advanced28 min studyLesson 341 in Mathematics

Functionals and first variations

Differentiate an objective over admissible functions while keeping the function space, boundary values, and topology explicit.

On this page

What you’ll understand

  • Specify a functional and its admissible perturbations.
  • Compute a directional first variation with justified differentiation.
  • Separate stationarity from local minimality and existence.

A stretched string is described by an entire shape, not by one adjustable number. Optimizing its energy therefore means comparing functions. Calculus of variations begins by asking how the objective changes along a small perturbation of a candidate shape. The calculation resembles an ordinary derivative, but the admissible perturbations depend on boundary conditions and on the chosen function space. Those choices determine what a local minimum means. A vanishing first variation identifies a candidate; it does not promise that the candidate minimizes, or even that a minimum exists somewhere.

The domain is part of the optimization problem

A functional maps a function to a number, often by integrating an energy density depending on its position and derivative. For a classical calculation take continuously differentiable real functions on a compact interval with fixed endpoint values. Their differences vanish at the endpoints. Alternatively the Sobolev space H1(a,b)H^1(a,b) consists of square-integrable functions with square-integrable weak derivative. In one dimension these functions have continuous representatives and endpoint traces, so prescribed endpoint values remain meaningful even when a classical derivative does not exist everywhere.

The admissible set is usually an affine set rather than a vector space: adding two functions with the same nonzero boundary value doubles that value. Perturbations instead lie in its associated homogeneous space. A fixed-endpoint perturbation hh must vanish at both ends. Additional inequalities can permit motion in only one direction, changing a zero-derivative condition into an inequality. A norm also matters: being close in maximum displacement allows sharp changes of slope, whereas being close in the continuously differentiable norm controls both shape and slope.

J[u]=∫abF(x,u(x),u′(x)) dx,A={u∈H1(a,b):u(a)=A, u(b)=B},h∈H01(a,b)J[u]=\int_a^b F(x,u(x),u'(x))\,dx,\qquad \mathcal A=\{u\in H^1(a,b):u(a)=A,\ u(b)=B\},\quad h\in H_0^1(a,b)
This Sobolev admissible set is suitable when the integrand has growth making the integral finite. The homogeneous perturbation space preserves its endpoint traces.↔ Scroll wide equations horizontally.

Differentiate along an admissible line

Fix a candidate and an admissible direction, then consider the scalar function of the perturbation parameter obtained by substituting u+εhu+\varepsilon h into the objective. Its derivative at zero is the first variation. In a classical setting, a continuously differentiable integrand on a neighborhood of the compact set traced by the curves supplies bounded derivatives and justifies differentiation under the integral. In a Sobolev setting, appropriate growth and integrability hypotheses are needed instead; formal differentiation alone does not establish them.

For smooth data the chain rule yields an integral linear in the direction and its derivative. Directional derivatives considered one direction at a time are weaker than Fréchet differentiability, which requires an error small relative to the chosen norm uniformly over nearby directions. The quadratic energy on H1H^1 has the stronger property directly from its exact expansion. For a general functional, existence of directional derivatives does not automatically supply a continuous linear derivative or a uniform approximation suitable for a numerical method.

δJ[u](h)=ddεJ[u+εh]∣ε=0=∫ab(Fuh+Fvh′) dx\delta J[u](h)=\left.\frac{d}{d\varepsilon}J[u+\varepsilon h]\right|_{\varepsilon=0}=\int_a^b(F_u h+F_v h')\,dx
Here v denotes the independent derivative argument of F. The last equality assumes the interchange of differentiation and integration is justified.↔ Scroll wide equations horizontally.

What a stationary function does and does not establish

At a local minimum with two-sided admissible line perturbations, the scalar function has a minimum at zero for each direction, so every first variation must vanish. Call this stationarity. A local maximum satisfies the same first-order condition, as can a saddle. The topology specifies which nearby functions are compared, and the phrase local minimum is incomplete without that context. For convex differentiable functionals on convex sets, first-order inequalities can become sufficient, but that requires a separate structural argument.

Existence is a different question again. On an open admissible interval even an ordinary smooth function can decrease toward an excluded boundary without attaining its infimum. In function spaces, minimizing sequences can oscillate or concentrate, leaving their limiting behavior outside a classical admissible class or defeating passage of the energy to the limit. Later lessons address these failures using weak compactness and lower semicontinuity. For now, keep a clear record of the space, constraints, derivative calculation, and conclusion actually proved.

An exact expansion around a straight shape

Let J[u]=12∫01(u′)2dxJ[u]=\tfrac12\int_0^1(u')^2dx with u(0)=0,u(1)=1u(0)=0,u(1)=1. Perturb u(x)=xu(x)=x by h(x)=x(1−x)h(x)=x(1-x) and compute the complete dependence on ε\varepsilon.

  1. The direction vanishes at both endpoints, so u+εhu+\varepsilon h preserves the boundary values. Its derivative is 1+ε(1−2x)1+\varepsilon(1-2x).

  2. Expand the square. The linear term integrates to ε∫01(1−2x)dx=0\varepsilon\int_0^1(1-2x)dx=0, while ∫01(1−2x)2dx=1/3\int_0^1(1-2x)^2dx=1/3.

  3. Therefore J[x+εh]=1/2+ε2/6J[x+\varepsilon h]=1/2+\varepsilon^2/6. Its first derivative at zero vanishes and its second derivative is 1/3>01/3>0 along this direction.

The result

This direction increases the energy quadratically. Establishing minimality against every admissible direction requires the corresponding general expansion or another global argument.

Explore the model

Interactive lab

Interactive lab · function

A stationary scalar restriction can bend either way

Reset and set both coefficient a=1a=1 and position x=0.5x=0.5; predict height 0.250.25 and slope 11. Reset and set both a=−1,x=0.5a=-1,x=0.5; predict height −0.25-0.25 and slope −1-1. Both parabolas have zero slope at the origin but opposite curvature. This scalar illustration explains why stationarity is insufficient; it does not evaluate the string functional or its admissible space.

A stationary scalar restriction can bend either wayThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forZero first-order change must be supplemented by curvature or convexity information.

Check your understanding

For fixed endpoint values u(0) = 0 and u(1) = 1, which perturbation preserves both values for every small positive and negative ε?

Try it yourself

For J[u]=12∫01[(u′)2+u2]dxJ[u]=\tfrac12\int_0^1[(u')^2+u^2]dx, endpoints zero and one, compute δJ[x](h)\delta J[x](h) for h=x(1−x)h=x(1-x). Is the straight shape stationary?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Separate the derivative-energy term from the displacement-energy term. The integral of h′ vanishes because h has equal endpoint values.

Compare with the solution

The variation is ∫01[h′+xh]dx=0+∫01(x2−x3)dx=1/12\int_0^1[h'+xh]dx=0+\int_0^1(x^2-x^3)dx=1/12. It is nonzero, so u=xu=x is not stationary for this functional. Choosing a small negative perturbation parameter lowers the objective to first order, despite the same straight shape being stationary for the derivative-only energy.

Take it with you

  • Admissible variations are determined by function spaces and constraints.
  • A first variation differentiates the objective along a permitted direction.
  • Stationarity, local minimality, and existence are separate claims.

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