In this section, we recall some basic results and notions which we consider helpful for readers to understand the present work. Proof of theorems are omitted as they can be proved by the following steps similar to the case of classical derivatives.
Theorem 3.1
If \(f \in C^{l} (\mathcalligra{R}^{+} \times \mathcalligra{R}^{+})\) and \(l=\max \{m_{1},m_{2}\}\), where \(m_{1},m_{2} \in \mathbb{Z}\). For \(i=1,2,\ldots , m_{1}\) and \(j=1,2,\ldots , m_{2}\), there exist \(k,\tau _{1},\tau _{2}>0\) such that \(\vert \frac{\partial ^{i+j}f(x,t)}{\partial x^{i}\partial t^{j}} \vert < ke^{x \tau _{1}+t\tau _{2}}\), then the “double Laplace transform” satisfies the following formulae [34]:
$$ \begin{aligned} &\mathcalligra{L}_{t}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{1}}f(x,t)}{\partial x^{m_{1}}} \biggr\} =s_{1}^{m_{1}} \mathcalligra{L}_{x}\mathcalligra{L}_{t} \bigl\{ f(x,t) \bigr\} - \sum_{i=0}^{{m_{1}}-1}s_{1}^{{m_{1}}-1-i} \mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i}f(0,t)}{\partial x^{i}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{2}}f(x,t)}{\partial t^{m_{2}}} \biggr\} =s_{2}^{m_{2}} \mathcalligra{L}_{x}\mathcalligra{L}_{t} \bigl\{ f(x,t) \bigr\} - \sum_{j=0}^{{m_{2}}-1}s_{1}^{{m_{2}}-1-j} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{j}f(0,t)}{\partial x^{j}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{1}+m_{2}}f(x,t)}{\partial t^{m_{1}}\partial x^{m_{2}}} \biggr\} \\&\quad =s_{1}^{m_{1}}s_{2}^{m_{2}} \Biggl[\mathcalligra{L}_{t}\mathcalligra{L}_{x} \bigl\{ f(x,t) \bigr\} -\sum_{j=0}^{m_{2}-1}s_{2}^{-j-1} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{j}f(x,0)}{\partial t^{j}} \biggr\} \\ &\qquad {}-\sum_{i=0}^{m_{1}-1}s_{1}^{-i-1} \mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i}f(0,t)}{\partial x^{i}} \biggr\} +\sum _{i=0}^{m_{2}-1} \sum_{j=0}^{m_{1}-1}s_{1}^{-i-1}s_{2}^{-j-1} \biggl\{ \frac{\partial ^{i+j}f(0,0)}{\partial x^{i}\partial t^{j}} \biggr\} \Biggr], \end{aligned} $$
(14)
where \(\frac{\partial ^{m_{1}+m_{2}}}{\partial t^{m_{1}}\partial x^{m_{2}}}f(x, t)\) denotes a mixed partial derivative at the point \((x, t)\).
Proof
The proof is similar to that of the Laplace transforms of the ordinary derivatives of functions of a single variable, see for more detail [35]. □
Theorem 3.2
Let \(f \in C^{l} (\mathcalligra{R}^{+} \times \mathcalligra{R}^{+}\times \mathcalligra{R}^{+})\), and \(l=\max \{m_{1},m_{2},m_{3}\}\), where \(m_{1},m_{2},m_{3} \in \mathbb{Z}\). For \(i_{1}=1,2,\ldots , m_{1}\), \(i_{2}=1,2,\ldots , m_{2}\), and \(i_{3}=1,2,\ldots , m_{3}\), there exist \(k,\tau _{1},\tau _{2},\tau _{3}>0\) such that \(\vert \frac{\partial ^{i_{1}+i_{2}+i_{3}}f(x,y,t)}{\partial x^{i_{1}}\partial y^{i_{2}}\partial t^{i_{3}}} \vert < ke^{x\tau _{1}+y\tau _{2}+t\tau _{3}}\), then the triple Laplace transform satisfies the following formulae [34]:
