# Convergent Finite Difference Scheme for Parabolic Equation

Consider the PDE $$u_t = b_{11}u_{xx} + 2b_{12}u_{xy} + b_{22}u_{yy},$$ where $$b_{11}, b_{22} > 0$$, and $$b_{12}^2 < b_1b_2$$. In Strikwerda's book, the ADI scehme \begin{align*} \left( 1 - \frac{k}{2} b_{11} \delta_x^2\right)v^{n+1/2} &= \left( 1 + \frac{k}{2} b_{22} \delta_y^2\right)v^{n} + kb_{12} \delta_{0x} \delta_{0y} v^n \\ \left( 1 - \frac{k}{2} b_{22} \delta_x^2\right)v^{n+1} &= \left( 1 + \frac{k}{2} b_{11} \delta_x^2\right)v^{n+1/2} + kb_{12} \delta_{0x} \delta_{0y} v^{n+1/2} \\ \end{align*} is stated to be convergent. I was able to show that it was consistent, but I am struggling with stability. Making the substitution $$v^{n} = g^n e^{il \theta}e^{i m \phi}$$, and $$v^{n+1/2} = \tilde{g}g^n e^{il \theta}e^{i m \phi}$$, \begin{align*} \left(1 + 2\mu b_{11} \sin^2{ (\theta/2 )} \right) \tilde{g} &= \left(1 - \mu( 2b_{22} \sin^2{ (\theta/2 )} + b_{12}\sin{\theta}\sin{\phi}) \right) \\ \left(1 + 2\mu b_{22} \sin^2{ (\theta/2 )} \right) g &= \left(1 - \mu( 2b_{11} \sin^2{ (\phi/2 )} + b_{12}\sin{\theta}\sin{\phi}) \right)\tilde{g}, \\ \end{align*} where $$\mu = \frac{k}{h^2}$$. This implies that $$g = \frac{\left(1 - \mu( 2b_{11} \sin^2{ (\theta/2 )} + b_{12}\sin{\theta}\sin{\phi}) \right)\left(1 - \mu( 2b_{22} \sin^2{ (\phi/2 )} + b_{12}\sin{\theta}\sin{\phi}) \right)}{\left(1 + 2\mu b_{11} \sin^2{ (\theta/2 )} \right)\left(1 + 2\mu b_{22} \sin^2{ (\phi/2 )} \right)}$$ Ultimately, I want to find some restriction on $$\mu$$ such that $$\left| g \right| \leq 1$$. Since most ADI schemes are unconditionally stable, I expect that this fraction is automatically bounded above by $$1$$, but I'm not sure how to show that. Any help would be much appreciated.