Functionals of Several Functions and Higher Derivatives

Ordinary calculus of variations tells one story: a single curve $y(x)$ threading two fixed endpoints so as to make $S[y]=\int_a^b F(x,y,y')\,\mathrm{d}x$ stationary, and the single Euler–Lagrange equation that story produces. Physics almost never presents itself so tidily. A double pendulum has two angles; a planet has three coordinates; and the energy stored in a bent ruler does not care about its slope at all — it cares about its curvature, the second derivative. Both departures from the textbook case — many functions, and derivatives beyond the first — are handled by the same variational reflex, and mastering them turns the calculus of variations from a geometry curiosity into the working language of mechanics, elasticity, and field theory. Two moves do all the work. When the integrand depends on several functions $y_1,\dots,y_n$ , each is perturbed by its own independent bump $\eta_i$ ; because the bumps are independent, the first variation splits into $n$ pieces that must each vanish on their own — one Euler–Lagrange equation per function, a coupled system. When the integrand depends on $y,y',\dots,y^{(m)}$ , the single bump $\eta$ appears differentiated up to $m$ times, and each derivative is integrated by parts back off the bump — $k$ times for the $k$ -th slot, flipping the sign each time. The alternating tower this builds is the Euler–Poisson equation, an ODE of order $2m$ . The leftover surface terms are not debris; they are the boundary physics — clamped, hinged, or free.

Key takeaways

A functional of several functions $y_1,\dots,y_n$ is varied one function at a time; independence of the variations yields one Euler–Lagrange equation per function — a coupled system of ODEs.
A functional with higher derivatives up to $y^{(m)}$ yields a single Euler–Poisson equation of order $2m$ , with the signature alternating sign $\sum_{k=0}^{m}(-1)^k\tfrac{\mathrm{d}^k}{\mathrm{d}x^k}\tfrac{\partial F}{\partial y^{(k)}}=0$ .
Higher-derivative problems demand more boundary data: an order- $2m$ ODE needs $2m$ conditions, supplied by fixing $y,y',\dots,y^{(m-1)}$ at each end or by natural boundary conditions on the conjugate quantities.
These two generalizations turn the calculus of variations into Lagrangian mechanics (several coordinates) and Euler–Bernoulli beam theory (curvature energy $\tfrac12 EI(y'')^2$ ).
Key idea: independent variations decouple — each independent function gives its own Euler–Lagrange equation, and each higher derivative contributes an integration-by-parts term with an alternating sign, $\sum_{k=0}^{m}(-1)^k\dfrac{\mathrm{d}^k}{\mathrm{d}x^k}\dfrac{\partial F}{\partial y^{(k)}}=0$ .

Where this lives today

The two branches are anything but historical. The several-functions branch is the skeleton of computational physics: molecular dynamics, robotics, and spacecraft trajectory optimization all integrate coupled Euler–Lagrange equations for thousands of generalized coordinates, and the same structure underlies every Lagrangian field theory in the Standard Model. The higher-derivative branch is just as alive. Euler–Bernoulli beam theory and its curvature functional govern everything from bridges to the silicon microcantilevers of atomic force microscopes, where a beam tens of micrometres long deflects by nanometres to read a single molecule; the same elastica energy models the buckling of DNA. And at the frontier, higher-derivative Lagrangians are a live and dangerous tool: $f(R)$ modified-gravity theories and the Pais–Uhlenbeck oscillator contain terms in $\ddot q$ or higher and run into the Ostrogradsky instability — the theorem that a non-degenerate higher-derivative theory has an energy unbounded below. Deciding when that "ghost" is fatal and when it can be tamed is an open problem, and it is nothing but the physics of equation (4) taken seriously.

