{00}\wedge B^{00}\wedge A\wedge B$
$\displaystyle\subseteq\\{0\\},$ $\displaystyle(A\wedge B)^{0}\wedge
A^{00}\wedge B^{00}\wedge A$ $\displaystyle\leq_{1}B^{0},$
$\displaystyle(A\wedge B)^{0}\wedge A^{00}\wedge B^{00}\wedge A$
$\displaystyle\leq_{1}B^{0}\wedge B^{00},$ $\displaystyle(A\wedge B)^{0}\wedge
A^{00}\wedge B^{00}\wedge A$ $\displaystyle\subseteq\\{0\\},$
$\displaystyle(A\wedge B)^{0}\wedge A^{00}\wedge B^{00}$
$\displaystyle\leq_{1}A^{0},$ $\displaystyle(A\wedge B)^{0}\wedge A^{00}\wedge
B^{00}$ $\displaystyle\leq_{1}A^{0}\wedge A^{00},$ $\displaystyle(A\wedge
B)^{0}\wedge A^{00}\wedge B^{00}$ $\displaystyle=0,$ $\displaystyle(A\wedge
B)^{0}$ $\displaystyle\leq_{1}(A^{00}\wedge B^{00})^{0},$
$\displaystyle(A^{00}\wedge B^{00})^{00}$ $\displaystyle\leq_{1}(A\wedge
B)^{00}.$

Together we obtain $(A\wedge B)^{00}=_{1}(A^{00}\wedge B^{00})^{00}$ and hence
$(A\wedge B)^{00}=(A^{00}\wedge B^{00})^{00}$ according to (vii). Now we have
$A^{00}\wedge B^{00}\leq_{1}A^{00},B^{00}$ and hence $(A^{00}\wedge
B^{00})^{00}\leq_{1}A^{00},B^{00}$ by (iv) and (viii) which implies
$(A^{00}\wedge B^{00})^{00}\leq_{1}A^{00}\wedge B^{00}$. Together with
$A^{00}\wedge B^{00}\leq_{1}(A^{00}\wedge B^{00})^{00}$ that follows from
(iii) this yields $(A\wedge B)^{00}=_{1}(A^{00}\wedge
B^{00})^{00}=_{1}A^{00}\wedge B^{00}$.

  11. (xi)

According to (iii), (iv), (ix), (viii) and the assumption we have

$A^{0}\wedge B^{0}\leq_{1}(A^{0}\wedge B^{0})^{00}\leq_{1}(A^{00}\vee
B^{00})^{0}\leq_{1}A^{000}\wedge B^{000}=A^{0}\wedge B^{0}.$

∎

The theorem just proved is very important because we will use it many times in
the proofs of most of the following statements.

######  Remark 2.3.

Assume $a,b\in L$ and $a^{00}\vee b^{00}\leq_{1}(a^{0}\wedge b^{0})^{0}$. Then
because of (xi) of Theorem 2.2 we have $(a^{00}\vee
b^{00})^{0}=_{1}a^{0}\wedge b^{0}$. Hence any of the following statements
implies the next one:

$\displaystyle 1$ $\displaystyle\in a^{00}\vee b^{00},$ $\displaystyle 1$
$\displaystyle\leq_{1}a^{00}\vee b^{00},$ $\displaystyle(a^{00}\vee
b^{00})^{0}$ $\displaystyle\leq_{1}0,$ $\displaystyle(a^{00}\vee b^{00})^{0}$
$\displaystyle=0,$ $\displaystyle a^{0}\wedge b^{0}$ $\displaystyle=0.$

Applying the previous results, we can show that the set of sharp elements of
$\mathbf{L}=(L,\vee,\wedge,0,1)$ forms a subsemilattice of $(L,\wedge)$.

######  Proposition 2.4.

Let $(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC and $a,b\in
S$. Then $a\wedge b\in S$.

###### Proof.

According to (x) of Theorem 2.2 we have $(a\wedge b)^{00}=_{1}a^{00}\wedge
b^{00}=a\wedge b$ and hence $(a\wedge b)^{00}=a\wedge b$ by (vi) of Theorem
2.2, i.e. $a\wedge b\in S$. ∎

######  Remark 2.5.

Unfortunately, $S$ need not be a sublattice of $\mathbf{L}$. Namely, consider
the lattice $\mathbf{L}$ visualized in Fig. 2. Here, obviously, $a,b\in S$ and
also $a\wedge b=0\in S$ in accordance with Proposition 2.4, but $a\vee
b=e\notin S$ since $e^{00}=1\neq e$.

Now we prove an expected statement about the set of dense elements.

######  Proposition 2.6.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC.
Then $D$ forms a $D$-filter of $\mathbf{L}$.

###### Proof.

Let $a,b\in D$ and $c\in L$. Then according to (x) of Theorem 2.2 we conclude
$(a\wedge b)^{00}=_{1}a^{00}\wedge b^{00}=1\wedge 1=1$ and hence $(a\wedge
b)^{00}=1$ according to (vi) of Theorem 2.2 proving $a\wedge b\in D$. If
$a\leq c$ then $c^{0}\leq_{1}a^{0}=0$ according to (iv) of Theorem 2.2 and
hence $c^{0}=0$, i.e. $c\in D$. ∎

It is well-known that the set $\mathcal{F}$ of all filters of $\mathbf{L}$
forms a complete lattice with respect to inclusion. Using Proposition 2.6 one
can recognize that the set of all $D$-filters of $\mathbf{L}$ forms a complete
sublattice of $\mathcal{F}$ with bottom element $D$.

