Question 10.8: CRANK–NICOLSON VERSUS FINITE-DIFFERENCE Consider the tempera...
CRANK-NICOLSON VERSUS FINITE-DIFFERENCE
Consider the temperature distribution problem studied in Examples 10.6 and 10.7. Assume h=0.25.
1. Use the \mathrm{CN} method with r=1 to find the temperature at the mesh points u_{11}, u_{21}, and u_{31} in the first time row.
2. Since r=1 does not satisfy the condition of r \leq \frac{1}{2}, pick r=0.25, for example, and h=0.25 as before, and apply the FD method to find the values at the points u_{11}, u_{21}, and u_{31} in (1). Considering the number of time steps has quadrupled, decide whether FD generates more accurate results than \mathrm{CN}.
1. We first note that
r=\frac{k \alpha^{2}}{h^{2}}=1 \underset{h=0.25}{\stackrel{\alpha=0.5}{\Rightarrow}} k=r\left(\frac{h}{\alpha}\right)^{2}=0.25
With r=1, 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)
becomes
4 u_{i, j+1}-u_{i-1, j+1}-u_{i+1, j+1}=u_{i-1, j}+u_{i+1, j}
Applying this equation with j=0, we find
\begin{gathered} 4 u_{11}-0-u_{21}=0+10 \\ 4 u_{21}-u_{11}-u_{31}=2(7.0711) \\ 4 u_{31}-u_{21}-0=10+0 \end{gathered} \Rightarrow\left[\begin{array}{ccc} 4 & -1 & 0 \\ -1 & 4 & -1 \\ 0 & -1 & 4 \end{array}\right]\left\{\begin{array}{l} u_{11} \\ u_{21} \\ u_{31} \end{array}\right\}=\left\{\begin{array}{c} 10 \\ 14.1422 \\ 10 \end{array}\right\} \Rightarrow \begin{aligned} & u_{11}=u_{31}=3.8673 \\ & u_{21}=5.4692 \end{aligned}
2. We note that
r=\frac{k \alpha^{2}}{h^{2}}=0.25 \underset{h=0.25}{\stackrel{\alpha=0.5}{\Rightarrow}} k=r\left(\frac{h}{\alpha}\right)^{2}=0.0625
Therefore, the step size along the t-axis has been reduced from k=0.25 to k=0.0625, implying that four time-step calculations are required to find the values of u_{11}, u_{21}, and u_{31} in (1). With r=0.25, Equation 10.28
u_{i, j+1}=(1-2 r) u_{i j}+r\left(u_{i-1, j}+u_{i+1, j}\right), \quad r=\frac{k \alpha^2}{h^2} (10.28)
reduces to
u_{i, j+1}=0.5 u_{i j}+0.25\left(u_{i-1, j}+u_{i+1, j}\right)
Proceeding as always, the solution estimates at the desired mesh points will be calculated. It should be mentioned that with the new, smaller step size, k=0.0625, the old u_{11}, u_{21}, and u_{31} in (1) are now labeled u_{14}, u_{24}, and u_{34}. The computed values are
\begin{aligned} & u_{14}=u_{34}=3.7533 \\ & u_{24}=5.3079 \end{aligned}
The numerical results obtained in (1) and (2) are summarized in Figure 10.14 where it is readily observed that although FD used four times as many time levels as \mathrm{CN}, the accuracy of the results by \mathrm{CN} is still superior.
