Assessment of an energy-based surface tension model for simulation of two-phase flows using second-order phase field methods

myaddress]Shahab Mirjalilimycorrespondingauthor
[mycorrespondingauthor]Corresponding author
myaddress]Makrand A. Khanwale
myaddress]Ali Mani
[myaddress]Center for Turbulence Research, Stanford University, Stanford, CA 94305, USA

Second-order phase field models have emerged as an attractive option for capturing the advection of interfaces in two-phase flows. Prior to these, state-of-the-art models based on the Cahn-Hilliard equation, which is a fourth-order equation, allowed for the derivation of surface tension models through thermodynamic arguments.

In contrast, the second-order phase field models do not follow a known energy law, and deriving a surface tension term for these models using thermodynamic arguments is not straightforward.
In this work, we justify that the energy-based surface tension model from the Cahn-Hilliard context can be adopted for second-order phase field models as well and assess its performance. We test the surface tension model on three different second-order phase field equations; the conservative diffuse interface model of Chiu and Lin, 2011, and two models based on the modified Allen-Cahn equation introduced by Sun and Beckermann, 2007.

Additionally, we draw the connection between the energy-based model with a localized variation of the continuum surface force (CSF) model.
Using canonical tests, we illustrate the lower magnitude of spurious currents, better accuracy, and superior convergence properties of the energy-based surface tension model compared to the CSF model, which is a popular choice used in conjunction with second-order phase field methods, and the localized CSF model. Importantly, in terms of computational expense and parallel efficiency, the energy-based model incurs no penalty compared to the CSF models.

two-phase flow, phase field, surface tension, diffuse interface, continuous surface force, Allen-Cahn

§ INTRODUCTION

In recent years, there has been considerable interest in second-order phase field equations that are suitable for modeling two-phase flows [Sun and Beckermann, 2007, Chiu and Lin, 2011, Mirjalili et al., 2020, Huang et al., 2020]. Unlike models based on the Cahn-Hilliard equation [Jacqmin, 1999, Abels et al., 2012], these equations are not derived from an underlying thermodynamic basis, and as a result, it is difficult to derive energy-based surface tension terms for these models. Based on the second law of thermodynamics, the Korteweg stress tensor has been used successfully to model surface tension effects in the context of Cahn-Hilliard models [Abels et al., 2012, Khanwale et al., 2020]. In this work, starting from the Korteweg stress tensor, we derive an energy-based surface tension force model for second-order phase field equations. Using canonical test problems on three different phase field equations, we show that the energy-based model has better properties than the continuum surface model (CSF), which is the highly explored alternative [Brackbill et al., 1992, Chiu and Lin, 2011, Mirjalili and Mani, 2021] and a localized variation of CSF. Specifically, the proposed model (1) is accurate and generates low spurious currents, (2) requires lower regularity compared to the direct implementation of the Korteweg stress tensor, (3) does not require computation of curvature, (4) is amenable to a wide range of spatio-temporal discretization schemes. We note regarding the third item that while the proposed surface tension model does not need curvature computation, some second-order phase field equations explicitly involve curvature fields.[color=ForestGreen]R2: #2.1

§ MODEL DERIVATION

To derive the energy-based surface tension model, let us consider a general phase field equation for incompressible flows in the form given by

\begin{equation}
\pd{\phi}{t} + \pd{}{x_j}\left(u_j \phi\right) = g(\phi, \mathbf{x}).
\end{equation}

The role of $g(\phi, \mathbf{x})$ is to regularize the interface. Traditional phase field models derive this right-hand side term such that it minimizes a free-energy functional. Most commonly, the Ginzburg-Landau free energy is utilized, corresponding to the chemical potential $\mu$  [Gurtin, 1996, Abels et al., 2012] given by

\begin{equation}
\mu = \frac{\hat{\sigma}}{\epsilon}\psi^\prime - \hat{\sigma}\epsilon\pd{}{x_j}\left({\pd{\phi}{x_j}}\right),\;\;\text{where}\;\;\psi =  \frac{1}{2}\left[\phi(1-\phi)\right]^2\;\;\text{and}\;\;\psi^\prime =  \phi\left(1-\phi\right)\left(1-2\phi\right).
\label{eqn:ac_mu}
\end{equation}

$\epsilon$ is the interfacial thickness parameter, and $\psi$ is the free energy for a phase field variable, $\phi\in [0,1]$. Additionally, $\hat{\sigma}$ is a constant related to surface energy, given by $\hat{\sigma}=6\sigma$[Note that for Cahn-Hilliard models, instead of the factor 6 we have the factor of $3/2\sqrt{2}\approx 1$. This distinction is due to the difference of the equilibrium profile of $\phi$ in the Cahn-Hilliard based models given by $\tanh{\left(s/\sqrt{2}\epsilon\right)}$ and $\phi \in [-1, 1]$, in contrast to $[1 + \tanh{\left(s/2\epsilon\right)}]/2$ and $\phi \in [0, 1]$ for phase field models considered here ($s$ is the signed distance from the interface).], where $\sigma$ is the surface tension coefficient.  In the special case of the Allen-Cahn model we would have $g(\phi, \mathbf{x}) = -\mu$. Note that the Allen-Cahn equation admits an energy dissipation law similar to the Cahn-Hilliard equation. However, the lack of mass conservation, in addition to the curvature-driven flow it generates, renders it unsuitable for simulation of two-phase flows, motivating the introduction of the alternative second-order phase field models we consider herein [Sun and Beckermann, 2007, Chiu and Lin, 2011, Mirjalili and Mani, 2021].

Historically, models based on the Cahn-Hilliard equation, where $g(\phi, \mathbf{x}) = \nabla^2\mu$, have been used for the simulation of two-phase flows. In this context, one can derive a surface tension model through the second law of thermodynamics. The surface tension force in this model is proportional to the divergence of the so-called Korteweg stress tensor,  $\pd{}{x_j}\left({\pd{\phi}{x_i}\pd{\phi}{x_j}}\right)$, which prescribes an energy dissipation law  [Shen and Yang, 2010, Abels et al., 2012, Guo et al., 2017, Khanwale et al., 2020]. While  this modeling choice for surface tension does not preserve total energy, it imparts stability and robustness to the Cahn-Hilliard-Navier-Stokes system[Jacqmin, 1999].[color=ForestGreen]R2: #2.2
However, the more recently introduced second-order phase field equations for simulating interfacial flows do not admit an energy dissipation law as they are not gradient flows to any known energy functional. As such, these equations have typically been coupled to the momentum transport equation via the CSF model, $F_i=\sigma\kappa\pd{\phi}{x_i}$, where $\kappa$ is the curvature  [Chiu and Lin, 2011, Mirjalili and Mani, 2021], defined as

\begin{equation}
\kappa=-\pd{}{x_j}\left(\frac{\pd{\phi}{x_j}}{\left(\pd{\phi}{x_k}\pd{\phi}{x_k}\right)^{1/2}}\right).
\label{eqn:curvature}
\end{equation}

In this work, we return to the Korteweg stress tensor to assess its suitability for second-order phase field equations. Due to better regularity, the flexibility offered in numerical discretization, and the well-balancedness of volumetric surface tension models compared to integral formulations  [Popinet, 2018], we utilize the equivalent  [Kim, 2005] surface tension force model,

\begin{equation}
\mu\frac{\partial\phi}{\partial{x}_{i}}.
\label{eqn:eb_st}
\end{equation}

While this surface tension force model has been used and derived in the literature for Cahn-Hilliard models [Jacqmin, 1999, Kim, 2005, Abels et al., 2012, Khanwale et al., 2020], for completeness of narrative, we reproduce its derivation here. Starting from the surface tension force based on the Korteweg stress tensor and using the product rule:

\begin{equation}
\begin{split}
\hat{\sigma} \epsilon \pd{}{x_j}\left({\pd{\phi}{x_i}\pd{\phi}{x_j}}\right) &=
\hat{\sigma} \epsilon \pd{\phi}{x_i}\left(\pd{}{x_j}\left({\pd{\phi}{x_j}}\right)\right)
+ \hat{\sigma} \epsilon \pd{\phi}{x_j}\left(\pd{}{x_j}\left({\pd{\phi}{x_i}}\right)\right) \\
& =
\hat{\sigma} \epsilon \pd{\phi}{x_i}\left(\pd{}{x_j}\left({\pd{\phi}{x_j}}\right)\right)
+\hat{\sigma} \epsilon \frac{1}{2}\pd{}{x_i}\left(  \pd{\phi}{x_j} \pd{\phi}{x_j} \right).
\label{eqn:forcing_expan}
\end{split}
\end{equation}

We manipulate this expression to write it in terms of the chemical potential:

\begin{equation}
\begin{split}
\hat{\sigma} \epsilon \pd{}{x_j}\left({\pd{\phi}{x_i}\pd{\phi}{x_j}}\right) &= \hat{\sigma} \epsilon \pd{\phi}{x_i}\left(\pd{}{x_j}\left({\pd{\phi}{x_j}}\right)\right)
+\hat{\sigma} \epsilon \frac{1}{2}\pd{}{x_i}\left(  \pd{\phi}{x_j} \pd{\phi}{x_j} \right)%\\
+ \hat{\sigma} \epsilon \frac{\psi^\prime}{\epsilon^2}\, \pd{\phi}{x_i}
- \hat{\sigma} \epsilon \frac{\psi^\prime}{\epsilon^2} \, \pd{\phi}{x_i},\\
&=   \pd{\phi}{x_i}\left(\frac{\hat{\sigma}}{\epsilon}\epsilon^2\pd{}{x_j}\left({\pd{\phi}{x_j}}\right) -  \frac{\hat{\sigma}}{\epsilon}\psi^\prime \, \right)  %\\
+\hat{\sigma} \epsilon\frac{1}{2}\pd{}{x_i}\left(  \pd{\phi}{x_j} \pd{\phi}{x_j} \right)
+ \hat{\sigma} \epsilon\frac{\psi^\prime}{\epsilon^2}  \, \pd{\phi}{x_i}.
\label{eqn:forcing_expan_f}
\end{split}
\end{equation}

The expression in the parenthesis in the first term can be replaced using the definition of $\mu$ <ref>, which leads to:

\begin{equation}
\begin{split}
\hat{\sigma} \epsilon\pd{}{x_j}\left({\pd{\phi}{x_i}\pd{\phi}{x_j}}\right) &=
+\hat{\sigma} \epsilon\frac{1}{2}\pd{}{x_i}\left(  \pd{\phi}{x_j} \pd{\phi}{x_j} \right)
+ \hat{\sigma} \epsilon\frac{\psi^\prime}{\epsilon^2}  \, \pd{\phi}{x_i},\\
+\pd{}{x_i}\left( \hat{\sigma} \epsilon \frac{1}{2}\pd{\phi}{x_j} \pd{\phi}{x_j}
+ \hat{\sigma} \frac{\psi}{\epsilon}  \,\right).
\end{split}
\label{eqn:forcing_expan_mu}
\end{equation}

Recognize that the last term in <ref> can absorb a redefined pressure.  Therefore, the first term in <ref> can be used as a surface tension force in the momentum equations.

