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\begin{document}
\title{Appendix 1}
\pacs{PACS}

\maketitle
\appendix
 
\section{Continuum model of the wing disc epithelium}
\subsection{Model definition}

We use a previously developed continuum model to describe tissue mechanics (Etournay et al., 2015; Popovic et al., 2017). Tissue shear flow $\tilde{v}_{ij}$ consists of convected co-rotational change of cell elongation $Q_{ij}$ and a contribution from cell rearrangements $R_{ij}$
\begin{align}
\label{eq:15}
\tilde{v}_{ij}&= \frac{D Q_{ij}}{D t} + R_{ij} \quad .
\end{align}
Tissue shear stress consists of an elastic part and a contribution from nematic cell polarity $q_{ij}$
\begin{align}
\label{eq:16}
\tilde{\sigma}_{ij}&= 2K Q_{ij} + \zeta q_{ij} \quad .
\end{align}
Shear flow due to cell rearrangements consists of term accounting for cell shape relaxation and a contribution due to nematic cell polarity $q_{ij}$
\begin{align}
\label{eq:17}
R_{ij}&= \frac{1}{\tau}Q_{ij} + \lambda q_{ij} \quad .
\end{align}
Force balance couples shear stress and area pressure
\begin{align}
\label{eq:20}
\partial_j \tilde{\sigma}_{ij} - \partial_i P&= 0
\end{align}
which will be reflected in cell area. We use a constitutive relation
\begin{align}
\label{eq:22}
P= - \overline{K} \ln{\left( \frac{a}{a_0} \right)} \quad ,
\end{align}
where $a$ represents cell area and $a_{0}$ corresponds to value of $a$ at $P= 0$.

%\subsection{Radially symmetric tissue}
We consider a simple model in which the tissue is a radially symmetric disc, see Fig. 3 C-E of the main text. We use polar coordinate system $(r, \varphi)$, where the radial force balance reads
\begin{align}
\label{eq:18}
\partial_rP= \partial_r\tilde{\sigma}_{rr}+ \frac{2}{r}\tilde{\sigma}_{rr}
\end{align}
Furthermore, in the model we consider a constrained tissue with negligible radial shear flow $\tilde{v}_{rr}= 0$, consistent with experimental observations, see Fig. 2 of the main text.

\subsection{Cell area profiles follow from cell elongation profiles}
Empirically, the radial cell elongation profile can be described by a power law
\begin{align}
\label{eq:24}
Q_{rr}&\simeq -\left( \frac{r}{r_Q} \right)^{\theta} \quad ,
\end{align}
with values for $r_Q= 140 \pm 20 \mu m$ and $\theta= 1.6\pm 0.1$ obtained by fitting the data on the last $\sim 5$ hours of the experiments, see red line in Appendix 1 - Figure \ref{fig:QrrChange} A. We find $r_Q= 84\pm 3 \mu m$ and $\theta= 2.8\pm 0.3$ by fitting the data from the first $\sim 5$ hours. Data is averaged over 5 experiments and errorbars represent a standard deviation among the mean values in each experiments.

\begin{figure}
\centering
\includegraphics[width= .45\textwidth]{Appendix1-figure1.pdf} 
\caption{Left: Radial cell elongation in the first half of the movies ('early') are shown in blue and in the second half of the movies ('late') are shown in red. Fits of power law are shown with dashed lines in corresponding colors, with fit parameters $r_Q=84\pm 3 \mu m$, $\theta= 2.8\pm 0.3$ and $r_Q= 140\pm 20 \mu m$, $\theta= 1.6\pm 0.1$ for 'early' and 'late' profiles, respectively.  Right: Radial cell elongation change suggests that there is no significant gradient of radial shear due to cell rearrangements $R_{rr}$ during the experiment.}
\label{fig:QrrChange}
\end{figure}

At this point it is useful to estimate the spatial profile of shear due to cell rearrangements $R_{rr}$. We calculate the difference between late and early cell elongation profiles, as shown in Appendix 1 - Figure \ref{fig:QrrChange} B. Since $\tilde{v}_{rr}\approx 2\cdot 10^{-3} h^{-1}$ we find $R_{rr}\approx (6\pm 7) \cdot 10^{-3} h^{-1}$ with no significant variation in $r$.

Now, we find an expression for cell area using Eqs. \ref{eq:16}, \ref{eq:17}, \ref{eq:22}, \ref{eq:18} and \ref{eq:24}
\begin{align}
\label{eq:41}
a= a_0e^{-\frac{c_a}{\overline{K}} + \frac{2K^{*}}{\overline{K}}\frac{\theta + 2}{\theta}\left(\frac{r}{r_{Q}}\right)^{\theta} - \frac{2K}{\overline{K}} \left( 1 - \frac{K^{*}}{K} \right) \tau R_{rr}\ln{\left(\frac{r}{r_{Q}}\right)}} ,
\end{align}
where 
\begin{align}
\label{eq:26}
K^{*}= \left( 1 - \frac{\zeta}{2 K \tau\lambda} \right)K 
\end{align}
and $c_a$ is an integration constant.
We fit the Eq. \ref{eq:41} to the experimental data as shown in Fig. 3 G of the main text, and we find parameter values
\begin{align}
  a_0e^{-\frac{c_a}{\overline{K}}}&= 3.3 \pm 0.3 \quad ,\\
  \label{eq:143}
  \frac{2K^{*}}{\overline{K}}\frac{\theta + 2}{\theta}&= 5.4 \pm 0.7 \quad ,\\
  \label{eq:155}
  \frac{2K}{\overline{K}} \left( 1 - \frac{K^{*}}{K} \right)\tau R_{rr}&= 0.00 \pm 0.07 \quad, 
\end{align}
where the reported uncertainties represent square roots of the fit covariance matrix and the parameter in Eq. \ref{eq:155} was constrained to be positive, motivated by the fact that we do not find $K^{*} > K$ (see Eq. \ref{eq:53} below).

