## 5 The Error Bound
 \tag{61}\]  
If we take \(\lambda_{1}^{*}=\max_{k\leq r\leq n}\left\{\left(A_{r,n}^{\sigma}\right)^{2}\right\}\), and \(\lambda_{2}^{*}=\max_{m\leq r\leq n}\left\{\left(A_{r,n}^{\sigma}\right)^{2}\right\}\), then we obtain the following:  
\[ \begin{split}\frac{\sum_{r=k}^{n}\frac{\left(A_{r,n}^{\sigma} \right)^{2}\Gamma^{2}(r+1)\,\mathcal{L}^{2r-2k}\,\Gamma\left(2\left(r-k \right)+\frac{1}{\sigma+2}\right)\,\Gamma\left(\frac{\sigma+1}{\sigma+2} \right)}{\Gamma^{2}(r-k+1)\,\Gamma(1-2+2\,r)}}{\Gamma^{2}(r-m+1)\,\Gamma(1-2 +2\,r)}\lesssim\frac{n^{2\,k}\,\mathcal{L}^{2n-2k}(2\left(n-k\right))^{-\left( \frac{\sigma+1}{\sigma+2}\right)}\left(n-k+1\right)\lambda_{1}^{*}}{n^{2\,m} \,\mathcal{L}^{2n-2m}(2\left(n-m\right))^{-\left(\frac{\sigma+1}{\sigma+2} \right)}\left(n-m+1\right)\lambda_{2}^{*}}\\ \lesssim\frac{n^{2\,k}\,\mathcal{L}^{2n-2k}(2\left(n-k\right))^{- \left(\frac{\sigma+1}{\sigma+2}\right)}\Gamma\left(n-k+2\right)\Gamma\left(n-m +1\right)}{n^{2\,m}\,\mathcal{L}^{2n-2m}(2\left(n-m\right))^{-\left(\frac{ \sigma+1}{\sigma+2}\right)}\Gamma\left(n-m+2\right)\Gamma\left(n-k+1\right)}\\ \lesssim n^{2\,\left(k-m\right)}\,\mathcal{L}^{2\,\left(m-k\right) }\left(\frac{n-k}{n-m}\right)^{\frac{1}{\sigma+2}}.\end{split}  \tag{62}\]Again, using the Stirling formula, that is,  
\[\left(\frac{n-k}{n-m}\right)^{\frac{1}{\sigma+2}}\lesssim n^{(m-k)\left(\frac{1}{ \sigma+2}\right)}, \tag{63}\]  
therefore, (62) gives the following:  
\[\frac{\sum_{r=k}^{n}\frac{\left(A_{r,n}^{\sigma}\right)^{2}\Gamma^{2}(r+1) \mathscr{L}^{2r-2k}\Gamma\left(2\left(r-k\right)+\frac{1}{\sigma+2}\right) \Gamma\left(\frac{\sigma+1}{\sigma+2}\right)}{\Gamma^{2}(r-k+1)\Gamma\left(1-2 k+2\tau\right)}\lesssim n^{(k-m)\left(\frac{2\tau+3}{\sigma+2}\right)}\mathscr{L}^{2 \left(m-k\right)}. \tag{64}\]  
Inserting (64) into (60) yields the following:  
\[||D_{z}^{k}\left(\mathcal{U}(z)-\widetilde{\mathcal{U}}(z)\right)||_{L_{2(z) }^{2}}^{2} \lesssim\mathscr{L}^{2\left(m-k\right)}\left(M+1\right)^{(k-m) \left(\frac{2\tau+3}{\sigma+2}\right)}\sum_{n=m}^{\infty}|u_{n}|^{2} \tag{65}\] \[\times\sum_{r=m}^{n}\frac{\left(A_{r,n}^{\sigma}\right)^{2}\Gamma ^{2}(r+1)\mathscr{L}^{2r-2m}\Gamma\left(2\left(r-m\right)+\frac{1}{\sigma+2} \right)\Gamma\left(\frac{\sigma+1}{\sigma+2}\right)}{\Gamma^{2}(r-m+1)\, \Gamma(1-2\,m+2\,r)}\] \[=\mathscr{L}^{2\left(m-k\right)}\left(M+1\right)^{(k-m)\left( \frac{2\tau+3}{\sigma+2}\right)}||D_{z}^{m}\,u(z)||_{L_{2(z)}^{2}}^{2}\] \[=\mathscr{L}^{2\left(m-k\right)}\left(\frac{\Gamma\left(M+2 \right)}{\Gamma\left(M+1\right)}\right)^{(k-m)\left(\frac{2\tau+3}{\sigma+2} \right)}||D_{z}^{m}\,u(z)||_{L_{2(z)}^{2}}^{2}\] \[\lesssim\mathscr{L}^{2\left(m-k\right)}\,M^{(k-m)\left(\frac{2 \tau+3}{\sigma+2}\right)}||D_{z}^{m}\,u(z)||_{L_{2(z)}^{2}}^{2}.\]  
Therefore, the result is achieved. Theorem 3.: _Given the following assumption: \(0\leq p\leq r\leq M+1\), and the approximation to \(\mathcal{U}(z,\tau)\in\mathbf{H}_{\omega(z,\tau)}^{r,s}(\Omega)\) is \(\widetilde{\mathcal{U}}(z,\tau)=\sum_{i=0}^{M}\sum_{j=0}^{N}u_{ij}G_{i}^{( \sigma)}(z)G_{j}^{(\tau)}(\tau)\). As a result, the following estimation is satisfied:_  
\[\left|\left|\frac{\partial^{p}}{\partial z^{p}}\left(\mathcal{U}(z,\tau)- \widetilde{\mathcal{U}}(z,\tau)\right)\right|\right|_{L_{\omega(z,\tau)}^{2} }\lesssim\mathscr{L}^{r-p}\,M^{\frac{1}{2}(r-p)\left(\frac{2\tau+3}{\sigma+2} \right)}\left|\mathcal{U}(z,\tau)\right|_{\mathbf{H}_{\omega(z,\tau)}^{r,0}}. \tag{66}\]  
Proof. : According to the definitions of \(\mathcal{U}(z,\tau)\) and \(\widetilde{\mathcal{U}}(z,\tau)\), one has the following:  
\[ \begin{split}\mathcal{U}(z,\tau)-\widetilde{\mathcal{U}}(z,\tau)& =\sum_{i=0}^{M}\sum_{j=N+1}^{\infty}u_{ij}G_{i}^{(\sigma)}(z)G_{j}^ {(\sigma)}(\tau)+\sum_{i=M+1}^{\infty}\sum_{j=0}^{\infty}u_{ij}G_{i}^{(\sigma) }(z)G_{j}^{(\sigma)}(\tau)\\ &\lesssim\sum_{i=0}^{M}\sum_{j=0}^{\infty}u_{ij}G_{i}^{(\sigma)}( z)G_{j}^{(\sigma)}(\tau)+\sum_{i=M+1}^{\infty}\sum_{j=0}^{\infty}u_{ij}G_{i}^{( \sigma)}(z)G_{j}^{(\sigma)}(\tau).\end{split}