$$ \begin{aligned} &\mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{1}}f(x,y,t)}{\partial x^{m_{1}}} \biggr\} \\ &\quad =s_{1}^{m_{1}} \mathcalligra{L}_{x} \mathcalligra{L}_{y}\mathcalligra{L}_{t} \bigl\{ f(x,y,t) \bigr\} -\sum_{i_{1}=0}^{{m_{1}}-1}s_{1}^{{m_{1}}-1-i_{1}} \mathcalligra{L}_{t} \mathcalligra{L}_{y} \biggl\{ \frac{\partial ^{i_{1}}f(0,y,t)}{\partial x^{i_{1}}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{2}}f(x,y,t)}{\partial y^{m_{2}}} \biggr\} \\ &\quad =s_{2}^{m_{2}} \mathcalligra{L}_{x}\mathcalligra{L}_{y}\mathcalligra{L}_{t} \bigl\{ f(x,y,t) \bigr\} -\sum_{i_{2}=0}^{{m_{2}}-1}s_{2}^{{m_{2}}-1-i_{2}} \mathcalligra{L}_{t} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}}f(x,0,t)}{\partial y^{i_{2}}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{3}}f(x,y,t)}{\partial t^{m_{3}}} \biggr\} \\ &\quad =s_{3}^{m_{3}} \mathcalligra{L}_{x}\mathcalligra{L}_{y}\mathcalligra{L}_{t} \bigl\{ f(x,y,t) \bigr\} -\sum_{i_{3}=0}^{{m_{3}}-1}s_{3}^{{m_{3}}-1-i_{3}} \mathcalligra{L}_{y} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{3}}f(x,y,0)}{\partial y^{i_{3}}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{1}+m_{2}+m_{3}}f(x,y,t)}{\partial t^{m_{1}}\partial y^{m_{2}}\partial x^{m_{3}}} \biggr\} \\ &\quad =s_{1}^{m_{1}}s_{2}^{m_{2}}s_{3}^{m_{3}} [\mathcalligra{L}_{t} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ f(x,y,t) \bigr\} \\ &\qquad {}-\sum_{i_{3}=0}^{m_{3}-1}\sum _{i_{2}=0}^{m_{2}-1}s^{m_{2}-i_{2}-1}_{2}s^{m_{3}-i_{3}-1}_{3} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}+i_{3} }f(x,0,0)}{\partial t^{i_{3}}\partial y^{i_{2}}} \biggr\} \\&\qquad {} -\sum _{i_{1}=0}^{m_{1}-1} \sum _{i_{3}=0}^{m_{3}-1}s^{m_{1}-i_{1}-1}_{1}s^{m_{3}-i_{3}-1}_{3} \mathcalligra{L}_{y} \biggl\{ \frac{\partial ^{i_{1}+i_{3} }f(0,y,0)}{\partial t^{i_{3}}\partial x^{i_{1}}} \biggr\} \\ &\qquad {}-\sum_{i_{1}=0}^{m_{1}-1}\sum _{i_{2}=0}^{m_{2}-1}s^{m_{1}-i_{1}-1}_{1}s^{m_{2}-i_{2}-1}_{2} \mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i_{1}+i_{2} }f(0,0,t)}{\partial x^{i_{1}}\partial y^{i_{2}}} \biggr\} \\&\qquad {}+\sum _{i_{1}=0}^{m_{1}-1} \sum _{i_{2}=0}^{m_{2}-1}\sum_{i_{3}=0}^{m_{3}-1}s^{-m_{1}-1}_{1}s^{-m_{2}-1}_{2}s_{3}^{-m_{3}-1} \biggl\{ \frac{\partial ^{i_{1}+i_{2}+i_{3} }f(0,0,0)}{\partial x^{i_{1}}\partial y^{i_{2}}\partial t^{i_{3}}} \biggr\} \end{aligned} $$
(15)
where \(\frac{\partial ^{i_{1}+i_{2}+i_{3}}}{\partial x^{i_{1}}\partial y^{i_{2}}\partial t^{i_{3}}}f(x,y, t)\) denotes a mixed partial derivative at the point \((x, t)\).
