TITLE: Find the number of real solutions of $x^3-3x-\sqrt{5}=0$
QUESTION [2 upvotes]: Find the number of real solutions of the equation  $x^3-3x-\sqrt{5}=0$
I don't know that how start for find !

REPLY [1 votes]: You can be sure that the equation has atleast one real solution since $f(1)<0$ and $f(3)>0$ and by Intermediate value theorem for continuous functions(where, $f(x)=x^3-3x-\sqrt5$ )there exists an $x_0 \in (1,3)$ such that $x_0^3 - 3x_0 -\sqrt5=0$. Moreover, you can also be certain that this is the only real solution of the equation since the function increases on $(3, \infty)$ (can be proven use second derivative test) and is negative on $(-\infty,1)$.
Given $x^3-3x-\sqrt5 = 0$
$$x^3-3x=\sqrt5$$
Squaring on both sides(since $\sqrt5$ is irrational and to use Rational root theorem I want all coefficients to be integers),
$$x^6-6x^4+9x^2-5=0$$
Using Rational root theorem the only possible rational solutions of the above equation are $1,-1,\frac{1}{5},-\frac{1}{5}$ and checking manually none of them are roots of the above equation. So the given equation has certainly no rational solutions.
Therefore $x^3-3x-\sqrt5$ has only one real solution(in addition you can also be certain that it is not a rational number).


text:
\section{GRU-MDN stochastic programming framework} \label{S:2}
To make data-driven SP that is capable of solving time-series problem, we propose a GRU-base mixture density network called GRU-MDN. The framework involves three components, GRU is in charge of predicting customer demands, GMM focuses on the probability distribution that are based on the outcomes from GRU, the SP is in charge of modeling uncertainty. We will elaborate our framework in this section.

\subsection{Gated Recurrent Unit}

Unlike the single and simple building block in RNNs, LSTM uses forgetting and gating mechanisms to select and filter information that is necessary for future computation. Figure \ref{fig:gru} shows the detail gating mechanism of GRU.

\begin{figure}
    \centering
    \includegraphics[scale = 0.3]{images/GRU.png}
    \caption{Gated Recurrent Unit}
    \label{fig:gru}
\end{figure}

The state of GRU for each time step $t$ is given by the following equations.

\begin{equation}
\begin{array}{l}
z_{t}=\sigma_{g}\left(W_{z} x_{t}+U_{z} h_{t-1}+b_{z}\right) \\
r_{t}=\sigma_{g}\left(W_{r} x_{t}+U_{r} h_{t-1}+b_{r}\right) \\
h_{t}=z_{t} \odot h_{t-1}+\left(1-z_{t}\right) \odot \phi_{h}\left(W_{h} x_{t}+U_{h}\left(r_{t} \odot h_{t-1}\right)+b_{h}\right)
\end{array}
\end{equation}
where $x_{t}$ is the input vector, $h_{t}$ is the output vecotr, $z_{t}$ is the update gate vector, $r_{t}$ is the reset gate vector, $W,U,b$ are the parameter matrices and vector. $\sigma_{g}$ and $\phi_{h}$ are activation functions in sigmoid and hyperbolic types, respectively.

\subsection{Mixture Density Network}
MDN is a variant of a neural network whose output is probability distribution(s) rather than single unit for most of neural networks. The basic idea of MDN is \textit{combining a deep neural network (DNN) and a (group of) mixture of distributions}.  Actually, most of the modern DNN architectures such as CNN, RNN and LSTM can be extended to become the special MDNs. The DNN provides the parameters for multiple distributions, which are then mixed by some weights. Also, These weights are provided by the DNN. Notice that the true distribution of the input data can be any type which cannot be described by the parametric methods (e.g. Gaussian distribution, Poisson distribution). Actually, the non-parametric method such as kernel density estimation (KDE) is able to handle with the arbitrary probability distribution learning problem. However, KDE method cannot be directly applied for probability distribution storage in MDN since KDE is a non-parametric approach which implies that it may contain infinite parameters that cannot be store by finite number of neurons. Therefore, the semi-parametric approach - Gaussian mixture model (GMM) is adopted to overcome the problem in MDN, which is formulated as follows.

$$p(\boldsymbol{X \big| \theta})=\sum_{i=1}^{K} w_{i} \mathcal{N}\left(\boldsymbol{X \big| \mu}_{i}, \boldsymbol{\Sigma}_{i}\right)$$
where $\theta = (W, \mu, \Sigma)$, $W = (w_{1}, w_{2}, \cdots, w_{K})$, $\mu = (\mu_{1}, \mu_{2}, \cdots, \mu_{K})$ and $\Sigma = (\Sigma_{1}, \Sigma_{2}, \cdots, \Sigma_{K})$. $K$ is the number of Gaussian distributions. Generally, GMM can be considered as a group of Gaussian distributions with different weights, where the $i-th$ Gaussian is determined by weight $w_{i}$, means $\mu_{i}$ and covariance matrix $\Sigma_{i}$ (variance for univariate Gaussian). Then the predicted probability distribution can be represented using GMM by adjusting the parameter $\theta$. In this work, we use expectation maximization (EM)\cite{dempster1977maximum} algorithm to determine the parameter of GMM, which is summarized as follows.

\begin{algorithm}
\caption{EM}\label{euclid}
\label{al}
\hspace*{\algorithmicindent}
\textbf{Input: The GMM $\sum_{k=1}^{K} \pi_{k} \mathcal{N}\left(\boldsymbol{x} | \boldsymbol{\mu}_{k}, \mathbf{\sigma}_{k}\right)$}
\\
\hspace*{\algorithmicindent}
\textbf{Output: GMM}

\begin{algorithmic}[1]

\State Initialize $\mu_{j}, \Sigma_{j}$ and $\pi_{j}$, $j=1, \cdots, K$
\State E-step. Compute $$
\gamma_{n j}=\frac{\pi_{j} \mathcal{N}\left(\xi_{n} | \mu_{j}, \Sigma_{j}\right)}{\sum_{i=1}^{K} \pi_{i} \mathcal{N}\left(\xi_{n} | \mu_{i}, \Sigma_{i}\right)}
$$
\State M-step. Re-estimate $$
\mu_{j}^{\mathrm{new}}=\frac{1}{N_{j}} \sum_{n=1}^{N} \gamma_{n j} \xi_{n}, \Sigma_{j}^{\mathrm{new}}=\frac{1}{N_{j}} \sum_{n=1}^{N} \gamma_{n j}\left(\xi_{n}-\mu_{j}^{\mathrm{new}}\right)\left(\xi_{n}-\mu_{j}^{\mathrm{new}}\right)^{\top}, \pi_{j}^{\mathrm{new}}=\frac{N_{j}}{N}
$$
where
$$
N_{j}=\sum_{n=1}^{N} \gamma_{n j}
$$
\State Check whether the convergence is satisfied. If not, return to step 2.

\end{algorithmic}
\end{algorithm}

The structure of MDN can be shown in the figure\ref{fig:mdn}.
\begin{figure}
    \centering
    \includegraphics[scale = 0.2]{images/MDN.png}
    \caption{Architecture of Mixture Density Network}
    \label{fig:mdn}
\end{figure}

One of the significant contribution of this work is to combine GRU and MDN, our proposed architecture of GRU-MDN can be shown in figure \ref{fig:GRU-MDN}.

\begin{figure}
    \centering
    \includegraphics[scale=0.4]{images/GRU-MDN.png}
    \caption{The Structure of GRU-MDN}
    \label{fig:GRU-MDN}
\end{figure}

\begin{comment}
\begin{algorithm}
\caption{Monte Carlo Sampling from GMM}\label{euclid}
\label{al}
\hspace*{\algorithmicindent}
\textbf{Input: The GMM $\sum_{k=1}^{K} \pi_{k} \mathcal{N}\left(\boldsymbol{x} | \boldsymbol{\mu}_{k}, \mathbf{\sigma}_{k}\right)$, number of data points $N$}
\\
\hspace*{\algorithmicindent}
\textbf{Output: the Monte Carlo sampling result}

\begin{algorithmic}[1]

\State $k \gets 0$
\State Repeat steps 1) and 2) until you have the desired amount of samples from the mixture distribution
\State Return the sampling result

\end{algorithmic}
\end{algorithm}
\end{comment}

\subsection{Data-driven dynamic stochastic programming framework}
Finally, we come to summarize our data-driven SP framework which is displayed in the figure \ref{fig:framework}.
\begin{figure}
    \centering
    \includegraphics[scale = 0.4]{images/framework.png}
    \caption{GRU-MDN SP Framework}
    \label{fig:framework}
\end{figure}

There are four major components involving in the proposed framework. Specifically, the time-series data is the input of the framework, then the predicted time-series probability distribution is obtained via GUR-MDN, the distribution is input as the parameters of SP, after that various model reformulation and decomposition algorithms can be applied to solve the SP model.

config: arxiv


text:
{\bf Problem.} Compute: $\left(\frac{1}{2} \right)^{3} \cdot \left(\frac{1}{7} \right)$.
{\bf Level.} Level 2
{\bf Type.} Prealgebra
{\bf Solution.} We have $\left(\frac{1}{2} \right)^{3}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}$, so \[\left(\frac{1}{2} \right)^{3} \cdot \left(\frac{1}{7} \right) = \frac18\cdot \frac17 = \boxed{\frac{1}{56}}.\]

set_name: MATH


TITLE: Rectangles in Hyperbolic geometry
QUESTION [3 upvotes]: This says rectangles don't exist in hyperbolic geometry. But according to this you can model the euclidean plane in Hyperbolic geometry. Wondering why you can't model the rectangle, I thought if you can do euclidean geometry in 2d plane you can get rectangle. The question is how to get (or introduce something to get) a rectangle in hyperbolic 2d plane.

REPLY [4 votes]: You have to be careful with your definitions. Yes, a Horosphere in Hyperbolic 3 space is an isometric model of the Euclidean plane, but the euclidean lines in this model are not hyperbolic lines. Thus, any model of a euclidean rectangle has four right angles but its sides are not hyperbolic lines, and therefore it is not a hyperbolic rectangle. See my answer to the recent MSE question 3119728 "Description of Model of Euclidean Geometry found in the Hyperbolic Plane".
The same kind of situation arises in using a sphere in Euclidean 3 space as a model of the elliptic plane. The elliptic lines in the model are great circles of the sphere and are not euclidean lines.


TITLE: Twisted torsions of reducible representations of knot groups
QUESTION [1 upvotes]: Let $L$ be a link in $S^3$ and $\rho : \pi_1(S^3 \setminus L) \to \operatorname{SL}_2(\mathbb C)$ be a representation of its knot group. If the twisted homology $H^\rho(S^3 \setminus L)$ is acyclic, we can obtain the twisted Reidemeister torsion $\tau(S^3 \setminus L, \rho)$, which is closely related to the twisted Alexander polynomial. (One can also do this for other matrix groups than $\operatorname{SL}_2(\mathbb C)$.)
Thinking of $\rho$ as a two-dimensional representation of $\pi_L := \pi_1(S^3 \setminus L)$, I am interested in the case where this representation is reducible, but possibly not indecomposable. After choosing the right basis this is the same as asking that the matrices of $\rho$ are always of the form
$$
\begin{pmatrix}
\kappa & \epsilon \\
0 & \kappa^{-1}
\end{pmatrix}
$$
for some $\kappa \ne 0$. If the $\epsilon$ can all be chosen to be zero, then the representation is decomposable and $\tau(S^3 \setminus L, \rho)$ is (a square of) the ordinary Reidemeister torsion.
There are examples where $\rho$ is reducible but not indecomposable. However, in those cases it seems that the indecomposability doesn't matter. More formally, one can obtain another representation $\bar \rho$ of $L$ by taking each meridian $x$ of $L$ and setting the upper-right entry of $\bar \rho(x)$ to be zero. (For a basis-independent definition, we are just replacing the meridians with diagonal matrices with the same eigenvalues.)
Then
$$
\tau(S^3 \setminus L, \rho) = \tau(S^3 \setminus L, \bar \rho)
$$
Is there a proof of this fact? Is it true?
This seems natural, because the torsion is a sort of determinant and the matrices above are upper-triangular. However, I am not aware of a proof or counterexample. (It's not totally obvious, because the twisted Burau representation used to define the torsion isn't necessarily upper-triangular even when $\rho$ is.)

REPLY [2 votes]: This should be true, here is a (hoperfully correct) sketch of proof: the Reidemeister torsion is a continuous function on the set of acyclic representations (as it can be computed from determinants of matrices whose coefficients vary continuously with the representation). On the other hand one can simultaneously conjugate any subset of upper triangular matrices arbitrarily close to diagonal matrices. So, if both the upper-triangular and diagonal representations are acyclic, the torsion is the same for them.


TITLE: computation of sum with natural numbers and roots
QUESTION [0 upvotes]: Let $n_1,...,n_m$ be non-negative natural numbers, such that their sum $n_1+\dots+n_m=2n$.
Fix $m$. I am wondering how to compute, or bound from above $$\sum_{i=1}^m \sqrt{1+\frac{1}{2n_i-1}}. $$
(One can assume that the $n_i$ are positive, since any term with $n_i=0$ doesn't contribute to the sum.)

REPLY [1 votes]: Here is an answer for the problem where $m$ is fixed. For any fixed positive integer $s$, note that the function
$$
f(t) = \sqrt{1+\frac1{2t-1}} + \sqrt{1+\frac1{2(s-t)-1}}
$$
is decreasing on the interval $[1,s/2]$ and increasing on the interval $[s/2,s-1]$ (this is a simple calculus exercise). In particular, if $n_1&ltn_2\le n_3&ltn_4$ are positive integers with $n_1+n_4 = s = n_2+n_3$, then
$$
\sqrt{1+\frac1{2n_1-1}} + \sqrt{1+\frac1{2n_4-1}} > \sqrt{1+\frac1{2n_2-1}} + \sqrt{1+\frac1{2n_3-1}}.
$$
This is enough to imply that the sum
$$
\sum_{i=1}^m \sqrt{1+\frac1{2n_i-1}}
$$
is maximized when $n_1=\cdots=n_{m-1}=1$ and $n_m = 2n-(m-1)$: for any other configuration, there are $n_i$ and $n_j$ both strictly larger than 1, in which case replacing them by $n_i-1$ and $n_j+1$ results in a larger sum.
By similar reasoning, it implies that the sum is minimized when the $n_i$ are all as equal as possible: specifically, with $k=[n/m]$ and $r=n\mod m$ (so that $0\le r&ltm$), set $n_1=\cdots=n_r=k+1$, $n_{r+1}=\cdots=n_m=k$.


TITLE: Triple integral bounds
QUESTION [1 upvotes]: Let $W$ be the region bounded by the planes
$x = 0$, $y = 0$, $z = 0$, $x + y = 1$, and $z = x + y$.
$(x^2 + y^2 + z^2)\, \mathrm dx\, \mathrm dy\, \mathrm dz$; $W$ is the region bounded by $x + y + z = a$ (where $a > 0$), $x = 0$, $y = 0$, and $z = 0$.

