Generalization of Fuzzy SHA-Transform with Medical Application

Definition 2.1 [7]

Parametrically speaking, a fuzzy number is a set of functions that meet the following criteria:

1. $\underline{\omega}(\vartheta)$ defines a left-continuous function in (0,1] continuing to the right and within bounds, and not diminishing 0.

2. $\bar{\omega}(\vartheta)$ defines a left-continuous function in (0,1] right continuous at infinity and bounded non-increasing 0.

3. $\underline\omega(\vartheta) \leq \bar{\omega}(\vartheta)$,0≤ϑ≤1.

For $\omega=\underline{\omega}(\vartheta), \bar{\omega}(\vartheta)$ and $v=\underline{v}(\vartheta), \bar{v}(\vartheta)$, the sum is defined by us $\omega \oplus v$ subtraction $\omega \ominus v$ and scalar multiplication by φ>0 as follows:

(a) Addition: $\omega \oplus v=\underline\omega(\vartheta)+\underline{v}(\vartheta), \bar{\omega}(\vartheta)+\bar{v}(\vartheta)$.

(b) Subtraction: $\omega \ominus v=\underline{\omega}(\vartheta)-\bar{v}(\vartheta), \bar{\omega}(\vartheta)-\underline{v}(\vartheta)$.

(c) Scalar multiplication: $\varphi \odot \omega=\left\{\begin{array}{ll}(\varphi \underline{\omega}, \varphi \bar{\omega}) & \varphi \geq 0 \\ (\varphi \bar{\omega}, \varphi \underline{\omega}) & \varphi<0\end{array}\right\}$.

Definition 2.2 [8]

Assume that $\omega, v \in \mathrm{E}$ (E: The set of all fuzzy numbers). If there exists $\rho \in \mathrm{E}$ such that ω+ν=ρ then ρ is called the Hukuhara difference of ω and νand it is identified by $\omega \ominus v$.

Definition 2.3 [9]

Let $\varphi(\varpi):(a, b) \rightarrow E$; be continuous fuzzy-valued function, φ is strongly generalized difference at $\varpi_0 \in(a, b)$. If there was an element $\varphi^{\prime}\left(\varpi_0\right) \in E$ then:

1. For all $\hbar>0$ small enough

$\exists \varphi\left(\varpi_0+\hbar\right) \ominus \varphi\left(\varpi_0\right), \exists \varphi\left(\varpi_0\right) \ominus \varphi\left(\varpi_0-\hbar\right)$ and the limit is $\varphi^{\prime}\left(\varpi_0\right)=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0+\hbar\right) \ominus \varphi\left(\varpi_0\right)}{\hbar}=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0\right) \ominus \varphi\left(\varpi_0-\hbar\right)}{\hbar}$.

Or

2. For all $\hbar>0$ small enough

$\exists \varphi\left(\varpi_0\right) \ominus \varphi\left(\varpi_0+\hbar\right), \exists \varphi\left(\varpi_0-\hbar\right) \ominus \varphi\left(\varpi_0\right)$ and the limit is $\varphi^{\prime}\left(\varpi_0\right)=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0\right) \ominus \varphi\left(\varpi_0+\hbar\right)}{-\hbar}=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0-\hbar\right) \ominus \varphi\left(\varpi_0\right)}{-\hbar}$.

Or

3. For all $\hbar>0$ small enough

$\exists \varphi\left(\varpi_0+\hbar\right) \ominus \varphi\left(\varpi_0\right), \exists \varphi\left(\varpi_0-\hbar\right) \ominus \varphi\left(\varpi_0\right)$ and the limit is $\varphi^{\prime}\left(\varpi_0\right)=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0+\hbar\right) \ominus \varphi\left(\varpi_0\right)}{\hbar}=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0-\hbar\right) \ominus \varphi\left(\varpi_0\right)}{-\hbar}$

Or

For all $\hbar>0$ small enough $\exists \varphi\left(\varpi_0\right) \ominus \varphi\left(\varpi_0+\hbar\right), \exists \varphi\left(\varpi_0-\hbar\right) \ominus \varphi\left(\varpi_0\right)$ and the limit is $\varphi^{\prime}\left(\varpi_0\right)=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0\right) \ominus \varphi\left(\varpi_0+\hbar\right)}{-h}=\lim _{\hbar \rightarrow 0^{+}} \frac{\varphi\left(\varpi_0-\hbar\right) \ominus \varphi\left(\varpi_0\right)}{\hbar}$.

Theorem 2.4 [10]

Assume that $\varphi: R \rightarrow E$ (E: the set of all fuzzy numbers) be a function and indicate $\varphi(\varpi)=(\underline{\varphi}(\varpi ; \vartheta), \bar{\varphi}(\varpi ; \vartheta))$ for each $\vartheta \in[0,1]$. Then:

1. If φ is the 1^st form, then $\underline\varphi(\varpi; \vartheta)$ and $\bar{\varphi}(\varpi ; \vartheta)$ can be differentiated and $\varphi^{\prime}(\varpi)=\underline{\varphi}(\varpi; \vartheta), \bar{\varphi}(\varpi; \vartheta)$.

2. If φ is the 2^nd form, then $\underline\varphi(\varpi; \vartheta)$ and $\bar{\varphi}(\varpi ; \vartheta)$ can be differentiated and $\varphi^{\prime}(\varpi)=\bar{\varphi}(\varpi;\vartheta), \underline{\varphi}(\varpi; \vartheta)$.

