Boundedness of Extremal

This question was asked in CSIR NET December 2017.

$$I[y]=\int_{0}^{1}\frac{1}{2}[(y')^{2}-4π^{2}(y)^{2}]dx$$

Let $$(P)m= \inf\{I[y]: y\in C^{1}[0,1], y(0)=0,y(1)=0\}$$ Let $$y_{0}$$ satisfy the Euler-Lagrange Equation associated with $$(P)$$. Then which of the following is /are true?

1. $$m= -\infty$$, $$I$$ is not bounded
1. $$m\in R$$ with $$I[y_{0}]=m$$
1. $$m\in R$$ with $$I[y_{0}]>m$$
1. $$m\in R$$ with $$I[y_{0}]

My Attempt

Here the Extremal function will be:

$$y_{0}=c\sin(2πx)$$

if we apply the boundary conditions. The answer should be the first option according to the answer key. But I am stuck around the boundedness of the extremal. Any help would be appreciated. Thank you.

Poincaré - Wirtinger Inequality $$\pi^2 \int_0^1 y^2 \leq \int_0^1 (y^\prime )^2$$, which is valid for any $$y \in C_0^1([0,1]):= C_0([0,1]) \cap C^1([0,1])$$, entails:

$$I[y] = \frac{1}{2}\ \int_0^1 (y^\prime)^2 - 4\pi^2 y^2 \geq \frac{1}{2}\ \int_0^1 \pi^2 y^2 - 4\pi^2 y^2 = -\frac{3}{2}\pi^2 \int_0^1 y^2$$

with equality iff $$y(x) = y_C(x) := C \sin(\pi x)$$; therefore:

$$I[y_C] = -\frac{3}{2}\pi^2 C^2 \int_0^1 \sin^2 (\pi x)\ \text{d}x \to -\infty \qquad \text{as } C \to +\infty$$

and $$I$$ is unbounded from below over $$C_0^1([0,1])$$.

If you are not familiar with inequalities like Poincaré - Wirtinger's, you can argue as follows.

Functions of the family $$y_C(x) := C \sin (\pi x)$$ belong to $$C_0^1([0,1])$$ for each $$C \in \mathbb{R}$$.
Evaluating $$I[\cdot]$$ on $$y_C$$ gives:

$$\begin{split} I[y_C] &= \frac{C^2}{2} \int_0^1 \Big[\pi^2 \cos^2(\pi x) - 4\pi^2 \sin^2(\pi x)\Big]\ \text{d} x\\ &= \frac{\pi^2 C^2}{2} \int_0^1 \Big[ 1 - 5 \sin^2 (\pi x)\Big]\ \text{d} x \\ &\stackrel{t=\pi x}{=} \frac{\pi C^2}{2} \underbrace{\int_0^\pi \Big[ 1 - 5 \sin^2 t\Big]\ \text{d} t}_{= - \frac{3\pi}{2}} \\ &= - \frac{3\pi^2}{4} C^2\end{split}$$

and the unboundedness of $$I[\cdot]$$ follows by letting $$C \to +\infty$$ as above.

• Thank you for your answer, but being a beginner I am unaware of the inequality you used. I had been following Calculus of Variations By I. M. Gelfand, S. V. Fomin. Can you provide me with some reference for the inequality and some related examples so that I can tackle when it pops up in future?
– S.S
Commented Aug 9, 2020 at 20:27
• Gel'fand & Fomin is a really good book, but is a bit outdated for it focus mostly on classical methods. There are many newer books about CoV you could refer to: for example, Dacorogna's Introduction to the Calculus of Variations is a good introductory book to the subject, focusing mostly on direct methods. Commented Aug 9, 2020 at 20:38
• You can find a proof of Poincaré - Wirtinger inequality here on MathSE: math.stackexchange.com/questions/702168/… Commented Aug 9, 2020 at 20:39
• Thank you for your suggestion and help.
– S.S
Commented Aug 9, 2020 at 20:40
• @ShatabdiSinha I edited my answer showing how to solve the problem also when you aren't familiar with P-W inequality. Commented Aug 10, 2020 at 0:27
1. FWIW, one way to physically understand the result, consider the Fourier series $$y(x)~=~\sum_{n\in\mathbb{N}}a_n \sin(\pi n x), \qquad a_n~\in~\mathbb{R}.\tag{A}$$ OP's action functional for the harmonic oscillator becomes $$I[y]~=~\frac{1}{2}\int_0^1 \! dx~(y^{\prime}(x)^2-4\pi^2 y(x)^2)~=~\ldots~=~\frac{\pi^2}{4}\sum_{n\in\mathbb{N}}(n^2-4)a^2_n.\tag{B}$$ We see that the only Fourier mode with negative contribution to the action (B) is the $$n=1$$ mode. By setting all the other modes to zero, and letting $$|a_1|\to\infty$$, we see that the action $$I[y]$$ is unbounded from below, i.e. option (1) is correct.

2. Fun fact: We can read off the classical solutions/natural frequency of the harmonic oscillator from formula (B). It is the mode $$n=2$$ with arbitrary amplitude $$a_2$$. This agrees with OP's last formula. It is a stationary but not a minimum configuration for the action. See also e.g. this related Phys.SE post.