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I have the following equation : $$ \phi(x) = \frac{x}{2} + \frac{\pi^2}{4}\int_{0}^{1}K(x,t)\phi(t)dt $$

where $$ K(x,t)= \left\{ \begin{array}{ll} \frac{x(2-t)}{2} & \mbox{if } 0 \leq x \leq t \\ \frac{t(2-x)}{2} & \mbox{if } t \leq x \leq1 \end{array} \right. $$

I tried differentiating twice to try and find the eigenfucntions of the homogenic equation but the ode is too hard to solve and I can't get a sturm liueville condition properly because phi on 0 is 0 but I there is not second condition.

** Edit : The question was made much simpler when a small change was made; probably due to a mistake in the exam that this question was taken out of.

Michael
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  • The kernel should be, right? $$ K(x,t)= \left{ \begin{array}{ll} \frac{x(2-t)}{2} & \mbox{if } 0 \leq x \leq t \ \frac{t(\color{red}{2-x})}{2} & \mbox{if } t \leq x \leq1 \end{array} \right. $$ – Math-fun Jul 13 '16 at 15:11
  • Most probably yeah, so there is a mistake in the exam form I took the question from. – Michael Jul 13 '16 at 15:46

1 Answers1

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If the correct Kernel is what I mentioned in the comment above - given that you mentioned about symmetry - then differenting twice we obtain $$\phi''(x)+\frac{\pi^2}{4}\phi(x)=0.$$Now this is standard, knowing that $\phi(0)=0$ and $\phi'(1)+\phi(1)=\frac12$.

Math-fun
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  • Well you are probably right, I got that question from an exam that took place a few years ago so I can't really ask anyone if that is a mistake.. so it is not just my mistake - this is indeed a nearly unsolvable question as posed. Thanks. – Michael Jul 13 '16 at 15:45
  • you are welcome :-) – Math-fun Jul 14 '16 at 07:18