Exercises — Calculus of variations — functionals, functional derivative

WHAT is $L$? The integrand: $L(x,y,y^{\prime })=3x{y}^{\prime 2}-y^2+5y$. The two partials treat $y$ and $y^{\prime }$ as independent slots (freeze one while differentiating the other): $\frac{\partial L}{\partial y}=-2y+5,\frac{\partial L}{\partial y^{\prime }}=6x{y}^{\prime }.$ Note $\partial L/\partial y^{\prime }$ still carries the $x$; that matters when we later take $\frac{d}{dx}$ of it.

Beltrami needs no explicit $x$ — meaning $x$ appears only through $y(x)$ and $y^{\prime }(x)$, never bare.

• (i) $\sqrt{1+y^{\prime 2}}$: no bare $x$ → applies.
• (ii) $x{y}^{\prime 2}$: bare $x$ out front → does not apply.
• (iii) $y^2+y^{\prime 2}$: no bare $x$ → applies.
• (iv) $y\sqrt{1+y^{\prime 2}}$: no bare $x$ → applies.

WHY E–L here: $L=\sqrt{1+y^{\prime 2}}$ has no bare $y$, so $\partial L/\partial y=0$; E–L reduces to $\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}=0$, i.e. $\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=\text{const}$. A constant of $\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}$ forces $y^{\prime }=\text{const}=m$ (the ratio is a monotonic function of the slope, so fixing it fixes the slope). Thus $y=mx+b$, a straight line — exactly the red line in Figure 2. Fit the endpoints: $y(0)=1\Rightarrow b=1$. $y(3)=7\Rightarrow 3m+1=7\Rightarrow m=2$. So $y=2x+1$. Length: slope $2$ over a run of $3$ gives rise $6$; length $=\sqrt{3^2+6^2}=\sqrt{45}=3\sqrt{5}\approx 6.708$.

$L=y^{\prime 2}+4y$. Partials: $\frac{\partial L}{\partial y}=4$, $\frac{\partial L}{\partial y^{\prime }}=2y^{\prime }$. E–L: $4-\frac{d}{dx}(2y^{\prime })=0\Rightarrow 4-2y^{\prime \prime }=0\Rightarrow y^{\prime \prime }=2$. Integrate twice: $y^{\prime }=2x+C_1$, then $y=x^2+C_{1}x+C_2$. Endpoints: $y(0)=0\Rightarrow C_2=0$. $y(1)=1\Rightarrow 1+C_1=1\Rightarrow C_1=0$. $\boxed{y=x^2}$.

WHAT changes: in the integration-by-parts step the boundary term $[\frac{\partial L}{\partial y^{\prime }}\eta {]}_0^1$ no longer dies at $x=1$, because $\eta (1)$ is not forced to $0$ (the endpoint is free to move).

• At $x=0$: $\eta (0)=0$ (that end is fixed) → term vanishes there.
• At $x=1$: $\eta (1)$ is arbitrary → to kill ${\Phi }^{\prime }(0)$ for all such $\eta$ we need $\frac{\partial L}{\partial y^{\prime }}{|}_{x=1}=0$. This is the natural boundary condition. Interior equation (E–L): $L=\frac{1}{2}{y}^{\prime 2}+y$, so $\partial L/\partial y=1$, $\partial L/\partial y^{\prime }=y^{\prime }$. E–L: $1-y^{\prime \prime }=0\Rightarrow y^{\prime \prime }=1$. Integrate: $y^{\prime }=x+C_1$, $y=\frac{1}{2}{x}^2+C_{1}x+C_2$. Apply conditions: fixed left end $y(0)=0\Rightarrow C_2=0$. Natural condition $y^{\prime }(1)=0\Rightarrow 1+C_1=0\Rightarrow C_1=-1$. $\boxed{y=\frac{1}{2}{x}^2-x}$, and indeed $y^{\prime }(1)=0$: the curve arrives horizontally at the free end.

Total $x$-derivative of $L(y,y^{\prime })$ (chain rule, no bare $x$): $\frac{dL}{dx}=L_y{y}^{\prime }+L_{y^{\prime }}{y}^{\prime \prime }$. Now differentiate the product $y^{\prime }{L}_{y^{\prime }}$: $\frac{d}{dx}(y^{\prime }{L}_{y^{\prime }})=y^{\prime \prime }{L}_{y^{\prime }}+y^{\prime }\frac{d}{dx}{L}_{y^{\prime }}$. Subtract: $\frac{d}{dx}(L-y^{\prime }{L}_{y^{\prime }})=\underset{dL/dx}{\underbrace{L_y{y}^{\prime }+L_{y^{\prime }}{y}^{\prime \prime }}}-\underset{\text{cancels}}{\underbrace{y^{\prime \prime }{L}_{y^{\prime }}}}-y^{\prime }\frac{d}{dx}{L}_{y^{\prime }}.$ The two $y^{\prime \prime }{L}_{y^{\prime }}$ terms cancel — one from $dL/dx$, one from the product rule — leaving $y^{\prime }(L_y-\frac{d}{dx}{L}_{y^{\prime }}).$ By E–L the bracket is $0$, so $L-y^{\prime }{L}_{y^{\prime }}=\text{const}$. The cancellation of the $y^{\prime \prime }$ term is exactly why Beltrami never needs the second derivative.

