# Finite Difference Method Stability with diffusion equation

The diffusion equation is:

$\frac{\partial T}{\partial t} = \alpha \left( \frac{\partial^2 T}{\partial x^2} \right)$

An explicit finite difference approach can be used to solve this, forward in time and central differences in space. Approximating the diffusion equation at a node i, yields,

$\frac{T_i^{n+1}-T_i^n}{\Delta t} = \alpha\frac{T_{i+1}^n-2T_i^n+T_{i-1}^n}{\Delta x^2}$

which gives

$T_i^{n+1} = T_i^n(1-\omega)+\omega(0.5T_{i+1}^n+0.5T_{i-1}^n)$

where $\omega = \frac{2\alpha \Delta t}{\Delta x^2}$

The book states that for stability condition, the coefficients of the right-hand side terms must be positive which implies that

$\Delta t \lt \frac{\Delta x^2}{2\alpha}$

My question is why should the coefficients of the right-hand side terms be positive for stability?

$$\frac{T_i^{n+1}-T_i^n}{\Delta t} = \alpha\frac{T_{i+1}^n-2T_i^n+T_{i-1}^n}{\Delta x^2}$$ Starting with the above and collect terms as you did $$T_i^{n+1} = T_i^n(1-\omega) + \omega(0.5T_{i+1}^n + 0.5T_{i-1}^n)$$ we use a plane wave solution for the stability of $T_i^n = T_0\mathrm{e}^{at+ikx}$ leads to $$T_0\mathrm{e}^{a(t + \Delta t)+ikx} = T_0\mathrm{e}^{at+ikx} (1-\omega) +\omega(0.5T_0\mathrm{e}^{at+ik(x+\Delta x)} + 0.5T_0\mathrm{e}^{at+ik(x-\Delta x)})$$ Dividing through by $T_i^n$ leads to  \mathrm{e}^{a\Delta t} = (1-\omega) + \omega(0.5\mathrm{e}^{ik\Delta x}+0.5\mathrm{e}^{-ik\Delta x}) = (1-\omega) +\omega\cos(k\Delta x) = 1 - \omega\left[2\cos^2\left(\frac{k\Delta x}{2}\right)\right] $$since we require the growth to be bounded i.e. \vert \mathrm{e}^{a\Delta t} \vert < 1 therefore$$ \vert 1 - \omega\left[2\cos^2\left(\frac{k\Delta x}{2}\right)\right]\vert < 1\implies 0< 2\omega < 2 $$finally reaching$$ 0 < \frac{2\alpha \Delta t}{(\Delta x)^2} < 1 $$and you finally reach your condition$$ 0 < \Delta t < \frac{(\Delta x)^2}{2\alpha}