**5. Examples**

In this section, we find the solution of partial convolution Volterra fuzzy integro-differential equation using DFST.

**Example 1.** *Consider the following PVFIDE*

$$\begin{aligned} &w\_{xx}^{\prime\prime}(\mathbf{x},y) \oplus w\_{yy}^{\prime\prime}(\mathbf{x},y) \oplus w(\mathbf{x},y) = \operatorname{g}(\mathbf{x},y) \oplus (FR) \stackrel{\scriptstyle\mathcal{Y}}{\underset{\scriptstyle\mathbf{0}}{\rightarrow}} \begin{smallmatrix} \stackrel{\scriptstyle\mathcal{Y}}{\underset{\scriptstyle\mathbf{0}}{\rightarrow}} \operatorname{\mathcal{F}} & \stackrel{\scriptstyle\mathcal{X}}{\underset{\scriptstyle\mathbf{0}}{\rightarrow}} \operatorname{\mathcal{F}} & \stackrel{\scriptstyle\mathcal{X}}{\underset{\scriptstyle\mathbf{0}}{\rightarrow}} \operatorname{\mathcal{F}}(\mathbf{a},\boldsymbol{\beta}) \operatorname{dad} \boldsymbol{\beta}, \end{smallmatrix} \\ & \begin{array}{ll} & (\mathbf{x},y) \in [0,1] \times [0,1] & r \in [0,1] \end{array} \end{aligned}$$

*with initial conditions*

$$w(\mathbf{x},0) = (\varepsilon^x(2+r), \varepsilon^x(4-r)), \; w\_y'(\mathbf{x},0) = (\varepsilon^x(2+r), \varepsilon^x(4-r)),$$

$$w(0,y) = (\varepsilon^y(2+r), \varepsilon^y(4-r)), \; w\_x'(0,y) = (\varepsilon^y(2+r), \varepsilon^y(4-r))$$

*and*

$$g(x,y) = (e^{x+y}(2+xy)(2+r), e^{x+y}(2+xy)(4-r)).$$

*In this case m* = *n* = 2*, a*1 = *b*1 = 0*, a*2 = *b*2 = 1*, c* = 1*, k*(*x* − *α*, *y* − *β*) = *ex*−*α*+*y*−*β* > 0 *for* 0 ≤ *α* ≤ *x* ≤ 1 *and* 0 ≤ *β* ≤ *y* ≤ 1*,*

$$\psi\_0(\mathbf{x},0) = (\varepsilon^\mathbf{x}(2+r), \varepsilon^\mathbf{x}(4-r)), \quad \varrho\_0(\mathbf{y}) = (\varepsilon^\mathbf{y}(2+r), \varepsilon^\mathbf{y}(4-r)).$$

$$\psi\_1(\mathbf{x},0) = (\varepsilon^\mathbf{x}(2+r), \varepsilon^\mathbf{x}(4-r)), \quad \varrho\_1(\mathbf{y}) = (\varepsilon^\mathbf{y}(2+r), \varepsilon^\mathbf{y}(4-r)).$$

*From* (10)*, we find*

$$s[k(\mathfrak{x}, y)] = s[e^{\mathfrak{x} + y}] = \frac{1}{(1 - \mathfrak{u})(1 - \mathfrak{v})} \check{\mathsf{x}}$$

$$S[\psi\_0(\mathbf{x})] = S[\psi\_1(\mathbf{x})] = (\mathbf{s}[e^{\mathbf{x}}(2+r)], \mathbf{s}[e^{\mathbf{x}}(4-r)]) = \left(\frac{1}{1-u}(2+r), \frac{1}{1-u}(4-r)\right),$$

$$S[\varphi\_0(\mathbf{y})] = S[\varphi\_1(\mathbf{y})] = (\mathbf{s}[e^{\mathbf{y}}(2+r)], \mathbf{s}[e^{\mathbf{y}}(4-r)]) = \left(\frac{1}{1-v}(2+r), \frac{1}{1-v}(4-r)\right)$$

*From Theorem* 3 *we obtain*

$$S[\lg(x,y)] = (s[(2+xy)e^{x+y}(2+r), s[(2+xy)e^{x+y}(4-r)]))$$

$$\mathcal{I} = \left( \left( 3 - \frac{uv}{(1-u)(1-v)} \right) \frac{(2+r)}{(1-u)(1-v)}, \left( 3 - \frac{uv}{(1-u)(1-v)} \right) \frac{(4-r)}{(1-u)(1-v)} \right).$$
 
$$\mathcal{I} = \left( \left( 3 - \frac{uv}{(1-u)(1-v)} \right), \left( 3 - \frac{uv}{(1-u)(1-v)} \right) \frac{(4-r)}{(1-u)(1-v)} \right).$$

*Then, of* (22) *and* (23) *for the solution of the equation we have*

$$s[\underline{w}(x,y,r)] = \frac{1}{(1-u)(1-v)}(2+r).$$

*and*

$$s\left[\underline{w}(x,y,r)\right] = \frac{1}{(1-u)(1-v)}(4-r).$$

*By inverse double Sumudu transform the solution of the equation is <sup>w</sup>*(*<sup>x</sup>*, *y*)=(*ex*+*y*(<sup>2</sup> + *<sup>r</sup>*),*ex*+*y*(<sup>4</sup> − *<sup>r</sup>*)).
