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How to solve this?

$$ g(t) = t \Theta (t) $$

$$ g * g(t)$$

I had hope to be able to use the $\delta $ function in some way to get eaiser calculations, but I can't see how. Is there any way to divide this into easier functions?

I would think that $(t*t)\Theta(t)$ would be easier to solve, but I can't find any rule that allows me to do that rewrite.

iveqy
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1 Answers1

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Suppose you have two (sufficiently regular) functions $a,b \colon \mathbb{R} \to \mathbb{R}$, and set $u(t) = a(t)\Theta(t)$, $v(t) = b(t) \Theta(t)$. Then for the convolution $u\ast v$ you get

\begin{align} (u\ast v)(t) &= \int_{\mathbb{R}} u(x)v(t-x)\,dx\\ &= \int_{\mathbb{R}} a(x)\Theta(x)v(t-x)\,dx\\ &= \int_{[0,\infty)} a(x)v(t-x)\,dx\\ &= \int_{[0,\infty)} a(x) b(t-x)\Theta(t-x)\,dx\\ &= \int_{[0,t]} a(x) b(t-x)\,dx, \end{align}

since $\Theta(y) = 0$ for $y < 0$.

Using that with $a = b = \operatorname{id}$, it is easy to compute $(g\ast g)(t)$.

Daniel Fischer
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  • This will give me a result $C$, however the solution should be $C\Theta(t)$. Following that reciepe seems to miss the $\Theta$ in the answer. (Talking numbers I get $t^3/6$ instead of $t^3/6\times \Theta$ – iveqy Jan 10 '15 at 12:01
  • Note that for $t < 0$, we have $[0,t] = \varnothing$, hence the convolution vanishes for $t \leqslant 0$. That is conveniently expressed by multiplying the expression you get for positive $t$ with the Heaviside function $\Theta$. – Daniel Fischer Jan 10 '15 at 12:04