No explicit $x$, so Beltrami applies. $L=\frac{1}{2}{y}^{\prime 2}-\frac{1}{2}{\omega }^2{y}^2$, $L_{y^{\prime }}=y^{\prime }$. $L-y^{\prime }{L}_{y^{\prime }}=(\frac{1}{2}{y}^{\prime 2}-\frac{1}{2}{\omega }^2{y}^2)-y^{\prime }\cdot y^{\prime }=-\frac{1}{2}{y}^{\prime 2}-\frac{1}{2}{\omega }^2{y}^2=\text{const}.$ Multiply by $-2$: $y^{\prime 2}+{\omega }^2{y}^2=\text{const}=E$. This is a conserved "energy". Consistency check: differentiate $y^{\prime 2}+{\omega }^2{y}^2=E$ w.r.t. $x$: $2y^{\prime }{y}^{\prime \prime }+2{\omega }^{2}y{y}^{\prime }=0\Rightarrow y^{\prime }(y^{\prime \prime }+{\omega }^{2}y)=0$. For non-constant $y$, $y^{\prime \prime }+{\omega }^{2}y=0$, i.e. $y^{\prime \prime }=-{\omega }^{2}y$ — exactly the E–L equation. ✓

WHY a multiplier: we minimise one functional subject to another being constant — the functional analogue of Lagrange multipliers. Form $\tilde{L}=y^{\prime 2}-\lambda y^2$ and apply E–L to $\tilde{L}$. $\frac{\partial \tilde{L}}{\partial y}=-2\lambda y$, $\frac{\partial \tilde{L}}{\partial y^{\prime }}=2y^{\prime }$. E–L: $-2\lambda y-\frac{d}{dx}(2y^{\prime })=0\Rightarrow y^{\prime \prime }=-\lambda y$. With $y(0)=y(\pi )=0$ this is the classic eigenvalue problem: $y=\sin (\sqrt{\lambda }x)$ needs $\sqrt{\lambda }\pi =n\pi$, so $\lambda =n^2$, $y_n=A\sin (nx)$. Smallest $\int y^{\prime 2}$ is the smallest eigenvalue: $n=1$, giving $y=A\sin x$. Fix $A$ from the constraint: ${\int }_0^{\pi }{A}^2{\sin }^{2}xdx=A^2\cdot \frac{\pi }{2}=1\Rightarrow A=\sqrt{2/\pi }$. Minimal value: ${\int }_0^{\pi }{y}^{\prime 2}dx={\int }_0^{\pi }{A}^2{\cos }^{2}xdx=A^2\cdot \frac{\pi }{2}=1$. So $\boxed{y=\sqrt{2/\pi }\sin x,\int y^{\prime 2}=1}$.

WHY a new equation: now $L$ depends on the second derivative, so the variation produces an ${\eta }^{\prime \prime }$ term needing integration by parts twice. Vary $y\to y+\epsilon \eta$ with $\eta (a)=\eta (b)={\eta }^{\prime }(a)={\eta }^{\prime }(b)=0$: ${\Phi }^{\prime }(0)={\int }_a^b(L_y\eta +L_{y^{\prime }}{\eta }^{\prime }+L_{y^{\prime \prime }}{\eta }^{\prime \prime })dx.$ IBP once on the ${\eta }^{\prime }$ term and twice on the ${\eta }^{\prime \prime }$ term; all boundary terms vanish because $\eta$ and ${\eta }^{\prime }$ are pinned at both ends. Extracting the common $\eta$: $\boxed{L_y-\frac{d}{dx}{L}_{y^{\prime }}+\frac{d^2}{d{x}^2}{L}_{y^{\prime \prime }}=0}(\text{Euler–Poisson}).$ Beam: $L=\frac{1}{2}{y}^{\prime \prime 2}$, so $L_y=0$, $L_{y^{\prime }}=0$, $L_{y^{\prime \prime }}=y^{\prime \prime }$. Equation: $\frac{d^2}{d{x}^2}(y^{\prime \prime })=0\Rightarrow y^{(4)}=0$. General solution: $y=c_0+c_{1}x+c_2{x}^2+c_3{x}^3$ — a cubic, the shape of an ideal bent elastic beam. ✓

The theorem in miniature: "$x$-independence of $L$" (a symmetry) forces $H\equiv y^{\prime }{L}_{y^{\prime }}-L$ to be constant along any extremal — this is the conserved Hamiltonian/energy, the Beltrami constant with a sign flip. Beltrami: $L-y^{\prime }{L}_{y^{\prime }}=\text{const}\iff H=y^{\prime }{L}_{y^{\prime }}-L=\text{const}$. Free particle $L=\frac{1}{2}{y}^{\prime 2}$: $L_{y^{\prime }}=y^{\prime }$, so $H=y^{\prime }\cdot y^{\prime }-\frac{1}{2}{y}^{\prime 2}=\frac{1}{2}{y}^{\prime 2}=\text{const}$. Constant $\frac{1}{2}{y}^{\prime 2}$ ⇒ constant $|y^{\prime }|$ ⇒ straight line. Symmetry (no preferred $x$) has produced a conservation law (constant "speed"), the pattern that in Lagrangian mechanics becomes energy conservation. ✓

Full E–L uses both $L_y$ AND $-\frac{d}{dx}{L}_{y^{\prime }}$ ::: yes — never just $L_y=0$. Beltrami condition ::: no explicit (bare) $x$ in $L$. Free endpoint gives ::: natural condition $\partial L/\partial y^{\prime }=0$ there. Constraint $\int g=$const handled by ::: functional Lagrange multiplier, extremise $L-\lambda g$. $L$ depends on $y^{\prime \prime }$ ⇒ equation ::: $L_y-\frac{d}{dx}{L}_{y^{\prime }}+\frac{d^2}{d{x}^2}{L}_{y^{\prime \prime }}=0$. $x$-independence of $L$ ⇒ ::: conserved $H=y^{\prime }{L}_{y^{\prime }}-L$ (Beltrami / baby Noether).