$D$-filters were described for distributive pseudocomplemented lattices by
M.S. Rao [6]. However, we are going to show that similar result can be stated
also for lattices being neither pseudocomplemented nor distributive, but
satisfying a weaker condition.

######  Theorem 2.7.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC
and $A$ a non-empty subset of $L$ such that for each $x,y\in\overline{A}$ the
following condition holds:

$\text{if }1\in(x^{00}\vee z^{00})\wedge(y^{00}\vee z^{00})\text{ for all
}z\in A\text{ then }1\in(x^{00}\wedge y^{00})\vee z^{00}\text{ for all }z\in
A.$

Then $\overline{A}$ is a $D$-filter of $\mathbf{L}$.

###### Proof.

Obviously, $D\subseteq\overline{A}$. Let $a,b\in\overline{A}$. Then
$1\in(a^{00}\vee z^{00})\wedge(b^{00}\vee z^{00})$ for all $z\in A$ and hence
$1\in(a^{00}\wedge b^{00})\vee z^{00}=(a\wedge b)^{00}\vee z^{00}$ for all
$z\in A$ according to (x) of Theorem 2.2, i.e. $a\wedge b\in\overline{A}$.
Further, if $c\in L$ and $a\leq c$ then $a^{00}\leq_{1}c^{00}$ according (iv)
of Theorem 2.2 and hence $1\in a^{00}\vee z^{00}\leq_{1}a^{00}\vee z^{00}$ for
all $z\in A$ showing $c\in\overline{A}$. Altogether, $\overline{A}$ is a
$D$-filter of $\mathbf{L}$. ∎

In the next example we show a lattice being neither distributive nor
pseudocomplemented, but satisfying the condition from Theorem 2.7.

######  Example 2.8.

Consider the non-distributive lattice $\mathbf{L}$ depicted in Fig. 2:

$0$$a$$b$$c$$d$$e$$f$$g$$1$Fig. 2Non-distributive non-pseudocomplemented
lattice

We have

$\begin{array}[]{r|r|r|r|r|r|r|r|r|r}x&0&a&b&c&d&e&f&g&1\\\ \hline\cr
x^{0}&1&bcd&acd&abd&abc&0&0&0&0\\\ \hline\cr x^{00}&0&a&b&c&d&1&1&1&1\\\
\hline\cr\overline{x}&F_{e}&F_{e}&F_{e}&F_{e}&F_{e}&F_{0}&F_{0}&F_{0}&F_{0}\end{array}$

$D=F_{e}$ and $S=\\{0,a,b,c,d,1\\}$. Hence $\mathbf{L}$ is not
pseudocomplemented. Further, $F_{a}$ is a $D$-filter of $\mathbf{L}$,
$\overline{\\{a,b\\}}=F_{e}$ is a $D$-filter of $\mathbf{L}$ in accordance
with Theorem 2.7 and $\\{a,b\\}^{D}=\\{c,d,e,f,g,1\\}$.

##  3 Stonean and $D$-Stonean lattices

The concept of a Stone lattice was introduced by R. Balbes and A. Horn [2],
see also the paper [7] by T.P. Speed. Recall from [2] that a bounded
pseudocomplemented lattice $(L,\vee,\wedge,0,1)$ is called Stone if

$x^{*}\vee x^{**}=1\text{ and }x^{*}\vee y^{*}=(x\wedge y)^{*}\text{ for all
}x,y\in L.$

In analogy to this definition we define for our sake the following two
concepts.

######  Definition 3.1.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC.
Then $\mathbf{L}$ is called Stonean if

$1\in x^{00}\vee y^{00}\text{ for every }x\in L\text{ and every }y\in x^{0}$
(1)

and $D$-Stonean if it is both Stonean and if

$\text{for all }x,y\in L,x\vee y\in D\text{ is equivalent to }1\in x^{00}\vee
y^{00}.$ (2)

Observe that (1) is equivalent to $x^{0}\subseteq\overline{x}$ for all $x\in
L$, and (2) is equivalent to $\overline{x}=x^{D}$ for all $x\in L$. Hence
$\mathbf{L}$ is Stonean if and only if $x^{0}\subseteq\overline{x}$ for all
$x\in L$, and $\mathbf{L}$ is $D$-Stonean if
$x^{0}\subseteq\overline{x}=x^{D}$ for all $x\in L$. The lattice visualized in
Fig. 2 is not Stonean since $b\in a^{0}$, but $1\notin e=a\vee b=a^{00}\vee
b^{00}$.

The following result relates the concept of a Stonean lattice with concepts
mentioned before.

######  Proposition 3.2.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC
and $a\in L$. Then the following holds:

  1. (i)

If $\mathbf{L}$ is Stonean and $a^{0}\leq_{2}\overline{a}$ then
$\overline{\overline{a}}=\overline{a^{0}}$.

  2. (ii)

$\mathbf{L}$ is Stonean if and only if
$\overline{\overline{x}}\subseteq\overline{x^{0}}$ for all $x\in L$.