Proof
In a similar fashion it can be proved as the Laplace transforms of the ordinary derivatives of functions of a single variable [35]. □
Theorem 3.3
Let \(f \in C^{l} (\mathcalligra{R}^{+} \times \mathcalligra{R}^{+}\times \mathcalligra{R}^{+}\times \mathcalligra{R}^{+})\) and \(l=\max \{m_{1},m_{2},m_{3}, m_{4}\}\), where \(m_{1},m_{2},m_{3},m_{4} \in \mathbb{Z}\). For \(i_{1}=1,2,\ldots , m_{1}\), \(i_{2}=1,2,\ldots , m_{2}\), \(i_{3}=1,2,\ldots , m_{3}\), and \(i_{4}=1,2,\ldots , m_{4}\), there exist \(k,\tau _{1},\tau _{2},\tau _{3},\tau _{4}>0\) such that \(\vert \frac{\partial ^{i_{1}+i_{2}+i_{3}+i_{4}}f(x,y,z,t)}{\partial x^{i_{1}}\partial y^{i_{2}}\partial z^{i_{3}}\partial t^{i_{4}}} \vert < ke^{x\tau _{1}+y\tau _{2}+z\tau _{3}+t\tau _{4}}\), then the fourth order Laplace transform satisfies the following formulae:
$$ \begin{aligned} &\mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{1}}f(x,y,z,t)}{\partial x^{m_{1}}} \biggr\} \\ &\quad =s_{1}^{m_{1}} \mathcalligra{L}_{x}\mathcalligra{L}_{y}\mathcalligra{L}_{z} \mathcalligra{L}_{t} \bigl\{ f(x,y,z,t) \bigr\} -\sum _{i_{1}=0}^{{m_{1}}-1}s_{1}^{{m_{1}}-1-i_{1}} \mathcalligra{L}_{t}\mathcalligra{L}_{z}\mathcalligra{L}_{y} \biggl\{ \frac{\partial ^{i_{1}}f(0,y,z,t)}{\partial x^{i_{1}}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{2}}f(x,y,z,t)}{\partial y^{m_{2}}} \biggr\} \\ &\quad =s_{2}^{m_{2}} \mathcalligra{L}_{x} \mathcalligra{L}_{y}\mathcalligra{L}_{z}\mathcalligra{L}_{t} \bigl\{ f(x,y,z,t) \bigr\} -\sum_{i_{2}=0}^{{m_{2}}-1}s_{2}^{{m_{2}}-1-i_{2}} \mathcalligra{L}_{t}\mathcalligra{L}_{x}\mathcalligra{L}_{z} \biggl\{ \frac{\partial ^{i_{2}}f(x,0,z,t)}{\partial y^{i_{2}}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{3}}f(x,y,z,t)}{\partial t^{m_{3}}} \biggr\} \\ &\quad =s_{3}^{m_{3}} \mathcalligra{L}_{x} \mathcalligra{L}_{y}\mathcalligra{L}_{z}\mathcalligra{L}_{t} \bigl\{ f(x,y,z,t) \bigr\} -\sum_{i_{3}=0}^{{m_{3}}-1}s_{3}^{{m_{3}}-1-i_{3}} \mathcalligra{L}_{z}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{3}}f(x,y,z,0)}{\partial y^{i_{3}}} \biggr\} , \\ &\mathcalligra{L}_{t}\mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{m_{1}+m_{2}+m_{3}}f(x,y,z,t)}{\partial t^{m_{1}}\partial y^{m_{2}}\partial x^{m_{3}}} \biggr\} \\ &\quad =s_{1}^{m_{1}}s_{2}^{m_{2}}s_{3}^{m_{3}}s_{4}^{m_{4}} [ \mathcalligra{L}_{t}\mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ f(x,y,z,t) \bigr\} \\ &\qquad {}-\sum_{i_{4}=0}^{m_{4}-1}\sum _{i_{3}=0}^{m_{3}-1}\sum_{i_{2}=0}^{m_{2}-1}s^{m_{2}-i_{2}-1}_{2}s^{m_{3}-i_{3}-1}_{3}s_{4}^{m_{4}-i_{4}-1} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}+i_{3}+i_{4} }f(x,0,0,0)}{\partial t^{i_{4}}\partial z^{i_{3}}\partial y^{i_{2}}} \biggr\} \\ &\qquad {}-\sum_{i_{1}=0}^{m_{1}-1}\sum _{i_{3}=0}^{m_{3}-1}\sum_{i_{4}=0}^{m_{4}-1}s^{m_{1}-i_{1}-1}_{1}s^{m_{3}-i_{3}-1}_{3}s_{4}^{m4-i_{4}-1} \mathcalligra{L}_{y} \biggl\{ \frac{\partial ^{i_{1}+i_{3}+i_{4} }f(0,y,0,0)}{\partial t^{i_{4}}\partial z^{i_{3}}\partial x^{i_{1}}} \biggr\} \\ &\qquad {}-\sum_{i_{1}=0}^{m_{1}-1}\sum _{i_{2}=0}^{m_{2}-1}\sum_{i_{4}=0}^{m_{4}-1}s^{m_{1}-i_{1}-1}_{1}s^{m_{2}-i_{2}-1}_{2}s_{4}^{m4-i_{4}-1} \mathcalligra{L}_{z} \biggl\{ \frac{\partial ^{i_{1}+i_{2}+i_{4} }f(0,0,z,0)}{\partial t^{i_{4}}\partial y^{i_{2}}\partial x^{i_{1}}} \biggr\} \\ &\qquad {}-\sum_{i_{1}=0}^{m_{1}-1}\sum _{i_{2}=0}^{m_{2}-1}\sum_{i_{3}=0}^{m_{3}-1}s^{m_{1}-i_{1}-1}_{1}s^{m_{2}-i_{2}-1}_{2}s_{3}^{m3-i_{3}-1} \mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i_{1}+i_{2}+i_{3} }f(0,0,0,t)}{\partial z^{i_{3}}\partial y^{i_{2}}\partial x^{i_{1}}} \biggr\} \\ &\qquad {}+\sum_{i_{1}=0}^{m_{1}-1}\sum _{i_{2}=0}^{m_{2}-1}\sum_{i_{3}=0}^{m_{3}-1} \sum_{i_{4}=0}^{m_{4}-1}s^{-m_{1}-1}_{1}s^{-m_{2}-1}_{2}s_{3}^{-m_{3}-1}s_{4}^{-m_{4}-1} \biggl\{ \frac{\partial ^{i_{1}+i_{2}+i_{3}+i_{4} }f(0,0,0,0)}{\partial x^{i_{1}}\partial y^{i_{2}}\partial z^{i_{3}}\partial t^{i_{4}}} \biggr\} , \end{aligned} $$
(16)
where \(\frac{\partial ^{i_{1}+i_{2}+i_{3}+i_{4}} }{\partial x^{i_{1}}\partial y^{i_{2}}\partial z^{i_{3}}\partial t^{i_{4}}}\) denotes a mixed partial derivative at the point \((x,y,z, t)\).
