Why is this unit step response continuous? This is from MIT 18.03SC Differential Equation, the lecture note on Initial Conditions enter link description here . In example 1, the IVP is given as
$\dot x = u(t), \quad x(0^-) = 0$
Apply Laplace transform and solve for $X(S)$
$ SX -Sx(0) = 1/S$
$ SX = 1/S $
$ X=1/S^2 $
Taking the inverse Laplace transform gives
$$ x = tu(t) $$
From previous lectures, the convention is that the unit step $u(t)$ is undefined at $t=0$. So, the result by cases is
$$ x = 0;\ t<0 \quad and \quad x = t;\ t>0$$
Note that the piecewise response doesn't include $t = 0$ in any of its cases. In the notes, it was written that because the pre-initial condition and post initial condition are equal $x(0^-) = x(0^+) = 0 $, so the response x(t) must be continuous. However, I think this contradicts the result where at t = 0 the response $x=tu(t)$ is undefined. Can anyone tell me how can the response be continuous at $t = 0 $ if $x(0)$ is undefined?
 A: As your right side is not continuous, you get not a simple ODE solution but some generalization. In putting together the parts for $t<0$ and $t>0$ where the equation is indeed a regular ODE the most simple continuation condition is continuity in $t=0$.
In short, this is not a result of some calculation but a demand on what a solution of this problem is.
A: Strictly speaking, if $x(0)$ is undefined, then $x$ can indeed not be continuous at $0$.
However, $u(t)$ is actually often conventionally defined at $0$, although possibly at different values. A common convention is to define it as $u(0)=\frac 12$, although $0$ or $1$ are also used.
Whichever finite value we pick, when we multiply it with $0$, it will be $0$. And if we define our solution to be $0$ at $0$, it is continuous after all. That is, we can make it continuous. It's called a removable discontinuity.
A: It is not just $u(0)$ which is not defined, but the whole differential equation, $\dot x(0)$ and $x(0)$.
Hence they solve left and right independently and impose that the solution be continuous at $0$, by requiring
$$\lim_{t\to 0^-}x(t)=\lim_{t\to 0^+}x(t)$$ and $$x(0):=\lim_{t\to 0}x(t).$$