Note that the sign of the surface tension force in <ref> is the opposite of the Korteweg stress tensor. Additionally, the sign of $\mu\pd{\phi}{x_i} $ would be positive on the right-hand side of the momentum equation. It is worth noting that by drawing analogies with Cahn-Hilliard models, a few prior studies have employed  <ref> in conjunction with second-order phase field methods[Abadi et al., 2018, Huang et al., 2020]. This work justifies why this approach is indeed accurate and advantageous compared to using alternative surface tension models.[color=cyan]R1: #1.1

§.§ Considered second-order phase field models

To ensure the generality of our conclusions, we examine the performance of the above energy-based surface tension force on three different second-order phase field equations. We consider (1) the conservative diffuse interface model, denoted by CDI [Chiu and Lin, 2011, Mirjalili et al., 2020], (2) the modified Allen-Cahn model of Sun and Beckermann, 2007, denoted by SB, and (3) an implicit-capable second-order diffuse interface model we denote by ISODI.  These equations are defined in <ref>.

Name     $g\left(\phi\right)$     Description

$\gamma\pd{}{x_j} \left[\epsilon\pd{\phi}{x_j} - \phi\left(1-\phi\right)n_j\right]$
Conservative Diffuse Interface (CDI) model [Chiu and Lin, 2011, Mirjalili et al., 2020]

Sun and Beckermann, 2007

$\gamma\left[ \epsilon\pdd{\phi}{x_j} - \frac{\phi\left(1 - \phi\right)\left(1 - 2\phi\right)}{\epsilon} + \epsilon\left(\pd{\phi}{x_k}\pd{\phi}{x_k}\right)^{1/2} \kappa \right]$
Curvature subtracted Allen-Cahn based model proposed in Sun and Beckermann, 2007

$\gamma\left[ \epsilon\pdd{\phi}{x_j} - \frac{\phi\left(1 - \phi\right)\left(1 - 2\phi\right)}{\epsilon} + \phi(1-\phi) \kappa\right]$
Implicit capable Second Order Diffuse Interface (ISODI) is a modification of model in  Sun and Beckermann, 2007 for full time-implicit schemes

Table describing the second order diffuse interface models under assessment. Note that the interface normal is $n_j = \left.{\pd{\phi}{x_j}}\middle/ {\left(\pd{\phi}{x_k}\pd{\phi}{x_k}\right)^{1/2}}\right.$. [color=ForestGreen, inline]R2: #2.3

Note that the ISODI model is derived by modifying the third term in the model by Sun and Beckermann, 2007 by invoking the equilibrium assumption, $\left({\partial\phi}/{\partial x_k}\;\;{\partial\phi}/{\partial x_k}\right)^{1/2}={\phi\left(1-\phi\right)}/{\epsilon}$.  This allows for an easier setup of Newton iteration for fully-implicit schemes. It is worth highlighting that since the interface is not necessarily at equilibrium in dynamic cases involving strong deformations, ISODI and the phase field model of Sun and Beckermann, 2007 can display different dynamics, which is why we examine both models here.[color=ForestGreen]R2: #2.4

§.§ Relation to a localized continuum surface force model

Before implementing the model proposed in <ref>, we present further analysis of its right-hand side to generate insight regarding its connection and contrast compared to CSF. For second-order phase field equations considered in this work, at equilibrium, the energy-based surface tension model (<ref>) is analytically equivalent to a modified CSF model given by $F_i=\hat{\sigma}\kappa\phi(1-\phi)\pd{\phi}{x_i}$. To prove this, note that at equilibrium, all three models in <ref> yield a hyperbolic tangent profile in the direction normal to the interface, whereby

\begin{equation}
\left(\pd{\phi}{x_k}\pd{\phi}{x_k}\right)^{1/2}=\frac{\phi\left(1-\phi\right)}{\epsilon}.
\label{eqn:tanh_identity}
\end{equation}

Note that we invoke this assumption just to establish a relationship between the energy-based model and the CSF formulations which are commonly used in the literature.
The requirement for <ref> to hold is that the interface should be in equilibrium, which is not necessarily true for dynamic systems.[color=ForestGreen]R2: #2.5

Also, note that, at equilibrium ($g(\phi) = 0$), for the three models presented in <ref>, using <ref>, we have the following equation[For CDI, here we expand the $g(\phi)$ to get $\epsilon\pdd{\phi}{x_j} - \left(\pd{\phi}{x_k}\pd{\phi}{x_k}\right)^{1/2}\frac{\left(1 - 2\phi\right)}{\epsilon} + \phi\left(1-\phi\right) \kappa$],

\begin{equation}
\epsilon\pdd{\phi}{x_j} - \frac{\phi\left(1 - \phi\right)\left(1 - 2\phi\right)}{\epsilon} = - \kappa \phi(1-\phi).
\label{eqn:rhs_eqmb}
\end{equation}

Now multiplying by $\hat{\sigma}$ and using the definition of $\mu$ (<ref>),

\begin{equation}
-\mu =  \hat{\sigma}\epsilon\pdd{\phi}{x_j} - \hat{\sigma}\frac{\phi\left(1 - \phi\right)\left(1 - 2\phi\right)}{\epsilon} = - \hat{\sigma}\kappa \phi(1-\phi).
\end{equation}

Now using <ref> and multiplying by $-\pd{\phi}{x_i}$ we get,

\begin{equation}
\mu \pd{\phi}{x_i} =  \hat{\sigma}\kappa \phi \left(1 - \phi\right)\pd{\phi}{x_i} = \hat{\sigma}\kappa \epsilon \left(\pd{\phi}{x_k}\pd{\phi}{x_k}\right)^{1/2} \pd{\phi}{x_i}.
\label{eqn:equivalence}
\end{equation}

It is important to note that in the right hand side of <ref>, the factor of $\phi(1-\phi)=\epsilon \left(\pd{\phi}{x_k}\pd{\phi}{x_k}\right)^{1/2}$ modifies the popular CSF model ($\hat{\sigma}\kappa\partial\phi/\partial x_i$) with a regularization kernel which is localized near the interface.
We call $\hat{\sigma}\kappa \phi \left(1 - \phi\right)\pd{\phi}{x_i}$, the localized CSF (LCSF) model.
The numerical results below confirm that this localization improves the accuracy of the surface tension model compared to the original CSF model.[color=ForestGreen]R2: #2.6

§ RESULTS AND DISCUSSION

In this section, using canonical tests, we numerically test the energy-based surface tension model by comparing its performance against CSF and LCSF. We use second-order finite differences on a staggered Cartesian grid in space and RK4 time-integration, as described in  Mirjalili and Mani, 2021.

§.§ Spurious currents

We consider a standard 2D benchmark for testing spurious currents generated by two-phase flow solvers [Williams et al., 1998, Popinet and Zaleski, 1999, François et al., 2006, Herrmann, 2008]. In a $1\times1$ domain with free-slip boundary conditions, at $t=0$, a circular drop with diameter $D=0.4$ is placed in the center of the domain with $\vec{u}=0$ everywhere. Note that all numbers and parameters reported in this work are non-dimensional[color=cyan]R2: #1.3. The density and viscosity of the two phases are equal, $\rho_1=\rho_2=300$ and $\mu_1=\mu_2=0.1$, the surface tension coefficient is $\sigma=1.0$, and the dimensionless parameter characterizing this problem is the Laplace number, $La=\sigma \rho D / { \mu }^{ 2 }=12000$. Any sustained flow in the numerical solution is due to errors in computing the surface tension force.

To understand how the spurious currents evolve, in <ref> we examine $Ca_{\infty}=({\norm{\vec{u}}_\infty\mu})/{\sigma}$ as a function of non dimensional time $t^*=\sigma t/( \mu D )$ using the CDI phase field model for all three (CSF, LCSF, and energy-based) surface tension formulations. We observe that after an initial transience, the energy-based model shows consistently lower spurious currents than the CSF and LCSF models. The transient higher values of spurious currents for the EB model at small times can be attributed to the interface being out of equilibrium. Based on these observations, and similar to  [Herrmann, 2008, Mirjalili et al., 2019], we choose to compare the spurious currents at $t^*=250$. Evidently, this is a sufficiently large time for transient effects to have passed and for spurious currents to be fully developed for all three surface tension models due to the balance of surface tension and viscous forces.[color=ForestGreen]R2: #2.7

Panel (a) of  <ref> shows the convergence of $Ca_{\infty}$ at $t^*=250$ as we increase the number of mesh points ($N\times N$) for a fixed interfacial thickness, $\epsilon=1/32$. By successively refining the mesh, this analysis illustrates the property of the models in the continuous limit, thereby teasing out the PDE errors. It is clear that for all three chosen phase field equations, the energy-based model (denoted with EB) outperforms CSF and LCSF due to a better convergence order (second order for EB). It is worth noting that LCSF consistently yields lower spurious currents than CSF. [color=ForestGreen]R2: #2.6

Panel (b) of  <ref> shows the convergence of $Ca_{\infty}$ at $t^*=250$ as we approach the sharp-interface limit. An important property of a physical model is to converge to true sharp interface physics as $\epsilon\rightarrow 0$. As we reduce $\epsilon$, we refine the mesh via ${\Delta x}={32}^{1/2}{\epsilon}^{3/2}$  [Jacqmin, 1999, Mirjalili et al., 2019]. In practice, to allow an integer number of mesh points, we successively refine the mesh by a factor of two while computing $\epsilon$ proportional to $\Delta x^{2/3}$. We observe that for all three phase field equations, the energy-based model converges faster (second order) to the sharp interface limit compared to the CSF and LCSF models. A faster convergence to the sharp interface limit has a practical benefit in terms of computational resource requirements, as it allows for the choice of larger $\epsilon$ values that usually require coarser spatio-temporal resolutions.  LCSF seems to consistently result in lower spurious currents than CSF for these simulations as well, although to a lesser degree than the fixed $\epsilon$ results shown in panel (a).[color=ForestGreen]R2: #2.6

§.§ Standing Wave

By examining surface oscillations of a standing wave, one can assess the accuracy of a two-phase flow solver in capturing interactions among inertial, capillary, and viscous forces [Gerlach et al., 2006, Herrmann, 2008, Mirjalili et al., 2019].