###### Proof.

  1. (i)

Assume that $\mathbf{L}$ is Stonean and $a^{0}\leq_{2}\overline{a}$. Then
$a^{0}\subseteq\overline{a}$ and hence
$\overline{\overline{a}}\subseteq\overline{a^{0}}$. Now let
$b\in\overline{a^{0}}$ and $c\in\overline{a}$. Since
$a^{0}\leq_{2}\overline{a}$ there exists some $d\in a^{0}$ with $d\leq c$. We
conclude $1\in b^{00}\vee d^{00}\leq_{1}b^{00}\vee c^{00}$ and hence $1\in
b^{00}\vee c^{00}$. This shows $b\in\overline{\overline{a}}$ and hence
$\overline{a^{0}}\subseteq\overline{\overline{a}}$. Together we obtain
$\overline{\overline{a}}=\overline{a^{0}}$.

  2. (ii)

According to the observation after Definition 3.1, $\mathbf{L}$ is Stonean if
and only if $x^{0}\subseteq\overline{x}$ for all $x\in L$. Now for every $x\in
L$ the inclusion $x^{0}\subseteq\overline{x}$ is equivalent to
$\overline{\overline{x}}\subseteq\overline{x^{0}}$.

∎

The next result is elementary but it will be used in the sequel.

######  Lemma 3.3.

Let $(L,\vee,\wedge,0,1)$ be a $D$-Stonean lattice and $a\in L$. Then $a\vee
a^{0}\subseteq D$.

###### Proof.

If $b\in a^{0}$ then $1\in a^{00}\vee b^{00}$ according to (1) which is
equivalent to $a\vee b\in D$ because of (2). ∎

######  Example 3.4.

  1. (i)

The lattice visualized in Fig. 2 satisfies neither (1) nor (2) since $b\in
a^{0}$ and $a\vee b=e\in D$, but $1\notin e=a\vee b=a^{00}\vee b^{00}$.

  2. (ii)

Consider the non-distributive lattice $\mathbf{L}$ depicted in Fig. 3:

$0$$a$$b$$c$$d$$e$$f$$g$$1$Fig. 3Non-distributive non-pseudocomplemented non-
Stonean lattice

We have

$\begin{array}[]{r|r|r|r|r|r|r|r|r|r}x&0&a&b&c&d&e&f&g&1\\\ \hline\cr
x^{0}&1&bcg&ceg&beg&bce&bcg&0&bce&0\\\ \hline\cr x^{00}&0&e&b&c&g&e&1&g&1\\\
\hline\cr\overline{x}&F_{f}&bcdfg1&adefg1&adefg1&abcef1&bcdfg1&L&abcef1&F_{0}\end{array}$

$D=F_{f}$ and $S=\\{0,b,c,e,g,1\\}$. Hence $\mathbf{L}$ is not
pseudocomplemented. Moreover, $\mathbf{L}$ satisfies neither (1) nor (2) since
$c\in b^{0}$ and $b\vee c=f\in D$, but $1\notin f=b\vee c=b^{00}\vee c^{00}$.
Therefore $\mathbf{L}$ is not Stonean.

######  Theorem 3.5.

Conditions (1) and (2) of Definition 3.1 are independent.

###### Proof.

The lattice visualized in Fig. 1 satisfies (1), but not (2) since $e\vee
g=1\in D$, but $1\notin g=g\vee g=e^{00}\vee g^{00}$. Hence it is Stonean, but
not $D$-Stonean. Now consider the non-distributive lattice $\mathbf{L}$
depicted in Fig. 4:

$0$$a$$b$$c$$d$$e$$f$$g$$h$$i$$j$$1$Fig. 4Non-distributive non-
pseudocomplemented non-Stonean lattice

We have

$\begin{array}[]{r|r|r|r|r|r|r|r|r|r|r|r|r}x&0&a&b&c&d&e&f&g&h&i&j&1\\\
\hline\cr x^{0}&1&j&dg&h&j&g&d&h&g&0&d&0\\\ \hline\cr
x^{00}&0&d&b&g&d&h&j&g&h&1&j&1\\\
\hline\cr\overline{x}&F_{i}&F_{c}&F_{i}&F_{a}&F_{c}&F_{c}&F_{a}&F_{a}&F_{c}&F_{0}&F_{a}&F_{0}\end{array}$

$D=F_{i}$ and $S=\\{0,b,d,g,h,j,1\\}$. Hence $\mathbf{L}$ is not
pseudocomplemented. The lattice $\mathbf{L}$ satisfies (2), but not (1) since
$d\in b^{0}$, but $1\notin h=b\vee d=b^{00}\vee d^{00}$. Therefore it is not
Stonean. ∎

How the concept of a Stonean lattice is related with its filters is shown in
the following theorem.

######  Theorem 3.6.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a Stonean lattice satisfying the ACC.
Then (2) is equivalent to any single of the following statements:

  1. (i)

$\overline{F}=F^{D}$ for all filters $F$ of $\mathbf{L}$,

  2. (ii)

$\overline{x}=x^{D}$ for all $x\in L$,

  3. (iii)

for every two filters $F,G$ of $\mathbf{L}$, $F\cap G\subseteq D$ is
equivalent to $F\subseteq\overline{G}$.