The physical picture

Picture the action as a machine with a bank of knobs — one knob for the value of each function at each point along the interval. A stationary configuration is one where nudging any single knob by an infinitesimal amount leaves the action unchanged to first order. If the problem has several functions, the knobs belonging to $y_1$ are physically independent of the knobs belonging to $y_2$ : you can wiggle one family while holding the other fixed. So the "no first-order change" demand must hold separately for each family, and that is precisely why $n$ independent functions give $n$ separate Euler–Lagrange equations rather than one blended condition. The equations may still be coupled — $F$ can mix the functions — but they are counted one per function.

Figure

Two independent functions, two independent bumps: wiggling y₁ with η₁ while holding y₂ fixed forces one stationarity condition; wiggling y₂ forces another, so several functions give one Euler–Lagrange equation each — ∂F/∂yᵢ − (d/dx)(∂F/∂yᵢ′) = 0

Now let the action feel curvature. A bent beam stores energy according to how sharply it bends — its second derivative — not merely how steeply it tilts. To extract a pointwise equation you must repeatedly push derivatives off the variation and onto the integrand, and each push through an integration by parts flips a sign; that alternating bookkeeping is the whole content of the Euler–Poisson equation. The price of curvature is paid at the ends: each integration by parts deposits a surface term, so a second-derivative problem leaves two boundary contributions — one proportional to $\delta y$ , one to $\delta y'$ — and whether you kill them by fixing the end or by cancelling the coefficient is the difference between a clamped beam and a free one.

Figure

A cantilever bent by its own uniform load q — clamped at the left (y = 0, y′ = 0), free at the right (y″ = 0, y‴ = 0). The higher-derivative Euler–Poisson equation gives EI y⁴ = q, whose tip droop is δ = qL⁴/(8EI) = 2.4 mm for E = 200 GPa, I = 4.17×10⁻⁶ m⁴, L = 2 m, q = 1000 N/m

The core equations

A functional of several functions is the natural home of a mechanical system with many coordinates. The integrand $F$ carries all $n$ functions and all their derivatives over one shared interval.

S[y_1,\dots,y_n]=\int_a^b F\!\left(x,y_1,\dots,y_n,y_1',\dots,y_n'\right)\mathrm{d}x \qquad (1)

where: $y_i(x)$ — the $i$ -th unknown function, $i=1,\dots,n$ ; $y_i'$ — its derivative $\mathrm{d}y_i/\mathrm{d}x$ ; $F$ — the integrand (a "Lagrangian density"), a function of $2n+1$ slots; $[a,b]$ — the interval with endpoints fixed. Reading: one scalar action assembled from $n$ competing profiles. Demanding stationarity under independent variations of each function produces not one equation but $n$ of them, each with the familiar Euler–Lagrange shape. They are coupled through $F$ but indexed one per function.

\frac{\partial F}{\partial y_i}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y_i'}=0,\qquad i=1,\dots,n \qquad (2)

where: $\partial F/\partial y_i$ — partial derivative holding every other slot fixed; $\mathrm{d}/\mathrm{d}x$ — the total derivative, through $x$ and every $y_j(x),y_j'(x)$ . Reading: $n$ second-order ODEs for $n$ unknowns — a well-posed system. The higher-derivative functional keeps one unknown but lets $F$ depend on derivatives up to order $m$ . Curvature energy ( $m=2$ ) is the important case, but the structure is general.

S[y]=\int_a^b F\!\left(x,y,y',y'',\dots,y^{(m)}\right)\mathrm{d}x \qquad (3)

where: $y^{(k)}$ — the $k$ -th derivative $\mathrm{d}^k y/\mathrm{d}x^k$ ; $m$ — the highest derivative present in $F$ . Reading: the action now senses slope, curvature, and beyond, not just position. Stationarity of (3) yields a single ordinary differential equation of order $2m$ : the Euler–Poisson equation. Each derivative slot is integrated by parts back onto $F$ , and the number of integrations is recorded as an alternating sign.

