Well, I have the advection equation (with $a>0$) $$\frac{\partial u}{\partial t} + a\frac{\partial u}{\partial x} = 0, \ \ \ \textrm{for} \ \ \left\{ \begin{array}{l} -\infty \leq x \leq \infty, \\ 0 <t, \end{array}\right.$$
The Fromm method for solving the advection equation is
$u_{i,j+1}= \frac{1}{4}\lambda(\lambda-1)u_{i-2,j}+ \frac{1}{4}\lambda(5-\lambda)u_{i-1,j}- \frac{1}{4}(\lambda-1)(\lambda+4)u_{i,j}+ \frac{1}{4}\lambda(\lambda-1)u_{i+1,j}$
Where $\lambda=\frac{ak}{h}$, $k$ is the time increment and $h$ the space increment.
To check if this method is stable or not, I use a solution like $u_{i,j} = w_je^{rx_iI}$ (where $I = \sqrt{-1}$)
And I'm doing the following $$w_{j+1}e^{rx_iI}= \frac{1}{4}\lambda(\lambda-1)w_je^{rx_{i-2}I}+ \frac{1}{4}\lambda(5-\lambda)w_je^{rx_{i-1}I} - \frac{1}{4}(\lambda-1)(\lambda+4)w_je^{rx_iI} + \frac{1}{4}\lambda(\lambda-1)w_je^{rx_{i+1}I}$$ $$w_{j+1}e^{rx_iI}=w_j\Bigg[ \frac{1}{4}\lambda(\lambda-1)e^{r(x_i-2h)I}+ \frac{1}{4}\lambda(5-\lambda)e^{r(x_i-h)I} - \frac{1}{4}(\lambda-1)(\lambda+4)e^{rx_iI} + \frac{1}{4}\lambda(\lambda-1)e^{r(x_i+h)I}\Bigg]$$ $$w_{j+1}=w_j\Bigg[ \frac{1}{4}\lambda(\lambda-1)e^{-2rhI}+ \frac{1}{4}\lambda(5-\lambda)e^{-rhI} - \frac{1}{4}(\lambda-1)(\lambda+4) + \frac{1}{4}\lambda(\lambda-1)e^{rhI}\Bigg]$$ $$4\frac{w_{j+1}}{w_j}=\lambda(\lambda-1)e^{-2rhI} + \lambda(5-\lambda)e^{-rhI} - (\lambda-1)(\lambda+4) + \lambda(\lambda-1)e^{rhI}$$
And I don't really know what should I do now... I have to check if $|\frac{w_{j+1}}{w_j}| \leq 1$, any idea how to process?