We compare the accuracy of the energy-based surface tension model against the CSF and LCSF models in capturing the amplitude of a 2D standing wave as a function of time. In a $2\pi\times 2\pi$ domain with periodic boundary conditions in the $x$ direction and free-slip on the top and bottom walls, a small-amplitude wave is placed between two immiscible phases at $t=0$. The initial wave height is $h_{\text{wave}}(x,t=0)=y-y_0+A_{0}\cos(x-\Delta x/2)$, where $y_0=\pi$ and $A_0=2\pi/100$. The wavelength is $\lambda=2\pi$. The surface tension is $\sigma=2$, density of the two phases are ${\rho}_{1}={\rho}_{2}=1$, kinematic viscosity of both phases are $\nu_1=\nu_2=0.064720863$, and the time step is $\Delta t=0.003$. We measure the amplitude of the standing wave at $x=\Delta x/2$ as a function of time.

Panels (a), (b), and (c) of  <ref> depict the normalized surface wave amplitude ($A/\lambda$) as a function of time for simulations performed
using CDI, ISODI, and SB phase field methods, respectively.

For panels (a), (b), and (c) of  <ref> we fix the value of $\epsilon=1/32$ and vary the mesh resolution.  We compare the performance of the energy-based surface tension model against CSF and the exact solution (see Prosperetti, 1981) for each resolution. Notice that for all three models (panels (a), (b), and (c)), the energy-based surface tension model (solid lines and denoted by EB) outperforms CSF (dashed lines and denoted by CSF) when compared to the exact solution (except for the lowest resolution simulations with $16\times16$ mesh points).
Additionally, it is important to note that as we refine the mesh, both CSF and the energy-based surface tension model converge to a solution, which implies a minimization of the spatial discretization error. Therefore, for the highest resolution, we predominantly observe a PDE error caused by choice of a finite $\epsilon$.  If we then inspect the converged solutions (purple solid line for the energy-based surface tension model and purple dashed line for the CSF model), it is clear that the PDE-level error committed using the energy-based surface tension model is much smaller than that of the CSF model for all three phase field methods (panels (a), (b), and (c)).  Therefore, we can infer from the results of panels (a), (b), and (c) that for a fixed $\epsilon$, the energy-based surface tension model has a smaller PDE error compared to CSF.

Panel (d) of   <ref> illustrates the convergence of the $l^2$ error computed between the surface wave amplitude and the theoretical solution [Prosperetti, 1981] as we approach the sharp interface limit (i.e., $\epsilon \rightarrow 0$).

While reducing $\epsilon$, we refine the mesh via ${\Delta x}={32}^{1/2}{\epsilon}^{3/2}$.

For all three phase field equations, we observe that the energy-based surface tension force model (symbols with solid lines) is more accurate than the CSF model (symbols with dashed lines) while maintaining its convergence order of one.

For panels (a), (b), and (c) of  <ref> we fix the value of $\epsilon=1/32$ while varying the mesh resolution to compare the performance of the energy-based surface tension model against LCSF and the exact solution (see Prosperetti, 1981). Similar to the comparison in <ref>, for all three phase field equations, the energy-based model (solid lines and denoted by EB) outperforms LCSF (dashed lines and denoted by LCSF) when compared against the exact solution. Furthermore, as we refine the mesh, LCSF and the energy-based surface tension model converge to a solution as the spatial discretization errors converge towards zero. The converged solutions show that the energy-based surface tension model (purple solid line) is much closer to the exact solution (solid black line) compared to the LCSF model (purple dashed line). This implies that the PDE-level error committed by the energy-based surface tension model is much smaller than the LCSF model for all three phase field methods.
[color=ForestGreen]R2: #2.6

Similar to panel (d) of  <ref>, panel (d) of   <ref> compares the convergence of the energy-based model and LCSF as we approach the sharp interface limit (i.e. $\epsilon \rightarrow 0$). We observe that the performance of the energy-based model (symbols with solid lines) is very close to the LCSF model, except for the CDI formulation.
To understand why we see higher errors for CDI for the energy-based model, let us revisit Panel (a); we observe from panel (a) that the energy-based surface tension model outperforms the LCSF model as the interface is progressively resolved with more mesh points, specifically for resolutions more than $64 \times 64$ which corresponds to $\epsilon/\Delta x\ge 2$. For panel (d), we have $\epsilon/\Delta x = \sqrt{1/\left(32\epsilon\right)}$. Therefore, it is only as we approach $\epsilon = 0.0124$ for which $\epsilon/\Delta x \approx 1.58$ that we begin to see the energy-based model accuracy catching up to LCSF. Indeed, at $\epsilon = 0.0124$, we observe that the LCSF model accuracy is flattening, whereas the energy-based model is still converging at the same first-order rate.

[color=ForestGreen]R2: #2.6

In  <ref>, we probe into the energetics of the standing wave problem. Namely, using CDI as the phase field model with $\epsilon=1/32$ on a $64\times64$ mesh, the amplitude evolution for the three different surface tension models is shown in panel (a). The time evolution of the kinetic energy (KE), surface energy (SE), and total energy (TE) for these simulations is shown in panel (b). SE is the system free energy given by $\frac{\hat{\sigma}}{\epsilon}\psi + \hat{\sigma}\epsilon\left({\pd{\phi}{x_j}}\right)\left({\pd{\phi}{x_j}}\right)$. Physically, energy is periodically exchanged between KE and SE while being dissipated by viscous dissipation. It is clear that during this process, compared to the energy-based surface tension model, the CSF and LCSF models exhibit unphysically high amplitude oscillations in  both KE and SE modes, which is correspondingly manifested by larger wave amplitudes in panel (a). The period of the oscillations seems to be also more erroneous for these surface tension models compared to the energy-based model. Overall, from panel (a) in conjunction with panel (b), it is evident that the energy-based model is more accurate than the CSF and LCSF models in flows involving dynamic interactions among surface energy, kinetic energy, and viscous dissipation.[color=ForestGreen]R2:#2.8

§ SUMMARY AND PERSPECTIVE

In this work, we provided theoretical justification for using  <ref> as an energy-based surface tension force model in conjunction with second-order phase field models. We described the connection of the energy-based surface tension model with a localized CSF model( <ref>), introduced in  <ref>. Through extensive testing on two canonical problems, we revealed the relative accuracy of the energy-based surface tension model compared to CSF and LCSF.  These two cases allowed us to assess the accuracy of the model in capturing static equilibrium solutions and dynamic interactions involving the interplay of surface tension forces with inertial and viscous forces. We observed from <ref> that the energy-based model leads to a faster rate of decay for spurious currents (second order) with respect to mesh resolution and interfacial thickness $\epsilon$ (the sharp interface limit) as compared to the CSF and LCSF models. This is analogous to observations from the Cahn-Hilliard model  [Jacqmin, 1999] and is intuitively expected since the equilibrium solution of the second-order phase field models is also a hyperbolic tangent. Additionally, from <ref>,<ref>, and <ref>, we observed that the energy-based surface tension model is more accurate compared to the CSF and LCSF models in dynamically active scenarios involving exchanges between various modes of energy.

Both of these properties are highly sought after for accurate two-phase flow simulations. While our tests were two-dimensional, our findings can be readily generalized to three-dimensional settings. With this short note, we propose the model presented in  <ref> as an alternative to the CSF model for surface tension forces in two-phase flow simulations that utilize second-order phase field equations.

§ ACKNOWLEDGEMENTS

We acknowledge financial support from the Office of Naval Research (Grant N00014-15-1-2523) and Palo Alto Research Center (Grant 249996). The computations in this note were performed on the Yellowstone cluster at the Stanford HPC Center, supported through awards from Intel, the National Science Foundation, DOD HPCMP, and the Office of Naval Research.

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# Identifying Self-Conjugate Partitions

by Rebecca Odom

## Abstract

A partition of a positive integer $n$ is defined as a non-increasing sequence
$P=[y_{0},y_{1},...,y_{m}]$ of positive integers which sum to $n$, where the
$y_{i}$ are called the parts of the partition. A Young diagram is a visual
representation of a partition using rows of boxes, where each row of boxes
corresponds to a part. The conjugate partition is similar to a transpose of a
matrix; we switch the rows with columns, or the index of a part with the part
itself. Self-conjugate partitions are partitions that are equal to their
conjugate; previously, the only known way to verify whether a partition is
self-conjugate was through the use of a Young diagram. In this research, by
proving preliminary lemmas and theorems about easily identifiable shapes which
are symmetric, we come to the main result: by simply adding the multiplicities
of parts appropriately, we can show whether or not a partition is self-
conjugate without the use of a Young diagram.

##  1 Introduction

The term “partition” usually means to parse out a mathematical object, such as
a set. This idea, however, of parsing an object can also be applied to
positive integers. The most natural way to do this is through integer
addition. For example, one can partition the integer $n=14$ as
$14=7+7=13+1=12+2$, and so on. In order to focus solely on the summands of an
integer partition and not the order in which they appear, we require the
summands to appear in descending order. The following is the formal definition
of an integer partition.

######  Definition 1.1.

A partition of a positive integer $n$ is a sequence

$P=[y_{0},y_{1},...,y_{m}]$

of positive integers where $m$ is a nonnegative integer,
$y_{0}+y_{1}+...+y_{m}=n$, and $y_{i}\geq y_{i+1}$ for all $0\leq i<m$. Each
$y_{i}$ is called a part of $P$, and $n=|P|:=y_{0}+y_{1}+...+y_{m}$ is called
the size of $P$.

Note that most textbooks use parentheses instead of brackets when describing a
partition. In this paper, we use brackets to improve the readability of our
proofs.

Below are some examples of partitions.

######  Example 1.1.

We can partition the integer $n=7$ in the following fifteen ways:

  1. $[7]:7=7$

  2. $[6,1]:7=6+1$

  3. $[5,2]:7=5+2$

  4. $[5,1,1]:7=5+1+1$

  5. $[4,3]:7=4+3$

  6. $[4,2,1]:7=4+2+1$

  7. $[4,1,1,1]:7=4+1+1+1$

  8. $[3,3,1]:7=3+3+1$

  9. $[3,2,2]:7=3+2+2$

  10. $[3,2,1,1]:7=3+2+1+1$

  11. $[3,1,1,1,1]:7=3+1+1+1+1$

  12. $[2,2,2,1]:7=2+2+2+1$

  13. $[2,2,1,1,1]:7=2+2+1+1+1$

  14. $[2,1,1,1,1,1]:7=2+1+1+1+1+1$

  15. $[1,1,1,1,1,1,1]:7=1+1+1+1+1+1+1$

There are two ways to visualize a partition that are quite common. The first
is a $Young\ diagram$ and the second is a $Ferrers\ diagram$. A Young diagram
maps a partition of $n$ to a series of rows of boxes, where each part of the
partition becomes a row. A Ferrers diagram is similar, but instead of mapping
to rows of boxes, the partition maps to rows of dots or circles. Let us see an
example. The graphics provided below (and every other figure to follow in the
paper) were created by using Geogebra [3].

