Question

The initial value problem \[ \frac{dy}{dt} + 2y = 0, y(0) = 1 \] is solved numerically using the forward Euler's method with a constant and positive time step of \(\Delta t\). Let \(y_n\) represent the numerical solution obtained after n steps. The condition \(|y_{n+1}| \le |y_n|\) is satisfied if and only if \(\Delta t\) does not exceed ................ (Answer in integer)

Hint:

For a general test equation \(\frac{dy}{dt} = -\lambda y\) with \(\lambda>0\), the stability limit for the forward Euler method is always \(|1 - \lambda \Delta t| \le 1\), which simplifies to \(\Delta t \le 2/\lambda\). In this problem, \(\lambda=2\), so the limit is \(\Delta t \le 2/2 = 1\). This is a useful formula to remember for stability analysis.

Answer 1

Step 1: Understanding the Concept:
The problem asks for the stability condition of the forward Euler method applied to a simple first-order ODE. The condition \(|y_{n+1}| \le |y_n|\) means that the numerical solution should not grow in magnitude from one step to the next, which is a condition for numerical stability, especially for an equation whose true solution decays.
Step 2: Key Formula or Approach:
1. Forward Euler's Method: For an ODE of the form \(\frac{dy}{dt} = f(t, y)\), the forward Euler update rule is: \[ y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n) \] 2. Apply to the given ODE: Rearrange the ODE to the standard form: \(\frac{dy}{dt} = -2y\). So, \(f(t, y) = -2y\).
3. Derive the update rule for this specific problem: Substitute \(f(t_n, y_n) = -2y_n\) into the Euler formula.
4. Apply the stability condition: Use the derived update rule to analyze the condition \(|y_{n+1}| \le |y_n|\) and find the constraint on \(\Delta t\).
Step 3: Detailed Calculation:
1. The ODE is \(\frac{dy}{dt} = -2y\).
2. The forward Euler update rule is: \[ y_{n+1} = y_n + \Delta t \cdot (-2y_n) \] \[ y_{n+1} = y_n (1 - 2\Delta t) \] 3. Now, apply the given stability condition: \[ |y_{n+1}| \le |y_n| \] Substitute the expression for \(y_{n+1}\): \[ |y_n (1 - 2\Delta t)| \le |y_n| \] \[ |y_n| |1 - 2\Delta t| \le |y_n| \] Assuming \(y_n \neq 0\), we can divide by \(|y_n|\): \[ |1 - 2\Delta t| \le 1 \] 4. Solve the inequality for \(\Delta t\). The inequality \(|x| \le a\) is equivalent to \(-a \le x \le a\). \[ -1 \le 1 - 2\Delta t \le 1 \] This gives two separate inequalities: - First inequality: \(-1 \le 1 - 2\Delta t\) \[ 2\Delta t \le 1 + 1 \] \[ 2\Delta t \le 2 \implies \Delta t \le 1 \] - Second inequality: \(1 - 2\Delta t \le 1\) \[ -2\Delta t \le 0 \implies 2\Delta t \ge 0 \implies \Delta t \ge 0 \] 5. Combine the results: We have \(0 \le \Delta t \le 1\).
The problem states that \(\Delta t\) is a positive time step, so \(\Delta t>0\).
The condition is satisfied if and only if \(0<\Delta t \le 1\).
Therefore, \(\Delta t\) must not exceed 1.
Step 4: Final Answer:
The condition is satisfied if and only if \(\Delta t\) does not exceed 1.
Step 5: Why This is Correct:
The solution correctly applies the forward Euler formula and the condition for numerical stability. The resulting inequality for \(\Delta t\) is solved correctly. For this specific decaying exponential problem, a time step greater than 1 would cause the numerical solution to oscillate with increasing amplitude, which is unstable and physically incorrect.