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Just to simplify the notation, I use that $u(s)$ and $f(t,s)$ vanish for $s<0$ or $t<0$, so I can remove the integration bounds and all integrals run from $-\infty$ to $\infty$. I might then as well take a Fourier transform instead of a Laplace transform.

$${\cal F}(\omega)=\int e^{i\omega t}F(t)dt=$$

$$\int e^{i\omega t}f(t-s,s-k)u(s)u(k)dkdsdt =$$

$$\int e^{i\omega \tau}e^{i\omega s}f(\tau,s-k)u(s)u(k)dkdsd\tau =$$

$$\int e^{i\omega \tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k)u(k)dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega\tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k){\cal U}(\omega')e^{-i\omega'k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega')u(\sigma+k)e^{i(\omega-\omega')k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega'){\cal U}(\omega-\omega')e^{-i(\omega-\omega')\sigma}d\omega'd\sigma d\tau =$$

$$(2\pi)^{-1}\int {\cal F}(\omega,\omega'){\cal U}(\omega'){\cal U}(\omega-\omega')d\omega'.$$

This is the desired relation between the Fourier (or Laplace) transforms ${\cal F}(\omega)$ of $F(t)$ and the transforms ${\cal F}(\omega,\omega')$ of $f(s,t)$ and ${\cal U}(\omega)$ of $u(t)$. You started out with a double convolution and upon transformation one convolution is left.

Just to simplify the notation, I use that $u(s)$ and $f(t,s)$ vanish for $s<0$ or $t<0$, so I can remove the integration bounds and all integrals run from $-\infty$ to $\infty$. I might then as well take a Fourier transform instead of a Laplace transform, ${\cal F}(\omega)=\int e^{i\omega t}F(t)dt$. The desired relation between the transforms ${\cal F}(\omega)$ of $F(t)$ and the transforms ${\cal F}(\omega,\omega')$ of $f(s,t)$ and ${\cal U}(\omega)$ of $u(t)$ is

$${\cal F}(\omega)=(2\pi)^{-1}\int {\cal F}(\omega,\omega'){\cal U}(\omega'){\cal U}(\omega-\omega')d\omega'.$$

You started out with a double convolution and upon transformation one convolution is left.

Derivation: $${\cal F}(\omega)=\int e^{i\omega t}f(t-s,s-k)u(s)u(k)dkdsdt =$$

$$\int e^{i\omega \tau}e^{i\omega s}f(\tau,s-k)u(s)u(k)dkdsd\tau =$$

$$\int e^{i\omega \tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k)u(k)dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega\tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k){\cal U}(\omega')e^{-i\omega'k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega')u(\sigma+k)e^{i(\omega-\omega')k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega'){\cal U}(\omega-\omega')e^{-i(\omega-\omega')\sigma}d\omega'd\sigma d\tau =$$

$$(2\pi)^{-1}\int {\cal F}(\omega,\omega'){\cal U}(\omega'){\cal U}(\omega-\omega')d\omega'.$$

Just to simplify the notation, I use that $u(s)$ and $f(t,s)$ vanish for $s<0$ or $t<0$, so I can remove the integration bounds and all integrals run from $-\infty$ to $\infty$. I might then as well take a Fourier transform instead of a Laplace transform.

$${\cal F}(\omega)=\int e^{i\omega t}F(t)dt=$$

$$\int e^{i\omega t}f(t-s,s-k)u(s)u(k)dkdsdt =$$

$$\int e^{i\omega \tau}e^{i\omega s}f(\tau,s-k)u(s)u(k)dkdsd\tau =$$

$$\int e^{i\omega \tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k)u(k)dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega\tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k){\cal U}(\omega')e^{-i\omega'k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega')u(\sigma+k)e^{i(\omega-\omega')k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega'){\cal U}(\omega-\omega')e^{-i(\omega-\omega')\sigma}d\omega'd\sigma d\tau =$$

$$(2\pi)^{-1}\int {\cal F}(\omega,\omega'){\cal U}(\omega'){\cal U}(\omega-\omega')d\omega'.$$

This is the desired relation between the Fourier (or Laplace) transforms ${\cal F}(\omega)$ of $F(t)$ and the transforms ${\cal F}(\omega,\omega')$ of $f(s,t)$ and ${\cal U}(\omega)$ of $u(t)$.

