# Closed-Form Analytical Solutions for the Deflection of Elastic Beams in a Peridynamic Framework
## 4 Analytical Solutions to Peridynamic Problems  
Let us recall the PD equation of motion (16):  
\[\rho A\bar{w}(x)=\frac{E\xi}{\beta}\int\limits_{-\delta}^{\delta}\frac{1}{\xi ^{2}}\bigg{[}\int\limits_{-\delta}^{\delta}\frac{w(x+\xi)-w(x)}{\xi^{2}}\text {d}\xi-\int\limits_{-\delta}^{\delta}\frac{w(x+\xi+\eta)-w(x+\xi)}{\eta^{2}} \text{d}\eta\bigg{]}\text{d}\xi+p(x). \tag{40}\]  
To formulate the analytical solutions to Equation (40), we apply the Laplace transform as follows:  
\[ \begin{split}\rho A\big{[}& s^{2}W(x,s)-sw_{0}(x)- v_{0}(x)\big{]}\\ &=c\int\limits_{-\delta}^{\delta}\frac{1}{\xi^{2}}\left[\begin{array} []{c}\int\limits_{-\delta}^{\delta}\frac{W(x+\zeta,s)-W(x,s)}{\xi^{2}}d\zeta \\ -\int\limits_{-\delta}^{\delta}\frac{W(x+\xi+\eta,s)-W(x+\xi,s)}{\eta^{2}}d \eta.\end{array}\right]d\xi+P(x,s),\end{split}  \tag{41}\]  
where  
\[W(x,s)=L[w(x,t)],\quad P(x,s)=L[p(x,t)],\quad w_{0}(x)=w(x,0),\quad v_{0}(x)= \psi(x,0). \tag{42}\]  
Consider a simply supported beam subjected to some arbitrary dynamic load. By analogy to Section 2.3.1, we can assume that the solution to Equation (41) has the following form:  
\[W(x,s)=\sum\limits_{n=1}^{\infty}B_{n}(s)\text{sin}\frac{n\pi tx}{L}. \tag{43}\]  
After plugging Equation (43) into Equation (41), we obtain  
\[\sum\limits_{n=1}^{\infty}B_{n}(s)\left\{c\left[\int\limits_{-\delta}^{\delta }\frac{1}{\xi^{2}}\bigg{(}1-\text{cos}\frac{n\pi\xi}{L}\bigg{)}d\xi\right]^{ 2}+\rho As^{2}\right\}\text{sin}\frac{n\pi tx}{L}=P(x,s)+s\rho Aw_{0}(x)+\rho Av _{0}(x). \tag{44}\]According to orthogonality, we obtain  
\[ \begin{split} B_{n}(s)&=\frac{2}{L}\frac{\int\limits_{0} ^{L}\left[P(x,s)+s\rho Aw_{0}(x)+\rho Av_{0}(x)\right]\text{sin}\frac{n\pi x}{L} dx}{\rho As^{2}+c\left[\int\limits_{-\delta}^{\delta}\frac{1}{\xi^{2}}\left(1-\text{ cos}\frac{n\pi\xi}{L}\right)d\xi\right]^{2}}\\ &=\frac{2}{L}\frac{\frac{1}{\rho A}\int\limits_{0}^{L}P(x,s)\text {sin}\frac{n\pi x}{L}dx+s\vartheta_{n}+\vartheta_{n}}{(s-i\omega_{n})(s+i \omega_{n})},\end{split}  \tag{45}\]  
where  
\[ \begin{split}\vartheta_{n}&=\int\limits_{0}^{L}w_{0} (x)\text{sin}\frac{n\pi x}{L}dx,\quad\vartheta_{n}=\int\limits_{0}^{L}v_{0}(x )\text{sin}\frac{n\pi x}{L}dx,\\ \omega_{n}&=\sqrt{\frac{c}{\rho A}}\int\limits_{- \delta}^{\delta}\frac{1}{\xi^{2}}\left(1-\text{cos}\frac{n\pi\xi}{L}\right)d \xi.\end{split}  \tag{46}\]  
Applying partial fractions to Equation (45) it follows that  
\[ \begin{split} B_{n}(s)&=\frac{2}{L}\left[\frac{1}{ \rho A}\frac{\int\limits_{0}^{L}P(x,s)\text{sin}\frac{n\pi x}{L}dx}{(s-i\omega_ {n})(s+i\omega_{n})}+\frac{i\omega_{n}\vartheta_{n}+\vartheta_{n}}{2i\omega_{ n}}\frac{1}{s-i\omega_{n}}+\frac{i\omega_{n}\vartheta_{n}-\vartheta_{n}}{2i \omega_{n}}\frac{1}{s+i\omega_{n}}\right].\end{split}  \tag{47}\]  
After performing the inverse Laplace transform on (47), we obtain  
\[ \begin{split} b_{n}(t)&=\frac{2}{L}\left\{\frac{1} {\rho A}\frac{1}{2i\omega_{n}}\int\limits_{0}^{t}\left(e^{i\omega_{n}u}-e^{-i \omega_{n}u}\right)\left[\int\limits_{0}^{L}b(x,t-u)\text{sin}\pi xdx\right]du \\ &+\frac{i\omega_{n}\vartheta_{n}+\vartheta_{n}}{2i\omega_{n}}e^ {i\omega_{n}t}+\frac{i\omega_{n}\vartheta_{n}-\vartheta_{n}}{2i\omega_{n}}e^ {-i\omega_{n}t}\right\}\\ &=\frac{2}{L}\left\{\vartheta_{n}\text{cos}\omega_{n}t+\frac{ \vartheta_{n}}{\omega_{n}}\text{sin}\omega_{n}t+\frac{1}{\rho A}\frac{1}{\omega _{n}}\int_{0}^{t}\text{sin}\omega_{n}u\left[\int\limits_{0}^{L}p(x,t-u)\text {sin}\frac{n\pi x}{L}dx\right]du\right\}.\end{split}  \tag{48}\]  
In this case, the solution is  
\[w(x,t)=L^{-1}[W(x,s)]=\sum\limits_{n=1}^{\infty}b_{n}(t)\text{sin}\frac{n\pi x }{L}. \tag{49}\]  
Coupling (48) with (49) yields the complete solution for a simply supported beam as follows:  
\[