# Need help solving functional analysis problem

hello everyone!
I had a stuck in solving problem for a week now, so need help.

the problem is as follows.

In a closed interval $I=[0,\pi]$, the 2-times continuously differentiable function $\phi(x)$ and $\psi(x)$ meet the following conditions (they're ranged in $\mathbb{R}$).

$\psi ''(x)+\psi(x)=\phi ''(x)+\phi(x)=0$

Assume $f(x)$ be a continuous function defined in $I$, and let $G(x,y), u(x)$ as followings.
$$G(x,y)=\Bigg\{\begin{array}{l}\phi(x)\psi(y)\quad (0\leq y\leq x\leq \pi) \\ \psi(x)\phi(y)\quad (0\leq x\leq y\leq \pi) \end{array} \\\\ u(x)=\int^\pi_0 G(x,y)f(y)dy$$

Now my question is, what would be a proof for the equation: $u''(x)+u(x)=Wf(x)$ (for $\exists W$ is a constant).

I found u''(x)+u(x) constantly becomes 0 (for reason that $u''(x)=\displaystyle{\int^\pi_0 \frac{\partial^2G(x,y)}{\partial{x}^2}f(y)dy} = -u(x)$).

Any hidden concept or my ignorance makes this so hard?

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$$u(x)=\int^\pi_0 G(x,y)f(y)dy = \int_0^x G(x, y) f(y) dy+ \int_x^{\pi} G(x, y) f(y) dy \\ = \int_0^x \phi(x) \psi(y) f(y) dy+ \int_x^{\pi} \psi(x) \phi(y) f(y) dy \\ = \phi(x) \int_0^x \psi(y) f(y) dy+ \psi(x)\int_x^{\pi} \phi(y) f(y) dy$$

Differentiate this carefully and you will see that your conclusion ## u''(x) = -u(x) ## is incorrect.

>voko
with your denotation, I found that $u''=-u+(\phi '\psi-\psi '\phi)f$ conveys the 1st differencial of $\phi '\psi-\psi '\phi$equals to 0 leads its constancy.
Thank you!

##\phi '\psi-\psi '\phi ## is not zero generally. It is a constant. The two functions are solutions of y'' + y = 0, and the general form of the solution is well known. Use it to prove the constancy of ##\phi '\psi-\psi '\phi ##.