## 2 Concentrated Couple in a Plane  
We consider a concentrated clockwise couple \(M\) applied at the origin of coordinates in a plane. In polar coordinates \((r,\theta)\), the components of classical (local) stress tensor \(\sigma\) are [44]  
\[\sigma_{rr}=0,\qquad\sigma_{\theta\theta}=0,\qquad\sigma_{r\theta}=\frac{A}{r^ {2}}, \tag{27}\]  
where  
\[A=\frac{M}{2\pi}. \tag{28}\]  
According to (27) and Equations (A1) and (A2) from Appendix A, we obtain that components of the nonlocal stress tensor  
\[t_{rr}=0,\qquad t_{\theta\theta}=0. \tag{29}\]  
From key equations of the theory (25) and (26), taking into account Equation (29) and Equations (A3) and (A4) from Appendix A in the axisymmetric case with \(\partial/\partial\theta=0\), it follows that the component \(t_{r\theta}\) is determined from the Cauchy problem \[\frac{\partial^{\alpha}t_{r\theta}}{\partial\tau^{\alpha}}=a\left(\frac{\partial^{2 }t_{r\theta}}{\partial r^{2}}+\frac{1}{r}\,\frac{\partial t_{r\theta}}{\partial r }-\frac{4}{r^{2}}\,t_{r\theta}\right), \tag{30}\]  
\[\tau=0:\ \ t_{r\theta}=\frac{A}{r^{2}}. \tag{31}\]  
The Laplace transform with respect to the nonlocality parameter \(\tau\) and the Hankel transform of the second order with respect to the radial coordinate \(r\) result in the solution in the transform domain  
\[\widehat{t}_{r\theta}^{*}(\xi,s)=\frac{A}{2}\,\frac{s^{\alpha-1}}{s^{\alpha}+ a_{\xi}^{2}}, \tag{32}\]  
where the hat marks the Hankel transform, \(\xi\) is the transform variable, and Equation (A5) from Appendix B has been used.  
Using the inverse Laplace transform formula [40,41]  
\[\mathcal{L}^{-1}\bigg{\{}\frac{s^{\alpha-1}}{s^{\alpha}+b}\bigg{\}}=E_{\alpha }(-b\tau^{\alpha}), \tag{33}\]  
where \(E_{\alpha}(z)\) is the ML function in one parameter \(\alpha\) (19), we get  
\[t_{r\theta}(r,\tau)=\frac{A}{2}\int_{0}^{\infty}E_{\alpha}\left(-a\xi^{2}\tau ^{\alpha}\right)I_{2}(r\xi)\xi\,\mathrm{d}\xi. \tag{34}\]  
For \(\alpha=1\),  
\[E_{1}\left(-z^{2}\right)=\mathrm{e}^{-z^{2}}, \tag{35}\]  
and taking into account Equation (A9) from Appendix B, we have  
\[t_{r\theta}(r,\tau)=\frac{A}{r^{2}}\bigg{[}1-\left(1+\frac{r^{2}}{4a\tau} \right)\exp\left(-\frac{r^{2}}{4a\tau}\right)\bigg{]}. \tag{36}\]  
The classical elasticity solution \(\sigma_{r\theta}=A/r^{2}\) (see Equation (27)) follows from Equation (36) when \(\tau\to 0\). As the ML function \(E_{\alpha}(0)=1\), in the general case described by the solution (34), the classical elasticity solution is obtained using the integral (A7) from Appendix B.  
Another particular case of the solution (34) is obtained for \(\alpha\to 0\) when  
\[E_{0}\left(-z^{2}\right)=\frac{1}{1+z^{2}}. \tag{37}\]  
Evaluating the corresponding integral (see Equation (A11) from Appendix B), we arrive at  
\[t_{r\theta}(r)=\frac{A}{2a}\left[\frac{2a}{r^{2}}-K_{2}\left(\frac{r}{\sqrt{a }}\right)\right]. \tag{38}\]  
Figure 1 shows the dependence of the nondimensional stress component \(\tilde{t}_{r\theta}\) on the nondimensional distance \(\tilde{r}\) for different values of the order of fractional derivative \(\alpha\). The following nondimensional quantities are introduced:  
\[\tilde{t}_{r\theta}=\frac{a\tau^{\alpha}}{A}\,t_{r\theta},\qquad\tilde{r}= \frac{r}{\sqrt{a}\tau^{\alpha/2}}. \tag{39}\]