\subsection{Laser ablation experiments provide an estimate of model parameters}
We write Eq. \ref{eq:16} in form 
\begin{align}
\label{eq:19}
\frac{\tilde{\sigma}_{rr}}{2K}&= \frac{K^{*}}{K}Q_{rr} +  \left(1 - \frac{K^{*}}{K}\right)\tau R_{rr} \quad .
\end{align}
We fit this equation to the experimentally obtained shear stress as a function of cell elongation in the laser ablation experiments. We find
\begin{align}
\label{eq:21}
  \frac{K^{*}}{K}&= 0.05\pm 0.02 \quad ,\\ 
  \left(1 - \frac{K^{*}}{K}\right)\tau R_{rr}&= 0.011\pm 0.002 \quad .
\end{align}
Using the estimate for $R_{rr}$ we find
\begin{align}
\label{eq:27}
\tau= (2 \pm 2) h \quad .
\end{align}
This value is consistent with the one found for the cell elongation relaxation time-scale in the pupal wing (Etournay et al., 2015).

In Eq. \ref{eq:17} we can now show that cell elongation profile in the wing pouch reflects the profile of nematic cell polarity. Namely, $|Q_{rr}| \gg \tau |R_{rr}| \approx 0.012$ in most of the tissue, except in the vicinity of the DV boundary, and thus
\begin{align}
\label{eq:28}
q_{rr} \approx -\tau \lambda Q_{rr} \quad .
\end{align}
\subsubsection*{Comparison of laser ablation and area profile fits}
It is interesting to notice that from Eq. \ref{eq:143} and the value for $2K/\overline{K}= 3.4 \pm 0.4$ obtained from the laser ablation experiments (main text and Methods), we can estimate
\begin{align}
\label{eq:53}
\frac{K^{*}}{K}= 0.6 \pm 0.1 \quad .
\end{align}
This value is significantly higher than the one obtained by a direct fit  to the laser ablation data (Eq. \ref{eq:21}). This apparent discrepancy stems in part from the fact that  the elastic constant $\overline{K}$ in Eq. \ref{eq:22} is a non-linear elastic modulus. It can only be related the one used in the laser ablation method by linearizing it around a typical value of $a$ in the data. When this value is not equal to $a_0$ the linearized elastic constant will differ from the original $\overline{K}$ in Eq. \ref{eq:22}. % Namely, in the laser ablation method, described in the next section, the elastic constitutive equation is linearised around a reference state that does not necessarily correspond to $a= a_0$ in Eq. \ref{eq:22} and therefore the effective value of the elastic constant will differ.
Furthermore, Eq. \ref{eq:22} does not take into account any adaptation of cell area to pressure or nematic cell polarity, which could influence the measured value $K^{*}/K$ when fitting the cell area profile. However interesting, these effects are beyond the scope of our current work. 

% TODO: what to do with area profile?
% Using Eqs.  \ref{eq:18}, \ref{eq:24} and \ref{eq:19}  we find
% \begin{align}
%   \label{eq:23}
%   \begin{split}
%   \frac{P}{\overline{K}}\approx c_P &- \frac{2K^{*}}{\overline{K}}\frac{\theta + 2}{\theta}\left( \frac{r}{r_Q} \right)^{\theta} \\
%   &- 2 \frac{2(K - K^{*}) }{\overline{K}}\tau R_{rr}\ln{\left(\frac{r}{r_{Q}}\right)} \quad ,
%   \end{split}
% \end{align}
% where $c_P$ is an integration constant, and expression for cell area profile can now be obtained using Eq. \ref{eq:22}.
% We use this expression in Eq. \ref{eq:22} to fit the 'late' cell area profile and find
% \begin{align}
% \label{eq:29}
%   a_0e^{c_P}&= (4.3\pm 0.5) \mu m^2\\
%   \frac{2K}{\overline{K}}\frac{K^{*}}{K} \frac{\theta + 2}{\theta}&= 5.5\pm 0.5\\
%   2 \tau R_{rr} \frac{2K}{\overline{K}}\left( 1 - \frac{K^{*}}{K} \right)&= 0.1 \pm 0.04 
% \end{align}

% As we have argued above $R_{rr}$ does not have a significant spatial profile and




\subsection{Mechanosensitivity leads to spontaneous cell polarity by self-organization}
\subsubsection{Homogeneous tissue}
We propose a dynamical equation of the polarity tensor $q_{ij}$
\begin{align}
\label{eq:polarityDynamicsLinear}
\frac{D q_{ij}}{Dt}&= -\frac{1}{\tau_q}  q_{ij} - \mu \tilde{\sigma}_{ij} \quad ,
\end{align}
where the parameter $\mu$ represents the mechanosensitive response of nematic cell polarity to shear stress. % The negative sign is introduced so that $\mu > 0$ when polarising component accumulates more along the bonds with increased tension.
We use Eqs. \ref{eq:15}, \ref{eq:16}, \ref{eq:17} and \ref{eq:polarityDynamicsLinear} to obtain a dynamic equation for shear stress
\begin{align}
  \label{eq:stressDynamicsLinear}
  \begin{split}
  \frac{\partial \tilde{\sigma}_{ij}}{\partial t}&= 2K \tilde{v}_{ij}-\left(\frac{1}{\tau} + \zeta \mu\right)\tilde{\sigma}_{ij} \\
  &- \left(2K\lambda  +  \zeta \left( \frac{1}{\tau_q} - \frac{1}{\tau} \right)\right)q_{ij}  \quad .
  \end{split}
\end{align}
The system of linear equations \ref{eq:polarityDynamicsLinear} and \ref{eq:stressDynamicsLinear} becomes unstable when
\begin{align}
\label{eq:linearStabilityCondition}
2K^{*} \tau_q \mu  > 1 \quad .
\end{align}
This instability will result in a finite value of nematic cell polarity in the tissue. To account for this polarised state we have to include a stabilising higher order term in Eq. \ref{eq:polarityDynamicsLinear}
\begin{align}
  \label{eq:stressDynamicsNonlinear}
  \begin{split}
  \frac{D q_{ij}}{Dt}= &-\frac{1}{\tau_q}q_{ij} - \mu\tilde{\sigma}_{ij} -  \alpha q^2 q_{ij}  \quad .
  \end{split}
\end{align}
% Therefore, when this condition is fulfilled, for example by increasing stiffness $K$ or cell shape relaxation timescale $\tau$ the isotropic state becomes unstable.
In the steady state the magnitude squared of the nematic cell polarity is 
\begin{align}
% \label{eq:30}
%   \tilde{\sigma}_{ij}&= - 2K^{*} q_{ij} \\
  \label{eq:polarisationSteadyState} 
q^2&= q_0^2 \equiv \frac{2K^{*}\tau_q\mu - 1}{\alpha\tau_q} \quad .
\end{align}
Here, we have introduced the strength of polarity $q_0$, corresponding to the magnitude of polarity in a homogeneous system. Note that the definition of $q_0^2$ allows for negative values, which would correspond to no spontaneous polarisation in the tissue.
% This result gives a uniform value of self-organised nematic cell polarity.
% Spatial variations of polarity, such as change of polarity angle that is present in wing tissue, will affect polarity dynamics.