$x,y,z$ being $0$ is throwing me off because I'm not sure how to graph it and get a bounded area; it seems like it would be infinite to me. What would the bounds for the triple integrals be?

REPLY [2 votes]: 1)

$$\int_{x=0}^{x=1}\int_{y=0}^{y=1-x}\int_{z=0}^{z=x+y}\ dzdydx $$
2)

$$\int_0^{x=a}\int_0^{y=a-x}\int_0^{a-x-y} (x^2 + y^2 + z^2) \;dz\;dy\;dx$$


TITLE: Stalk of a generic point of the special fiber
QUESTION [0 upvotes]: Im trying to understand the paper Weight functions on non-archimedean analytic spaces and the kontsevich-soibelman skeleton by Johannes Nicaise.
Let $R$ be a discrete valuation ring and $K=Frac(R)$. Let $X$ be a normal integral separated $K-$scheme of finite type and $\mathscr{X}$ a normal $R-$model.
An $R-$model $\mathscr{X}$ of $X$ is a normal flat separated $R-$scheme of finite type endowed with an isomorphism of $K-$schemes $\mathscr{X}_{K} \longrightarrow X$.
Why if we consider $E=\overline{\{\xi\}}$ an irreducible component of the special fiber $\mathscr{X}_{k}$ ( where $\xi$ is the generic point of $E$), then $\mathcal{O}_{\mathscr{X,\xi}}$ is a discrete valuation ring with fraction field $K(X)=\mathcal{O}_{X,\eta}$ (where $\eta$ is the generic point of $X$)?

REPLY [0 votes]: Let $\pi : X \rightarrow Spec(R)$ is the structure morphisms. Since $X$ is normal, then irreducible and connected components of $X$ are the same.
Let $X = \cup X_i$, where $X_i$ are irreducible components. Thus each of the $X_i$ are connected components and hence open. Now, since $X$ is flat, we get each of the $X_i$ is flat over $Spec(R)$, hence $\pi(X_i)$ is also open and hence contains the generic point. It follows that the generic fiber is not irreducible, a contradiction. Thus we get $\mathscr{X}$ is irreducible.
Now since $\pi$ is flat and $X$ is irreducible and thus equi-dimensional, each of the irreducible components of all of the fibers, we get $E$ is one dimensional. Hence $\xi$ is a codimension one point in $X$ and $X$ is normal, hence the local ring $\mathcal{O}_{\mathscr{X},\xi}$ is a discrete valuation ring.


TITLE: Symplectic basis $(A_i,B_i)$ such that $S= $ span$(A_1,B_1,...,A_k,B_k)$ for some $k$ when $S$ is symplectic
QUESTION [2 upvotes]: Let $(V,\omega)$ be a symplectic vector space of dimension $2n$. How can I show that for a symplectic subspace $S \subset V$, there exists a symplectic basis $(A_i,B_i)$ such that $S= $ span$(A_1,B_1,...,A_k,B_k)$ for some $k$.
I know that since $S$ is symplectic, $S \cap S^\perp= \lbrace0 \rbrace$ and this is true iff $\omega|_S$ is nondegenerate. Let $(\alpha^1, \beta^1,...,\alpha^n, \beta^n)$ be the corresponding dual basis for $V^\ast$. Then $\omega$ is defined by $\omega=\sum_{i=1}^n \alpha^i \wedge \beta^i$. I do not know how to continue.

REPLY [4 votes]: S is a symplectic space so it has a symplectic basis. Its orthogonal subspace is also a symplectic space, so it has a symplectic base. Now consider the union of these two bases.


TITLE: Fractional power of self adjoint operators.
QUESTION [1 upvotes]: I want to know that $$\langle T^*Tz, z\rangle  = \langle(T^*T)^{\frac {1}{2}}z, (T^*T)^{\frac {1}{2}}z\rangle $$ holds or not? What is the meaning of $$(T^*T)^{\frac{1}{2 }} $$

REPLY [1 votes]: I'll assume that $z \in H$, where $H$ is a Hilbert space and $T$ is a bounded operator on $H$. It is a well-known result that a positive bounded operator $S$ has a unique positive square root, i.e., there is a unique positive operator $S^{1/2}$ that satisfies $(S^{1/2})^2 = S$. Note that $T^* T$ is positive since $\langle T^* T x, x \rangle = \langle T x, T x \rangle = \| T x \|^2 \geq 0$. So $T^* T$ has a unique positive square root $(T^* T)^{1/2}$. Since positive operators are self-adjoint it follows that
$$
\langle T^* T x, y \rangle = \langle ((T^* T)^{1/2})^2 x, y \rangle = \langle (T^* T)^{1/2} x, (T^* T)^{1/2} y \rangle
$$
for all $x, y \in H$.


text:
\begin{document}

\title{Asymptotic Analysis of a Two-Phase Model of Solid Tumour Growth}
\author{{\sc Andrea Genovese de Oliveira}\thanks{Departamento de Matem\'atica, Universidade de Bras\'ilia, Brazil, email: {\tt andreagenovese@unb.br}.}
\quad
{\sc John R. King}\thanks{School of Mathematical Sciences, University of Nottingham, United Kingdom, email: {\tt John.King@nottingham.ac.uk}.}}

\pagestyle{headings}
\markboth{A. GENOVESE DE OLIVEIRA}{\rm ASYMPTOTIC ANALYSIS OF TUMOUR MODEL}
\maketitle

\begin{abstract}
{\noindent We investigate avascular tumour growth as a two-phase process consisting of cells and liquid.  Based on the one-dimensional continuum moving-boundary model formulated by (\textit{Byrne, King, McElwain, Preziosi, Applied Mathematics Letters, 2003, 16, 567-573}), we defined boundary conditions for the analogous model of tumour growth in two dimensions.  We investigate linear stability of one dimensional time-dependent solution profiles in the moving-boundary formulation of a limit case (with negligible nutrient consumption and cell drag).  For this, we obtain an asymptotic limit of the two-dimensional perturbations for large time (in the case where the tumour is growing) by using the method of matched asymptotic approximations.  Having characterised an asymptotic limit of the perturbations, we compare it to the time-dependent solution profile in order to analytically obtain a condition for instability.  Numerical simulations are mentioned.

\vspace{.3cm}

\noindent\textit{Keywords}:  Avascular tumour, tumour growth model, multiphase model, two-phase model, linear stability, asymptotic approximations, matched asymptotic approximation, moving boundary.}
\end{abstract}

\section{Introduction}

We explore the two-phase model of avascular tumour growth proposed by Byrne, King, McElwain and Preziosi in 2003 \cite{byrne2003two}.  This is one of the seminal multiphase models of tumour growth.  By considering two limit cases of this model, the authors were able to draw parallels to previous models of tumour growth such as \cite{adam1986simplified}, \cite{gatenby1996reaction}, \cite{greenspan1972models}, and \cite{sherratt1992oncogenes}, .  On the other hand, it has been extended to three and four phase models of vascular tumour growth in \cite{breward2003multiphase} and \cite{hubbard2013multiphase}, respectively, using similar arguments as the ones used in this paper.

Multiphase models consider the simultaneous movement of materials in different states (liquids, gases and solids) or with different chemical properties, such as viscosity and heterogeneity.  In tumour modelling, these phases can be, among others, healthy and cancer cells, extra-cellular material, and blood vessels, which will clearly have different properties.  Additionally, these phases may not be all included simultaneously in every model, depending on the complexity needed.  Indeed, this makes multiphase formulations conveniently constructive.

We will approach this model through the study of stability, which is important in tumour growth models.  Stability can represent a non-malignant dormant tumour that stabilises at a certain size and can potentially be moved by redirecting its nutrient supply.  On the other hand, instability may represent a malignant tumour that grows uncontrollably, disintegrates or even breaks into pieces.  In the study of avascular tumours, particularly, the distinction between dormant and malignant tumours is a central point of research and this motivates the use of linear stability and perturbation methods to study the model proposed here.

More models of tumour growth can be found in the following reviews: \cite{araujo2004history,roose2007mathematical}

\section{Model formulation}

The equations we analyse are those of the two phase model of tumour growth defined in \cite{byrne2003two}, namely

\begin{alignat}{2}[left={\empheqlbrace}]
\dfrac{\partial \alpha}{\partial t}  & = S_c(\alpha,C)- \nabla \cdot (\mathbf{v}_c \alpha), \label{eq:2.1}
\\
0 & = \nabla \cdot(\alpha\mathbf{v}_c+(1-\alpha)\mathbf{v}_w),\label{eq:2.2}
\\
\mathbf{0} & = \nabla \cdot(-p \mathbf{I} -\alpha \Sigma_c(\alpha)\mathbf{I} + \mu_c \alpha ( \nabla \mathbf{v}_c + \nabla \mathbf{v}_c^T) +  \lambda_c \alpha (\nabla \cdot \mathbf{v}_c)\mathbf{I}),\label{eq:2.3}
\\
\mathbf{0} & = (1-\alpha) \nabla p + k(\alpha)(\mathbf{v}_w- \mathbf{v}_c),\label{eq:2.4}
\\
0 & =\nabla^2 C - Q_c(\alpha, C),\label{eq:2.5}
\end{alignat}

wherein \eqref{eq:2.1} represents the conservation of mass equation for the cell phase (with cell volume fraction $\alpha$, cell velocity $\mathbf{v}_c$, net birth rate $S_c$), \eqref{eq:2.2} represents overall mass conservation (with water volume fraction $1-\alpha$, water velocity $\mathbf{v}_w$), \eqref{eq:2.3} is the overall momentum conservation equation (with water pressure $p$, cell extra pressure $\Sigma_c$, cell shear viscosity $\mu_c$, and cell bulk viscosity $\lambda_c $), \eqref{eq:2.4} is the water momentum equation (with interphase drag coefficient $k$) and \eqref{eq:2.5} is the (quasi-steady) nutrient consumption equation (with consumption rate $Q_c$).  We focus here on the two-dimentional case, with spatial coordinate $(x,y)$ and time $t$.

The following expressions were used for the functions $S_c(\alpha,C)$, $k(\alpha)$, and $Q_c(\alpha,C)$:

\begin{align*}
S_c(\alpha, C)  = \Big(\dfrac{s_0C}{1+s_1C}\Big)\alpha(1-\alpha) - \Big(\dfrac{s_2+s_3C}{1+s_4C}\Big)\alpha, \quad s_i>0,\quad  i=0\ldots 4 \\
k(\alpha)  = k_0 \alpha(1-\alpha), \quad k_0 >0, \qquad Q_c(\alpha, C)  = \dfrac{Q_0 C \alpha}{1+Q_1C}, \quad Q_0, Q_1 >0.
\end{align*}

The function $\Sigma_c(\alpha)$ is defined as the difference between the pressures of the two phases, chosen as in \cite{byrne2003two}:

$$

\Sigma_c(\alpha) = p_c - p_w =\dfrac{\hat{\Sigma}_c |\alpha - \alpha^\ast|^{r-1}}{(1-\alpha)^q}(\alpha-\alpha^\ast) H(\alpha-\alpha_{min}),$$

where $H$ is the Heaviside function and $q,r,\hat\Sigma_c>0$.  Additionally, $0<\alpha_{min}<\alpha^\ast<1$ with $\alpha^\ast$ being the natural cell density. Here, the cell pressure $p_c$ represents the isotropic stresses associated with cell-cell interactions and the fluid pressure $p_w$ is the hydrodynamic pressure in the water.  In summary, the equation above states that the difference in pressure between the two phases may depend on the cell concentration.

Additionally, we define the outer boundary of the tumour at time $t$ by

\begin{equation*}
\Gamma(t) :=: \{(x,y)\in \mathcal{R}^2: x=R(y,t)\}
\end{equation*}

such that the kinematic condition reads

\begin{equation} \label{eq:2.6}
\dfrac{\partial \mathbf{R}}{\partial t}(y,t) \cdot \mathbf{n}(y,t) = \mathbf{v}_c(\mathbf{R}(y,t),t)\cdot \mathbf{n}(y,t),
\end{equation}

where $\mathbf{n}$ is the outward unit normal to the curve $\mathbf{R}(y,t)=(R(y,t),y)$.  At the moving boundary $\Gamma(t)$ we also impose the following boundary conditions:

\begin{equation} \label{eq:2.7}
[- \Sigma_c(\alpha)\mathbf{I} + \mu_c  ( \nabla \mathbf{v}_c + \nabla \mathbf{v}_c^T) +  \lambda_c  (\nabla \cdot \mathbf{v}_c)\mathbf{I})]\cdot \mathbf{n} = \mathbf{0}, \qquad p=0,

\qquad C=C_\infty.
\end{equation}

These represent zero stress in each phase and fixed nutrient concentration at the free surface.  Additionally, at the inner boundary $x=0$ we impose the following conditions:

\begin{equation} \label{eq:2.8}
\mathbf{v}_c=0, \qquad v_w^1 :=\mathbf{v}_w \cdot \mathbf{e}_1=0, \qquad \dfrac{\partial C}{\partial x} = 0,
\end{equation}

where $\mathbf{e}_1$ is the unit vector in the $x$ direction. This corresponds to assuming that $x=0$ is a rigid impermeable boundary.

Notice that the boundary conditions imposed on the moving boundary can be reduced to the conditions required in the one dimensional model in \cite{byrne2003two} by taking

$$\mathbf{R}(y,t)=(R(t),y), \qquad   \mathbf{n}=\mathbf{e}_1.

$$

Additionally, recall that, in the one dimensional model, the outer boundary conditions on $v_c$ and $p$ were considered to be derived from $1$-D versions of

$$\boldsymbol\sigma_c \cdot \mathbf{n} = 0, \qquad\boldsymbol\sigma_w \cdot \mathbf{n} = 0,$$

which are equivalent to the first two conditions defined on the two dimensional moving boundary.  We observe that \eqref{eq:2.4} and \eqref{eq:2.7} imply that

\begin{equation} \label{eq:2.9}
(\mathbf{v}_w-\mathbf{v}_c) \cdot \boldsymbol{\tau}=0
\end{equation}
on $\Gamma(t).$

To complete the necessary notation, let

\begin{equation*}
\Omega(t) :=: \{(x,y) \in \mathcal{R}^2: 0<x<R(y,t)\}
\end{equation*}

denote our domain of interest for each $t>0$.

\section{Limit case}\label{sec:3}

Motivated by results obtained in the simpler one dimensional case, we will analyse a specific limit case of this system in which nutrient consumption and cell drag are taken to be negligible.  This will allow us to obtain some analytically tractable simplifications of the full model.