######  Example 1.2.

Suppose we have the partition $P=[4,3,2,2,1]$ of $n=12$. Below are the two
visual representations of $P$: the left diagram is a Young diagram, and the
right is a Ferrers diagram.

Notice how each row’s number of boxes or dots corresponds to each part in the
partition $P$ of $n$.

##

There are quite a few things that we can do with partitions. To begin, let us
define the partition function $p(n)$.

######  Definition 1.2.

The partition function $p(n)$ is the number of partitions of $n$.

Now, $p(n)$ can be defined to be zero when $n<0$, and $p(0)=1$, which
references the empty partition with no parts. The next six values of $p(n)$
are as follows, where we will temporarily use an exponent to show repeated
values:

  1. $p(1)=1\ \to\ 1=[1].$

  2. $p(2)=2\ \to\ 2=[2],\ 1+1=[1^{2}].$

  3. $p(3)=3\ \to\ 3=[3],\ 2+1=[2,1],\ 1+1+1=[1^{3}]$

  4. $p(4)=5\ \to\ 4=[4],\ 3+1=[3,1],\ 2+2=[2^{2}],\ 2+1+1=[2,1^{2}],\ 1+1+1+1=[1^{4}].$

  5. $p(5)=7\ \to\ 5=[5],\ 4+1=[4,1],\ 3+2=[3,2],\ 3+1+1=[3,1^{2}],\ 2+2+1=[2^{2},1],$

  6. $\ 2+1+1+1=[2,1^{3}],\ 1+1+1+1+1=[1^{5}].$

  7. $p(6)=11\ \to\ 6=[6],\ 5+1=[5,1],\ 4+2=[4,2],\ 4+1+1=[4,1^{2}],\ 3+3=[3^{2}],$

  8. $3+2+1=[3,2,1],\ 3+1+1+1=[3,1^{3}],\ 2+2+2=[2^{3}],$

  9. $2+2+1+1=[2^{2},1^{2}],\ 2+1+1+1+1=[2,1^{4}],\ 1+1+1+1+1+1=[1^{6}].$

This function grows rapidly in value; for example [1], “$p(10)=42$,
$p(20)=627$, $p(50)=204226$, $p(100)=190569292$, and $p(200)=3972999029388$.”
In fact, this leads us to the historical collaboration between G. H. Hardy and
S. Ramanujan, who together showed what the value of $p(n)$ would be for any
$n$ [2]:

######  Theorem 1.3.

The value of $p(n)$ is the integer nearest

$\frac{1}{2\sqrt{2}}\sum_{q=1}^{v}\sqrt{q}A_{q}(n)\psi_{q}(n),$

where $A_{q}(n)=\sum\omega_{p,q}e^{-2n\rho\pi i/q}$, the sum is over $p<q$
with $\gcd(p,q)=1$, $\omega_{p,q}$ is a certain 24$q$th root of unity, $v$ is
of the order of $\sqrt{n}$, and

$\psi_{q}(n)=\frac{d}{dn}\left(\emph{exp}\left(\frac{C\sqrt{\left(n-\frac{1}{24}\right)}}{q}\right)\right),\
C=\pi\sqrt{\frac{2}{3}}$

This formula was later refined by H. Rademacher to be [4]:

######  Theorem 1.4.

$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}A_{k}(n)k^{\frac{1}{2}}\left[\frac{d}{dx}\left(\frac{\sinh((\pi/k)(2/3\left(x-1/24\right))^{\frac{1}{2}})}{\left(x-1/24\right)^{\frac{1}{2}}}\right)\right]_{x=n},$

where

$A_{k}(n)=\sum_{h\ \emph{mod}\ k}^{\ }\omega_{h,k}e^{-2\pi inh/k},\
\gcd(h,k)=1\ for\ h\ \emph{mod}\ k$

An example of using this second formula for $p(200)$ is as follows, where we
compute the first eight terms:

  1. +3,972,998,993,185.896

  2. +36,282.978

  3. -87.555

  4. +5.147

  5. +1.424

  6. +0.071

  7. +0.000

  8. +0.043

  9. ——————————

  10. 3,972,99,029,388.004

which is, within .004, the correct value for $p(200)$ [1]. However, the proofs
for both formulas are quite lengthy, so we will omit them. Continuing on, the
same Ramanujan and Hardy also discovered divisibility properties concerning
$p(n)$. Ten of these divisibility properties are listed below [1]:

  1. (1)

$p(4),\ p(9),\ p(14),\ p(19),...\equiv 0\ ($mod$\ 5)$

  2. (2)

$p(5),\ p(12),\ p(19),\ p(26),...\equiv 0\ ($mod$\ 7)$

  3. (3)

$p(6),\ p(17),\ p(28),\ p(39),...\equiv 0\ ($mod$\ 11)$

  4. (4)

$p(24),\ p(49),\ p(74),\ p(99),...\equiv 0\ ($mod$\ 25)$

  5. (5)

$p(19),\ p(54),\ p(89),\ p(124),...\equiv 0\ ($mod$\ 35)$

  6. (6)

$p(47),\ p(96),\ p(145),\ p(194),...\equiv 0\ ($mod$\ 49)$

  7. (7)

$p(39),\ p(94),\ p(149),...\equiv 0\ ($mod$\ 55)$

  8. (8)

$p(61),\ p(138),...\equiv 0\ ($mod$\ 77)$

  9. (9)

$p(116),...\equiv 0\ ($mod$\ 121)$

  10. (10)

$p(99),...\equiv 0\ ($mod$\ 125)$

This leads to Ramanujan’s conjecture of the following theorem:

######  Theorem 1.5.

If $\delta=5^{a}7^{b}11^{c}$ and $24\lambda\ \equiv\ 1\ (\emph{mod}\ \delta)$,
then

$p(\lambda),\ p(\lambda+\delta),\ p(\lambda+2\delta),\ ...\ \equiv\ 0\
(\emph{mod}\ \delta).$

However, a counterexample to this conjecture was made by S. Chowla in the
1930s [1]. Specifically, he noticed that

$p(243)=133978259344888\ \not\equiv\ 0\ (\textrm{mod}\ 7^{3}).$

Though his conjecture was incorrect, Ramanujan did provide proofs (or sketches
of proofs) for

$p(5n+4)\ \equiv\ 0\ (\textrm{mod}\ 5),$ $p(7n+5)\ \equiv\ 0\ (\textrm{mod}\
7),$ $p(25n+24)\ \equiv\ 0\ (\textrm{mod}\ 25),$ $p(49n+47)\ \equiv\ 0\
(\textrm{mod}\ 49).$

Furthermore, in 1938, G. N. Watson showed that a “modified version of the
conjecture was true for all powers of 5 and 7” [1]. Later, in 1967 A. O. L.
Atkin gave proof for the true, full modified conjecture:

######  Theorem 1.6.

If $\delta=5^{a}7^{b}11^{c}$ and $24\lambda\ \equiv\ 1\ (\emph{mod}\ \delta)$,
then $p(\lambda)\ \equiv\ 0\ (\emph{mod}\ 5^{a}7^{[(b+2)/2]}11^{c})$.

Clearly, the value of $p(n)$ is more complicated than it first appears. Let us
now consider what happens when we consider subsets of partitions [1]:

Let $\varphi$ denote the set of all partitions, and let $p(S,n)$ denote the
number of partitions of $n$ that belong to a subset $S\subseteq\varphi$.
Furthermore, let $\mathcal{O}$ denote the set of all partitions with only odd
parts and $\mathcal{D}$ denote the set of all partitions with distinct parts.
Below we list partitions in $\mathcal{O}$ and $\mathcal{D}$ where we will,
again, let an exponential represent the multiplicity of a part.

  1. $p(\mathcal{O},1)=1\ \to\ 1=[1].$

  2. $p(\mathcal{O},2)=1\ \to\ 1+1=[1^{2}].$

  3. $p(\mathcal{O},3)=2\ \to\ 3=[3],\ 1+1+1=[1^{3}].$

  4. $p(\mathcal{O},4)=2\ \to\ 3+1=[3,1],\ 1+1+1+1=[1^{4}].$

  5. $p(\mathcal{O},5)=3\ \to\ 5=[5],\ 3+1+1=[3,1^{2}],\ 1+1+1+1+1=[1^{5}].$

  6. $p(\mathcal{O},6)=4\ \to\ 5+1=[5,1],\ 3+3=[3^{2}],\ 3+1+1+1=[3,1^{3}],\ 1+1+1+1+1+1=[1^{6}].$

  7. $p(\mathcal{O},7)=5\ \to\ 7=[7],\ 5+1+1=[5,1^{2}],\ 3+3+1=[3^{2},1],\ 1+1+1+1+1+1+1=[1^{7}].$

  8. $p(\mathcal{D},1)=1\ \to\ 1=[1].$

  9. $p(\mathcal{D},2)=1\ \to\ 2=[2].$

  10. $p(\mathcal{D},3)=2\ \to\ 3=[3],\ 2+1=[2,1].$

  11. $p(\mathcal{D},4)=2\ \to\ 4=[4],\ 3+1=[3,1].$

  12. $p(\mathcal{D},5)=3\ \to\ 5=[5],\ 4+1=[4,1],\ 3+2=[3,2].$

  13. $p(\mathcal{D},6)=4\ \to\ 6=[6],\ 5+1=[5,1],\ 4+2=[4,2],\ 3+2+1=[3,2,1].$

  14. $p(\mathcal{D},7)=5\ \to\ 7=[7],\ 6+1=[6,1],\ 5+2=[5,2],\ 4+3=[4,3],\ 4+2+1=[4,2,1].$

Consider the fact that, for $n\leq 7$, $p(\mathcal{O},n)=p(\mathcal{D},n)$. In
fact, let us prove that this is true for all $n$. First, let us define
generating functions for sequences and notation for subsets of positive
integers:

######  Definition 1.7.

The generating function $f(q)$ for the sequence $a_{0}$, $a_{1}$, $a_{2}$,
$a_{3}$, … is the power series $f(q)=\sum_{n\geq 0}a_{n}q^{n}$.