\subsubsection{Polarisation of a radially symmetric tissue}
To account for spatial variations of nematic cell polarity, we introduce a Laplacian term in Eq. \ref{eq:stressDynamicsNonlinear}
\begin{align}
\label{eq:stressDynamicsNonlinearLaplacian}
  \frac{D q_{ij}}{Dt}= &-\frac{1}{\tau_q}q_{ij} - \mu\tilde{\sigma}_{ij} -  \alpha q^2 q_{ij}  + D \nabla^2 q_{ij} 
\end{align}
% Tissue shear flow equation (together with constitutive equations for shear stress and shear due to cell rearrangements) can be written as dynamical equation for shear stress
% \begin{align}
%   \label{eq:stressDynamicsLinear}
%   \begin{split}
%   \frac{\partial \boldsymbol{\tilde{\sigma}}}{\partial t}&= 2K \mathbf{\tilde{v}}-\left(\frac{1}{\tau} + \zeta \mu\right)\boldsymbol{\tilde{\sigma}} \\
%   &- \left(2K\lambda  +  \zeta \left( \frac{1}{\tau_q} - \frac{1}{\tau} \right)\right)\mathbf{q} -\zeta a |\mathbf{q}|^2 \mathbf{q} \quad .
%   \end{split}
% \end{align}

% We proceed to fit the In general this equation has to be solved numerically but we can discuss dependence of the solution on $q_0$ by rewriting Eq. \ref{eq:32} in form
% \begin{align}
% \label{eq:36}
% \overline{q}^3_{rr} - \overline{q}_{rr}&= \partial_\rho^2\overline{q}_{rr} + \frac{1}{\rho}\partial_\rho \overline{q}_{rr} - \frac{\overline{q}_{rr}}{\rho^2} \quad ,
% \end{align}
% where
% \begin{align}
% \label{eq:37}
%   \overline{q}_{rr}&= \frac{q_{rr}}{q_0} \\
%   \rho&= \frac{r}{L} \\
%   L^2&= \frac{D}{aq_0^2}\quad .
% \end{align}
% Therefore, any solution will be of the form
% \begin{align}
% \label{eq:38}
% q_{rr}(r; q_0)&= q_0 f \left( \frac{r}{L} \right) \quad,
% \end{align}
% where $f$ is a solution of Eq. \ref{eq:36}.

% % If we consider a solution describing a wild-type wing
% % \begin{align}
% % \label{eq:39}
% % q_{rr}^{WT}&= q^{WT}_0 f \left( \frac{r}{L^{WT}} \right) \quad,
% % \end{align}
% % then by decreasing $q_0$ in myoVI mutant wings we expect to find
% % \begin{align}
% % \label{eq:40}
% % q_{rr}^{myoVI}&= q^{myoVI}_0 f \left( \frac{r}{L^{myoVI}} \right) \quad .
% % \end{align}
% % Therefore, in the steady state, for a function $f$ growing with radius we expect to find both reduced magnitude and reduced spatial gradient of the nematic cell polarity in myoVI mutant wings. Since in the steady state $Q_{rr}=-\lambda\tau q_{rr}$ this prediction is consistent with the experimental observation observation of reduced magnitude and radial gradient of the tangential cell elongation in \textit{myoVI} mutant wings.

To fit the radial profile of $Q_{rr}$ observed in time-lapse experiments we account for non-vanishing $R_{rr}$ as well. Consistent with the observation that during the experiments $R_{rr}$ does not vary significantly in time (Fig. 2 B main text) we use approximation $\tau \partial_tR_{rr} \ll R_{rr}$. Then $q_{rr}$ satisfies
\begin{align}
  \label{eq:25}
  \begin{split}
  \partial_r^2q_{rr} + \frac{1}{r}\partial_r q_{rr} - \frac{q_{rr}}{r^2}&=  \frac{\alpha}{D}\left(q_{rr}^3 - q_0^2q_{rr}\right)\\
  &+\frac{1}{D}\left(\frac{1}{\tau \lambda} +2K\mu\tau\right)R_{rr} \quad ,
  \end{split}
\end{align}
where the left hand side is the radial component of the Laplacian of $q_{ij}$ in polar coordinate system. From solution of this equation the $Q_{rr}$ profile is obtained as $Q_{rr} \approx -\tau \lambda q_{rr}$. This quantity is fitted to the experimentally measured tangential cell elongation profile by optimizing 4 tissue parameters, as well as two boundary conditions $q_{rr}(r_{in})$ and $\partial_r q_{rr}(r_{in})$ required to solve Eq. \ref{eq:25}. Here $r_{in}= 10 \mu m$ is the smallest radius at which we measure the tangential cell elongation profile in Fig. 6 C of the main text.
Parameters obtained by the fit are reported in Table 1 of the main text and the fitted $Q_{rr}$ profile is shown in Fig. 6 of the main text.
% \begin{align}
% \label{eq:31}
%   q_0^2&= 0.01,\\
%   \frac{D}{a}&= 0.48 \mu m^2\quad ,\\
%   \frac{1}{D}\left(\frac{1}{\tau \lambda} +2K\mu\tau\right)R_{rr}&= 9\cdot 10^{-4} \mu m^{-2} \quad ,\\
%   \lambda\tau&= 1.0 \quad .
% \end{align}
Parameter uncertainties were estimated as the 10 and 90 percentile intervals obtained by fitting uniformly sampled profiles of $Q_{rr}$ from the interval $Q_{rr}(r) + \Delta Q_{rr}(r)$ and $Q_{rr}(r) - \Delta Q_{rr}(r)$, see Fig. 3 of the main text, with 101 realisations. Obtained uncertainty intervals are reported in the Table 1 of the main text.