In the one dimensional formulation this limit case is obtained by taking $Q_c(\alpha,C) \equiv 0$ and $k(\alpha)\equiv 0$ in the original formulation.  The former raises no difficulties, simply implying that $C \equiv C_\infty$, (i.e. nutrient-rich behaviour) throughout $\Omega(t)$, but the latter requires more care.   If $k(\alpha) \equiv 0$ in the two dimensional formulation, then \eqref{eq:2.4} and \eqref{eq:2.7} imply

$p\equiv 0$, only one equation, from the vector system \eqref{eq:2.4}.  To avoid this loss of information, we must reinstate the small $k(\alpha)$ taking $k(\alpha) = k_0 \alpha(1-\alpha)$ as in \cite{byrne2003two}, with the constant $k_0>0$ small.  Then,

\begin{equation*}
0 = \nabla p + k_0 \alpha  (\mathbf{v}_w-\mathbf{v}_c).
\end{equation*}

and hence

\begin{equation*}
0 = \nabla \times (\alpha( \mathbf{v}_w-\mathbf{v}_c))
\end{equation*}

and thus we have obtained the second equation (and extra two equations in three dimensions) required.

Therefore,  $p\equiv 0$ and $(\alpha, C, \mathbf{v}_c, \mathbf{v}_w, \mathbf{R})$ satisfy

\begin{alignat}{2}[left={\empheqlbrace}]
\dfrac{\partial \alpha}{\partial t} & = S_c(\alpha,C) - \nabla \cdot (\mathbf{v}_c \alpha),
\\
0 & = \nabla \cdot(\alpha\mathbf{v}_c+(1-\alpha)\mathbf{v}_w),
\\
0 & = \nabla \cdot(-\alpha \Sigma_c(\alpha)\mathbf{I} + \mu_c \alpha ( \nabla \mathbf{v}_c + \nabla \mathbf{v}_c^T) +  \lambda_c \alpha (\nabla \cdot \mathbf{v}_c)\mathbf{I}),
\\
0 &= \nabla \times [\alpha( \mathbf{v}_w-\mathbf{v}_c)]
\\
0 & = \nabla^2 C, \qquad\qquad\qquad\qquad (x,y)\in \Omega(t), \quad t\in(0,\infty),
\end{alignat}

with kinematic condition \eqref{eq:2.6} and other boundary conditions as in \eqref{eq:2.7} and \eqref{eq:2.8}.

Motivated by results obtained in the one dimensional case, described in \cite{genovese2017asymptotic}, we consider the stability of the following non-homogeneous solution of the limit case defined in the previous section.  Let the constants $\alpha=\alpha_h\in (0,1)$ and $\lambda_2$ be defined by

\begin{equation*}
\lambda_2 = \dfrac{\Sigma_c(\alpha_h)}{\hat\mu_c} =  \dfrac{ S_c(\alpha_h, C_\infty)}{\alpha_h}
\end{equation*}

and

$$\mathbf{v}_{c}(x) = \lambda_2 x \hspace{.05cm} \mathbf{e_1}\qquad
\mathbf{v}_{w}(x) = -\dfrac{\lambda_2 \alpha_h}{1-\alpha_h}x \hspace{.05cm} \mathbf{e_1}, \qquad C=C_\infty.$$

We can also define the two-dimensional moving boundary $\Gamma_\ast (t)$ and domain $\Omega_\ast(t)$, via

\begin{equation*}
\mathbf{R}_\ast(t,y) = (R_\ast(t),y) = (R_0 e^{\lambda_2 t},y).
\end{equation*}

\section{Linear stability analysis}

We shall consider two dimensional perturbations on this one dimensional solution.  For this purpose, we linearise about this one dimensional solution and obtain a system for the respective two-dimensional perturbations

$$
\begin{array}{c}
\hat\alpha(x,y,t), \qquad \hat C(x,y,t), \qquad \hat p(x,y,t),\qquad \mathbf{\hat R}(y,t) =(\hat R(y,t),0)\\
\mathbf{\hat{v}}_{c}(x,y,t)=(\hat v^1_{c}(x,y,t),\hat v^2_{c}(x,y,t)) ,\quad \mathbf{\hat{v}}_{w}(x,y,t)=(\hat v^1_{w}(x,y,t),\hat v^2_{w}(x,y,t)).
\end{array}
$$

Then, we fix the boundary.  For this, we make the following change of variable:

\begin{equation*}
x =  R_\ast(t)\xi
\end{equation*}

and therefore

\begin{align*}
\dfrac{\partial}{\partial t} & \longrightarrow  \dfrac{\partial}{\partial t} - \dfrac{\xi}{R_\ast(t)} \dfrac{dR_\ast}{dt} \dfrac{\partial}{\partial \xi}
\\[.1cm]

\dfrac{\partial}{\partial x} & \longrightarrow  \dfrac{1}{R_\ast(t)} \dfrac{\partial}{\partial \xi}
\\[.1cm]
\dfrac{\partial}{\partial y} & \longrightarrow  \dfrac{\partial}{\partial y}.
\end{align*}

Having done this, we noticed that it is not possible to separate variables in $\xi$ or $t$ as both $\xi$ and $t$ are present in the coefficients in ways that they cannot be extracted by simple exponentiation.  However, it is possible to separate variables in $y$.

Indeed, consider now

\begin{equation*}
\hat\alpha(\xi,y,t)= e^{i\kappa y}\tilde{\alpha}(\xi,t)+ c.c.
\end{equation*}

\begin{equation*}
\hat C(\xi,y,t)= e^{i\kappa y}\tilde{C}(\xi,t)+ c.c.\qquad\hat p(\xi,y,t)= e^{i\kappa y}\tilde{p}(\xi,t)+ c.c.,
\end{equation*}

\begin{equation*}
\mathbf{\hat{v}}_{c}(\xi,y,t)= e^{i\kappa y}\mathbf{\tilde{v}}_c(\xi,t) + c.c.= e^{i\kappa y}(\tilde v_c^1(\xi,t),\tilde v_c^2(\xi,t))+ c.c.,
\end{equation*}

\begin{equation*}
\mathbf{\hat v}_{w}(\xi,y,t)= e^{i\kappa y}\mathbf{\tilde{v}}_w(\xi,t)+ c.c.= e^{i \kappa y}(\tilde v_w^1(\xi,t),\tilde v_w^2(\xi,t))+ c.c.
\end{equation*}

and

$$ \mathbf{\hat R}(y,t) = (\hat R(y,t), 0)+ c.c.=e^{i\kappa y} (\tilde R(t),0)+ c.c,$$

where $c.c.$ stands for the complex conjugate of its preceding term and $\kappa\in \mathcal{R}$ is arbitrary corresponding to the wavelength of the perturbation in question.

This reduces the system to

\begin{align}\begin{cases}
\label{eq:limit2d}
\dfrac{\partial \tilde \alpha }{\partial t} &=  \Big[\dfrac{\partial S_c}{\partial \alpha}(\alpha_h, C_\infty) -\lambda_2 \Big]\tilde \alpha  - \dfrac{\alpha_h }{R_\ast} \dfrac{\partial \tilde v^1_{c}}{\partial \xi} -\alpha_h i\kappa \tilde v_c^2
\\[.3cm]

0 &= \dfrac{\lambda_2}{1-\alpha_h}\Big(1+\xi \dfrac{\partial }{\partial \xi}\Big) \tilde  \alpha  +  \dfrac{\alpha_h}{R_\ast} \dfrac{\partial  \tilde  v_c^1}{\partial \xi }  +\dfrac{ 1-\alpha_h}{R_\ast} \dfrac{\partial \tilde  v_w^1 }{\partial \xi }\\[.3cm]
& +\alpha_h i\kappa \tilde  v_c^2 + (1-\alpha_h) i\kappa  \tilde  v_w^2
\\[.3cm]

0 & = - \dfrac{\Sigma_c'(\alpha_h)}{R_\ast }   \dfrac{\partial \tilde  \alpha }{\partial \xi} + \dfrac{\hat \mu_c }{R_\ast^2} \dfrac{\partial^2 \tilde  v_c^1}{\partial \xi^2}+\dfrac{(\lambda_c+ \mu_c )}{ R_\ast } i\kappa \dfrac{\partial \tilde  v^2_c}{\partial \xi}   - \mu_c \kappa^2  \tilde  v_c^1
\\[.3cm]

0 & =  \Big( - \Sigma_c(\alpha_h)-\alpha_h \Sigma_c'(\alpha_h) + \lambda_c  \lambda_2  \Big)i\kappa \tilde \alpha
+ \dfrac{\mu_c \alpha_h}{R_\ast^2} \dfrac{\partial^2 \tilde  v_c^2}{\partial \xi^2}
\\
& + (\mu_c+\lambda_c)\dfrac{\alpha_h i\kappa}{R_\ast}\dfrac{\partial \tilde v_c^1}{\partial \xi}
- \alpha_h \kappa^2 \hat \mu_c  \tilde v_c^2
\\[.3cm]

0 &= \dfrac{\alpha_h }{R_\ast^2} \dfrac{\partial}{\partial \xi}( \tilde v_w^2 -  \tilde v_c^2) - \dfrac{i\kappa \alpha_h}{R_\ast} (\tilde v_w^1 - \tilde v_c^1)  +\dfrac{i\kappa \lambda_2 \xi}{1-\alpha_h}\tilde \alpha

\end{cases}
\end{align}

for $(\xi,t)\in (0,1) \times (0,\infty)$, with

$$\dfrac{d \tilde  R}{d t}(t) = \lambda_2 \tilde R(t)+\tilde  v_c^1(1,t),\qquad t\in (0,\infty),$$

subject to the boundary conditions:

\begin{align}
\begin{cases}
\tilde v_c^1\hspace{4pt}=\tilde v_c^2=\tilde v_w^1=0, \qquad &\mbox{ at } \xi=0,
\\%[.1cm]

0 \hspace{8pt}= - \Sigma_c'(\alpha_h) \tilde \alpha + \dfrac{\hat \mu_c }{R_\ast } \dfrac{\partial \tilde v_c^1}{\partial \xi}+\lambda_c  i\kappa \tilde v_c^2,\qquad &\mbox{ at } \xi= 1,
\\[.2cm]

0  \hspace{8pt}= i\kappa \Big( \tilde v_c^1 + 2 \lambda_2  \tilde R \Big) + \dfrac{1}{R_\ast }\dfrac{\partial \tilde v_c^2}{\partial \xi}, \qquad &\mbox{ at } \xi = 1,
\\[.2cm]

\tilde v_w^2  = \dfrac{i \kappa \lambda_2 R_\ast}{1-\alpha_h} \tilde R+\tilde v_c^2, \qquad &\mbox{ at } \xi = 1,
\end{cases}
\end{align}

for $t\in (0,\infty)$.

\section{Asymptotic results}

Since this system cannot be solved analytically, we obtain asymptotic expressions for large $t$ to classify the stability of the one dimensional solution.  For that, we  restrict ourselves to the case where the base state solution is one of a growing tumour, i.e. $\lambda_2>0$ and therefore

$$\epsilon = \epsilon(t) :=: \dfrac{1}{R_\ast(t)} = \dfrac{1}{R_0 e^{\lambda_2 t}} \longrightarrow 0, \qquad \text{ when } \qquad t\rightarrow \infty.$$

Motivated by this limit, $\epsilon(t)$ will be used as a small parameter for $t$ large in the asymptotic approximations that follow.

Once this asymptotic behaviour is characterised we can compare these to the one dimensional base state solution in order to, ultimately, reach a condition for instability of the growing solutions of the limit case with negligible nutrient uptake and cell drag.  Details of this derivation can be found in \cite{genovese2017asymptotic}.  The large-time limit is of singular perturbation type, necessitating the application of the method of matched asymptotic expansions.

\subsection{Outer solution}

We first characterised the long time behaviour of the solution to the system \eqref{eq:limit2d} at a position $\xi\in [0,1]$ away from any possible boundary layers.  Notice that the first three equations of this system pertain only to $\{\tilde \alpha, \tilde v_c^1, \tilde v_c^2\}$.  Thus, we may decouple these equations to solve for $\{\tilde \alpha, \tilde v_c^1, \tilde v_c^2\}$ first and then for $\{\tilde v_w^1, \tilde v_w^2\}$.

Suppose that $\{ a_0, a_1, a_2 \}$ are decay rates of $\tilde \alpha$, $\tilde v_c^1$, and $\tilde v_c^2$, respectively, i.e.

$$\tilde \alpha(\xi,t) = e^{a_0t} \overline \alpha(\xi), \qquad

\tilde v_c^1 (\xi,t) = e^{a_1t} \overline v_c^1 (\xi), \qquad

\tilde v_c^2 (\xi,t) = e^{a_2t} \overline v_c^2 (\xi),$$

with

$$\overline \alpha(\xi) =O(1), \qquad \overline v_c^1(\xi) =O(1), \qquad \overline v_c^2(\xi) =O(1), \qquad t\rightarrow 0,$$

for all relevant $\xi$ (away from any possible boundary layers).  By substituting these into the first three equations in \eqref{eq:limit2d}, we may prove via case-by-case logic, that there is only one set of values $\{a_0,a_1,a_2\}$ that balances these equations for large time and yields non-trivial solutions for $\{\overline \alpha, \overline v_c^1, \overline v_c^2\}$.  Then, using these results we applied the same reasoning to find $\{\tilde v_w^1, \tilde v_w^2\}$ using the last two equations of the system.

