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The values of $\lambda$ for which the following equation has a non-trivial solution $\phi(x)=\lambda\int_0^{\pi} K(x,t)\phi(t)dt$ where $0\leq x\leq \pi$ and $K(x,t)= \begin{cases} \sin x\cos t & 0\leq x\leq t \\ \cos x\sin t & t\leq x\leq \pi \end{cases} $ are

$(a)$ $\bigg(n+\frac{1}{2}\bigg)^2-1$, $n\in \mathbb{N}$.

$(b)$ $n^2-1$, $n\in \mathbb{N}$.

$(c)$ $\frac{1}{2}(n+1)^2-1$, $n\in \mathbb{N}$.

$(d)$ $\frac{1}{2}(2n+1)^2-1$, $n\in \mathbb{N}$.

I converted this integral equation into its equivalent ordinary second order linear differential equation $\phi''(x)+2\lambda\cos^2(x)\phi(x)=0$ with mixed boundary conditions $\phi(0)=0$ and $\phi'(\pi)=0$. Next I need to find the eigenvalues of this ODE to find the required answer. $\lambda=0$ will give us trivial solution, so this possibility is discarded. But cannot check for the other two cases $\lambda=\mu^2>0$ and $\lambda=-\mu^2<0$. Any help will be appreciated in this regard.

am_11235...
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1 Answers1

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Unfortunately there seems to be a mistake in the above. Substituting the kernel in and given it's form, it is easy to prove that:

$$\phi(x)=\lambda\sin x\int_{x}^\pi dt \cos t~\phi (t)+\lambda\cos x\int_{0}^x dt \sin t~\phi (t)$$

Then applying derivatives successively we can also prove that:

$$\phi'(x)=\lambda\cos x\int_{x}^\pi dt \cos t~\phi (t)-\lambda\sin x\int_{0}^x dt \sin t~\phi (t)$$

$$\phi''(x)=-\lambda\sin x\int_{x}^\pi dt \cos t~\phi (t)-\lambda\cos x\int_{0}^x dt \sin t~\phi (t)-\lambda\cos^2 x~\phi(x)-\lambda\sin^2x~\phi(x)=-(1+\lambda)\phi(x)$$

We also obtain the constraints $\phi(0)=0,\phi'(\pi)=0$ as mentioned in the question.

and now we can solve the equation setting $\omega=\sqrt{1+\lambda}$:

$$\phi(x)=A\cos(\omega t)+B\sin(\omega t)$$

and applying the constraints we obtain

$$A=0~~,~~\omega\pi=(n+\frac{1}{2})\pi$$

and therefore the eigenvalues that produce legitimate solutions to the equation are given by

$$\lambda_n=(n+\frac{1}{2})^2-1$$

DinosaurEgg
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