## 3 Problem Formulation and a First Solution  
Now, we are going into the solution of the fractional Ambartsumian Equation (3). For stability reasons, the derivative order (\(\alpha\)) can only assume any real value in the interval \((0,2)\). The equation, as written above, corresponds to the representation of a system with null input, but having a non-null initial-condition (IC). Traditionally, the IC problems have been treated with the unilateral Laplace transform. However, this has proven to be inconsistent [35]. On the other hand, when used in connection with RL or C derivatives, it inserts wrong ICs [36]. In [36], an alternative approach into the IC problem is described. It is based on the fractional jump formula [44]. Basically, the procedure consists of modifying the original equation so that the initial-condition appears explicitly. In this particular case, we get the equation  
\[D^{\alpha}x(t)+x(t)-x_{0}D^{\alpha-1}\delta(t)=\frac{1}{q}x\bigg{(}\frac{t}{q }\bigg{)},\quad t\geq 0,\ q\geq 1, \tag{12}\]  
where \(\delta(t)\) is the Dirac impulse. This relation can be rewritten as  
\[D^{\alpha}x(t)+x(t)-x_{0}D^{\alpha}\varepsilon(t)=\frac{1}{q}x\bigg{(}\frac{t }{q}\bigg{)},\quad t\geq 0,\ q\geq 1,\]since \(\delta(t)=D\epsilon(t)\). We modify it into  
\[D^{\alpha}[x(t)-x_{0}\epsilon(t)]+x(t)=\frac{1}{q}x\bigg{(}\frac{t}{q}\bigg{)} \quad t\geq 0,\ q\geq\]  
that highlights the role of IC: to make continuous the function at the origin, before computing the derivative. **Remark 1**. : _The brightness, \(x(t)\), is dimensionless. Therefore, having in mind that a derivative of order \(\alpha\) introduces a division by a power of time, we must adjust the equation dimensions. Consequently, \(q\) must have units \([s^{\alpha}]\)._  
Now, we are in a condition of looking for the solution of our problem. The case of \(q=1\) presents no particular difficulty. Therefore, we will not consider it. **Theorem 1**. : _Let \(q>1\) and \(0<\alpha<2\). The solution of (3), under the IC \(x_{0}\), is given by:_  
\[x(t)=\sum_{n=0}^{\infty}C_{n}\frac{t^{n\alpha}}{\Gamma(n\alpha+1)},\quad t>0, \tag{13}\]  
_where \(C_{0}=x_{0}\) and_  
\[C_{n}=(-1)^{n}\prod_{k=0}^{n-1}\Bigl{[}1-q^{-k\alpha-1}\Bigr{]}C_{0}, \tag{14}\]  
_for \(n=1,2,\dots\)._  
Proof. : A somehow obvious approach to solve (12) is a search for a fractional Taylor solution as in [2,5] which corresponds, in the BLT domain, to assume that \(X(s)=\mathcal{L}x(t)\) has the Laurent form  
\[X(s)=\sum_{n=0}^{\infty}C_{n}s^{-n\alpha-1}, \tag{15}\]  
where \(C_{n},\ n=0,1,2\cdots\) are real parameters to be computed. In terms of the bilateral Laplace transform [35], Equation (12) becomes  
\[s^{\alpha}X(s)-x_{0}s^{\alpha-1}+X(s)=X(qs),\quad Re(s)>0, \tag{16}\]  
where the scale property of the BLT was used. We have,  
\[s^{\alpha}\sum_{0}^{\infty}C_{n}s^{-n\alpha-1}+\sum_{0}^{\infty}C_{n}s^{-n \alpha-1}=x_{0}s^{\alpha-1}+\sum_{0}^{\infty}C_{n}q^{-n\alpha-1}s^{-n\alpha- 1},\]  
that leads to  
\[\sum_{n=-1}^{\infty}C_{n+1}s^{-n\alpha}+\sum_{0}^{\infty}C_{n}s^{-n\alpha}=x_ {0}s^{\alpha}+\sum_{0}^{\infty}C_{n}q^{-n\alpha-1}s^{-n\alpha}.\]  
From this equation, we conclude immediately that  
\[C_{n+1}=-C_{n}+C_{n}q^{-n\alpha-1}=-\Bigl{[}1-q^{-n\alpha-1}\Bigr{]}C_{n}, \tag{17}\]  
with \(C_{0}=x_{0}\). The solution of this difference equation is the q-shifted factorial (14) [41,45]. These coefficients constitute a decreasing sequence as \(n\to\infty\). In fact, from (17) and, as \(q>1\),  
\[D^{\alpha}x(t)+x(t)-x_{0}D^{\alpha}\varepsilon(t)=\frac{1}{q}x\bigg{(}\frac{t }{q}\bigg{)},\quad t\geq 0,\ q\geq 1,\]0On the other hand, from (14) and using well-known results, we obtain  
\[D^{\alpha}x(t)+x(t)-x_{0}D^{\alpha}\varepsilon(t)=\frac{1}{q}x\bigg{(}\frac{t }{q}\bigg{)},\quad t\geq 0,\ q\geq 1,\]1  
showing the convergence of the infinite product in (14).