Professor told me that solving:
$\ddot{\theta}+\dot{\theta} +f(\theta) = c_0 \times \delta(t)$
can be done by taking a limit
$\lim_{e \to 0} \int_{-e}^{e} \left ( \ddot{\theta}+\dot{\theta} +f(\theta) \right) dt = c_0\lim_{e \to 0} \delta(t)$, $$\lim_{e \to 0} \int_{-e}^{e} \left ( \ddot{\theta}+\dot{\theta} +f(\theta) \right) dt = c_0,$$ (where in the right part dirac - delta function goes to 1.)
But how can I receive on the left part of the equation?:
$[\dot\theta]$, so that
$[\dot\theta] = c_0.$ <- it was said that this is the answer.
Here is professors handwriting:
Only thing I can guess is that, and it is only a guess, that
$\lim_{e \to 0} \int_{-e}^{e} f(\theta) = 0$
Dear grdgfgr Thank you. Still some unclear questions: 1) what do you mean by writing wrt t? 2) what is the dirac delta derivative? Does it exist ? If I look on the definition, I see that it equals infinity at 0, and 0 else, that is why I suppose it is not defined.. 3) I did not understand how you say that the jump can be only in $\ddot{\theta}$ case. BUT! While writing this question I think, I have realized, could you be pleased, to say am I right?
I got that $\dot{\theta(e)} - \dot{\theta(-e)}$ or $\theta(e) - > \theta(-e)$ or $F(\theta(e)) - F(\theta(-e))$ should have somewhere the value $c_0$ to be consistent with LHS.
If I look at the condition that $e \to 0$, I understand that all this 3 differences are going to be equal 0. But I go from the last one. I assume that function is smooth enough therfore $F(\theta(e)) - > F(\theta(-e)) = 0$, as well as $\theta(e) - \theta(-e) = 0 $, however, what I think is:
differnce $\dot{\theta(e)} - \dot{\theta(-e)}$ can be the jump, because, when I have a saddle point, or a peak, the values of the derivatives of some variable at $e$ and $-e$ positions can be very different especially in 2 close positions. Is this the logic, why you say that the jump can be only in $\dot{\theta}$?
I don't like this question but I think this is what is supposed to happen:
$$\ddot{\theta}(t)+\dot{\theta}(t) +f(\theta(t)) = c_0 \times \delta(t)$$
Integrate both sides wrt t
$$\int^{\epsilon}_{-\epsilon}\bigg(\ddot{\theta}(t)+\dot{\theta} (t)+f(\theta(t))\bigg) dt=\int^{\epsilon}_{-\epsilon} c_0 \times \delta(t)dt$$
Integral of $\delta$ is $1$. $F(\theta(t))$ is the antiderivative of $f(\theta(t))$. We get:
$$\dot{\theta}(\epsilon)-\dot{\theta}(-\epsilon)+\theta(\epsilon)-\theta(-\epsilon)+F(\theta(\epsilon))-F(\theta(-\epsilon))=c_0$$
Now since $\epsilon$ is so small, normally the left side should have been 0. But this indicates one of those functions must have a jump. Jump has to be caused by a delta in $\ddot{\theta}(t)$ since if it were caused by, for example, $\dot{\theta}(t)$, then $\ddot{\theta}(t)$ would contain a dirac delta derivative, which we dont see on RHS.
Update:
wrt : with respect to.
dirac delta derivative is complicated to explain, but if you had the derivative of dirac delta, you would certainly need to take that into account. But we don't see such a think on the RHS (right hand side) so we can say that neither $\dot{\theta}(t)$ nor $\theta (t)$ may contain dirac delta.
you can consider delta to be the derivative of this step function. And the integral of delta to be this step function.
so $\dot \theta$ in this example looks like the step function above.
This is a 1D function, it can't have a saddle point. We also are not concerned by peaks.