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Article

Solution of Linear Caputo Fractional Differential Equations with Fractional Initial Conditions

by
Aghalaya S. Vatsala
1,* and
Govinda Pageni
2
1
Department of Mathematics, University of Louisiana at Lafayette, Lafayette, LA 70504, USA
2
Department of Mathematics, Florida State University Panama City, Panama City, FL 32405, USA
*
Author to whom correspondence should be addressed.
AppliedMath 2025, 5(4), 157; https://doi.org/10.3390/appliedmath5040157
Submission received: 28 August 2025 / Revised: 16 October 2025 / Accepted: 23 October 2025 / Published: 7 November 2025

Abstract

The computation of solutions of Caputo fractional differential equations is paramount in modeling to establish its benefits over the corresponding integer order models. In the literature so far, in order to compute the solution of Caputo fractional differential equations, the solution is typically assumed to be a C n function, which is a sufficient condition for the Caputo derivative to exist. In this work, we assume the necessary condition for the Caputo derivative of order n q , ( n 1 ) < n q < n , to exist, which means that we assume it to be a C n q function. Recently, it has been established that the Caputo derivative of order n q is sequential of order q . As such, we assume the fractional initial conditions. In our work, we have obtained an analytical solution for the Caputo fractional differential equation of order n q with fractional initial conditions by two different methods. Namely, the approximation method and the Laplace transform method. The application of our main results is illustrated with examples.

1. Introduction

In the past few decades, the qualitative and quantitative study of fractional differential equations has grown enormously due to its applications. See [1,2,3] for more on fractional differential equations with applications. Among the many types of fractional differential equations, the Caputo fractional differential equation has received special attention, particularly because it is more closely related to integer-order differential equations and is more practical for initial value problems. In [4], the author has used the value of q as a parameter to enhance the mathematical model of realistic data. In addition, the Caputo fractional derivative is in the convolution integral form. As such, the Laplace transform method is an elegant method to solve Caputo fractional initial value problems even with fractional initial conditions. In order to use q, the order of the Caputo derivative, as a parameter, we need to compute the solution of Caputo fractional differential equations either analytically or numerically.
In this work, we focus on solving the linear Caputo fractional differential equation of order n q , where ( n 1 ) < n q < n , with initial conditions. Traditionally, initial conditions are taken as in the integer-order case. Therefore, to compute the solution, we normally assume that it is a C n function on its interval of existence. The reason behind this assumption is that the Caputo derivative of any C n function exists. This is sufficient to ensure that the Caputo derivative of order n q exists and is well-defined. As a result, to compute the solution of n q order Caputo fractional differential equations using the approximation method, the authors in [5] have used the basis as
1 , ( t a ) , ( t a ) 2 , ( t a ) ( n 1 ) ,
and chosen the initial approximation as
b 0 + b 1 ( t a ) + b 1 ( t a ) 2 + + b n 1 ( t a ) ( n 1 ) ,
where the initial conditions are specified as the integer order of the form
u ( k ) ( a ) = b k ,
for k = 1 , 2 , ( n 1 ) .
This choice of initial approximation ensures that the initial conditions are exactly satisfied by the initial approximation. For more details, see page 230 of [5]. Additional studies on Caputo fractional differential equations and their applications can be found in [3,5,6,7,8,9,10,11,12,13]. No doubt that the solution yields the integer result as a special case. However, the solution that is obtained by starting with a C n function as the initial approximation is only a C n q function. As an example, when n = 1 , the solution is E q , 1 ( λ ( t a ) q ) , which is not a C 1 function, but only a C q function. See [5,10,11,12] for the solutions of linear Caputo fractional differential equations in terms of Mittag–Leffler functions. Throughout the analysis, we assume that the solution we plan to obtain is a C n q function. According to the results presented in [14], the Caputo fractional derivative of order n q exists for functions of the form ( t a ) ω , provided ω n q . Furthermore, the Caputo derivative of order n q , where ( n 1 ) < n q < n , is sequential of order q . In this work, we provide an explicit formula to find the solution of the linear nonhomogeneous Caputo fractional differential equations using two methods.