Theorem 2.5 [9]

Let $\varphi(\varpi ; \vartheta): R \rightarrow E$  (E: the set of all fuzzy numbers) and it is represented by $[\underline{\varphi}(\varpi ; \vartheta), \bar{\varphi}(\varpi ; \vartheta)]$. For any fixed $\vartheta \in(0,1]$ assume that $\underline\varphi(\varpi ; \vartheta)$ and $\bar{\varphi}(\varpi ; \vartheta)$ “ are Riemann-integrable functions on [a, b] for every b≥a, there are two positive functions $\underline{M}_{\vartheta}$ and $\bar{M}_{\vartheta}$ such that $\int_a^b|\underline{f}(\varpi ; \vartheta)| d \varpi \leq \underline M_{\vartheta}$ and $\int_a^b|\bar{\varphi}(\varpi ; \vartheta)| d \varpi \leq \overline{M_{\vartheta}}$. Then, $f(\varpi)$ is an improper fuzzy Riemann-integrable function on [a,∞). Furthermore, we have:

$\int_a^{\infty} \varphi(\varpi) d \varpi=\left[\int_a^{\infty} \underline{\varphi}(\varpi ; \vartheta) d \varpi, \int_a^{\infty} \bar{\varphi}(\varpi ; \vartheta) d \varpi\right]$

Definition 2.6 [5]

Let $\varphi(\sigma)$ exist as a continuous fuzzy-valued function, let's say that $\varepsilon \int_0^{\infty} e^{-\left(\mathrm{i} \sqrt[2 n]{\varepsilon}+\varepsilon\right) \sigma} \varphi(\sigma) d \sigma$ is an improper fuzzy Riemannintegrable on $[0, \infty)$, then $\varepsilon \int_0^{\infty} e^{-\left(i \sqrt[2 n]{\varepsilon}+\varepsilon\right) \sigma} \varphi(\sigma) d \sigma$ is called $\widehat{S H A}$-transform and is denoted as: ${S H A}[\varphi(\sigma)]={S H A}(\varepsilon)=\varepsilon \int_0^{\infty} e^{-\left(i \sqrt[2 n]{\varepsilon}+\varepsilon\right) \sigma} \varphi(\sigma) d \sigma \mathrm{n} \geq 1$.

From Theorem 2:

$\varepsilon \int_0^{\infty} e^{-\left(i^2 \sqrt[n]{\varepsilon}+\varepsilon\right) \sigma} \varphi(\sigma) d \sigma=\varepsilon \int_0^{\infty} e^{-\left(i^2 \sqrt[n]{\varepsilon}+\varepsilon\right) \sigma} \underline{\varphi}(\sigma ; \phi) d \sigma$,

$\varepsilon \int_0^{\infty} e^{-\left(\mathrm{i}^2 \sqrt[n]{\varepsilon}+\varepsilon\right) \sigma} \bar{\varphi}(\sigma ; \phi) d \sigma$

Also by the definition of classic SHA-transform:

${S H A}[\underline{\varphi}(\sigma ; \phi)]=\varepsilon \int_0^{\infty} e^{-\left(i \sqrt[2n]{\varepsilon}+\varepsilon\right) \sigma} \underline{\varphi}(\sigma ; \phi) d \sigma$,

${S H A}[\bar{\varphi}(\sigma ; \phi)]=\varepsilon \int_0^{\infty} e^{-\left(i\sqrt[2n]{\varepsilon} \sqrt[n]{\varepsilon}+\varepsilon\right) \sigma} \bar{\varphi}(\sigma ; \phi) d \sigma$

So:

${S H A}[\varphi(\sigma ; \phi)]=S H A[\underline{\varphi}(\sigma ; \phi)], \operatorname{SHA}[\bar{\varphi}(\sigma ; \phi)]$.

Theorem 2.7 [8]

Duality Between Fuzzy Laplace –$\widehat{S H A}$ transforms. If F(p) is fuzzy Laplace transform of φ(σ) and SHA(ε) is $\widehat{S H A}$ -transform of φ(σ) then ${S H A}(\varepsilon)=\varepsilon F(\sqrt[2 n]{\varepsilon}+\varepsilon)$.

Theorem 2.8

Let $\mathfrak{J}(\delta)$ by fuzzy function $\delta \geq 0, \eta(\varepsilon)=\varepsilon, \varepsilon \neq 0$ be positive real function and $\beta(\varepsilon)=i \sqrt[2 n]{\varepsilon}+\varepsilon, \varepsilon \neq 0$ be positive complex function then the derivatives of $\mathfrak{I}(\delta)$ for $\mathrm{n}^{\text {th }}$ - order will be as following:

1. $\operatorname{SHA}\{\delta \mathfrak{\Im}(\delta)\}=-\frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{\operatorname{SHA}(\mathfrak{\Im}(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}$

2. $\operatorname{SHA}\left\{\delta^2 \Im(\delta)\right\}=(-1)^2 \frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{\operatorname{SHA}(\Im(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}\right)^{\prime}$

3. $S H A\left\{\delta^n \mathfrak{I}(\delta)\right\}=(-1)^n \frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}$

$\left.\left(\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\cdots\left(\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{\operatorname{SHA}(\mathfrak{I}(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}\right)^{\prime}\right)^{\prime}\right)^{\prime}\right)^{\prime} \ldots\right)^{\prime}$

Proof:

1. Since

$\operatorname{SHA}\{\mathfrak{J}(\delta), \varepsilon\}=\varepsilon \int_0^{\infty} \underline{\mathfrak{J}}(\delta ; \varepsilon) e^{-\left(i^2 \sqrt[n]{\varepsilon}-\varepsilon\right) \delta} d \delta, \varepsilon \int_0^{\infty} \overline{{\mathfrak{J}}}(\delta ; \varepsilon) e^{-\left(i{ \sqrt[2 n]{\varepsilon}}+\varepsilon\right) \delta} d \delta$

$\Rightarrow \frac{\operatorname{SHA}\{\mathfrak{J}(\delta), \varepsilon\}}{\varepsilon}=\int_0^{\infty} \underline{\mathfrak{J}}(\delta ; \varepsilon) e^{-\left(i{\sqrt[2 n]{\varepsilon}}+\varepsilon\right) \delta} d \delta, \varepsilon \int_0^{\infty} {\overline{\mathfrak{J}}}(\delta ; \varepsilon) e^{-\left(i\sqrt[2 n]{\varepsilon}+\varepsilon\right) \delta} d \delta$

Derivative above equation with respect ε, to get:

$\left(\frac{S H A\{\mathfrak{J}(\delta), \varepsilon\}}{\varepsilon}\right)^{\prime}=\frac{d}{d \varepsilon}\left[\begin{array}{l}\int_0^{\infty} \underline{\mathfrak{J}}(\delta ; \varepsilon) e^{-\left(i^{2 \sqrt[n]{\varepsilon}+\varepsilon) \delta}\right.} d \delta, \\ \int_0^{\infty} \overline{\mathfrak{J}}(\delta ; \varepsilon) e^{-\left(i^{2 \sqrt [n]{\varepsilon}}+\varepsilon\right) \delta} d \delta\end{array}\right]$