######  Definition 1.8.

Let $\mathcal{H}$ be a set of positive integers. We let “$\mathcal{H}$” denote
the set of all partitions whose parts lie in $\mathcal{H}$. Consequently,
$p(``\mathcal{H}",n)$ is the number of partitions of $n$ that have all their
parts in $\mathcal{H}$. Thus, if $\mathcal{H}_{0}$ is the set of all odd
positive integers, then “$\mathcal{H}_{0}$”$=\mathcal{O}$. As in,
$p($“$\mathcal{H}_{0}$”$,n)=p(\mathcal{O},n)$.

######  Definition 1.9.

Let $\mathcal{H}$ be a set of positive integers. We let “$\mathcal{H}$”$(\leq
d)$ denote the set of all partitions in which no part appears more than $d$
times and each part is in $\mathcal{H}$. Thus, if $\mathbb{N}$ is the set of
all positive integers, then $p($“$\mathbb{N}$”$(\leq 1),n)=p(\mathcal{D},n)$.

This leads us to a theorem and its proof [1].

######  Theorem 1.10.

Let $\mathcal{H}$ be a set of positive integers, and let

$f(q)=\sum_{n\geq 0}p(``H",n)q^{n},$ $f_{d}(q)=\sum_{n\geq 0}p(``H"(\leq
d),n)q^{n}.$

Then for $|q|<1$

$f(q)=\prod_{n\in\mathcal{H}}(1-q^{n})^{-1}$ (1)
$f_{d}(q)=\prod_{n\in\mathcal{H}}(1+q^{n}+\dots+q^{dn})$ $\ \ \ \ \ \ \ \ \ \
=\prod_{n\in\mathcal{H}}(1-q^{(d+1)n})(1-q^{n})^{-1}$ (2)

Note that the two forms for $f_{d}(q)$ came from the sum of a finite geometric
series:

$1+x+x^{2}+\dots+x^{r}=\frac{1-x^{r+1}}{1-x}\ \
\cite[cite]{[\@@bibref{}{Book}{}{}]}.$

###### Proof.

[1] We wish to prove (1) and (2). First, let us index all elements of
$\mathcal{H}$ such that $\mathcal{H}={h_{1},h_{2},h_{3},h_{4},\dots}$; then we
have

$\prod_{n\in\mathcal{H}}(1-q^{n})^{-1}=\prod_{n\in\mathcal{H}}(1+q^{n}+q^{2n}+q^{3n}+\cdots)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
=(1+q^{h_{1}}+q^{2h_{1}}+q^{3h_{1}}+\cdots)$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\ (1+q^{h_{2}}+q^{2h_{2}}+q^{3h_{2}})$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \times\ (1+q^{h_{3}}+q^{2h_{3}}+q^{3h_{3}}+\cdots)$
$\times\cdots\ \ \ \ \ \ \ \ \ \ $ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ =\sum_{a_{1}\geq 0}\sum_{a_{2}\geq 0}\sum_{a_{3}\geq 0}\cdots
q^{a_{1}h_{1}+a_{2}h_{2}+a_{3}h_{3}+\cdots}$

and the exponent of $q$ is the size of the partition
$[\dots,h_{3}^{a_{3}},h_{2}^{a_{2}},h_{1}^{a_{1}}]$. Therefore, the $q^{n}$
term appears $p(``\mathcal{H}",n)$ times for each $n$ [1]. Thus,

$\prod_{n\in\mathcal{H}}(1-q^{n})^{-1}=\sum_{n\geq
0}p(``\mathcal{H}",n)q^{n}.$

Furthermore, the proof of (2) is identical, except we have a finite geometric
series instead of an infinite one:

$\prod_{n\in\mathcal{H}}(1+q^{n}+q^{2n}+\dots+q^{dn})$ $\ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{d\geq a_{1}\geq 0}\ \sum_{d\geq
a_{2}\geq 0}\ \sum_{d\geq a_{3}\geq 0}\cdots
q^{a_{1}h_{1}+a_{2}h_{2}+a_{3}h_{3}+\cdots}$ $\ \ \ =\sum_{n\geq
0}p(``\mathcal{H}"(\leq d),n)q^{n}.$

∎

######  Corollary 1.11.

_(Euler)._ $p(\mathcal{O},n)=p(\mathcal{D},n)$ for all $n$.

It is evident that the study of partitions is quite diverse, but the focus of
this paper is the conjugate of a partition. If the reader is familiar with
matrix methods or linear algebra, the concept will be familiar since it is
similar to the transpose of a matrix. Taking the conjugate of a partition is
when every part of the partition is switched with its index. This is similar
to finding a transpose of a matrix, since we make every row of the Young
diagram (or Ferrers diagram) a column and every column a row. This is
equivalent to reflecting a partition diagram about its main diagonal through
the first part’s first box or dot. We see this visually in the next example.

######  Example 1.3.

Suppose we are given the same partition $P=[4,3,2,2,1]$ as in Example 1.2. We
want to find its conjugate. We label each column below by color. The first
column of $P$ is purple, and this will become the first part of the conjugate
$P^{\prime}$ of some partition $P$ of $n$. The second column of $P$ is green,
and this will become the second part of $P^{\prime}$. We continue this process
with every column, flipping the Young diagram of $P$ about its diagonal
through the first box of the first part of $P$. Thus, we are left with
$P^{\prime}=[5,4,2,1]$.

Figure 1: $P=[4,3,2,2,1]$ to the left; $P^{\prime}$ to the right.

Suppose we were to find the conjugate $P^{\prime}$ of some partition $P$ of
$n$. If $P=P^{\prime}$, then we say $P$ is self-conjugate. Let us see an
example below:

######  Example 1.4.

Suppose we have the partition $P=[4,3,2,1]$ of $n=10$. We find the conjugate
of $P$ below, using the same method as before, and see that $P=P^{\prime}$.

Figure 2: $P=[4,3,2,1]$ to the left; $P^{\prime}$ to the right.

This poses the question that this paper will answer: is it possible to
determine if a partition $P$ is self-conjugate without the use of a Young or
Ferrers diagram as a visual determinant? In particular, is there something
about the parts themselves, a pattern to be found, that would let us know if
$P$ is self-conjugate?

In Section 2, we prove some preliminary lemmas to begin our investigation of
this question. In Section 3, we prove an intermediate theorem regarding
symmetry of Young diagrams. In Section 4, we introduce the idea of partition
addition and sub-partitions which will be essential in the proof of our main
theorem. In Section 5, we discuss the idea of nesting and prove a few more
lemmas about symmetry of Young diagrams. Finally, in Section 6, we state and
prove our main theorem, which gives a necessary and sufficient condition for a
partition to be self-conjugate based only on the parts and not on the Young
diagram.

## Acknowledgements

Let it be known that without Dr. Madeline Dawsey from the University of Texas
at Tyler, this paper and research would not have been possible. Dr. Dawsey is
the one who started me on this journey through partitions and she has been my
rock and foundation throughout all of this; I am forever grateful and indebted
to her and the time she has given me. Furthermore, I would not be where I am
today without several of the faculty of the mathematics department: Dr. Kassie
Archer, Dr. Scott LaLonde, Dr. Katie Anders, Dr. Joseph Vandehey, Dr. Debbie
Koslover, Dr. Sheldon Davis, Dr. Alex Bearden, Dr. Regan Beckham, Dr. David
Milan, Dr. Christy Graves, and Dr. Steve Graves. Each of these professors are
masters of their craft and have imbued me with the knowledge and skills to be
successful both in mathematics and the world. I am far from the easiest
student to instruct, and despite that, they all made my cup of success
overflow. The mathematics department of the University of Texas at Tyler has
my full gratitude and recommendation; there is no better place nor better
group of fantastic people from which to learn the terrifying beauty that is
the world of mathematics. Last, but not least, I would like to thank my
grandparents and dedicate this paper to them: Benny and Linda Huff. In a full
evaluation, it would be most obvious my life has been more full of turmoil
than of joy, and yet, my grandparents have seen me through each dark moment; I
owe every drop of success and goodness in me to the two of them.

##  2 Simple Lemmas and Counting

Before we introduce our lemmas and their proofs, let us see some examples and
non-examples of self-conjugate partitions with their respective Young
diagrams.

######  Example 2.1.

$P=P^{\prime}=[5,5,5,5,5]$ versus
$Q=[6,5,5,5,5]\not=Q^{\prime}=[5,5,5,5,5,1]$.

Figure 3: Left:$P=P^{\prime}$, Center: $Q$, Right: $Q^{\prime}$.

######  Example 2.2.

$P=P^{\prime}=[5,4,3,2,1]$ versus $Q=[4,4,3,2,1]\not=Q^{\prime}=[5,4,3,2]$.

Figure 4: Left:$P=P^{\prime}$, Center: $Q$, Right: $Q^{\prime}$.

######  Example 2.3.

$P=P^{\prime}=[5,1,1,1,1]$ versus $Q=[5,1,1,1]\not=Q^{\prime}=[4,1,1,1,1]$.

Figure 5: Left:$P=P^{\prime}$, Center: $Q$, Right: $Q^{\prime}$.

In all of these examples, it is easy to see that the first part of $P$ must
equal the number of parts of $P$ when $P=P^{\prime}$; otherwise, the conjugate
of our partition will not have the correct number of parts. This leads us to
our first lemma.

###  2.1 Size Equality

######  Lemma 2.1.

If a partition is self-conjugate, then the first part is equal to the number
of parts.

###### Proof.

Suppose $m,\ n\in\mathbb{N}$. Let $P=[n,\ a,\ b,...,\ c]$ be a partition that
is self-conjugate. Let $P$ have length $\ell(P)=m$ (as in, $P$ has $m$ parts).

  1. 1.

Suppose $m>n$. Then the first row of the Young diagram will be $n$-long for
the original partition $P$ and $m$-long for the conjugate partition
$P^{\prime}$. But, since $m>n$, this contradicts $P$ being self-conjugate
since the first parts of the original and conjugate partition are not equal.

  2. 2.

Suppose $m<n$. We get a similar contradiction as before since the first row of
the Young diagram will be $n$-long for the original partition $P$ and $m$-long
for the conjugate partition $P^{\prime}$, and thus, the first parts are not
equal.

