On several papers, I found the following model for a multiple integral equation:
$$g(s)=\int\limits_{\Omega} h(s,t)f(t)\,\mathrm{d}t$$
where $s,t \in \mathbb{R}^3$, and $\Omega \subseteq \mathbb{R}^3$.
I would like to know wether it is possible to express the above equation as:
$$g(x,y,z)=\iiint\limits_{\Omega} h\left(x,y,z,u,v,w\right) f\left(u,v,w\right) \, \mathrm{d}u\, \mathrm{d}v\, \mathrm{d}w$$
or are two different problems.
In the first equation it seems to say that $g$, a function of $s$, is the time integral, over the (probably) closed region $\Omega$, of the product of a function of $t$, and a function of both $t$ and $s$. Now, stop and think here. When I integrate a function over a region I don't keep the variable that I integrated over; I don't keep $t$. The thing that I bought by spending $t$ is a function of the boundary alone, the fundamental theorem of calculus says so.
$$\int_{a}^b f(x) dx = F(b) - F(a)$$
See, no $x$s. Here $F$ is the antiderivative of $f$, so for us that means that the boundaries of $\Omega$ are functions of $s$. I'm going to assume that by $s$ you mean $\{x,y,z\}$, i.e., flat 3D space. If that's right then the left side of your equations agree.
$$ s: \{x,y,z\} \rightarrow s$$ $$g: s \rightarrow g $$ $$\Rightarrow g: \{x,y,z\} \rightarrow s \rightarrow g $$ $$\Rightarrow g: \{x,y,z\} \rightarrow g $$
As for the right hand, what you are saying is that you're going to try to find relations between $t$, and a new 3D space, $\{v, u, w\}$, such that you can integrate the same product of functions over the same region, only now with three new variables which give you the same information as $t$ used to. This task is probably impossible if you were looking for a symbolic answer.
On the other hand, what I think you meant is this:
$$\int_{\omega}\,h(x,y,z,t)\,f(t)\,dt= \iiint_{\Omega}\,H (x,y,z,u,v,w)\,F (u,v,w)\,du\,dv\,dw$$ Or maybe: $$\int_{\omega} h(x,y,z,t)\,f(t)\,dt= \iiint_{\Omega} h (x,y,z,t(u,v,w))\,f (t(u,v,w))\, du\,dv\,dw$$ This one seams the most pheasable: $$\int_{\omega} h(x(t),y(t),z(t),t)\,f(t)\,dt$$ $$= \iiint_{\Omega} h (x(t(u,v,w)),y(t(u,v,w)),z(t(u,v,w)),t(u,v,w))\,f (t(u,v,w))\, du\,dv\,dw$$
Which is just substituting an expression in for $t$ and finding an $\Omega$.
Like I said, this is probably a computer problem, but there is still something to be gotten from experimenting.
The question if these are two different problems or not, will depend upon whether and how $s$ and $t$ might be dependent on $\Bbb R^3$. There is no dependency between your expressions yet if you dont clearly express a dependency between $s$ and $t$.
Yes. All you did was to rename the components of $s,t$ as $x,y,z$ and $u,v,w$. Therefore the two expressions are completely identical, differing only in that you define $g(x,y,z):=g(\,(x,y,z)\,)$ and so on.