$\left(\frac{S H A\{\mathfrak{J}(\delta), \varepsilon\}}{\varepsilon}\right)^{\prime}=-\left(i^{2} \sqrt[n]{\varepsilon}+\varepsilon\right) \int_0^{\infty} \underline{\mathfrak{J}}(\delta ; \varepsilon) e^{-\left(i^2 \sqrt[n]{\varepsilon}+\varepsilon\right) \delta} d \delta$,

$-\left(i^{2} \sqrt[n]{\varepsilon}+\varepsilon\right) \int_0^{\infty} \overline{\mathfrak{I}}(\delta ; \varepsilon) e^{-\left(i^2 \sqrt[n]{\varepsilon}+\varepsilon\right) \delta} d \delta$          (1)

From Eq. (1), to get:

$\left(\frac{S H A\{\mathfrak{I}(\delta), \varepsilon\}}{\varepsilon}\right)^{\prime}=-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \frac{S H A\{\underline{\mathfrak{I}}(\delta ; \varepsilon), \varepsilon\}}{\varepsilon}$,

$-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \frac{S H A\{\overline{\mathfrak{J}}(\delta ; \varepsilon), \varepsilon\}}{\varepsilon}$

$\left(\frac{\operatorname{SHA}(\mathfrak{\Im}(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}=-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \frac{\operatorname{SHA}\{\delta \mathfrak{\Im}(\delta), \varepsilon\}}{\varepsilon}$

Then, to get:

$\operatorname{SHA}\{\delta \mathfrak{I}(\delta)\}=-\frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{\operatorname{SHA}(\mathfrak{\Im}(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}$

2. Since from the first part, we have:

$\operatorname{SHA}\{\delta \mathfrak{J}(\delta), \varepsilon\}=-\frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{\operatorname{SHA}(\mathfrak{J}(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}$

Taking the derivative for both sides of the above equation:

$-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \frac{1}{\varepsilon} \int_0^{\infty} \delta^2 \underline{\mathfrak{J}}(\delta ; \varepsilon) e^{-\left(i \sqrt[2n]{\varepsilon}+\varepsilon\right) \delta} d \delta$,

$-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \int_0^{\infty} \delta^2 \bar{\Im}(\delta ; \varepsilon) e^{-(i 2 \sqrt[2 n]{\varepsilon}+\varepsilon) \delta} d \delta=\left(-\frac{\varepsilon}{(i \sqrt[2 n]{\varepsilon}+\varepsilon)}\left(\frac{\operatorname{SHA}\{\Im(\delta), \varepsilon\}}{\varepsilon}\right)^{\prime}\right)^{\prime}$

$-\frac{(i \sqrt[2 n]{\varepsilon}+\varepsilon)}{\varepsilon} S H A\left\{\delta^2 \underline{\mathfrak{J}}(\delta ; \vartheta)\right\}$,

$\frac{(i \sqrt[2 n]{\varepsilon}+\varepsilon)}{\varepsilon} S H A\left\{\delta^2 \bar{\mathfrak{J}}(\delta ; \vartheta)\right\}=\left(-\frac{\varepsilon}{(i \sqrt[2 n]{\varepsilon}+\varepsilon)}\left(\frac{S H A\{{\mathfrak{J}}(\delta), \varepsilon\}}{\varepsilon}\right)^{\prime}\right)^{\prime}$

Thus:

$S H A\left\{\delta^2 \mathfrak{I}(\delta)\right\}=(-1)^2 \frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(-\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{S H A(\Im(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}\right)^{\prime}$.

3. In similar way, we can prove the third part

$\operatorname{SHA}\left\{\delta^2 {\Im}(\delta)\right\}=(-1)^2 \frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(-\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{\operatorname{SHA}({{\Im}}(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}\right)^{\prime}$

derivative both side of above equation (n-2)-Times, we get:

$S H A\left\{\delta^n \mathfrak{I}(\delta)\right\}=(-1)^n \frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon}$

$\left(\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\ldots\left(\frac{1}{i \sqrt[2 n]{\varepsilon}+\varepsilon}\left(\frac{\operatorname{SHA}(\Im(\delta), \varepsilon)}{\varepsilon}\right)^{\prime}\right)^{\prime}\right)^{\prime}\right)^{\prime} \ldots\right)$.

Theorem 2.9

Let $\eta(\varepsilon)=\varepsilon$ and $\beta(\varepsilon)=i \sqrt[2 n ]{\varepsilon+\varepsilon}$ are differentiable functions such that $\Im(\delta)$ be fuzzy function, then:

$\operatorname{SHA}\left\{\delta \mathfrak{J}^{(n)}(\delta)\right\}=-\frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon} \frac{d}{d \varepsilon}\left(\frac{\operatorname{SHA}\left(\mathfrak{J}^{(n)}(\delta)\right)}{\varepsilon}\right)$

Proof:

Since

$S H A\left\{\mathfrak{J}^{(n)}(\delta), \varepsilon\right\}=\varepsilon \int_0^{\infty} \underline{\mathfrak{I}}^{(n)}(\delta ; \vartheta) e^{-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \delta} d \delta, \varepsilon \int_0^{\infty} \overline{\mathfrak{J}}^{(n)}(\delta ; \vartheta) e^{-\left(i^{2 n} \sqrt{\varepsilon}+\varepsilon\right) \delta} d \delta$

$\frac{\operatorname{SHA}\left\{\mathfrak{J}^{(n)}(\delta), \varepsilon\right\}}{\varepsilon}=\int_0^{\infty} \mathfrak{J}^{(n)}(\delta ; \vartheta) e^{-\left(i^2 \sqrt{\varepsilon}+\varepsilon\right) \delta} d \delta, \int_0^{\infty} \overline{\mathfrak{J}}^{(n)}(\delta ; \vartheta) e^{-\left(i^{ \sqrt[2n]{\varepsilon}}+\varepsilon\right) \delta} d \delta$

By derivative above equation respect to ε, then:

