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

[itex]\psi ''(x)+\psi(x)=\phi ''(x)+\phi(x)=0[/itex]

Assume [itex]f(x)[/itex] be a continuous function defined in [itex]I[/itex], and let [itex]G(x,y), u(x)[/itex] as followings.
[tex]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[/tex]

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

I found u''(x)+u(x) constantly becomes 0 (for reason that [itex]u''(x)=\displaystyle{\int^\pi_0 \frac{\partial^2G(x,y)}{\partial{x}^2}f(y)dy} = -u(x)[/itex]).Any hidden concept or my ignorance makes this so hard?

 

[voko]

[tex]
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
[/tex]

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

 

[BaitiTamam]

>voko
with your denotation, I found that [itex]u''=-u+(\phi '\psi-\psi '\phi)f[/itex] conveys the 1st differencial of [itex]\phi '\psi-\psi '\phi[/itex]equals 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 ##.

 

[blue_raver22]

Hello,

It seems like you are trying to solve a functional analysis problem involving continuous functions and integration. This type of problem can be quite challenging, so it is understandable that you have been stuck for a week.

Firstly, let me clarify some notation for the benefit of those reading this response. In the given problem, \phi(x) and \psi(x) are two continuous, twice continuously differentiable functions defined on the closed interval I=[0,\pi]. They satisfy the following conditions:

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

Furthermore, we have a continuous function f(x) defined on the same interval I. We define two new functions G(x,y) and u(x) as follows:

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, the question is asking for a proof of the equation:

u''(x)+u(x)=Wf(x) (for \exists W is a constant).

In other words, we need to show that there exists a constant W such that the given equation holds for any choice of f(x).

You have already made some progress in your attempt to solve this problem. You have shown that u''(x)+u(x) is always equal to 0. This is because u''(x) can be written as an integral involving the second derivative of G(x,y) with respect to x, and this derivative is equal to -u(x) (as you have correctly shown).

However, this does not necessarily imply that u''(x)+u(x)=Wf(x) for some constant W. In fact, it is not true in general. To prove this, we can consider a counterexample. Let us choose a specific function f(x) and see if we can find a constant W such that the equation holds.

Let f(x) = x. Then, u(x) = \int^\pi_0 G(x,y)ydy = \frac{1}{2}x^2\psi(x).

Now, let us assume that there exists