课本中ADI格式(2.4.27)等价于如下单步格式
$$ \begin{aligned} & \left(1-\frac{1}{2}\mu_x\delta_x^2\right) \left(1-\frac{1}{2}\mu_y\delta_y^2\right) \left(1-\frac{1}{2}\mu_z\delta_z^2\right)U_{j,k,l}^{m+1} =\\ & \left(1+\frac{1}{2}\mu_x\delta_x^2\right) \left(1+\frac{1}{2}\mu_y\delta_y^2\right) \left(1+\frac{1}{2}\mu_z\delta_z^2\right)U_{j,k,l}^{m} -\frac{1}{4}\mu_x\mu_y\mu_z\delta_x^2\delta_y^2\delta_z^2U_{j,k,l}^{m} \end{aligned} $$
使用Fourier方法，代入波形 $$U_{j,k,l}^{m} = \lambda_{s}^m \exp\left[i\pi \left(\frac{s_x j}{N_x} + \frac{s_y k}{N_y} + \frac{s_z l}{N_z} \right)\right], $$ 注意到 $$ \begin{aligned} \delta_A^2 U_{j,k,l}^{m} &= U_{j,k,l}^{m}\left( \exp\left(i\frac{\pi s_A}{N_A}\right) -2 + \exp\left(-i\frac{\pi s_A}{N_A}\right) \right)\\ &=U_{j,k,l}^m \left(-4\sin^2\left(\frac{\pi s_A}{2N_A}\right)\right), \qquad A = x,y,z. \end{aligned} $$
所以， $$ \begin{aligned} & \left(1 + 2\mu_x \sin^2\left(\frac{\pi s_x}{2N_x}\right)\right) \left(1 + 2\mu_y \sin^2\left(\frac{\pi s_y}{2N_y}\right)\right) \left(1 + 2\mu_z \sin^2\left(\frac{\pi s_z}{2N_z}\right)\right) \lambda_s \\ =& \left(1 - 2\mu_x \sin^2\left(\frac{\pi s_x}{2N_x}\right)\right) \left(1 - 2\mu_y \sin^2\left(\frac{\pi s_y}{2N_y}\right)\right) \left(1 - 2\mu_z \sin^2\left(\frac{\pi s_z}{2N_z}\right)\right)\\ & + 16\mu_x\mu_y\mu_z \sin^2\left(\frac{\pi s_x}{2N_x}\right) \sin^2\left(\frac{\pi s_y}{2N_y}\right) \sin^2\left(\frac{\pi s_z}{2N_z}\right) \end{aligned} $$
记 \(S_A = \mu_A \sin^2\left(\frac{\pi s_A}{2N_A}\right), A=x,y,z\), 有 $$ \lambda_s = \frac{(1-2S_x)(1-2S_y)(1-2S_z) + 16S_xS_yS_z}{(1+2S_x)(1+2S_y)(1+2S_z)} $$
由于\(S_A \ge 0, A=x,y,z\), 可以验证 \(-1\le \lambda_s \le 1\). 故格式无条件 \(L_2\) 稳定.