This way, we obtained that the non-trivial outer solution for $\xi$ away from any possible boundary layers is, at first order,

\begin{align}\label{eq:outersolutiondecay1}
\tilde \alpha (\xi,t) & = e^{(\gamma_0 -\lambda_2) t}  \overline \alpha (\xi),\\

\tilde v_c^1 (\xi,t) & = \dfrac{\gamma_1}{R_0 }   e^{(\gamma_0 -2\lambda_2) t} \dfrac{d \overline \alpha }{d \xi}(\xi),\label{eq:outersolutiondecay2}\\

\tilde v_c^2 (\xi,t) & = \gamma_3 e^{(\gamma_0-\lambda_2) t}  \overline \alpha(\xi),\label{eq:outersolutiondecay3}\\

\tilde v_w^1(\xi,t) & =  \dfrac{\lambda_2 R_0}{\alpha_h ( 1-\alpha_h)} e^{\gamma_0 t}\xi \overline \alpha (\xi),\label{eq:outersolutiondecay4}\\

\tilde v_w^2(\xi,t)  & = -  e^{(\gamma_0-\lambda_2) t} \Bigg [ \dfrac{\lambda_2}{i \kappa \alpha_h ( 1- \alpha_h)^2}\bigg(1+\xi \dfrac{d}{d\xi} \bigg) \overline \alpha(\xi) + \dfrac{\alpha_h \gamma_3 }{1-\alpha_h} \overline \alpha(\xi)  \Bigg ]\label{eq:outersolutiondecay5}
\end{align}

and it satisfies

\begin{align*}
\begin{cases}
\dfrac{\partial \tilde \alpha }{\partial t} &=  \Big[\dfrac{\partial S_c}{\partial \alpha}(\alpha_h, C_\infty) -\lambda_2 \Big]\tilde \alpha  -\alpha_h i\kappa \tilde v_c^2  \\[.3cm]

0 &= \dfrac{\lambda_2}{1-\alpha_h}\Big(1+\xi \dfrac{\partial }{\partial \xi}\Big) \tilde  \alpha  +\dfrac{ 1-\alpha_h}{R_\ast} \dfrac{\partial \tilde  v_w^1 }{\partial \xi } +\alpha_h i\kappa \tilde  v_c^2 + (1-\alpha_h) i\kappa  \tilde  v_w^2\\[.3cm]

0 & = - \Sigma_c'(\alpha_h)   \dfrac{\partial \tilde  \alpha }{\partial \xi} +(\lambda_c+ \mu_c ) i\kappa \dfrac{\partial \tilde  v^2_c}{\partial \xi}   - \mu_c \kappa^2 R_\ast   \tilde  v_c^1 \\[.3cm]

0 & =  \Big( - \Sigma_c(\alpha_h)-\alpha_h \Sigma_c'(\alpha_h) + \lambda_c  \lambda_2  \Big)i\kappa\tilde \alpha - \alpha_h \kappa^2 \hat \mu_c  \tilde v_c^2. \\[.3cm]

0 &=  - \dfrac{i\kappa \alpha_h}{R_\ast} \tilde v_w^1 +\dfrac{i\kappa \lambda_2 \xi}{1-\alpha_h}\tilde \alpha,

\end{cases}
\end{align*}

where

$$
\begin{array}{c}

\gamma_0  = \dfrac{\partial S_c}{\partial \alpha}(\alpha_h,C_\infty) - \dfrac{\alpha_h\Sigma_c'(\alpha_h)}{\hat \mu_c} +\lambda_2 \Big(\dfrac{\lambda_c}{\hat \mu_c} -1\Big),\\[.5cm]

\gamma_1  = -\dfrac{\Sigma_c'(\alpha_h)}{\mu_c \kappa^2}-\dfrac{ (\lambda_c+\mu_c)( - \Sigma_c(\alpha_h)-\alpha_h \Sigma_c'(\alpha_h) + \lambda_c  \lambda_2  )}{\mu_c \kappa^2 \hat \mu_c \alpha_h},\\[.5cm]

\gamma_3  = -  \dfrac{( - \Sigma_c(\alpha_h)-\alpha_h \Sigma_c'(\alpha_h) + \lambda_c  \lambda_2 )}{\hat \mu_c \alpha_h i \kappa}.

\end{array}
$$

Notice that $\gamma_0$, $\lambda_2$, and therefore all the exponential decay rates specified for the outer solution, interestingly enough, do not depend on the wavelength $\kappa$ of the perturbation in question.

\subsection{Boundary layers}

Now, in order to find the location of possible boundary layers, we must be able to find if there are any regions where rapid change may be observed.  For that, we tested different time frames of our solution numerically and discovered the following behaviour in $\tilde \alpha$ (see figure \ref{fig:1}).

The graph on the right represents the same data as the figure on the left, except that the figure on the right shows the behaviour for $\tilde \alpha$ for $t>5$ rather than $t>0$, as seen on the left.  If we focus our attention on the graph on the right, we see that there are regions of rapid change near $\xi=0$ and $\xi=1$.  This, in turn, suggests that there is a boundary layer near each of these boundaries.

\begin{figure}[H]
\centering
\begin{minipage}{.49\textwidth}
\centering
\includegraphics[width=\textwidth]{FIG1.pdf}
\end{minipage}

\hfill

\begin{minipage}{.49\textwidth}
\centering
\includegraphics[width=\textwidth]{FIG2.pdf}
\end{minipage}
\caption{Perturbation of the cell concentration $\tilde \alpha(\xi,t)$ for $t\in [0,20]$ (on the left) and $t\in[5,20]$ (on the right) showing rapid change near inner and outer boundaries $\xi=0$ and $\xi=1$, indicating the position of boundary layers.}
\label{fig:1}
\end{figure}

\subsection{Inner solution near free boundary}

Recall that, in the previous section, we have seen numerical results that suggest that there might be boundary layers near boundaries at $\xi=0$ and $\xi=1$, which we will call the inner and outer boundary, respectively.  Having said that, in this section, we will analytically characterise an inner solution near the outer boundary. Indeed, by considering all possible asymptotic balances for our equations near $\xi=1$, we obtain four possible non-trivial balances, which may be narrowed down to only one possibility that is non-trivial and allows for matching to be performed between the outer solution and the inner solution.

For this, consider the following change of variable

$$\xi =1 - \dfrac{X}{R_\ast(t)^\beta}, \qquad \beta>0, \qquad X=O(1)\quad\text{ as }\quad t\rightarrow \infty.$$

This will allow us to focus within the boundary layer, a thin strip near the outer boundary $\xi=1$.  By the product rule, we then have

\begin{align*}
\dfrac{\partial}{\partial t} & \longrightarrow \dfrac{\partial}{\partial t} + \beta \lambda_2 X \dfrac{\partial}{\partial X} \\[.1cm]
\dfrac{\partial}{\partial \xi} & \longrightarrow -R_\ast(t)^\beta \dfrac{\partial}{\partial X}\\[.1cm]
\dfrac{\partial^2}{\partial \xi^2} & \longrightarrow R_\ast(t)^{2\beta} \dfrac{\partial^2}{\partial X^2}.
\end{align*}

Recall also that the outer solutions all had a common coefficient: $e^{\gamma_0 t}$.   Thus, to facilitate matching between the outer and inner solutions, we will choose to assume that our inner solution is of the form

$$\tilde \alpha(X,t) = e^{(\gamma_0 + a_0) t} A(X),\quad \tilde v_c^1(X,t) = e^{(\gamma_0 +a_1) t} V_c^1(X),\quad \tilde v_c^2(X,t) = e^{(\gamma_0+ a_2) t} V_c^2(X),$$

$$\tilde v_w^1(X,t) = e^{(\gamma_0+b_1) t} V_w^1(X), \quad \tilde v_w^2(X,t) = e^{(\gamma_0 +b_2 )t} V_w^2(X).$$

By substituting these into the system above and exploring the exponential pre-factors of each of its terms and their behaviour when $t\rightarrow \infty$, it is possible to prove through a case-by-case logic (similar to the one used before) that the only possible non-trivial asymptotic balances happen when

$$a_0= a_1 =a_2, \qquad b_1-a_0 - \lambda_2 = b_2 - a_0 - \lambda_2=0,$$

where the asymptotic balance is given by

\begin{empheq}[left=\empheqlbrace]{align}
0&=  \Big[\dfrac{\partial S_c}{\partial \alpha}(\alpha_h, C_\infty)-\gamma_0 -a_0 -\lambda_2 \Big( 1+ X\dfrac{d}{dX} \Big)\Big]A \nonumber \\
&+ \alpha_h \dfrac{ d V^1_{c}}{d X} -\alpha_h i\kappa V_c^2  \label{eq:AAI4_1} \\[.1cm]

0 &= - \dfrac{\lambda_2 R_0 }{(1-\alpha_h)^2}\dfrac{d A}{d  X}  - \dfrac{d  V_w^1 }{d  X } + i\kappa V_w^2 \label{eq:AAI4_4}\\[.1cm]

0 & = \Sigma_c'(\alpha_h) \dfrac{d A}{d X} + \hat \mu_c \dfrac{d^2 V_c^1}{d X^2}- (\lambda_c+ \mu_c ) i\kappa \dfrac{d V^2_c}{d X}   - \mu_c \kappa^2  V_c^1 \label{eq:AAI4_2}\\[.1cm]

0 & =  \Big( - \Sigma_c(\alpha_h)-\alpha_h \Sigma_c'(\alpha_h) + \lambda_c  \lambda_2  \Big)  i\kappa A
+  \mu_c \alpha_h  \dfrac{d^2 V_c^2}{d X^2} \nonumber \\
  & - (\mu_c+\lambda_c) \alpha_h i\kappa \dfrac{d V_c^1}{d X}
- \alpha_h \kappa^2 \hat \mu_c  V_c^2  \label{eq:AAI4_3}\\[.1cm]

0 &= -\dfrac{d V_w^2}{d  X}   - i\kappa  V_w^1 +\dfrac{i\kappa\lambda_2 R_0 }{\alpha_h(1-\alpha_h)} A, \label{eq:AAI4_5}

\end{empheq}

with conditions at $X=0$:

\begin{align*}
\begin{cases}
0  &= - \Sigma_c'(\alpha_h) A - \hat \mu_c \dfrac{d V_c^1}{d X}+\lambda_c  i\kappa V_c^2,\\[.3cm]

0 &= \dfrac{(\lambda_2-\gamma_0-a_0)}{i\kappa } \dfrac{d V_c^2}{d X}+ (\lambda_2 +\gamma_0+a_0)V_c^1\\[.3cm]

V_w^2 & =\dfrac{R_0}{2(1-\alpha_h)}\Big[ \dfrac{d V_c^2}{dX}- i\kappa V_c^1\Big].
\end{cases}
\end{align*}

Consider initially the first three equations \eqref{eq:AAI4_1}-\eqref{eq:AAI4_3} and the corresponding set $\{ A, V_c^1, V_c^2\}$.  Notice  that the asymptotic balance of these equations preserves all the terms in the original system.  Therefore, it is no surprise that we cannot solve this system by using basic ordinary differential equation methods as we have done in previous cases.  However, we can begin to tackle equations \eqref{eq:AAI4_1}-\eqref{eq:AAI4_3} by defining $\{F_1, F_2\}$ as

$$F_1 = \dfrac{d V_c^1}{dX} -i\kappa V_c^2, \qquad F_2 =\dfrac{d V_c^2}{dX} +i\kappa V_c^1.$$

By substituting these into \eqref{eq:AAI4_1} and \eqref{eq:AAI4_2}, we can find expressions for $\{F_1, F_2\}$ in terms of $A$ which we can then substitute into \eqref{eq:AAI4_3} to obtain a third order ordinary differential equation for $A$:

\begin{equation}\label{eq:AAI4_7}
0 =  \lambda_2 X\dfrac{d^3 A}{dX^3}+\Big[a_0+\lambda_2(\dfrac{\lambda_c}{\hat \mu_c}+2)\Big]\dfrac{d^2 A}{dX^2}-\lambda_2 \kappa^2 X \dfrac{dA}{dX} - \kappa^2(a_0+\lambda_2)A.
\end{equation}

Then, through some algebraic manipulations, the expressions for $\{F_1, F_2\}$ in terms of $A$ allow us to recover $\{V_c^1,V_c^2\}$ by solving:

\begin{align}
\begin{cases}\label{eq:AAI4_8}
\dfrac{d^2 V_c^1}{d X^2}-\kappa^2 V_c^1 & = H_1\\[.1cm]

V_c^2 -\dfrac{1}{i\kappa} \dfrac{dV_c^1}{dX} & = H_2,
\end{cases}
\end{align}

where $H_1$ and $H_2$ depend only on $A$:

\begin{align*}
H_1& =\dfrac{\mu_c + \lambda_c}{\mu_c \alpha_h} \bigg[ \Big(\dfrac{\partial S_c}{\partial \alpha}(\alpha_h, C_\infty)-\gamma_0-a_0 -2 \lambda_2 -\dfrac{\alpha_h \Sigma_c'(\alpha_h)}{\mu_c + \lambda_c}  \Big) \dfrac{dA}{dX}  - \lambda_2 X \dfrac{d^2 A}{d X^2}  \bigg]  \\[.2cm]

H_2& = \dfrac{1}{\alpha_h i\kappa} \Big(\dfrac{\partial S_c}{\partial \alpha}(\alpha_h, C_\infty)-\gamma_0 -a_0-\lambda_2  \Big)A - \dfrac{\lambda_2}{\alpha_h i \kappa } X \dfrac{dA}{dX}.
\end{align*}

However, this whole simplification process relies on solving $A$ from a third order ordinary differential equation, namely \eqref{eq:AAI4_7}, which cannot be easily solved analytically.  Therefore, we will approach the problem differently and find the asymptotic behaviour of $\{A,V_c^1,V_c^2\}$ when $X\rightarrow \infty$ in order to be able to perform matching between the outer and inner solutions to obtain a value for $a_0$ and consequently for $\{a_1,a_2,b_1,b_2\}$.

For this, consider $\hat X = \epsilon X$, with $0<\epsilon\ll 1$.  Thus, equations \eqref{eq:AAI4_7} and \eqref{eq:AAI4_8} above may be found to be, at first order:

$$0=\lambda_2 \hat X \dfrac{dA}{d \hat X}(\hat X) + (a_0+\lambda_2)A(\hat X),\quad

V_c^1(\hat X) = 0,$$

$$V_c^2(\hat X)   = \dfrac{1}{\alpha_h i\kappa} \Big[\dfrac{\partial S_c}{\partial \alpha}(\alpha_h, C_\infty)-\gamma_0 +\lambda_2 \Big(1-\hat X \dfrac{d }{d \hat X}\Big)\Big] A(\hat X).$$

Through traditional ordinary differential equation methods, these can be solved to find that

$$A(\hat X) = C_1 \hat X^{-\dfrac{a_0+\lambda_2}{\lambda_2}}, \qquad V_c^1(\hat X) = 0, \qquad   V_c^2(\hat X)= \gamma_3 C_1 \hat X^{-\dfrac{a_0+\lambda_2}{\lambda_2}}.$$

Thus, as $X\rightarrow +\infty$,

$$A(X) \sim C_1 X^{-\dfrac{a_0+\lambda_2}{\lambda_2}}, \qquad V_c^1(X) \rightarrow 0, \qquad V_c^2(X) \sim \gamma_3 C_1 X^{-\dfrac{a_0+\lambda_2}{\lambda_2}}.$$

Now, by using this asymptotic behaviour and recasting it in terms of $(\xi,t)$ using $X=(1-\xi)R_\ast(t)$, we see that as $X\rightarrow +\infty$,

\begin{align*}
e^{a_0 t} A(X) & \sim e^{a_0 t} C_1 X^{-\dfrac{a_0+\lambda_2}{\lambda_2}}

= C_1 [R_0(1-\xi)]^{-\dfrac{a_0+\lambda_2}{\lambda_2}} e^{-\lambda_2 t},\\

e^{a_0 t} V_c^1(X) & \rightarrow 0,\\

e^{a_0 t} V_c^2(X) & \sim e^{a_0 t} C_1 \gamma_3 X^{-\dfrac{a_0+\lambda_2}{\lambda_2}}

 = \gamma_3 C_1 [R_0(1-\xi)]^{-\dfrac{a_0+\lambda_2}{\lambda_2}} e^{-\lambda_2 t}.
\end{align*}

By matching these to the outer solutions, we see that $a_0 = -2\lambda_2$ and

$$\overline \alpha(\xi) \sim C_1 R_0(1-\xi), \qquad \text{ and } \qquad \dfrac{d \overline \alpha}{d \xi}(\xi) \rightarrow 0$$

when $\xi \rightarrow 1$.  Furthermore,

we see that

$$a_0=a_1=a_2=-2\lambda_2, \qquad b_1=b_2 = a_0+\lambda_2 =-\lambda_2,$$

and thus we have characterised the decay rates of the inner solution near the outer boundary.