\subsubsection{Predictions  for experiments}
Spontaneous cell polarity in our model depends on mechanosensitive feedback characterized by the
feedback strength $\mu$. The characteristic polarity strength $q_0$ grows monotonically with $\mu$.
A prediction from our model is therefore that when $\mu$ is reduced, cell elongation
will be reduced because it is mediated by cell polarity. In order to test this prediction, we performed
MyoVI knockdown experiments which indeed show a decrease in cell elongation.

Our qualitative prediction is rather robust and does not depend on details of the system such as the
influence of compartment boundaries which disrupt the radial symmetry. However boundary conditions,
which we do not know, could have a significant influence on the system. Therefore a full quantitative
prediction of experiments is difficult. 
%The qualitative prediction we make in the main text is justified by the fact that $q_0$, which is the characteristic self-organised polarity scale, grows monotonically with $\mu$. Although $q_0$ is not the only quantity determining the polarity pattern, as long there are no unexpected changes of other contributing factors, such as boundary conditions, we expect to find a clear qualitative change of the cell elongation magnitude. 
To show the self-consistency of our prediction,  we test if the value of $q_0$ that we use
to describe the data is consistent with the actual cell elongations observed. Indeed we have
$q_0\approx 0.1$, see Table 1 which predicts a typical strength of cell elongation
 $Q_0=-\lambda\tau q_0 \approx -0.1$, consistent with experimentally measured values,
 see Fig. 6 C of the main text. This suggests that our qualitative prediction is robust given the uncertainties
 in the system. 
% FinaExact boundary conditions imposed on the tissue boundary are at the moment unknown. We expect that the effective boundary values of polarity appropriate for our model will change with the polarity scale $q_0$, if polarity in the wing is self-organised. This is confirmed in $myoVI$ RNAi experiments.


% \subsection{Steady state}
% In steady state, for a radially symmetric tissue we find:
% \begin{align}
% \label{eq:1}
% q^3_{rr}= q_0^2q_{rr}+ \frac{D}{a} \left(\partial_r^2q_{rr} + \frac{1}{r}\partial_r q_{rr} - \frac{q_{rr}}{r^2} \right) \quad ,
% \end{align}
% where bracket on the right hand side is tensor Laplacian of a traceless symmetric tensor in polar coordinates and  $q_0$ is polarity in homogeneous system defined above in Eq. \ref{eq:polarisationSteadyState}.

% \end{document}

% \subsubsection{Fitting theory to the time-lapse data}
% In order to fit the theory to the measured cell elongation profile in time-lapse experiments we have to account for the fact that tissue is not in the steady state. To fully solve the problem one would have to solve the PDE system of Eqs. \ref{eq:stressDynamicsLinear} and \ref{eq:stressDynamicsNonlinearLaplacian}. Here, we instead use the fact that we measure $R_{rr}$  to be approximately constant and we can express all time derivatives in terms of $R_{rr}$
% \begin{align}
%   \label{eq:2}
%   \begin{split}
% \partial_r^2q_{rr} + \frac{1}{r}\partial_r q_{rr} - \frac{q_{rr}}{r^2}=&       \frac{a}{D} \left(q^2_{rr} - q_0^2\right)q_{rr}\\
%     +&\frac{1 + \tau \zeta\mu}{D \tau\lambda}R_{rr} \quad .
%   \end{split}
% \end{align}
% To compare this result with experimental data we use Eq. \ref{eq:28} 
% \begin{align}
% \label{eq:35}
%   \begin{split}
% \partial_r^2Q_{rr} + \frac{1}{r}\partial_r Q_{rr} - \frac{Q_{rr}}{r^2}=&       \frac{a}{D (\tau\lambda)^2} \left(Q^2_{rr} - Q_0^2\right)Q_{rr}\\
%     -&\frac{1 + \tau \zeta\mu}{D}R_{rr} \quad .
%   \end{split}
% \end{align}

% \end{document}

\section{Theory of circular laser ablations}
We developed a method to infer the stresses in an epithelial tissue from the tissue boundary shape after a circular ablation. % We describe an epithelial tissue as a homogeneous two-dimensional linearly elastic sheet. 
Here, we first briefly review linear elasticity in polar coordinates. Then, we derive equations relating shape of the  inner and outer boundaries obtained by the ablation of an infinite elastic sheet in the small deformation regime. The solution to this problem is by no means new, see for example (Muskhelishvili, 1977). Here we present our derivation of the solution. The application of the method and the obtained parameter estimates are described in Methods section.

