# 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 ?

• It just means that using the recurrence relation to solve for the PDF is expected to be numerically stable. Commented Oct 15, 2018 at 20:17

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.