## 2 Modified Nonlocal Strain Gradient Law for Rods
\begin{array}{c}\left.\partial_{x}N\left(x,L_{c},l\right)\right|_{x =0}=\frac{1}{L_{c}}N\left(0,L_{c},l\right)+\frac{l^{2}}{L_{c}^{2}}\left. \partial_{x}\left(A_{E}\ \varepsilon_{el}\right)\left(x\right)\right|_{x=0}\\ \left.\partial_{x}N\left(x,L_{c},l\right)\right|_{x=L}=-\frac{1}{L_{c}}N\left( L,L_{c},l\right)+\frac{l^{2}}{L_{c}^{2}}\left.\partial_{x}\left(A_{E}\ \varepsilon_{el}\right)\left(x\right)\right|_{x=L}.\end{array} \right. \tag{7}\]  
Proof. : Since the bi-exponential averaging function is given by  
\[\phi(x,L_{c})=\frac{1}{2L_{c}}\exp\left(-\frac{\left|x\right|}{L_{c}}\right), \tag{8}\]a direct evaluation provides the first derivative of the convolutions (Equation (3))  
\[ \begin{split}\partial_{x}N_{0}\left(x,L_{c}\right)=\frac{1}{L_{c}} \left[\int_{x}^{L}\phi\left(x-\xi,L_{c}\right)\ \left(A_{E}\,\varepsilon_{el}\right)\left(\xi\right)d\xi+\right.\\ \left.-\int_{0}^{x}\phi\left(x-\xi,L_{c}\right)\ \left(A_{E}\, \varepsilon_{el}\right)\left(\xi\right)d\xi\right]\\ \partial_{x}N_{1}\left(x,L_{c},I\right)=\frac{l^{2}}{L_{c}} \left[\int_{x}^{L}\phi\left(x-\xi,L_{c}\right)\ \partial_{\xi}\left(A_{E}\, \varepsilon_{el}\right)\left(\xi\right)d\xi+\right.\\ \left.-\int_{0}^{x}\phi\left(x-\xi,L_{c}\right)\ \partial_{\xi}\left(A_{E}\, \varepsilon_{el}\right)\left(\xi\right)d\xi\right].\end{split}  \tag{9}\]  
The second derivative of the convolutions (Equation (3)) follows from Equation (9) using Equation (3) to get  
\[ \begin{split}\partial_{x}^{2}N_{0}\left(x,L_{c}\right)=\frac{1}{ L_{c}^{2}}\left[l^{2}\int_{0}^{L}\phi\left(x-\xi,L_{c}\right)\ \left(A_{E}\,\varepsilon_{el}\right)\left(\xi\right)d\xi+\right.\\ \left.-\left(A_{E}\,\varepsilon_{el}\right)\left(x\right)\right] =\frac{1}{L_{c}^{2}}\left[N_{0}\left(x,L_{c}\right)-\left(A_{E}\, \varepsilon_{el}\right)\left(x\right)\right]\\ \partial_{x}^{2}N_{1}\left(x,L_{c},I\right)=\frac{l^{2}}{L_{c}^{ 2}}\left[\int_{0}^{L}\phi\left(x-\xi,L_{c}\right)\ \partial_{\xi}\left(A_{E}\, \varepsilon_{el}\right)\left(\xi\right)d\xi+\right.\\ \left.-\partial_{x}\left(A_{E}\,\varepsilon_{el}\right)\left(x \right)\right]=\\ \left.=\frac{1}{L_{c}^{2}}\left[N_{1}\left(x,L_{c},I\right)-l^{2} \partial_{x}\left(A_{E}\,\varepsilon_{el}\right)\left(x\right)\right].\right. \end{split}  \tag{10}\]  
Subtracting the third derivative of Equation (9)\({}_{2}\) from Equation (9)\({}_{1}\), it turns out to be  
\[ \begin{split}\partial_{x}^{2}N_{0}\left(x,L_{c}\right)-\partial_ {x}^{3}N_{1}\left(x,L_{c},I\right)\\ =\frac{1}{L_{c}^{2}}\left[N_{0}\left(x,L_{c}\right)-\left(A_{E} \,\varepsilon_{el}\right)\left(x\right)\right]+\\ -\frac{1}{L_{c}^{2}}\left[\partial_{x}N_{1}\left(x,L_{c},I\right) -l^{2}\partial_{x}^{2}\left(A_{E}\,\varepsilon_{el}\right)\left(x\right) \right]\end{split}  \tag{11}\]  
so that, recalling Equation (5) and rearranging the terms, we have  
\[ \begin{split}\partial_{x}^{2}N&\left(x,L_{c},I\right) =\partial_{x}^{2}N_{0}\left(x,L_{c}\right)-\partial_{x}^{3}N_{1}\left(x,L_{c},I\right)=\\ =\frac{1}{L_{c}^{2}}\left[N_{0}\left(x,L_{c}\right)-\partial_{x} N_{1}\left(x,L_{c},I\right)\right]-\frac{1}{L_{c}^{2}}\left(A_{E}\, \varepsilon_{el}\right)\left(x\right)+\frac{l^{2}}{L_{c}^{2}}\partial_{x}^{2} \left(A_{E}\,\varepsilon_{el}\right)\left(x\right)\end{split}  \tag{12}\]  
and Equation (6) is recovered. The CBC in Equation (7) can be recovered as follows. Using Equation (9)\({}_{2}\), Equation (5) can be rewritten in the form  
\[ \begin{split} N&\left(x,L_{c},I\right)=N_{0}\left(x,L _{c}\right)-\frac{l^{2}}{L_{c}}\left[\int_{x}^{L}\phi\left(x-\xi,L_{c}\right)\ \partial_{\xi}\left(A_{E}\, \varepsilon_{el}\right)\left(\xi\right)d\xi+\right.\\ \left.-\int_{0}^{x}\phi\left(x-\xi,L_{c}\right)\ \partial_{\xi} \left(A_{E}\,\varepsilon_{el}\right)\left(\xi\right)d\xi\right]\end{split}