If we apply the same process used above for $\{A,V_c^1,V_c^2\}$ now to $\{V_w^1,V_w^2\}$, we see that equations \eqref{eq:AAI4_4} and \eqref{eq:AAI4_5} imply that, at first order,

$$V_w^1(X) \sim \dfrac{\lambda_2 R_0}{\alpha_h (1-\alpha_h)} C_1 X \qquad \text{ and }\qquad V_w^2(X) \rightarrow 0,$$

when $X\rightarrow \infty$ and matching can be performed due to values of $b_1$ and $b_2$ and the conditions previously derived on $\overline \alpha(\xi)$, namely

$$\overline \alpha(\xi) \sim C_1 R_0(1-\xi), \qquad \text{ and } \qquad \dfrac{d \overline \alpha}{d \xi}(\xi) \rightarrow 0,$$
when $\xi \rightarrow 1$.

Now, notice that the two conditions above together imply that $C_1=0$ (via L'Hopital's Rule) and therefore that the asymptotic behaviour obtained here when $X\rightarrow \infty$ is actually trivial.  In order to remedy this, we must revisit equation \eqref{eq:AAI4_7}, now with $a_0=-2\lambda_2$, namely

$$0 =  X\dfrac{d^3 A}{dX^3}+\dfrac{\lambda_c}{\hat \mu_c}\dfrac{d^2 A}{dX^2}-\kappa^2 X \dfrac{dA}{dX} + \kappa^2 A.$$

Notice that this is a third order ordinary differential equation, so we expect three linearly independent solutions and not just one, as we obtained previously with the first order approximation.  Therefore, we will extract the other two solutions by applying a WKBJ expansion to $A(X)$.

Initially, consider $\hat X = \epsilon X,$ with $0<\epsilon\ll 1.$  Then, the equation above becomes:

$$0 = \epsilon^2 \Big (\hat X \dfrac{d^3 A}{d \hat X^3} + \dfrac{\lambda_c}{\hat \mu_c}\dfrac{d^2 A}{d \hat X^2}\Big) -\kappa^2 \Big( \hat X \dfrac{dA}{d \hat X}-A\Big).$$

Note that, at first order, we can retrieve the solution obtained in the previous section:  $A(\hat X)= C_1 \hat X$. Now, as usual in a WKBJ approximation, let

$$A(\hat X, \epsilon ) = \exp\Big\{\dfrac{G(\hat X,\epsilon)}{\epsilon}\Big\},\quad \text{ where } \quad G(\hat X,\epsilon)= G_0(\hat X) + \epsilon G_1(\hat X) + O( \epsilon^2).$$

Through substitution into the equation above, we obtain

\begin{align*}
0 &= \hat X  \Bigg[  \Bigg(\dfrac{d G_0}{d \hat X} \Bigg)^3 - \kappa^2 \dfrac{d G_0}{d \hat X} \Bigg]+ \epsilon \Bigg[\kappa^2 + \dfrac{\lambda_c}{\hat \mu_c}\Bigg(\dfrac{d G_0}{d \hat X} \Bigg)^2 + 3\hat X \dfrac{d G_0}{d \hat X} \dfrac{d^2 G_0}{d \hat X^2} \\
& + 3 \hat X \Bigg(\dfrac{d G_0}{d \hat X} \Bigg)^2 \dfrac{d G_1}{d \hat X} - \hat X \kappa^2 \dfrac{d G_1}{d \hat X}  \Bigg]+O(\epsilon^2).
\end{align*}

Equating terms in $O(1)$ and then $O(\epsilon)$, we see that either

$$\{G_0(\hat X) = \tilde C_0, \quad G_1(\hat X) = \ln(\hat X) + \tilde C_1\},$$
or
$$\{G_0(\hat X) = \pm \kappa \hat X + \tilde C_0, \quad G_1(\hat X) = \omega \ln(\hat X) + \tilde C_1\},$$

where $\omega = -\dfrac{1}{2}\big(1+\dfrac{\lambda_c}{\hat \mu_c}\big)$. Therefore, grouping the constants, we have

$$A(\hat X, \epsilon ) = \exp \Big\{ \frac{G(\hat X,\epsilon)}{\epsilon} \Big\} = \exp \Big\{\frac{G_0(\hat X)}{\epsilon}+G_1(\hat X) \Big\} = C_1 \hat X,$$

or

$$A(\hat X, \epsilon ) = \exp \Big\{ \frac{G(\hat X,\epsilon)}{\epsilon} \Big\} = \exp \Big\{\frac{G_0(\hat X)}{\epsilon}+G_1(\hat X) \Big\}  = C_0 \hat X^{\omega} \exp \Big\{\dfrac{\pm \kappa\hat X}{\epsilon}\Big\}.$$

Notice that the first solution is the same as found previously, by using a first order approximation.  Thus, the latter equation corresponds to the other two solutions for which we searched.  Therefore, since the equation for $A$ is a third order ordinary differential equation, we thus conclude that $A(X)$ satisfies

\begin{equation}\label{eq:AAI4_9}
A(X) \sim C_1 X + C_2 X^{\omega}e^{\kappa X} + C_3 X^{\omega}e^{-\kappa X}, \qquad \omega = -\dfrac{1}{2}\big(1+\dfrac{\lambda_c}{\hat \mu_c}\big)
\end{equation}

when $X\rightarrow \infty$ for some constants $C_1, C_2, C_3$.  However, for matching to be possible, we must have $C_2=0$.

Recall now the equations in \eqref{eq:AAI4_8} for $\{V_c^1, V_c^2\}$ were obtained via algebraic manipulation and the definition of two auxiliary functions $\{ F_1, F_2\}$ that depended on $\{V_c^1, V_c^2\}$.  Similarly, we take an analogous approach to recast equations \eqref{eq:AAI4_4} and \eqref{eq:AAI4_5} to find $\{V_w^1,V_w^2\}$.  To be more specific, we define

$$F_3 = \dfrac{d V_w^1}{dX} -i\kappa V_w^2, \qquad F_4 =\dfrac{d V_w^2}{dX} +i\kappa V_w^1,$$

and, through some algebraic manipulation, recast equations \eqref{eq:AAI4_4} and \eqref{eq:AAI4_5} as

\begin{align}
\begin{cases}\label{eq:AAI4_10}
\dfrac{d^2 V_w^2}{d X^2}-\kappa^2 V_w^2 & = \dfrac{i \kappa R_0 \lambda_2}{\alpha_h (1-\alpha_h)^2} \dfrac{dA}{dX}\\[.1cm]

V_w^1 +\dfrac{1}{i\kappa} \dfrac{d V_w^2}{dX} & = \dfrac{\lambda_2 R_0}{\alpha_h (1-\alpha_h)}A .
\end{cases}
\end{align}

Therefore, using the expression for $A$ in \eqref{eq:AAI4_9}, we can find the asymptotic behaviour of $\{V_c^1, V_c^2, V_w^1, V_w^2\}$ via equations \eqref{eq:AAI4_8} and \eqref{eq:AAI4_10}.  For this, we must solve two second order ordinary differential equations for $\{V_c^1,V_w^2\}$ and obtain $\{V_c^2,V_w^1\}$ by substitution of $\{A,V_c^1,V_w^2\}$ into the remaining equations.

Note also that the solution to the homogeneous version of the ordinary differential equations for $\{V_c^1,V_w^2\}$ will have both decaying and growing exponential terms of the type $e^{-\kappa X}$ and $e^{\kappa X}$, respectively.  However, for matching to be possible, we cannot allow the existence of the growing exponential term of the type $e^{\kappa X}$.  If this is taken into consideration, we can simplify our calculations by assuming any pre-factor of $e^{\kappa X}$ is null.  Thus, by inserting the expression for $A(X)$ into the equations above, we obtained that

\begin{align*}
A(X) & \sim  C_1 X +C_2 X^{\omega} e^{\kappa X}+C_3 X^{\omega} e^{-\kappa X}\\
& \Longrightarrow C_2 = 0\\

V_c^1(X) & \sim -C_1 \gamma_1 + C_4 e^{\kappa X}+C_5 e^{-\kappa X}+C_3 X^{\omega} e^{-\kappa X} p_1^2(X)\\
& \Longrightarrow C_4 = 0\\

V_c^2(X) & \sim \gamma_3 C_1 X + C_5 i e^{-\kappa X} +C_3 X^{\omega-1} e^{-\kappa X} p_2^3(X)\\

V_w^1(X) & \sim \dfrac{\lambda_2 R_0}{\alpha_h ( 1-\alpha_h)} C_1 X + C_6 e^{-\kappa X}+C_3 X^{\omega-1} e^{-\kappa X} p_3^2(X) \\[.1cm]

V_w^2(X) & \sim \dfrac{\lambda_2 R_0 C_1}{i \kappa \alpha_h ( 1-\alpha_h)^2}  +C_6 i e^{-\kappa X}+C_7 e^{\kappa X}+C_3 X^{\omega} e^{-\kappa X}p_4^1(X)\\
& \Longrightarrow C_7 = 0,
\end{align*}

when $X\rightarrow \infty$.  Here we used the following notation:

$$p_i^n (X) = A_{i,0}+A_{i,1}X+A_{i,2}X^2 ... + A_{i,n}X^n, \qquad i=1,2,3,4, \qquad n\in \mathbf{N},$$

where every $A_{i,j}$ is a constant that may be found explicitly in terms of the parameters of our system.  Because these polynomials appear in terms that decay exponentially when $X\rightarrow \infty$, there is no need to make the expressions for $A_{i,j}$ explicit here as these are negligible terms for the purpose of matching.

In the process used to derive the expressions above, we also used the first order term of the following asymptotic expansion:

\begin{align*}
\int^X e^{-2\kappa s} s^\eta ds & \sim - \dfrac{X^\eta  e^{-2\kappa X}}{2\kappa} \Big( 1+ \dfrac{\eta }{2\kappa}\dfrac{1}{X}+\dfrac{\eta (\eta -1)}{(2\kappa)^2}\dfrac{1}{X^2}\\
 & +\dfrac{\eta (\eta -1)(\eta -2)}{(2\kappa)^3}\dfrac{1}{X^3} + O\Big(\Big(\dfrac{1}{X}\Big)^4\Big) \Big),
\end{align*}

when $X\rightarrow \infty$, which can be proven by integrating by parts repeatedly (See \cite{temme2014asymptotic}).

The terms that grow exponentially must be discarded in order to allow us to perform matching when $X\rightarrow \infty$, and therefore $C_2=C_4=C_7=0$.  This way, no exponentially growing term is carried along in the derivation of $V_c^1$, $V_c^2$, $V_w^1$, and $V_w^2$ and we may obtain $3$ boundary conditions for the inner solution when $X\rightarrow \infty$.  Since there are $3$ boundary conditions already at $X=0$, this implies that we have a total of $6$ boundary conditions that may be applied to the inner solution at $X=0$ and when $X\rightarrow \infty$ in order to perform numerical simulations.

Notice that thus we have $4$ degrees of freedom, corresponding to the constants $\{ C_1,C_3,C_5,C_6 \}$.  However, $\{ C_3,C_5,C_6 \}$ do not play a role in matching, since these are the coefficients of terms that decay exponentially when $X\rightarrow \infty$.

That being said, we will now find a condition for $C_1$ by matching the inner solution to the outer solution when $X\rightarrow \infty$.  Indeed, from the expressions above, we see that when $X\rightarrow \infty$, at first order

\begin{align*}
A(X) & \sim  C_1 X \\

V_c^1(X) & \sim - C_1 \gamma_1 \\

V_c^2(X) & \sim C_1 \gamma_3 X \\

V_w^1(X) & \sim \dfrac{\lambda_2 R_0}{\alpha_h ( 1-\alpha_h)} C_1 X \\[.1cm]

V_w^2(X) & \sim \dfrac{\lambda_2 R_0 C_1}{i\kappa \alpha_h ( 1-\alpha_h)^2}.
\end{align*}

Notice that the asymptotic far-field behaviour found for $A$, $V_c^2$, and $V_w^1$ is the same as found previously, by using a first order approximation.  However, instead of having

$$V_c^1(X) \rightarrow 0, \qquad \text{ and } \qquad V_w^2(X) \rightarrow 0$$

when $X\rightarrow \infty$, we now have

$$V_c^1(X) \sim - C_1 \gamma_1, \qquad \text{ and } \qquad V_w^2(X) \sim \dfrac{\lambda_2 R_0 C_1}{i\kappa \alpha_h ( 1-\alpha_h)^2}$$

when $X\rightarrow \infty.$  If we match these to the outer solution when $\xi\rightarrow 1$, we obtain that for matching to be possible we must have

$$C_1 = -\dfrac{1}{R_0} \lim_{\xi \rightarrow 1} \dfrac{d \overline \alpha}{d \xi}(\xi), \qquad \text{ and } \qquad \overline \alpha(\xi) \sim C_1 R_0 (1-\xi),$$

when $\xi\rightarrow 1$, where $\overline \alpha(\xi)$ is part of the outer solution.

As in the previous section, notice again that all the exponential decay rates specified for the inner solution near the outer boundary, interestingly enough, do not depend on the wavelength $\kappa$ of the perturbation in question.

\subsection{Inner solution near fixed boundary}

Finally, we characterised the inner solution in the boundary layer near the inner boundary $\xi=0$.  For that, we have performed a process very similar to the one applied to the characterisation of the inner solution near the outer boundary.  Most changes stem from different rescaling on $\xi$, namely

$$\xi =\dfrac{x}{R_\ast(t)^\beta}, \quad x=O(1)\text{ as }t\rightarrow \infty, \quad \beta>0.$$

Indeed, we now want to consider a thin strip near $\xi=0$, rather than $\xi=1$, which justifies the rescaling above.

