## 5 Consistency, Stability, and Convergence
We defined  
\[+K_{1}\frac{u(x_{j+1},t^{n})-u(x_{j-1},t^{n})}{2\Delta x}-K_{2} \frac{u(x_{j+1},t^{n})-2u(x_{j},t^{n})+u(x_{j-1},t^{n})}{\Delta x^{2}}-f(x_{j},t^{n})\]3  
and  
\[+K_{1}\frac{u(x_{j+1},t^{n})-u(x_{j-1},t^{n})}{2\Delta x}-K_{2} \frac{u(x_{j+1},t^{n})-2u(x_{j},t^{n})+u(x_{j-1},t^{n})}{\Delta x^{2}}-f(x_{j},t^{n})\]4  
Equation (16) can be written as  
\[+K_{1}\frac{u(x_{j+1},t^{n})-u(x_{j-1},t^{n})}{2\Delta x}-K_{2} \frac{u(x_{j+1},t^{n})-2u(x_{j},t^{n})+u(x_{j-1},t^{n})}{\Delta x^{2}}-f(x_{j},t^{n})\]5  
From (17), and taking into account that \(T_{n,k+1}-T_{n,k}>0\), we obtained  
\[+K_{1}\frac{u(x_{j+1},t^{n})-u(x_{j-1},t^{n})}{2\Delta x}-K_{2} \frac{u(x_{j+1},t^{n})-2u(x_{j},t^{n})+u(x_{j-1},t^{n})}{\Delta x^{2}}-f(x_{j},t^{n})\]6  
where we defined  
\[+K_{1}\frac{u(x_{j+1},t^{n})-u(x_{j-1},t^{n})}{2\Delta x}-K_{2} \frac{u(x_{j+1},t^{n})-2u(x_{j},t^{n})+u(x_{j-1},t^{n})}{\Delta x^{2}}-f(x_{j},t^{n})\]7  
But, since \(\sum_{k=0}^{n-1}(T_{n,k+1}-T_{n,k})=(\Delta t^{n})^{-\alpha}\) and recalling that \(T_{n,n}=(\Delta t^{n})^{-\alpha}\) and \(T_{n,0}=0\),  
\[+K_{1}\frac{u(x_{j+1},t^{n})-u(x_{j-1},t^{n})}{2\Delta x}-K_{2} \frac{u(x_{j+1},t^{n})-2u(x_{j},t^{n})+u(x_{j-1},t^{n})}{\Delta x^{2}}-f(x_{j},t^{n})\]8  
Thus, we concluded that  
\[+K_{1}\frac{u(x_{j+1},t^{n})-u(x_{j-1},t^{n})}{2\Delta x}-K_{2} \frac{u(x_{j+1},t^{n})-2u(x_{j},t^{n})+u(x_{j-1},t^{n})}{\Delta x^{2}}-f(x_{j},t^{n})\]9 \[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]0i.e.,  
\[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]1  
for \(1\leq n\leq N\). This means that the present method is unconditionally-stable. **Convergence. ** We let \(u(x_{j},t_{n})\) be the exact solution of Equation (1) at mesh point \((x_{j},t^{n})\) for \(j=0,1,\cdots,J\) and \(n=0,1,\cdots,N\). Denoting \(\epsilon_{j}^{n}=u(x_{j},t^{n})-U_{j}^{n}\), we obtained the error equations  
\[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]2  
with \(\epsilon_{j}^{0}=0\), for \(j=0,1,\cdots,J\) and \(\epsilon_{j}^{n}=0\), for \(n=1,2,\cdots,N\). \(R_{j}^{n}\) is the local truncation error. We introduced the following norm  
\[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]3  
then, we obtained  
\[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]4 \[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]5  
where \(R_{max}=\max_{n,i}|R_{i}^{n}|\). For \(n=1\), we obtained  
\[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]6  
Then,  
\[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]7  
To obtain the error estimates of the numerical solutions, we needed the uniform error bounds on all time levels. Then, applying the first of the estimates (5), we were able to consider  
\[= \frac{1}{\Gamma(2-\alpha)}\sum_{k=1}^{n}T_{n,k}(u(x_{j},t^{k})-u (x_{j},t^{k-1}))-\frac{\partial^{\alpha}}{\partial t^{\alpha}}u(x_{j},t^{n})\]8  
where \(C_{R}\) is a positive constant that is dependent on \(T\), \(\alpha\) and the exact solution \(u(x,t)\), but is independent of \(N\) and \(\Delta x\). Thus, we proved that the solution of the difference method (11), with initial and boundary conditions given by (10), is convergent.