$\frac{d}{d \varepsilon}\left[\frac{S H A\left\{\mathfrak{J}^{(n)}(\delta), \varepsilon\right\}}{\varepsilon}\right]=\left[\int_0^{\infty} \underline{\mathfrak{I}}^{(n)}(\delta ; \vartheta) e^{-\left(i^\sqrt[2n]{\varepsilon}+\varepsilon\right) \delta} d \delta, \int_0^{\infty} \overline{\mathfrak{J}}^{(n)}(\delta ; \vartheta) e^{-\left(i^{2 n} \sqrt{\varepsilon}+\varepsilon\right) \delta} d \delta\right]$

$\frac{d}{d \varepsilon}\left[\frac{\operatorname{SHA}\left\{\mathfrak{J}^{(n)}(\delta), \varepsilon\right\}}{\varepsilon}\right]=-\left(i \sqrt[2n]{\varepsilon}+\varepsilon\right) \int_0^{\infty} \underline{\mathfrak{J}}^{(n)}(\delta ; \vartheta) e^{-\left(i \sqrt[2n]{\varepsilon}+\varepsilon\right) \delta} d \delta$

$-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \int_0^{\infty} \overline{\mathfrak{J}}^{(n)}(\delta ; \vartheta) e^{-\left(i \sqrt[2n]{\varepsilon}+\varepsilon\right) \delta} d \delta$          (2)

From Eq. (2):

$\frac{d}{d \varepsilon}\left[\frac{\operatorname{SHA}\left\{\mathfrak{J}^{(n)}(\delta), \varepsilon\right\}}{\varepsilon}\right]=-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \frac{\operatorname{SHA}\left\{\delta \underline {\mathfrak{I}}^{(n)}(\delta ; \vartheta), \varepsilon\right\}}{\varepsilon}$,

$-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \frac{S H A\left\{\delta \overline{\mathfrak{J}}^{(n)}(\delta ; \vartheta), \varepsilon\right\}}{\varepsilon}$

$\frac{d}{d \varepsilon}\left[\frac{S H A\left\{\mathfrak{J}^{(n)}(\delta), \varepsilon\right\}}{\varepsilon}\right]=-(i \sqrt[2 n]{\varepsilon}+\varepsilon) \frac{S H A\left\{\delta \mathfrak{J}^{(n)}(\delta ; \vartheta), \varepsilon\right\}}{\varepsilon}$

Then: $S H A\left\{\delta \mathfrak{J}^{(n)}(\delta)\right\}=-\frac{\varepsilon}{i \sqrt[2 n]{\varepsilon}+\varepsilon} \frac{d}{d \varepsilon}\left(\frac{S H A\left(\mathfrak{J}^{(n)}(\delta)\right)}{\varepsilon}\right)$

Theorem 2.10

Assume that $\varphi^\backslash(\sigma)$ be continuous fuzzy-valued function and $\varphi(\sigma)$ the primitive of $\varphi^\backslash(\sigma)$ on $[0, \infty)$, we have:

1. $S H A\left[\varphi^{\backslash}(\sigma)\right]=(i \sqrt[2 n]{\varepsilon}+\varepsilon) S H A[\varphi(\sigma)] \ominus \varepsilon \varphi(0)$, where, φ is the first form differentiable

2. $\operatorname{SHA}\left[\varphi^{\backslash \backslash }(\sigma)\right]=-\varepsilon \varphi(0) \ominus(-i \sqrt[2 n]{\varepsilon}+\varepsilon) \operatorname{SHA}[\varphi(\sigma)]$, where, φ is the second form differentiable.

Theorem 2.11 [6]

Assume that, $\varphi(\sigma), \varphi^{\backslash}(\sigma)$ are continuous fuzzy-valued function on $[0, \infty)$, fuzzy derivative of FSHA-Transform about second order will be as following:

1. If $\varphi, \varphi^\backslash$ are first form then:

$S H A\left[\varphi^{\backslash\backslash}(\sigma)\right]=(i \sqrt[2 n]{\varepsilon}+\varepsilon)^2 S H A[\varphi(\sigma)] \ominus \varepsilon(i \sqrt[2 n]{\varepsilon}+\varepsilon) \varphi(0) \ominus \varepsilon \varphi^{\backslash}(0)$

2. If φ first form and $\varphi^\backslash $ then the second form:

$\operatorname{SHA}\left[\varphi^{\backslash \backslash}(\sigma)\right]=-\varepsilon(i \sqrt[2 n]{\varepsilon}+\varepsilon) \varphi(0) \ominus(i \sqrt[2 n]{\varepsilon}+\varepsilon)^2 \operatorname{SHA}[\varphi(\sigma)] \ominus \varepsilon \varphi^{\backslash}(0)$

3. If φ consists of a second form and $\varphi^{\backslash}$ initial form followed by:

$S H A\left[\varphi^{\backslash \backslash \backslash}(\sigma)\right]=-\varepsilon(i \sqrt[2 n]{\varepsilon}+\varepsilon) \varphi(0) \ominus\left(-i^{2 n} \sqrt{\varepsilon}+\varepsilon\right)^2 S H A[\varphi(\sigma)]-\varepsilon \varphi^{\backslash}(0)$

4. If $\varphi, \varphi^{\backslash}$ are second form then:

$S H A\left[\varphi^{\backslash \backslash}(\sigma)\right]=(i \sqrt[2 n]{\varepsilon}+\varepsilon)^2 S H A[\varphi(\sigma)] \ominus \varepsilon(-i \sqrt[2 n]{\varepsilon}+\varepsilon) \varphi(0)-\varepsilon \varphi^{\backslash}(0)$

Theorem 2.12 [8]

Assume that $\mathfrak{I}(\delta), \mathfrak{I}^\backslash(\delta), \ldots, \mathfrak{I}^{n-1}(\delta)$ be from continuous fuzzy-valued functions on $[0, \infty)$. Let $\mathfrak{J}^{\left(i_1\right)}(\delta), \mathfrak{J}^{\left.(i_2\right)}(\delta), \ldots, \mathfrak{J}^{\left(i_m\right)}(\delta)$ are differentiable functions $2^{s t}$ from functions for $0 \leq i_1 \leq i_2 \ldots \leq i_m \leq n-1$ and $\mathfrak{J}^{(p)}$ is the $1^{s t}$ from differentiable function for p≠i_j, j=1,2,...,m, then:

