J. R. Solvability of a problem for a time fractional m atematika Instituti Byulleteni 2021, Vol. 4, №4, 9-18 b. Bulletin of the Institute of Mathematics 2021, Vol. 4, №4, pp. 9-18 Бюллетень

Preliminaries

Fractional derivatives and integrals

Definition 1. The Riemann-Liouville (R-L) fractional integral of a function f(x) of order α is defined by[1]
.
Definition 2. The Riemann-Liouville (R-L) left-sided fractional derivative of order α of a function f(x) is defined by [1]

Definition 3. The Caputo-Gerasimov left-sided fractional derivative of order α is defined by [1]

where n ∈ N and n = [α] + 1.
Definition 4. [11] We consider the weighted spaces of continuous functions
C_γ[a,b] = {f : (a,b] → R : (x − a)^γf(x) ∈ C[a,b]}, 0 ≤ γ < 1,
and
C_γⁿ[a,b] = {f ∈ Cⁿ⁻¹[a,b] : f⁽ⁿ⁾ ∈ C_γ,n ∈ N,
C_γ⁰[a,b] = C_γ[a,b],
with the norms
kfk_Cγ = k(x − a)^γf(x))k_C

and	n−1 kfkCγn = X kf(k)kC + kf(n)kCγ. k=0

These spaces satisfy the following properties.

1. 
   .
   
2. 
   .
   
3. 
   C_γ1[a,b] ⊂ C_γ2[a,b], 0 ≤ γ₁ < γ₂ < 1.
   

Lemma 1. , then the fractional derivatives in Definition 1 and Definition 2 exist on (a,b] and
.
Definition 5. Hilfer fractional derivative of order α and type µ with respect to t is defined by [23]

whenever the right-hand side exists.
The derivative is considered as an interpolation between the Riemann-Liouville and Caputo derivative:
.
Lemma 2. ([1] ) Let for x ∈ (a,b] and when f(x) ∈ C[a,b], the equality holds at any point x ∈ (a,b].
Lemma 3. ([1] ) Let α > 0, 0 ≤ γ < 1. If f(x) ∈ C_γ[a,b], then

for x ∈ (a,b] and when f(x) ∈ C[a,b], the equality holds at any point x ∈ (a,b].
Lemma 4. ([1]) Let α > 0, 0 ≤ γ < 1, n = [α] + 1 and and
, then
,
where and f_n−α(x) ∈ Cⁿ[a,b] then the equality holds at any point x ∈ (a,b].

Mittag-Leffler function

A two-parametr of the Mittag-Leffler function at α > 0, for all β ∈ C and z ∈ C, represented as follows [4]
.
Lemma 5. (see[4]Theorem 1.6, p.35) if α < 2, β is arbitrary real number, µ is such that πα/2 < µ < min{π,πα} and C₁ is a real constant, then
.
Definition 6. We introduce the inner scalar product and the norm on graph as follows:
,
Here we understand the functions f(x) and g(x) in the following forms:
 f(1)(x1)   g(1)(x1) 
f(x) =  f(2)(x2) , g(x) =  g(2)(x2) 
 ...   ...  f(k)(xk) g(k)(xk)

Formulation of a problem

Let us consider simple star graph Γ with three semi-finite bonds connected at the point O. The point O is the vertex of the graph. We label bonds of the graph as B_k, k = 1,2,3 . Let us define coordinate x_k on the bond B_k,k = 1,2,3, and x_k ∈ (0,M_k). At each bond the coordinate of the vertex point O is equal to zero. Further, we will use x instead of x_k.
On the each edges of the over defined graph, we consider fractional differential equations
, (1)
where is Hilfer operator, 0 < α < 1, , 0 ≤ µ ≤ 1, f^(k)(x,t) (k = 1,2,3) are known functions. We will study the following problem for equation (1) in Γ.
Problem. To find functions u^(k)(x,t) in the domain B_k ×(0,T), satisfy an equation (1) for 0 < α < 1 with the following properties:

1. 
   t1−α−µ+αµu(k)(x,t) ∈ C([0,Mk] × [0,T]), u⁽_xx^k)(x,t),D₀₊^α,µu^(k) (x,t) ∈ C ((0,M_k) × (0,T));
   
2. 
   local conditions:
   

; (2)

3. vertex conditions
u^(1)(0,t) = u^(2)(0,t) = u^(3)(0,t),t ∈ [0,T],	(3)
u^(1)x (0,t) + u^(2)x (0,t) + u^(3)x (0,t) = 0, t ∈ [0,T], k = 1,2,3 and boundary conditions	(4)
u^(^k^)(Mk,t) = 0, t ∈ [0,T], k = 1,2,3. where ϕ^(^k^)(x) are sufficiently smooth given functions, moreover	(5)

(6)

ϕ^(^k^)(Mk) = 0, k = 1,2,3.

