## 4 Heat Conduction Models with Mittag-Leffler-Type Memories: Examples
### Heat Flux with Mittag-Leffler (One-Parameter) Memory Kernel  
Now, using the relation (26) we constitute a new relaxation (memory) function for the heat flux [86]  
\[R_{\rm ML}=k_{1}\delta(t)+k_{2}\lambda\frac{E_{\alpha,0}(-\lambda t^{\alpha})} {t}=k_{1}\delta(t)+k_{2}\frac{d}{dt}[E_{\alpha,1}(-\lambda t^{\alpha})] \tag{97}\]  
The second term in (97) is a monotonically decaying function, singular at \(t\to 0\), and thus, obeys all necessary properties to be used as a memory kernel. Then, the heat flux with memory can be expressed as [86]  
\[q(x,t)=-k_{1}\frac{\partial T(x,t)}{\partial x}-k_{2}\lambda\int\limits_{- \infty}^{t}\frac{\partial T(x,s)}{\partial x}\frac{d}{ds}\big{[}E_{\alpha,1} \big{(}\lambda(t-s)^{\alpha}\big{)}\big{]}ds \tag{98}\]  
Now, the integration by parts of the integral (denoting \(F(x,t)=\partial T(x,t)/\partial x\) ) in (98) yields  
\[\begin{split}&\int\limits_{-\infty}^{t}E_{\alpha,1}\big{(}\lambda (t-s)^{\alpha}\big{)}F(x,s)d\lambda s=\left.E_{\alpha,1}\big{(}(t-s)^{\alpha} \big{)}\big{[}F(x,s)-F(x,t)\big{]}\right|_{-\infty}^{t}+\\ &+\int\limits_{-\infty}^{t}E_{\alpha,1}\big{(}\lambda(t-s)^{ \alpha}\big{)}\frac{\partial F(x,s)}{\partial s}d\lambda s\end{split} \tag{99}\]  
The first term in the RHS of (99) is zero, while the second one matches the definition of the Atangana-Baleanu derivative fractional derivative (56).  
The coefficient \(\lambda\in(\infty,0)\) can be mapped by (see the same relations in [66] (see (93) used earlier)\[\lambda=\frac{\alpha}{1-\alpha}\in[0,\infty],\quad\alpha=\frac{1}{1+1/\lambda}= \frac{\lambda}{1+\lambda}\in[0,1] \tag{100}\]  
Then, assuming the lower limit in the Stieltjes integral at zero (the causality principle), we have a new expression for the second term of the heat flux with a memory [86]  
\[q_{e}=-a_{2}\int\limits_{0}^{t}E_{\alpha,1}\biggl{[}-\frac{\alpha}{1-\alpha}(t- s)^{\alpha}\biggr{]}\frac{d}{ds}\biggl{[}\frac{\partial T(x,s)}{\partial x} \biggr{]}ds \tag{101}\]  
We can recast the integral in (101) as the elastic component of the flux \(q_{e}\) because it resembles how the ABC derivative is constructed. [86].  
\[\begin{split}& q_{e}=-a_{2}(1-\alpha)\left\{\frac{1}{1-\alpha} \int\limits_{0}^{t}E_{\alpha,1}\biggl{[}-\frac{\alpha}{1-\alpha}(t-s)^{\alpha} \biggr{]}\frac{d}{ds}\biggl{[}\frac{\partial T(x,s)}{\partial x}\biggr{]}ds \right\}=\\ &=-a_{2}(1-\alpha)^{ABC}D_{t}^{\alpha}\left[\frac{\partial T(x,t) }{\partial x}\right]\end{split} \tag{102}\]  
Consequently  
\[\begin{split}&-\frac{\partial q_{e}}{\partial x}=a_{2}(1-\alpha) \int\limits_{0}^{t}E_{\alpha,1}\biggl{[}-\frac{\alpha}{1-\alpha}(t-s)^{\alpha} \biggr{]}\frac{d}{ds}\biggl{[}\frac{\partial^{2}T(x,s)}{\partial x^{2}} \biggr{]}ds=\\ &=-a_{2}(1-\alpha)^{ABC}D_{t}^{\alpha}\left[\frac{\partial^{2}T( x,t)}{\partial x^{2}}\right]\end{split} \tag{103}\]  
Finally, the energy conservation equation yields [86]  
\[\frac{\partial T(x,t)}{\partial t}=a_{1}\frac{\partial^{2}T(x,t)}{\partial x^ {2}}+a_{2}(1-\alpha)^{ABC}D_{t}^{\alpha}\biggl{[}\frac{\partial^{2}T(x,t)}{ \partial x^{2}}\biggr{]} \tag{104}\]  
For \(\alpha=1\), it reduces to the Fourier equation.  
Based on the fading memory formalisms for the heat flux, we found that the ABC derivative spontaneously manifests itself when a suitable memory kernel (97) is incorporated into the constitutive equation.