Thus, since $m\not>n$ and $m\not<n$, and $m,\ n\in\mathbb{N}$, then we have
$m=n$ when $P$ is a self-conjugate.
∎

From now on, we only want to consider partitions of dimension
$d\in\mathbb{N}$, where we say a partition has $dimension$ $d$ if the first
part equals the number of parts equals $d$. Lemma 2.1 is not an if and only if
statement, but we do not want to consider partitions where the number of parts
does not equal the first part, because those partitions clearly cannot be
self-conjugate. All partitions of dimension $d$ that are self-conjugate fit
into a square, $d\times d$ Young diagram. In fact, the following section
provides a formula for the number of dimension $d$ partitions for any positive
integer $d$.

###  2.2 Counting

In the world of mathematics, we often concern ourselves with how many
mathematical objects we have of certain types. We now wish to see how many
self-conjugate partitions of given a dimension $d$ there are. First, we count
(where the main color represents the partition and the secondary color
represents what we have taken away from the square $d\times d$ Young diagram
to obtain the partition):

  1. $d=1:$ Figure 6: Self-conjugate partition of dimension 1.

There are $1=2^{0}=2^{1-1}$ self-conjugate partitions (red).

  2. $d=2:$

Figure 7: Self-conjugate partitions of dimension 2.

There are $2=2^{1}=2^{2-1}$ self-conjugate partitions (orange).

  3. $d=3:$

Figure 8: Self-conjugate partitions of dimension 3.

There are $4=2^{2}=2^{3-1}$ self-conjugate partitions (blue).

  4. $d=4:$

Figure 9: Self-conjugate partitions of dimension 4.

There are $8=2^{3}=2^{4-1}$ self-conjugate partitions (pink).

The following theorem provides a formula for the number of sel-conjugate
partitions of dimension $d\geq 3$.

######  Theorem 2.2.

There are

$2^{d-1}=2+\sum_{i=1}^{d-2}2^{i}$

self-conjugate partitions of dimension $d$, for $d\geq 3$.

###### Proof.

First, let us prove that $2^{k}=2+\sum_{i=1}^{k-1}2^{i}$. Notice:

  1. $2+\sum_{i=1}^{k-1}2^{i}$

  2. $=$

$2+2+\sum_{i=2}^{k-1}2^{i}$

  3. $=$

$4+\sum_{i=2}^{k-1}2^{i}=2^{2}+4+\sum_{i=3}^{k-1}2^{i}$

  4. $=$

$8+\sum_{i=3}^{k-1}2^{i}=2^{3}+\sum_{i=3}^{k-1}2^{i}$

  5. $\vdots$

  6. $=$

$2^{k-2}+\sum_{i=k-2}^{k-1}2^{i}=2^{k-2}+2^{k-2}+\sum_{i=k-1}^{k-1}2^{i}$

  7. $=$

$2(2^{k-2})+2^{k-1}=2^{k-1}+2^{k-1}=2(2^{k-1})$

  8. $=$

$2^{k}$

Thus, the claim that $2^{k}=2+\sum_{i=1}^{k-1}2^{i}$ holds true (for $k\geq
2$). Now, we must show that for dimension $d$, there are $2^{d-1}$ self-
conjugate partitions. From the pictures above, there are two things that are
readily clear: for every dimension $d\geq 2$, there will be a partition that
has no squares removed and a partition that has only one square removed from
its Young diagram. In other words, for dimension $d\geq 2$, we will have
partitions of the integers $d^{2}$ and $d^{2}-1$, which each have unique
representations in a given dimension $d$. The rest of the self-conjugate
partitions relate to the rest of the previous dimensions, where we remove
squares in the pattern of the previous self-conjugate partitions. Thus, let us
say $x_{d}$ is the number of self-conjugate partitions of dimension $d$; we
have $x_{d}=x_{d-1}+x_{d-2}+x_{d-3}+...+x_{d-(d-3)}+x_{d-(d-2)}+2$. So, our
proof becomes a proof by induction.

  1. Base case: Suppose $d=3$. From the work before, we have that there are $2^{2}=2+\sum_{i=1}^{2-1}2^{i}=2+2=4$ self-conjugate partitions.

  2. Induction: Suppose for all $d=j\leq k$, there are $2^{j-1}$ self-conjugate partitions of dimension $d=j$. We wish to find how many self-conjugate partitions there are for $d=k+1$. From our work before, we know $2^{k-1}=2+\sum_{i=1}^{k-2}2^{i}$. We also know that the number of self-conjugate partitions of dimension $d=k+1$ is $x_{k+1}=2+\sum_{j=2}^{k}x_{j}$. Thus, we write:

    1. $x_{k+1}$

$=2+\sum_{j=2}^{k}x_{j}$

    2. $=2+x_{k}+\sum_{j=2}^{k-1}x_{j}$

    3. $=2+2^{k-1}+\sum_{j=2}^{k-1}x_{j}$

    4. $=2+2^{k-1}+x_{k-1}+\sum_{j=2}^{k-2}x_{j}$

    5. $=2+2^{k-1}+2^{k-2}+\sum_{j=2}^{k-2}x_{j}$

    6. $\vdots$

    7. $=2+2^{k-1}+2^{k-2}+2^{k-3}+...+2^{k-(k-2)}+2^{k-(k-1)}$

    8. $=2+\sum_{i=1}^{k-1}2^{i}$

    9. $=2^{k}$

This proves the theorem. ∎

One special type of partition which we noticed above and will discuss next is
a partition whose Young diagram is a perfect square. This type of partition is
related to the Durfee square of a partition, the largest perfect square that
can fit inside the Young diagram of the partition. In particular, we will
focus next on partitions whose Young diagrams are equal to their own Durfee
squares. We will refer to this type of partition as a “pure Durfee square”,
and we will show that pure Durfee squares of any dimension are self-conjugate.

###  2.3 Durfee Square

######  Lemma 2.3.

Suppose the first part of a partition $P$ is equal to the number of parts of
$P$, and all parts of $P$ are equal; namely, $P$ is a pure Durfee square. Then
$P$ is self-conjugate.

###### Proof.

Suppose $P$ is a pure Durfee square. Then we have that $P=[n,\ n,...,\ n]$
with $n$ parts. This means all parts of the conjugate partition $P^{\prime}$
will be of size $n$, since every column and every row equals $n$. Thus,
$P=P^{\prime}=[n,\ n,...,\ n]$, so $P$ is a self-conjugate partition.
∎

The self-conjugate partition $P$ from Example 2.2 is also a special type of
self-conjugate partition: a “fancy triangle”. In the pure Durfee square case,
$P=[n,\ n,...,\ n]$ with $n$ parts is self-conjugate, whereas in the fancy
triangle case, the pattern that holds is a series of perfectly descending
parts. Explicitly, a partition $P$ is a fancy triangle if
$P=[n,n-1,n-2,...,2,1]$. We now show this type of partition is always self-
conjugate as well.

###  2.4 Fancy Triangle

######  Lemma 2.4.

Suppose the first part of a partition $P$ is equal to the number of parts, and
each subsequent part is exactly one less than the previous part; namely, $P$
is a fancy triangle. Then P is self-conjugate.

###### Proof.

Suppose $P$ is a fancy triangle:

$P=[n,\ n-1,\ n-2,...,\ n-(n-1)].$

The first row of the Young diagrams of both $P$ and its conjugate $P^{\prime}$
are $n$-long by assumption. Thus, there is symmetry along the first (outer)
row and column of the Young diagram. Now, remove the first row and column of
$P$ to form the new partition

$Q=[(n-1),\ (n-1)-1,\ (n-1)-2,...,\ (n-1)-((n-1)-1)].$

Let $m=n-1$. Then

$Q=[m,\ m-1,\ m-2,...,\ m-(m-1)].$

So, $Q$ has $m=n-1$ parts and each part is exactly one less than the previous.
Thus, $Q$ is another fancy triangle, and the first row of the Young diagrams
of both $Q$ and its conjugate $Q^{\prime}$ are $m$-long, so $P$ has symmetry
along the first (outer) two rows and columns. Continue inductively until we
get to one of two cases depending on whether $n$ is even or odd: a partition
$Z$ that either is $Z=[1]$ or $Z=[2,1]$. Both cases are trivially self-
conjugate, since $Z=[1]=Z^{\prime}$ is symmetric by Lemma 2.3 and
$Z=[2,1]=Z^{\prime}$ is the base case of a symmetric fancy triangle. Thus, the
claim of Lemma 2.4 holds true.
∎

There is one last special self-conjugate partition we wish to discuss: the
“fancy L”. This partition only has one row and one column in its Young
diagram, and both have equal length. Thus, we will have a partition with the
first part $n$ and all other parts 1. We saw an example of this in Example
2.3.

###  2.5 Fancy L

######  Lemma 2.5.

Suppose $P=[n,1,1,...,1]$ with $n$ parts; namely, $P$ is a fancy L. Then $P$
is self-conjugate.

###### Proof.

Suppose $P=[n,1,1,...,1]$ with $n$ parts. The first row of the Young diagram
of the conjugate $P^{\prime}$ is $n$-long because $P$ has $n$ parts.
Furthermore, $P^{\prime}$ has $n$ parts, with one part of size $n$ followed by
$n-1$ parts of size 1, because only the first row of the Young diagram of $P$
is $n$-long and the rest of the parts are of size 1. Thus, the first (outer)
row and column of $P$ is symmetric. Since there are no other boxes to consider
in the Young diagrams of either $P$ or $P^{\prime}$, our claim for Lemma 2.5
holds true.
∎

##  3 Piecewise Theorem

We now change direction slightly and prove a seemingly unrelated theorem. The
idea of this section is that if a partition is self-conjugate, then the area
on either side of the diagonal in a Young diagram is equal; this theorem and
its proof will help us in the proof of the main theorem later.

######  Theorem 3.1.

Let $\phi$ be a map on the set of all partitions such that if
$P=[z_{0},z_{1},...,z_{m}]$, then $\phi(P)$ is given by the piecewise function

$\phi:\ P\to f(x)=\begin{cases}z_{0}&\mathrm{if\ }0\leq x<1\\\
z_{1}&\mathrm{if\ }1\leq x<2\\\ \vdots&\\\ z_{m}&\mathrm{if\ }m-1\leq
x<m.\end{cases}$

Let $g(x)=x$. If $P$ is self-conjugate, then the area of the region bounded by
$f$, $g$, and the $x$-axis is equal to the area of the region bounded by $f$,
$g$, and the $y$-axis.

###### Proof.

Suppose $P$ is a self-conjugate partition, and suppose you calculated the
areas of the regions defined above by integration.

Figure 10: Superimposing areas.

Assume by way of contradiction that when you superimpose one region on the
other, by reflecting one across $g$, one has more area. Without loss of
generality, assume the region bounded by the x-axis has more area.

Figure 11: Comparing bounds.

