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Consider the following integral equation: $$ \int_0^r f(t) \arcsin \left( \frac{t}{r} \right) \, \mathrm{d}t + \frac{\pi}{2} \int_r^R f(t) \, \mathrm{d} t = r \, \qquad (0<r<R) \, , $$ where $f(t)$ is the unknown function. By differentiating both sides of this equation with respect to $r$, 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 $$ f(t) = -1 \, . $$

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

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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).

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  • $\begingroup$ In symbols, you have $f = g \Rightarrow f' = g'$, but not the other way ($\Leftarrow$)! If you check your solutions at the end, you effectively check which ones of them could still be pulled through the "missing" $\Leftarrow$ back to get $f = g$. $\endgroup$ – ComFreek Sep 22 at 16:19
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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

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

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  • $\begingroup$ Thanks for the answer. Your solution does not seem to be defined for the second term on the left hand side of the original equation since $t>r$. Yet, again, the original problem is not solved using your $f(t)$ $\endgroup$ – Volterra Sep 21 at 16:37
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    $\begingroup$ @Volterra - you’re right. For an alternative approach, I provided one particular solution in the answer. $\endgroup$ – Quanto Sep 21 at 17:50
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    $\begingroup$ Note that $r$ is not a constant but a variable here, so it cannot be used in the definition of $f$. $\endgroup$ – ComplexYetTrivial Sep 21 at 19:47
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    $\begingroup$ r is a varying constant: the equation must hold for any r in the range. I dont see any steps in the solution that restrict the range or r in the definition of; therefore, it ought to be valid. For precision, we could write $f(t) = \dots, 0 \lt r \lt R$ @ComplexYetTrivial $\endgroup$ – D. Ben Knoble Sep 22 at 4:13
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    $\begingroup$ @D.BenKnoble I think that the point of these equations is to find one function $f$ which solves them for every $r \in (0,R)$. If $f$ depends on $r$, we instead have a family of functions, each of which solves the equation for one value of $r$ only. $\endgroup$ – ComplexYetTrivial Sep 22 at 8:50
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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).

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  • $\begingroup$ hi, I can easily understand the other solution but yours seems dull to me. Why are you saying there isn't any solution since the author of the question is saying he just have to add a constant to get the good result? $\endgroup$ – Marine Galantin Sep 22 at 11:10
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    $\begingroup$ @MarineGalantin If the right-hand side of the original equation was $r - \frac{\pi}{2}R$ instead of $r$, $f \equiv -1$ would be a solution. But since this additional constant is not present, this is not the case. However, the differentiated equation shows that $f \equiv -1$ is the only possible solution. This contradiction is why the first equation cannot be solved. $\endgroup$ – ComplexYetTrivial Sep 22 at 11:57
  • $\begingroup$ oh thank you :) $\endgroup$ – Marine Galantin Sep 22 at 12:02
  • $\begingroup$ Thanks for the hint. Just a quick question please. Suppose one has to solve an Abel integral equation with a vanishing right hand side, is $f \equiv 0$ the only unique possible solution? Thanks $\endgroup$ – Volterra Sep 23 at 12:47
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    $\begingroup$ @Volterra Yes! You can multiply the equation $\int_0^u \frac{f(t)}{\sqrt{u-t}} \, \mathrm{d}t = 0$ by $\frac{1}{\sqrt{v-u}}$ and integrate from $0$ to $v$ with respect to $u$. Then interchange the order of integration to find that $f$ has a vanishing antiderivative and therefore must be equal to zero itself. $\endgroup$ – ComplexYetTrivial Sep 23 at 15:49
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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.

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    $\begingroup$ Thanks for the comment. Yes, this approach solves my original problem involving a more complicated right hand side in the starting equation. $\endgroup$ – Volterra Sep 23 at 9:45

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