(7)

Main Result

Thorem. If and
and absolutely integrable functions in (0,M_k) and (B_k × (0,T)) such that ϕ⁽¹⁾ (0) = ϕ⁽²⁾ (0) = ϕ⁽³⁾ (0) , and
, then the solution of the investigated problem exists and unique.
Proof: Using by the method separations of variables for the homogeneous equation we will get integer order differential equations
(8)
and fractional order differential equations
,
moreover, from the conditions (3)-(5), we obtain

X(1)(0) = X(2)(0) = X(3)(0),

(9)

, (10)

X^(^k^)(Mk) = 0.k = 1,2,3.

(11)

By virtue conditions (9)-(11) from the general solution of equation (8), we can find eigenfunction and eigenvalues:
X^(k)(x) = a_k cosλx + b_k sinλx; x ∈ B_k. (12)
We assume that f^(k)(x,t) ∈ L₂[0;M_k] and then we expand into the Fourier series in terms of eigenfunctions, i.e.
, (13)
where f_n(t) is the coefficients of the Fourier series (13). Further, we search a solution of the equation (1) in the form
. (14)
Substituting (14) into the equation (1), we obtain
.
Consequently
. (15)
General solution of the Eq.(15) has a form (see, [3], Lemma 2):
(16)
where γ = α + µ − αµ. Considering (14) and (16) we can write the general solution of equation (1) in the following form:
(17)
Applying the operator to the (17) and considering Definition 1, we have
(18)

Assuming ϕ^(k)(x) ∈ L₂[0;M_k] and expanding functions ϕ^(k)(x) into a Fourier series:
, (19)
where, ϕ_n are the coefficients of the Fourier series (19).
;

By virtue (2) from (19) we find that
An = ϕn. where		(20)
	(1)  An (2) An  An(3)
Further, integrating by parts two times the functions ϕ^(^k^)(x) and considering (6)-(7), we get:

(21)
It is required to prove the convergence of functions in the domain B_k ×(0,T).
We use Lemma 5 and the following inequalities
, (22)
owing to (20)-(21), we find
.
where C₁ = const > 0. From (13) and based on the conditions of the Theorem,

taking into account (22),
. (23)
where C₂ is positive const. From (17) and (21)-(23) we obtain



where C_i = const > 0,i = 3,5 and C₅ ≥ C₃ + C₄. So u^(k)(x,t) are uniform convergent. Taking into account , we can write

We infer, that the function be uniformly convergent due to the Definition 4.



Now consider for convergence.




where m_i = const > 0, (i = 1,4) and m₁ + m₂ + m₃ ≤ m₄. According to the asymtotes of λ_n ∼ cn (c = const) (see[11]) we can conclude that series of is uniformly convergent. The operator which is defined in definition 4 can be written as
µ(1−α) (1−µ)(1−α) (k) − u(xxk)(x,t) = f(k)(x,t),
I0+ DI0+ u (x,t)
I0+γ−αDI0+1−γu(k)(x,t) = u(xxk)(x,t) + f(k)(x,t),γ = α + µ − αµ.
Introducing notation
(24)
equation (1) we will rewrite as follows:

Further applying and considering Lemma 2, we deduce
.
Using Definition 3 we obtain
(25)
where . Considering (2)-(5), and from (24) we deduce

v^(^k^)(x,0) = ϕ^(^k^)(x),k = 1,2,3, x ∈ Bk, and vertex conditions	(26)
v^(1)(0,t) = v^(2)(0,t) = v^(3)(0,t),t ∈ [0,T],	(27)

(28)

boundary conditions
v^(^k^)(Lk,t) = 0, t ∈ [0,T], k = 1,2,3.	(29)

Using standart scheme, a solution the problem (25)-(29)(see[7],[10],[6] ) is unique. If andϕ^(k)(x) ≡ 0 , then v^(k)(x,t) ≡ 0. Based on (24) we get u^(k)(x,t) ≡ 0. Therefore, we can say that the solution to problem
(1)-(5) is unique.