  • Approximation Method: This method is based on representing the solution in terms of the basis functions
    1 , ( t a ) q , ( t a ) 2 q , , ( t a ) ( n 1 ) q .
  • Laplace Transform Method: This method is helpful because the Caputo derivative can be expressed in convolution integral form, making it suitable for solution via Laplace transform methods.
Remark 1.
In the approximation method, we have used basis functions
1 , ( t a ) q , ( t a ) 2 q , , ( t a ) ( n 1 ) q ,
using the fact that the Caputo fractional derivative of order n q is a sequential Caputo fractional derivative of order q.
In both methods, we have used the Caputo fractional initial conditions and the solutions are expressed in terms of appropriate Mittag–Leffler functions, which are determined by the values q , n q , and λ . In short, the solution can be determined directly, once the initial conditions are known. A key advantage of our solution approach methods, compared to standard methods, is that it avoids the need to find the roots of the characteristic polynomial r n = λ . Additionally, it eliminates the necessity of solving a nonhomogeneous system of n equations, which is typically required in the standard method.
We provide two illustrative examples demonstrating that the solution obtained using our method matches those derived via the standard method, which is similar to the integer-order method. In particular, when q = 1 , our solution reduces to the integer-order result as a special case.
An important question is whether our method can be extended to cases where the equation includes Caputo fractional derivatives with lower-order terms, along with fractional initial conditions. The answer is affirmative. Such a scalar differential equation of order n q with fractional initial conditions can be rewritten as a system of Caputo fractional differential equations of order q with initial conditions. This is a special case of the general linear Caputo fractional system of order q with initial conditions and a constant matrix A. The method of successive approximations, previously used for solving our linear Caputo fractional differential equations, can be applied here, starting from an initial vector derived from the initial conditions provided. The solution takes the following form:
u = E q , 1 ( A ( t a ) q ) u 0 .
The fundamental matrix E q , 1 ( A ( t a ) q ) can be computed using the eigenvalue method. A key advantage of our solution method for the Caputo fractional differential system with initial conditions is that the value of q can be chosen as a parameter to enhance the mathematical modeling.
The structure of our research article is as follows. In Section 2, we present the necessary background, including relevant definitions and known results, and some new results related to the Laplace transform method that support our main findings. In Section 3, we derive the solution of linear Caputo fractional differential equations of order n q , where n 1 < n q < n , subject to fractional initial conditions. Two distinct approaches are employed: the method of successive approximations and the Laplace transform technique. We also provide some examples to demonstrate that the solutions obtained via Theorems 1 and 2 are in perfect agreement with those derived through standard analytical methods. We can extend our results to the Caputo fractional differential equation with fractional order n q and initial conditions having lower-order fractional derivatives with constant coefficients. This is achieved by reducing the fractional differential equation with fractional order n q to an n-system of Caputo fractional differential equations of order q with initial conditions. As an application of our results, we have provided an example.

