Question 10.7: CRANK–NICOLSON METHOD; 1D HEAT EQUATION Consider the tempera...

With r=0.4, Equation 10.32

2(1+r) u_{i, j+1}-r\left(u_{i-1, j+1}+u_{i+1, j+1}\right)=2(1-r) u_{i j}+r\left(u_{i-1, j}+u_{i+1, j}\right), \quad r=\frac{k \alpha^2}{h^2}     (10.32)

is written as

2.8 u_{i, j+1}-0.4\left(u_{i-1, j+1}+u_{i+1, j+1}\right)=1.2 u_{i j}+0.4\left(u_{i-1, j}+u_{i+1, j}\right)       (10.33)

Applying Equation 10.33 at the j = 0 level, we find

\begin{aligned} & 2.8 u_{11}-0.4\left(u_{01}+u_{21}\right)=1.2 u_{10}+0.4\left(u_{00}+u_{20}\right) \\ & 2.8 u_{21}-0.4\left(u_{11}+u_{31}\right)=1.2 u_{20}+0.4\left(u_{10}+u_{30}\right) \\ & 2.8 u_{31}-0.4\left(u_{21}+u_{41}\right)=1.2 u_{30}+0.4\left(u_{20}+u_{40}\right) \end{aligned}

Substituting the values from boundary and initial conditions, yields

\left[\begin{array}{ccc} 2.8 & -0.4 & 0 \\ -0.4 & 2.8 & -0.4 \\ 0 & -0.4 & 2.8 \end{array}\right]\left\{\begin{array}{l} u_{11} \\ u_{21} \\ u_{31} \end{array}\right\}=\left\{\begin{array}{l} 12.4853 \\ 17.6569 \\ 12.4853 \end{array}\right\} \quad \begin{array}{ll} \stackrel{\stackrel{\text { Solve }}{\Rightarrow}}{\text { tridiagonal system }} & u_{11}=5.5880=u_{31} \\ & u_{21}=7.9026 \end{array}

Next, Equation 10.33 is applied at the j=1 level:

\left[\begin{array}{ccc} 2.8 & -0.4 & 0 \\ -0.4 & 2.8 & -0.4 \\ 0 & -0.4 & 2.8 \end{array}\right]\left\{\begin{array}{l} u_{12} \\ u_{22} \\ u_{32} \end{array}\right\}=\left\{\begin{array}{c} 9.8666 \\ 13.9535 \\ 9.8666 \end{array}\right\} \quad \underset{\text{tridiagonal system }}{\stackrel{ \text { Solve }}{\Rightarrow}} \quad \begin{array}{l} u_{12}=4.4159=u_{32} \\ u_{22}=6.2451\end{array}

This procedure is repeated until the approximate values of u at the mesh points along the j=5 row are calculated. Execution of the user-defined function Heat1DCN yields