1. If m is an even number, we have

$L\left(\mathfrak{J}^{(n)}(\delta)\right)=p^n L(\mathfrak{J}(\delta)) \ominus p^{n-1} \mathfrak{J}(0) \otimes \sum_{k=1}^{n-1} p^{n-(k+1)} \mathfrak{J}^{(k)}(0)$

such that

$\otimes=\left\{\begin{array}{l} \ominus, \text{In case the second order repetition count is equal to the first} \\ \text { order repetition count } 2 \text { among } i_1, \ldots, i_k, \text {, is an even number } \\ -, \text { If the frequency of occurrence of the number } 2 \text { among } i_1, \ldots, i_k \\ \text { is an odd number }\end{array}\right.$

2. If m is an odd number, we have

$L\left(\mathfrak{J}^{(n)}(\delta)\right)=-p^{n-1} \mathfrak{J}(0) \ominus\left(-p^n\right) L(\mathfrak{J}(\delta)) \otimes \sum_{k=1}^{n-1} p^{n-(k+1)} \mathfrak{J}^{(k)}(0)$

such that

$\otimes=\left\{\begin{array}{l}\ominus, \text { If the frequency of occurrence of the number } 2 \text { among } i_1, \ldots, i_k \\ \text { is an odd number } \\ -, \text { If the frequency of occurrence of the number } 2 \text { among } i_1, \ldots, i_k \\ \text { is an even number }\end{array}\right.$

Theorem 2.13

Suppose that, $\mathfrak{J}(\delta), \mathfrak{J}^\backslash(\delta)$ and $\mathfrak{J}^{\backslash \backslash}(\delta)$ are continuous fuzzy-valued function on [0,∞), fuzzy derivative of fuzzy SHA-transform about third order it will be:

1. If $\mathfrak{J}, \mathfrak{J}^\backslash, \mathfrak{J}^{\backslash\backslash}$ are first form then:

$\operatorname{SHA}\left[\Im^{\backslash \backslash \backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\Im(\delta)]$

$\ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \Im(0) \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^{\backslash}(0) \ominus \varepsilon \widetilde{\Im}^{\backslash \backslash}(0)$

2. If $\mathfrak{J}^{\backslash}, \mathfrak{J}^{\backslash\backslash}$ are first form and $\mathfrak{J}$ second form then:

$S H A\left[\mathfrak{J}^{\backslash \backslash \backslash}(\delta)\right]=\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \mathfrak{\Im}(0)(-\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[{\Im}(\delta)]$

$-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^\backslash(0)-{\varepsilon}\mathfrak{J}^{\backslash\backslash}(0)$

3. If $\mathfrak{J}, \mathfrak{J}^{\backslash\backslash}$ are first form and $\mathfrak{J}^{\backslash }$ second form then:

$S H A\left[\mathfrak{J}^{\backslash \backslash}(\delta)\right]=-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \Im(0) \ominus(-\sqrt[2 n]{\varepsilon}+\varepsilon)^3$

$\operatorname{SHA}[\mathfrak{\Im}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^{\backslash}(0) \ominus \varepsilon {\Im}^{\backslash\backslash}(0)$

4. If $\mathfrak{J}, \mathfrak{J}^{\backslash}$ are first form and $\mathfrak{J}^{{\backslash \backslash}}$ second form then:

$S H A\left[\mathfrak{J}^{\backslash \backslash\backslash}(\delta)\right]=-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \mathfrak{J}(0) \ominus(-\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\mathfrak{J}(\delta)]$

$-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^\backslash(0) \ominus {\varepsilon\mathfrak{J}}^{\backslash\backslash}(0)$

5. If $\mathfrak{J}, \mathfrak{J}^\backslash$ are second form and $\mathfrak{J}^\backslash\backslash $ first form then:

$S H A\left[\mathfrak{J}^{\backslash \backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\mathfrak{J}(\delta)] \ominus\left(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \mathfrak{J}(0)\right.$

$-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^\backslash(0)-\varepsilon{\mathfrak{J}}^{\backslash\backslash}(0)$

6. If  $\mathfrak{J}, \mathfrak{J}^{\backslash\backslash}$ are second form and ${\mathfrak{J}}^{\backslash }$ first form then:

$\operatorname{SHA}\left[{\mathfrak{J}}^{\backslash \backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[{\mathfrak{J}}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

${\mathfrak{J}}(0) \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) {\mathfrak{J}^\backslash}(0)-\varepsilon{{\mathfrak{J}}}^{\backslash \backslash}(0)$

7. If $\mathfrak{J}^{\backslash}, \mathfrak{J}^{\backslash\backslash}$ are second form and $\mathfrak{J}$ first form then:

$\operatorname{SHA}\left[{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[{\mathfrak{J}}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\mathfrak{J}(0)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) {\mathfrak{J}}^{\backslash}(0) \ominus {\varepsilon}{\mathfrak{J}}^{\backslash\backslash}(0)$

8. If $\mathfrak{J}, \mathfrak{J}^{\backslash}, \mathfrak{J}^{\backslash\backslash}$ are second form then:

$S H A\left[\mathfrak{J}^{\backslash \backslash \backslash}(\delta)\right]=-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \mathfrak{J}(0) \ominus(-\sqrt[2 n]{\varepsilon}+\varepsilon)^3$

$S H A[\mathfrak{J}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^\backslash(0)-\varepsilon\mathfrak{J}^{\backslash\backslash}(0)$

Proof: We will prove four cases:

Case 1

1. $\mathfrak{J}, \mathfrak{J}^{\backslash}, \mathfrak{J}^{\backslash\backslash}$ are first form and for any arbitral $\phi \in$ [0, 1], then:

$S H A[\mathfrak{J}$ "' $(\delta)]=S H A\left[\underline{\mathfrak{J}}^{\prime \prime \prime}(\delta, \phi)\right], \mathrm{SHA}\left[\overline{\mathfrak{J}}^{\prime \prime \prime}(\delta, \phi)\right]$

By definition of SHA derivative of second order we get:

$\operatorname{SHA}[{\underline{\mathfrak{J}}^{\backslash\backslash\backslash}}(\delta)]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\underline{\mathfrak{ J }}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline{{\mathfrak{J}}^\backslash(0, \phi)}-\varepsilon \underline{{\mathfrak { J }}^{\backslash\backslash}(0, \phi)}$           (3)

