Question

The ordinary differential equation \( \frac{dy}{dt} = -\pi y \) subject to an initial condition \( y(0) = 1 \) is solved numerically using the following scheme:
\[ \frac{y(t_{n+1}) - y(t_n)}{h} = -\pi y(t_n) \] \text{where } \( h \) is the time step, \( t_n = nh \), and \( n = 0, 1, 2, \dots \). \text{This numerical scheme is stable for all values of } h \text{ in the interval} ________.

• A. \( 0<h<\frac{2}{\pi} \)
• B. \( 0<h<1 \)
• C. \( 0<h<\frac{\pi}{2} \)
• D. for all \( h>0 \)

Hint:

When using the explicit Euler method for solving differential equations, ensure the step size \( h \) satisfies the stability condition to avoid numerical instability.

Answer 1

The Correct Option is A

The given ordinary differential equation \( \frac{dy}{dt} = -\pi y \) is solved numerically using an explicit Euler method: \[ y(t_{n+1}) = y(t_n) + h \cdot (-\pi y(t_n)) = y(t_n) - h \pi y(t_n) \] The stability of the numerical scheme depends on the choice of the time step \( h \). For this scheme to be stable, we require that the numerical solution does not grow unbounded as the iterations proceed. The stability condition for the explicit Euler method is given by: \[ |1 - h\pi|<1 \] Solving this inequality, we get: \[ -1<1 - h\pi<1 \quad \Rightarrow \quad 0<h<\frac{2}{\pi} \] Thus, the scheme is stable for \( h \) in the interval \( 0<h<\frac{2}{\pi} \). Final Answer: (A)