Proof
The proof is similar to that of the Laplace transforms of the ordinary derivatives of functions of a single variable, and therefore the readers are suggested to see [35] and the references therein. □
In the following theorems we define the double, triple, and fourth order Laplace transform of fractional integrals.
Theorem 3.4
Let \(\alpha , \beta \in \mathbb{C} \) such that \(\mathcalligra{R}(\alpha ), \mathcalligra{R}( \beta ) \geq 0\). Let \(a, b \in \mathcalligra{R}\) with \(a,b > 0\) and \(f \in \mathcalligra{L}_{1}[(0,a)\times (0,b)]\). Further assume that for \(x>a\), \(t>b\) and constants \(k,\ \tau _{1},\ \tau _{2} > 0\) the inequality \(\vert f(x,t) \vert \leq ke^{x\tau _{1}+t\tau _{2}}\) holds. Then the double Laplace transform of fractional integral is given by [34]
$$\begin{aligned}& \mathcalligra{L}_{t}\mathcalligra{L}_{x}\bigl\{ _{0}{I}_{x}^{\alpha }f(x,t)\bigr\} (s_{1},s_{2})= \frac{1}{s_{1}^{\alpha }} \mathcalligra{L}_{t}\mathcalligra{L}_{x}\bigl\{ f(x,t)\bigr\} (s_{1},s_{2}), \end{aligned}$$
(17)
$$\begin{aligned}& \mathcalligra{L}_{t}\mathcalligra{L}_{x}\bigl\{ _{0}{I}_{t}^{\beta }f(x,t)\bigr\} (s_{1},s_{2})= \frac{1}{s_{2}^{\beta }} \mathcalligra{L}_{t}\mathcalligra{L}_{x}\bigl\{ f(x,t)\bigr\} (s_{1},s_{2}), \end{aligned}$$
(18)
and
$$ \mathcalligra{L}_{t}\mathcalligra{L}_{x}\bigl\{ {_{0}I_{t}}^{\beta }{_{0}I_{x}}^{\alpha }f(x,t) \bigr\} (s_{1},s_{2})=\frac{1}{s_{1}^{\alpha }s_{2}^{\beta }} \mathcalligra{L}_{t} \mathcalligra{L}_{x}{f(x,t)}(s_{1},s_{2}), $$
(19)
where \(s_{1}\) and \(s_{2}\) are parameters of Laplace transforms of x and t respectively.
