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Eigenvalues of an operator

Find all the functions $f \in C([0,\frac{\pi}{2}])$ which are solutions of $$ f(x) = \lambda \int_0^{\pi/2} \cos({x-y)}f(y)\,dy, \qquad \lambda \in \mathbb R. $$

I just tried some algebraic manipulations and calculations, but I do not know ho to face this problem. Any help or hint is welcomed. Thanks a lot.

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marked as duplicate by J. M., GEdgar, sdcvvc, William, Rudy the Reindeer Sep 3 '12 at 10:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
Differentiate twice under the integral sign. –  Jayesh Badwaik Aug 24 '12 at 13:39

3 Answers 3

up vote 5 down vote accepted

Expand $\cos(x-y)$ into $\cos(x)\cos(y)+\sin(x)\sin(y)$ to show that $f(x)=a\cos(x)+b\sin(x)$ with $$ a=\lambda\int_0^{\pi/2}\cos(y)f(y)dy,\qquad b=\lambda\int_0^{\pi/2}\sin(y)f(y)dy. $$ Since $$ \int_0^{\pi/2}\cos^2(y)dy=\int_0^{\pi/2}\sin^2(y)dy=\frac{\pi}4,\qquad \int_0^{\pi/2}\cos(y)f(y)dy=\frac12, $$ this yields $$ 4a=\lambda(a\pi+2b),\qquad 4b=\lambda(2a+\pi b). $$ If $\lambda\pi-4\ne\pm2\lambda$, the only solution is $a=b=0$, hence $f=0$. If $\lambda\pi-4=\pm2\lambda$, then $\lambda\ne0$ hence every $a=\mp b$ is a solution. For $\lambda=4/(\pi\pm2)$, $f(x)=a(\cos x\pm\sin x)$, or, equivalently, $f(x)=c\cos(x\mp\pi/4)$.

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A different solution from the one by nbubis is differentiation under the integral sign. Hence we get, \begin{equation} \frac{df(x)}{dx} = - \lambda \int_0^{\pi/2} \sin({x-y)}f(y)\,dy, \qquad \lambda \in \mathbb R. \end{equation}

and then differentiating again we get \begin{equation} \frac{d^{2}f(x)}{dx^{2}} = - \lambda \int_0^{\pi/2} \cos({x-y)}f(y)\,dy, \qquad \lambda \in \mathbb R. \end{equation}

which is basically \begin{equation} \frac{d^{2}f(x)}{dx^{2}} = - f(x), \qquad \lambda \in \mathbb R. \end{equation}

Hence, \begin{equation} f(x) = a\cos(x) + b \sin(x) \end{equation}

now you can put this in the integral and you will get appropriate values of $a,b$

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You can try to expand $f(x)$ as a Fourier series, and then compare the result, term by term. You will then see that the all higher orders $(n > 1)$ must be zero, since the integral only yields terms of order $1$.

So we now know that $f(x)$ must be of the form: $$f(x) = a \sin(x) + b \cos(x)$$ And inserting this leads to the following equations for $a,b$: $$ a = \lambda\left(\frac{\pi a}{4} + \frac{b}{2}\right)\\ b = \lambda\left(\frac{\pi b}{4} + \frac{a}{2}\right)$$ Unless $\lambda$ is one of two values, $a,b$ must be zero, and so for almost all values of lambda, the only function satisfying your equation is $f(x) = 0$.

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