# New Fuzzy Numerical Methods for Solving Cauchy Problems
## 2 Basic Concepts
### Numeric Scheme II: Modified 2-Step Adams Moulton Method Based on FT and NIM for Cauchy Problem  
The Scheme I uses 1-step method for solving Cauchy problem (15). In this subsection, we improve 2-step Adams Moulton method using FT and NIM for solving Cauchy problem (15). The 2-step Adams Moulton method can be improved to effectively approximate the solution of (15) by the FT components, \(Y_{k}\), \(k=2,\ldots,n-1\), of the unknown function \(y(t)\) with respect to choose some power of  
where  
\[F_{k}^{e} = \frac{\int_{a}^{b}f(t,y_{k})B_{k}(t)dt}{\int_{a}^{b}B_{k}(t)dt},\ \ \ F_{k+1}^{e*}\ \ =\ \ \ \frac{\int_{a}^{b}f(t,y_{k+1}^{*})B_{k+1}(t)dt}{\int_{a}^{b}B_{k+1}(t)dt},\ \ \ F_{k+1}^{e**}\ \ =\ \ \ \frac{\int_{a}^{b}f(t,y_{k+1}^{**})B_{k+1}(t)dt}{\int_{a}^{b}B_{k+1}(t)dt},\]  
and the truncation error \(T_{k}\) of the Scheme I is given by  
\[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]  
Rearranging (16), we get  
\[0=\frac{Y_{k+1}-Y_{k}}{h}-\frac{1}{2}\left(F_{k}+F_{k+1}^{**}\right). \tag{24}\]  
If we denote the error \(e_{k+1}=Y_{k+1}-y_{k+1}\) and subtracting (24) from (23), so:  
\[T_{k}h=e_{k+1}-e_{k}-\frac{h}{2}\left(F_{k}-F_{k}^{e}\right)-\frac{h}{2}\left( F_{k+1}^{**}-F_{k+1}^{e**}\right). \tag{25}\]  
**Lemma 8**. : _Let \(f\) is assumed to be sufficiently smooth function of its arguments on \([a,b]\) and satisfies the Lipschitz condition with the constant \(L\) with respect to \(y\), then we get for \(k=1,\ldots,n\),_  
\[\left|e_{k+1}\right|\leq \left|e_{k}\right|(1+c)+Th\ \ \ \text{ and }\ \ \left|F_{k}^{e}-F_{k+1}^{e**}\right|\leq LhM_{2}\]  
_where \(c=hL+\frac{h^{2}L^{2}}{2}+\frac{h^{2}L^{3}}{8}\), \(T=\max\limits_{1\leq k\leq n}\left|T_{k}\right|\), \(M_{2}\) is upper bound for \(f\), and \(F_{k}^{e}\), \(F_{k+1}^{e**}\) are determined by Formula (22)._  
Proof. : By hypothesis, \(f\) satisfies the Lipschitz condition and using Lemma 5, Formulas (16), (17), (21) and (22), we get  
\[\left|F_{k}-F_{k}^{e}\right| \leq\frac{1}{h}\left|\int_{a}^{b}f(t,Y_{k})B_{k}(t)dt-\int_{a}^{b }f(t,y_{k})B_{k}(t)dt\right|\leq L\left|e_{k}\right|\] \[\left|F_{k+1}^{*}-F_{k+1}^{e**}\right| \leq\frac{1}{h}\left|\int_{a}^{b}f(t,Y_{k+1}^{*})B_{k+1}(t)dt-\int _{a}^{b}f(t,y_{k+1}^{*})B_{k+1}(t)dt\right|\leq L\left|Y_{k+1}^{*}-y_{k+1}^{*}\right|\] \[\left|F_{k+1}^{**}-F_{k+1}^{e**}\right| \leq\frac{1}{h}\left|\int_{a}^{b}f(t,Y_{k+1}^{**})B_{k+1}(t)dt-\int _{a}^{b}f(t,y_{k+1}^{**})B_{k+1}(t)dt\right|\leq L\left|Y_{k+1}^{**}-y_{k+1}^{ **}\right|\] \[\left|Y_{k+1}^{*}-y_{k+1}^{*}\right| \leq\left|(Y_{k}+hF_{k}/2)-(y_{k}+hF_{k}^{e}/2)\right|\leq\left| e_{k}\right|\left(1+\frac{hL}{2}\right)\] \[\left|Y_{k+1}^{**}-y_{k+1}^{**}\right| \leq\left|\left(Y_{k+1}^{*}+hF_{k+1}^{*}/2\right)-\left(y_{k+1}^{ *}+hF_{k+1}^{e*}/2\right)\right|\leq\left|e_{k}\right|\left(1+\frac{hL}{2} \right)^{2}\] \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]0 \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]1 \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]2Furthermore, by using \(\left|f(t,y(t))\right|\leq M_{2}\), we get  
\[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]3 \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]4 \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]5 \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]6 \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]7 \[T_{k}=\frac{y_{k+1}-y_{k}}{h}-\frac{1}{2}\left(F_{k}^{e}+F_{k+1}^{e**}\right). \tag{23}\]8  
This completes the proof. **Theorem 2**. : _Consider the the numeric Scheme I (16), where \(f\in C^{2}\left[a,b\right]\) and satisfies the Lipschitz condition with the constant \(L\) with respect to \(y\).