# ILC with Initial State Learning for Fractional Order Linear Distributed Parameter Systems
## 4 Numerical Example  
In this section, two numerical examples are presented to demonstrate the validity of the design method. Consider a tubular flow reactor with a cooling jacket where an irreversible exothermic zeroth order reaction takes place. The reaction rate is considered to be temperature dependent, then it varies with time and the location of the reactor. The cooling jacket and feed temperatures are chosen as the manipulated variables. The thermal spatiotemporal dynamics can be derived from the energy balance inside the reactor and can be presented as the following distributed parameter systems [40]:  
\[ \begin{array}{l}\frac{\partial T}{\partial t}=\frac{k}{\rho C_{p}}\frac{ \partial^{2}T}{\partial z^{2}}-u\frac{\partial T}{\partial z}\\ \quad-\frac{hA_{c}}{\rho C_{p}}T+\frac{(-\Delta H)}{\rho C_{p}}r_{0}e^{(\frac{ -E}{RT})},\end{array}  \tag{65}\]  
where \(T\) is the stream temperature inside the reactor, \(t\) is time and \(z\in[0,L]\) denotes the spatial coordinate, \(L\) is the reactor length. The terms \(k,\ \rho,\ C_{p},\ u,\ (-\Delta H),\ r_{0},\ E\) and \(h\) are used for thermal conductivity, density, heat capacity, axial velocity, heat of reaction, pre-exponential reaction constant, activation energy and the heat transfer coefficient between the reactor and cooling jacket, respectively. \(A_{c}\) presents the cooling jacket temperature. In the vicinity of the steady-state working point, nonlinear Model (65) can be linearized as [41]:  
\[\frac{\partial T(z,t)}{\partial t}=\theta_{1}\frac{\partial^{2}T}{\partial z^{ 2}}+\theta_{2}(t)T(z,t)+\theta_{3}(t)U(z,t), \tag{66}\]  
where \(\theta_{2}(t)\) and \(\theta_{3}(t)\) vary nonlinearly with temperature during the entire heating process. Define the measured value of the thermocouple as:  
\[Y(z,t)=\theta_{4}(t)T(z,t)+\theta_{5}(t)U(z,t). \tag{67}\]  
In order to simplify the example, set the above distributed parameter system with fractional order \(\alpha=0.9\) and:  
\[\theta_{1}=I_{2},\ \theta_{2}(t)=\left[ \begin{array}{cc}e^{-2t}&0 \\ 0&e^{-3t}\end{array} \right],\] \[\theta_{3}(t)=\left[ \begin{array}{cc}e^{-t}&0\\ 0.5&e^{-1.5t}\end{array} \right],\] \[\theta_{4}(t)=\left[ \begin{array}{cc}0.5&1\\ 0.9&2\end{array} \right],\ \theta_{5}(t)=\left[ \begin{array}{cc}0.98&0\\ 1.80&1.5\end{array} \right].\]  
where \((z,t)\in[0,1]\times[0,0.8]\). Let the reference be \(y_{1d}=(1-\cos z)t\), \(y_{2d}=\sin(2\pi^{2}t)\cos(2\pi z)\). For the above system, we apply the following P-type ILC updating law:  
\[u_{k+1}(z,t)=u_{k}(z,t)+\Gamma e_{k+1}(z,t), \tag{68}\]  
where \(\Gamma=\left[ \begin{array}{cc}0.98&0\\ 1.80&1.5\end{array} \right].\) The initial state value is \(q_{0}(z,0)=\sin z\), while the control input value at the beginning of learning is set to zero. By some straightforward calculations, we can find:  
\[\|(I+G(t)\Gamma)^{-1}\|^{2}=0.81>0.50.\]  
That is, the convergence Condition (64) is not satisfied (Theorem 1 in [19] is invalid). However,  
\[\rho\{(I+G(t)\Gamma)^{-1}\}=0.51<1.\]  
From Corollary 3, it can be concluded that the tracking error is convergent. To simulate the fractional order operator, we use the following expression: [42]  
\[\Delta^{\alpha}y(k)=\frac{1}{h^{\alpha}}\sum_{j=0}^{L}(-1)^{j}\binom{\alpha}{ j}y(k-j) \tag{69}\]  
where \(h\) represents the sampling interval and the number of samples is denoted as \(k\). \(\binom{\alpha}{j}\) is the notation of the binomial coefficients. To conserve computing resources, the length of the memory \(L\) is set as a finite recording number [42], usually \(L=20\). To obtain the numerical solution, we use the two order explicit center difference scheme, in which the sampling periods are taken as \(\Delta z=0.1,\ \Delta t=0.001\). From the viewpoint of viscoelastic theory, the initial history (terminal) should be the state over a time-interval, not a state at a given time [43; 44]. To consider the physically consistent initial conditions, a simple solution could be obtained by the procedure proposed in [43; 45].