2. Preliminary Results

In this section, we provide basic definitions of fractional derivatives and integrals, describe their properties, and highlight important mathematical results.
Definition 1.
The Gamma function is defined by the integral
Γ ( q ) = 0 t q 1 e t d t ,
where 0 < q 1 , which converges in the right half of the complex plane when R e ( q ) > 0 .
Definition 2.
The β function is defined by the integral
β ( z , w ) = 0 1 t z 1 ( 1 t ) w 1 d t ,
where R e ( z ) , R e ( w ) > 0 .
The Γ and β functions are related by the relation
β ( p , q ) = Γ ( p ) Γ ( q ) Γ ( p + q ) .
Definition 3.
The Riemann–Liouville (left-sided) fractional derivative of u ( t ) of order n q , where ( n 1 ) < n q < n , is defined by
D a + n q u ( t ) = 1 Γ ( n n q ) d n d t n a t ( t s ) n n q 1 u ( s ) d s , t > a .
Definition 4.
The Caputo (left-sided) fractional derivative of u ( t ) of order n q , where n 1 n q < n , is defined by
D a + n q c u ( t ) = 1 Γ ( n n q ) a t ( t s ) n n q 1 u ( n ) ( s ) d s , t > a ,
where u ( n ) ( t ) = d n u d t n .
If we replace n by 1 in the above definition, we obtain the following special case of the Caputo fractional derivative of order q.
Definition 5.
The Caputo (left) fractional derivative of u ( t ) of order q, where 0 < q < 1 , is defined as
D a + q c u ( t ) = 1 Γ ( 1 q ) a t ( t s ) q u ( s ) d s , t > a .
See [5,10,12,15] for more on all the above definitions and other details.
The next definition is useful in our basic Caputo fractional differential inequalities.
Definition 6.
If the Caputo derivative of a function u ( t ) of order q , 0 < q 1 , exists on an interval J = [ a , T ] , T > a , then we say u C q [ J , R ] , where J = [ a , T ] , T > a .
If u C 1 [ a , T ] , then certainly D a + q c u ( t ) exists on [ a , T ] . Note that all C 1 functions on [ a , T ] are C q functions on [ a , T ] . However, the converse need not be true. For example, the function f ( t ) = ( t a ) ω for any ω is a C q function when q ω < 1 . However, it is easy to see f ( t ) does not exist at t = a .
Next, we define the two-parameter Mittag–Leffler function, which will be useful in solving systems of linear Caputo fractional differential equations.
Definition 7.
The two-parameter Mittag–Leffler function is defined as
E q , r ( λ t q ) = k = 0 ( λ t q ) k Γ ( q k + r ) ,
where q , r > 0 , and λ is a constant. Furthermore, for r = q , (5) reduces to
E q , q ( λ t q ) = k = 0 ( λ t q ) k Γ ( q k + q ) , q > 0 .
If r = 1 in (5), then
E q , 1 ( λ t q ) = k = 0 ( λ t q ) k Γ ( q k + 1 ) , q > 0 .
If q = 1 and r = 1 in (5), then we obtain
E 1 , 1 ( λ t ) = k = 0 ( λ t ) k Γ ( k + 1 ) = e λ t ,
where e λ t is the usual exponential function.
See [15,16] for more details on the Mittag–Leffler functions. If the scalar λ is replaced by a square matrix A in the above definition, the corresponding Mittag–Leffler function is referred to as the matrix Mittag–Leffler function.
Definition 8.
The matrix Mittag–Leffler function E q , 1 ( A t q ) is defined by the series
E q , 1 ( A t q ) = k = 0 ( A t q ) k Γ ( k q + 1 ) , q > 0 ,
where A is a square matrix.
Note: In the context of solving systems of ordinary differential equations, the expression e A t is commonly used to represent the solution operator, and the general solution takes the form e A t c , where c is a constant column vector. Such a result is not yet available for a fractional differential system. See [17] for more details on the matrix Mittag–Leffler function.
Next, we will define fractional trigonometric functions and generalized fractional trigonometric functions of order q, which will be required in our main results.
Definition 9.
The fractional trigonometric functions sin q , 1 ( λ t q ) and cos q , 1 ( λ t q ) are given by
sin q , 1 ( λ t q ) = 1 2 i [ E q , 1 ( i λ t q ) E q , 1 ( i λ t q ) ] ,
and
cos q , 1 ( λ t q ) = 1 2 [ E q , 1 ( i λ t q ) + E q , 1 ( i λ t q ) ] ,
respectively.
We can also define sin q , q ( λ t q ) and cos q , q ( λ t q ) in a similar way using E q , q ( λ t q ) in place of E q , 1 ( λ t q ) .
Consider the scalar Caputo fractional linear initial value problem
D a + q c u ( t ) = λ u ( t ) + h ( t ) , u ( a ) = u 0 ,