\subsection{Linear elasticity in polar coordinates}\label{app:Theory}
 
\subsubsection{Displacement gradient}
The deformation of an elastic object is described by the displacement vector of each point on the sheet
\begin{align}\label{eq:defDisplacement}
u_i(x_j) &= x'_i(x_j) - x_i \quad 
\end{align}
where $x_i$ is the intial position of a point on the elastic sheet and $x'_i$ is the position of that point after the deformation. Shape changes are described by the symmetric traceless gradient of displacement
\begin{align}
u_{ij}&= \frac{1}{2}\left(\partial_j u_i  + \partial_i u_j\right) \quad ,
\end{align}
which can be decomposed into isotropic part $u= u_{kk}$ and shear part $\tilde{u}_{ij}= u_{ij} -\delta_{ij} u/2 $. Note that we use summation convention over repeated indices.
In polar coordinates displacement gradient is given by 
\begin{align}\label{eq:defPolar}
u_{rr} &= \partial_r u_r\\
u_{\varphi\varphi} &= \frac{1}{r}\partial_\varphi u_\varphi + \frac{1}{r}u_r\\
u_{r\varphi} &= \frac{1}{2}\left(\partial_r u_\varphi - \frac{1}{r}u_\varphi + \frac{1}{r}\partial_\varphi u_r\right)
\end{align}
% \subsubsection{Stress}
% Polar components of the stress tensor expressed in terms of the cartesian components, assuming $\sigma_{xy} = 0$
% \begin{align}\label{eq:stressPolar}
% \sigma_{rr}&= \frac{\sigma_{xx} + \sigma_{yy}}{2} + \frac{\sigma_{xx} - \sigma_{yy}}{2}\cos{2\varphi}\\
% \sigma_{r\varphi}&= -\frac{\sigma_{xx} - \sigma_{yy}}{2}\sin{2\varphi}\\
% \sigma_{\varphi\varphi}&= \frac{\sigma_{xx} + \sigma_{yy}}{2} - \frac{\sigma_{xx} - \sigma_{yy}}{2}\cos{2\varphi}
% \end{align}
\subsubsection{Linear elastic constitutive relation}
We are considering the linear elastic constitutive relation
\begin{align}
\label{eq:13}
  \sigma_{ij}&= 2K \tilde{u}_{ij}+ \zeta q_{ij}  + \overline{K}u \delta_{ij}\quad , 
\end{align}
where $\tilde{\sigma}_{ij}$ is traceless symmetric component of shear stress, % , $\tilde{u}_{ij}$ is tracelles symmetric component of displacement gradient,
with shear elastic modulus $K$ and bulk elastic modulus $\overline{K}$. Here, we allow for the presence of active stresses due to a nematic field $q_{ij}$, corresponding to the nematic cell polarity in the wing disc. However, a constant active stress does not influence our results in the small deformation limit, since it simply redefines the reference configuration.
We can therefore write this constitutive equation as
\begin{align}\label{eq:Hooke}
u_{ij} &= \frac{1}{2K}\tilde{\sigma}_{ij} +\frac{1}{4\overline{K}}\sigma_{kk}\delta_{ij}\quad .
\end{align}

% In a linear elastic problem a uniform active stress contribution in constitutive equation is equivalent to a shift of reference strain. Assuming that the ablation does not perturb the source of active stress on the relevant time-scale of the ablation of $\sim 1 \text{min}$ this does not affect results below.

\subsubsection{Force balance and compatibilty: Airy stress function}
Force balance imposes
\begin{align}
  \label{eq:14}
  \partial_j \sigma_{ij}&= 0 \quad .
\end{align}
In two dimensions, force balance is satisfied by all stress fields whose components are derived from the Airy stress function $\Phi$ 
\begin{align}\label{eq:defAiry}
\begin{split}
\sigma_{xx} &= \partial^2_y \Phi \quad ,\\
\sigma_{xy} &= -\partial_x\partial_y \Phi \quad ,\\
\sigma_{yy} &= \partial^2_x \Phi \quad ,
\end{split}
\end{align}
which can be expressed in polar coordinates 
\begin{align}\label{eq:airyPolar}
\sigma_{rr} &= \frac{1}{r}\partial_r\Phi + \frac{1}{r^2}\partial^2_\varphi\Phi \quad ,\\
\sigma_{r\varphi} &= -\partial_r\left(\frac{1}{r}\partial_\varphi\Phi\right) \quad ,\\
\sigma_{\varphi\varphi} &= \partial^2_r \Phi \quad .
\end{align}
Now, a compatibility condition, which follows from requirement of continuity of displacement field, can be written as 
\begin{align}
\nabla^4 \Phi &= 0 \quad. 
\end{align}
The general solution of this equation in polar coordinates, also called Michell solution, is given by
\begin{align}\label{Airy}
\begin{split}
\Phi&= \left[A_0 r^2 + B_0 r^2\ln{r} + C_0 \ln{r}\right]\\
  &+ \left[I_0 r^2 + I_1 r^2 \ln{r} + I_2 \ln{r} + I_3\right]\varphi\\
  &+ \left[A_1 r  + B_1 r^{-1} + \overline{B}_1 r \varphi + C_1 r^3 + D_1 r \ln{r}\right]\cos{\varphi}\\
  &+ \left[E_1 r  + F_1 r^{-1} + \overline{F}_1 r \varphi + G_1 r^3 + H_1 r \ln{r}\right]\sin{\varphi}\\
  &+ \sum\limits_{n> 1}\left[A_nr^n + B_nr^{-n} + C_n r^{n+2} + D_n r^{-n+2}\right] \cos{n\varphi}\\
  &+ \sum\limits_{n> 1}\left[E_nr^n + F_nr^{-n} + G_n r^{n+2} + H_n r^{-n+2}\right] \sin{n\varphi} \quad .
\end{split}
\end{align}
Inferring the stresses from laser ablation experiments consists of solving two elastic problems using the Michell solution as we now show.

\subsection{Circular ablation of an elastic sheet under stress}

Laser ablation splits the elastic sheet into inner and outer pieces. We now show that the boundary shapes of the two pieces are ellipses by solving the corresponding elastic problems. 

\subsubsection{Inner piece}
We consider a piece of elastic sheet stretched by stress $\sigma_{xx}= \sigma_{xx}^0$, $\sigma_{yy}= \sigma^0_{yy}$, $\sigma_{xy} = 0$. After a circular laser ablation the inner piece is under no stress and to infer the original stresses in the sheet we determine the shape of the sheet boundary assuming the stresses are known. Then, given the boundary shape we will be able to infer the stresses. % we can find the shape of the sheet when the stress is relaxed. , if we know the relaxed shape, we would like to infer the original stresses $\sigma^0_{xx}$ and $\sigma^0_{yy}$ in the elastic sheet.

