## 2 Exact Differential Equation of the Deflection of an Elastic Beam
To make the equation more solvable with a simple form, a differentiation of Equation (1) with respect to \(x\) will change it to  
\[\frac{w^{\prime\prime\prime}(x)}{\sqrt{\left[1+w^{\prime 2}(x)\right]^{3}}}-3 \frac{w^{\prime\prime}2(x)w^{\prime}(x)}{\sqrt{\left[1+w^{\prime 2}(x) \right]^{5}}}=-\frac{p}{EI}. \tag{3}\]  
Figure 1: An elastic cantilever beam with essential properties and deformation variables under a point load. With a substitution of  
\[\frac{dw(x)}{dx}=g(x), \tag{4}\]  
Equation (3) will be changed to a new form as  
\[\frac{g^{\prime\prime}(x)}{\sqrt{\left[1+g^{2}(x)\right]^{3}}}-3\frac{g^{\prime 2 }(x)g(x)}{\sqrt{\left[1+g^{2}(x)\right]^{5}}}=-\frac{P}{EI}. \tag{5}\]  
By further introducing a dimensionless variable \(\xi=x/l\), Equation (5) will be changed to  
\[\frac{g^{\prime\prime}(\xi)}{\sqrt{\left[1+g^{2}(\xi)\right]^{3}}}-3\frac{g^{ \prime 2}(\xi)g(\xi)}{\sqrt{\left[1+g^{2}(\xi)\right]^{5}}}+\alpha=0, \tag{6}\]  
where \(\alpha=\frac{P^{2}}{EI}\) is a parameter used in the differential equation of an elastica with the primary variable of the rotation of the cross-section [46]. Now let the solution of Equation (6) is a trigonometric series in the form of  
\[g(\xi)=\sum_{n=1}^{\infty}A_{n}\sin n\xi, \tag{7}\]  
satisfying boundary conditions \(g(0)=0\) and \(g^{\prime}(1)=0\). Apparently, by letting the series as  
\[g(\xi)=\sum_{n=1}^{N}A_{n}\sin\frac{(2n-1)\pi}{2}\xi,n=1,\;2,\;3,\cdots \tag{8}\]  
it will be a possible combination for the optimal solution with fewer terms. With the solution assumption of Equation (8), applying the [PERSON] method to Equation (6) as  
\[\int_{0}^{1}\Biggl{\{}\frac{g^{\prime\prime}(\xi)}{\sqrt{\left[1+g^{2}(\xi) \right]^{3}}}-3\frac{g^{\prime 2}(\xi)g(\xi)}{\sqrt{\left[1+g^{2}(\xi)\right]^{5}}}+ \alpha\Biggr{\}}\delta gdz=0 \tag{9}\]  
with the approximate expansion of \(\frac{1}{\sqrt{\left[1+g^{2}(\xi)\right]^{3}}}\approx 1,\frac{1}{\sqrt{\left[1+g^{2}( \xi)\right]^{3}}}\approx 1-\frac{5}{2}g^{2}(\xi)+\frac{35}{8}g^{4}(\xi)\) and Equation (8), Equation (9) will be written as  
\[\begin{split}&\int_{0  
equation for the coefficient from Equation (8), it is easily obtained that the coefficient \(A_{1}\) is the solution from the nonlinear equation  
\[\frac{2\,\alpha}{\pi}-\frac{A_{1}\,\left(512+384\,\,A_{1}^{2}-480\,\,A_{1}^{4}+5 25\,\,A_{1}^{6}\right)\pi}{4096}=0. \tag{11}\]  
The equation above has no analytical solution with parameter \(\alpha\), and the implicit solution can be obtained with the aid of a Chebyshev series using symbolic mathematical software such as Matlab(r) or Mathematica(r),  
\[ \begin{array}{c}A_{n}(\alpha)=\sum\limits_{k=0}^{\infty}C_{k}T_{k}\big{(} \frac{\pi}{5}-1\big{)},\\ C_{k}=\frac{2}{5\pi}\int_{0}^{10}A_{n}(\alpha)^{\frac{T_{k}\big{(}\frac{\pi}{5 }-1\big{)}}{\sqrt{1-\big{(}\frac{\pi}{5}-1\big{)}^{2}}}}d\alpha,\,\,(k=1,\,2, \,3,\,\cdots,\infty),\\ C_{0}=\frac{1}{5\pi}\int_{0}^{10}A_{0}(\alpha)^{\frac{T_{0}\big{(}\frac{\pi}{ 5}-1\big{)}}{\sqrt{1-\big{(}\frac{\pi}{5}-1\big{)}^{2}}}}d\alpha.\end{array}  \tag{12}\]  
By evaluating with symbolic software tools, it is easily obtained that the coefficient as a polynomial  
\[ \begin{array}{rl}A_{1}&=-0.00045679290254130\\ &+0.5457933917923579\alpha-0.14674177515953207\alpha^{2}\\ &+0.13351098076023404\alpha^{3}-0.11194284067204267\alpha^{4}\\ &+0.051839004053003124\alpha^{5}-0.01485367508499742\alpha^{6}\\ &+0.0028596124004239727\alpha^{7}-0.0003868993474699787\alpha^{8}\\ &+0.00003761922105859693\alpha^{9}-0.000002645845880748856\alpha^{10}\\ &+1.335599198474874\times 10^{-7}\alpha^{11}-4.719951997798395\times 10^{-9} \alpha^{12}\\ &+1.108925046190397\times 10^{-10}\alpha^{13}-1.556059378808374\times 10^{-12} \alpha^{14}\\ &+9.868406118277864\times 10^{-15}\alpha^{15}.\end{array}  \tag{13}\]  
Then, the rotation angle of the beam is obtained and should be convergent for a reasonable parameter \(\alpha\).