Need help solving functional analysis problem

[BaitiTamam]

hello everyone!
I had a stuck in solving problem for a week now, so need help.
Please 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}<br /> \\\\<br /> u(x)=\int^\pi_0 G(x,y)f(y)dy

Now my question is, what would be a proof for the equation: u&#039;&#039;(x)+u(x)=Wf(x) (for \exists W is a constant).

I found u''(x)+u(x) constantly becomes 0 (for reason that u&#039;&#039;(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?

 

[voko]

<br /> 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<br /> \\ = \int_0^x \phi(x) \psi(y) f(y) dy+ \int_x^{\pi} \psi(x) \phi(y) f(y) dy<br /> \\ = \phi(x) \int_0^x \psi(y) f(y) dy+ \psi(x)\int_x^{\pi} \phi(y) f(y) dy<br />

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

 

[BaitiTamam]

>voko
with your denotation, I found that u&#039;&#039;=-u+(\phi &#039;\psi-\psi &#039;\phi)f conveys the 1st differencial of \phi &#039;\psi-\psi &#039;\phiequals to 0 leads its constancy.
Thank you!

 

[voko]

$\phi '\psi-\psi '\phi$ 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 $\phi '\psi-\psi '\phi$.