where t J and u C q ( J ) . Here, h ( t ) C ( J , R ) denotes a continuous function from J to R . As seen in [1,5,12], if u C q ( J ) , then the solution of (11) can be written as
u ( t ) = u 0 E q , 1 ( λ ( t a ) q ) + a t ( t s ) ( q 1 ) E q , q ( λ ( t s ) q ) h ( s ) d s .
Note that this is a C q solution on the interval [ a , T ] , for any T > a .
Next, consider the linear Cauchy problem for the Caputo fractional differential equation
D a + n q c u ( t ) = λ u ( t ) + f ( t ) , t > a , ( n 1 ) < n q < n , n N , λ R ,
with initial conditions
u ( k ) ( a ) = b k , b k R , k = 0 , 1 , , n 1 .
Note that the initial conditions correspond to those of an integer order n. This is because, assuming u C n , the Caputo derivative of order n q exists. Accordingly, the basis functions are taken to be those of the integer order, namely,
1 , ( t a ) q , ( t a ) 2 q , , ( t a ) ( n 1 ) q .
Then the solution of (13) and (14) can be written as
u ( t ) = j = 0 n 1 b j ( t a ) j E n q , j + 1 λ ( t a ) n q + a t ( t s ) n q 1 E n q , n q λ ( t s ) n q f ( s ) d s .
Note that the solution provided in (15) is only of class C n q on [ a , T ] , and not of class C n on [ a , T ] .
Definition 10.
If the Caputo derivative of a function u ( t ) of order n q , where ( n 1 ) < n q n , exists on an interval J = [ a , T ] , with T > a , then we say that u C n q [ J , R ) .
If u C n [ a , T ] , then the Caputo fractional derivative D a + n q c u ( t ) certainly exists on [ a , T ] . Note that all C n functions on [ a , T ] are also in C n q [ a , T ] . However, the converse need not be true. For example, the function f ( t ) = ( t a ) ω is a C n q function, when n q ω < n . But it is easy to see that the integer-order derivative f ( n ) ( t ) does not exist at t = a if q < 1 .
Definition 11.
The Laplace transform F(s) of a function f(t) is
L [ f ( t ) ] = F ( s ) = 0 e s t f ( t ) d t ,
defined for all s such that the integral converges.
Since D 0 + q c f ( t ) is in the convolution integral form, the Laplace transform of D 0 + q c f ( t ) is given by
L [ c D 0 + q f ( t ) ] = s q F ( s ) s q 1 f ( 0 ) , 0 < q 1 ,
where F ( s ) = L ( f ( t ) ) .
See [11], for the initial work on the Laplace transform for fractional differential equations.
Lemma 1.
The inverse Laplace transform of
s n q r q 1 s n q λ ,
is given by
L 1 s n q r q 1 s n q λ = t r q E n q , r q + 1 ( λ t n q ) ,
for r = 0 , 1 , 2 , , n 1 .
Proof. 
To prove the lemma, consider the series expansion of t r q E n q , r q + 1 ( λ t n q ) :
t r q E n q , r q + 1 ( λ t n q ) = t r q 1 Γ ( r q + 1 ) + λ t n q Γ ( n q + r q + 1 ) + + λ k t k n q Γ ( k n q + r q + 1 ) + .
Taking the Laplace transform of both sides, we obtain
L t r q E n q , r q + 1 ( λ t n q ) = 1 s r q + 1 + λ s n q + r q + 1 + + λ k s k n q + r q + 1 + .
The right-hand side can be factored as
1 s r q + 1 1 + λ s n q + λ 2 s 2 n q + + λ k s k n q + = 1 s r q + 1 1 1 λ s n q ,
= s n q r q 1 s n q λ , for | λ | < s n q .
Thus, we have shown that
L t r q E n q , r q + 1 ( λ t n q ) = s n q r q 1 s n q λ ,
for r = 0 , 1 , 2 , , n 1 .
The conclusion of the lemma follows by taking the inverse Laplace transform of (16) on both sides. □
Next, we provide the definition of the sequential Caputo fractional derivative of order n q , which is the sequential Caputo derivative of order q.
Definition 12.
The Caputo left fractional derivative of a function u ( t ) of order n q , where ( n 1 ) < n q < n , is called the sequential left Caputo fractional derivative of order q if the following relation holds:
D a + n q c u ( t ) = D a + q c D a + ( n 1 ) q c u ( t ) ,
for all integers n = 2 , 3 , .
Remark 2.
In [14], it was established that the left Caputo derivative of order n q of a function f ( t ) , starting from t = a , can be computed only when f ( t ) can be expressed as a function of ( t a ) . Moreover, the exponent ω in ( t a ) ω must satisfy ω n q in order for the derivative to exist. Additionally, it was shown that the Caputo derivative of order n q is always a sequential Caputo derivative of order q.
Using Remark 2 and observing that the Caputo derivative is a convolution integral, we can show that
L D 0 + n q c u ( t ) = s n q U ( s ) s n q 1 u ( 0 ) s ( n 1 ) q 1 D 0 + q c u ( t ) t = 0 s q 1 D 0 + ( n 1 ) q c u ( t ) t = 0 .