Stress in a sheet that is in mechanical equilibrium, under no external body force, is uniform throughout the sheet. Therefore, using Eqs. \ref{eq:defPolar} we can integrate Eq. \ref{eq:Hooke} to find the components of the displacement vector $u_i$
\begin{align}
u_r&= \frac{r}{2K}\left[\tilde{\sigma}\cos{2\varphi} + \frac{K}{2\overline{K}}\sigma\right]\\
u_{\varphi}&= -\frac{r}{2K}\tilde{\sigma}\sin{2\varphi} \quad ,
\end{align}
where we have introduced 
\begin{align}
\sigma &= \sigma^0_{xx} + \sigma^0_{yy}\\
\tilde{\sigma} &= (\sigma^0_{xx} - \sigma^0_{yy})/2 \quad .
\end{align}
To relate the displacement to the shape of the inner piece boundary, we have to solve Eq. \ref{eq:defDisplacement} for $r(\varphi)$, in the configuration illustrated in Appendix 1 - Figure \ref{fig:cutSchematic}
\begin{align}\label{eq:circleInner}
r\hat{r} + u_r \hat{r} + u_\varphi \hat{\varphi} &= R \hat{r}' \quad . 
\end{align}
Using the fact that 
\begin{align}
\hat{r}' \cdot \hat{r} &= \cos\left(\varphi' - \varphi\right)\\
\hat{r}' \cdot \hat{\varphi}&= \sin\left(\varphi' - \varphi\right)
\end{align}
we obtain
\begin{align}\label{eq:innerSolution}
r\left(\varphi \middle| \frac{\sigma}{\overline{K}}, \frac{\tilde{\sigma}}{K}\right) &= \frac{R}{\sqrt{\left[1 + \frac{\sigma}{4\overline{K}}+ \frac{\tilde{\sigma}}{2K}\cos(2\varphi)\right]^2 + \left[\frac{\tilde{\sigma}}{2K}\sin(2\varphi)\right]^2}} \quad .
\end{align}
To simplify the notation we introduce
\begin{align}
\label{eq:sSimple}
  \tilde{s}&= \frac{\tilde{\sigma}}{2K} \\ \label{eq:tildeSSimple}
  s&= \frac{\sigma}{\overline{K}}  \quad, 
\end{align}
and we write the solution in the form
\begin{align}\label{eq:innerSolutionSimple}
r\left(\varphi \middle| s, \tilde{s}\right) &= \frac{R}{\sqrt{\left[1 + \frac{1}{4}s+ \tilde{s}\cos(2\varphi)\right]^2 + \left[\tilde{s}\sin(2\varphi)\right]^{2}}} \quad .
\end{align}
We denote semiaxes of this ellipse along the $x$ and $y$ axes, as defined in Appendix 1 - Figure \ref{fig:cutSchematic},  by $a_{\text{in}}$ and $b_{\text{in}}$, respectively. We find that 
\begin{align}
\label{eq:innerShort}
  a_{\text{in}}&= \frac{R}{\left| 1 + \frac{1}{4}s + \tilde{s} \right|}\\
  \label{eq:innerLong}
  b_{\text{in}}&= \frac{R}{\left| 1 + \frac{1}{4}s - \tilde{s} \right|}
\end{align}


\begin{figure}
\centering
\includegraphics[width= .45\textwidth]{Appendix1-figure2.pdf} 
\caption{A. Inner piece problem: Blue ellipse shape of the boundary after the ablation is the reference shape deformed by uniform stress $\sigma_{xx}= \sigma_{xx}^0$, $\sigma_{yy} = \sigma_{yy}^0$ to the red circle boundary shape before the ablation. B) Outer piece problem: The red circle boundary shape before the ablation is the reference in this case. The final boundary shape after the ablation is blue ellipse.}
\label{fig:cutSchematic}
\end{figure}


\subsubsection{Outer piece}
% Now we solve the elastic problem to find  for the outside piece of membrane. %
% One could solve the elastic problem using the state in Eq. \ref{eq:innerSolution} as reference, however, since this solution is an ellipse applying stress free boundary conditions would either mix different modes in polar coordinates or we would have to use elliptical coordinates. Therefore, 
As in the case of the inner piece we need to find the boundary shape as a function of the original stresses in the sheet. In this case, the normal stress at the ablation boundary is zero, but the sheet is not stress-free. To simplify the problem, we consider initially stress-free sheet with a circular hole to which boundary stresses $-\sigma^0_{ij}$ are applied. For small deformations, where elastic problem is linear and solutions of the problems commute, the boundary shape is equivalent to the one created by the laser ablation of a sheet under stress $\sigma^0_{ij}$. % since this is equivalent to substraction of a homogeneous solutions of the linear elastic problem.

Now we have solve the full elastic problem since stresses are not uniform. We keep terms of the Airy stress function $\Phi$, which obey nematic symmetry of the problem and relax sufficiently fast at infinity. We also assume that the polar axis is oriented along one of the principal axes of stress tensor such that the off-diagonal component of stress vanishes. We obtain
\begin{align}
\Phi&= C_0 \ln{r} + \left(D_2 + \frac{B_2}{r^2}\right)\cos(2\varphi)
\end{align}
and for stress components, using Eq. \ref{eq:airyPolar}
\begin{align}\label{eq:outerStressGeneral}
\sigma_{rr} &= \frac{C_0}{r^2} - 2\left(2\frac{D_2}{r^2} + 3\frac{B_2}{r^4} \right)\cos(2\varphi)\\
\sigma_{r\varphi}&= -2\left(\frac{D_2}{r^2} + 3 \frac{B_2}{r^4}\right)\sin(2\varphi)\\
\sigma_{\varphi\varphi}&= -\frac{C_0}{r^2} + 6\frac{B_2}{r^4}\cos(2\varphi)
\end{align}

