# Annex: Derivatives of Airy function

$$\Phi = \frac{\left(n+1\right)^2}{n\left(2n+1\right)} r^{\frac{2n+1}{n+1}} f\left(\theta\right)$$ $$\Phi_{,r} = \frac{n+1}{n} r^{\frac{n}{n+1}} f\left(\theta\right)$$ $$\Phi_{,rr} = r^{\frac{-1}{n+1}} f\left(\theta\right)$$ $$\Phi_{,\theta} = \frac{\left(n+1\right)^2}{n\left(2n+1\right)} r^{\frac{2n+1}{n+1}} f'\left(\theta\right)$$ $$\Phi_{,\theta\theta} = \frac{\left(n+1\right)^2}{n\left(2n+1\right)} r^{\frac{2n+1}{n+1}} f''\left(\theta\right)$$ $$\Phi_{,r\theta} = \Phi_{,\theta r} = \frac{n+1}{n} r^{\frac{n}{n+1}} f'\left(\theta\right)$$

# Annex: Derivatives of stream function

$$\Psi = - r^{\frac{n+2}{n+1}} g\left(\theta\right)$$ $$\Psi_{,r} = -\frac{n+2}{n+1} r^{\frac{1}{n+1}} g\left(\theta\right)$$ $$\Psi_{,rr} = -\frac{n+2}{\left(n+1\right)^2} r^{\frac{-n}{n+1}} g\left(\theta\right)$$ $$\Psi_{,\theta} = -r^{\frac{n+2}{n+1}} g'\left(\theta\right)$$ $$\Psi_{,\theta\theta} = -r^{\frac{n+2}{n+1}} g''\left(\theta\right)$$ $$\Psi_{,r\theta} = -\frac{n+2}{n+1} r^{\frac{1}{n+1}} g'\left(\theta\right)$$

# Annex: Power law and normal flow

We have defined the power law to be

$$\bar{\varepsilon} = \frac{\alpha \sigma_p^0}{E} \left( \frac{\sigma_e}{\sigma_p^0} \right)^n\,,\label{eq:powerapp}$$

with the equivalent von Mises stress being

$$\label{eq:vonMises} \sigma_e = \sqrt{\frac{3}{2} \mathbf{s}:\mathbf{s}} \, \text{.}$$

Since we have assumed plane-$\varepsilon$ state and the material to be incompressible, the strain tensor $\strain$ can be expressed in terms of the deviatoric stress tensor $\mathbf{s}$, the equivalent strain $\bar{\varepsilon}$ and the equivalent stress through the normal flow condition

$$\label{eq:normality1} \strain = \bar{\varepsilon} \sqrt{\frac{3}{2}} \frac{\mathbf{s}}{\sqrt{\mathbf{s}: \mathbf{s}}} = \frac{3\bar{\varepsilon}}{2\sigma_e} \mathbf{s}\,.$$

Therefore, since $\strain_{zz}=0$, one has $\mathbf{s}_{zz}=0$ and

$$\cauchy_{zz} - \frac{\cauchy_{rr} + \cauchy_{\theta\theta} + \cauchy_{zz}}{3} = 0 \, \text{,} \label{eq:stressTensor_zz}$$

due to the plane-$\varepsilon$ assumption. Using (\ref{eq:stressTensor_zz}), the out of plane stress can be expressed by

$$\cauchy_{zz} = \frac{\cauchy_{rr} + \cauchy_{\theta\theta}}{2} \, \text{,}$$

and one has thus

$$\text{tr}{\cauchy} = \frac{3}{2}\left(\cauchy_{rr}+\cauchy_{\theta\theta}\right) \, \text{,}$$

and the non-zero components of the deviatoric stress tensor are

\begin{eqnarray}\label{eq:stressTensor_nonZero} \uT{s}=\left( \begin{array}{cc} {\frac{\cauchy_{rr}-\cauchy_{\theta\theta}}{2}} & \cauchy_{r\theta}\\ \cauchy_{r\theta} & \frac{\cauchy_{\theta\theta}-\cauchy_{rr}}{2} \end{array}\right) \, \text{.} \end{eqnarray}

Now, inserting (\ref{eq:stressTensor_nonZero}) into (\ref{eq:vonMises}) and resolving the double contractions, $\mathbf{A}:\mathbf{B}=A_{ij}B_{ij}$, results in

$$\sigma_e =\sqrt{\frac{3}{2}} \sqrt{\uT{s}:\uT{s}} = \sqrt{\frac{3}{4}\left(\cauchy_{rr}-\cauchy_{\theta\theta}\right)^2+3\cauchy_{r\theta}^2 }\,.$$