3. Main Results

In this section, we derive the solution form of a nonhomogeneous linear Caputo fractional differential equation of order n q , where ( n 1 ) < n q < n . The equation includes a linear term and is subject to fractional initial conditions.
The parameter q is not only the order of the derivative but also a flexible modeling tool. Since it affects the initial conditions, it helps to enhance the model. This makes the role of q especially valuable when using the method of successive approximations to find the solution.
Consider the linear nonhomogeneous Caputo fractional differential equation of order n q , where ( n 1 ) < n q < n :
( c D a + n q ) u ( t ) = λ u ( t ) + f ( t ) , λ R ,
along with the following fractional initial conditions:
( c D a + k q ) u ( t ) | t = a = a k , for k = 0 , 1 , 2 , , ( n 1 ) .
We assume that f ( t ) C [ a , T ) , where T > a . The solution to the above equation, along with the initial conditions, is similar to the solution of an integer-order differential equation of order n. However, in this case, the initial conditions involve fractional derivatives. Since the Caputo fractional derivative of order n q is a sequential Caputo derivative of order q, our initial conditions are given in terms of the Caputo fractional derivative of order k q , where k = 0 , 1 , 2 , , ( n 1 ) .
Additionally, the basis solutions for solving ( c D a + n q ) u = 0 are given by
1 , ( t a ) q Γ ( q + 1 ) , , ( t a ) ( n 1 ) q Γ ( ( n 1 ) q + 1 ) .
Theorem 1.
Let u ( t ) C n q [ a , T ] be the solution of (19) and (20) on the interval [ a , T ] . Then the solution u ( t ) is given by
u ( t ) = k = 0 n 1 a k ( t a ) k q E n q , k q + 1 λ ( t a ) n q + a t ( t s ) n q 1 E n q , n q λ ( t s ) n q f ( s ) d s .
Proof. 
The solution of (19) and (20) can be written as the solution of the homogeneous equation and the solution of the nonhomogeneous terms, which are λ u + f ( t ) . It is easy to see that the solution of the homogeneous equation ( c D a + n q ) u ( t ) = 0 satisfying the fractional initial condition (20) is given by
u ( t ) = k = 0 n 1 a k Γ ( k n q + k q + 1 ) ( t a ) k n q + k q .
Using the integral representation for the nonhomogeneous part, we can write the solution of (19) and (20) as
u ( t ) = k = 0 n 1 a k Γ ( k n q + k q + 1 ) ( t a ) k n q + k q + λ Γ ( n q ) a t ( t s ) n q 1 u ( s ) d s + 1 Γ ( n q ) a t ( t s ) n q 1 f ( s ) d s .
We will find this solution by the method of successive approximation, starting with
u 0 ( t ) = k = 0 n 1 a k Γ ( k q + 1 ) ( t a ) k q + 1 Γ ( n q ) a t ( t s ) n q 1 f ( s ) d s ,
and
u m ( t ) = k = 0 n 1 a k Γ ( k q + 1 ) ( t a ) k q + λ Γ ( n q ) a t ( t s ) n q 1 u m 1 ( s ) d s + 1 Γ ( n q ) a t ( t s ) n q 1 f ( s ) d s ,
where m N .
For m = 1 , we obtain
u 1 ( t ) = k = 0 n 1 a k Γ ( k q + 1 ) ( t a ) k q + λ Γ ( n q ) a t ( t s ) n q 1 u 0 ( s ) d s + 1 Γ ( n q ) a t ( t s ) n q 1 f ( s ) d s .
Substituting for u 0 ( s ) from (23) into the formula for u 1 ( t ) and simplifying, we get
u 1 ( t ) = k = 0 n 1 a k 1 Γ ( k q + 1 ) + λ ( t a ) n q Γ ( n q + k q + 1 ) ( t a ) k q + a t ( t s ) n q 1 1 Γ ( n q ) + λ ( t s ) n q Γ ( 2 n q ) f ( s ) d s .
Continuing this process, we obtain the following general form:
u m ( t ) = k = 0 n 1 a k i = 0 m λ i ( t a ) i n q Γ ( i n q + k q + 1 ) ( t a ) k q + a t ( t s ) n q 1 i = 0 m λ i ( t s ) i n q Γ ( ( i + 1 ) n q ) f ( s ) d s .
Taking the limit as m , we get the solution
u ( t ) = k = 0 n 1 a k ( t a ) k q E n q , k q + 1 λ ( t a ) n q + a t ( t s ) n q 1 E n q , n q λ ( t s ) n q f ( s ) d s .
Next, we prove the same result using the Laplace transform method. For convenience, we set a = 0 in the definition of the Caputo derivative, and the fractional derivatives are taken at a + 0 .
Theorem 2.
Let u ( t ) C n q [ 0 , T ] be the solution of (19) and (20) on the interval [ 0 , T ] . Then the solution u ( t ) is given by
u ( t ) = k = 0 n 1 a k t k q E n q , k q + 1 λ t n q + 0 t ( t s ) n q 1 E n q , n q λ ( t s ) n q f ( s ) d s .
Proof. 
