Stability of y(t) = x(t²)

y(t) = x(t²)

Linearity

The response to a·x₁ + b·x₂ is a·x₁(t²) + b·x₂(t²) = a·y₁ + b·y₂. The system is linear.

Time invariance

Delay the input: x₁(t) = x(t - t₀) gives y₁(t) = x₁(t²) = x(t² - t₀). Delay the output: y(t - t₀) = x((t - t₀)²). Since t² - t₀ ≠ (t - t₀)², the system is time-varying.

So the system is not LTI, and the LTI stability test through h(t) cannot be applied.

BIBO stability

If |x(t)| ≤ M for all t, then |y(t)| = |x(t²)| ≤ M. A bounded input always gives a bounded output, so the system is BIBO stable.

Causality (for completeness)

For 0 < t < 1, t² < t (a past value is used), but for t > 1, t² > t, so y(t) uses a future value of x; for example, y(2) = x(4). For negative t, t² > t as well, e.g. y(-1) = x(1). So the system is non-causal.

Answer: y(t) = x(t²) is stable (BIBO), but it is not an LTI system (it is linear but time-varying), and it is non-causal.