###### Proof.

(2) $\Rightarrow$ (i):
This follows from

$\displaystyle\overline{F}$ $\displaystyle=\\{x\in L\mid 1\in x^{00}\vee
y^{00}\text{ for all }y\in F\\},$ $\displaystyle F^{D}$ $\displaystyle=\\{x\in
L\mid x\vee y\in D\text{ for all }y\in F\\}$

for all filters $F$ of $\mathbf{L}$.
(i) $\Rightarrow$ (ii):
For all $x\in L$ we have $\overline{x}=\overline{F_{x}}=F_{x}^{D}=x^{D}$.
(ii) $\Rightarrow$ (iii):
Let $F,G$ be filters of $\mathbf{L}$. If $a\in F\cap G$ then $a=a\vee a\in
F\vee G$. If, conversely, $a\in F\vee G$ then there exists some $b\in F$ and
some $c\in G$ with $b\vee c=a$ and hence $a\in F\cap G$. This shows $F\cap
G=F\vee G$. Now the following are equivalent:

$\displaystyle F\cap G$ $\displaystyle\subseteq D,$ $\displaystyle F\vee G$
$\displaystyle\subseteq D,$ $\displaystyle x\vee y$ $\displaystyle\in D\text{
for all }x\in F\text{ and all }y\in G,$ $\displaystyle y$ $\displaystyle\in
x^{D}\text{ for all }x\in F\text{ and all }y\in G,$ $\displaystyle y$
$\displaystyle\in\overline{x}\text{ for all }x\in F\text{ and all }y\in G,$
$\displaystyle 1$ $\displaystyle\in x^{00}\vee y^{00}\text{ for all }x\in
F\text{ and all }y\in G,$ $\displaystyle F$
$\displaystyle\subseteq\overline{G}.$

(iii) $\Rightarrow$ (2):
For all $x,y\in L$ the following are equivalent: $x\vee y\in D$; $F_{x\vee
y}\subseteq D$; $F_{x}\cap F_{y}\subseteq D$;
$F_{x}\subseteq\overline{F_{y}}$; $x\in\overline{y}$; $1\in x^{00}\vee
y^{00}$. ∎

The results of the previous theorem can be checked in the following example.

######  Example 3.7.

Consider the non-distributive lattice $\mathbf{L}$ depicted in Fig. 5:

$0$$a$$b$$c$$d$$1$Fig. 5Non-distributive non-pseudocomplemented $D$-Stonean
lattice

We have

$\begin{array}[]{r|r|r|r|r|r|r}x&0&a&b&c&d&1\\\ \hline\cr
x^{0}&1&bc&cd&bd&bc&0\\\ \hline\cr x^{00}&0&d&b&c&d&1\\\
\hline\cr\overline{F_{x}}=F_{x}^{D}=\overline{x}=x^{D}&F_{1}&bc1&acd1&abd1&bc1&F_{0}\end{array}$

$D=F_{1}$ and $S=\\{0,b,c,d,1\\}$. Hence $\mathbf{L}$ is not
pseudocomplemented, but it is $D$-Stonean. Moreover, for $x,y\in L$ both
$F_{x}\cap F_{y}\subseteq D$ and $F_{x}\subseteq\overline{F_{y}}$ are
equivalent to

$1\in\\{x,y\\}\text{ or }(x,y\in\\{a,b,c,d\\}\text{ and }x\neq y\text{ and
}\\{x,y\\}\neq\\{a,d\\})$

in accordance with Theorem 3.6.

##  4 Coherent and closed filters

We are going to introduce the so-called coherent filters. For this, let us
define the operator $c$ on filters of $\mathbf{L}$ as follows:

$c(F):=\\{x\in L\mid\overline{x}\wedge F=L\\}.$

Now we define

######  Definition 4.1.

A filter $F$ of a bounded lattice $\mathbf{L}=(L,\vee,\wedge,0,1)$ satisfying
the ACC is called coherent if $c(F)=F$.

One can easily show that if $F$ and $G$ are filters of $\mathbf{L}$ with
$F\subseteq G$ then $c(F)\subseteq c(G)$. Hence $c(F\cap G)\subseteq c(F)\cap
c(G)$ for all filters $F,G$ of $\mathbf{L}$.

######  Example 4.2.

Consider the lattice from Fig. 2. Then we have

$\begin{array}[]{r|r|r|r|r|r|r|r|r|r}x&0&a&b&c&d&e&f&g&1\\\ \hline\cr
c(F_{x})&F_{0}&F_{e}&F_{e}&F_{e}&F_{e}&F_{e}&F_{e}&F_{e}&F_{e}\end{array}$

and hence $F_{0}$ and $F_{e}$ are the only coherent filters. The filters
$F_{a}$ and $F_{f}$ of the lattice depicted in Fig. 3 are coherent, but the
filter $F_{b}$ is not since $b\in F_{b}\setminus c(F_{b})$.

######  Lemma 4.3.

If $\mathbf{L}=(L,\vee,\wedge,0,1)$ is a bounded lattice satisfying the ACC
and $F$ a filter of $\mathbf{L}$ satisfying $\overline{x}\cup F=L$ for all
$x\in F$ then $F\subseteq c(F)$.