$S H A\left[\overline{\mathfrak{J}}^{\backslash \backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\overline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{\mathfrak{J}^{\backslash}(0, \phi)}-{\varepsilon}\overline{{\mathfrak{J}^{\backslash\backslash}(0, \phi)}}$          (4)

If we substitute (3) in (4) and since $\mathfrak{I}^{\prime}(\delta)$, $\mathfrak{I}^{\prime \prime}(\delta)$ are first form we get:

SHA $\left[\underline{\mathfrak{J}}^{\backslash\backslash \backslash }(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[ \underline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline{\mathfrak{I}^{\backslash }(0, \phi)}-\varepsilon \underline{\mathfrak{J}^{\backslash \backslash }(0, \phi)}$

,$(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\widetilde{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \overline{\mathfrak{J}^{\backslash\backslash}(0, \phi)}$

by Theorem (2) we get:

$S H A\left[\mathfrak{J}^{\backslash \backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[{\mathfrak{J}}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\mathfrak{J}(0) \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) {\mathfrak{J}^\backslash}(0) \ominus\varepsilon {\mathfrak{J}}^{\backslash\backslash}(0)$

2. If $\mathfrak{J}^\backslash, \mathfrak{J}^{\backslash\backslash}$ are first form and f second form then:

$\operatorname{SHA}\left[\mathfrak{J}^{\prime \prime}(\delta)\right]=S H A\left[\underline {\mathfrak{J}}^{\prime \prime \prime}(\delta, \phi)\right], \operatorname{SHA}[\overline{\mathfrak{J}}$ "' $(\delta, \phi)]$

By definition of SHA derivative of second order we get:

$S H A\left[\underline{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\underline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline{{\mathfrak{J}^{\backslash}}(0, \phi)}-\varepsilon \underline{\mathfrak{{J}}^{\backslash \backslash}(0, \phi)}$           (5)

$S H A\left[\overline{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\overline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \overline{\mathfrak{I}^{\backslash\backslash}(0, \phi)}$          (6)

If we substitute (5) in (6) and since $\mathfrak{I}^{\prime}(\delta)$, $\mathfrak{I}^{\prime \prime}(\delta)$ are first form we get:

$\operatorname{SHA}[\underline{{\mathfrak{J}}}^{\backslash\backslash\backslash}(\delta)]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\overline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{{\mathfrak{J}}^\backslash(0, \phi)}-\varepsilon \overline{{\mathfrak{J}}^{\backslash\backslash}(0, \phi)}$

,$(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\underline{\mathfrak{I}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline{{\mathfrak{J}^\backslash}(0, \phi)}-\varepsilon \underline{{\mathfrak{J}}^{\backslash\backslash}(0, \phi)}$

by Theorem (2) we get:

$\operatorname{SHA}\left[\mathfrak{J}^{\backslash\backslash\backslash}(\delta)\right]=\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \mathfrak{J}(0) \ominus(-\sqrt[2 n]{\varepsilon}+\varepsilon)^3$

$\operatorname{SHA}[\mathfrak{J}(\delta)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^{\backslash}(0)-\varepsilon{\mathfrak { J }}^{\backslash\backslash}(0)$

3. If $\mathfrak{J}, \mathfrak{J}^{\backslash}$ are second form and $\mathfrak{J}^{\backslash\backslash }$ first form then:

$S H A\left[\mathfrak{J}^{\prime \prime}(\delta)\right]=S H A\left[\underline{\mathfrak{J}}^{\prime \prime \prime}(\delta, \phi)\right], \mathrm{SHA}\left[\overline{\mathfrak{J}}^{\prime \prime \prime}(\delta, \phi)\right]$

By definition of SHA derivative of second order we get:

$S H A\left[\underline {\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\underline{\mathfrak { J }}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline {\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline {\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \underline{\mathfrak{J}^{\backslash\backslash}(0, \phi)}$

$S H A\left[\overline{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\overline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \overline{\mathfrak{J}^{\backslash\backslash}(0, \phi)}$

If we substitute (4) in (3) and since $\mathfrak{J}^{\prime}(\delta)$  second and $\mathfrak{J}^{\prime \prime}(\delta)$  first form we get:

$S H A\left[\underline{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\underline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline{\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \underline{\mathfrak{ { J }}^{\backslash\backslash}(0, \phi)}$

,$(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\overline{\mathfrak{I}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{{ J }}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \overline{\mathfrak{I}^{\backslash\backslash}(0, \phi)}$

by Theorem (2) we get:

$\operatorname{SHA}\left[\mathfrak{J}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\mathfrak{J}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\mathfrak{J}(0)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon){\mathfrak{J}}^{\backslash}(0)-\varepsilon {\mathfrak{J}}^{\backslash\backslash}(0)$

4. If $\mathfrak{I}, \mathfrak{I}^{\backslash}, \mathfrak{I}^{\backslash\backslash}$ are second form then:

$S H A\left[\mathfrak{I}^{\prime \prime}(\delta)\right]=S H A\left[\underline{\mathfrak{I}}^{\prime \prime \prime}(\delta, \phi)\right], \mathrm{SHA}\left[\overline{\mathfrak{I}}^{\prime \prime \prime}(\delta, \phi)\right]$

By definition of SHA derivative of second order we get:

$S H A\left[\underline{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\mathfrak{\underline { J }}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline{\mathfrak { J }}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline{\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \underline{{\mathfrak{J}}^{\backslash\backslash}(0, \phi)}$

$S H A\left[\overline{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\overline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{\mathfrak{J}^{\backslash}(0, \phi)}-{\varepsilon} \overline{\mathfrak{J}^{\backslash\backslash}(0, \phi)}$

If we substitute (6) in (5) and since $\mathfrak{J}^{\prime}(\delta)$, $\mathfrak{J}^{\prime \prime}(\delta)$ are first form we get:

$\operatorname{SHA}\left[\underline{\mathfrak{J}}^{\backslash\backslash\backslash}(\delta)\right]=(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 \operatorname{SHA}[\overline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\overline{\mathfrak{T}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \overline{\mathfrak{J}^{\backslash}(0, \phi)}-\varepsilon \overline{{\mathfrak { J }}^{\backslash\backslash}(0, \phi)}$

