Why solving a differentiated integral equation might eventually lead to erroneous solutions of the original problem?

Question

Consider the following integral equation araising in a mathematical-physical problem: $$ \int_0^r f(t) \arcsin \left( \frac{t}{r} \right) \, \mathrm{d}t + \frac{\pi}{2} \int_r^R f(t) \, \mathrm{d} t = r \, \qquad (0<r<R) \, , $$ where $f(t)$ is the unknown function, and $R$ is a positive real number. By differentiating both sides of this equation with respect to the variable $r$, one obtains $$ -\frac{1}{r} \int_0^r \frac{f(t)t \, \mathrm{d}t}{\sqrt{r^2-t^2}} = 1 \, , $$ the solution of which can readily be obtained as $$ f(t) = -1 \, . $$

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

i was wondering whether some math details or assumptions are overlooked here 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

Answer 1

Why solving a differentiated integral equation might eventually lead to erroneous solutions of the original problem?

The reason is that taking a derivative is not an invertible operation. So the new equation you obtain is true, but not equivalent to the original one -- the set of solutions has increased.

The simplest example is trying to solve an ordinary equation, say, $$ x=1 $$ The obvious solution is $x=1$. But if you square both sides, you obtain $x^2=1$, which now has two solutions, $x=\pm 1$. The new "wrong" solution appeared because taking a square is not invertible (the kernel is the negatives).

Similarly, taking a derivative is not invertible. Consider the equation $$ f(t)=t $$ The obvious solution is $f(t)=t$. But if you take a derivative, you get $f'(t)=1$, whose general solution is $f(t)=t+c$, for an arbitrary $c$. The "wrong" solutions, those with $c\neq0$, appeared because taking a derivative is not invertible (the kernel is the constants).

Answer 2

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

$$ -\frac{1}{r} \int_0^r \frac{f(t)t \, \mathrm{d}t}{\sqrt{r^2-t^2}} = 1 \, , $$

For example,

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

Edit: an alternative approach

Similarly, the original integral equation also admits multiple solutions. One particular solution can be derived by assuming that $f(t)=a$ is a simple flat function, where $a$ is a constant. Then, we have

$$a\int_0^r \arcsin\left( \frac tr \right)dt +\frac{\pi}{2}a(R-r)=r$$

or,

$$a\left[ \left( \frac {\pi}{2} -1 \right)r+ \frac{\pi}{2}(R-r)\right]=r$$

and the flat function solution is

$$f(t)=a= \frac{1}{ \frac{\pi}{2}\frac Rr -1}$$

Answer 3

Your derivation seems perfectly fine to me. Letting $s = t^2$ and $u = r^2$ you can rewrite the second integral equation as $$ \int \limits_0^u \frac{f(\sqrt{s})}{\sqrt{u - s}} \, \mathrm{d} s = - 2 \sqrt{u} \, , \, u \in [0,R^2] . $$ 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).

Answer 4

As pointed out by @AccidentalFourierTransform, taking the derivative will eventually lead to another integral equation. Therefore, the solution of the differentiated equation may not be necessarily the same as the original equation.

In fact, a problem occur in the second integral at the upper integration limit $r=R$ giving raise to the additional term $-\pi R/2$.

So, in order to get rid of this undesired term, it suffices to consider a solution of the form $$ f(t) = -1 + c \delta(t-R) , $$ where $c$ is a constant to be chosen so as to satisfy the original equation. Upon substitution, it follows readily that $c=2R$.

@Quanto $f(t)$ should not be a function of $r$!

Hope this help.