Proof
Formula (17) can be derived by taking the double Laplace transform of the convolution with respect to x. By taking the double Laplace transform of the convolution with respect to t, one can easily prove formula (18). For the proof of formula (19), one may consider the double Laplace transform of the double convolution. For further details on the double Laplace transforms, see [3, 36] and the references therein. □
Theorem 3.5
Let \(\alpha , \beta , \gamma \in \mathbb{C} \) such that \(\mathcalligra{R}(\alpha ), \mathcalligra{R}( \beta ), \mathcalligra{R}(\gamma ) \geq 0\). Let \(a, b, c \in \mathcalligra{R}\) with \(a,b ,c> 0\) and \(f \in \mathcalligra{L}_{1}[(0,a)\times (0,b)\times (0,c)]\). Further assume that for \(x>a\), \(y>b\), \(t>c\) and constants \(k,\ \tau _{1},\ \tau _{2},\tau _{3} > 0\) the inequality \(\vert f(x,y,t) \vert \leq ke^{x\tau _{1}+y\tau _{2}+t\tau _{3}}\) holds. Then the triple Laplace transform of fractional integrals is given by [34]
$$\begin{aligned}& \mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ _{0}{I^{\alpha }_{x}} f(x,y,t) \bigr\} (s_{1},s_{2},s_{3})= \frac{1}{s_{1}^{\alpha }}\mathcalligra{L}_{t} \mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ f(x,y,t)\bigr\} (s_{1},s_{2},s_{3}), \end{aligned}$$
(20)
$$\begin{aligned}& \mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ _{0}{I^{\beta }_{y}} f(x,y,t) \bigr\} (s_{1},s_{2},s_{3})= \frac{1}{s_{2}^{\beta }}\mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ f(x,y,t)\bigr\} (s_{1},s_{2},s_{3}), \end{aligned}$$
(21)
$$\begin{aligned}& \mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ _{0}{I}^{\gamma }_{t} f(x,y,t) \bigr\} (s_{1},s_{2},s_{3})= \frac{1}{s_{3}^{\gamma }}\mathcalligra{L}_{t} \mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ f(x,y,t)\bigr\} (s_{1},s_{2},s_{3}), \end{aligned}$$
(22)
and
$$ \mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ {_{0}I}^{\gamma }_{t}{_{0}I}^{\beta }_{t} {_{0}I}^{\alpha }_{x} f(x,y,t)\bigr\} (s_{1},s_{2},s_{3})= \frac{1}{s_{1}^{\alpha }s_{2}^{\beta }s_{3}^{\gamma }} \mathcalligra{L}_{t} \mathcalligra{L}_{y}\mathcalligra{L}_{x}{f(x,y,t)}(s_{1},s_{2},s_{3}), $$
(23)
where \(s_{1}\), \(s_{2}\), and \(s_{3}\) are the parameters of Laplace transforms of x, y, and t respectively.
Theorem 3.6
Let \(\alpha , \beta , \gamma , \sigma \in \mathbb{C} \) such that \(\mathcalligra{R}(\alpha ), \mathcalligra{R}( \beta ), \mathcalligra{R}(\gamma ), \mathcalligra{R}(\sigma ) \geq 0\). Let \(a, b, c, d \in \mathcalligra{R}\) with \(a,b ,c,d> 0\) and \(f \in L^{1}[(0,a)\times (0,b)\times (0,c)\times (0,d)]\). Further assume that for \(x>a\), \(y>b\), \(z>c\), \(t>d\) and constants \(k,\ \tau _{1},\ \tau _{2},\tau _{3},\tau _{4} > 0\) the inequality \(\vert f(x,y,z,t) \vert \leq ke^{x\tau _{1}+y\tau _{2}+z\tau _{3}+t \tau _{4}}\) holds. Then the triple Laplace transform of fractional integrals is given by
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x}\bigl\{ _{0}{I}^{\alpha }_{x} f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4})\\ \quad = \frac{1}{s_{1}^{\alpha }}\mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4}), \end{gathered} \end{aligned}$$
(24)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x}\bigl\{ _{0}{I}^{\beta }_{y} f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4})\\ \quad = \frac{1}{s_{2}^{\beta }}\mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4}), \end{gathered} \end{aligned}$$
(25)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x}\bigl\{ _{0}{I}^{\gamma }_{z} f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4})\\ \quad = \frac{1}{s_{3}^{\gamma }}\mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4}), \end{gathered} \end{aligned}$$
(26)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x}\bigl\{ _{0}{I}^{\sigma }_{t} f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4})\\ \quad = \frac{1}{s_{3}^{\sigma }}\mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y} \mathcalligra{L}_{x}\bigl\{ f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4}), \end{gathered} \end{aligned}$$
(27)
and
$$ \begin{gathered}[b] \mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ {_{0}I}^{\sigma }_{t}{_{0}I}^{\gamma }_{z}{_{y}I}^{\beta }_{0^{+}} {_{0}I}^{\alpha }_{x} f(x,y,z,t)\bigr\} (s_{1},s_{2},s_{3},s_{4}) \\ \quad =\frac{1}{s_{1}^{\alpha }s_{2}^{\beta }s_{3}^{\gamma }s_{4}^{\sigma }} \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x}{f(x,y,z,t)}(s_{1},s_{2},s_{3},s_{4}), \end{gathered} $$
(28)
where \(s_{1}\), \(s_{2}\), \(s_{3}\), and \(s_{4}\) are the parameters of Laplace transforms of x, y, z, and t respectively.
