## 4 Numerical Examples
### 1-D Bratu Problem  
The 1-D Bratu problem [8] is given by  
\[\frac{d^{2}U}{dx^{2}}+\lambda\exp U(x)=0,\,\lambda>0,\,0<x<1, \tag{26}\]  
with the boundary conditions \(U(0)=U(1)=0\). The 1-D planar Bratu problem has two known, bifurcated, exact solutions for values of \(\lambda<\lambda_{c}\), one solution for \(\lambda=\lambda_{c}\) and no solution for \(\lambda>\lambda_{c}\). \begin{table}
\begin{tabular}{c c c c c c c c c} \hline \hline
**Methods** & \multicolumn{2}{c}{**TP1**} & \multicolumn{2}{c}{**TP2**} & \multicolumn{2}{c}{**TP3**} \\ \hline  & \(M\) & \(err_{min}\) & \(p_{c}\) & \(M\) & \(err_{min}\) & \(p_{c}\) & \(M\) & \(err_{min}\) & \(p_{c}\) \\ \hline \(2^{nd}NM\) Equation (1) & 7 & 4.6e\(-\)114 & 2.00 & 9 & 1.7e\(-\)107 & 2.00 & 8 & 3.9e\(-\)145 & 2.02 \\ \(3^{rd}HM\) Equation (2) & 5 & 1.4e\(-\)174 & 2.99 & 6 & 4.5e\(-\)139 & 3.00 & 5 & 2.9e\(-\)291 & 4.10 \\ \(4^{th}NR\) Equation (3) & 4 & 4.6e\(-\)114 & 4.02 & 5 & 1.7e\(-\)107 & 4.00 & 5 & 2.9e\(-\)291 & 4.11 \\ \(4^{th}SGS\) Equation (4) & 4 & 7.1\(-\)108 & 3.99 & 6 & 0 & 3.99 & 5 & 8.8e\(-\)257 & 4.03 \\ \(4^{th}HM\) Equation (6) & 4 & 1.4e\(-\)105 & 3.99 & 6 & 0 & 4.00 & 5 & 5.5e\(-\)247 & 4.12 \\ \(6^{th}CHMT\) Equation (5) & 4 & 0 & 5.91 & 5 & 0 & 5.98 & 4 & 4.6e\(-\)199 & 6.12 \\ \(6^{th}HM\) Equation (7) & 4 & 0 & 5.90 & 5 & 0 & 5.98 & 4 & 6.1e\(-\)194 & 6.13 \\ \(8^{th}HM\) Equation (7) & 4 & 0 & 7.90 & 4 & 1.9e\(-\)133 & 7.99 & 4 & 0 & 8.64 \\ \(10^{th}HM\) Equation (7) & 3 & 1.1e\(-\)154 & 9.90 & 4 & 2.2e\(-\)248 & 9.99 & 4 & 0 & 10.76 \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Comparison of different methods for system of nonlinear equations. \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Methods** & **TP1** & **TP2** & **TP3** \\ \hline \(2^{nd}NM\) & 1.161405 & 1.734549 & 1.758380 \\ \(3^{rd}HM\) & 0.950678 & 2.445676 & 1.969176 \\ \(4^{th}NR\) & 0.808851 & 1.569021 & 1.452089 \\ \(4^{th}SGS\) & 1.052950 & 2.649530 & 2.571427 \\ \(4^{th}HM\) & 1.001148 & 2.170088 & 2.456138 \\ \(6^{th}CHMT\) & 1.132364 & 2.117847 & 2.405149 \\ \(6^{th}HM\) & 0.944062 & 2.137319 & 2.528262 \\ \(8^{th}HM\) & 0.986300 & 2.328460 & 2.071641 \\ \(10^{th}HM\) & 1.029707 & 2.482167 & 2.213744 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Comparison of CPU time (s). The critical value of \(\lambda_{c}\) is simply \(8(\eta^{2}-1)\), where \(\eta\) is the fixed point of the hyperbolic cotangent function \(\coth\left(x\right)\). The exact solution to Equation (26) is known and can be presented here as  
\[U(x)=-2\ln\Bigg{[}\frac{\cosh\left(x-\frac{1}{2}\right)\frac{\theta}{2}}{\cosh \left(\frac{\theta}{4}\right)}\Bigg{]}, \tag{27}\]  
where \(\theta\) is a constant to be determined, which satisfies the boundary conditions and is carefully chosen and assumed to be the solution of the differential Equation (26). Using a similar procedure as in [14], we show how to obtain the critical value of \(\lambda\). Substitute Equation (27) in Equation (26), simplify and collocate at the point \(x=\dfrac{1}{2}\) because it is the midpoint of the interval. Another point could be chosen, but low order approximations are likely to be better if the collocation points are distributed somewhat evenly throughout the region. Then, we have  
\[\theta^{2}=2\lambda\cosh^{2}\bigg{(}\frac{\theta}{4}\bigg{)}. \tag{28}\]  
Differentiating Equation (28) with respect to \(\theta\) and setting \(\dfrac{d\lambda}{d\theta}=0\), the critical value \(\lambda_{c}\) satisfies  
\[\theta=\frac{1}{2}\lambda_{c}\cosh\bigg{(}\frac{\theta}{4}\bigg{)}\sinh\bigg{(} \frac{\theta}{4}\bigg{)}. \tag{29}\]  
By eliminating \(\lambda\) from Equations (28) and (29), we have the value of \(\theta_{c}\) for the critical \(\lambda_{c}\) satisfying  
\[\frac{\theta_{c}}{4}=\coth\bigg{(}\frac{\theta_{c}}{4}\bigg{)} \tag{30}\]  
for which \(\theta_{c}=4.798714560\) can be obtained using an iterative method. We then get \(\lambda_{c}=3.513830720\) from Equation (28). Figure 1 illustrates this critical value of \(\lambda\).