The stress boundary condition reads% of (relative to the reference stress)
\begin{align}
\sigma_{rr}(r= R)&= -\frac{\sigma}{2} - \tilde{\sigma}\cos(2\varphi)\\
\sigma_{r\varphi}(r= R)&= \tilde{\sigma}\sin(2\varphi)
\end{align}
allows us to determine coefficients in Eq. \ref{eq:outerStressGeneral}
\begin{align}
C_0&= -\frac{1}{2}\sigma R^2\\
B_2&= -\frac{1}{2}\tilde{\sigma}R^4\\
D_2&= \tilde{\sigma}R^2 \quad .
\end{align}
Now we can calculate and then integrate the deformation gradient components to obtain
\begin{align}
u_r&= R\left[\frac{\overline{K}}{4K}\frac{\sigma}{\overline{K}}\frac{R}{r} + \left[2\left(1 + \frac{K}{\overline{K}}\right)\frac{R}{r} -\frac{R^3}{r^3}\right]\frac{\tilde{\sigma}}{2K}\cos(2\varphi)\right]\\
u_\varphi&= -R\left[2\frac{K}{\overline{K}}\frac{R}{r} + \frac{R^3}{r^3}\right]\frac{\tilde{\sigma}}{2K}\sin(2\varphi) \quad .
\end{align}
Finally, as for the inner cut, we have to solve 
\begin{align}\label{eq:circleOuter}
R\hat{r} + u_r \hat{r} + u_\varphi \hat{\varphi} &= r' \hat{r}'
\end{align}
for $r'(\varphi')$. We express the solution using the angle $\varphi$ as a parameter
\begin{widetext}
\begin{align}\label{eq:phiPrime}
\varphi'\left(\varphi \middle| \frac{\sigma}{\overline{K}}, \frac{\tilde{\sigma}}{2K}, \frac{K}{\overline{K}}\right)  &= \varphi - \arctan{\left(\frac{\left(1 + 2\frac{K}{\overline{K}}\right)\frac{\tilde{\sigma}}{2K}\sin(2\varphi)}{1 + \frac{\overline{K}}{4K}\frac{\sigma}{\overline{K}} + \left(1 + 2\frac{K}{\overline{K}}\right) \frac{\tilde{\sigma}}{2K}\cos(2\varphi)}\right)}\\ \label{eq:rPrime}
r'\left(\varphi \middle| \frac{\sigma}{\overline{K}}, \frac{\tilde{\sigma}}{2K}, \frac{K}{\overline{K}}\right) &= R\sqrt{\left[1 + \frac{\overline{K}}{4K}\frac{\sigma}{\overline{K}} + \left(1 + 2\frac{K}{\overline{K}}\right)\frac{\tilde{\sigma}}{2K}\cos(2\varphi)\right]^2 + \left[\left(1 + 2 \frac{K}{\overline{K}}\right)\frac{\tilde{\sigma}}{2K}\sin(2\varphi)\right]^2} \quad .
\end{align}
\end{widetext}
Note ratio of elastic constants appears as a parameter of shape. % This result is due to the fact that stress is not uniform and therefore isotropic and shear stress components mix in the force balance equation. The ratio of elastic constants describes how deformations arising from these different components add to the total deformation.

Now we show that the shape defined by Eqs. \ref{eq:phiPrime} and \ref{eq:rPrime} is an ellipse. To this end, we first define
\begin{align}
\label{eq:1}
  A&= \left(1 + \frac{2K}{\overline{K}}\right)\frac{\tilde{\sigma}}{2K}\\
  B&= 1 + \frac{\overline{K}}{4K}\frac{\sigma}{\overline{K}} \quad. 
\end{align}
This allows us to write Eqs. \ref{eq:phiPrime} and \ref{eq:rPrime} as
\begin{align}
\label{eq:phiPrimeSimple}
\varphi'\left(\varphi \right)  &= \varphi - \arctan{\left(\frac{A \sin{\left(2 \varphi \right)}}{B + A \cos{\left( 2\varphi \right)}} \right)}\\ \label{eq:rPrimeSimple}
r'\left(\varphi \right) &= R\sqrt{A^{2} + B^{2} + 2 AB\cos 2\varphi} \quad .
\end{align}
Now we show that $r'(\varphi')$ is an ellpise. An ellipse equation can be written as
\begin{align}
\label{eq:3}
r'(\varphi')&= \frac{a_{\text{out}}b_{\text{out}}}{\sqrt{b_{\text{out}}^{2}\cos^{2}\varphi' + a_{\text{out}}^{2}\sin^{2}\varphi'}} \quad ,
\end{align}
where $a_{\text{out}}$ and $b_{\text{out}}$ are semiaxes of the ellipse oriented along $x$ and $y$ axes, respectively.
Starting from Eq. \ref{eq:3} and expressing $\varphi'$ from Eq. \ref{eq:phiPrimeSimple} we will show that we can recover Eq. \ref{eq:rPrimeSimple} for a particular choice of parameters $a$ and $b$.

