# 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)
For a kernel function of the type (15), the operator \(B_{2}\) in (22) by removing the modulus in the integrand can be written equivalently as the Volterra-Fredholm integro-differential operator  
\[\mathcal{D}(B) = \Big{\{}w(x)\in C^{4}[0,L]:w(0)=w^{\prime}(0)=0,\]1 \[\mathcal{D}(B) = \Big{\{}w(x)\in C^{4}[0,L]:w(0)=w^{\prime}(0)=0,\]2  
see [39] for details, and as a result, the FIDBVP in (25) degenerates to the Volterra-Fredholm integro-differential boundary value problem  
\[\mathcal{D}(B) = \Big{\{}w(x)\in C^{4}[0,L]:w(0)=w^{\prime}(0)=0,\]3After substituting (26) into (28), the exact solution of VFIDBVP can be obtained by applying Theorem 1. Comparing (28) with (9), we take \(n=2\), \(m=1\),  
\[\mathcal{D}(B) = \Big{\{}w(x)\in C^{4}[0,L]:w(0)=w^{\prime}(0)=0,\]4 \[\Phi(w) = \left( \begin{array}{c}\Phi_{1}(w)\\ \Phi_{2}(w)\end{array} \right)=\left( \begin{array}{c}w(0)\\ w^{\prime}(0)\end{array} \right),\] \[Kw(x) = -\frac{\xi_{2}}{\tau\xi_{1}}\int_{0}^{x}\sinh\biggl{(}\frac{x-t} {\tau}\biggr{)}\frac{d^{2}w(t)}{dt^{2}}dt,\] \[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\] \[\Psi(Aw) = \biggl{(}\int_{0}^{L}e^{-\frac{t}{\tau}}\frac{d^{2}w(t)}{dt^{2}} dt\biggr{)},\] \[f(x) = \frac{1}{EI_{0}^{2}}\biggl{[}\int_{0}^{x}(x-t)q(t)dt-x\int_{0}^{ L}q(t)dt+\int_{0}^{L}tq(t)dt\biggr{]}.\]  
In addition, we have  
\[Dz(x) = (A+K)z(x)=\frac{d^{2}z(x)}{dx^{2}}-\frac{\xi_{2}}{\tau\xi_{1}} \int_{0}^{x}\sinh\biggl{(}\frac{x-t}{\tau}\biggr{)}\frac{d^{2}z(t)}{dt^{2}}dt,\] \[\mathcal{D}(D) = \mathcal{D}(A)=\{z(x)\in C^{2}[0,L]:\Phi(z)=\mathbf{0}\}.\]  
First, we find the inverse operator \(D^{-1}\) by solving the boundary value problem \(Dz(x)=f(x)\) via the Laplace transform method. By applying the Laplace transform operator on both sides, using the convolution theorem and utilizing the boundary conditions \(z(x)=z^{\prime}(x)=0\), we get  
\[\mathcal{L}\{Dz(x)\} = \mathcal{L}\biggl{\{}\frac{d^{2}z(x)}{dx^{2}}\biggr{\}}-\frac{ \xi_{2}}{\tau\xi_{1}}\mathcal{L}\Bigl{\{}\sinh\Bigl{(}\frac{x}{\tau}\Bigr{)} \Bigr{\}}\mathcal{L}\biggl{\{}\frac{d^{2}z(x)}{dx^{2}}\biggr{\}}\] \[= \biggl{[}1-\frac{\xi_{2}}{\tau\xi_{1}}\biggl{(}\frac{\frac{1}{ \tau}}{s^{2}-\frac{1}{\tau^{2}}}\biggr{)}\biggr{]}s^{2}Z(s)=F(s),\]  
from where it follows that  
\[Z(s)=F(s)Q(s), \tag{29}\]  
where  
\[Z(s)=\mathcal{L}\{z(x)\},\quad F(s)=\mathcal{L}\{f(x)\},\quad Q(s)=\frac{\xi_ {1}(\tau^{2}s^{2}-1)}{s^{2}(\xi_{1}\tau^{2}s^{2}-1)}.\]  
Taking the inverse Laplace transform of (29), we obtain  
\[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]0  
Since Equation (30) holds for every \(f(x)\in X\), it is implied that the operator \(D\) is bijective. Next, we compute  
\[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]1  
where \(G(s)=\mathcal{L}\{g(x)\}\), and subsequently  
\[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]2 \[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]3If  
\[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]4  
then from Theorem 1, it follows that the operator \(B_{2}\) is bijective and the problem (28) admits a unique solution. To find the solution, we further compute  
\[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]5 \[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]6  
Finally, by substituting (30)-(32) and \(W^{-1}\) into (13), we obtain the exact solution in the closed form of VFIDBV in (28), viz. \[g(x) = \left(-\frac{\xi_{2}}{2\tau\xi_{1}}e^{\frac{x}{\tau}}\right),\]7  
This is the solution of the boundary value problem (23) and so the solution to the nonlocal Euler-Bernulli Equation (16) subject to the boundary conditions (20).