Following the same process as in the previous section, we found that the inner solution near the inner boundary $\xi=0$ satisfies at order one:

\begin{align*}
\tilde \alpha (x,t) & = e^{(\gamma_0 -2 \lambda_2) t}  A(x)\\
\tilde v_c^1 (x,t) & = e^{(\gamma_0 -2 \lambda_2) t}  V_c^1(x)\\
\tilde v_c^2 (x,t) & = e^{(\gamma_0 -2 \lambda_2) t}  V_c^2(x)\\
\tilde v_w^1 (x,t) & = e^{(\gamma_0 - 2\lambda_2) t}  V_w^1(x)\\
\tilde v_w^2 (x,t) & = e^{(\gamma_0 - 2\lambda_2) t}  V_w^2(x)
\end{align*}

where

$$\xi=\dfrac{x}{R_\ast(t)}$$

and $A(x),$ $V_c^1(x)$, $V_c^2(x)$, $V_w^1(x)$, and $V_w^2(x)$ satisfy

\begin{align*}
\begin{cases}
0&=  \Big[\dfrac{\partial S_c}{\partial \alpha}(\alpha_h, C_\infty)-\gamma_0 +\lambda_2 \Big( 1- x\dfrac{d}{dx} \Big)\Big]A- \alpha_h \dfrac{ d V^1_{c}}{d x} -\alpha_h i \kappa  V_c^2  \\[.2cm]

0 & = -\Sigma_c'(\alpha_h) \dfrac{d A}{d x} + \hat \mu_c \dfrac{d^2 V_c^1}{d x^2}+ (\lambda_c+ \mu_c ) i \kappa   \dfrac{d V^2_c}{d x}   - \mu_c \kappa^2  V_c^1 \\[.2cm]

0 & =  \Big( - \Sigma_c(\alpha_h)-\alpha_h \Sigma_c'(\alpha_h) + \lambda_c  \lambda_2  \Big)  i \kappa  A
+  \mu_c \alpha_h  \dfrac{d^2 V_c^2}{d x^2}\\%[.1cm]
& + (\mu_c+\lambda_c) \alpha_h i \kappa  \dfrac{d V_c^1}{d x}
- \alpha_h \kappa^2 \hat \mu_c  V_c^2  \\[.2cm]

0 &=  \dfrac{\lambda_2}{1-\alpha_h} \big( 1+x \dfrac{d}{dx} \big) A  +\alpha_h \dfrac{d V_c^1 }{d x }+\alpha_h i \kappa V_c^2 \\%[.1cm]
& + (1-\alpha_h) \dfrac{d  V_w^1 }{d  x } + (1-\alpha_h) i \kappa   V_w^2\\[.2cm]

0 &= \dfrac{\lambda_2 i \kappa }{1-\alpha_h}x A +\alpha_h i \kappa  V_c^1-\alpha_h \dfrac{d V_c^2 }{d x }  - \alpha_h i \kappa   V_w^1+ \alpha_h \dfrac{d V_w^2}{d  x},

\end{cases}
\end{align*}

and $V_c^1(0)=V_c^2(0)=V_w^1(0)=0$.

Far field behaviour when $x\rightarrow \infty$ is, at first order

\begin{align*}
A(x) & \sim  D_1 x \\

V_c^1(x) & \sim D_1 \gamma_1 \\

V_c^2(x) & \sim D_1 \gamma_3 x \\

V_w^1(x) & \sim \dfrac{\lambda_2 }{\alpha_h ( 1-\alpha_h)} D_1 x^2\\% +D_1 B_0 \\

V_w^2(x) & \sim D_1 B_1 x,
\end{align*}

where matching yields

$$D_1 = \dfrac{1}{R_0} \lim_{\xi \rightarrow 0} \dfrac{d \overline \alpha}{d \xi}(\xi), \qquad \text{ and } \qquad \overline \alpha(\xi) \sim D_1 R_0 \xi,$$

when $\xi \rightarrow 0$.  Details of this derivation can be found in \cite{genovese2017asymptotic}.

As in the previous sections, notice again that all the exponential decay rates specified for the inner solution near the inner boundary, interestingly enough, do not depend on the wavelength $\kappa$ of the perturbation in question.

At this point, notice that the asymptotic characterisation of the outer and inner solutions has been obtained.

Additionally, the decays found in this characterisation were validated numerically by simulating the system satisfied by the perturbations and comparing decays to those characterised in the previous sections, for various regions of the tumour.  The interested reader may find more details in \cite{genovese2017asymptotic}.

\section{Summary}

The non-homogeneous base state solutions are unstable to two dimensional perturbations if one of the following has an infinite limit for some $(x,y)\in (0, R_\ast(t))\times \mathcal{R}$ when $t\rightarrow \infty$:

$$\Bigg|\dfrac{\hat \alpha(x,y,t)}{\alpha_h}\Bigg|, \qquad  \dfrac{ || \mathbf{\hat  v}_c(x,y,t)|| }{||\mathbf{v}_{cs}(x)||},\qquad  \dfrac{|| \mathbf{\hat v}_w(x,y,t)|| }{||\mathbf{v}_{ws}(x)||}, \qquad  \Bigg|\dfrac{\hat R(y,t)}{R_\ast(t)}\Bigg|,$$

where $||\cdot || $ is the euclidean norm.  Equivalently, since the base state solution is one dimensional, it is linearly unstable if one of the following is infinite for some $(x,y)\in (0, R_\ast(t))\times \mathcal{R}$  when $t\rightarrow \infty$:

$$\Bigg|\dfrac{\hat \alpha(x,y,t)}{\alpha_h}\Bigg|, \qquad  \Bigg|\dfrac{\hat v_c^j(x,y,t)}{v_{cs}(x)}\Bigg|, \qquad  \Bigg|\dfrac{\hat v_w^j(x,y,t)}{v_{ws}(x)}\Bigg|, \qquad \Bigg|\dfrac{\hat R(y,t)}{R_\ast(t)}\Bigg|, \qquad j=1,2.$$

By inserting the definition of the base state solutions into the ratios above and using the $(\xi,t)$, $(X,t)$ and $(x,t)$ formulations defined in the previous section, through some algebraic manipulation, it is thus possible to prove that the base state solution is linearly unstable to two dimensional perturbations if one of the following is infinite for some $\xi\in (0,1)$, $X,x \in [0,\infty)$  when $t\rightarrow \infty$:

\begin{table}[H]
\centering
\begin{tabular}{p{2.5cm} p{3cm} p{3cm} }

$|\tilde \alpha(\xi,t)|$ & $|\tilde v_c^j(\xi,t)| e^{-\lambda_2 t}$ & $|\tilde v_w^j(\xi,t)|e^{-\lambda_2 t}$ \\

$|\tilde \alpha(X,t)|$ & $|\tilde v_c^j(X,t)| e^{-\lambda_2 t}$ & $|\tilde v_w^j(X,t)| e^{-\lambda_2 t}$ \\

 $|\tilde \alpha(x,t)|$ & $|\tilde v_c^j(x,t)|$ & $|\tilde v_w^j(x,t)|$

\end{tabular}
\end{table}

\noindent for $j=1,2$ or

$$\lim_{t\rightarrow \infty} |\tilde R(t)| e^{-\lambda_2 t} = \infty.$$

By inserting the asymptotic characterisation of the perturbations obtained in the previous section into the expressions above, we can find a condition for instability of the growing base state solutions in terms of $\gamma_0$ and $\lambda_2$.  By doing this, we may conclude that for the base state solution to be unstable, one of the following exponentials must tend to $\infty$ when $t\rightarrow \infty$:

$$e^{(\gamma_0-\lambda_2)t}, \qquad e^{(\gamma_0-2\lambda_2)t}, \qquad e^{(\gamma_0-3\lambda_2)t}.$$

Thus, since $\lambda_2>0$, for instability we must have $\gamma_0-\lambda_2>0$.
\vspace{.5cm}

\subsection{Conclusion}

The growing base state solutions $(\alpha_h, \mathbf{v}_{cs}, \mathbf{v}_{ws}, \mathbf{R}_\ast)$ of the two dimensional limit case system with negligible nutrient uptake and cell drag obtained in section \ref{sec:3} are unstable to two dimensional perturbations when

$$\gamma_0  -\lambda_2 = \dfrac{\partial S_c}{\partial \alpha}(\alpha_h,C_\infty) - \dfrac{\alpha_h\Sigma_c'(\alpha_h)}{\hat \mu_c} +\lambda_2 \Big(\dfrac{\lambda_c}{\hat \mu_c} -2\Big) >0,$$

where $\alpha_h$ and $\lambda_2$ are defined by

$$\hat\mu_c S_c(\alpha_h, C_\infty) = \alpha_h \Sigma_c(\alpha_h), \qquad \lambda_2 =\dfrac{\Sigma_c(\alpha_h)}{\hat\mu_c}.$$

\vspace*{6pt}

\noindent\textbf{Acknowledgements}\vspace*{3pt}

\noindent This work was produced with the financial support of the University of Nottingham, \textit{Coordena\c{c}\~ao de Aperfei\c{c}oamento de Pessoal de N\'ivel Superior} (CAPES), \textit{Funda\c{c}\~ao de Apoio a Pesquisa do Distrito Federal} (FAPDF), \textit{Conselho Nacional de Desenvolvimento Cient\'ifico e Tecnol\'ogico} (CNPq) and the University of Brasilia along with the collaboration of Prof. Daniele Avitabile from the Department of Mathematics, Vrije Universiteit Amsterdam, Netherlands.


TITLE: Showing that $\{ n\mid \nu(n) \equiv 0 \pmod e\}$ is independent of primitive root
QUESTION [1 upvotes]: Let us suppose we have a prime $p > 3$ and two primitive roots $g_1, g_2$ of $p$ with index functions $\nu_1$ and $\nu_2$ (so that $g_1^{\nu_1(n)} = n$). Also suppose that $ef = p-1$ is a (nontrivial) factorization.

In short, my question is: Why is it 'obvious' that the set $\{n \mid \nu_1(n) \equiv 0 \pmod e \}$ is the same as the set $\{ n \mid \nu_2(n) \equiv 0 \pmod e\}$?

The rest is just for context and motivation.
This comes up in a standard proof of the prime-modulus-only case of Dirichlet's Theorem on the density of primes in arithmetic progressions, but my justification of why this fact is true has always bugged me. Like many number theoretic arguments, it's sort of a trick. In particular, we need that $\sum_{(\nu(n) \equiv 0 \mod e)} \zeta^n$ is independent of the primitive root we use, where $\zeta = e^{2\pi i / p}$ is the first primitive $p$th root of unity and $\nu(n)$ denotes the index of $n$ relative to whatever primitive root we're using. One way to show that this is true is to just realize that the sum $\displaystyle \frac{1}{e}\sum_{x = 0}^{p-1} \zeta^{x^e}$ gives us the sum we want and is independent of $\zeta$. But there's no intuition behind this statement to me, really.
It's not true that shifted equivalence classes are also permutations of each other here, i.e. that the set $\{ n | \nu(n) \equiv j \pmod e \}$ for $j \neq 0$ will be independent of the primitive root in general. But the sort of trick that I know above doesn't shed light onto this. (Why doesn't the trick work here? If you boil it down, the trick relies on the fact that $\nu(x^e) = e \nu(x) \equiv 0 \pmod e$, and this doesn't work so well with $\nu(x^{e + 1})$.)

REPLY [2 votes]: Another way to put the answer by André Nicolas: the cyclic group $(\mathbf Z/p \mathbf Z)^\times$ only has a single (cyclic) subgroup of order $f=(p-1)/e$, and both congruences you gave necessarily describe this subgroup.


text:
\begin{document}

\begin{abstract}
A \textit{curious number} is a palindromic number whose base ten representation has the form \linebreak $a \ldots a b \ldots b a \ldots a$. In this paper, we  determine all curious numbers that are perfect squares. Our proof involves reducing the search for such numbers to several single variable families. From here, we complete the proof in two different ways. The first approach is elementary, though somewhat ad hoc. The second entails studying integral points on elliptic curves and is more systematic.
\end{abstract}

\maketitle

\vspace{-.5cm}

\section{introduction}

Ian Stewart begins his popular recreational mathematics book \textit{Professor Stewart's Hoard of Mathematical Treasures} \cite{St2010} with the following \quotes{calculator curiosity}:
\begin{align*}
(8 \times 8) + 13 &= 77 \\
(8 \times 88) + 13 &= 717 \\
(8 \times 888) + 13 &= 7117 \\
(8 \times 8888) + 13 &= 71117 \\
(8 \times 88888) + 13 &= 711117 \\
(8 \times 888888) + 13 &= 7111117 \\
(8 \times 8888888) + 13 &= 71111117 \\
(8 \times 88888888) + 13 &= 711111117.
\end{align*}
The numbers on the right hand sides of the equalities above are examples of what we call \textit{curious} numbers.

\begin{defn} For nonnegative integers $m,n$, an integer is \textit{$(m,n)$-curious} if its base ten representation is
\[ \underbrace{a \ldots a}_{m} \underbrace{b \ldots b}_{n} \underbrace{a \ldots a}_{m} \]
for some integers $1 \leq a \leq 9$ and $0 \leq b \leq 9$. A number is called \textit{curious} if it is $(m,n)$-curious for some $m,n$.
\end{defn}

We see that none of the numbers in Stewart's \quotes{calculator curiosity} are perfect squares. This follows from the observation that each number in the family $71\ldots17$ is congruent to $2$ modulo $5$, a quadratic non-residue.

The question of which repdigits (i.e. $(0,n)$-curious numbers) are perfect squares is a pleasant exercise in elementary number theory. There are 28 repdigits of length at most three. Among those, only $0,1,4,$ and $9$ are perfect squares. No repdigit of length at least four is a perfect square since one may verify that each of
\[ 1111, \, 2222, \, 3333, \, 4444, \, 5555, \, 6666, \, 7777, \, 8888, \, \text{and } 9999 \]
is a quadratic non-residue modulo 10000.

The general question of which repdigits are perfect powers is known as \textit{Obl\'ath's problem}. After R. Obl\'ath's  \cite{Ob1956} partial solution, the problem was fully solved by Y. Bugeaud and M. Mignotte \cite{BM19999} using bounds on $p$-adic logarithms. Variations on Obl\'ath's problem have been studied by several authors. For instance, A. Gica and L. Panaitopol \cite{GP2003} determined all perfect squares among the \textit{near repdigits}, which are integers for which all but a single digit are equal. Recently, B. Goddard and J. Rouse \cite{GR2017}  determined the perfect squares that may be written as the sum of two repdigits. Whereas other authors relied on techniques related to Pell equations to settle difficult cases, Goddard and Rouse used a powerful technique involving elliptic curves.

In this paper, we determine which curious numbers are perfect squares. Specifically, we prove the following.
\begin{thrm} \label{main-thrm} The curious numbers that are perfect squares are  $0, \, 1,\, 4,\, 9,\, 121,\, 484,\, 676,$ and $44944.$
\end{thrm}
Broadly speaking, we prove Theorem \ref{main-thrm} as follows. First, in \S \ref{alg-reform}, we recast the problem in algebraic terms. Next, in \S \ref{mod107}, we reduce modulo $10^7$ to narrow down our search for curious squares to thirteen single variable families. From here, we complete the proof in two distinct ways. The first approach, given in \S \ref{modfamilies}, proceeds by studying the families via modular arithmetic. This approach is somewhat ad hoc, whereas our second approach, given in \S \ref{elliptic},  is more  systematic. Here we study the families via integral points on elliptic curves, along the lines of Goddard and Rouse \cite[\S4]{GR2017}.

