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Let $X=\left\{f\in C^1[0,1]: f(0)=0=f(1)\right\}$.

Define $J:X \to \Bbb R$ by $J(f)=\int\limits^1_0 \left(f^\prime (x)^2-4\pi^2 f(x)^2\right)dx.$

Does $\inf\limits_{f\in X}J(f)$ exist?

If $F(x,f,f^\prime)=\left(f^\prime(x)^2-4\pi^2 f(x)^2\right)$, then to minimize $J(f)$, the function $F$ must satisfy the Euler-Lagrange equation: $F_{f}-\frac{d}{dx}\left(F_{f^\prime}\right)=0$. Let me know how to approach this? I came to know that $\inf\limits_{f\in X}J(f)=-\infty$. How?

Edit: I have corrected the definition of $J(f)$.

Community
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user149418
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2 Answers2

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Take $f_n(x)=n\sin(\pi x)$. Then $$ J(f_n)=\int_0^1(n^2\pi^2\cos^2(\pi x)-4\pi^2n^2\sin^2(\pi x))\,dx=-\frac{3\pi^2}{2}n^2\to -\infty,\quad n\to+\infty $$

A.Γ.
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  • Though I agree this is correct, how would one arrive at this without guessing? What if the boundary conditions didn't necessarily admit an obvious guess? – Gregory Jun 23 '18 at 21:28
  • @Gregory Yeah, some kind of guessing is always necessary in counter-examples, but here trial and error is quite intuitively obvious. I think even a simple triangle shape like $f(x)=x$ first and $1-x$ at the end may work here, since the coefficient $4\pi^2$ is large. Once you found a function with negative $J$ the rest is just to amplify it because $J$ is homogeneous. – A.Γ. Jun 23 '18 at 21:50
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The E-L equation yields $f'' + 4 \pi^2 f = 0$. The result is $f(x) = A \sin (2\pi x) + B \cos (2 \pi x)$. The boundary conditions imply $f(x) = A \sin (2\pi x)$. Plugging this into $J$ \begin{align*} J(f) & = 4 \pi ^2 A^2 \int_0^1 \cos^2(2 \pi x ) - \sin^2(2\pi x) \ dx \\ & = 4 \pi^2 A \int_0^1 \cos(4\pi x) \ dx \\ & = \pi A \sin (4\pi x) \big|_0^1 \\ & = 0. \end{align*} The minimum of the function should be 0. In fact if we try the Ansatz $f_n = \sin (2 \pi n x)$ we also see that it takes a minimum value at $0$.

Gregory
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