Need help solving functional analysis problem

  • Thread starter BaitiTamam
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  • #1
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?
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Answers and Replies

  • #2
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.
  • #3
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!
  • #4
##\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 ##.