\sum_{k=0}^{m}(-1)^{k}\,\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}\!\left(\frac{\partial F}{\partial y^{(k)}}\right)=0 \qquad (4)

where: $(-1)^k$ — the sign left by $k$ integrations by parts; $\partial F/\partial y^{(k)}$ — derivative of $F$ with respect to its $k$ -th derivative slot. Reading: for $m=1$ this is Euler–Lagrange; for $m=2$ it is $\partial_y F-\tfrac{\mathrm{d}}{\mathrm{d}x}\partial_{y'}F+\tfrac{\mathrm{d}^2}{\mathrm{d}x^2}\partial_{y''}F=0$ . Integrating a second-derivative term by parts twice leaves two surface terms behind. Their vanishing is a separate demand from the field equation, and it encodes the endpoint physics.

\Big[\,\delta y\Big(\frac{\partial F}{\partial y'}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y''}\Big)+\delta y'\,\frac{\partial F}{\partial y''}\Big]_a^b=0 \qquad (5)

where: $\delta y$ — variation of the deflection at an endpoint; $\delta y'$ — variation of the slope there. Reading: at each end, fix $y$ and $y'$ (essential/clamped), or annihilate the coefficients (natural/free). Feeding the curvature energy of a slender rod into the Euler–Poisson equation collapses everything to the workhorse of structural engineering — the fourth-order Euler–Bernoulli beam equation.

\frac{\mathrm{d}^2}{\mathrm{d}x^2}\!\left(EI\,\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right)=q(x) \qquad (6)

where: $E$ — Young's modulus; $I$ — second moment of area; $q(x)$ — transverse load per unit length; $y(x)$ — transverse deflection. Reading: the fourth derivative of the deflection, weighted by the stiffness $EI$ , balances the load. Finally, choosing time as the variable and generalized coordinates as the functions turns (2) into the central equation of analytical mechanics.

\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=0,\qquad i=1,\dots,n \qquad (7)

where: $L=T-V$ — the Lagrangian; $q_i$ — generalized coordinate; $\dot q_i$ — generalized velocity. Reading: several-functions Euler–Lagrange is Lagrange's equations of motion.

Derivation ACoupled Euler-Lagrange equations for several functions

The goal is to show that a functional built from $n$ functions splits under variation into $n$ separate stationarity conditions. We perturb every function by its own independent bump, expand the action to first order, integrate each velocity term by parts to strip the derivative off the bump, then switch the bumps on one at a time. The fundamental lemma does the final work, converting "the integral vanishes for every admissible set of bumps" into "each coefficient vanishes pointwise" — which is $n$ equations, one per function.

Step 1. Perturb each function independently, $y_i(x)\to y_i(x)+\varepsilon\,\eta_i(x)$ , with smooth comparison functions obeying $\eta_i(a)=\eta_i(b)=0$ . Why: pinning every $\eta_i$ at the endpoints keeps the competitors inside the same fixed-endpoint class, so the boundary terms will later drop and only the bulk equation survives.

Step 2. Form the first variation by differentiating under the integral and setting $\varepsilon=0$ , giving $\displaystyle\delta S=\int_a^b\sum_{i=1}^n\Big(\frac{\partial F}{\partial y_i}\eta_i+\frac{\partial F}{\partial y_i'}\eta_i'\Big)\mathrm{d}x$ . Why: to first order the change in the action is linear in the bumps, and the chain rule distributes it across every function and every velocity slot.

Step 3. Integrate each velocity term by parts, $\displaystyle\int_a^b\frac{\partial F}{\partial y_i'}\eta_i'\,\mathrm{d}x=\Big[\frac{\partial F}{\partial y_i'}\eta_i\Big]_a^b-\int_a^b\Big(\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y_i'}\Big)\eta_i\,\mathrm{d}x$ , and drop the boundary term since $\eta_i$ vanishes at both ends. Why: this moves the derivative off the arbitrary $\eta_i'$ and onto $F$ , so every term now multiplies the undifferentiated bump $\eta_i$ — the only form the fundamental lemma can act on.

