Identifying extrema of a functional

Question

I'm new on Calculus of variations and I don't figure out how to find a minimum (or maximum) for the following functional

$$ J(f) = \int_{-3}^{-2}(f^2(t)+f'(t)) ~dt . \tag{1}$$

I have tried to use the Euler-Lagrange equation

$$\frac{\partial\mathcal{L}}{\partial f}-\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial f'}\right) = 0, \tag{2}$$ where $\mathcal{L}\left(f(t),f^\prime(t),t\right) = f^2(t)+f'(t).$ However, the only solution that I have found was $f(t) =0$, but I don't know if $f(t)=0$ is a local minimum, local maximum or a saddle path of $J$.

Question: If $f$ satisfies the Euler-Lagrange equation, how to test if $f$ is a minimum, maximum or saddle path ? In ordinary calculus, we can compute the Hessian, but I don't know how to proceed in calculus of variations.

Thanks in advance for any help!

Answer 1

$f(t)=0$ is a saddle, indeed there exist

$f_1(t)=-\frac{3}{38}t$ and $f_2(t)=t$ such that

$J(f_1)=-\frac{3}{76}<0=J(f)<\frac{22}{3}=J(f_2)$.

Since from $g(t)=kt$ it follows that $J(g)=\frac{19}{3}k^2+k$, it means that the functional $J$ is unbounded from above, so it does not have maximum value and supremum of $J$ is $+\infty$.

Moreover there exists a sequence of functions $\{h_n(t)\}_{n\in\mathbb{N}-\{0\}}$ defined as $h_n(t)=\frac{1}{2\left(t+2-\frac{1}{n}\right)}$ such that

$J(h_n)=-\frac{n}{4}\left(1-\frac{1}{n+1}\right)$.

It means that the functional $J$ is also unbounded from below, so it does not have minimum value and infimum of $J$ is $-\infty$.

Answer 2

OP's functional is of the form $$J[f]~=~f(b) -f(a) +\int_a^b \mathrm{d}t~f(t)^2. \tag{A}$$ An infinitesimal variation reads $$\delta J[f]~=~\delta f(b) -\delta f(a) +2\int_a^b \mathrm{d}t~f(t)~\delta f(t). \tag{B}$$ If we consider a test function of the form $$ f(t)~=~|t-c|^{-1/3}, \tag{C}$$ then the integral (A) is bounded but we can make $f(a)$ or $f(b)$ go to $+\infty$ by letting $c\to a$ or $c\to b$, respectively. This shows that OP's functional (A) is unbounded from both above and below.

Answer 3

To find the extrema of $$ J(f)=\int_{-3}^{-2}\left(f^2(t)+f'(t)\right)\mathrm{d}t $$ note that $$ \begin{align} \delta J(f) &=\int_{-3}^{-2}\left(2f(t)\,\delta f+\delta f'(t)\right)\mathrm{d}t\\ &=\int_{-3}^{-2}\left(2f(t)+\delta_0(t+2)-\delta_0(t+3)\right)\delta f\,\mathrm{d}t \end{align} $$ which indicates a critical function would look like $$ f(t)=\tfrac12\delta_0(t+3)-\tfrac12\delta_0(t+2) $$ where $\delta_0$ is the usual Dirac delta function.

To approximate this function, we can test $$ f_n(x)=\left\{\begin{array}{} n-n^2(x+3)&\text{if }x\in\left[-3,-3+\frac1n\right]\\ 0&\text{if }x\in\left[-3+\frac1n,-2-\frac1n\right]\\ -n-n^2(x+2)&\text{if }x\in\left[-2-\frac1n,-2\right] \end{array}\right. $$

Computing, we get $$ J(f_n)=-\frac43n $$ We can also test $$ g_n(x)=\left\{\begin{array}{} n^2(x+3)&\text{if }x\in\left[-3,-3+\frac1n\right]\\ 2n-n^2(x+3)&\text{if }x\in\left[-3+\frac1n,-3+\frac2n\right]\\ 0&\text{if }x\in\left[-3+\frac2n,-2-\frac2n\right]\\ -2n-n^2(x+2)&\text{if }x\in\left[-2-\frac2n,-2-\frac1n\right]\\ n^2(x+2)&\text{if }x\in\left[-2-\frac1n,-2\right] \end{array}\right. $$

Computing, we get $$ J(g_n)=\frac43n $$ Thus, there are no extremal functions since functions can be found that go to both $\pm\infty$.

This is not a good example to show how to determine maxima and minima in Calculus of Variations since extremal functions do not exist.