# About the Use of Generalized Forms of Derivatives in the Study of Electromagnetic Problems
## 4 Fractional RLC Circuit  
In the previous section, we dealt with first-order ODE and we managed to obtain the corresponding non-homogeneous contributions. Now we provide a natural further extension by studying an RLC circuit [46; 47; 48; 49; 50; 51]. **Problem 3** (**RLC Circuit**). : _Regarding a second-order ODE of the type_  
\[ \begin{cases} L\dfrac{d^{2}}{dt^{2}}i(t)+R\dfrac{d}{dt}i(t)+\dfrac{1}{C}i(t)=V^ {\prime}(t)\\ V^{\prime}(t)=V_{0}f(t)\end{cases} , \tag{62}\]  
_we can proceed by just extending the outlined procedure and writing_  
Figure 11: Equation (59) for different values of \(\gamma\). \[ \begin{split}\frac{d^{2}}{d\tau^{2}}\tilde{i}(\tau)+\frac{1}{\omega_{0} \tau_{L}}\frac{d}{d\tau}\tilde{i}(\tau)+\tilde{i}(\tau)=f^{*}(\tau),\\ \tilde{i}=\frac{i}{i_{0}},\qquad i_{0}=\frac{V_{0}}{L\omega_{0}^{ 2}},\qquad\omega_{0}^{2}=\frac{1}{LC},\qquad\tau=\omega_{0}t,\qquad\tau_{L}= \frac{L}{R},\qquad f^{*}(\tau)=\left(\frac{\tau}{\omega_{0}}\right)\end{split}  \tag{63}\]  
with \(L\) being the circuit inductance. The solution of Equation (63) for vanishing initial conditions \(\tilde{i}_{0}\), \(\frac{d}{d\tau}\tilde{i}(\tau)\mid_{\tau=0}\) can be written as  
\[ \begin{split}\tilde{i}(\tau)&=\left(\frac{d^{2}}{d \tau^{2}}+\frac{1}{\omega_{0}\tau_{L}}\frac{d}{d\tau}+1\right)^{-1}f^{*}(\tau) =\int_{0}^{\infty}e^{-s}\exp\biggl{[}-s\biggl{(}\frac{d^{2}}{d\tau^{2}}+\frac {1}{\omega_{0}\tau_{L}}\frac{d}{d\tau}\biggr{)}\biggr{]}ds\ f^{*}(\tau)=\\ &=\int_{0}^{\infty}e^{-s}\exp\biggl{[}-s\biggl{(}\frac{d^{2}}{d \tau^{2}}\biggr{)}\biggr{]}f^{*}\biggl{(}\tau-\frac{s}{\omega_{0}\tau_{L}} \biggr{)}ds.\end{split}  \tag{64}\]  
In order to make the previous formal solution effective, we use the identity  
\[e^{-s\frac{d^{2}}{d\tau^{2}}}=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}\exp \left(-\xi^{2}+2i\sqrt{s}\frac{d}{d\tau}\xi\right)d\xi \tag{65}\]  
and eventually end up with  
\[\tilde{i}(\tau)=\frac{1}{\sqrt{\pi}}\int_{0}^{\infty}e^{-s}\biggl{(}\int_{- \infty}^{\infty}e^{-\xi^{2}}f^{*}\biggl{(}\tau-\frac{s}{\omega_{0}\tau_{L}}+2 i\sqrt{s}\xi\biggr{)}d\xi\biggr{)}ds \tag{66}\]  
which is easily integrated and the relevant solutions are reported in Figure 12. Alternatively, when the previous solution displays convergence problems, the use of the Fourier-Transform method yields  
\[\tilde{i}(\tau)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\tilde{f}^ {*}(k)e^{ik\tau}}{-k^{2}+\frac{ik}{\omega_{0}\tau_{L}}+1}dk. \tag{67}\]  
In Figure 12 we report the current wave form for \(V^{\prime}(\tau)\propto e^{-\tau^{2}}\) and different values of \(A=\omega_{0}\tau_{L}\). Small \(A\) values determine a strongly damped behavior. For future convenience, we consider a capacitor defined as an operator specified in terms of fractional derivatives as  
\[\hat{C}=C_{0}P^{(\alpha,\gamma)},\qquad\qquad P^{(\alpha,\gamma)}\left(\frac{ d}{dt}\right)=\left(1+\frac{\kappa}{\left(1+\left(\tau\frac{d}{dt}\right)^{1- \alpha}\right)^{1-\gamma}}\right),\qquad\qquad\kappa\ \ constant. \tag{68}\]  
whose physical meaning will be discussed in the forthcoming section and providing  
\[\tilde{i}_{P}(\tau)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\tilde{ f}^{*}(k)e^{ik\tau}}{-k^{2}+\frac{ik}{\omega_{0}\tau_{L}}+\left(P^{(\alpha, \gamma)}(ik)\right)^{-1}}dk. \tag{69}\]  
The effect of this new term on the current wave-form is provided in Figure 13.  
The behavior of the current, generally oscillating, exhibits a critical dependence on the various parameters defining the operator function \(P^{(\alpha,\gamma)}(. )\), whose physical meaning will be discussed in the forthcoming section. In the previous sections we provided a solution to fractional ODE using a technique privileging the FT method. The procedure we have envisaged is useful when non-homogeneous terms are present and does not apply for homogeneous cases. We would like however to underline that the search for a solution of fractional ODE is not an easy task and a general theory is not yet available.