Step 4. Collect to get $\displaystyle\delta S=\int_a^b\sum_i\Big(\frac{\partial F}{\partial y_i}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial F}{\partial y_i'}\Big)\eta_i\,\mathrm{d}x=0$ , then switch on one bump at a time (set all $\eta_j=0$ except $\eta_i$ ). Why: with only $\eta_i$ active the sum reduces to a single integral, and the fundamental lemma forces its bracket to vanish for all $x$ ; sweeping $i$ over $1,\dots,n$ delivers all $n$ equations. *For $n=1$ this is the ordinary Euler–Lagrange equation; for $F=L=T-V$ in time $t$ it is Lagrange's equations of motion.*

Go deeper · Derivation B · 1 minThe Euler-Poisson equation for higher derivatives

Derivation BThe Euler-Poisson equation for higher derivatives

The aim is to establish the alternating-sign tower for a functional depending on derivatives up to order $m$ , and to see clearly where the signs come from. A single bump $\eta=\delta y$ enters the first variation differentiated up to $m$ times, and each differentiated copy must be integrated by parts enough times to land back on a bare $\eta$ . The count is exact: the $k$ -th slot needs $k$ integrations by parts, each contributing a factor $-1$ , so the $k$ -th term arrives with sign $(-1)^k$ . Assembling all $m+1$ terms and invoking the fundamental lemma produces one ODE of order $2m$ .

Step 1. Write the first variation, $\displaystyle\delta S=\int_a^b\sum_{k=0}^m\frac{\partial F}{\partial y^{(k)}}\,\eta^{(k)}\,\mathrm{d}x$ , where $\eta^{(k)}$ is the $k$ -th derivative of the bump. Why: because $F$ lists $y,y',\dots,y^{(m)}$ as separate arguments, the chain rule attaches one term to each, and the $k$ -th carries $\eta$ differentiated $k$ times.

Step 2. Integrate the $k$ -th term by parts $k$ times to reach $\displaystyle\int_a^b\frac{\partial F}{\partial y^{(k)}}\eta^{(k)}\,\mathrm{d}x=(-1)^k\int_a^b\Big(\frac{\mathrm{d}^k}{\mathrm{d}x^k}\frac{\partial F}{\partial y^{(k)}}\Big)\eta\,\mathrm{d}x+\text{boundary}$ . Why: each integration by parts trades one derivative on $\eta$ for one derivative on $\partial F/\partial y^{(k)}$ and flips the sign; after $k$ of them the bump is bare and the accumulated sign is $(-1)^k$ .

Step 3. Impose $\eta(a)=\eta(b)=\dots=\eta^{(m-1)}(a)=\eta^{(m-1)}(b)=0$ so every surface term vanishes, leaving $\displaystyle\delta S=\int_a^b\Big(\sum_{k=0}^m(-1)^k\frac{\mathrm{d}^k}{\mathrm{d}x^k}\frac{\partial F}{\partial y^{(k)}}\Big)\eta\,\mathrm{d}x=0$ . Why: fixing the function and its first $m-1$ derivatives at both ends is exactly the data that annihilates all the boundary terms generated by the repeated parts, isolating the bulk integrand.

Step 4. Apply the fundamental lemma to the arbitrary $\eta$ , forcing the whole bracket to vanish pointwise — the Euler–Poisson equation, an ODE of order $2m$ . Why: the lemma converts "vanishes against every admissible $\eta$ " into "vanishes identically", and the highest derivative present, $\mathrm{d}^m/\mathrm{d}x^m$ of $\partial F/\partial y^{(m)}$ , makes it order $2m$ . *For $m=1$ the Euler–Poisson equation collapses to Euler–Lagrange; for $m=2$ and $F=\tfrac12 EI(y'')^2-q\,y$ it yields $\tfrac{\mathrm{d}^2}{\mathrm{d}x^2}(EI\,y'')=q$ , the Euler–Bernoulli equation.*

Go deeper · Derivation C · 1 minNatural boundary conditions for a free beam end

Derivation CNatural boundary conditions for a free beam end

The purpose here is to show that the surface terms left over from the higher-derivative integration by parts are the source of the physical boundary conditions at any end you do not clamp. We keep the $m=2$ boundary bracket, refuse to prescribe $y$ or $y'$ at the right end, and let the variational principle decide what holds there. Because $\delta y(b)$ and $\delta y'(b)$ are then free and independent, each coefficient must vanish separately, and inserting the beam energy turns those into "zero bending moment" and "zero shear force" — the free-end conditions, derived rather than assumed.

