I need to solve, in $C[0,1]$, the equation $\displaystyle x(t) - \lambda \int_{0}^{\pi}(\sin t \cos s)x(s) ds = \sin t$.

Adding the integral part to both sides, I obtain $x(t) = \sin t + \lambda \int_{0}^{\pi}(\sin t \cos s)x(s) ds$, which is, I believe, a Fredholm Integral Equation of the second kind (yay, Wikipedia!).

However, other than briefly mentioning in class what they are, our professor never really went over how to solve them.

I feel kind of bad asking for a crash course in how to solve Fredholm Integral Equations of the Second Kind (sounds like a bad scifi movie from the early 80s) in $C[0,1]$, but that's essentially what I'm doing...in the context of this particular integral equation of course.

Thank you for your time & patience! :)


Since $$\int_0^\pi (\sin t \cos s) \sin s \, ds = \sin t \int_0^\pi \sin s \cos s \, ds = 0$$ $x(t) = \sin t$ is a solution. Are you worried about uniqueness?

OK, suppose you want the general solution to $$x(t) - \lambda \int_{0}^{\pi}(\sin t \cos s)x(s) ds = \sin t.$$ Rearrange this as $$x(t) = \sin t \left( 1 + \lambda \int_0^\pi \cos s x(s) \, ds\right)$$ to get that $x(t) = A \sin t$, where $A$ is the number in the parentheses. Plug this back into the original equation to find $A =1$.

  • $\begingroup$ you know, I'm not sure...let's say that I am. Then what? In any case, $x(t) = \sin t $ is a solution in $C[0,1]$? It just seems too easy... $\endgroup$ – ALannister May 16 '16 at 13:58
  • $\begingroup$ I added a few lines to the answer. $\endgroup$ – Umberto P. May 16 '16 at 14:11
  • $\begingroup$ I saw that. Thank you! That helps a lot. Just checking though, it is in $C[0,1]$? $\endgroup$ – ALannister May 16 '16 at 14:17
  • $\begingroup$ That depends on what you mean by $C[0,1]$. $\endgroup$ – Umberto P. May 16 '16 at 14:18
  • $\begingroup$ I suppose the set of all continuous functions on $[0,1]$ equipped with the uniform (sup) norm $\Vert \cdot \Vert_{\infty}$ $\endgroup$ – ALannister May 16 '16 at 14:53

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