Note that the Caputo left derivative is in a convolution integral form, which simplifies taking the Laplace transform of the left Caputo derivative of a function u ( t ) . For further details, see [11]. Additionally, it is well-established that the Caputo left derivative of order n q , where ( n 1 ) < n q < n , is a sequential Caputo derivative of order q.
Taking the Laplace transform of both sides of Equation (19), we get
s n q U ( s ) s n q 1 ( c D 0 + q ) u ( t ) | t = 0 s ( n 1 ) q 1 ( c D 0 + q ) u ( t ) | t = 0 s q 1 ( c D 0 + ( n 1 ) q ) u ( t ) | t = 0 = λ U ( s ) + F ( s ) ,
where L [ u ( t ) ] = U ( s ) and L [ f ( t ) ] = F ( s ) .
Using the initial conditions ( c D 0 + k q ) u ( t ) | t = 0 = a k for k = 0 , 1 , 2 , , ( n 1 ) , and solving for U ( s ) , we obtain
U ( s ) = s n q 1 s n q λ a 0 + s ( n 1 ) q 1 s n q λ a 1 + + s q 1 s n q λ a n 1 + F ( s ) s n q λ .
Note that the inverse Laplace transform of the last term is a convolution integral of the functions f ( t ) and E n q , n q ( λ t n q ) .
Next, we take the inverse Laplace transform of both sides and apply Lemma 1 for r = 0 , 1 , 2 , , ( n 1 ) , which yields the coefficients a i , for i = 0 , 1 , 2 , , ( n 1 ) .
As a result, we obtain the following solution:
u ( t ) = k = 0 n 1 a k t k q E n q , k q + 1 λ t n q + 0 t ( t s ) n q 1 E n q , n q λ ( t s ) n q f ( s ) d s .
Remark 3.
For any λ R , the solution to the initial value problem (19) and (20) on the interval [ a , T ] is given by
u ( t ) = k = 0 n 1 a k ( t a ) k q E n q , k q + 1 λ ( t a ) n q + a t ( t s ) n q 1 E n q , n q λ ( t s ) n q f ( s ) d s .
See [18] for more details on the convergence of Mittag–Leffler functions. This result reduces to the solution of the n th order differential equation with initial conditions as a special case. However, the standard method to solve the above equation requires finding the roots of the corresponding n th -degree polynomial.
Next, we provide some examples to demonstrate that the solution obtained by Theorems 1 and 2 is exactly the same as the solution obtained by the standard method, which requires the computation of the n roots of the polynomial m n = λ .
Example 1.
Consider the Caputo fractional differential equation
( D 0 + 2 q c ) u ( t ) = λ u ( t ) , λ R ,
with the fractional initial conditions
( D 0 + k q c ) u ( t ) | t = 0 = a k , for k = 0 , 1 .
Then, using Theorem 1, the solution is given by
u ( t ) = a 0 E 2 q , 1 ( λ t 2 q ) + a 1 t q E 2 q , q + 1 ( λ t 2 q ) .
In particular, if λ = 1 , the solution becomes
u ( t ) = a 0 E 2 q , 1 ( t 2 q ) + a 1 t q E 2 q , q + 1 ( t 2 q ) .
Alternatively, using standard methods for solving fractional differential equations of the form ( D 0 + 2 q c ) u ( t ) = λ u ( t ) , one obtains the two linearly independent solutions:
E q , 1 ( t q ) and E q , 1 ( t q ) ,
which serve as fundamental solutions of the equation. The general solution satisfying the given initial conditions is then expressed as
u ( t ) = a 0 + a 1 2 E q , 1 ( t q ) + a 0 a 1 2 E q , 1 ( t q ) .
It is straightforward to verify that the solutions given in Equations (32) and (33) are equivalent. In particular, both are linear combinations of the same fundamental solutions, and both satisfy the initial conditions.
Example 2.
Consider the Caputo fractional differential equation
( D 0 + 2 q c ) u ( t ) = λ u ( t ) , λ R ,
subject to the fractional initial conditions
( D 0 + k q c ) u ( t ) | t = 0 = a k , for k = 0 , 1 .
Then, using Theorem 1, the solution is given by
u ( t ) = a 0 E 2 q , 1 ( λ t 2 q ) + a 1 t q E 2 q , q + 1 ( λ t 2 q ) .
In particular, if λ = 1 , the solution becomes
u ( t ) = a 0 E 2 q , 1 ( t 2 q ) + a 1 t q E 2 q , q + 1 ( t 2 q ) .
Alternatively, by using standard techniques, we can express the two linearly independent solutions as the generalized cosine and sine functions:
cos q , 1 ( t q ) = E 2 q , 1 ( t 2 q ) , sin q , 1 ( t q ) = t q E 2 q , q + 1 ( t 2 q ) .
Hence, the solution satisfying the same initial conditions can also be written as
u ( t ) = a 0 cos q , 1 ( t q ) + a 1 sin q , 1 ( t q ) .
