How to calculate the convolution product of $(H_0 e^{\alpha t}) * (H_0 e^{-\alpha t})$?

Question

everyone!

I am doing an exercise concerning the resolution of differential eaqtion in the sense of distributions.

Let $\alpha \in \mathbb{R}_+^*$. Let be the differential equation, for $f \in \mathrm{C}^0(\mathbb{R}, \mathbb{R})$ and $t \ge 0$ :

\begin{align} \left\{ \begin{array}{c1} u''(t) - \alpha^2u(t) = f(t)\\ u(0) = u_0, \ \ \ \ u'(0) = u_1 \end{array} \right. \end{align}

I have shown that $\varphi_\pm (t) = e^{\pm \alpha t}$ is a solution.

I have solved $w''(t) = \alpha^2w(t)$ in $\mathbb{R}$ with $w(0) = 0$ and $w'(0) = 1$. Which gives us : \begin{align} w(t) = \frac{1}{2\alpha}(e^{\alpha t} - e^{-\alpha t}) \end{align}

Now I'm trying to calculate $(H_0 e^{\alpha t}) * (H_0 e^{-\alpha t})$ but I don't see how to do it at all...(The answer is $H_0 w$)

Where $H_0$ is the Heaviside function : https://en.wikipedia.org/wiki/Heaviside_step_function

I tried this: \begin{align} \displaystyle \begin{array}{rcl} \displaystyle (H_0 e^{\alpha t}) * (H_0 e^{-\alpha t}) & = & f * g \\ \displaystyle & = & \int_{\mathbb{R}} e^{\alpha (t -x)}1_{[0, +\infty[}(t - x) \ \ e^{-\alpha x}1_{[0, +\infty[}(x) dx\\ \displaystyle & = & e^{\alpha (t)}\int_{\mathbb{R}} e^{-\alpha x}1_{[0, +\infty[}(t - x) \ \ e^{-\alpha x}1_{[0, +\infty[}(x) dx \displaystyle \end{array} \end{align}

But I don't think I'm off to a good start... If you have any ideas, I'd be happy to hear them.

Thanks for your help

Answer 1

You want the integral $$ J = \int_{-\infty}^\infty dt' H(t-t')e^{\alpha[t-t']}H(t')e^{-\alpha t'} = H(t) e^{\alpha t}\int_{0}^t dt' e^{- 2 \alpha t'}.$$ Note I have put in the second $H(t)$ by hand: the integrand vanishes if $t$ is not positive. This evaluates to $$J = \frac{H(t)e^{\alpha t}}{2\alpha}\left( 1-e^{-2\alpha t} \right)\\ J= H(t)\frac{\sinh(\alpha t)}{\alpha}$$ as given by 138 Aspen from mathematica. Switching back to your notation, this is $J = H_0 w$ as in the problem set. Your approach is perfectly good, you just didn't follow through.