,$(\sqrt[2 n]{\varepsilon}+\varepsilon)^3 S H A[\underline{\mathfrak{J}}(\delta, \phi)]-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2$

$\underline{\mathfrak{J}}(0, \phi)-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \underline{{\mathfrak{J}^\backslash}(0, \phi)}-\varepsilon \underline{{\mathfrak { J }}^{\backslash\backslash}(0, \phi)}$

by Theorem (2) we get:

$S H A\left[\mathfrak{J}^{\backslash\backslash\backslash}(\delta)\right]=-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^2 \mathfrak{J}(0) \ominus(-\sqrt[2 n]{\varepsilon}+\varepsilon)^3$

$\operatorname{SHA}[\mathfrak{\Im}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon) \mathfrak{J}^{\backslash}(0)-\varepsilon{\mathfrak {J}}^{\backslash\backslash}(0)$

5. Fuzzy SHA-Transform for Fuzzy nth- Order Derivative

Theorem 2.14

Assume that $\mathfrak{J}(\delta), \mathfrak{J}^\backslash(\delta), \ldots, \mathfrak{J}^{n-1}(\delta), \mathfrak{J}^{(n)}(\delta)$ are continuous fuzzy-valued functions on [0,∞) and of exponential order and that $\mathfrak{J}^{(n)}(\delta)$ be pricewise continuous fuzzy-valued on [0,∞). Let $\mathfrak{J}^{\left(i_1\right)}(\delta), \mathfrak{J}^{\left.(i_2\right)}(\delta), \ldots, \mathfrak{J}^{\left(i_{\varphi}\right)}(\delta)$ are the second form differentiable functions for $0 \leq i_1 \leq i_2 \ldots \leq i_{\varphi} \leq n-1$ and $\mathfrak{J}^{(p)}$ be first form differentiable function for p≠i_j, j=1, 2,..., φ, then:

(1) If φ is an even number, we have

$M\left(\mathfrak{J}^{(n)}(\delta)\right)=(\sqrt[2 n]{\varepsilon}+\varepsilon)^n M(\mathfrak{J}(\delta)) \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)$

$\mathfrak{J}(0) \circledR \sum_{\kappa=1}^{n-1} \varepsilon^{n-\kappa+1} \mathfrak{J}^{(\kappa)}(0)$

such that

$\circledR=\left\{\begin{array}{l} \ominus,\text {If there is a detectable difference between the } \\ \text { occurrences of the second form and the first, } \\ \text { then } i_1, \ldots, i_\kappa \text { is an even number } \\ -, \text { If there is a detectable difference between the occurrences } \\ \text { of the second form and the first, then } i_1, \ldots, i_\kappa \\ \text { is an odd number } \end{array}\right.$

(2) If φ is an odd number, we have

$M\left(\mathfrak{J}^{(n)}(\delta)\right)=-\varepsilon^{n+1} \mathfrak{J}(0) \ominus\left(-\varepsilon^n\right) M(\mathfrak{J}(\delta)) \circledR \sum_{\kappa=1}^{n-1} \varepsilon^{n-\kappa+1} \mathfrak{J}^{(\kappa)}(0)$,

such that

$\circledR=\left\{\begin{array}{l} \ominus \text {  iIf there is a detectable difference between }\\ \text {  the occurrences of the second form and the first,}\\\text { then } i_1, \ldots, i_\kappa \text { is an odd number }\\\text {-, If there is a detectable difference between }\\\text { the occurrences of the second form and the first,}  \\  \text { then} i_1, \ldots, i_\kappa \text { is an even number}\end{array}\right.$

Proof: The proof depends on the duality between fuzzy Laplace –SHA-Transforms from as follows:

$\operatorname{SHA}(\varepsilon)=\operatorname{SHA}[\mathfrak{J}(\varepsilon)] F(p)=L[\mathfrak{J}(\mathfrak{\aleph})]$

$\operatorname{SHA}_n(\varepsilon)=\operatorname{SHA}\left[\mathfrak{J}^{(n)}(\delta)\right]$ and $F_n(p)=L\left[\mathfrak{J}^{(n)}(\mathfrak{\aleph})\right]$

From duality relation (1), we have

$S H A_n(\delta)=S H A\left[\mathfrak{J}^{(n)}(\delta)\right]=\varepsilon F_n(\sqrt[2 n]{\varepsilon}+\varepsilon)$

Let φ is an even number. Then from Theorem 4 when φ is an even number, Eq. (2) becomes:

$S H A_n(\delta)=\varepsilon\left[\begin{array}{l}(\sqrt[2 n]{\varepsilon}+\varepsilon)^n F(\sqrt[2 n]{\varepsilon}+\varepsilon) \ominus(\sqrt[2 n]{\varepsilon}+\varepsilon)^{(n-1)} \mathfrak{J}(0) \\ \left({ }^{\circledR} \sum_{\kappa=1}^{n-1}(\sqrt[2 n]{\varepsilon}+\varepsilon)^{n-(\kappa+1)} \mathfrak{J}^{(\kappa)}(0)\right.\end{array}\right]$

$=(\sqrt[2 n]{\varepsilon}+\varepsilon)^n[\varepsilon F(\sqrt[2 n]{\varepsilon}+\varepsilon)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^{(n-1)}$

$\mathfrak{J}(0) \circledR \varepsilon \sum_{\kappa=1}^{n-1}(\sqrt[2 n]{\varepsilon}+\varepsilon)^{n-(\kappa+1)} \mathfrak{J}^{(\kappa)}(0)$

$=(\sqrt[2 n]{\varepsilon}+\varepsilon)^n S H A[\mathfrak{J}(\delta)] \ominus \varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^{(n-1)}$

$\mathfrak{J}(0) \circledR \varepsilon \sum_{\kappa=1}^{n-1}(\sqrt[2 n]{\varepsilon}+\varepsilon)^{n-(\kappa+1)} \mathfrak{J}^{(\kappa)}(0)$

Let φ is an odd number. Then from Theorem 4, Eq. (3) becomes:

