# Convolution of convolution

Let us write a convolution $\int_{0}^{t} A(t-\tau) \mathrm{d}x(\tau)$ as $A \star \mathrm{d}x$ I would like to write down the expression for the double convolution $A \star \mathrm{d}x \star \mathrm{d}x$ Following the definition I obtain $\int_{0}^{t} \int_{0} ^{t-\tau} A(t-\tau-s) \mathrm{d}x(s) \mathrm{d}x(\tau)$ Can this be given a more compact form, especially in reference to the upper limit of integration in the inner integral? I would like to perform the change of variable $t-\tau = w$ but unsure as tyo how to proceed, any hint would be the most appreciated, thanks

• It is possible in the case where $x$ is differentiable. In this case $A\star dx= A\ast \dot x$ and you can use the fact that $A\ast \dot x \ast \dot x = A \ast (\dot x\ast\dot x)$. Commented Feb 18, 2014 at 10:12
• Yes x is differentiable, but I am unsure I got your point, could you please expand a bit on your reply, thank you very much
– Buco
Commented Feb 18, 2014 at 10:27

When the function is differentiable and you can write the operation as a regular convolution, you can use the fact that $\dot x\ast \dot x$ makes sense, differently from $dx\star dx$, which is not defined.

In this case you would have $A\star dx\star dx = A\ast \dot x \ast \dot x = A\ast (\dot x \ast \dot x)$: $$\int_0^t A(t-u) \int_0^u \dot x(u-s)\dot x(s)\,ds\,du.$$

If you want to rewrite it as before: $$\int_0^t A(t-u) \int_0^u \dot x(u-s)\,dx(s)\,du.$$

Otherwise, you can change the limits, but at the cost of defining another function $x^t(w)=x(t-w)$ when changing $t-\tau=w$, in this case $dx^t(w)=-dx(t-w)$:

$$\int_0^t \int_0^{t-\tau} A(t-\tau-s)\,dx(s)\,dx(\tau)=\int_0^t \int_0^{w} A(w-s)\,dx(s)\,dx^t(w)$$

• I see, thank you very much. So I cannot get away with changing the limits and maintaining the same functions, much appreciated
– Buco
Commented Feb 18, 2014 at 11:42

The result above can be obtained directly by a straight forward calculation. Suppose you have three Schwartz function $f,g,h$ (in particular: continuous, smooth and decaying fast enough at infinity). Then $$m(t) = (f*g)(t) = \int_0^t f(t-s_1) g(s_1) ds_1$$ and $$(f*g*h)(t) = (f*g)*h = (m*h)(t) = \int_0^t m(t-s_2) h(s_2) ds_2$$ putting the explicit expression for $m(t-s_2)$ we obtain $$= \int_0^t ds_2 \int_0^{t-s_2} f(t-s_2-s_1) g(s_1) h(s_2) ds_2$$ so $$(f*g*h)(t) = \int_0^t \int_0^{t-s_2} f(t-(s_1+s_2))g(s_1)h(s_2)ds_1 ds_2$$ and without further assumptions this as far as you can go.