# Laplace transform having this unusual property in convolution?

Here is the problem

Solve $y'(t) = 1 - \int_{0}^{t} y(t - v)e^{-2v}dv$

The solution sets $\mathcal{L}(y) = Y(s)$ and does the following

Notice that in step 1, they have $$Y(s)\dfrac{1}{s+2}$$

Are they implying $$\mathcal{L}(y(t) * e^{-2t}) = \mathcal{L}(y(t)) \mathcal{L}(e^{-2t}) = Y(s)\mathcal{L}(e^{-2t}) = Y(s)\dfrac{1}{s+2}$$

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There must be some mistake - inside the integral you have $y(t-v)$ and $v$ is not there in the later equations. Also, if the integral is with respect to $dt$, why is $t$ in the limits of the integral? is the integral w.r.t. $dv$? – svenkatr Jun 14 '12 at 4:31
fixed mistake on exp(-2t). Thank you for catching that svenkatr – Hawk Jun 14 '12 at 4:38

Yes they are implying that. Applying the Laplace transform to a convolution gives a product. Write

$$\mathcal{L}\{f*g\}(s):=\int_0^\infty \left(\int_0^x f(u)g(x-u)du\right)e^{-sx}dx=\iint_D f(u)g(x-u)e^{-sx}dudx$$

The region of integration in the $xu$-plane is $D=\{(x,u):0\le u\le x\}$, an infinite triangle. Writing out the substitution $v=x-u$ our region of integration in the $uv$-plane is simply $(0,\infty)\times(0,\infty)$, so

$$=\iint_{(0,\infty)^2} f(u)g(v)e^{-s(u+v)}dudv=\int_0^\infty f(u)e^{-su}du\int_0^\infty g(v)e^{-sv}dv=\mathcal{L}\{f\}(s)\cdot \mathcal{L}\{g\}(s).$$

Note the Jacobian determinant of the transformation $(x,u)\mapsto(u,x-u)$ is simply $1$ (in abs. value).

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Also note this rule is not unusual. It is listed on Wikipedia for example. – anon Jun 14 '12 at 5:42
Will the region always be a rectangle so that this property holds in all cases? – Hawk Jun 14 '12 at 17:51
@jak: What cases are you talking about? The triangular region comes from the definition of convolution and the Laplace transform being put together, and the infinite square comes from the change of variables: none of that has anything to do with choice of $f$ and $g$ (except as far as regularity and therefore convergence is concerned). – anon Jun 14 '12 at 19:17