Definition 1. Given a function f : R⁺ × R⁺ × R⁺⟶R, the conformable partial fractional derivatives (CPFDs) of orders α, β, and γ of the function f(x, y, t) are defined as follows:

where 0 < α, β, γ ≤ 1, x, y, t > 0, and $$ and $$ are called the fractional partial derivatives of orders α, β, and γ, respectively.

We prove the basic Theorem 1 and the relation between the CPFDs and partial derivatives as follows.

Theorem 1. Let α, β, γ ∈ (0,1] and f(x, y, t) be a differentiable at a point for x, y, t > 0. Then,

• (i)

  $$

• (ii)

  $$

• (iii)

  $$

Proof. By the definition of CFPD,

Similarity, we can prove the results (ii) and (iii).

In the next preposition, we mention the conformable partial fractional derivative of some functions. By using Theorem 1, it can be verified easily.

Proposition 1. Let α, β, γ ∈ (0,1] and a, b ∈ R, l, m, n ∈ N. Then, we have the following:

• (1)

  (∂^α/∂x^α)(au(x, y, t) + bv(x, y, t)) = a(∂^α/∂x^α)u(x, y, t) + b(∂^α/∂x^α)v(x, y, t)

• (2)

  (∂^α+β+γ/∂x^α∂y^β∂t^γ)(x^l y^m tⁿ) = lmn x^l−α y^m−β t^n−γ

• (3)

  $$, $$

• (4)

  (∂^α/∂x^α)(sin(x^α/α)cos(t^γ/γ)) = cos(x^α/α)cos(t^γ/γ)

• (5)

  (∂^β/∂y^β)(sin(x^α/α)cos(y^β/β)cos(t^γ/γ)) = −sin(x^α/α)sin(y^β/β)cos(t^γ/γ)

Definition 2. Let the function u : (0, ∞)⟶R and 0 < α ≤ 1 be the piecewise continuous function. Then, the conformable Laplace transform (CLT) of function u of exponential of order α is defined and denoted by

Definition 3. Let u(x, y) be a piecewise continuous function on the domain D of R⁺ × R⁺ of exponential order α and β. Then conformable double Laplace Transform (CDLT) of u(x, y) is defined and denoted by

where x, y > 0,  p, q ∈ ∁, α, β ∈ (0,1].

Now, we define the conformable triple Laplace transform, for α, β, γ ∈ (0,1], and p, q, s ∈ ∁ are the Laplace variables.

Definition 4. Let u(x, y, t) be a real valued piecewise continuous function of x, y,  and t defined on the domain D of R⁺ × R⁺ × R⁺ of exponential order α, β, and γ, respectively. Then, the conformable triple Laplace transform (CTLT) of u(x, y, t) is defined as follows:

where p, q, s ∈ ∁  are Laplace variables and α, β, γ ∈ (0,1].

The conformable inverse triple Laplace transform, denoted by u(x, y, t), is defined by

Definition 5. A unit step or Heaviside unit step function is defined as follows:

Theorem 2. If $$, $$, and A, B, and C are constants, then

• (a)

  Linearity property:

  $$ ()

• (b)

  $$, where C is the constant.

• (c)

  $$, where Γ(⋅) is the gamma function. Note that Γ(n + 1) = n!, for n = 0, 1, 2, 3, …

• (d)

  The first shifting theorem for conformable triple Laplace transform:

•  

  If $$, then

  $$ ()

• (e)

  $$, then

  $$ ()

• (f)
  $$ ()

Proof. From results (a)–(d) and (f), it can be easily proved by using the definition of conformable triple Laplace transform (CTLT). Here only we see the proof of result (e).

So, by the definition of CTLT (equation (6)), we have

By differentiating with respect to p, l-times, we get

It reduces to

Now, we again differentiate with respect to q and s, m- and n-times, respectively, and we obtain the simplification as follows:

which implies

Now, we multiply (−1)^l+m+n, on both sides, and we get the required result:

Theorem 3. If $$, then

where H(x, y, t) is a Heaviside unit step function as defined in equation (8).

