Shifted Lagrangian Jacobi collocation scheme for numerical solution of a model of HIV infection

Abstract

In this paper, a system of nonlinear ordinary differential equations (NODEs), namely the equation model of the human immunodeficiency virus (HIV) infection of $CD4^+ T$ cells, is studied. Our approach is implemented by using the Shifted-Lagrangian Jacobi (SLJ) polynomials formed by Shifted-Jacobi-Gauss-Radau (SJ-GR) points. In a new insight, by applying Quasilinearization method (QLM) the system of NODE’s is simplified and changed into a system of Linear ordinary differential equations (LODE’s) and instead of working on a system of NODE’s, all processes and works are done on a system of LODE’s. Therefore, unlike the most of the current studies working on nonlinear algebraic equations, the problem is reduced to a system of linear algebraic equations. Then, to solve the problem and find the unknown approximation coefficients, a system of $Ax=b$ has been solved. At the end, the accuracy and reliability of this method are shown and comparisons with the other current work’s results are made.

Appendix

In this Appendix we are going to apply QLM to equations in (1). In first equation in (1), we are coping with

$$\begin{aligned} \frac{dT}{dx}= & {} c_1-c_2T(x)+c_3T(x)\left( 1-\frac{T(x)+I(x)}{T_{Max}}\right) -c_4V(x)T(x),\\ \frac{dT}{dx}= & {} c_1-c_2T(x)+c_3T(x) -c_3\frac{T^2(x)}{T_{Max}}-c_3\frac{T(x)I(x)}{T_{Max}}-c_4V(x)T(x). \end{aligned}$$

Applying the QLM, we will have

$$\begin{aligned} \frac{dT_{i+1}(x)}{dx}= & {} \underbrace{c_1-c_2T_{i}(x)+c_3T_{i}(x) -c_3(x)\frac{T_{i}^2(x)}{T_{Max}}-c_3\frac{T_{i}(x)I_{i}(x)}{T_{Max}}-c_4V_{i}(x)T_{i}(x)}_{A_1(x)}\nonumber \\&+\big (T_{i+1(x)}-T_{i}(x)\big ).\left( -c_2+c_3-c_3\frac{2T_{i}(x)}{T_{Max}}-c_3\frac{I_{i}(x)}{T_{Max}}-c_4V_{i}(x)\right) .\nonumber \\ \end{aligned}$$

(25)

All the i-th step forms are known. So then we can show these forms as

$$\begin{aligned} \frac{dT_{i+1}(x)}{dx}= & {} A_1(x)+\big (T_{i+1}(x)-T_{i}(x)\big ).\left( -c_2+c_3-c3\frac{2T_{i}(x)}{T_{Max}}-c_3\frac{I_{i}(x)}{T_{Max}}-c_4V_{i}(x)\right) ,\\ \frac{dT_{i+1}(x)}{dx}= & {} A_1(x)+T_{i+1}(x)\left( -c_2+c_3-c_3\frac{2T_{i}(x)}{T_{Max}}-c_3\frac{I_{i}(x)}{T_{Max}}-c_4V_{i}(x)\right) \\&\quad -\underbrace{T_{i}(x).\left( -c_2+c_3-c3\frac{2T_{i}(x)}{T_{Max}}-c_3\frac{I_{i}(x)}{T_{Max}}-c_4V_{i}(x)\right) }_{A_2(x)},\\ \frac{dT_{i+1}(x)}{dx}= & {} A_1(x)+T_{i+1}(x)\underbrace{\left( -c_2+c_3-c3\frac{2T_{i}(x)}{T_{Max}}-c_3\frac{I_{i}(x)}{T_{Max}}-c_4V_{i}(x)\right) }_{A_3(x)}+A_2(x),\\ \frac{dT_{i+1}(x)}{dx}= & {} A_3(x)T_{i+1}(x)+\underbrace{A_1(x)+A_2(x)}_{A_4(x)},\\ \frac{dT_{i+1}(x)}{dx}= & {} A_3(x)T_{i+1}(x)+A_4(x). \end{aligned}$$

Thus, at the end

$$\begin{aligned} A_3T_{i+1}(x)+A_4(x)-\frac{dT_{i+1}(x)}{dx}=0. \end{aligned}$$

(26)

Similarly, for the second and third equations in (1), we do the same

$$\begin{aligned} \frac{dI(x)}{dx}= & {} c_4V(x)T(x)-c_5I(x), \nonumber \\ \frac{dI_{i+1}(x)}{dx}= & {} \underbrace{ c_4V_{i}(x)T_{i}(x)-c_5I_{i}(x)}_{B_1(x)}+\big (I_{i+1}(x)-I_{i}(x)\big )(-c5), \nonumber \\ \frac{dI_{i+1}(x)}{dx}= & {} B_1(x)+(I_{i+1}(x)-I_{i}(x))(-c5), \nonumber \\ \frac{dI_{i+1}(x)}{dx}= & {} B_1(x)+I_{i+1}(x)(-c5)-\underbrace{I_{i}(x)(-c5)}_{B_2(x)}, \nonumber \\ \frac{dI_{i+1}(x)}{dx}= & {} B_1(x)+I_{i+1}(x)\underbrace{(-c5)}_{B_3(x)}+B_2(x), \nonumber \\ \frac{dI_{i+1}(x)}{dx}= & {} \underbrace{B_1(x)+B_2(x)}_{B_4(x)}+B_3(x)I_{i+1}(x), \nonumber \\ \frac{dI_{i+1}(x)}{dx}= & {} B_3(x)I_{i+1}(x)+B_4(x). \end{aligned}$$

$$\begin{aligned} B_3(x)I_{i+1}(x)+B_4(x)-\frac{dI_{i+1}(x)}{dx}=0. \end{aligned}$$

(27)

For the third equation,

$$\begin{aligned} \frac{dV}{dx}= & {} Nc_5I(x)-c_6V(x), \\ \frac{dV_{i+1}(x)}{dx}= & {} \underbrace{Nc_5I_{i}(x)-c_6V_{i}(x)}_{C_1(x)}+(V_{i+1}(x)-V_{i}(x))(-c_6), \\ \frac{dV_{i+1}(x)}{dx}= & {} C_1(x)+(V_{i+1}(x)-V_{i}(x))(-c_6), \\ \frac{dV_{i+1}(x)}{dx}= & {} C_1(x)+V_{i+1}(x)(-c_6)-\underbrace{V_{i}(x)(-c_6)}_{C_2(x)}, \\ \frac{dV_{i+1}(x)}{dx}= & {} C_1(x)+V_{i+1}(x)\underbrace{(-c_6)}_{C_3(x)}+C_2(x), \\ \frac{dV_{i+1}(x)}{dx}= & {} \underbrace{C_1(x)+C_2(x)}_{C_4(x)}+C_3(x)V_{i+1}(x), \\ \frac{dV_{i+1}(x)}{dx}= & {} C_3(x)V_{i+1}(x)+C_4(x). \end{aligned}$$

$$\begin{aligned} C_3(x)V_{i+1}(x)+C_4(x)-\frac{dV_{i+1}(x)}{dx}=0. \end{aligned}$$

(28)