It is straightforward to verify that the solutions given in Equations (37) and (38) are indeed equivalent.
We now extend the result of Theorem 1 to a linear system of Caputo fractional differential equations of order q, subject to initial conditions.
Consider the system
( D 0 + q c ) u ( t ) = A u ( t ) + f ( t ) , 0 < q < 1 ,
where u ( t ) is an N × 1 column vector, A is an N × N constant nonsingular matrix, and f ( t ) is a given N × 1 column vector-valued function.
The initial condition is specified as
u ( 0 ) = u 0 ,
where u 0 is a known N × 1 constant vector.
The proof follows along the same lines as in Theorem 1, with the scalar λ replaced by the matrix A. See [19] for more details.
The solution to the system (39) is given by
u ( t ) = E q , 1 ( A t q ) u 0 + 0 t ( t s ) q 1 E q , q ( A ( t s ) q ) f ( s ) d s .
The details of the proof of (41) will be presented elsewhere.
Example 3.
Consider the 2 q order Caputo fractional differential equation with lower-order terms
( D 0 + 2 q c ) u ( t ) 3 D 0 + q c u ( t ) + 2 u ( t ) = 0 ,
with fractional initial conditions
u 1 ( 0 ) = u 01 , D 0 + q c u 1 ( 0 ) = u 02 .
Then, (42) and (43) can be reduced to the linear two-system of Caputo fractional differential equations of order q:
D 0 + q c u 1 ( t ) D 0 + q c u 2 ( t ) = 0 1 2 3 u 1 ( t ) u 2 ( t ) ,
with initial conditions
u 1 ( 0 ) u 2 ( 0 ) = u 01 u 02 .
Then, the solution is given by
u ( t ) = E q , 1 ( A t q ) u 0 ,
where the matrix A and vector u 0 are defined as
A = 0 1 2 3 , u 0 = u 01 u 02 .
If q = 1 , the solution reduces to the following classical case:
u ( t ) = e A t u 0 .
The two distinct eigenvalues of the matrix A can be easily obtained as λ 1 = 1 and λ 2 = 2 .
Using the eigenvalues of the matrix A, the solution to (44) can be expressed explicitly as
u ( t ) = E q , 1 ( t q ) E q , 1 ( 2 t q ) E q , 1 ( t q ) 2 E q , 1 ( 2 t q ) 2 u 01 u 02 u 02 u 01 .
Component-wise, this gives:
u 1 ( t ) = ( 2 u 01 u 02 ) E q , 1 ( t q ) + ( u 02 u 01 ) E q , 1 ( 2 t q ) , u 2 ( t ) = ( 2 u 01 u 02 ) E q , 1 ( t q ) + ( u 02 u 01 ) 2 E q , 1 ( 2 t q ) .
Notice that the solution can also be expressed in the matrix form as
u 1 ( t ) u 2 ( t ) = 2 E q , 1 ( t q ) E q , 1 ( 2 t q ) E q , 1 ( 2 t q ) E q , 1 ( t q ) 2 E q , 1 ( t q ) 2 E q , 1 ( 2 t q ) 2 E q , 1 ( 2 t q ) E q , 1 ( t q ) u 01 u 02 .
This is precisely what we wanted to achieve. For any change in the initial conditions, the solution can be computed immediately once the fundamental matrix solution has been determined.
Remark 4.
In the scalar linear fractional differential equation of order n q with fractional initial conditions, we can express the solution directly by using the initial conditions without solving the polynomial of degree n, namely, r n + λ = 0 . The solutions are expressed using the Caputo fractional initial conditions and appropriate Mittag–Leffler functions, which are determined by the values q , n q , and λ . However, if the linear fractional differential equation includes lower-order derivative terms, such as ( c D 0 + k q ) u ( t ) for 1 < k < n , our method with fractional initial conditions, as presented in Theorem 1, will not work.
Nevertheless, when fractional initial conditions are provided, we can reduce the n q -order fractional differential equation to a system of n fractional differential equations with initial conditions. Note that this reduction is not possible if the initial conditions are of integer order.
Thus, having fractional initial conditions is both appropriate and useful when we want to treat q as a parameter. Theorems 1 and 2 provide the solution, which coincides with the standard solution for the integer-order case.
The Laplace transform method is an indirect method of obtaining the forward solution (19) and (20). However, the approximation method is a more direct method of obtaining the forward solution of (19) and (20). The advantage of the approximation method is that the backward solution of the linear Caputo fractional differential equations with fractional initial conditions can also be obtained.