Either the first part does not equal the number of parts in $P$, failing Lemma
2.1; or, for some $i\in\\{1,\ 2,...,\ m\\}$, and for $(i-1)\leq x<i$, we have
$f(x)=j$, and for some $(j-1)\leq x\textrm{*}<j$ and $j\leq(i-1)$, we have
$f(x\textrm{*})=k$, where $k<i$ (see Figure 11 for an illustration).

This implies the conjugate of $P$, $P^{\prime}$, will not equal $P$, since
$f(x\textrm{*})=k$ for $P$, but $f(x\textrm{*})=i$ for $P^{\prime}$ and $k<i$.
Thus, we have have a contradiction of our assumption that $P=P^{\prime}$.

Similarly, if the region bounded by the $y$-axis had more area, we would draw
a similar conclusion. Thus, if the areas are unequal, $P$ cannot be a self-
conjugate. ∎

The following is another symmetry property shared by all self-conjugate
partitions.

######  Proposition 3.2.

Let $\phi$ be the map defined in Theorem 3.1. If $P$ is self-conjugate, then
$f(x)=f^{-1}(x)$.

Note that the converse of Theorem 3.1 is not true. Having equal area on either
side of the diagonal does not imply that the partition is self-conjugate.
Consider the counterexample below.

######  Counterexample 3.1.

Suppose we have the partition $P=[3,3,1,1]$. We graph $\phi(P)$ below on the
Cartesian plane.

Notice that the areas on either side of the diagonal $g(x)=x$ (orange and
teal) are equal, but the partition is not self-conjugate. Thus, equal areas on
either side of the diagonal cannot imply a self-conjugate partition, so the
relationship is not “if and only if”.

##  4 Addition and Sub-Partitions

We now introduce two definitions with some examples. Being able to add
partitions and to sub-partition as defined below will be necessary to later
proofs.

######  Definition 4.1.

Suppose $P$ and $Q$ are partitions. To add the partitions $P+Q=R$, combine the
parts of $P$ and $Q$ into one partition R, reordering the parts as necessary
to satisfy Definition 1.1.

This operation on partitions has been previously defined by Schneider as the
product of two partitions [5]. Note that addition of partitions is both
associative and commutative. For example, given the partitions $A=[4,3,2,1]$
and $B=[3,2,2]$, we have $A+B=[4,3,3,2,2,2,1]=B+A$.

######  Definition 4.2.

A sub-partition $Q$ of a partition $P$ is a set of sequential parts of $P$
which is itself a partition and satisfies Definition 1.1.

This sub-partition $Q$, when correctly added to another sub-partition, or
series of sub-partitions, will yield $P$ itself. To sub-partition $P$ is to
partition $P$ into sub-partitions. Thus, “sub-partition” is both a noun and a
verb. Sub-partition is the inverse operation of addition.

Furthermore, the empty partition $E=[\ ]$ will serve as an identity, so
$E+P=P+E=P$ for any partition $P$.

For example, given $P=[a,b,c,d]$, we could sub-partition $P$ in some of the
ways below (and yield the following valid sub-partitions):

  1. 1.

$P=[a,b,c,d]+[\ ]=[\ ]+[a,b,c,d]$

  2. 2.

$P=[a,b,c]+[d]$

  3. 3.

$P=[a,b]+[c,d]$

  4. 4.

$P=[a]+[b,c,d]$

  5. 5.

$P=[a,b]+[c]+[d]$

  6. 6.

$P=[a]+[b,c]+[d]$

  7. 7.

$P=[a]+[b]+[c,d]$

  8. 8.

$P=[a]+[b]+[c]+[d]$

  9. 9.

$P=[a,d]+[b,c]$

  10. 10.

$P=[a,c]+[b,d]$

##  5 Nesting

Figure 12: Nested Diagram of some partition $P$.

When looking at the Young diagram of some symmetric partition $P$, we can see
a “nest” of fancy L(s) and an “egg” that is either a pure Durfee square or a
fancy triangle (see Figure 12 above for an illustration).

######  Example 5.1.

Consider the self-conjugate partition $P=[7,7,6,4,3,3,1]$. Its Young diagram
is given below.

The blue, green, and orange areas are fancy Ls, and the purple area is a
(trivial) Durfee square. The fancy Ls combined are the nest and the Durfee
square is the egg. Now, we can combine the blue and green area to be a fancy L
of width 2, as opposed to the fancy Ls of width 1 in Lemma 2.4. These higher
width fancy Ls are still symmetric.

We want to show that if both the nest and the egg are truly symmetric, then we
will have a self-conjugate partition. We first prove the trivial cases, where
we only have a single fancy L of width 1 nesting a symmetric egg (first, a
pure Durfee square, and second, a fancy triangle).

###  5.1 The Trivial Cases

######  Lemma 5.1.

Suppose the first part of a partition $P$ is equal to the number of parts of
$P$, and

$P=[n,\ m,\ m,...,\ m,\ 1,\ 1,...,1].$

Sub-partition $P$ as such:

$P=:[n]+[m,\ m,...,\ m]+[1,\ 1,...,1]=X+Y+Z,$

respectively. Suppose $Y$ has $m-1$ parts and $Z$ has $n-(m-1)-1$ parts. Then,
$P$ is self-conjugate.

###### Proof.

Let $P$ be as above. Remove the first (outer) row and column. Then we are left
with the partition

$Q=:[(n-n)]+[(m)-1,\ (m)-1,...,\ (m)-1]+[(1-1),\ (1-1),...,\ (1-1)]$ $=[\
]+[(m-1),\ (m-1),...,\ (m-1)]+[\ ]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $=E+[(m-1),\ (m-1),...,\ (m-1)]+E\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ $=[m-1,\ m-1,...,\
m-1].\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $

So, $Q$ is a pure Durfee square, of dimension $m-1$, and is self-conjugate by
Lemma 2.3. The first (outer) row and column was a fancy L, which is self-
conjugate by Lemma 2.5. Thus, the Young diagram of $P$ is symmetric, so $P$ is
self-conjugate.
∎

######  Lemma 5.2.

Suppose the first part of a partition $P$ is equal to the number of parts of
$P$ and

$P=[n,\ m,\ (m-1),...,\ (m-(m-2)),\ \ 1,\ 1,...,1].$

Sub-partition $P$ as such:

$P=:[n]+[m,\ (m-1),...,\ (m-(m-2))]+[1,\ 1,...,1]=X+Y+Z,$

respectively. Suppose $Y$ has $m-1$ parts and $Z$ has $n-(m-1)-1$ parts. Then
$P$ is self-conjugate.

###### Proof.

Let $P$ be as above. Remove the first (outer) row and column. Then we are left
with the partition

$Q=:[(n-n)]+[m-1,\ (m-1)-1,...,\ (m-(m-2))-1]+[(1-1),...,\ (1-1)]$ $=[\
]+[m-1,\ (m-1)-1,...,\ (m-(m-2))-1]+[\ ]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
$=E+[m-1,\ (m-1)-1,...,\ (m-(m-2))-1]+E\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
$=[m-1,\ (m-1)-1,...,\ (m-1)-((m-1)-1)].\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ $

So, $Q$ is a fancy triangle and is self-conjugate by Lemma 2.4. The first
(outer) row and column was a fancy L, which is self-conjugate by Lemma 2.5.
Thus, the Young diagram of $P$ is symmetric, so $P$ is self-conjugate.
∎

We now want to show that if both the nest and the egg are symmetric, then we
will have a self-conjugate partition in general, where our nests are non-
trivial multiple fancy Ls. We will show this is true for eggs that are pure
Durfee squares and fancy triangles.

###  5.2 The Non-Trivial Cases

######  Lemma 5.3.

Suppose the first part of a partition $P$ is equal to the number of parts of
$P$ and

$P=[n,\ n,...,\ n,\ m,\ m,...,\ m,\ j,\ j,...,\ j].$

Sub-partition $P$ as such:

$P=:[n,\ n,...,\ n]+[m,\ m,...,\ m]+[j,\ j,...,\ j]=X+Y+Z,$

respectively. Suppose $X$ has $j$ parts, $Y$ has $m-j$ parts, and $Z$ has
$n-(m-j)-j$ parts. Then $P$ is self-conjugate.

###### Proof.

Let $P$ be as above. Remove the first (outer) $j$ rows and $j$ columns. You
will be left with:

  1. $Q=:$

$[(n-n),\ (n-n),...,\ (n-n)]+[m-j,\ m-j,...,\ m-j]\\\ +[(j-j),\ (j-j),...,\
(j-j)]$

  2. $=[\ ]+[m-j,\ m-j,...,\ m-j]+[\ ]$

  3. $=E+[m-j,\ m-j,...,\ m-j]+E$

  4. $=[m-j,\ m-j,...,\ m-j]$

So, $Q$ is a Durfee square and is self-conjugate by Lemma 2.3. The first
(outer) $j$ rows and $j$ columns were a fancy L of width $j$, which is self-
conjugate by iterating Lemma 2.5 $j$ times. Thus, the Young diagram of $P$ is
symmetric, so $P$ is self-conjugate.
∎

######  Lemma 5.4.

Suppose the first part of a partition $P$ is equal to the number of parts of
$P$ and

$P=[n,\ n,...,\ n,\ m,\ m-1,...,\ m-(m-j)+1,\ j,\ j,...,\ j].$

Sub-partition $P$ as such:

$P=:[n,\ n,...,\ n]+[m,\ m-1,...,\ m-(m-j)+1]+[j,\ j,...,\ j]=X+Y+Z,$

respectively. Suppose $X$ has $j$ parts, $Y$ has $m-j$ parts, and $Z$ has
$n-(m-j)-j$ parts. Then $P$ is self-conjugate.

###### Proof.

Let $P$ be as above. Remove the first (outer) $j$ rows and $j$ columns. Then
we are left with the partition

  1. $Q$

$=[(n-n),\ (n-n),...,\ (n-n)]$

  2. $\ \ \ +[m-j,\ (m-1)-j,...,\ (m-(m-j)+1)-j]$

  3. $\ \ \ +[(j-j),\ (j-j),...,\ (j-j)]$

  4. $=[\ ]+[(m-j),\ (m-j)-1,...,\ ((m-j)-(m-j))+1]+[\ ]$

  5. $=E+[(m-j),\ (m-j)-1,...,\ ((m-j)-(m-j))+1]+E$

  6. $=[(m-j),\ (m-j)-1,...,\ (m-j)-((m-j)-1)]$

So, $Q$ is a fancy triangle and is self-conjugate by Lemma 2.4, and the
(outer) $j$ rows and $j$ columns were a fancy L of width $j$, which is self-
conjugate by iterating Lemma 2.5 $j$ times. Thus, the Young diagram of $P$ is
symmetric, so $P$ is self-conjugate.
∎

###  5.3 The Nest-And-Egg Theorem

######  Theorem 5.5.

Given a nest, which is a series of fancy Ls of any width, and an egg that is
either a Durfee square or a fancy triangle of any size $($or an empty egg,
$E=[\ ])$, a partition composed of this nest and egg will always be self-
conjugate.

