MIT微分方程公开课：单位阶跃响应各阶连续性疑问

Great question! Let’s break down why the unit step response ( x(t) ) for linear constant-coefficient differential equation (LCCDE) systems has continuous derivatives, and walk through the full proof step by step — matching the context from your MIT OCW notes.

First, let’s set the stage: we’re looking at the zero-state response (meaning all initial conditions are zero: ( x(0^-) = x'(0^-) = \dots = x^{(n-1)}(0-) = 0 )) of an nth-order LCCDE to a unit step input ( u(t) ):
[
a_n x^{(n)}(t) + a_{n-1} x^{(n-1)}(t) + \dots + a_1 x'(t) + a_0 x(t) = u(t)
]

1. Continuity for ( t \neq 0 ) #

• For ( t < 0 ): The unit step ( u(t) = 0 ), so our equation becomes a homogeneous LCCDE with zero initial conditions. The only solution here is ( x(t) = 0 ) — a constant function where every derivative is also 0, so all derivatives are trivially continuous.
• For ( t > 0 ): ( u(t) = 1 ), a constant. The solution ( x(t) ) is a combination of a constant particular solution (to match the input) and homogeneous solutions (exponential terms, assuming the system is stable with poles in the left half-plane). All these functions are smooth (infinitely differentiable), so every derivative of ( x(t) ) is continuous on ( t > 0 ).

2. Continuity at ( t = 0 ) (First ( n-1 ) Derivatives) #

We can prove this using induction and contradiction:

• Base case (the function itself, ( k=0 )): Suppose ( x(t) ) was discontinuous at ( t=0 ), meaning ( x(0^+) \neq x(0^-) = 0 ). This would make ( x'(t) ) include an impulse function ( (x(0^+) - x(0^-))\delta(t) ). Plugging this into the LCCDE, the left-hand side would have an nth-order impulse term — but the right-hand side ( u(t) ) only has a step (no impulses). This is impossible, so ( x(0^+) = x(0^-) = 0 ), making ( x(t) ) continuous at ( t=0 ).
• Inductive step: Assume all derivatives up to the ( m )-th order (where ( 0 \leq m \leq n-2 )) are continuous at ( t=0 ). Now consider the ( (m+1) )-th derivative: if it were discontinuous, the ( (m+2) )-th derivative would contain an impulse. Again, substituting into the LCCDE would create an impulse term on the left that can’t be canceled by the step input on the right. Thus ( x^{(m+1)}(0+) = x^{(m+1)}(0-) = 0 ), so this derivative is also continuous at ( t=0 ).

By induction, every derivative from the 0th (the function itself) up to the ( (n-1) )-th order is continuous at ( t=0 ).

3. A Quick Note on the nth Derivative #

The nth derivative will have a jump discontinuity at ( t=0 ). To confirm, plug ( t=0^+ ) into the LCCDE:
[
a_n x^{(n)}(0+) + \sum_{k=0}^{n-1} a_k x^{(k)}(0+) = 1
]
Since all lower derivatives are 0 at ( 0^+ ), this simplifies to ( a_n x^{(n)}(0+) = 1 ), so ( x^{(n)}(0+) = 1/a_n ). At ( t=0^- ), ( x^{(n)}(0-) = 0 ), hence the jump.

Your notes mention "all derivatives are continuous" — this is likely a shorthand for all derivatives on ( t \neq 0 ) (where they’re perfectly smooth) or continuity of all derivatives except the nth (which matches examples like second-order systems, where the first derivative is continuous, and the second has a jump).

内容的提问来源于stack exchange，提问作者molecularlionel