Just to simplify the notation, I use that $u(s)$ and $f(t,s)$ vanish for $s<0$ or $t<0$, so I can remove the integration bounds and all integrals run from $-\infty$ to $\infty$. I might then as well take a Fourier transform instead of a Laplace transform.

$${\cal F}(\omega)=\int e^{i\omega t}F(t)dt=$$

$$\int e^{i\omega t}f(t-s,s-k)u(s)u(k)dkdsdt =$$

$$\int e^{i\omega \tau}e^{i\omega s}f(\tau,s-k)u(s)u(k)dkdsd\tau =$$

$$\int e^{i\omega \tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k)u(k)dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega\tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k){\cal U}(\omega')e^{-i\omega'k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega')u(\sigma+k)e^{i(\omega-\omega')k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega'){\cal U}(\omega-\omega')e^{-i(\omega-\omega')\sigma}d\omega'd\sigma d\tau =$$

$$(2\pi)^{-1}\int {\cal F}(\omega,\omega'){\cal U}(\omega'){\cal U}(\omega-\omega')d\omega'.$$

This is the desired relation between the Fourier (or Laplace) transforms ${\cal F}(\omega)$ of $F(t)$ and the transforms ${\cal F}(\omega,\omega')$ of $f(s,t)$ and ${\cal U}(\omega)$ of $u(t)$. You started out with a double convolution and upon transformation one convolution is left.

Just to simplify the notation, I use that $u(s)$ and $f(t,s)$ vanish for $s<0$ or $t<0$, so I can remove the integration bounds and all integrals run from $-\infty$ to $\infty$. I might then as well take a Fourier transform instead of a Laplace transform.

$${\cal F}(\omega)=\int e^{i\omega t}F(t)dt=$$

$$\int e^{i\omega t}f(t-s,s-k)u(s)u(k)dkdsdt =$$

$$\int e^{i\omega \tau}e^{i\omega s}f(\tau,s-k)u(s)u(k)dkdsd\tau =$$

$$\int e^{i\omega \tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k)u(k)dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega\tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k)∫{\cal U}(\omega')e^{-i\omega'k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma)u(\sigma+k){\cal U}(\omega')e^{-i(\omega-\omega')k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega')u(\sigma+k)e^{i(\omega'-\omega)k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\tau){\cal U}(\omega'){\cal U}(\omega'-\omega)e^{-i(\omega'-\omega)\sigma}d\omega'd\sigma d\tau =$$

$$(2\pi)^{-1}\int {\cal F}(\omega,2\omega-\omega'){\cal U}(\omega'){\cal U}(\omega'-\omega)d\omega'.$$

This is the desired relation between the Fourier transform ${\cal F}(\omega)$ of $F(t)$ and the Fourier transforms ${\cal F}(\omega,\omega')$ of $f(s,t)$ and ${\cal U}(\omega)$ of $u(t)$.

Just to simplify the notation, I use that $u(s)$ and $f(t,s)$ vanish for $s<0$ or $t<0$, so I can remove the integration bounds and all integrals run from $-\infty$ to $\infty$. I might then as well take a Fourier transform instead of a Laplace transform.

$${\cal F}(\omega)=\int e^{i\omega t}F(t)dt=$$

$$\int e^{i\omega t}f(t-s,s-k)u(s)u(k)dkdsdt =$$

$$\int e^{i\omega \tau}e^{i\omega s}f(\tau,s-k)u(s)u(k)dkdsd\tau =$$

$$\int e^{i\omega \tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k)u(k)dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega\tau}e^{i\omega\sigma}e^{i\omega k}f(\tau,\sigma)u(\sigma+k){\cal U}(\omega')e^{-i\omega'k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega')u(\sigma+k)e^{i(\omega-\omega')k}d\omega'dkd\sigma d\tau =$$

$$(2\pi)^{-1}\int e^{i\omega \tau}e^{i\omega \sigma}f(\tau,\sigma){\cal U}(\omega'){\cal U}(\omega-\omega')e^{-i(\omega-\omega')\sigma}d\omega'd\sigma d\tau =$$

$$(2\pi)^{-1}\int {\cal F}(\omega,\omega'){\cal U}(\omega'){\cal U}(\omega-\omega')d\omega'.$$

This is the desired relation between the Fourier (or Laplace) transforms ${\cal F}(\omega)$ of $F(t)$ and the transforms ${\cal F}(\omega,\omega')$ of $f(s,t)$ and ${\cal U}(\omega)$ of $u(t)$.