Stability Analysis
I28) and (29)0)ividing the whole equation by \(e^{\text{iIϕ}}\)
\begin{equation}
\frac{\phi^{k+1}-\phi^{k+1}e^{-I\phi}}{h}=\frac{A}{2}(\phi^{k}+\phi^{k+1}e^{-I\phi})\nonumber \\
\end{equation}\begin{equation}
\left(1.D-e^{-I\phi}1.D-\frac{h}{2}A\right)\phi^{k+1}=\frac{h}{2}A\phi^{k}\nonumber \\
\end{equation}
Where 1.D is an identity matrix of order 5 by 5.
\(G=\frac{\phi^{k+1}}{\phi^{k}}=\left(1.D-e^{-I\phi}1.D-\frac{h}{2}A\right)^{-1}(\frac{h}{2}A)\)
Then the above equation can be expressed by
\(G=\left(1-e^{-I\phi}-\frac{h}{2}\lambda_{\max}\right)^{-1}(\frac{h}{2}\lambda_{\max})\)
Where \(\lambda_{\max}\)is the maximum Eigenvalue of A. By
employing stability condition \(\left|G\right|\leq 1,\) the resulting
inequalities are given by
\(\left|\frac{h}{2}{(1-e^{-I\phi}-\frac{h}{2}\lambda_{\max})}^{-1}\lambda_{\max}\right|\leq 1\)
\(\frac{h}{2}\left|\lambda_{\max}\right|\leq\left|1-e^{-I\phi}-\frac{h}{2}\lambda_{\max}\right|\leq 1+1+\frac{h}{2}\left|\lambda_{\max}\right|\)
\(\frac{h}{2}\left|\lambda_{\max}\right|\leq 2+\frac{h}{2}\left|\lambda_{\max}\right|\)
\(0\leq 2\)
Which is always right. This the Adams-Moulton second-order method using
Gauss-Seidel iterative method is unconditionally stable for the
considered equations,
For the case of Adams-Moulton’s third-order using Gauss-Seidel iterative
process, applying the third-order Adams-Moulton’s method on the equation
is given by
\(\frac{X_{i}-X_{i-1}}{h}=\frac{A}{12}(5X_{i}+8X_{i-1}-X_{i-2})\)(31)
By applying the Gauss-Seidel iterative procedure on equation (31) the
resulting equation is
\(\frac{X_{i}^{k+1}-X_{i-1}^{k+1}}{h}=\frac{A}{12}(5X_{i}^{k}+8X_{i-1}^{k+1}-X_{i-2}^{k+1})\)(32)
By employing the Van-Neuman stability criteria and using
\(X_{i}^{k+1}=\phi^{k+1}e^{\text{iIϕ}},\ X_{i}^{k}=\phi^{k}e^{\text{iIϕ}},\ X_{i-1}^{k+1}=\phi^{k+1}e^{(i-1)I\phi},\ \ X_{i-2}^{k+1}=\phi^{k+1}e^{(i-2)I\phi}\)in to equation (32) yields
\(\frac{\phi^{k+1}e^{\text{iIϕ}}-\phi^{k+1}e^{(i-1)I\phi}}{h}=\frac{A}{12}(5\phi^{k}e^{\text{iIϕ}}+8\phi^{k+1}e^{\left(i-1\right)\text{Iϕ}}-\phi^{k+1}e^{(i-2)I\phi})\)(33)
Dividing the whole equation by \(e^{\text{iIϕ}}\) on both side of the
equation (33) yields
\begin{equation}
\frac{\phi^{k+1}-\phi^{k+1}e^{-I\phi}}{h}=\frac{A}{12}(5\phi^{k}+{8\phi}^{k+1}e^{-I\phi}-\phi^{k+1}e^{-2I\phi})\nonumber \\
\end{equation}\begin{equation}
\left(1.D-e^{-I\phi}1.D-\frac{2h}{3}Ae^{-I\phi}+\frac{h}{12}Ae^{-2I\phi}\right)\phi^{k+1}=\frac{5h}{12}A\phi^{k}\nonumber \\
\end{equation}
Where 1.D is an identity matrix of order 5 by 5.
\(G=\frac{\phi^{k+1}}{\phi^{k}}=\left(1.D-e^{-I\phi}1.D-\frac{2h}{3}e^{-I\phi}A+\frac{h}{12}e^{-2I\phi}A\right)^{-1}(\frac{5h}{12}A)\)
By applying the stability condition, inequality can be expressed as
\begin{equation}
\left|{(1-e^{-I\phi}-\frac{2}{3}he^{-I\phi}\lambda_{\max}+\frac{h}{12}e^{-2I\phi}\lambda_{\max})}^{-1}\frac{5h}{12}\lambda_{\max}\right|\leq 1\nonumber \\
\end{equation}
\(\frac{5h}{2}\left|\lambda_{\max}\right|\leq\left|1-e^{-I\phi}-\frac{2}{3}he^{-I\phi}\lambda_{max+}\frac{h}{12}e^{-2I\phi}\lambda_{max+}\right|\leq 1+1+\frac{2}{3}h\left|\lambda_{\max}\right|+\frac{h}{12}\left|\lambda_{\max}\right|\)
\(\frac{h}{12}(5-8-1)\left|\lambda_{\max}\right|\leq 2\)
\(\left|\lambda_{\max}\right|\geq-\frac{6}{h}\)