## 3 Theoretical Analysis and Proof
### Robustness Analysis  
When the machine is working, the noise from the surroundings or the machine itself has a great impact on the data collection. Based on this, a noise perturbation item can be defined as \(\psi(t)\). Thus, EAEND model (10) with a noise disturbance item is rewritten as  
\[ \begin{split}\mathbf{x}_{k+1}=& 1.5\mathbf{x}_{k}-\mathbf{x}_{k-1}+0.5 \mathbf{x}_{k-2}+N_{k}^{-1}\Big{(}(\mathbf{g}_{k}-\mathbf{g}_{k-1})-\\ &(N_{k}-N_{k-1})\mathbf{x}_{k}-\mathrm{Y}(N_{k}\mathbf{x}_{k}- \mathbf{g}_{k})-\Omega\sum_{i=0}^{k}(N_{i}\mathbf{x}_{i}-\mathbf{g}_{i})+\psi( t)\Big{)}.\end{split}  \tag{19}\]  
Then, in order to illustrate the robustness of EAEND model (19) in the aforementioned circumstance, the following theorem is presented. Theorem 3.1: _Assuming that EAEND model (19) aiding CEM method (1) for Arctic sea ice identification is affected by the constant noise \(\psi(t)=\mu\), the steady-state computing error of EAEND model (19) is \(O(\sigma^{3})\), regardless of the magnitude of \(\psi(t)=\mu\)._Proof. : EAEND model (10) is converted into a linear system (14), and thus the steady-state computing error of the model is the superposition effects of the constant noise \(\psi(t)=\mu\) and \(O(\sigma^{3})\). Thus, from (14), the \(\theta\)th subsystem with the constant noise term can be regarded as  
\[\varepsilon_{k+1}^{\theta}=(1.5-\mathrm{Y})\varepsilon_{k}^{\theta}- \varepsilon_{k-1}^{\theta}+0.5\varepsilon_{k-2}^{\theta}-\Gamma\sum_{i=0}^{k} \varepsilon_{i}+\mu^{\theta}. \tag{20}\]  
The aforementioned equation can be converted via the Z-transform into the following form:  
\[ \begin{split}& z\varepsilon^{\theta}(z)-z\varepsilon^{\theta}(0) \\ &=(1.5-\mathrm{Y})\varepsilon^{\theta}(z)-z^{-1}(\varepsilon^{ \theta}(z)+z\varepsilon^{\theta}(-1))+0.5z^{-2}(\varepsilon^{\theta}(z)+\\ & z^{2}\varepsilon^{\theta}(-2)+z\varepsilon^{\theta}(-1))- \Gamma\frac{\varepsilon^{\theta}(z)}{1-z^{-1}}+\frac{z\mu^{\theta}}{z-1}, \end{split}  \tag{21}\]  
where \(\varepsilon^{\theta}(0)\) denotes the initial value of \(\varepsilon^{\theta}\). The above equation can be further expressed as  
\[ \begin{split}&(z\varepsilon^{\theta}(0)-(0.5z^{-1}-1) \varepsilon^{\theta}(-1)+0.5\varepsilon^{\theta}(-2)+\frac{z\mu^{\theta}}{z-1 })\varepsilon^{\theta}(z)\\ &=z-1.5+\mathrm{Y}+z^{-1}-0.5z^{-2}+\frac{z\Gamma}{z-1}.\end{split}  \tag{22}\]  
Therefore, the expression about \(\varepsilon^{\theta}(z)\) is  
\[ \begin{split}&\varepsilon^{\theta}(z)=\\ &\frac{z^{3}(z-1)(\varepsilon^{\theta}(0)+z^{-2}(0.5-z) \varepsilon^{\theta}(-1))}{z^{4}+(2.5+\mathrm{Y}+\Gamma)z^{3}+(2.5-\mathrm{Y} )z^{2}-1.5z+0.5}+\\ &\frac{z^{3}(z-1)(0.5z^{-1}\varepsilon^{\theta}(-2)+(z-1)^{-1} \mu^{\theta})}{z^{4}+(2.5+\mathrm{Y}+\Gamma)z^{3}+(2.5-\mathrm{Y})z^{2}-1.5z+0.5}.\end{split}  \tag{23}\]  
Evidently, two of the four roots have real parts whose absolute value is less than 1, both within the unit circle. That is, model (20) is stable. Based on the final value theorem, we have  
\[ \begin{split}&\lim_{k\to\infty}\varepsilon_{k}^{\theta}\\ &=\lim_{z\to 1}(z-1)\varepsilon^{\theta}(z)\\ &=\lim_{z\to 1}\frac{z^{3}(z-1)^{2}(\varepsilon^{\theta}(0)+z^{-2}(0.5-z) \varepsilon^{\theta}(-1))}{z^{4}+(2.5+\mathrm{Y}+\Gamma)z^{3}+(2.5-\mathrm{Y} )z^{2}-1.5z+0.5}+\\ &\lim_{z\to 1}\frac{z^{3}(z-1)^{2}(0.5z^{-1}\varepsilon^{\theta}(-2)+(z-1 )^{-1}\mu^{\theta})}{z^{4}+(2.5+\mathrm{Y}+\Gamma)z^{3}+(2.5-\mathrm{Y})z^{2} -1.5z+0.5}\\ &=0.\end{split}  \tag{24}\]  
Thus, the steady-state computing error for subsystem \(\theta\) can be further obtained as \(\lim_{k\to\infty}\varepsilon_{k}=0\). Since the truncation error of (15) is \(O(\sigma^{3})\), the steady-state error \(\lim_{k\to\infty}\lVert\varepsilon_{k}\rVert_{2}\) of EAEND model (19) is also \(O(\sigma^{3})\), regardless of the variation of the constant noise \(\psi(t)=\mu\). The proof is complete. In addition, considering the effect of the random noise, Theorem 4.1 is given. **Theorem 4.1**.