4. Conclusions

In this work, we have presented a methodology for computing the solution of linear, nonhomogeneous Caputo (left) fractional differential equations of order n q , where ( n 1 ) < n q < n , with a linear term λ u and fractional initial conditions. This approach had an advantage over the standard method commonly used for integer-order differential equations, which requires additional computations to solve a system of nonhomogeneous equations.
To establish our results, we have presented two distinct methods, namely the successive approximation method and the Laplace transform method. Moreover, the successive approximation method can also be applied to compute the backward solution of the Caputo (right) fractional differential equations of order n q , starting from t = b with fractional initial conditions specified at t = b .
Furthermore, the successive approximation method can be applied to obtain solutions of linear nonhomogeneous systems of Caputo fractional differential equations of order q, with 0 < q < 1 , using fractional initial conditions expressed in terms of the matrix Mittag–Leffler function. This formulation naturally recovers the classical integer-order results as a special case.
In our future work:
  • We plan to use the value of q as a parameter to enhance the mathematical model with realistic data.
  • We aim to develop iterative methods to compute the solution of nonlinear Caputo fractional differential equations with fractional initial conditions and fractional boundary conditions.

Author Contributions

Conceptualization, A.S.V. and G.P.; methodology, A.S.V. and G.P.; validation, A.S.V. and G.P.; formal analysis, A.S.V.; investigation, G.P.; writing—original draft preparation, A.S.V.; writing—review and editing, G.P.; supervision, A.S.V. All authors have read and agreed to the published version of the manuscript.

Funding

No funding was received to assist with the preparation of this manuscript.

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The authors declare no conflicts of interest.

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Vatsala, A.S.; Pageni, G. Solution of Linear Caputo Fractional Differential Equations with Fractional Initial Conditions. AppliedMath 2025, 5, 157. https://doi.org/10.3390/appliedmath5040157

AMA Style

Vatsala AS, Pageni G. Solution of Linear Caputo Fractional Differential Equations with Fractional Initial Conditions. AppliedMath. 2025; 5(4):157. https://doi.org/10.3390/appliedmath5040157

Chicago/Turabian Style

Vatsala, Aghalaya S., and Govinda Pageni. 2025. "Solution of Linear Caputo Fractional Differential Equations with Fractional Initial Conditions" AppliedMath 5, no. 4: 157. https://doi.org/10.3390/appliedmath5040157

APA Style

Vatsala, A. S., & Pageni, G. (2025). Solution of Linear Caputo Fractional Differential Equations with Fractional Initial Conditions. AppliedMath, 5(4), 157. https://doi.org/10.3390/appliedmath5040157

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