Expressing the stress components in terms of $r$ and $f$ as was derived previously,

$$\begin{cases} \cauchy_{rr}= r^{\frac{-1}{n+1}}\left(\frac{n+1}{n} f+\frac{\left(n+1\right)^2}{n\left(2n+1\right)}f'' \right)\,, \\ \cauchy_{\theta\theta}= r^{\frac{-1}{n+1}} f\,, \\ \cauchy_{r\theta}= r^{\frac{-1}{n+1}}\left(- \frac{n+1}{n} + \frac{\left(n+1\right)^2}{n\left(2n+1\right)} \right)f'=r^{\frac{-1}{n+1}} \frac{-\left(n+1\right)} {\left(2n+1\right)}f' \,,\end{cases}$$

$$\sigma_e^2 =r^{\frac{-2}{n+1}}\left[ \frac{3}{4}\left(\frac{1}{n}f +\frac{\left(n+1\right)^2}{n\left(2n+1\right)}f''\right)^2+3\left(\frac{n+1}{2n+1}\right)^2f'^2\right] \, \text{.}$$

Similarly, we can obtain an expression for the equivalent strain by inserting the individual components which we derived in terms of $r$ and $f$ as

$$\begin{cases} \strain_{rr}=\frac{\Psi_{,r\theta}}{r}-\frac{\Psi_{,\theta}}{r^2} = r^{\frac{-n}{n+1}} \left(-\frac{n+2}{n+1}g'+g'\right)=-\frac{1}{n+1} r^{\frac{-n}{n+1}} g'\,, \\ \strain_{\theta\theta}=\frac{1}{n+1} r^{\frac{-n}{n+1}} g' \,,\\ \strain_{r\theta}=\frac{1}{2}\left(\frac{\Psi_{,\theta\theta}}{r^2}-\Psi_{,rr}+\frac{\Psi_{,r}}{r}\right) = \frac{r^{\frac{-n}{n+1}}}{2} \left(-g''+\frac{n+2}{\left(n+1\right)^2}g-\frac{n+2}{n+1}g \right) = - \frac{r^{\frac{-n}{n+1}}}{2} \left(g''+ n \frac{n+2}{\left(n+1\right)^2}g \right) \, \text{,} \end{cases} \label{eq:strainFromF}$$

$$\bar{\varepsilon}^2 = \frac{2}{3} \strain : \strain = \frac{2}{3}\left(\strain_{rr}^2+\strain_{\theta\theta}^2+2\strain_{r\theta}^2\right) = \frac{2}{3}r^{\frac{-2n}{n+1}} \left[\frac{2}{\left(1+n\right)^2}g'^2+\frac{1}{2}\left(g''+n\frac{n+2}{\left(n+1\right)^2}g\right)^2\right] \,.$$

In order to derive the first $2^{\text{nd}}$-order ODE, we introduce these expressions in the power law (\ref{eq:powerapp}) rewritten as

$$\bar{\varepsilon}^2=\left(\frac{\alpha\sigma_p^0}{E}\right)^2\left(\frac{\sigma_e}{\sigma_p^0}\right)^{2n}\,,$$

which becomes

$$\begin{gathered} \left[\frac{4}{3\left(1+n\right)^2}g'^2+\frac{1}{3}\left(g''+n\frac{n+2}{\left(n+1\right)^2}g\right)^2\right] = \left(\frac{\alpha\sigma_p^0}{E\left(\sigma_p^0\right)^n}\right)^2 \\ \left[ \frac{3}{4}\left(\frac{1}{n} +\frac{\left(n+1\right)^2}{n\left(2n+1\right)}f''\right)^2+3\left(\frac{n+1}{2n+1}\right)^2f'^2\right]^n\,. \end{gathered}$$

In order to derive the first $2^{\text{nd}}$-order ODE, we introduce these expressions in the normal flow condition (\ref{eq:normality1}), leading to

$$\strain = \bar{\varepsilon} \sqrt{\frac{3}{2}} \frac{\uT{s}}{\sqrt{\uT{s}:\uT{s}}} = \frac{3 \bar{\varepsilon} }{2\sigma_e } \uT{s} = \frac{3 \bar{\varepsilon} }{2\sigma_e }\left( \begin{array}{cc} {\frac{\cauchy_{rr}-\cauchy_{\theta\theta}}{2}} & \cauchy_{r\theta}\\ \cauchy_{r\theta} & \frac{\cauchy_{\theta\theta}-\cauchy_{rr}}{2} \end{array}\right) \, \text{.}$$

From the latter result, the following strain-stress relations arise

$$\begin{gathered} \strain_{rr}-\strain_{\theta\theta} = \frac{3\bar{\varepsilon}}{2\sigma_e}\left(\cauchy_{rr}-\cauchy_{\theta\theta}\right) \\ \strain_{r\theta} = \frac{3\bar{\varepsilon}}{2\sigma_e} \cauchy_{r\theta} \end{gathered}\,,$$

which implies

$$\left(\strain_{rr}-\strain_{\theta\theta}\right)\cauchy_{r\theta} = \left(\cauchy_{rr}-\cauchy_{\theta\theta}\right)\strain_{r\theta} \, \text{,}$$

or in terms of $f\left(\theta\right)$ and $g\left(\theta\right)$

$$\frac{2}{2n+1}g'f' = -\frac{1}{2} \left[\frac{1}{n}f+\frac{\left(n+1\right)^2}{n\left(2n+1\right)}f''\right] \left[g''+n\frac{n+2}{\left(n+1\right)^2}g\right]\,.$$

