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Solve the integral equation $$ f(x) = \lambda\int_0^\pi(x+y)f(y)\mathrm dy $$

with $f(x)$ integrable in [0,π].

From my comment below: ...I think that a non trivial solution is $f(x)=Ax+B$ with $A$ and $B$ constant, but i can't check it.

(see the comments for others attempts)

Thanks

Alex
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  • Are you assuming anything else about function $f$? – welshman500 Feb 19 '18 at 17:41
  • Integrable in $[0,\pi]$.I think that a non trivial solution is $f(x)=Ax+B$ with $A$ and $B$ constant, but i can't check it.... – Alex Feb 19 '18 at 17:46
  • Alex: I get the system $A=\lambda\left[\frac{A\pi^2}{2}+B\pi\right], ; B=\lambda\left[\frac{a\pi^3}{3}+\frac{B\pi^2}{2}\right]$. That looks like an invertible system to me. [EDIT] Yeah, thanks copper.hat. Also, Wolfram Dev Platform yields only the trivial solution. So, back to square one. – Adrian Keister Feb 19 '18 at 18:10
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    @AdrianKeister: You have a small typo., $a \to A$. – copper.hat Feb 19 '18 at 18:14
  • @AdrianKeister Yes i get the same system. I tried to solve it and I found something like $ B_i = c_i(\lambda_i)*A_i$ with $\lambda_1=\frac{(4\sqrt(3)-6)}{\pi^2}$ and $\lambda_2=-\frac{(4\sqrt(3)+6)}{\pi^2}$ and $c(\lambda_i)$ a rational function, but when I substituted this solution in the equation I get an incorrect result, perhaps due to calculation errors I do not know.... – Alex Feb 19 '18 at 18:26
  • Wolfram Dev Platform doesn't help much - the DSolveValue command doesn't yield a non-trivial solution. – Adrian Keister Feb 19 '18 at 18:30
  • @Alex: the system is invertible and homogeneous, which means it has only the trivial solution. – Adrian Keister Feb 19 '18 at 18:31
  • @AdrianKeister but is it invertible for all $\lambda$? The determinant is always non zero? For example for $\lambda=\frac{4\sqrt(3)-6}{\pi^2}$ the solution seems to be $B= \frac{\pi(2-\sqrt(3))A}{2\sqrt(3)-3}$. But you're probably right and I made mistakes! Sorry and thanks! – Alex Feb 19 '18 at 19:28
  • Ah, I see what you are about. Yes, $\lambda=\frac{-6\pm 4\sqrt{3}}{\pi^2}$ will give you non-trivial solutions for $A$ and $B$. I haven't worked out what $A$ and $B$ are. Lemma type this all up into a solution for you, if you don't mind. – Adrian Keister Feb 19 '18 at 19:39

1 Answers1

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It is clear that $f(x)=0$ is a trivial solution to this integral equation. However, in separating out the original IE like this: $$f(x)=x\left(\lambda \int_0^{\pi}f(y) \, dy\right)+\lambda\int_0^{\pi}y \, f(y) \, dy,$$ the ansatz $f(x)=Ax+B$ suggests itself. Plugging this into the equation yields the equation $$Ax+B=x\lambda\left[\frac{A\pi^2}{2}+B\pi\right] + \lambda\left[\frac{A\pi^3}{3}+\frac{B\pi^2}{2}\right].$$ Setting the coefficients of like powers of $x$ equal yields the (mostly) invertible (and definitely) homogeneous system of equations: $$A=\lambda\left[\frac{A\pi^2}{2}+B\pi\right]$$ $$B=\lambda\left[\frac{A\pi^3}{3}+\frac{B\pi^2}{2}\right].$$ For most values of $\lambda$, the system is invertible, which implies the trivial solution. (So we see that the trivial solution is a special case of our ansatz, though we could have seen that from the get-go.) However, if we cast this into standard $\mathbf{A}x=0$ matrix form, we can take the determinant of the matrix $\mathbf{A}$ and set it equal to zero to find if there are non-trivial solutions. We obtain the equivalent system $$0=A\left[\frac{\lambda\pi^2}{2}-1\right]+B\lambda\pi$$ $$0=A\,\frac{\lambda\pi^3}{3}+B\left[\frac{\lambda\pi^2}{2}-1\right].$$ Taking the determinant yields the quadratic $$\frac{\lambda^2\pi^4}{12}+\lambda\pi^2-1=0,$$ with solutions $$\lambda=\frac{-6\pm 4\sqrt{3}}{\pi^2}.$$ We can plug this into one of the relationships between $A$ and $B$ to get a parametrized family of solutions. We obtain $$B=\frac{A\,(-2\pm\sqrt{3}\,)\,\pi}{3\pm 2\sqrt{3}}.$$ The final solution, therefore, to the integral equation is $$f(x)=tx+\frac{t\,(-2\pm\sqrt{3}\,)\,\pi}{3\pm 2\sqrt{3}},$$ where $t$ is a parameter.

Adrian Keister
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  • Nice answer, +1. OTOH, it's better not to answer a question that shows no effort whatsoever. – Xam Feb 19 '18 at 21:13
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    Well, he did put forth a significant effort in the comments. – Adrian Keister Feb 19 '18 at 21:13
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    Then, he should have included such effort in the body of his question. It's not a good practice to give further clarification in comments because not everyone read them. – Xam Feb 19 '18 at 21:16