Convolution of $te^{2t}$ and $\delta_1-\delta_2$?

Question

I seek to find $f*g$ where $f=te^{2t}$ and $g=\delta_1-\delta_2$ and $\delta_a(t)= \displaystyle \lim_{\epsilon \to 0^+}d_{a,\epsilon}(t)$; i.e. $\delta$ is the Dirac Delta function.

We have learned and proved two theorems regarding the Dirac Delta function, including

(1) $\mathcal{L}\{\delta_a\} = e^{-as}$ where $\mathcal{L}$ is notation for the Laplace transform, and

(2) $(f*\delta_a)(t) = f(t-a)\mathcal{U}(t-a)$ where $\mathcal{U}$ denotes the unit step function.

However, the convolution theorem states that $\mathcal{L}^{-1}\{f*g\}=F(s)G(s)$ where $F = \mathcal{L}(f)$ and $G = \mathcal{L}(g)$.

So my question is this: Is it possible to use the convolution theorem in this case and take $\mathcal{L}\{\delta_1-\delta_2\}$ to obtain $G$ using (1)? Or is the solution obtained by messing with (2)? I suspect that we learned (2) because the Dirac Delta function is "bad" and does not satisfy the conditions of the convolution theorem, which requires that $f,g \in \mathcal{H}$, where $\mathcal{H}$ denotes the Heaviside class.

EDIT: Here's what I did. Using the convolution theorem and (1) above, we have

$f*g = \mathcal{L}^{-1}\{FG\}$, and $F = \mathcal{L}\{te^{2t}\} = \frac{1}{(s-2)^2}$, $G=\mathcal{L}\{\delta_1 - \delta_2\} = e^{-s} - e^{-2s}$.

Thus,

\begin{equation} \begin{split} f*g &= \mathcal{L}^{-1}\left\{\frac{e^{-s}}{(s-2)^2} - \frac{e^{-2s}}{(s-2)^2}\right\} \\ &= \mathcal{U}(t-1)\mathcal{L}^{-1}\left\{\frac{1}{(s-2)^2}\right\} - \mathcal{U}(t-2)\mathcal{L}^{-1}\left\{\frac{1}{(s-2)^2}\right\}\\ &= \mathcal{U}(t-1)(t-1)e^{2(t-1)} - \mathcal{U}(t-2)(t-2)e^{2(t-2)}\\ &= \begin{cases} 0 &\mbox{if } 0 \leq t \lt 1 \\ (t-1)e^{2(t-1)} & \mbox{if } 1 \leq t \lt 2 \\ (t-1)e^{2(t-1)} - (t-2)e^{2(t-2)} & \mbox{if } t\geq 2 \end{cases} \end{split} \end{equation}

Answer 1

Why don't simply use the definition of the convolution of two distributions, $$(T\star S) (\varphi) = T_x (S_y (\varphi(x+y)))$$

Hence, take $S = \delta_1-\delta_2$ and you got $S_y(\varphi(x+y))=\varphi(x+1)-\varphi(x+2)$. Hence \begin{align*} (te^{2t} \star (\delta_1-\delta_2))(\varphi)& = \int_{-\infty}^{+\infty}te^{2t}(\varphi(t+1)-\varphi(t+2))dt\\ & = \int_{-\infty}^{+\infty}te^{2t}\varphi(t+1)dt - \int_{-\infty}^{+\infty}te^{2t}\varphi(t+2) dt \\ & = \int_{-\infty}^{+\infty}(t-1)e^{2t-2}\varphi(t)dt - \int_{-\infty}^{+\infty}(t-2)e^{2t-4}\varphi(t) dt. \end{align*}

Finally we get $$te^{2t} \star (\delta_1-\delta_2) =(t-1)e^{2t-2}-(t-2)e^{2t-4}.$$

Answer 2

I'll be a little bit straightforward. We have \begin{align} f(t) &= t e^{2 t} \theta_0(t),\\ g(t) &= \delta_1(t) - \delta_2(t), \end{align} where $\theta_x(t) = \mathbb{1}_{\left\{t\geq x\right\}}$ and $\delta_x(t) = \mathbb{1}_{\left\{t=x\right\}}$.

Given that both functions are supported on $\mathbb{R}_{+}$, we define $$h(t) = f(t) \ast g(t),$$ and we can, for some $s\in \mathbb{R}_{+}^{*}$, use the Laplace Transformation to determine $h(t)$.

$$\mathcal{L}\{h(t)\} = \mathcal{L}\{f(t) \ast g(t)\}$$ Using the Convolution theorem, $$\mathcal{L}\{h(t)\} = \mathcal{L}\{f(t)\} \times \mathcal{L}\{g(t)\} = \frac{1}{\left(s-2\right)^2} \times \left(e^{-s}-e^{-2 s} \right)$$ thus $$h(t) = \mathcal{L}^{-1} \left( \frac{e^{-s}-e^{-2 s} }{\left(s-2\right)^2} \right) = e^{2 t-2} (t-1) \theta_1(t)- e^{2 t-4}(t-2)\theta_2(t) $$ or $$ h(t) = \begin{cases} 0 \qquad & t<1 ,\\ e^{2 t-2} (t-1) \qquad & 1 \leq t <2 ,\\ e^{2 t-2} (t-1)+e^{2 t-4} (2-t)\qquad & t \geq 2. \end{cases} $$ Your answer is correct.

Answer 3

Observe that $$ \big(f(t) * \delta(t-T)\big)= \int_{-\infty}^\infty f(\tau) \delta(t-T-\tau) \,\mathrm d\tau= \int_{-\infty}^\infty f(\tau) \delta(\tau-(t-T)) \,\mathrm d\tau = f(t-T) $$ So you have $$f(t) * \big(\delta(t-1)-\delta(t-1)\big)=f(t) * \delta(t-1)-f(t) * \delta(t-2)=f(t-1)-f(t-2)$$ where $f(t)=t\mathrm e^{2t}u(t)$ and then $$\color{blue}{ f*(\delta_1-\delta_2)=(t-1)\mathrm e^{2(t-1)}u(t-1)-(t-2)\mathrm e^{2(t-2)}u(t-2)} $$ that is $$ f*(\delta_1-\delta_2)=\begin{cases} 0 \qquad & t<1 ,\\ e^{2 t-2} (t-1) \qquad & 1 \leq t <2 ,\\ e^{2 t-2} (t-1)+e^{2 t-4} (2-t)\qquad & t \geq 2. \end{cases} $$