###### Proof.

Let $a\in F$ and $b\in L$. Then $1\in F\cap\overline{a}$ and either
$b\in\overline{a}$ and hence $b=b\wedge 1\in\overline{a}\wedge F$ or $b\in F$
and hence $b=1\wedge b\in\overline{a}\wedge F$. This shows $a\in c(F)$ and
therefore $F\subseteq c(F)$. ∎

Since $\overline{x}=L$ for all $x\in D$ we have $F\subseteq c(F)$ for all
filters $F$ of $\mathbf{L}$ being contained in $D$.

It is easy to see that if a lattice $\mathbf{L}$ is distributive then $c(F)$
is closed under $\wedge$ for every filter $F$ of $\mathbf{L}$. However, we can
show that this holds for not necessarily distributive lattices under a weak
condition.

######  Proposition 4.4.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC
and $F$ a proper filter of $\mathbf{L}$ and $a,b\in L$. Then the following
holds:

  1. (i)

If $c(F)$ is closed with respect to $\wedge$ then $c(F)$ is a $D$-filter of
$F$,

  2. (ii)

the inclusion $\overline{a}\wedge\overline{b}\subseteq\overline{a\wedge b}$
holds if and only if $x,y\in L$ and $1\in(x^{00}\vee a^{00})\wedge(y^{00}\vee
b^{00})$ together imply $1\in(x\wedge y)^{00}\vee(a\wedge b)^{00}$,

  3. (iii)

if $\overline{x}\wedge\overline{y}\subseteq\overline{x\wedge y}$ for all
$x,y\in c(F)$ then $c(F)$ is closed with respect to $\wedge$ and hence a
$D$-filter of $\mathbf{L}$.

###### Proof.

  1. (i)

If $a\in D$ then $a^{00}=1$ and hence $\overline{a}=L$ whence $L=L\wedge
1\subseteq L\wedge F=\overline{a}\wedge F\subseteq L$, i.e.
$\overline{a}\wedge F=L$ showing $a\in c(F)$. Therefore $D\subseteq c(F)$. If
$b\in c(F)$, $c\in L$ and $b\leq c$ then $b^{00}\leq_{1}c^{00}$ and hence
$L=\overline{b}\wedge F\subseteq\overline{c}\wedge F\subseteq L$ which implies
$\overline{c}\wedge F=L$, i.e. $c\in c(F)$.

  2. (ii)

We have

$\displaystyle\overline{a}\wedge\overline{b}$ $\displaystyle=\\{x\wedge y\mid
x,y\in L\text{ and }1\in(x^{00}\vee a^{00})\wedge(y^{00}\vee b^{00})\\},$
$\displaystyle\overline{a\wedge b}$ $\displaystyle=\\{x\in L\mid 1\in
x^{00}\vee(a\wedge b)^{00}\\}.$

  3. (iii)

if $\overline{x}\wedge\overline{y}\subseteq\overline{x\wedge y}$ for all
$x,y\in c(F)$ then for all $x,y\in c(F)$ we have

$L=L\wedge L=(\overline{x}\wedge F)\wedge(\overline{y}\wedge
F)=(\overline{x}\wedge\overline{y})\wedge(F\wedge
F)=(\overline{x}\wedge\overline{y})\wedge F\subseteq\overline{x\wedge y}\wedge
F\subseteq L$

and hence $\overline{x\wedge y}\wedge F=L$ which means nothing else than
$x\wedge y\in c(F)$.

∎

The condition in (ii) of Proposition 4.4 holds for all proper filters of the
lattice from Fig. 2.

For $D$-Stonean lattices we can prove the following result.

######  Theorem 4.5.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a $D$-Stonean lattice and $F$ a
$D$-filter of $\mathbf{L}$. Then $c(F)\subseteq F$.

###### Proof.

Let $a\in c(F)$. Then $\overline{a}\wedge F=L$ and hence there exists some
$b\in\overline{a}$ and some $c\in F$ with $b\wedge c=a$. Now

$\overline{a}=\\{x\in L\mid 1\in x^{00}\vee a^{00}\\}=\\{x\in L\mid x\vee a\in
D\\}$

and hence $b=b\vee(b\wedge c)=b\vee a\in D\subseteq F$ which implies
$a=b\wedge c\in F$. ∎

Let us note that the condition that $\mathbf{L}$ is $D$-Stonean is only
sufficient but not necessary.

######  Example 4.6.

The filter $F_{a}$ of the non-Stonean lattice visualized in Fig. 2 is not
coherent since $a\in F_{a}\setminus c(F_{a})$, but $c(F_{a})=F_{e}\subseteq
F_{a}$.

Combining Lemma 4.3 and Theorem 4.5 we obtain

######  Corollary 4.7.

If $\mathbf{L}=(L,\vee,\wedge,0,1)$ is a $D$-Stonean lattice and $F$ a
$D$-filter of $\mathbf{L}$ satisfying $\overline{x}\cup F=L$ for all $x\in F$
then $F$ is coherent.

Now we turn our attention to the so-called closed filters. A filter $F$ of
$\mathbf{L}$ is called closed if it is a closed subset of $L$ as defined in
the introduction, i.e. if $\overline{\overline{F}}=F$. Of course,
$F\subseteq\overline{\overline{F}}$ holds for every filter $F$ of
$\mathbf{L}$.