Step 1. From the $m=2$ analysis, the surviving right-end terms are $\Big[\big(\tfrac{\partial F}{\partial y'}-\tfrac{\mathrm{d}}{\mathrm{d}x}\tfrac{\partial F}{\partial y''}\big)\delta y+\tfrac{\partial F}{\partial y''}\,\delta y'\Big]_{x=b}=0$ . Why: at a free end nothing pins the deflection or slope, so unlike the clamped case we cannot set $\delta y(b)=\delta y'(b)=0$ — the terms stay and must be dealt with honestly.

Step 2. Because $\delta y(b)$ and $\delta y'(b)$ are arbitrary and independent, set each coefficient to zero separately: $\tfrac{\partial F}{\partial y''}\big|_b=0$ and $\big(\tfrac{\partial F}{\partial y'}-\tfrac{\mathrm{d}}{\mathrm{d}x}\tfrac{\partial F}{\partial y''}\big)\big|_b=0$ . Why: a sum $A\,\delta y+B\,\delta y'$ can vanish for all independent choices of the two variations only if $A=B=0$ — the same independence argument that gave one equation per function now gives one condition per free variation.

Step 3. Insert the pure curvature energy $F=\tfrac12 EI(y'')^2$ , so $\partial F/\partial y''=EI\,y''$ and $\partial F/\partial y'=0$ , yielding $EI\,y''(b)=0$ and $\tfrac{\mathrm{d}}{\mathrm{d}x}\big(EI\,y''\big)\big|_b=0$ . Why: $EI\,y''$ is the bending moment and its derivative is the shear force, so the natural conditions read directly as an unloaded free end carrying neither moment nor shear. *These are the standard free-end conditions $M=EI\,y''=0$ , $V=(EI\,y'')'=0$ ; a clamped end instead prescribes $y=y'=0$ , and a simple (hinged) support gives $y=0$ , $EI\,y''=0$ .*

Limiting cases

The machinery reduces to a chain of familiar special cases as you dial down the functions, the highest derivative, or the coupling:

Case	Result	Meaning
Single function, first derivative	$\dfrac{\partial F}{\partial y}-\dfrac{\mathrm{d}}{\mathrm{d}x}\dfrac{\partial F}{\partial y'}=0$	recovers ordinary Euler–Lagrange when $n=1$ , $m=1$
$n$ functions, no cross-terms	$F=\textstyle\sum_i F_i(x,y_i,y_i')$	the system separates into $n$ independent Euler–Lagrange equations
No explicit $x$ (higher order)	$\displaystyle\sum_{k\ge 1}\Big(\sum_{j\ge k}(-1)^{j-k}\tfrac{\mathrm{d}^{j-k}}{\mathrm{d}x^{j-k}}\tfrac{\partial F}{\partial y^{(j)}}\Big)y^{(k)}-F=\text{const}$	a Beltrami-type first integral — the higher-order energy is conserved
Pure curvature energy	$F=\tfrac12 EI(y'')^2$	Euler–Poisson gives $EI\,y^{(4)}=0$ , the unloaded Euler–Bernoulli beam
Time as variable, $F=T-V$	$\tfrac{\mathrm{d}}{\mathrm{d}t}\partial_{\dot q_i}L-\partial_{q_i}L=0$	Lagrange's equations for $n$ generalized coordinates

Worked examples

1 · Deflection of a loaded cantilever (fully worked)

A steel cantilever of length $L=2\ \mathrm{m}$ is clamped at $x=0$ and free at $x=L$ , and carries a uniform transverse load $q=1000\ \mathrm{N/m}$ . Its rectangular cross-section is $b=0.05\ \mathrm{m}$ wide and $h=0.10\ \mathrm{m}$ deep, with $E=200\ \mathrm{GPa}$ . Find the stiffness $EI$ , the deflection profile, and the tip droop.

