2
$\begingroup$

\begin{align*} y(t) &= e^{2t} - e^t + \big(e^{2(t - 2)} - e^{t - 2}\big) u(t - 2), \\ y'(t) &= 2e^{2t} - e^t + \big(2e^{2(t - 2)} - e^{t - 2}\big) u(t - 2), \\ y''(t) &= 4e^{2t} - e^t + \big(4e^{2(t - 2)} - e^{t - 2}\big) u(t - 2) + \delta(t - 2). \end{align*}
Why is there a $\delta(t − 2)$ in the equation for $y''(t)$? In which $u$ is the unit step function. When $y'(t)$ is calculated there's no $\delta(t − 2)$ added, why do you need to add an $\delta(t − 2)$ for $y''(t)$?

$\endgroup$
2
  • $\begingroup$ Could you add a little more context? What is $u$? Is it a step function? What is $y$ and where is it coming from? $\endgroup$
    – user223391
    Jan 5 '16 at 3:50
  • 6
    $\begingroup$ $y'$ has $\delta(t-2)$ but it comes multiplied by $f$ s.t. $f(2)=0$, so it drops out. Not so for $y''$. $\endgroup$
    – A.S.
    Jan 5 '16 at 3:51
0
$\begingroup$

First: my background is physics, so this will be characteristically unrigorous.

Three important facts:

  1. $\delta(u)$ is the derivative of $u(t)$ in the sense that \begin{align} u(t) = \int_{-\infty}^t \delta(s) ds. \end{align}
  2. The derivative of $\delta(t)$ satisfies $\delta(t) = - t \delta(t)$. (This can be demonstrated with integration by parts.)
  3. The Dirac delta $\delta(t-a)$ is zero everywhere except at $t=a$, so $f(t)\delta(t-a)=f(a) \delta(t-a)$ ... the $\delta$ pulls out one particular function value.

Step 1: Let's write your $y(t)$ as \begin{align} y(t) = e^{2 t}-e^t + f(t) u(t-2) \end{align} Step 2: Then using fact 1: \begin{align} y'(t) = 2 e^{2 t} -e^t + \delta (t-2) f(t) + \theta (t-2) f'(t) \end{align} (Do not use Fact 3 to simplify $f(t)\delta(t-2)$ term yet.)

Step 3: Differentiating again: \begin{align} y''(t) =4 e^{2 t} -e^t + 2 \delta (t-2) f'(t) + \delta'(t-2) f(t) + \theta (t-2) f''(t) \end{align} Step 4a: Near $t=2$, $f(t) \approx t-2$ (do a Taylor approx). So it seems safe to replace $\delta' (t-2) f(t)$ by $\delta' (t-2) (t-2)$: \begin{align} y''(t) =4 e^{2 t} -e^t + 2 \delta (t-2) f'(t) + \delta'(t-2) (t-2) + \theta (t-2) f''(t) \end{align} Step 4b: Now use Fact 2 above to replace $\delta'(t-2) (t-2)$ by $-\delta(t-2)$: \begin{align} y''(t) =4 e^{2 t} -e^t + 2 \delta (t-2) f'(t) - \delta(t-2) + \theta (t-2) f''(t) \end{align} Step 5: Now that we're done differentiating, let's clean things up with Fact 3. Since $f'(2)=1$, we get: \begin{align} y''(t) = & 4 e^{2 t} -e^t + 2 \delta (t-2) - \delta(t-2) + \theta (t-2) f''(t)\\ = & 4 e^{2 t} -e^t + \delta (t-2) + \theta (t-2) f''(t) \end{align} Which is the second derivative that you stated.


Again, my background is physics, and I do not have a super rigorous knowledge of delta distributions. I suspect all the equalities above really only hold underneath integral signs ... or something.

In particular, I'd be interested if someone could comment on why using Fact 3 to simplify in the second step will eventually give us the wrong answer.

$\endgroup$
1
  • $\begingroup$ Everything here is correct in the sense of distributions. Note that the step function is not differentiable in the classical sense, but we could make use of the definition of weak derivatives. $\endgroup$
    – Chee Han
    Jan 5 '16 at 8:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.