The resolution of an elasticity exercise has lead to the following integral equation $$ \int_r^R \frac{F_1(t) \, \mathrm{d}t}{(t^2-r^2)^{1/2}} + \int_0^r \frac{F_2(t) \, \mathrm{d}t}{(r^2-t^2)^{1/2}} = 0 \quad\quad (0<r<R) \, , $$ where $R>0$. I was wondering whether a relation between $F_1$ and $F_2$ could be established.
Any help is highly appreciated.
Thanks
R
You can simplify one integral at the expense of making the other one more complicated, but I am not sure whether this is waht you want. Multiplying both sides with $r/\sqrt{s^2-r^2}$ and integrating $r$ from $0$ to $s$, $$ \int_0^s \frac{rdr}{(s^2-r^2)^{1/2}}\int_r^R \frac{F_1(t)dt}{(t^2-r^2)^{1/2}} +\int_0^s \frac{rdr}{(s^2-r^2)^{1/2}}\int_0^r \frac{F_2(t)dt}{(r^2-t^2)^{1/2}}=0. $$ Changing the order of integration in the second expression, $$ \int_0^s \frac{rdr}{(s^2-r^2)^{1/2}}\int_0^r \frac{F_2(t)dt}{(r^2-t^2)^{1/2}} = \int_0^s F_2(t)dt\int_{t}^s \frac{rdr}{(s^2-r^2)^{1/2}(r^2-t^2)^{1/2}}. $$ The inner integration is exactly $\pi/2$, and we get $$ \int_0^s \frac{rdr}{(s^2-r^2)^{1/2}}\int_r^R \frac{F_1(t)dt}{(t^2-r^2)^{1/2}} + \frac{\pi}{2}\int_0^s F_2(t)dt=0. $$ Taking derivative on both sides, $$ F_2(s) = -\frac{2}{\pi}\frac{d}{ds}\int_0^s \frac{rdr}{(s^2-r^2)^{1/2}}\int_r^R \frac{F_1(t)dt}{(t^2-r^2)^{1/2}}. $$ Similarly, we can express $F_1$ in terms of $F_2$, by multiplying both sides with $r/\sqrt{s^2-r^2}$ and integrating $r$ from $s$ to $R$