In the theorems given below, we give the double, triple, and fourth order Laplace transforms of the fractional Caputo derivatives.
Theorem 3.7
Let \(m_{1},m_{2} \in \mathbb{N}\) and \(\alpha , \beta > 0\) such that \(m_{2}-1 < \alpha \leq m_{2}\), \(m_{1}-1 < \beta \leq m_{1}\). Let us choose \(l=\max \{m_{1},m_{2}\}\) and let \(f \in C^{l}(\mathcalligra{R}^{+} \times \mathcalligra{R}^{+})\). Assume further that for \(a,b >0\) we have \(f^{(l)} \in \mathcalligra{L}_{1}[(0,a)\times (0,b)]\). Further assume that for \(x>a\), \(t>b\) and constants \(k,\ \tau _{1},\ \tau _{2} > 0\) the inequality \(\vert f(x,t) \vert \leq ke^{x\tau _{1}+t\tau _{2}}\) holds. The double Laplace transforms of the partial fractional Caputo derivatives can be defined as
$$\begin{aligned}& \mathcalligra{L}_{t}\mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\alpha }_{x} f(x,t) \bigr\} =s_{1}^{\alpha } \Biggl[\mathcalligra{L}_{t} \mathcalligra{L}_{x} \bigl\{ f(x,t) \bigr\} -\sum _{i_{1}=0}^{m_{1}-1}s_{1}^{-1-{i_{1}}} \mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i_{1}}f(0,t)}{\partial x^{i_{1}}} \biggr\} \Biggr], \end{aligned}$$
(29)
$$\begin{aligned}& \mathcalligra{L}_{t}\mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\beta }_{t} f(x,t) \bigr\} =s_{2}^{\beta } \Biggl[\mathcalligra{L}_{t} \mathcalligra{L}_{x} \bigl\{ f(x,t) \bigr\} -\sum _{i_{2}=0}^{m_{2}-1}s_{2}^{-1-i_{2}} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}}f(0,t)}{\partial t^{i_{2}}} \biggr\} \Biggr], \end{aligned}$$
(30)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t} \mathcalligra{L}_{x} \bigl\{ {_{0}{^{c}D}^{\alpha }_{x}} {_{0}{^{c}D}^{\beta }_{t}} f(x,t) \bigr\} \\ \quad =s_{1}^{\alpha }s_{2}^{\beta } \Biggl[ \mathcalligra{L}_{t}\mathcalligra{L}_{x} \bigl\{ f(x,t) \bigr\} -\sum_{i_{1}=0}^{m_{2}-1}s_{1}^{-1-{i_{1}}} \mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i_{1}}f(0,t)}{\partial x^{i_{1}}} \biggr\} \\ \qquad {}-\sum_{i_{2}=0}^{m_{1}-1}s_{2}^{-1-i_{2}} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}}f(x,0)}{\partial t^{i_{2}}} \biggr\} +\sum _{{i_{2}}=0}^{m_{1}-1} \sum _{{i_{1}}=0}^{m_{2}-1}s_{1}^{-{i_{1}}-1}s_{2}^{-{i_{2}}-1} \frac{\partial ^{{i_{1}}+{i_{2}}}f(0,0)}{\partial x^{i_{1}}\partial t^{i_{2}}} \Biggr]. \end{gathered} \end{aligned}$$
(31)
Theorem 3.8
Let \(m_{1},m_{2},m_{3} \in \mathbb{N}\) and \(\alpha , \beta ,\gamma > 0\) such that \(m_{3}-1 < \gamma \leq m_{3}\), \(m_{2}-1 < \beta \leq m_{2}\), \(m_{1}-1 < \alpha \leq m_{1}\). Let us choose \(l=\max \{m_{1},m_{2},m_{3}\}\) and let \(f\in C^{l}(\mathcalligra{R}^{+} \times \mathcalligra{R}^{+}\times \mathcalligra{R}^{+})\). Assume further that for \(a,b,c >0\) we have \(f^{(l)} \in \mathcalligra{L}_{1}[(0,a)\times (0,b)\times (0,c)]\). Further assume that for \(x>a\), \(y>b\), \(t>c\) and constants \(k,\ \tau _{1},\ \tau _{2},\tau _{3} > 0\) the inequality \(\vert f(x,y,t) \vert \leq ke^{x\tau _{1}+y\tau _{2}+t\tau _{3}}\) holds. The triple Laplace transforms of the partial fractional Caputo derivatives can be defined as [17]