######  Example 4.8.

The filter $F_{a}$ of the $D$-Stonean lattice visualized in Fig. 5 is closed
and coherent since $\overline{\overline{F_{a}}}=\overline{\\{b,c,1\\}}=F_{a}$
and $c(F_{a})=F_{a}$.

######  Corollary 4.9.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC
and $A\subseteq L$. Then the following holds:

  1. (i)

$D\subseteq\overline{A}$,

  2. (ii)

$\overline{A}=L$ if and only if $A\subseteq D$,

  3. (iii)

every closed filter of $\mathbf{L}$ is a $D$-filter.

###### Proof.

  1. (i)

and (ii) We use Lemma 1.1.

  2. (iii)

This follows from (i).

∎

######  Example 4.10.

The filter $F_{e}$ of the lattice depicted in Fig. 2 is both coherent and
closed since $c(F_{e})=F_{e}$ and
$\overline{\overline{F_{e}}}=\overline{F_{0}}=F_{e}$, but its subfilter
$F_{f}$ is neither coherent nor closed since $e\in c(F_{f})\setminus F_{f}$
and $\overline{\overline{F_{f}}}=\overline{F_{0}}=F_{e}\neq F_{f}$.

######  Remark 4.11.

Since the set of closed subsets of $L$ is closed under arbitrary intersection,
the same is true for the set of closed filters and the latter forms a complete
lattice with respect to inclusion. Moreover, for every filter $F$ of
$\mathbf{L}$ there is a smallest closed filter of $\mathbf{L}$ including it.

##  5 Maximal, prime and median filters

First let us recall some well-known classes of lattice filters. We say that a
filter $F$ of a lattice $(L,\vee,\wedge)$ is

  * •

proper if $F\neq L$,

  * •

maximal if it is a maximal proper filter,

  * •

prime if $x,y\in L$ and $x\vee y\in F$ imply $x\in F$ or $y\in F$.

It is well-known (see e.g. [8]) that every maximal filter of a distributive
lattice is prime. Unfortunately, this does not hold for non-distributive
lattices. For example, the filter $F_{a}$ of the lattice in Fig. 5 is maximal,
but it is not prime since $b\vee c=1\in F_{a}$, but $b,c\notin F_{a}$.
However, we can prove the following:

######  Theorem 5.1.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC
and $F$ a proper filter of $\mathbf{L}$. Then the following holds:

  1. (i)

$F$ is maximal if and only if $x^{0}\cap F\neq\emptyset$ for all $x\in
L\setminus F$,

  2. (ii)

if $F$ is maximal then $F$ is a $D$-filter,

  3. (iii)

if $F$ is maximal and $(x\vee y)\wedge z=0$ for all $x,y\in L\setminus F$ and
all $z\in F$ with $x\wedge z=y\wedge z=0$ then $F$ is prime.

###### Proof.

  1. (i)

Assume $F$ to be maximal and $a\in L\setminus F$. Then $G:=\\{x\in
L\mid\text{there exists some }f\in F\text{ with }f\wedge a\leq x\\}$ is a
filter of $\mathbf{L}$ strictly including $F$. Since $F$ is maximal we
conclude $G=L$. Hence $0\in G$, i.e. there exists some $g\in F$ with $g\wedge
a\leq 0$, it means $g\wedge a=0$. Therefore there exists some $b\in a^{0}$
with $g\leq b$. Since $F$ is a filter we conclude $b\in F$ and hence $b\in
a^{0}\cap F$ whence $a^{0}\cap F\neq\emptyset$. Conversely, assume $x^{0}\cap
F\neq\emptyset$ for all $x\in L\setminus F$. Suppose $F$ not to be maximal.
Then there exists a proper filter $H$ of $\mathbf{L}$ strictly including $F$.
Let $c\in H\setminus F$. Then $c^{0}\cap F\neq\emptyset$. Let $d\in c^{0}\cap
F$. Then $c,d\in H$. Since $H$ is a filter of $\mathbf{L}$ we have $0=c\wedge
d\in H$ and hence $H=L$, a contradiction. This shows that $F$ is maximal.

  2. (ii)

Assume $F$ to be maximal, but not being a $D$-filter. Then there exists some
$a\in D\setminus F$. According to (i) $0\cap F=a^{0}\cap F\neq\emptyset$ and
hence $0\in F$ which implies $F=L$, a contradiction. Therefore $F$ is a
$D$-filter.

  3. (iii)

Assume $F$ to be maximal and $(x\vee y)\wedge z=0$ for all $x,y\in L\setminus
F$ and all $z\in F$ with $x\wedge z=y\wedge z=0$. Suppose $F$ to be not prime.
Then there exist some $a,b\in L\setminus F$ with $a\vee b\in F$. According to
(i) we have $a^{0}\cap F,b^{0}\cap F\neq\emptyset$. Let $f\in a^{0}\cap F$ and
$g\in b^{0}\cap F$ and put $h:=f\wedge g$. Then $h\in F$ and $a\wedge
f=b\wedge g=0$ and hence $a\wedge h=b\wedge h=0$. By the above assumption we
conclude $0=(a\vee b)\wedge h\in F$ which implies that $F$ is not proper which
is a contradiction. Hence $F$ is prime.