We first calculate
\begin{align}
\label{eq:4}
  \cos \varphi'&= \frac{\left( A+B \right)\cos\varphi}{\sqrt{A^{2}+ B^{2}+ 2AB\cos 2\varphi}}\\
  \sin \varphi'&= \frac{\left( B-A \right)\sin\varphi }{\sqrt{A^{2}+ B^{2}+ 2AB\cos 2\varphi}} \quad .
\end{align}
Inserting these identities into Eq. \ref{eq:3} we find
\begin{align}
\label{eq:2}
r'(\varphi')&= \frac{a_{\text{out}}b_{\text{out}} \sqrt{A^{2}+ B^{2}+ 2AB\cos 2\varphi}}{\sqrt{b_{\text{out}}^{2}\left( A + B \right)^{2}\cos^{2}\varphi + a_{\text{out}}^{2}\left( B-A \right)^{2}\sin^{2}\varphi}} \quad .
\end{align}
Now, we only need to find $a_{\text{out}}$ and $b_{\text{out}}$ for which
\begin{align}
  \label{eq:5}
  \begin{split}
    \frac{a_{\text{out}}^{2}b_{\text{out}}^{2}}{R^{2}} &=  b_{\text{out}}^{2}\left( A + B \right)^{2}\cos^{2}\varphi\\
    &+ a_{\text{out}}^{2}\left( B-A \right)^{2}\left( 1 - \cos^{2}\varphi \right) \quad ,
    \end{split}
\end{align}
for all $\varphi$. The solutions are
\begin{align}
  \label{eq:6}
  a_{\text{out}}&= R \left| A + B \right| \\
  b_{\text{out}}&= R\left| B-A \right| \quad ,
\end{align}
or in terms of the original parameters
\begin{align}
\label{eq:7}
  a_{\text{out}}&= R \left| 1 + \frac{\overline{K}}{4K}\frac{\sigma}{\overline{K}}+ \left(1 + \frac{2K}{\overline{K}}\right)\frac{\tilde{\sigma}}{2K} \right| \\
  b_{\text{out}}&= R\left| 1 + \frac{\overline{K}}{4K}\frac{\sigma}{\overline{K}}-\left(1 + \frac{2K}{\overline{K}}\right)\frac{\tilde{\sigma}}{2K} \right| \quad .
\end{align}
If, in addition to Eqs. \ref{eq:sSimple} and \ref{eq:tildeSSimple} we define
\begin{align}
\label{eq:10}
\xi&= \frac{2K}{\overline{K}} \quad ,
\end{align}
we can write
\begin{align}
\label{eq:outerLong}
  a_{\text{out}}&= R \left| 1 + \frac{1}{2\xi}s+ \left(1 + \xi\right)\tilde{s} \right| \\
  \label{eq:outerShort}
  b_{\text{out}}&= R \left| 1 + \frac{1}{2\xi}s- \left(1 + \xi\right)\tilde{s} \right|
\end{align}

Equations \ref{eq:innerShort}, \ref{eq:innerLong}, \ref{eq:outerLong} and \ref{eq:outerShort} impose 4 constraints on 3 parameters: $s$, $\tilde{s}$ and $\xi$. When determining these parameters from experimental data we are in general not be able to satisfy all 4 equations simultaneously. Instead, we find the 3 parameters that best fit the segmented outlines of tissue boundaries, see Methods.

\section{Correlation between cell elongation and cell movement contributes to the tissue radial shear flow}\label{sec:correlation_term}

In this work we use the previously developed method to define and calculate cell elongation tensor and tissue shear due to T1 transitions, cell divisions and extrusions (Etournay et al., 2015; Merkel et al., 2015). However, we apply the method in the polar coordinate system and to make it fully consistent we have to introduce a new correlation term $S_{ij}$, which arises in the polar coordinate system when  translation and elongation are correlated. To demonstrate the origin and derive an expression for $S_{ij}$ we consider a tissue that translates over the time interval $\Delta t$ but does not change otherwise. The tissue is triangulated by connecting the neighboring cell centers, see (Merkel et al., 2017). Initially, the angle of a triangle $k$ with with respect to a center point is  $\varphi$, blue triangle in Fig. Appendix 1 - Figure \ref{fig:are_x}. During the translation by $\Delta \vec{r}= \vec{r}(t + \Delta t) - \vec{r}(t)$ the radial angle of the triangles changes by a small angle $\Delta \varphi^k$. Therefore, the components of triangle elongation tensor $\tilde{q}^t_{ij}$ in polar coordinate system change by
\begin{align}
  \Delta \tilde{q}^t_{ij}&= - 2 \epsilon_{ik} \tilde{q}^t_{kj} \Delta \varphi \quad .
\end{align}

 \begin{figure}[ht!]
  \centering
  \includegraphics[width= .35\textwidth]{Appendix1-figure3.pdf}
  \caption{Triangle is translated by $\Delta \vec{r}= \vec{r}(t + \Delta t) - \vec{r}(t)$. Although the triangle elongation did not change during the translation, the radial component of the triangle elongation changes due to the change of radial direction by $\Delta \varphi$.}
  \label{fig:are_x}
\end{figure}
 \begin{figure}[ht!]
  \centering
  \includegraphics[width= .45\textwidth]{Appendix1-figure4.pdf}
  \caption{The three correlation terms contributing to the radial tissue shear flow measured in wing disc pouch, averaged over 5 time-lapses. Shaded regions correspond to one standard deviation of that sample. Blue: Correlation term stemming from correlated fluctuations of rotation and elongation of triangles. Green: Correlation term stemming from correlated fluctuations of area and elongation of triangles. Black: The new correlation term stemming from translation and elongation of triangles. The new correlation term (black) is very small in comparison to the other two.}
  \label{fig:correlations}
\end{figure}
However, since the triangles are only translating, there is no shear flow. Therefore, the average radial shear flow also vanishes and an additional term is necessary to account for the change in the average radial cell elongation in a translating tissue
\begin{align}
  \tilde{v}_{ij}&= 0 = \frac{\Delta Q_{ij}}{\Delta t} + S_{ij} \quad .
\end{align}
Therefore, the correlation term is given by
\begin{align}
  S_{ij}= \frac{1}{\Delta t}\left\langle 2 \epsilon_{ik} \tilde{q}^t_{kj} \Delta \varphi  \right\rangle \quad .
\end{align}
In Appendix 1 - Figure \ref{fig:correlations} we show the new correlation term in black measured in the pouch of the wing disc, averaged over 5 time-lapse experiments. For comparison we show the correlation term stemming from correlated fluctuations of triangle rotation and elongation in blue, and correlation term stemming from correlated fluctuations of triangle area and elongation in green. For definitions and discussion of the other two correlation terms see (Etournay et al., 2015; Merkel et al., 2017). We find that the contribution of the new correlation term to the radial shear flow is much smaller that the other two correlation terms. Therefore, contribution to the radial tissue shear flow of correlation between translation and triangle elongation is a negligible in the wing disc.


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