Discussion
In order to numerically solve the ordinary differential equation \(\dfrac{dy}{dt} = -y\) for \(t \gt 0\), with an initial condition \(y(0) = 1\), the following scheme is employed
\[ \begin{aligned} \frac{y_{n+1} - y_{n}}{\Delta t} \\ &= -\frac{1}{2}(y_{n+1} + y_{n}). \end{aligned} \]
Here, \(\Delta t\) is the time step and \(y_{n} = y(n\Delta t)\) for \(n = 0, 1, 2, \ldots\). This numerical scheme will yield a solution with non-physical oscillations for \(\Delta t \gt h\). The value of \(h\) is
Answer: D
The value of h is 2.
The scheme is the trapezoidal (Crank-Nicolson) rule. Solve it for \(y_{n+1}\):
\[ \begin{aligned} y_{n+1}\left(1 + \frac{\Delta t}{2}\right) &= y_{n}\left(1 - \frac{\Delta t}{2}\right) \\ y_{n+1} &= \frac{1 - \Delta t/2}{1 + \Delta t/2}\,y_{n} \end{aligned} \]
The exact solution \(e^{-t}\) decays without changing sign. The amplification factor is always smaller than 1 in size, so the scheme is stable, but it becomes negative when \(1 - \Delta t/2 \lt 0\), that is for \(\Delta t \gt 2\). The numerical solution then alternates in sign from step to step, a non-physical oscillation. So \(h = 2\); 1/2, 1 and 3/2 all still give a positive factor.
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