I could not prove the following:

A function $f \in \mathscr{C}^2([0, \pi])$, such that $$f(0) = f(\pi) = 0,\\ \int_0^{\pi} (f'(x))^2dx = 1,\\ \text{and }\int_0^{\pi} (f(x))^2dx = 2$$ Then such function does not exist.

I think that I have to use the Rayleigh quotient and have a contradiction for the eigenvalue $\frac{1}{2}$. Thanks in advance.

  • 1
    $\begingroup$ To @Mårten W who edited this question earlier, pde tag was relevant, because this question can be related to the eigenvalue problem of the minus Laplacian $-\Delta u = \lambda u$. $\endgroup$ – Shuhao Cao Jul 3 '13 at 22:26
  • $\begingroup$ Did you mean $\int_0^\pi f^2=2$ and $\int_0^\pi(f')^2=1$? If that the case you can use Fourier analysis! $\endgroup$ – Mercy King Jul 3 '13 at 22:43
  • $\begingroup$ @ShuhaoCao I had this question in a pde test yesterday and it was written $f'$, maybe that was mistyped by my professor. $\endgroup$ – user40276 Jul 3 '13 at 22:56
  • $\begingroup$ @user40276 It should be a typo as you said, because the Reyleigh quotient is for $-f'' = \lambda f$. $\endgroup$ – Shuhao Cao Jul 3 '13 at 23:01
  • $\begingroup$ @ShuhaoCao The question is not necessarily related to the Rayleigh quotient, this was just what I have tried (I tried Bessel inequality and a Fourier expansion too). Furthermore the Rayleigh quotient is not used only for the Laplacian, it's used in other symmetric (hermitian) operators. $\endgroup$ – user40276 Jul 3 '13 at 23:08

It does exist. For example, $$f(x)=a x(\pi-x)+bx^2(\pi-x)$$ with $a\approx-0.718151$, $b\approx 0.290939$ (exact values are too long to type but can be found easily).

Another example: $$f(x)=2\sqrt{\frac{2}{\pi}} \sin^2x-\frac{32\sqrt{2}+\sqrt{2048-144\pi^2}}{6\pi^{3/2}}\sin x.$$

Maybe you have forgotten to mention some further condition on $f$.

  • $\begingroup$ There are actually four solutions of the form $f(x)=(a+bx)x(\pi-x)$. The LaTeX was too long, so I posted it as another answer. $\endgroup$ – robjohn Jul 3 '13 at 22:39
  • $\begingroup$ I think OP means to prove that the function doesn't exist for eigenvalue problem $-f'' = \lambda f$, it is a standard exercise in pde class. $\endgroup$ – Shuhao Cao Jul 3 '13 at 22:41

This was meant as a comment to O.L.'s answer, but it was too long:

Here is an array of the four values of $(a,b)$ for $f(x)=(a+bx)x(\pi-x)$ that satisfy the given conditions:

$$ \left( \begin{array}{cc} \frac{\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}-\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi ^5} , -\frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}} \\[6pt] \frac{\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}+\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi ^5} , -\frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}} \\[6pt] \frac{-\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}-\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi ^5} , \frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}} \\[6pt] \frac{-\pi ^{5/2} \sqrt{210 \left(-5+\pi ^2\right)}+\sqrt{630 \pi ^5-30 \pi ^7}}{4 \pi ^5} , \frac{\sqrt{\frac{1}{2} \left(-525+105 \pi ^2\right)}}{\pi ^{7/2}} \end{array} \right) $$


I think for user40276 mentioned the Rayleigh quotient, he meant to prove that

There does not exist such $f\in C^2$ solving the Dirichlet eigenvalue problem for $$-\Delta f= -f''= \lambda f\tag{1}$$ on the interval satisfying those conditions.

Notice the condition means that the $f$ is an eigenfunction of $-\Delta$ with eigenvalue $2$. All the eigenfunction have the form $$v(x) = A\sin(\sqrt{\lambda}x) + B\cos(\sqrt{\lambda}x)$$ which solves problem (1). Now for the boundary condition: $$ v(0) =0 \implies B=0, $$ so $v(x) = A\sin(\sqrt{\lambda}x)$, now apply the other boundary condition: $$ v(\pi) =0 \implies A\sin(\sqrt{\lambda}\pi) = 0\implies \sqrt{\lambda} = k\in \mathbb{Z}. $$ This says that the eigenvalue for problem (1) can only be complete squares, $1,4,9,\ldots$. Hence $2$ can not be one.

UPDATE: OP updated that the eigenvalue problem should be $$f' + \lambda f = 0,\tag{2}$$ then the first integration condition is $$ \int_0^{\pi} \big(f'(x)\big)^2dx = \int_0^{\pi} \big(\lambda f(x)\big)^2dx= 2, $$ together with the second it implies $\lambda = \pm\sqrt{2}$. The solution to $(2)$ is $f = Ce^{\pm \sqrt{2} x}$. Notice here it is not $f =c_1 e^{\sqrt{2} x} + c_2 e^{-\sqrt{2} x}$, $f$ can be just one, not a linear combination of both. So there is no way $f(0) = f(\pi) = 0$.


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