show that the integral equation $$\phi(x)- \lambda\int^{\pi}_0 \sin x \sin 2t\phi(t)\,dt=0 , 0 \leq x \leq \pi$$ has no eigenvalue.
can anyone help how can I able to solve this problem please.thanks for your time.
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almost same as this:[http://math.stackexchange.com/questions/349133/solution-of-an-integral-equation-phix-int1-0-xtxt-phit-dt-x-0/349224#349224] – ABC Apr 02 '13 at 17:24
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We have $$\phi(x) = \overbrace{\left(\lambda \int_0^{\pi} \sin(2t) \phi(t) dt\right)}^k \sin(x)$$ Hence, $\phi(x) = k \sin(x)$. This means $$\int_0^{\pi} \sin(2t) \phi(t) dt = k \int_0^{\pi} \sin(2t) \sin(t)dt = 0$$ This gives us $$\phi(x) = 0$$ is the only possible solution to the equation. Hence, $\ldots$