###### Proof.

Suppose we have a nest and a trivial egg; the egg $K$ is either $K=[\ ]=E$,
$K=[1]$, $K=[2,1]$, or $K=[2,2]$, and the rest of our Young diagram belongs to
the nest (and thus, is some fancy L or a series of fancy Ls of any width).
Since all Ls of the nest are symmetric by Lemma 2.5 and all of these egg cases
are trivially symmetric, the claim holds true. Suppose we have a nest and a
non-trivial egg, such as a Durfee square or fancy triangle of dimension 3 or
more. By Lemma 5.3 and Lemma 5.4, the claim holds true, since removing enough
outer fancy Ls from our Young diagram will leave us with one of these cases.
Thus, the theorem holds true in all cases.
∎

##  6 The Main Theorem

With all of the aforementioned lemmas and theorems, we finally wish to prove
the main theorem: the Self-Conjugate Partition Theorem. This theorem will
allow us to take any partition, without its Young diagram, and determine if it
is self-conjugate, even if it does not follow any of the forms or patterns
above. Although the statement of the theorem and its proof may appear
complicated, it will show us a simple way to determine whether a partition is
self-conjugate by only using clever counting skills and summation.

However, we note that the $x_{i}$-values in the next theorem represent the
multiplicity of parts (how many times a part repeats in the partition) and not
the value or index of the parts themselves. Figure 13 will be a useful
illustration for the proof of the theorem.

Figure 13: The Young diagram of a self-conjugate partition $P$.

######  Theorem 6.1 (The Self-Conjugate Partition Theorem).

A partition $P$ is self-conjugate if and only if it can be written as the
following:

$P=\left[\sum_{i=0}^{n}x_{i},\ \sum_{i=0}^{n}x_{i},...,\sum_{i=0}^{n}x_{i},\
\sum_{i=0}^{n-1}x_{i},\
\sum_{i=0}^{n-1}x_{i},...,\sum_{i=0}^{n-1}x_{i},...,\sum_{i=0}^{0}x_{i},\
\sum_{i=0}^{0}x_{i},...,\sum_{i=0}^{0}x_{i},\right]$
$=\left[\sum_{i=0}^{n}x_{i},\
\sum_{i=0}^{n}x_{i},...,\sum_{i=0}^{n}x_{i}\right]+\left[\sum_{i=0}^{n-1}x_{i},\
\sum_{i=0}^{n-1}x_{i},...,\sum_{i=0}^{n-1}x_{i}\right]+...+\left[\sum_{i=0}^{0}x_{i},\
\sum_{i=0}^{0}x_{i},...,\sum_{i=0}^{0}x_{i}\right]$
$=:Y_{n}+Y_{n-1}+\cdots+Y_{0},$

respectively, where $P$ has $\sum_{i=0}^{n}x_{i}$ parts, $Y_{n}$ has $x_{0}$
parts, $Y_{n-1}$ has $x_{1}$ parts, …, $Y_{0}$ has $x_{n}$ parts, and all
$x_{i}\in\mathbb{N}$.

###### Proof.

  1. Suppose a partition $P$ is self-conjugate. We write

$P=[z_{n},\ z_{n},\ ...,\ z_{n},\ z_{n-1},\ z_{n-1},\ ...,\ z_{n-1},\ ...,\
z_{0},\ z_{0},\ ...,\ z_{0}]$ $=[z_{n},\ z_{n},\ ...,\ z_{n}]+[z_{n-1},\
z_{n-1},\ ...,\ z_{n-1}]+...+[z_{0},\ z_{0},\ ...,\ z_{0}]$
$=:Y_{n}+Y_{n-1}+\cdots+Y_{0},$

where $P$ has $z_{n}$ parts. It must be that $Y_{n}$ has $z_{0}$ parts, lest
we contradict Theorem 3.1, since $P$ is self-conjugate, and the boundaries of
our piecewise function have to be equal on either side of $g(x)=x$ (see Figure
13). Likewise, $Y_{n-1}$ has $z_{1}-z_{0}$ parts, lest we contradict Theorem
3.1. Continuing inductively, we find that

$\ell(Y_{n})=z_{0},\ \ell(Y_{n-1})=z_{1}-z_{0},\ \ell(Y_{n-2})=z_{2}-z_{1},\
...,\ \ell(Y_{0})=z_{n}-z_{n-1}.\ $

Let $x_{i}$ denote the number of parts of size $z_{n-1}$ in $P$, for all
$0\leq i\leq n$. Again, by Theorem 3.1, since the boundaries must match, we
have that

$z_{n}=\sum_{i=0}^{n}x_{i},\ z_{n-1}=\sum_{i=0}^{n-1}x_{i},\ ...,\
z_{0}=\sum_{i=0}^{0}x_{i}.$

Thus, if $P$ is a self-conjugate partition, it can be written as the above
partition.

  2. Conversely, suppose we have the partition

$P=\left[\sum_{i=0}^{n}x_{i},\ \sum_{i=0}^{n}x_{i},...,\sum_{i=0}^{n}x_{i},\
\sum_{i=0}^{n-1}x_{i},\
\sum_{i=0}^{n-1}x_{i},...,\sum_{i=0}^{n-1}x_{i},...,\sum_{i=0}^{0}x_{i},\
\sum_{i=0}^{0}x_{i},...,\sum_{i=0}^{0}x_{i},\right].$

Remove the first $x_{0}$ (outer) rows and columns. We are left with:

$Q=\left[\left(\sum_{i=0}^{n}x_{i}\right)-\left(\sum_{i=0}^{n}x_{i}\right),\left(\sum_{i=0}^{n}x_{i}\right)-\left(\sum_{i=0}^{n}x_{i}\right),...,\left(\sum_{i=0}^{n}x_{i}\right)-\left(\sum_{i=0}^{n}x_{i}\right)\right]$

$+\left[\left(\sum_{i=0}^{n-1}x_{i}\right)-x_{0},\
\left(\sum_{i=0}^{n-1}x_{i}\right)-x_{0},...,\left(\sum_{i=0}^{n-1}x_{i}\right)-x_{0}\right]+\cdots$

$=[\
]+\left[\sum_{i=1}^{n-1}x_{i},\sum_{i=1}^{n-1}x_{i},...,\sum_{i=1}^{n-1}x_{i}\right]+\cdots+\left[\sum_{i=1}^{1}x_{i},\sum_{i=1}^{1}x_{i},...,\sum_{i=1}^{1}x_{i}\right]+[\
]$

$=\left[\sum_{i=1}^{n-1}x_{i},\sum_{i=1}^{n-1}x_{i},...,\sum_{i=1}^{n-1}x_{i}\right]+\left[\sum_{i=1}^{n-2}x_{i},\sum_{i=1}^{n-2}x_{i},...,\sum_{i=1}^{n-2}x_{i}\right]+\cdots+\left[\sum_{i=1}^{1}x_{i},\sum_{i=1}^{1}x_{i},...,\sum_{i=1}^{1}x_{i}\right].$

Let $j=i-1$ and $m=n-1$. Then,

$Q=\left[\sum_{j=0}^{m}x_{j},\sum_{j=0}^{m}x_{j},...,\sum_{j=0}^{m}x_{j}\right]+\left[\sum_{j=0}^{m-1}x_{j},\sum_{j=0}^{m-1}x_{j},...,\sum_{j=0}^{m-1}x_{j}\right]+\cdots+\left[\sum_{j=0}^{0}x_{j},\sum_{j=0}^{0}x_{j},...,\sum_{j=0}^{0}x_{j}\right].$

We are now back where we started. Continue this process inductively, removing
every next outer set of $x_{k}$ rows and columns, for $k\in\\{1,2,...,n\\}$.
This will peel back our nest, and by Theorem 5.5, we will eventually get a
symmetric egg. This is because every layer we peel back is a fancy L, and the
egg itself is either empty or made of fancy Ls (Durfee square, fancy
triangle). Alternatively, assume the supposition above and one or more parts
of the partition makes the partition a non-self-conjugate; this will lead to a
contradiction because of our supposition on the multiplicity of parts. Thus,
$P$ is a self-conjugate partition.

∎

Now, let us go through a few examples and apply this theorem to see if various
partitions are self-conjugate.

######  Example 6.1.

Let $P=[5,5,3,2,2]$. We see that $x_{0}=2$, $x_{1}=1$, $x_{2}=2$, since 5
appears twice, 3 once, and 2 twice. We have

$5=\sum_{i=0}^{2}x_{i}=2+1+2,\ \ 3=\sum_{i=0}^{1}x_{i}=2+1,\ \
2=\sum_{i=0}^{0}x_{i}=2,$

so $P$ is self-conjugate.

######  Example 6.2.

Let $Q=[20,7,7,7,7,7,6,6,5,5,5,5,4,4,2,2,1,1,1,1]$. We see that $x_{0}=1$,
$x_{1}=5$, $x_{2}=2$, $x_{3}=4$, $x_{4}=2$, $x_{5}=2$, and $x_{6}=4$. We have

$\sum_{i=0}^{6}x_{i}=1+5+2+4+2+2+4=20,\ \
\sum_{i=0}^{5}x_{i}=1+5+2+4+2+2=16\neq 7,$

so $Q$ is not self-conjugate.

######  Example 6.3.

Let $R=[9,9,7,7,4,4,4,2,2]$. We see that $x_{0}=2$, $x_{1}=2$, $x_{2}=3$,
$x_{3}=2$. We have

$\sum_{i=0}^{3}x_{i}=2+2+3+2=9,\ \ \sum_{i=0}^{2}x_{i}=2+2+3=7,\ \
\sum_{i=0}^{1}x_{i}=2+2=4,\ \ \sum_{i=0}^{0}x_{i}=2,$

so $R$ is self-conjugate.

## References

  * [1] G. E. Andrews, The Theory of Partitions. Cambridge University Press, Cambridge (1984).
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https://www.geogebra.org/calculator
https://www.maa.org/external_archive/joma/Volume7/Hohenwarter/History.html

  * [4] H. Rademacher, On the Partition Function $p(n)$. Proceedings of the London Mathematical Society, Second Series, 43(4):241-254 (1937).
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https://cocalc.com/