Question bank — Calculus of variations — functionals, functional derivative

False. It only enforces $\delta J=0$ (a stationary point); the solution could be a maximum or a saddle, just as $f^{\prime }(x)=0$ finds maxima and inflections too. You need a second-variation test to confirm a minimum.

False. If $L$ has no bare $y$, then $\partial L/\partial y=0$, so E–L reduces to $\frac{d}{dx}(\partial L/\partial y^{\prime })=0$, meaning $\partial L/\partial y^{\prime }$ is constant — not $y$. For $L=\sqrt{1+y^{\prime 2}}$ this gives a straight line with nonzero slope.

False. It is a whole function of $x$ — one "gradient component" per point, defined by $\delta J=\int \frac{\delta J}{\delta y(x)}\delta y(x)dx$. Only after evaluating at a specific $x$ do you get a number.

Partly false. The E–L interior equation holds either way, but with free endpoints the boundary term $[L_{y^{\prime }}\eta {]}_a^b$ survives and adds the natural boundary conditions $L_{y^{\prime }}=0$ at the free end. So the full stationarity condition genuinely changes.

False. It is a first integral of E–L, valid only when $\partial L/\partial x=0$. It contains E–L (differentiate it and you recover $y^{\prime }(L_y-\frac{d}{dx}{L}_{y^{\prime }})=0$), so it carries the same information with one fewer derivative.

False. $\eta$ is arbitrary in the interior but is forced to satisfy $\eta (a)=\eta (b)=0$, because the trial curve $y+\epsilon \eta$ must keep the endpoints pinned. That very constraint is what kills the boundary term.

False. The outer $\frac{d}{dx}$ in E–L is a total derivative that also sees $y(x)$ and $y^{\prime }(x)$ through the chain rule, producing $L_{y^{\prime }x}+L_{y^{\prime }y}{y}^{\prime }+L_{y^{\prime }{y}^{\prime }}{y}^{\prime \prime }$. The two orders of operation do not commute here.

Error: $\partial L/\partial y=0$ only kills the first term. E–L still demands $\frac{d}{dx}(y^{\prime }/\sqrt{1+y^{\prime 2}})=0$, forcing constant slope. Not every curve qualifies — only straight lines.

Error: $y$ and $y^{\prime }$ are coupled — perturbing $y(x)$ automatically perturbs $y^{\prime }(x)$. Dropping the $-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}$ term throws away half the condition (and, in mechanics, all the dynamics).

Error: The boundary term is not "small," it is exactly zero here — and only because $\eta (a)=\eta (b)=0$. Calling it "small" hides the fact that with free endpoints it is a genuine, nonzero condition you must keep.

Error: The Fundamental Lemma needs the integral to vanish for every smooth $\eta$ vanishing at the ends, not one. One $\eta$ tells you almost nothing; the "all $\eta$" is what pins $f$ down pointwise.

Error: The second $L$ contains an explicit $x$, so $\partial L/\partial x\ne 0$ and Beltrami does not apply. Beltrami is licensed only when $x$ is absent from $L$.

Error: A minimum forces ${\Phi }^{\prime }(0)=0$ but ${\Phi }^{\prime \prime }(0)\ge 0$, not $=0$. Setting ${\Phi }^{\prime \prime }(0)=0$ would be the condition for a degenerate/inflection case, not a generic minimum.

Because the unknown is a function $y(x)$, so stationarity must hold against wiggles at every point $x$; that infinite family of conditions collapses (via the Fundamental Lemma) into E–L holding at each $x$ — a differential equation.

$\Phi (\epsilon )$ is an ordinary function of the single number $\epsilon$, so we can use ordinary calculus; moving $\partial /\partial \epsilon$ inside the $x$-integral (legal for smooth integrands) turns a hard variational problem into a familiar chain-rule computation.

We want to extract a single common factor $\eta$ so the Fundamental Lemma can act. As long as $\eta$ and ${\eta }^{\prime }$ both appear, we cannot factor and conclude the bracket is zero; IBP moves the derivative off ${\eta }^{\prime }$ onto $L_{y^{\prime }}$.

The Fundamental Lemma needs a "bump" concentrated wherever $f\ne 0$, and the derivation used ${\eta }^{\prime }$ inside an integral — both require enough differentiability. Smoothness also matches the smoothness assumed of $y$.

It is a first-order relation (a first integral), so it already used up one integration; you avoid computing $y^{\prime \prime }$ and the messy total derivative, turning a 2nd-order ODE into a 1st-order one.

In finite dimensions the gradient has one component per coordinate $x_i$; here the "coordinates" are the values $y(x)$ at each point $x$, so the gradient has one component per point — a function rather than a vector.

At a fixed endpoint $\eta$ is forced to zero, killing the boundary term automatically. At a free endpoint $\eta$ may be nonzero there, so for $\delta J=0$ to still hold the coefficient $L_{y^{\prime }}$ must vanish instead.

The $y^{\prime \prime }$-term coefficient $L_{y^{\prime }{y}^{\prime }}$ vanishes, so E–L degenerates into an algebraic (non-differential) constraint $P_y-Q_x=0$; there is generally no free curve to solve for — a sign the problem is ill-posed or has a hidden divergence structure.

Then $\tilde{y}=y$ and ${\Phi }^{\prime }(0)$ trivially gives $0=0$ — no information. This is exactly why we demand the condition for all nonzero admissible $\eta$; the zero perturbation is the useless degenerate case.

The catenary solution $y=c\cosh (\frac{x-x_0}{c})$ has minimum value $c>0$; forcing endpoints too close to the axis (small $y$) can make no smooth solution exist, and the true minimiser degenerates to a "Goldschmidt" discontinuous solution — a genuine failure of the smooth theory.

Yes — the E–L solution is $y=mx+b$ with slope fixed by the endpoints; equal endpoints just give $m=0$, a horizontal segment, still the shortest path. No degeneracy occurs.

Then one integration by parts is not enough; you integrate by parts twice, producing the higher Euler–Lagrange equation $L_y-\frac{d}{dx}{L}_{y^{\prime }}+\frac{d^2}{d{x}^2}{L}_{y^{\prime \prime }}=0$ and two boundary terms at each end, requiring extra endpoint data.

$L=y$ gives $\partial L/\partial y=1$ and $\partial L/\partial y^{\prime }=0$, so E–L reads $1=0$ — no stationary curve exists. The functional simply has no interior extremum; it is pushed to the boundary of whatever constraints you add.

All partials vanish, so E–L is $0=0$: every admissible curve is stationary because $J$ is the same constant $\times (b-a)$ for all of them. The variational problem is degenerate — nothing to optimise.