# Closed-Form Solution of the Bending Two-Phase Integral Model of Euler-Bernoulli Nanobeams
## 3 Closed-Form Solution of [PERSON]'s Two-Phase Integral Model Equations
### Cantilever Beam (CF)  
Let us consider the case of a cantilever beam subject to the following boundary conditions  
\[w(0)=w^{\prime}(0)=0,\quad M(L)=M^{\prime}(L)=0. \tag{20}\]  
Let \(X=C[0,L]\). Taking into account the equilibrium Equation (16) and the definition (14), we define the operator \(B:X\to X\) as  
\[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\] \[\mathcal{D}(B) = \Big{\{}w(x)\in C^{4}[0,L]:w(0)=w^{\prime}(0)=0,\] \[\bigg{[}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{2}}{\xi_{1}}\int_{0}^ {x}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{]}_{x=L}=0,\] \[\bigg{[}\frac{d}{dx}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{x}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)}\bigg{]}_{x=L} =0\bigg{\}},\]  
and write the boundary value problem (16), (20) in the compact form  
\[Bw(x)=\frac{1}{EI\xi_{1}}q(x),\quad 0<x<L,\quad q(x)\in X. \tag{21}\]  
Let the Fredholm integro-differential operator \(B_{2}:X\to X\) be defined by  
\[B_{2}w(x) = \frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{2}}{\xi_{1}}\int_{0}^{L}k(x, t)\frac{d^{2}w(t)}{dt^{2}}dt,\] \[\mathcal{D}(B_{2}) = \Big{\{}w(x)\in C^{2}[0,L]:\,w(0)=w^{\prime}(0)=0\Big{\}}. \tag{22}\]  
Furthermore, let the differential operator \(B_{1}:X\to X\) be  
\[B_{1}y(x) = \frac{d^{2}y(x)}{dx^{2}},\] \[\mathcal{D}(B_{1}) = \Big{\{}y(x)\in C^{2}[0,L]:y(L)=y^{\prime}(L)=0\Big{\}}.\]If we take \(y(x)=B_{2}w(x)\), then  
\[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]0 \[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]1 \[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]2 \[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]3 \[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]4 \[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]5  
That is, the operator \(B\) can be factorized as \(B=B_{1}B_{2}\), and therefore, the boundary value problem (21) is carried to  
\[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]6  
The solution of (23) can now be obtained by solving the following two boundary value problems, namely the differential boundary value problem  
\[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]7  
and the Fredholm integro-differential boundary value problem  
\[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]8  
The solution of DBVP in (24) in closed form for any \(q(x)\in X\) is given by  
\[Bw(x) = \frac{d^{2}}{dx^{2}}\bigg{(}\frac{d^{2}w(x)}{dx^{2}}+\frac{\xi_{ 2}}{\xi_{1}}\int_{0}^{L}k(x,t)\frac{d^{2}w(t)}{dt^{2}}dt\bigg{)},\]9 \[\mathcal{D}(B) = \Big{\{}w(x)\in C^{4}[0,L]:w(0)=w^{\prime}(0)=0,\]0  
see, for example, in [40]. The solution procedure for the FIDBVP in (25) is determined by the type of the kernel \(k(x,t)\).