My question here regards these course notes from MIT’s open courseware program on differential equations: https://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iii-fourier-series-and-laplace-transform/unit-step-and-unit-impulse-response/MIT18_03SCF11_s25_1text.pdf
I'm confused as to why it's stated that the response x(t) to the unit step response u(t) is continuous at all derivatives of x(t). This is explained in examples 1 and 3, but as far as I can tell the proofs assume the very thing that's supposed to be proved.
In example 1, dx/dt = u(t), and therefore it is stated that x(t) = t. But isn’t this leaving out the constant of integration, C1? If we include this constant, and state that x(t)=t+c1, then the value of C1 depends on the initial condition at x(0+). If we assume that x(0+)=0, then of course C1 = 0 and x(0+) = x(0-) = 0, and x(t) is continuous.
But why must we assume that x(0+)=0? Wouldn’t it be just as reasonable to use x(0+)=1? And if we make this assumption, then x(0+)=1 and x(0-)=0, and x(t) is not continuous.
So it seems to me that there is a sleight of hand taking place here.
The same basic argument would apply to the second order differential equation in example 3 as well — if the initial value of x(0+) = 1 (which seems just as reasonable as using x(0+)=0 ), then x(t) is not continuous as there is a jump at x(0), since x(0-)=0 and x(0+)=1.
I can understand why, in a physical/engineering sense, it would make more sense to assume that both x(0+) and x(0-) = 0, since there is a finite rise time for any physical system. But strictly in regards to this math, is this necessary?
As an aside, I have no problems with examples 2 and 4 in the texts, discussing the same thing but with the unit impulse response as the system input, because the way the unit impulse response is defined it doesn’t make sense to add a constant of integration to it. But I don't see why this should be the case with the unit step response.
These kinds of questions have as background that the system is "switched on" at $t=0$, that is, $x(t)=0$ for all $t<0$ which implies that also all derivatives are zero for negative times.
Then $\dot x=u$ has the solution $x(t)=\max(0,t)$ which is indeed continuous and piecewise differentiable.