\section{Algebraic reformulation} \label{alg-reform}

In this section, we recast our problem in algebraic terms. To do so, we start by developing an algebraic expression for curious numbers. Note the following standard expression for repdigits,
\begin{equation} \label{repdigit} a_m \coloneqq \underbrace{a \ldots a}_m = a \cdot \frac{10^m-1}{10-1}. \end{equation}
Now observe that
\begin{equation} \label{cur-notation}  a_mb_na_m \coloneqq
\underbrace{a \ldots a}_{m} \underbrace{b \ldots b}_{n} \underbrace{a \ldots a}_{m} = 10^{m+n} \cdot a_m + 10^m \cdot b_n + a_m.
\end{equation}
In the shorthand $ a_mb_na_m$, juxtaposition denotes concatenation rather than multiplication. Though this notation is more compact, it may cause confusion. In such instances, we opt to use the longhand notation instead. Now, upon combining equations (\ref{repdigit}) and (\ref{cur-notation}), we obtain an expression for curious numbers,
\begin{equation} \label{exp1}
a_mb_na_m
= a \cdot 10^{m+n} \cdot \frac{10^m-1}{10-1} + b \cdot 10^m \cdot \frac{10^n-1}{10-1} + a\cdot \frac{10^m-1}{10-1}. \\
\end{equation}

To streamline matters, we define the integers
\begin{equation} \label{MN}
    M_{a,b,m} \coloneqq 10^m \cdot (a-b) - a \quad \text{ and } \quad
    N_{a,b,m} \coloneqq  10^m (a \cdot 10^m + b - a).
\end{equation}
After some regrouping, (\ref{exp1}) becomes
\begin{equation} \label{cur-alg-clean}
a_m b_n a_m
= \frac{1}{9} \p{N_{a,b,m} \cdot 10^n + M_{a,b,m}}\!.
\end{equation}

We're interested in determining the collection of $a_mb_na_m$ that are perfect squares. Thus, by multiplying (\ref{cur-alg-clean}) through by 9, we record that our problem reduces to solving the equation
\begin{equation} \label{alg-eq} (3y)^2 = N_{a,b,m} \cdot 10^n + M_{a,b,m}  \end{equation}
in the nonnegative integers $a,b,m,n,y$ with the restrictions that $1 \leq a \leq 9$ and $0 \leq b \leq 9$.

\section{Narrowing search to several single variable families} \label{mod107}

Let $\mathcal{C}_{m,n}$ denote the set of $(m,n)$-curious numbers and write $\mathcal{C} = \bigcup_{m,n \geq 0} \mathcal{C}_{m,n}$ for the set of all curious numbers. Let $\mathcal{S}$ denote the set of all perfect squares. For a positive integer $k$, define the set
\[ \mathcal{S}_k \coloneqq \set{s^2 \in \mathcal{S} : s \leq k}\!. \]
Finally, we denote the reduction modulo $10^7$ map by
\[\pi: \Z \longrightarrow \Z/10^7\Z.\]

In the notation above, our ultimate objective is to determine the intersection $\mathcal{C} \cap \mathcal{S}$. We start by computing the intersection $\pi(\mathcal{C}) \cap \pi(\mathcal{S})$. Here, our choice to reduce modulo $10^7$ is somewhat arbitrary. In fact, what follows works provided that we reduce modulo $10^k$ for any $k \geq 4$. However, we choose $k = 7$ because this is the smallest exponent for which the number of families that we will need to consider in the next section is minimal.

Observe that $\pi(\mathcal{C})$ and $\pi(\mathcal{S})$ may be realized as
\[ \pi(\mathcal{C}) = \bigcup_{\substack{(m,n) \\ 0 \leq m \leq 7 \\ 0 \leq n \leq 7-m}} \pi(\mathcal{C}_{m,n}) \quad \text{and} \quad \pi(\mathcal{S}) = \pi(\mathcal{S}_{10^{7}}). \]
As the sets $\mathcal{C}_{m,n}$ and $\mathcal{S}_{10^{7}}$ are finite and of reasonable size, it is straightforward to determine $\pi(\mathcal{C})$ and $\pi(\mathcal{S})$ via machine computation. We do so using SageMath \cite{SAGE}, and find that they have the intersection
\begin{align*}
\pi(\mathcal{C}) \cap \pi(\mathcal{S}) =
\{
&0,
 1,
 4,
 9,
 121,
 161,
 484,
 656,
 676,
 929,
 969,
 1001,
 1441,
 1881,
 4004,
 4224,
 5225,
 6116, \\
& 6336,
 9009,
 9449,
 9889,
 10001,
 14441,
 18881,
 40004,
 44544,
 44644,
 44944,
 52225,
 67776, \\
& 90009,
 94449,
 98889,
 100001,
 144441,
 188881,
 400004,
 442244,
 447744,
 522225,
 655556, \\
& 677776,
 900009,
 944449,
 988889,
 1000001,
 1444441,
 1888881,
 2222224,
 2222225,
 2222244, \\
& 3333444,
 4000004,
 4222224,
 4222244,
 4333444,
 4422244,
 4433344,
 4433444,
 4441444, \\
& 4444441,
 4444449,
 4445444,
 4449444,
 4477444,
 4777444,
 4777744,
 5222225,
 5555556, \\
& 6555556,
 7777444,
 8888881,
 8888889,
 9000009,
 9444449,
 9888889
\}.
\end{align*}

We have that $\pi(\mathcal{C} \cap \mathcal{S}) \subseteq \pi(\mathcal{C}) \cap \pi(\mathcal{S})$. Those elements of $\pi(\mathcal{C}) \cap \pi(\mathcal{S})$ whose preimage under the map $\mathcal{C} \overset{\pi}{\longrightarrow} \Z/10^7\Z$ consists of a single non-square integer are not members of $\pi(\mathcal{C} \cap \mathcal{S})$. For instance, the preimage of $4333444$ under $\mathcal{C} \overset{\pi}{\longrightarrow} \Z/10^7\Z$ is the singleton $\set{444333444}$. Since $444333444$ is not a perfect square, we ascertain that $44333444 \not\in \pi(\mathcal{C} \cap \mathcal{S})$. Proceeding in this way, we find that
\begin{align*}
\pi(\mathcal{C} \cap \mathcal{S}) \subseteq
\{
& 0,
 1,
 4,
 9,
 121,
 484,
 676,
 44944,
 2222224,
 2222225,
 2222244,
 3333444, \\
& 4444441,
 4444449,
 5555556,
 7777444,
 8888881,
 8888889
\}.
\end{align*}
Taking the preimage of the above set inclusion under the map $\mathcal{C} \overset{\pi}{\longrightarrow} \Z/10^7\Z$, we deduce that
\begin{align} \label{CSpos}
\mathcal{C} \cap \mathcal{S} \subseteq
\{
&0,
1,
4,
9,
121,
 484,
 676,
 44944,
 10 \ldots 01,
 14\ldots41,
 18\ldots81,
 40 \ldots 04,
 42\ldots24,
 442\ldots244, \\
& 4443\ldots3444,
 4447 \ldots 7444,
 52\ldots25,
 65\ldots56,
 90 \ldots 09,
 94\ldots49,
 98\ldots89 \nonumber
\}.
\end{align}
The elements above that are listed with an ellipsis are placeholders for the appropriately corresponding single variable family of curious numbers. For instance, we write $4443\ldots3444$ to denote the family
\[ 4443\ldots3444 \coloneqq \{444\underbrace{3\ldots3}_n444 : n \geq 0\}. \]

The assertion of Theorem \ref{main-thrm} is that $\mathcal{C} \cap \mathcal{S} = \set{0, \, 1, \, 4, \, 9, \, 121, \, 484, \, 676, \, 44944}$. Thus, by (\ref{CSpos}), it suffices to prove that there does not exist a perfect square in any of the following thirteen single variable families,
\begin{align*} \mathcal{F} \coloneqq \{
 &10 \ldots 01, \,
 14\ldots41, \,
 18\ldots81, \,
 40 \ldots 04, \,
 42\ldots24, \,
 442\ldots244, \,
 4443\ldots3444, \, \\
& 4447 \ldots 7444, \,
 52\ldots25, \,
 65\ldots56, \,
 90 \ldots 09, \,
 94\ldots49, \,
 98\ldots89 \}.
 \end{align*}

\section{Considering families via modular arithmetic} \label{modfamilies}

In this section, we use modular arithmetic considerations to prove that none of the families listed in $\mathcal{F}$ contain a square. Let's begin by considering $10\ldots01$. Observe that each number in this family is congruent to 2 modulo 3. As 2 is a quadratic non-residue modulo 3, it follows at once that no number in this family is a square. In fact, since each of the numbers in $40\ldots04$ and $90\ldots09$ is a square multiple of a number in $10\ldots01$, we deduce that these two families contain no squares as well.

Given the above, our problem is reduced to proving that the ten remaining families of $\mathcal{F}$ contain no squares. For each of these families, we record below the coefficients  $M_{a,b,m}$ and $N_{a,b,m}$, as defined in (\ref{MN}).
\renewcommand{\arraystretch}{1.20}
\begin{table}[H]
\begin{tabular}{|lcccrr|} \hline
Family           & $a$ & $b$ & $m$ & $M_{a,b,m}$ & $N_{a,b,m}$ \\ \hline
$14\ldots41$     & 1   & 4   & 1   & $-31$       & 130         \\
$18\ldots81$     & 1   & 8   & 1   & $-71$       & 170         \\
$42\ldots24$     & 4   & 2   & 1   & 16          & 380         \\
$442\ldots244$   & 4   & 2   & 2   & 196         & 39800       \\
$4443\ldots3444$ & 4   & 3   & 3   & 996         & 3999000     \\
$4447\ldots7444$ & 4   & 7   & 3   & $-3004$     & 4003000     \\
$52\ldots25$     & 5   & 2   & 1   & 25          & 470         \\
$65\ldots56$     & 6   & 5   & 1   & 4           & 590         \\
$94\ldots49$     & 9   & 4   & 1   & 41          & 850         \\
$98\ldots89$     & 9   & 8   & 1   & 1           & 890 \\ \hline
\end{tabular}
\caption{} \label{MN_table}
\end{table}
\vspace{-.4cm}

We rule out the possibility of perfect squares in each of the above families via two elementary lemmas.

\begin{lemm} \label{mod-lem} If $M_{a,b,m}$ is a quadratic non-residue modulo $N_{a,b,m}$, then $a_mb_na_m$ is not a square for each $n \geq 0$.
\end{lemm}
\begin{proof} We prove the contrapositive. Suppose that $n_0$ is a nonnegative integer for which $a_mb_{n_0}a_m$ is a square. Then $a,b,m,n_0,$ and $y \coloneqq \sqrt{a_m b_{n_0}a_m}$ give a solution to the equation (\ref{alg-eq}). Reducing this equation modulo $N_{a,b,m}$, we find that
\[ M_{a,b,m} \equiv (3y)^2 \pmod{N_{a,b,m}}. \]
Hence, $M_{a,b,m}$ is a quadratic residue modulo $N_{a,b,m}$.
\end{proof}

For instance, let's consider the family $14\ldots41$ in view of the above lemma. From Table \ref{MN_table}, we read that $(a,b,m) = (1,4,1)$,  $M_{a,b,m} = -31$, and $N_{a,b,m} = 130$. Note that $-31$ is a quadratic non-residue modulo $130$. Thus, we deduce that the family $14\ldots41$ contains no perfect squares. Applying Lemma \ref{mod-lem} to the data from Table \ref{MN_table} in this way for each of the ten families, we conclude that none of the following five families contain a perfect square:
\[ 14\ldots41, \, 18\ldots81, \, 4443\ldots3444, \, 4447\ldots7444, \, \text{and } 94\ldots49. \]

This leaves the five other families to consider. We do so via our next lemma, which requires some notation from elementary number theory. Let $M,N$ be  nonzero integers with $\gcd(M,N) = 1$. Then $M$ is invertible modulo $N$. Hence, we may speak of its multiplicative order modulo $N$, which is denoted by  $\o{ord}_N(M)$. Explicitly, $\o{ord}_N(M)$ is the least positive integer with the property that $M^{\o{ord}_N(M)} \equiv 1 \pmod{N}$.

\begin{lemm} \label{mod-lem2} Let $N$ be a positive integer with $\gcd(N,10) = 1$. If for each integer $k$ with $0 \leq k < \o{ord}_N(10)$, we have that $N_{a,b,m} \cdot 10^k + M_{a,b,m}$ is a quadratic non-residue modulo $N$, then $a_m b_n a_m$ is not a square for each $n \geq 0$.
\end{lemm}
\begin{proof} We prove the contrapositive. Suppose  that $n_0$ is such that $a_m b_{n_0} a_m$ is a square. Then $a,b,m,n_0,$ and $y \coloneqq \sqrt{a_m b_{n_0}a_m}$ give a solution to the equation (\ref{alg-eq}). Write $k_0$ to denote the integer for which $0 \leq k_0 < \o{ord}_N(10)$ and $k_0 \equiv n_0 \pmod{\o{ord}_N(10)}$. Then $10^{n_0} \equiv 10^{k_0} \pmod{N}$, so upon reducing (\ref{alg-eq}) modulo $N$, we find that
\[ N_{a,b,m} \cdot 10^{k_0} + M_{a,b,m} \equiv N_{a,b,m} \cdot 10^{n_0} + M_{a,b,m} \equiv (3y)^2 \pmod{N}. \]
Hence, $N_{a,b,m} \cdot 10^{k_0} + M_{a,b,m}$ is a quadratic residue modulo $N$.
\end{proof}

Using SageMath, we search for (and find) appropriate $N$ as in Lemma \ref{mod-lem2} for each of five remaining families. The relevant data is tabulated below.
\renewcommand{\arraystretch}{1.20}
\begin{table}[H]
\begin{tabular}{|llll|}
\hline
Family         & $N$     & $\o{ord}_N(10)$ & $N_{a,b,m} \cdot 10^k + M_{a,b,m} \mod N$ \, for\, $0 \leq k < \o{ord}_N(10)$ \\ \hline
$42\ldots24$   & $999$   & $3$             & $396, \, 819, \, 54$                                                                \\
$442\ldots244$ & $77$    & $6$             & $33, \, 29, \, 66, \, 51, \, 55, \, 18$                                                      \\
$52\ldots25$   & $91$    & $6$             & $40, \, 84, \, 69, \, 10, \, 57, \, 72$                                                      \\
$65\ldots56$   & $13837$ & $8$             & $594, \, 5904, \, 3656, \, 8850, \, 5442, \, 12873, \, 4161, \, 63$                                \\
$98\ldots89$   & $1001$  & $6$             & $891, \, 893, \, 913, \, 112, \, 110, \, 90$                                                 \\ \hline
\end{tabular}
\caption{}
\end{table}
\vspace{-.4cm}

Together with Lemma \ref{mod-lem2}, this data  proves that none of the five remaining families contain a perfect square. To highlight an example, let's consider the family $42\ldots24$. We read the values $396,\, 819,$ and $54$ from the fourth column of the first row of data. Because each of these is a quadratic non-residue modulo $N = 999$, Lemma \ref{mod-lem2} implies that the family in question contains no perfect squares.

