<Forward Euler method> gives
$$
u_m^{n+1}=u_m^n
+r(u_{m-1}^n-2u_m^n+u_{m+1}^n)
+s(u_{m+1}^n-u_{m-1}^n),
$$
where
$$
r=\frac{\Delta t}{\Delta x^2},
\qquad
s=\frac{\alpha\Delta t}{2\Delta x}.
$$
Its <amplification factor> is
$$
G(\vartheta)=1-4r\sin^2\frac\vartheta2
+i\frac{\alpha\Delta t}{\Delta x}\sin\vartheta.
$$
Writing $X=\sin^2(\vartheta/2)$, the condition $|G|^2\leq1$ for every $0\leq X\leq1$ is equivalent to
$$
r\leq\frac12,
\qquad
\left(\frac{\alpha\Delta t}{\Delta x}\right)^2\leq2r.
$$
Therefore
$$
\boxed{
0\leq\Delta t\leq
\min\left\{\frac{\Delta x^2}{2},\frac2{\alpha^2}\right\}
}
$$
with the second bound omitted when $\alpha=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 341 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-341/4/c/solution