∎

Of course, condition (iii) of Theorem 5.1 holds for any distributive lattice.
However, the following example shows that it may be satisfied also in a non-
distributive lattice.

######  Example 5.2.

The non-distributive lattice visualized in Fig. 6

$0$$a$$b$$c$$d$$1$Fig. 6Non-distributive lattice

satisfies the condition in (iii) of Theorem 5.1 for the maximal proper filter
$F_{a}$. In accordance with this theorem, this filter is prime.

In the following we adopt a certain modification of the concept of a median
filter defined in [6].

######  Definition 5.3.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC.
The filter $F$ of $\mathbf{L}$ is called median if it is maximal and if for
each $x\in F$ there exists some $y\in L\setminus F$ with $1\in x^{00}\vee
y^{00}$.

######  Example 5.4.

The filter $F_{a}$ of the non-Stonean lattice $(L,\vee,\wedge,0,1)$ visualized
in Fig. 3 is median, closed and coherent since $b\in L\setminus F_{a}$, $1\in
x^{00}\vee b^{00}$ for every $x\in F_{a}$ and
$\overline{\overline{F_{a}}}=\overline{\\{b,c,d,f,g,1\\}}=F_{a}$. The filter
$F_{a}$ of the non-Stonean lattice $(L,\vee,\wedge,0,1)$ depicted in Fig. 4 is
prime, median, closed and coherent since $c\in L\setminus F_{a}$, $1\in
x^{00}\vee c^{00}$ for all $x\in F_{a}$ and
$\overline{\overline{F_{a}}}=\overline{F_{c}}=F_{a}$. We list all closed
filters of the lattices from Fig. 1 to Fig. 5:

$\begin{array}[]{c|l}\text{Fig.}&\text{closed filters}\\\ \hline\cr
1&F_{0},F_{d},F_{f},F_{1}\\\ \hline\cr 2&F_{0},F_{e}\\\ \hline\cr
3&F_{0},F_{a},F_{d},F_{f}\\\ \hline\cr 4&F_{0},F_{a},F_{c},F_{i}\\\ \hline\cr
5&F_{0},F_{b},F_{c},F_{d},F_{1}\end{array}$

Now we present several basic properties of proper, prime and median filters.

######  Theorem 5.5.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC,
$F$ a proper filter of $\mathbf{L}$ and $a\in L$. Then the following holds:

  1. (i)

If $x^{00}\vee y^{00}\leq_{1}(x\vee y)^{00}$ for all $x,y\in L$, $F$ is a
prime $D$-filter and $a\in L\setminus F$ then ${\overline{a}}\subseteq F$,

  2. (ii)

if $x^{00}\vee y^{00}\leq_{1}(x\vee y)^{00}$ for all $x,y\in L$, $F$ is a
median $D$-filter and $a\in F$ then $\overline{\overline{a}}\subseteq F$,

  3. (iii)

If $a\in F$ then $a^{0}\not\subseteq F$,

  4. (iv)

if $\mathbf{L}$ is $D$-Stonean and $F$ a prime $D$-filter of $\mathbf{L}$ then
$a\in F$ if and only if $a^{0}\not\subseteq F$.

###### Proof.

  1. (i)

Assume $x^{00}\vee y^{00}\leq_{1}(x\vee y)^{00}$ for all $x,y\in L$ and let
$F$ be a prime $D$-filter of $\mathbf{L}$, $a\in L\setminus F$ and
$b\in{\overline{a}}$. Then $1\in a^{00}\vee b^{00}\leq_{1}(a\vee b)^{00}$.
Since $(a\vee b)^{00}$ is an antichain we have $(a\vee b)^{00}=1$ which
implies $(a\vee b)^{0}=(a\vee b)^{000}=0$, i.e. $a\vee b\in D$. Since $F$ is a
$D$-filter, we have $D\subseteq F$ and therefore $a\vee b\in F$. Because
$a\notin F$ and $F$ is prime we conclude $b\in F$ proving
${\overline{a}}\subseteq F$.

  2. (ii)

Assume $x^{00}\vee y^{00}\leq_{1}(x\vee y)^{00}$ for all $x,y\in L$ and let
$F$ be a median $D$-filter of $\mathbf{L}$, $a\in F$ and
$b\in\overline{\overline{a}}$. Since $F$ is median there exists some $c\in
L\setminus F$ with $1\in a^{00}\vee c^{00}$. Hence
$c\in{\overline{a}}=\overline{\overline{\overline{a}}}\subseteq{\overline{b}}$.
Since $c\notin F$ we have ${\overline{c}}\subseteq F$ according to (i) and we
obtain $b\in\overline{\overline{b}}\subseteq{\overline{c}}\subseteq F$ proving
$\overline{\overline{a}}\subseteq F$.

  3. (iii)

If we would have $a^{0}\subseteq F$ for some $a\in F$ then $0=a\wedge
a^{0}\subseteq F$ and hence $F=L$ contradicting the assumption that $F$ is
proper.

  4. (iv)

Suppose $\mathbf{L}$ to be $D$-Stonean, $F$ to be a prime $D$-filter of
$\mathbf{L}$ and $a^{0}\not\subseteq F$. Then there exists some $b\in
a^{0}\setminus F$. Since $\mathbf{L}$ is $D$-Stonean we have $1\in a^{00}\vee
b^{00}$ and hence $a\vee b\in D\subseteq F$ according to the assumption that
$F$ is a $D$-filter. Because $F$ is prime and $b\notin F$ we obtain $a\in F$.
The converse implication follows from (i).

