Solving for integrand from integrated quantities.

Question

Given equations of the form:

$A(r) = \int_{t_{1}}^{t_{2}}F(r,t)dt$

$B(t) = \int_a^b F(r,t)r^2dr$

where $A(r)$, $B(t)$, and all of the limits on the integrals are known, is there enough information to solve for F(r,t)? If so how would one do this?

For more context this is a scenario where there is a quantity, $F(r,t)$, that varies in space and time but is measured as only a function of time and a function of space separately. I am trying to figure out if the full space and time dependence can be reconstructed from these two measurements alone.

EDIT: Perhaps a better way of stating the question.

Is $F(r,t)$ uniquely constrained given $A(r)$ and $B(t)$?

Answer 1

Let's look at the problem $$ A(r) = \int_{t_{1}}^{t_{2}}F(r,t)dt\\ B(t) = \int_a^b F(r,t)r^2dr $$ in the particular case where $t_1 = -1, t_2 = 1, a = -1, b = 1$.

Then if $F$ is any odd function, (odd in $r$, I mean: $F(r, t) = -F(-r, t)$), then $B$ will be the constant function $0$. So the question becomes: are there odd functions $F_1$ and $F_2$ such that $$ A_1(r) = \int_{-1}^{1}F_1(r,t)dt\\ A_2(r) = \int_{-1}^{1}F_2(r,t)dt\\ $$ are the same function, even though $F_1$ and $F_2$ are different? But once again, if $F_1$ and $F_2$ are both odd (in $t$) then both integrals will be everywhere zero.

So let's pick $F_1(r, t) = 0$ everywhere, and $F_2(r, t) = tr$. Then the functions $A(r)$ and $B(t)$, computed for either $F_1$ or $F_2$, will be everywhere zero.

In short: there's no uniqueness result to be had.

Answer 2

What I would do is to find $$A(r,t) = \int_{t_1}^t F(r,t) dt$$ so that $$A'_t(r,t) = F(r,t)$$ Hence, $$A'_t(r,t_1)=F(r,t_1), \ A'_t(r,t_2)=F(r,t_2), \ A'_t(r,t_3)=F(r,t_3), \dots$$ which means you should be able reconstruct numerically the function $F$ by finding $A$ as a function of $r$ and different time $t_i$ endpoints if you have enough data taken and are able to find a time gradient of $A$. We can also apply an analogous trick to $B$ and make sure the two reconstructions match.