\section{Considering families via elliptic curves} \label{elliptic}
In this section, we give a method that determines the squares in a given single variable family  (as in \S\ref{modfamilies}). As we'll see, the squares are in one-to-one correspondence with the integral points of a specific form on certain elliptic curves. An \textit{elliptic curve} $E$ (defined over the rationals) is a projective curve given by an equation
\[ E : y^2 = x^3 + ax + b \]
for some $a,b \in \Q$ with nonzero discriminant $\Delta \coloneqq -16(4a^3 + 27b^2)$. The set of \textit{integral points} of $E$ is
\[ E(\mathbb{Z}) \coloneqq \set{(x,y) \in \Z \times \Z : y^2 = x^3 + ax + b}\!. \]
This set is finite and fairly computable. See \cite{Si2009} for a thorough treatment of the theory of elliptic curves and specifically Chapter IX for a treatment of integral points. In what follows, we use the \texttt{IntegralPoints} command in Magma \cite{MAGM} to rigorously compute integral points on various elliptic curves.

We define some notation. Let $a,b,m$ be nonnegative integers with $1 \leq a \leq 9$, $0 \leq b \leq 9$, and $m\geq 1$. Then
\[ \set{a_m b_n a_m : n \geq 0} \]
is a family of curious numbers of the sort that we considered in \S \ref{modfamilies}. We're interested in determining the set
\[
Q_{a,b,m} \coloneqq \set{n : a_mb_na_m \text{is a perfect square}}\!.
\]
For each $j \in \set{0,1,2}$, we set
\[ B_{a,b,m,j} \coloneqq  N_{a,b,m}^{2}\cdot10^{2j}\cdot M_{a,b,m} \]
and consider the elliptic curve
\[E_{a,b,m,j} : y^2 = x^3 + B_{a,b,m,j}. \]
Indeed, the above equation defines an elliptic curve since
\[    \Delta  =   -16(4\cdot0 +27\cdot B_{a,b,m,j}^{2}) = -16\cdot27\cdot B_{a,b,m,j}^{2} \neq 0.
\]
We're interested in the integral points of $E_{a,b,m,j}$ and, more  specifically, the subset
\[
L_{a,b,m,j} \coloneqq \set{(X,Y) \in  E_{a,b,m,j}(\Z) : (X,Y) =  ( N_{a,b,m}\cdot10^{j+k} , N_{a,b,m}\cdot10^{j}\cdot3y) \text{ for some integers } k,y \geq 0 }\!.
\]

\begin{prop}\label{ell} In the notation above, we have a bijection
\begin{align*} \Phi: Q_{a,b,m} &\longrightarrow L_{a,b,m,0} \cup L_{a,b,m,1} \cup L_{a,b,m,2}  \\
 n = 3k + j &\longmapsto  ( N_{a,b,m}\cdot10^{j+k} , N_{a,b,m}\cdot10^{j}\cdot 3\sqrt{a_m b_n a_m}),
\end{align*}
where we write $n = 3k + j$ with $j \in \set{0,1,2}$.
\end{prop}
\begin{proof} The map $\Phi$ is well-defined. Indeed, the decomposition $n = 3k +j$ with $j \in \set{0,1,2}$ is unique and we now verify that the image of $\Phi$ is contained in the stated codomain. If $n = 3k+j \in Q_{a,b,m}$, then
\[
(3 \sqrt{a_m b_n a_m})^2  = N_{a,b,m}\cdot10^{3k+j} + M_{a,b,m}
\]
holds by (\ref{alg-eq}). Multiplying through  by $N_{a,b,m}^{2}\cdot10^{2j}$ and regrouping, we find that
\begin{equation*}
(N_{a,b,m}\cdot10^{j}\cdot3\sqrt{a_mb_na_m})^{2} = (N_{a,b,m}\cdot10^{j+k})^{3} + N_{a,b,m}^{2}\cdot10^{2j}\cdot M_{a,b,m}.
\end{equation*}
Consequently, $\Phi(n) \in L_{a,b,m,j}$, establishing that the image of $\Phi$ is contained in the stated codomain. On reading this argument backwards, we deduce that $\Phi$ is surjective. What remains is to show that $\Phi$ is injective.

 For this, suppose that $n = 3k+j \in Q_{a,b,m}$ and $n'=3k'+j' \in Q_{a,b,m}$  are such that $\Phi(n) = \Phi(n')$. Then
\begin{equation} \label{ell_inj}
( N_{a,b,m}\cdot10^{j+k} , N_{a,b,m}\cdot10^{j}\cdot 3\sqrt{a_m b_n a_m}) = ( N_{a,b,m}\cdot10^{j'+k'} , N_{a,b,m}\cdot10^{j'}\cdot 3\sqrt{a_m b_{n'} a_m}).
\end{equation}
By comparing the first coordinates, we find that $j+k = j' + k'$. Note that since $a,m$ are nonzero, $a_mb_na_m$ is not divisible by $10$. Thus $\sqrt{a_mb_na_m}$ is not divisible by 10 (and nor is  $\sqrt{a_mb_{n'}a_m}$). Thus comparing the second coordinates of (\ref{ell_inj}), we find that $j = j'$. Since $j + k = j' + k'$ and $j = j'$, we have that $n = n'$.
\end{proof}

The above proposition, along with the  data from Table \ref{elliptic_table} (Appendix \ref{app}), gives an alternate proof that none of the thirteen families in
$\mathcal{F}$ contain a perfect square.
To illustrate, let's consider the family
$42\ldots24$. Here, $(a,b,m) = (4,2,1)$ and the  corresponding elliptic curves for $j \in \{0,1,2\}$ are
\begin{align*}
     E_{4,2,1,0}: y^2 &= x^3 + 23104\cdot 10^{2}\\
     E_{4,2,1,1}: y^2 &= x^3 + 23104\cdot 10^{4}\\
     E_{4,2,1,2}: y^2 &= x^3 + 23104\cdot 10^{6}.
\end{align*}
Using Magma \cite{MAGM}, we compute their integral points:
\begin{align*}
     E_{4,2,1,0}(\Z) &= \{(80, \pm 1680), (0, \pm1520), (1520, \pm59280), (-76, \pm1368)\} \\
     E_{4,2,1,1}(\Z) &=  \{(0, \pm15200)\} \\
     E_{4,2,1,2}(\Z) &= \{(0, \pm152000)\}.
\end{align*}
None of these points are of the form $( N_{a,b,m}\cdot10^{j+k} , N_{a,b,m}\cdot10^{j}\cdot3y)$ for  nonnegative integers $y,k$. Hence,
\[ L_{4,2,1,0} = L_{4,2,1,1} = L_{4,2,1,2} = \emptyset. \]
So, by Proposition \ref{ell}, we conclude that the family $42\ldots24$ contains no perfect squares.

\bibliographystyle{amsplain}
\bibliography{Curious_Squares}

\newpage

\appendix

\section{} \label{app}

\renewcommand{\arraystretch}{1.08}
\begin{table}[H]
\begin{tabular}{|lllp{95mm}|} \hline
Family            & $j$ & $B_{a,b,m,j}$                                                                  & Integral points of $E_{a,b,m,j}: y^2 = x^3 + B_{a,b,m,j}$ up to sign     \\ \hline
$10\ldots01$      & 0   & $ 729 \cdot 10^2$                                                            & \{($-36$, 162), (0, 270), (40, 370), (45, 405), (180, 2430), (216, 3186), (23940, 3704130)\}                                                                                \\
                   & 1   & $ 729\cdot 10^4$                                                          & \{(0, 2700)\}                                                                                \\
                  & 2   & $ 729\cdot 10^6$                                                        & \{($-900$, 0), (1800, 81000), (0, 27000)\}                                                                                \\ \hline
$14\ldots41$     & 0   & $ -5239\cdot 10^2$                & \{(100, 690), (140, 1490), (160, 1890), (1589, 63337), (28261, 4750959)\}                                                                                \\
                    & 1   & $ -5239\cdot 10^4$                                                         & \{(376, 876)\}                                                                                \\
                    & 2   & $ -5239\cdot 10^6$                                                       & \{(3500, 194000)\}                                                                                \\ \hline
$18\ldots81$       & 0   & $ -20519\cdot 10^2$                                                          & \{(960, 29710)\}                                                                                \\
                  & 1   & $ -20519\cdot 10^4$                                                        &$\emptyset$                                                                                \\
                   & 2   & $ -20519\cdot 10^6$                                                      &$\emptyset$                                                                                \\ \hline
$40\ldots04$      & 0   & $ 46656\cdot 10^2$                                                          & \{($-144$, 1296), ($-135$, 1485), (0, 2160), (160, 2960), (180, 3240),  (720, 19440), (864, 25488), (95760, 29633040)\}                                                             \\
                   & 1   & $ 46656\cdot 10^4$                                                        & \{(0, 21600)\}                                                                                \\
                   & 2   & $ 46656\cdot 10^6$                                                      & \{ ($-3600$, 0), (0, 216000), (7200, 648000)\}                                                                                \\ \hline
$42\ldots24$       & 0   & $ 23104\cdot 10^2$                                                          & \{($-76$, 1368), (0, 1520), (80, 1680), (1520, 59280)\}                                                                                \\
                     & 1   & $ 23104\cdot 10^4$                                                        & \{(0, 15200)\}                                                                                \\
                    & 2   & $ 23104\cdot 10^6$                                                      & \{(0, 152000)\}                                                                                \\ \hline
$442\ldots244$   & 0   & $ 31047184\cdot 10^4$                                                     & \{($-4975$, 432825), (0, 557200), (5600, 697200), (44576, 9427824)\}                                                                                \\
                    & 1   & $ 31047184\cdot 10^6$                                                   & \{(0, 5572000), (8959776, 26819194976)\}                                                                                \\
                      & 2   & $ 31047184 \cdot 10^8$                                                 & \{($-84000$, 50120000), (0, 55720000), (1671600, 2161936000)\}                                                                                \\ \hline
$4443\ldots3444$  & 0   & $ 15928032996\cdot 10^6$                                                & \{(198400, 154070000)\}                                                                                \\
                     & 1   & $ 15928032996\cdot 10^8$                                              & \{($-356000$, 1244060000)\}                                                                                \\
                     & 2   & $ 15928032996\cdot 10^{10}$ &$\emptyset$                                                                                \\ \hline
$4447\ldots7444$    & 0   & $-48136123036\cdot 10^6$                                                 &$\emptyset$                                                                                \\
                      & 1   & $-48136123036\cdot 10^8$   &$\emptyset$                                                                                \\
                     & 2   & $-48136123036\cdot 10^{10}$ &$\emptyset$                                                                                \\ \hline
$52\ldots25$    & 0   & $  55225\cdot 10^2$                                                         & \{(0, 2350)\}                                                                                \\
                    & 1   & $ 55225\cdot 10^4$                                                        & \{(0, 23500)\}                                                                                \\
                   & 2   & $ 55225\cdot 10^6$                                                      & \{(0, 235000)\}                                                                                \\ \hline
$65\ldots56$    & 0   & $  13924\cdot 10^2 $           & \{(0, 1180), (80, 1380), (944, 29028)\}                                                                                \\
                     & 1   & $ 13924\cdot 10^4 $            & \{(0, 11800)\}                                                                                \\
                      & 2   & $ 13924\cdot 10^6 $        & \{($-2400$, 10000), (0, 118000), (4425, 317125), (751296, 651203344)\}                                                                                \\ \hline
$90\ldots09$    & 0   & $ 531441\cdot 10^2$                                                         & \{ ($-324$, 4374),  (0, 7290), (360, 9990), (405, 10935), (1620, 65610), (1944, 86022),  (215460, 100011510)\}                                                                      \\
                      & 1   & $ 531441\cdot 10^4 $        & \{(0, 72900)\}                                                                                \\
                    & 2   & $ 531441\cdot 10^6$         & \{($-8100$, 0), (0, 729000), (16200, 2187000)\}                                                                                \\ \hline
$94\ldots49$    & 0   & $ 296225\cdot 10^2$             & \{($-200$, 4650), ($-100$, 5350), (349, 8493), (10300, 1045350)\}                                                                                \\
                    & 1   & $ 296225\cdot 10^4$                                                       & \{($-800$, 49500),  (200, 54500), (625, 56625), (11416, 1220964)\}                                                                                \\
                    & 2   & $ 296225\cdot 10^6 $        &$\emptyset$                                                                                \\ \hline
$98\ldots89$    & 0   & $  7921\cdot 10^2 $            & \{(0, 890)\}                                                                                \\
                   & 1   & $ 7921\cdot 10^4$                                                         & \{($-400$, 3900), (0, 8900), (1424, 54468)\}                                                                                \\
                  & 2   & $ 7921\cdot 10^6$            & \{(0, 89000), (8400, $-775000$)\}  \\ \hline
\end{tabular}
\caption{. Integral points data for \S \ref{elliptic} } \label{elliptic_table}
\end{table}

\end{document}


TITLE: Prove $a_n = \lfloor ni \rfloor$ for some irrational $i$ has no pattern
QUESTION [4 upvotes]: I have a sequence $a_n = \lfloor ni \rfloor$ for an irrational $i$ and I want to prove that there is no 'pattern' to these terms.
In particular, I have $1 < i <2$ and I want to prove that $\forall a,b \in \mathbb{N}, \exists k \in \mathbb{N}$ such that $a +bk \neq a_n$ for any $n$.
This makes sense to me intuitively because $a_n - a_{n-1}$ will be 1 or 2 'randomly', so the sequence must eventually 'skip over' one of the terms of $a+bk$.
My approach has been to assume that it holds, then show that $i$ must be rational for a contradiction. However, the floor is messing up the simple divisibility techniques I would use.

REPLY [4 votes]: This is a case of the equidistribution theorem.  If the sequence repeats, subdivide $[0,1]$ finely enough and some pieces will get nothing.


TITLE: Summation Notation of a Linear Combination Over a Set
QUESTION [0 upvotes]: Suppose I have a set $S = \{\bar{u_1},\bar{u_2},...,\bar{u_{|S|}}\}$ where $|S|$ is the cardinality of $S$. I want to express a linear combination of all vectors in the set using a summation like so:
$$ \sum_{u_i \in S} a_i \bar{u_i} $$
Is this valid?
(This question is an extension of a pre-existing question: Notation of the summation of a set of numbers )

REPLY [0 votes]: I would write $S = \{\bar u_i: i ∈ I\}$. And then $∑_{i ∈ I} a_i \bar u_i$. But note that this is an actual linear combination only if $\{i ∈ I: a_i ≠ 0\}$ is finite.
If you want the index set to be well-ordered, you may write $S = \{\bar u_α: α < |S|\}$ or $\{\bar u_α: α < κ\}$ and then $∑_{α < κ} a_α \bar u_α$. If $S$ is finite, you may use $n$ instead of $κ$ and $i$, $j$, or $k$ instead of $α$.