# Annex: 4$^{\text{th}}$-order ODE

We aim to reduce the following two 2$^{\text{nd}}$-order differential equations

$$\label{eq:ODEs_1} \begin{cases} \left[\tilde{f}''+\frac{2n+1}{\left(n+1\right)^2} \tilde{f}\right]\left[\tilde{g}''+n\frac{n+2}{\left(n+1\right)^2}\tilde{g}\right] = - \frac{4n}{\left(n+1\right)^2}\tilde{g}'\tilde{f}' \,,\\ \left[\frac{4}{\left(1+n\right)^2}\tilde{g}'^2+\left(\tilde{g}''+n\frac{n+2}{\left(n+1\right)^2}\tilde{g}\right)^2\right]^{\frac{1}{n}} = \frac{3^{\frac{n+1}{n}}}{4}\frac{\left(n+1\right)^4}{n^2\left(2n+1\right)^2} \left[\left(\frac{2n+1}{\left(n+1\right)^2}\tilde{f} + \tilde{f}''\right)^2+\frac{4n^2}{\left(n+1\right)^2}\tilde{f}'^2\right]\,, \end{cases}$$

to a single 4$^{\text{th}}$-order ordinary differential equation. To do that, let us define $\tilde{l}$ as follows

$$\label{eq:defineL} \tilde{l}\left(\theta\right) = \frac{1}{n}\left[\left(n+1\right)\tilde{f}''+\frac{2n+1}{n+1}\tilde{f}\right] \, \text{.}$$

Introducing this new function $\tilde{l}$ allows us to redefine the two ODEs in (\ref{eq:ODEs_1})

$$\label{eq:ODEs_2} \begin{cases} \left[\frac{4}{\left(1+n\right)^2}\tilde{g}'^2+\left(\tilde{g}''+n\frac{n+2}{\left(n+1\right)^2}\tilde{g}\right)^2\right]^{\frac{1}{n}} = \frac{3^{\frac{n+1}{n}}}{4}\frac{\left(n+1\right)^2}{\left(2n+1\right)^2} \left[\tilde{l}^2+4\tilde{f}'^2\right]\,, \\ \tilde{l} \left[\tilde{g}''+n\frac{n+2}{\left(n+1\right)^2}\tilde{g}\right] = - \frac{4}{\left(n+1\right)}\tilde{g}'\tilde{f}' \, \text{.} \end{cases}$$

These equations allow expressing $\tilde{g}$ in terms of $\tilde{f}$ as

$$\left[\frac{4}{\left(1+n\right)^2}+\frac{16}{\left(n+1\right)^2\tilde{l}^2}\tilde{f}'^2\right]^{\frac{1}{n}}\tilde{g}'^{\frac{2}{n}} = \frac{3^{\frac{n+1}{n}}}{4}\frac{\left(n+1\right)^2}{\left(2n+1\right)^2} \left[\tilde{l}^2+4\tilde{f}'^2\right]\,,$$

or again

$$\tilde{g}' = \sqrt{\left[\frac{\frac{3^{n+1}}{4^n}\frac{\left(n+1\right)^{2n}}{\left(2n+1\right)^{2n}} \left[\tilde{l}^2+4\tilde{f}'^2\right]^n}{\left[\frac{4}{\left(1+n\right)^2}+\frac{16}{\left(n+1\right)^2\tilde{l}^2}\tilde{f}'^2\right]}\right]} \, \text{.} \label{eq:gfromf}$$

Differentiating the second equation of (\ref{eq:ODEs_2}) and substituting $\tilde{g}'$ and its derivatives using (\ref{eq:gfromf}), yield a $4^{\text{th}}$-order ODE

\begin{eqnarray} &&\tilde{f}'''' +\frac{2n+1}{\left(1+n\right)^2}\tilde{f}'' + \frac{1}{n\tilde{l}^2+4\tilde{f}'^2}\left\{ \frac{n^2}{\left(1+n\right)^2}\left[\frac{4}{n}\tilde{f}''+\frac{n+2}{1+n}\tilde{l}\right]\left[\tilde{l}^2+4\tilde{f}'^2\right]+\right.\nonumber\\ &&\left.\frac{4n\left(n-1\right)}{\left(n+1\right)^2}\tilde{f}'\left[\tilde{l}\tilde{l'}+4\tilde{f}'\tilde{f}''\right] + \frac{n\left(n-1\right)}{n+1}\left[3\tilde{l}\tilde{l}'^2+8\tilde{l}'\tilde{f}'\tilde{f}''+4\tilde{l}\left(\tilde{f}''^2+\tilde{f}'\tilde{f}'''\right)\right]\right.\nonumber\\ &&\left.+\frac{n\left(n-1\right)\left(n-3\right)}{n+1}\frac{\tilde{l}}{\tilde{l}^2+4\tilde{f}'^2}\left[\tilde{l}\tilde{l}' +4 \tilde{f}'\tilde{f}''\right]^2\right\}=0 \, \text{.} \end{eqnarray}