$S H A_n(\varepsilon)=\left[\begin{array}{l}-(\sqrt[2 n]{\varepsilon}+\varepsilon)^{(n-1)} \mathfrak{J}(0) \ominus\left(-(\sqrt[2 n]{\varepsilon}+\varepsilon)^n\right) F(\sqrt[2 n]{\varepsilon}+\varepsilon) \\ \ \circledR\sum_{\kappa=1}^{n-1}(\sqrt[2 n]{\varepsilon}+\varepsilon)^{n-(\kappa+1)} \mathfrak{J}^{(\kappa)}(0)\end{array}\right]$

$=-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^{(n-1)} \mathfrak{J}(0) \ominus\left(-(\sqrt[2 n]{\varepsilon}+\varepsilon)^n\right)[\varepsilon F(\sqrt[2 n]{\varepsilon}+\varepsilon)]$

$\circledR\varepsilon \sum_{\kappa=1}^{n-1}(\sqrt[2 n]{\varepsilon}+\varepsilon)^{n-(\kappa+1)} \mathfrak{J}^{(\kappa)}(0)$

$=-\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)^{(n-1)} \mathfrak{I}(0) \ominus\left(-(\sqrt[2 n]{\varepsilon}+\varepsilon)^n\right) S H A[\mathfrak{J}(\delta)]$

$\circledR\varepsilon \sum_{\kappa=1}^{n-1}(\sqrt[2 n]{\varepsilon}+\varepsilon)^{n-(\kappa+1)} \mathfrak{J}^{(\kappa)}(0)$

Example 2.15

The concentration of a drug in an organ. If blood transports a specific drug into an organ at a variable rate of rI(t) cm³/s and out of the organ at a variable rate of rO(t) cm³/s, and the organ has an initial blood volume of V cm³, the following equation can be used: For c(t)g/cm³ of blood entering the organ, compute an ODE for the amount of drug present in the organ at time t following the partnership.

$\mathfrak{I}^{\backslash}(t)=\Psi \Upsilon-\hbar \mathfrak{J}(t), \quad \mathfrak{J}(0)=(\mathfrak{J}(0, r), \mathfrak{J}(r, 0))$

such that $\Psi=1, \Upsilon=c(t)=1, \hbar=v(t)=1$

To solve the equation, we have two cases:

Case 1

By using FSHA-Transform for both sides of the original equation we get:

$(\sqrt[2 n]{\varepsilon}+\varepsilon) S H A[\mathfrak{J}(\delta)] \ominus \varepsilon \mathfrak{J}(0)=-S H A[\mathfrak{J}(\delta)]$

Eq. (4) becomes:

$(\sqrt[2 n]{\varepsilon}+\varepsilon) S H A[\underline{\mathfrak{J}}(\delta)]-\varepsilon \underline{\mathfrak{J}}(0)=-S H A[\mathfrak{J}(\delta)]$,

$(\sqrt[2 n]{\varepsilon}+\varepsilon) S H A[\overline{\mathfrak{J}}(\delta)]-\varepsilon \overline{\mathfrak{J}}(0)=-S H A[\overline{\mathfrak{J}}(\delta)]$

By solve Eq. (5) we get:

$S H A[\underline{\mathfrak{J}}(\delta ; \vartheta)]=\frac{\varepsilon}{(\sqrt[2 n]{\varepsilon}+\varepsilon)+1} \underline{\mathfrak { J }}(0 ; \delta)$,

$\operatorname{SHA}[\overline{\mathfrak{J}}(\delta ; \vartheta)]=\frac{\varepsilon}{(\sqrt[2 n]{\varepsilon}+\varepsilon)+1} \overline{\mathfrak{J}}(0 ; \delta)$

Now we use inverse fuzzy SHA-Transform for above equation:

$\underline{\mathfrak{J}}(\delta ; \vartheta)=e^{-\delta} \underline{\mathfrak{J}}(0 ; \delta), \overline{\mathfrak{I}}(\delta ; \vartheta)=e^{-\delta} \overline{\mathfrak{I}}(0 ; \delta)$

Case 2

By using fuzzy SHA-Transform for both sides of the original equation we get:

$-\varepsilon \mathfrak{I}(0) \ominus(-(\sqrt[2 n]{\varepsilon}+\varepsilon)) S H A[\mathfrak{J}(\delta)]=-S H A[\mathfrak{J}(\delta)](5)$

Eq. (4) becomes:

$(\sqrt[2 n]{\varepsilon}+\varepsilon) S H A[\overline{\mathfrak{J}}(\delta)]-\varepsilon \overline{\mathfrak{J}}(0)=-S H A[\underline{\mathfrak{J}}(\delta)]$,

$(\sqrt[2 n]{\varepsilon}+\varepsilon) S H A[\underline{\mathfrak{J}}(\delta)]-\varepsilon \underline{\mathfrak{J}}(0)=-S H A[\overline{\mathfrak{J}}(\delta)]$

By solve Eq. (5) we get:

$S H A[\underline{\mathfrak{J}}(\delta ; \vartheta)]=\frac{\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)}{(\sqrt[2 n]{\varepsilon}+\varepsilon)^2-1} \overline{\mathfrak{J}}(0 ; \delta)-\frac{\varepsilon}{(\sqrt[2 n]{\varepsilon}+\varepsilon)^2-1} \underline{\mathfrak{J}}(0 ; \delta)$

$S H A[\overline{\mathfrak{J}}(\delta ; \vartheta)]=\frac{\varepsilon(\sqrt[2 n]{\varepsilon}+\varepsilon)}{(\sqrt[2 n]{\varepsilon}+\varepsilon)^2-1} \underline{\mathfrak{J}}(0 ; \delta)-\frac{\varepsilon}{(\sqrt[2 n]{\varepsilon}+\varepsilon)^2-1} \overline{\mathfrak{J}}(0 ; \delta)$

Now we use inverse fuzzy SHA-transform for above equation:

$\underline{\mathfrak{J}}(\delta ; \vartheta)=\cosh \varpi \overline{\mathfrak{J}}(0 ; \delta)-\sinh \varpi \mathfrak{J}(0 ; \delta)$,

$\overline{\mathfrak{I}}(\delta ; \vartheta)=\cosh \varpi \underline{\mathfrak{J}}(0 ; \delta)-\sinh \varpi \overline{\mathfrak{J}}(0 ; \delta)$