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\alpha }_{x} f(x,y,t) \bigr\} \\ \quad =s_{1}^{\alpha } \Biggl[\mathcalligra{L}_{t} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ f(x,y,t) \bigr\} -\sum_{i_{1}=0}^{m_{1}-1}s_{1}^{-1-{i_{1}}} \mathcalligra{L}_{y}\mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i_{1}}f(0,y,t)}{\partial x^{i_{1}}} \biggr\} \Biggr], \end{gathered} \end{aligned}$$
(32)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\beta }_{y} f(x,y,t) \bigr\} \\ \quad =s_{2}^{\beta } \Biggl[\mathcalligra{L}_{t} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ f(x,y,t) \bigr\} -\sum_{i_{2}=0}^{m_{2}-1}s_{2}^{-1-i_{2}} \mathcalligra{L}_{t}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}}f(x,0,t)}{\partial y^{i_{2}}} \biggr\} \Biggr], \end{gathered} \end{aligned}$$
(33)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\gamma }_{t} f(x,y,t) \bigr\} \\ \quad =s_{3}^{\gamma } \Biggl[ \mathcalligra{L}_{t} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ f(x,y,t) \bigr\} -\sum_{i_{3}=0}^{m_{3}-1}s_{3}^{-1-i_{3}} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{3}}f(x,y,0)}{\partial t^{i_{2}}} \biggr\} \Biggr], \end{gathered} \end{aligned}$$
(34)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ {_{0}{^{c}D}^{\alpha }_{x}} {_{0}{^{c}D}^{\beta }_{y}} {_{0}{^{c}D}^{\gamma }_{t}} f(x,y,t) \bigr\} \\ \quad =s_{1}^{\alpha }s_{2}^{\beta }s_{3}^{\gamma } \Biggl[\mathcalligra{L}_{t} \mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ f(x,y,t) \bigr\} -\sum_{i_{1}=0}^{m_{2}-1}s_{1}^{-1-{i_{1}}} \mathcalligra{L}_{t} \mathcalligra{L}_{y} \biggl\{ \frac{\partial ^{i_{1}}f(0,y,t)}{\partial x^{i_{1}}} \biggr\} \\ \qquad {} -\sum_{i_{2}=0}^{m_{1}-1}s_{2}^{-1-i_{2}} \mathcalligra{L}_{t}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}}f(x,0,t)}{\partial y^{i_{2}}} \biggr\} -\sum_{i_{3}=0}^{m_{3}-1}s_{3}^{-1-i_{3}} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{3}}f(x,y,0)}{\partial t^{i_{3}}} \biggr\} \\ \qquad {}+ \sum_{i_{3}=0}^{m_{3}-1} \sum_{i_{2}=0}^{m_{2}-1}\sum _{{i_{1}}=0}^{m_{1}-1}s_{1}^{-{i_{1}}-1}s_{2}^{-{i_{2}}-1}s_{3}^{-{i_{3}}-1} \frac{\partial ^{{i_{1}}+{i_{2}}+{i_{3}}}f(0,0,0)}{\partial x^{i_{1}}\partial y^{i_{2}}\partial t^{i_{3}}} \Biggr]. \end{gathered} \end{aligned}$$
(35)
Theorem 3.9
Let \(m_{1},m_{2},m_{3},m_{4} \in \mathbb{N}\) and \(\alpha , \beta ,\gamma ,\sigma > 0\) such that \(m_{4}-1 < \sigma \leq m_{4}\), \(m_{3}-1 < \gamma \leq m_{3}\), \(m_{2}-1 < \beta \leq m_{2}\), \(m_{1}-1 < \alpha \leq m_{1}\). Let us choose \(l=\max \{m_{1},m_{2},m_{3},m_{4}\}\) and let \(f \in C^{l}(R^{+} \times \mathcalligra{R}^{+} \times \mathcalligra{R}^{+}\times \mathcalligra{R}^{+})\). Assume further that for \(a,b,c,d >0\) we have \(f^{(l)} \in \mathcalligra{L}_{1}[(0,a)\times (0,b)\times (0,c)\times (0,d)]\). Further assume that for \(x>a\), \(y>b\), \(z>c\), \(t>d\) and constants \(k,\ \tau _{1},\ \tau _{2},\tau _{3},\tau _{4} > 0\) the inequality \(\vert f(x,y,z,t) \vert \leq ke^{x\tau _{1}+y\tau _{2}+z\tau _{3}+t \tau _{4}}\) holds. The fourth order Laplace transforms of the partial fractional Caputo derivatives can be defined as
