heat equation stability condition random walk I consider the heat equation : $$\frac{\partial f}{\partial t}(t,x) = \frac{\partial^2f}{\partial x^2}(t,x)$$ on ${\Bbb R}_+^* \times [0,1]$.
One can proove that the explicit Euler scheme is stable iff $\frac{\tau}{\delta^2} \le \frac{1}{2}$ with $\tau =$ time step  and $\delta=$ spatial step.
We can modelize the heat equation by the deplacement of a particule on ${\Bbb Z}$. 
Let $(X_n)_{n\ge 1}$ random variables with $P(X_n=-1)=1/2$ and $P(X_n=1)=1/2$.
Define $S_n = \sum_{k=1}^nX_k$. 
$\delta S_n$ is the position of the particule after time $n\tau$. 
Let $p_{n}(k) = P(S_n=k)$.
We can proove $$\frac{p_{n+1}(k)-p_n(k)}{\tau} = \frac{\delta^2}{2\tau}\frac{p_{n}(k+1)-2p_n(k)+p_n(k-1)}{\delta^2}$$
(We recognize the Euler explicit scheme)
Question : what does the stability condition ($\frac{\tau}{\delta^2} \le \frac{1}{2}$) mean in the context of the random walk ?
 A: I don't know how your equation relates to probability but when you use the mesh the parameter $s$ you get
$$ s =\frac{k \Delta t}{(\Delta x)^{2}}  \tag{1}$$
is the same parameter you are talking about. Now if you use matrix notation when you solve this, you end up with a tridiagonal matrix. 
$$ A = \begin{bmatrix} 1-2s & s & 0 & 0 & 0 & 0 & 0 \\   s & 1-2s & s & 0 & 0 & 0 & 0 \\ 0 & s & 1-2s & 0 & 0 & 0 & 0 \\ 0  & 0 & \cdots & \cdots & \cdots & 0 & 0 \\ 0 & 0 & 0 & s & 1-2s & s & 0 \\ 0 & 0 & 0 & 0 & 0 & s & 1-2s \\\end{bmatrix} \tag{2}$$
which you can write like this
$$ u^{m+1} = Au^{m} \\ u^{1} = Au^{0} \\ u^{2} = Au^{1}  = A^{2}u^{0} \tag{3} $$
right then you can work out this 
$$ u^{m} = \sum_{n=1}^{N-1} c_{n}^{0} (\mu_{n})^{m} \xi_{n} \tag{4} $$
then the change of the solution depends on $\mu_{n}$ 
$$ (\mu_{n})^{m} =\begin{align}\begin{cases} \textrm{ explosive growth } &  \mu_{n} > 1 \\ \textrm{ exponential decay }  &  0 < \mu_{n} < 1   \\ \textrm{ convergent oscillation }  &-1 < \mu_{n}  < 0  \\ \textrm{divergent oscillation }  & \mu_{n} < -1 \end{cases} \end{align} \tag{5}$$
where $m = \frac{t}{\Delta t}$
So this is unstable if $\mu_{n} > 1 $ or $\mu_{n} < -1$  going further.
$$ A \xi = \mu \xi \tag{6} $$
$$ s \xi_{j+1} + (1-2s)\xi_{j} + s\xi_{j-1} = \mu\xi_{j} \tag{7}  $$
$$ \xi_{j+1} + \xi_{j-1} = \bigg( \frac{\mu+2s -1}{s}\bigg)\xi_{j} \tag{8} $$
We then see the eigenvalues $\mu$ of $A$ are the eigenvalues $\lambda$ obtained from the following equation
$$ \mu = 1-2s(1-\cos(\alpha \Delta x)) \tag{9} $$
where $\alpha = \frac{n\pi}{L}$
This is Gerschgorins circle theorem
$$| \mu - a_{ii} |\leq \sum_{j=1}^{N-1} |a_{ij} |  \tag{10}$$
$$ | \mu -(1-2s) | < 2s \tag{11}  $$
then the eigenvalues lie in the resulting region
$$ 1-4s \leq \mu \leq 1 \tag{12} $$
which gives stability when 
$$ -1 \leq \mu \leq 1 \tag{13} $$
but this is only stable if $s \leq \frac{1}{2}$. If $s >\frac{1}{2}$ it doesn't imply it is unstable. 
