# New Approximation Methods Based on Fuzzy Transform for Solving SODEs: [PERSON]*, [PERSON], [PERSON], [PERSON]
## 3 FzT for Solving SODEs
### Numerical Scheme I for SODEs
\begin{array}{ccc}2\int_{t_{1}}^{t_{n}}f(t,X_{k},Y_{k} )A_{1}(t)dt&\text{if}&k=1,\\ &\\ \int_{t_{1}}^{t_{n}}f(t,X_{k},Y_{k})A_{k}(t)dt&\text{if}&k=2,\ldots,n-1,\\ &\\ Y_{k+1}-Y_{k}=\left\{\begin{array}{ccc}2\int_{t_{1}}^{t_{n}}g(t,X_{k},Y_{k} )A_{1}(t)dt&\text{if}&k=1,\\ &\\ \int_{t_{1}}^{t_{n}}g(t,X_{k},Y_{k})A_{k}(t)dt&\text{if}&k=2,\ldots,n-1,\\ \end{array}\right.\end{array} \right. \tag{24}\]  
Proof. : The proof can be obtained from Remark 1; in particular, by the properties of the uniform fuzzy partition \(\int_{t_{1}}^{t_{n}}A_{k}(t)dt=\
icefrac{{h}}{{2}}\) for \(k=1\) and \(\int_{t_{1}}^{t_{n}}A_{k}(t)dt=h\) for \(k=2,\ldots,n-1\) and after substituting this into (14)-(15). **Lemma 3**. : _Suppose that \(f,g\) are continuous and bounded on \(I=[t_{1},t_{n}]\), and consider that Scheme I (14)-(15) with respect to the basic functions forms a uniform fuzzy partition of \(I\). Then, we have for fixed \(k=1,\ldots,n-1\):_  
\[|X_{k+1}-X_{k}|\leq M_{1}h,\ \ |Y_{k+1}-Y_{k}|\leq M_{2}h,\]  
_where \(M_{1}=\underset{t\in[t_{1},t_{n}]}{\max}\ |f(t,x,y)|\) and \(M_{2}=\underset{t\in[t_{1},t_{n}]}{\max}\ |g(t,x,y)|\)._  
Proof. : Let us choose a value of \(k\) in the range \(1\leq k\leq n-1\). By using Lemma 2, we get:  
\[|X_{2}-X_{1}| =\left|2\int_{t_{1}}^{t_{2}}f(t,X_{k},Y_{k})A_{1}(t)dt\right|,\] \[\leq 2\int_{t_{1}}^{t_{2}}|f(t,X_{k},Y_{k})A_{1}(t)|\,dt \leq 2M_{1} \int_{t_{k-1}}^{t_{k+1}}A_{1}(t)dt=M_{1}h,\] \[|X_{k+1}-X_{k}| =\left|\int_{t_{k-1}}^{t_{k+1}}f(t,X_{k},Y_{k})A_{k}(t)dt\right|,\] \[\leq\int_{t_{k-1}}^{t_{k+1}}|f(t,X_{k},Y_{k})A_{k}(t)|\,dt \leq M_{1} \int_{t_{k-1}}^{t_{k+1}}A_{k}(t)dt=M_{1}h.\]  
Similarly, \(|Y_{2}-Y_{1}|\leq M_{2}h\) and \(|Y_{k+1}-Y_{k}|\leq M_{2}h\). Throughout the assumptions of Theorem 3, we consider \((t,x_{1},x_{2})\) instead \((t,x,y)\). **Theorem 3**. : _Suppose that \(f(t,x_{1},x_{2}),\)\(g(t,x_{1},x_{2})\in C^{2}\left[t_{1},t_{2}\right]\). Let \(\left|\frac{\partial g}{\partial x_{i}}\right|\leq L_{f}\left(\left|\frac{ \partial g}{\partial x_{i}}\right|\leq L_{g}\right),\)\(i=1,2\), and \(|f(t,x_{1},x_{2})|\leq M_{1}\)\((|g(t,x_{1},x_{2})|\leq M_{2})\). Consider Scheme I (14)-(15) for some positive integer \(k\), and \(\left\{A_{k}\ \mid k=1,\ldots,n-1\right\},\)\(n\geq 2,\) is a uniform fuzzy partition of \([t_{1},t_{n}]\), then the following hold true:__for a value of \(k\) in the range \(1\leq k\leq n-1\):_ \[\left|\hat{F}_{k}-\hat{F}_{k-1}\right|\leq L_{f}hU_{k,k-1}\,,\,\,\,\,\left|\hat{G }_{k}-\hat{G}_{k-1}\right|\leq L_{g}hU_{k,k-1}\,,\] _where_ \(U_{k,k-1}=\left|X_{k}-X_{k-1}\right|+\left|Y_{k}-Y_{k-1}\right|\)_._
2. _for all_ \(k=1,\ldots,n-1\)_:_ \[\left|X_{n+1}-X_{n}\right|\leq\frac{Mh}{2}e^{n}(2 Lh^{2}),\,\,\,\,\left|Y_{n+1} -Y_{n}\right|\leq\frac{Mh}{2}e^{n}(2 Lh^{2})\,,\] _where_ \(L=L_{f}+L_{g}\) _and_ \(M=\sum_{i}^{2}M_{i}\)_._  
Proof. : 1. Using (6), we can get for each \(k=2,\ldots,n-1\) and \(t\in I\cap[t_{k},t_{k+1}]\): \[\left|\hat{F}_{k}-\hat{F}_{k-1}\right| = \left|\frac{\int_{t_{k-1}}^{t_{k+1}}f(t,X_{k},Y_{k})A_{k}(t)dt}{ \int_{t_{k-1}}^{t_{k+1}}A_{k}(t)dt}-\frac{\int_{t_{k-2}}^{t_{k}}f(t,X_{k-1},Y _{k-1})A_{k-1}(t)dt}{\int_{t_{k-2}}^{t_{k}}A_{k-1}(t)dt}\right|.\] Based on Remark 1 and Definition 1, the properties of the uniform fuzzy partition, we replace \(t\) by \(t-h\) and then \(A_{k-1}(t-h)\) by \(A_{k}(t)\). Thus, \[\left|\hat{F}_{k}-\hat{F}_{k-1}\right| = \left|\frac{\int_{t_{k-1}}^{t_{k+1}}f(t,X_{k},Y_{k})A_{k}(t)dt}{ \int_{t_{k-1}}^{t_{k+1}}A_{k}(t)dt}-\frac{\int_{t_{k-1}}^{t_{k+1}}f(t-h,X_{k-1 },Y_{k-1})A_{k}(t)dt}{\int_{t_{k-1}}^{t_{k+1}}A_{k}(t)dt}\right|,\] (25) \[\leq L_{f}\left(\left|X_{k}-X_{k-1}\right|+\left|Y_{k}-Y_{k-1}\right| \right)\int_{t_{k-1}}^{t_{k+1}}A_{k}(t)dt,\] \[|X_{2}-X_{1}| =\left|2\int_{t_{1}}^{t_{2}}f(t,X_{k},Y_{k})A_{1}(t)dt\right|,\]0 In a similar way, \(\left|\hat{G}_{k}-\hat{G}_{k-1}\right|\leq L_{g}hU_{k,k-1}\). 2. We first prove the estimate for \(k=1\).