∎

The next result illuminates the relationship between coherent and median
filters.

######  Theorem 5.6.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a bounded lattice satisfying the ACC
and $F$ a maximal filter of $\mathbf{L}$. Then the following holds:

  1. (i)

If $F$ is coherent then it is median,

  2. (ii)

if $\overline{F}\not\subseteq F$ then $F$ is median,

  3. (iii)

if $F=\overline{L\setminus F}$ then $F$ is median.

###### Proof.

Let $a\in F$.

  1. (i)

Assume $F$ to be coherent. Then $a\in c(F)$, i.e. $\overline{a}\wedge F=L$ and
hence there exists some $b\in\overline{a}$ and some $c\in F$ with $b\wedge
c=0$. Because of $b\in\overline{a}$ we have $1\in a^{00}\vee b^{00}$. Now
$b\in F$ would imply $0=b\wedge c\in F$ and hence $F=L$ contradicting the
maximality of $F$. Hence $b\notin F$ showing $F$ to be median.

  2. (ii)

Assume $\overline{F}\not\subseteq F$. Then there exists some
$b\in\overline{F}\setminus F$. Hence $1\in a^{00}\vee b^{00}$ and $b\in
L\setminus F$ proving $F$ to be median.

  3. (iii)

Since $a\in\overline{L\setminus F}$ we have $1\in a^{00}\vee x^{00}$ for all
$x\in L\setminus F$. Because $F$ is proper there exists such an $x$.

∎

An interesting property of filters being both median and prime shown for
distributive pseudocomplemented lattices by M.S. Rao [6] can be proved also in
a more general case.

######  Proposition 5.7.

Let $\mathbf{L}=(L,\vee,\wedge,0,1)$ be a $D$-Stonean lattice, $F$ a median
prime filter of $\mathbf{L}$ and $a,b\in L$. Then the following holds:

  1. (i)

If $F$ is a $D$-filter of $\mathbf{L}$, $a\in F$ and
$\overline{a}=\overline{b}$ then $b\in F$,

  2. (ii)

if $a\vee b\in F$ then there exists some $c\in L\setminus F$ such that
$\\{a\vee c,b\vee c\\}\cap D\neq\emptyset$.

###### Proof.

  1. (i)

Assume $F$ to be a $D$-filter of $\mathbf{L}$, $a\in F$ and
$\overline{a}=\overline{b}$. Since $F$ is median there exists some $c\in
L\setminus F$ with $1\in a^{00}\vee c^{00}$. Hence
$c\in\overline{a}=\overline{b}$, i.e. $1\in b^{00}\vee c^{00}$. Since
$\mathbf{L}$ is $D$-Stonean and $F$ is a $D$-filter we have $b\vee c\in
D\subseteq F$. Because $F$ is prime we conclude $b\in F$.

  2. (ii)

Suppose $a\vee b\in F$. Since $F$ is prime we have $a\in F$ or $b\in F$.
Because $F$ is median, in the first case we see that there exists some $c\in
L\setminus F$ with $1\in a^{00}\vee c^{00}$. Since $\mathbf{L}$ is $D$-Stonean
we obtain $a\vee c\in D$. The case $b\in F$ can be treated analogously.

∎

Data availability statement No datasets were generated or analyzed during the
current study.

## References

  * [1] 9
  * [2] R. Balbes and A. Horn, Stone lattices. Duke Math. J. 37 (1970), 537–545.
  * [3] I. Chajda and H. Länger H, Algebraic structures formalizing the logic with unsharp implication and negation. Logic J. IGPL (submitted).
  * [4] O. Frink, Pseudo-complements in semi-lattices. Duke Math. J. 29 (1962), 505–514.
  * [5] M.S. Rao, Median prime ideals of pseudo-complemented distributive lattices. Arch. Math. (Brno) 58 (2022), 213–226.
  * [6] M.S. Rao, Median filters of pseudo-complemented distributive lattices. Discuss. Math. General Algebra Appl. (to appear).
  * [7] T.P. Speed, On Stone lattices. J. Austral. Math. Soc. 9 (1969), 297–307.
  * [8] G. Szász, Introduction to Lattice Theory. Akadémiai Kiadó, Budapest 1963.

Authors’ addresses:

Ivan Chajda
Palacký University Olomouc
Faculty of Science
Department of Algebra and Geometry
17\. listopadu 12
771 46 Olomouc
Czech Republic
<EMAIL_ADDRESS>
Miroslav Kolařík
Palacký University Olomouc
Faculty of Science
Department of Computer Science
17\. listopadu 12
771 46 Olomouc
Czech Republic
<EMAIL_ADDRESS>
Helmut Länger
TU Wien
Faculty of Mathematics and Geoinformation
Institute of Discrete Mathematics and Geometry
Wiedner Hauptstraße 8-10
1040 Vienna
Austria, and
Palacký University Olomouc
Faculty of Science
Department of Algebra and Geometry
17\. listopadu 12
771 46 Olomouc
Czech Republic
<EMAIL_ADDRESS>