Xjyuk

Consider the following integral equation:
$$
int_0^r f(t) arcsin left( fractr right) , mathrmdt
+ fracpi2 int_r^R f(t) , mathrmd t = r , qquad
(0<r<R) , ,
$$
where $f(t)$ is the unknown function.
By differentiating both sides of this equation with respect to $r$, one obtains
$$
-frac1r int_0^r fracf(t)t , mathrmdtsqrtr^2-t^2 = 1 , ,
$$
the solution of which can readily be obtained as
$$
f(t) = -1 , .
$$

However, upon substitution of this solution into the original integral equation above, one rather gets an additional $-pi R/2$ term on the left hand side.

i was wondering whether some math details are overlooked during this resolution. Any help would be highly appreciated.

An alternative resolution approach that leads to the desired solution is also most welcome.

Thank you

The problem is that $f(t) =-1$ is not the unique solution of the integral equation below

$$
-frac1r int_0^r fracf(t)t , mathrmdtsqrtr^2-t^2 = 1 , ,
$$

For example,

$$ f(t) =- frac 2r sqrtr^2-t^2$$ would be another valid solution. In fact, there are numerous number of functions $f(t)$ that satisfy the integral equation above.

Edit:

Similarly, the original integral also admits multiple solutions. One particular solution is to assume that $f(t)=a$ is a flat function, where $a$ is a constant. Then, we have

$$aint_0^r arcsinleft( frac tr right)dt +a fracpi2(R-r)=r$$

or,

$$aleft[ left( frac pi2 -1 right)r+ fracpi2(R-r)right]=r$$

and the solution is

$$f(t)=a= frac1 fracpi2frac Rr -1$$

Your derivation seems perfectly fine to me. Letting $s = t^2$ and $u = r^2$ you can rewrite the second integral equation as
$$ int limits_0^u fracf(sqrts)sqrtu - s , mathrmd s = - 2 sqrtu , , , u in [0,R^2] . $$
This is an Abel integral equation, which admits a unique solution (note that the right-hand side is absolutely continuous). The formula yields $f(t) = -1$ for $t in [0,R]$ in agreement with your result. Therefore, the only possible solution to your original equation is $f equiv -1$. Since this leads to a contradiction, we conclude that it simply does not have a solution at all. This is not uncommon for such integral equations (Fredholm equations of the first kind).