Step 1. The bending stiffness needs the second moment of area of a rectangle, $I=bh^3/12=(0.05)(0.10)^3/12=5\times10^{-5}/12=\mathbf{4.17\times10^{-6}\ m^4}$ , so $EI=(200\times10^9)(4.17\times10^{-6})=\mathbf{8.33\times10^{5}\ N\,m^2}$ . This is the coefficient multiplying $y^{(4)}$ in the field equation.

Step 2. With constant $EI$ and constant $q$ , the Euler–Bernoulli equation (6) is $EI\,y^{(4)}=q$ , integrated four times to $y(x)=\dfrac{q}{24EI}x^4+\dfrac{c_3}{6}x^3+\dfrac{c_2}{2}x^2+c_1x+c_0$ . The four constants are exactly the four boundary conditions the fourth-order equation demands.

Step 3. Apply the clamped end at $x=0$ (essential conditions $y(0)=0$ , $y'(0)=0$ ): these force $c_0=0$ and $c_1=0$ . Apply the free end at $x=L$ (natural conditions $y''(L)=0$ , $y'''(L)=0$ ): from $y'''=\tfrac{q}{EI}x+c_3$ we get $c_3=-qL/EI$ , and from $y''=\tfrac{q}{2EI}x^2+c_3x+c_2$ we get $c_2=qL^2/2EI$ . The free-end conditions are the natural boundary conditions derived above — zero moment and zero shear.

Step 4. Assemble the profile $y(x)=\dfrac{q}{24EI}\big(x^4-4Lx^3+6L^2x^2\big)$ and evaluate at the tip $x=L$ : the bracket is $L^4-4L^4+6L^4=3L^4$ , so $y(L)=\dfrac{q\,(3L^4)}{24EI}=\dfrac{qL^4}{8EI}$ . Numerically $y(L)=\dfrac{(1000)(2)^4}{8(8.33\times10^5)}=\dfrac{16000}{6.67\times10^6}=\mathbf{2.4\times10^{-3}\ m}$ .

Step 5. The tip droops by $\mathbf{\delta=2.4\ mm}$ , about $L/830$ — comfortably small, so the linear small-slope theory is self-consistent. The calculation used both the higher-derivative field equation and its natural boundary conditions; drop either and the answer is wrong. δ = qL⁴/(8EI) = 2.4 mm.

2 · A central-force orbit as a two-function system

A particle of mass $m$ moves in a plane under a central potential $V(r)$ . Using polar coordinates $r(t)$ and $\phi(t)$ as the two functions, derive its equations of motion and identify the conserved quantity that the $\phi$ -equation hands you.

Step 1. The Lagrangian in polar coordinates is $L=\tfrac12 m(\dot r^2+r^2\dot\phi^2)-V(r)$ , a functional of two independent functions $r(t)$ and $\phi(t)$ — the several-functions setting with $n=2$ and time as the variable.

Step 2. The $r$ -equation is $\tfrac{\mathrm{d}}{\mathrm{d}t}\partial_{\dot r}L-\partial_r L=0$ : with $\partial_{\dot r}L=m\dot r$ and $\partial_r L=m r\dot\phi^2-V'(r)$ , this gives $\mathbf{m\ddot r-m r\dot\phi^2+V'(r)=0}$ — Newton's radial law with the centrifugal term restored.