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\alpha }_{x} f(x,y,z,t) \bigr\} \\ \quad =s_{1}^{\alpha } \Biggl[ \mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ f(x,y,z,t) \bigr\} -\sum _{i_{1}=0}^{m_{1}-1}s_{1}^{-1-{i_{1}}} \mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{t} \biggl\{ \frac{\partial ^{i_{1}}f(0,y,z,t)}{\partial x^{i_{1}}} \biggr\} \Biggr], \end{gathered} \end{aligned}$$
(36)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\beta }_{y} f(x,y,z,t) \bigr\} \\ \quad =s_{2}^{\beta } \Biggl[ \mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ f(x,y,z,t) \bigr\} -\sum _{i_{2}=0}^{m_{2}-1}s_{2}^{-1-i_{2}} \mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}}f(x,0,z,t)}{\partial y^{i_{2}}} \biggr\} \Biggr], \end{gathered} \end{aligned}$$
(37)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\gamma }_{z} f(x,y,z,t) \bigr\} \\ \quad =s_{3}^{\gamma } \Biggl[ \mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ f(x,y,z,t) \bigr\} -\sum _{i_{3}=0}^{m_{3}-1}s_{3}^{-1-i_{3}} \mathcalligra{L}_{y} \mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{3}}f(x,y,0,t)}{\partial z^{i_{3}}} \biggr\} \Biggr], \end{gathered} \end{aligned}$$
(38)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t}\mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ _{0}{^{c}D}^{\sigma }_{t} f(x,y,z,t) \bigr\} \\ \quad =s_{4}^{\sigma } \Biggl[\mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ f(x,y,z,t) \bigr\} -\sum_{i_{4}=0}^{m_{4}-1}s_{4}^{-1-i_{4}} \mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{4}}f(x,y,z,0)}{\partial t^{i_{4}}} \biggr\} \Biggr], \end{gathered} \end{aligned}$$
(39)
$$\begin{aligned}& \begin{gathered}[b] \mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{y} \mathcalligra{L}_{x} \bigl\{ {_{0}{^{c}D}^{\alpha }_{x}} {_{0}{^{c}D}^{\beta }_{y}} {_{0}{^{c}D}^{\gamma }_{z}{_{0}{^{c}D}^{\sigma }_{t}}} f(x,y,z,t) \bigr\} \\ \quad =s_{1}^{\alpha }s_{2}^{\beta }s_{3}^{\gamma } \Biggl[\mathcalligra{L}_{t} \mathcalligra{L}_{z} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \bigl\{ f(x,y,z,t) \bigr\} \\ \qquad {}-\sum_{i_{1}=0}^{m_{2}-1}s_{1}^{-1-{i_{1}}} \mathcalligra{L}_{t} \mathcalligra{L}_{z}\mathcalligra{L}_{y} \biggl\{ \frac{\partial ^{i_{1}}f(0,y,z,t)}{\partial x^{i_{1}}} \biggr\} -\sum_{i_{2}=0}^{m_{2}-1}s_{2}^{-1-i_{2}} \mathcalligra{L}_{t}\mathcalligra{L}_{z}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{2}}f(x,0,z,t)}{\partial y^{i_{2}}} \biggr\} \\ \qquad {}-\sum_{i_{3}=0}^{m_{3}-1}s_{3}^{-1-{i_{3}}} \mathcalligra{L}_{t} \mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{3}}f(x,y,0,t)}{\partial z^{i_{3}}} \biggr\} -\sum_{i_{3}=0}^{m_{3}-1}s_{4}^{-1-i_{4}} \mathcalligra{L}_{t}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{4}}f(x,y,z,0)}{\partial t^{i_{4}}} \biggr\} \\ \qquad {}-\sum_{i_{4}=0}^{m_{4}-1}\sum _{i_{3}=0}^{m_{3}-1}s_{4}^{-1-i_{4}} \mathcalligra{L}_{z}\mathcalligra{L}_{y}\mathcalligra{L}_{x} \biggl\{ \frac{\partial ^{i_{4}}f(x,y,z,0)}{\partial t^{i_{4}}} \biggr\} \\ \qquad {}+\sum_{i_{4}=0}^{m_{4}-1}\sum _{i_{3}=0}^{m_{3}-1}\sum_{i_{2}=0}^{m_{2}-1} \sum_{{i_{1}}=0}^{m_{1}-1}s_{1}^{-{i_{1}}-1}s_{2}^{-{i_{2}}-1}s_{3}^{-{i_{3}}-1}s_{4}^{-{i_{4}}-1} \frac{\partial ^{{i_{1}}+{i_{2}}+{i_{3}}+{i_{4}}}f(0,0,0,0)}{\partial x^{i_{1}}\partial y^{i_{2}}\partial z^{i_{3}}t^{i_{4}}} \Biggr]. \end{gathered} \end{aligned}$$
(40)