Proof. By applying the definition of CTLT, we have

Now, using the definition of Heaviside unit step function H(x, y, t), we have

By putting $$, we have

Theorem 4. The conformable triple Laplace transform of the function $$, $$ and (∂^β+γ/∂y^β∂t^γ)(u(x, y, t)) is given by

• (a)

  $$

• (b)

  $$

• (c)

  $$

• (d)

  $$

Proof. By applying the definition of CTLT and Theorem 2 (e), till equation (15) can be the required result (a). Remaining results (b)–(d) can be obtained via the same process.

Theorem 5. For α, β, γ ∈ (0,1]. Let u(x, y, t) = u(x^α/α, y^β/β, t^γ/γ) be the real valued piecewise continuous function x, y,  and t of the domain D on (0, ∞) × (0, ∞) × (0, ∞). The CTLT (conformable triple Laplace transform) of the conformable partial fractional derivatives of order α,  β, and γ is given by

• (a)

  $$

• (b)

  $$

• (c)

  $$

• (d)

  $$

• (e)

  $$

• (f)

  $$

• (g)

  $$

Proof. Here, we go for proof of result (a), and the remaining results (b)–(g) can be proved. To obtain conformable triple Laplace transform of the fractional partial derivatives, we use integration by parts and Theorem 1.

By applying the definition of CTLT, we have

Since we have Theorem 1, (∂^α(u)/∂x^α) = x^1−α (∂u/∂x). We use this result in equation (23).

Therefore, equation (23) becomes

The integral inside the bracket is given by

By substituting equation (25) in equation (24), and simplifying, we get the required result (a), that is

In general, the above results in Theorem 5 can be extended as follows:

• (a)
  $$ ()

• (b)

  If $$, then

  $$ ()

Example 1. To illustrate the proposed method, let us consider the following nonlinear partial fractional differential equation:

with initial condition u(x, y, 0) = xy, and where α, β, γ ∈ (0,1], x, y, t∈[0, ∞).

Solution 1. Rewrite equation (41) as

Taking the conformable triple Laplace transform on both sides of equation (42), we have

Recalling Theorem 5 (c), $$.

So, equation (43) reduces to

Since u(x, y, 0) = xy, we have

Now, by applying the conformable inverse triple Laplace transform to equation (44) and using initial condition equation (45), we obtain from equation (44)

By applying the proposed method, in particular, equations (35)–(37), let u₀(x, y, t) = xy, and the recursive relation is given by

where A_n is the Adomian polynomial to decompose the nonlinear terms by using the relation

Let the nonlinear term be represented as

Substituting equation (49) in equation (48), and also calculating (∂^2αu_n/∂x^2α), the resulting expression of equation (47) reduces to the following.

For n = 0,

We have

For n = 1,

Using equation (51), u₀,  and u₁ and simplifying, equation (52) becomes

We have

Simplifying equation (54), we have

For n = 2,

Substituting u₀, u₁,  and u₂ in equation (56) and simplifying, we obtain

Therefore, we have

Simplifying equation (58), we have

and so on….

The approximate series solution is expressed as

From equation (60), if we consider α = β = γ = 1, then the solution of equation (41) reduces to

Figures 1 and 2 show the 3D graphical representations of equation (60) with various values of α and β.

Example 2. Consider a nonlinear nonhomogeneous partial fractional differential equation, for α, β, γ ∈ (0,1], x, y, t∈[0, ∞).

with initial conditions

Solution 2. Rewrite equation (62) as

Imposing the conformable triple Laplace transform to both sides,

Recalling Theorem 5 (d),

Equation (66) becomes

Using initial condition (equation (63)) and taking the inverse triple Laplace transform on equation (67), we obtain

By applying the proposed method, we have the following.

Let u₀ = yt − x, and the recursive relation is

where A_n is the Adomian polynomial to decompose the nonlinear terms by using the following relation:

Let the nonlinear term be represented as

Note that, in Adomian relation, the linear term can be considered, in place of nonlinear. So, you may take f(u) = u, and it leads to the same answer. So, this method is also valid for linear partial fractional differential equation.

For n = 0,

and so on…

The approximate series solution is written as

For α, β, γ ∈ ( 0,1, [x, y, t∈]0, ∞ ).

From equation (73), note that for α, β, γ = 1, the solution of equation (63) reduces to

Figures 3 and 4 show the 3D graphical representations of equation (74) with various values of γ and β.