Step 3. The $\phi$ -equation is $\tfrac{\mathrm{d}}{\mathrm{d}t}\partial_{\dot\phi}L-\partial_\phi L=0$ : since $\phi$ is cyclic ( $L$ has no bare $\phi$ ), $\partial_\phi L=0$ and $\tfrac{\mathrm{d}}{\mathrm{d}t}(m r^2\dot\phi)=0$ . The bracket is constant.

Step 4. That constant is the angular momentum $\ell=m r^2\dot\phi=\text{const}$ — conservation falls out directly because one function was cyclic. Two functions give two coupled equations; the cyclic one delivers a conservation law $\ell = m r^2\dot\phi$ .

Physics of the portrait

The generalization to several functions and higher derivatives is where the calculus of variations grew up into physics — carrying the names of Euler, who first minimized a bending energy; Lagrange, who made "several functions" the coordinates of mechanics; and Poisson, whose name rides the higher-derivative equation and whose elasticity work is exactly where curvature functionals live.

Origins · 1 minSiméon Denis Poisson and how this was discovered

Siméon Denis Poisson

1781–1840 · elasticity and the higher-derivative variational calculus

Siméon Denis Poisson was Laplace and Lagrange's most prodigious student at the École Polytechnique, and he spent a career turning their methods into equations that still bear his name — the Poisson equation of potential theory, the Poisson brackets of mechanics, the Poisson distribution, and, in the theory of stressed solids, Poisson's ratio. That last thread is why his name attaches to this subject. When a variational integrand depends on curvature or higher derivatives, the stationarity condition generalizes Euler–Lagrange into the alternating-sign tower now called the Euler–Poisson equation, and Poisson's memoirs on the calculus of variations wrestled directly with such higher-order integrands and the extra boundary conditions they force. His work on the bending and vibration of elastic plates and rods — the first serious continuum theories of curvature energy $\int(y'')^2$ — is the physical shadow of that mathematics: the plate and beam equations are exactly the Euler–Poisson equations of an elastic energy that cares about the second derivative. This variational calculus is his natural monument.

History

1744

— Euler's Methodus inveniendi founds the calculus of variations and, following Daniel Bernoulli, solves the elastica by minimizing bending energy

1755

— the nineteen-year-old Lagrange sends Euler his δ-algorithm, mechanizing the variation of several coordinates at once

1788

— Lagrange's Mécanique analytique recasts mechanics as a system of coupled Euler-Lagrange equations in generalized coordinates

1833

— Poisson's memoir on the calculus of variations treats integrands containing higher derivatives and their boundary terms

1850

— Ostrogradsky's canonical formalism handles higher-derivative Lagrangians, and exposes the instability now named for him

Going further

Texts. Gelfand & Fomin, Calculus of Variations, §§10–12 (several functions and higher derivatives) · Stone & Goldbart, Mathematics for Physics, ch. 1 · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., ch. 22 · Lanczos, The Variational Principles of Mechanics, ch. 2 · Landau & Lifshitz, Theory of Elasticity, §§17–20. The $(-1)^k$ sign structure follows by induction on the integrations by parts. The $m=2$ case reduces to $\tfrac{\mathrm{d}^2}{\mathrm{d}x^2}(EI\,y'')=q$ , and the cantilever tip deflection is $\delta=qL^4/(8EI)=2.4\ \mathrm{mm}$ with $EI=8.33\times10^5\ \mathrm{N\,m^2}$ ; $E=200\ \mathrm{GPa}$ for steel is standard.

Prerequisites. First Variation and Stationarity · Euler–Lagrange Equation · Fundamental Lemma of the Calculus of Variations · Functionals and Function Spaces. Leads to. Lagrangian Mechanics · Hamilton's Principle of Stationary Action · Natural and Essential Boundary Conditions · Geodesics as Variational Problems.

Ask the tutor: Want to see the clamped, simply-supported, and free versions of the same beam solved side by side, or how the Ostrogradsky ghost emerges from a concrete $\ddot q^2$ Lagrangian?

Why does having several functions give a coupled system rather than one big equation?

Because you vary the functions independently. When you perturb $y_i\to y_i+\varepsilon\eta_i$ , the bump for one function has nothing to do with the bump for another — you can switch one on while holding the rest at zero. The first variation is a sum of integrals, one per function, and setting all bumps but one to zero isolates a single integral whose bracket the fundamental lemma forces to vanish. Doing this for each function produces $n$ equations. They can still be coupled — $F$ may mix $y_1$ and $y_2$ , so each equation contains the others — but there are exactly $n$ of them, which is what makes the system well-posed.

Where exactly does the alternating sign in the Euler-Poisson equation come from?

It is pure integration-by-parts bookkeeping. In the first variation the $k$ -th derivative slot contributes $\partial F/\partial y^{(k)}$ times $\eta^{(k)}$ , the bump differentiated $k$ times. To apply the fundamental lemma you need the bump bare, so you integrate by parts $k$ times, each move shifting one derivative from $\eta$ onto $\partial F/\partial y^{(k)}$ and introducing one factor $-1$ . After $k$ moves the accumulated sign is $(-1)^k$ : position ( $k=0$ ) is $+$ , slope ( $k=1$ ) is $-$ , curvature ( $k=2$ ) is $+$ , and so on. That is the entire origin of the $\sum_k(-1)^k$ pattern — no deeper principle, just counting sign flips.

How many boundary conditions do I actually need, and why do people get it wrong?

An Euler–Poisson equation from an integrand of highest derivative $m$ is an ODE of order $2m$ , so it needs $2m$ boundary conditions — for a beam, $m=2$ and you need four. People get it wrong by importing the habit from first-order problems, where two conditions (one per end) suffice, and simply prescribing $y$ at both ends of a beam. That leaves two constants undetermined. The fix is to supply $m$ conditions at each end: fix or free the whole set $y,y',\dots,y^{(m-1)}$ . For a beam that means specifying, at each end, either the deflection or the shear, and either the slope or the bending moment — the essential/natural pairing.

What is the difference between essential and natural boundary conditions?

Essential (or "geometric") conditions are ones you impose by hand because the setup fixes them: a clamped beam end has $y=0$ and $y'=0$ because the wall holds it there, and these are built into the space of admissible functions. Natural conditions are ones the variational principle derives: at any end you do not constrain, the leftover boundary terms must vanish on their own, and since the variations there are free, their coefficients must be zero — for a beam, $EI\,y''=0$ (no moment) and $(EI\,y'')'=0$ (no shear) at a free end. Essential conditions constrain the trial functions; natural conditions come out of the mathematics, and together they always add up to $2m$ .

Is a functional of several functions the same thing as a Lagrangian in mechanics?

Structurally, yes — that is the point. Take the variable to be time $t$ , the functions to be generalized coordinates $q_1(t),\dots,q_n(t)$ , and the integrand to be $L=T-V$ . Then the several-functions Euler–Lagrange system is precisely Lagrange's equations of motion, $\tfrac{\mathrm{d}}{\mathrm{d}t}\partial L/\partial\dot q_i-\partial L/\partial q_i=0$ . The abstract "coupled stationarity conditions of a multi-function action" and the concrete "equations of motion" are the same equations in different clothes — which is exactly why Hamilton's principle can serve as the foundation of mechanics.

Why are higher-derivative theories considered dangerous in modern physics?

Because of the Ostrogradsky instability. When a Lagrangian depends non-trivially on accelerations or higher derivatives (and is "non-degenerate"), Ostrogradsky's 1850 canonical analysis shows the Hamiltonian is linear in one conjugate momentum, so the energy is unbounded below — the system can lower its energy without limit by exciting a "ghost" degree of freedom, and coupling to matter makes it unstable. This is why fundamental theories are kept first-order in time, and why higher-derivative modified-gravity models must be built with special degeneracies to dodge the ghost. The threat is a direct consequence of the order- $2m$ structure the Euler–Poisson equation makes explicit.