Extremal 2026-10-31
A sufficiently regular admissible function at which the first variation of a functional vanishes for every permitted variation. An extremal is a stationary candidate; it need not be a local maximum or minimum. For a first-order integral functional with fixed endpoint values, it satisfies the Euler-Lagrange equation under the usual regularity assumptions.
With the curvature sign convention in which , the inverse-metric variation of is plus a boundary term, for a scalar multiplier held fixed in that variation. The Palatini identity expresses the curvature variation as divergences of . Two integrations by parts move the derivatives onto , yielding the displayed terms. For constant they vanish. The opposite curvature convention instead has .
Put , the interval of imaginary time, and assume . For a variation with , the first variation of the Euclidean action isThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationIts sign differs from the real-time harmonic oscillator equation because the Euclidean action contains positive kinetic energy and positive potential energy. The general solution is . Applying the two endpoint conditions gives and . Equivalently, the equal-endpoint Euclidean oscillator path isThis form displays reflection about the midpoint and satisfies both endpoints immediately. On this path,Thus the Euclidean action reduces to a boundary term. Since and ,The positive quadratic form is the second variation, so this stationary path is the unique minimum, rather than merely an arbitrary stationary point.
Past exam of the elasticity and plasticity course of the University of Oxford 2008 1 b Solution 2026-10-31
Choose zero energy in the unstrained state. The required strain energy density isIts variation is , giving the prescribed stress as the tensor gradient. This variational notation handles the symmetric off-diagonal components without a spurious factor of two.
Decompose into orthogonal deviatoric and spherical parts. ThenThe two squares vanish together only for . Thus and are sufficient for positive definiteness. They are necessary by testing a nonzero trace-free strain and a pure dilation separately. Equivalently both shear modulus and bulk modulus must be positive.
Past exam of the elasticity and plasticity course of the University of Oxford 2008 3 a Solution 2026-10-31
For an admissible variation with ,Stationarity for every variation therefore gives the Euler-Lagrange equationThe first energy term is bending energy: the curvature of an inextensible planar beam is . The shortening relative to the straight reference is . A prescribed compressive dead load does positive work on that shortening, so its potential-energy contribution is minus times that integral. A minimizer must satisfy the stationary equation, but a stationary solution need not be a minimizer.
Past exam of the elasticity and plasticity course of the University of Oxford 2008 3 c Solution 2026-10-31
Linearization gives . The two homogeneous Dirichlet conditions select . For any small trial amplitude in this mode, , soFor every nonzero quadratic variation is positive by the sharp Poincaré inequality. For , the first mode gives negative energy at sufficiently small amplitude, so zero cannot be a local minimum.
The printed strict “only if” conclusion omits the critical case. At the quadratic test is degenerate, not negative. In fact zero is still the unique minimum of the full displayed functional: use and to get , with strict inequality for a nonzero continuous function because when . In particular on the critical first mode,Thus strict positivity of the second variation holds for , while local minimality of the full energy holds also at . This is the distinction captured by critical-load minimality of the angular Euler strut.
Charles Darwin was neither the first nor the only pre-1859 advocate of changing species. The claim needs a second qualification: believing that species have histories is not the same as identifying natural selection, and neither entails that every lineage changes continuously at a constant rate. Several rival explanations were available before the publication of On the Origin of Species by Charles Darwin.
Jean-Baptiste Lamarck's 1809 programme is an unambiguous earlier example. In Zoological Philosophy by Jean-Baptiste Lamarck, forms which appear stable over a short interval can change through the accumulated effects of circumstances, altered habits, use and disuse, and transmission of acquired modifications. His separate account of increasing organizational complexity also mattered. This was a substantial causal programme, rather than a passing acknowledgment that individuals differ. It was not, however, an anticipation of every feature of Charles Darwin's explanation: an internally ordered tendency toward complexity and a response produced through an organism's activities differ from the sorting of heritable differences through unequal reproductive success.[ref]
Zoological Philosophy by Jean-Baptiste Lamarck. 1809, Part I, chapters III, VI and VII: species, organization and the influence of circumstances
Robert Chambers's anonymously published Vestiges of the Natural History of Creation by Robert Chambers supplied another public alternative in 1844. It placed the succession of living forms within a law-governed history extending from the formation of the heavens to human beings. Its treatment of development was speculative and did not contain the developed selection mechanism of On the Origin of Species by Charles Darwin. Nevertheless, its very existence rules out an intellectual landscape in which an isolated Charles Darwin first introduced the possibility of organic change to an otherwise uniformly fixist audience. The difference between a privately developed theory and a book circulating publicly is especially important here.[ref]
Vestiges of the Natural History of Creation by Robert Chambers. 1844 edition, General Considerations respecting the Origin of the Animated Tribes and Hypothesis of the Development of the Vegetable and Animal Kingdoms
Alfred Russel Wallace provides the decisive counterexample even to the narrower claim of unique possession of natural selection. His 1858 essay argued that varieties could persist and depart progressively from their original type through differences in their capacity to survive under particular conditions. The materials jointly communicated to the Linnean Society on 1 July 1858 included his essay, extracts from Charles Darwin's earlier work and an account of his thinking in correspondence. The introductory communication explicitly presented the two as independent originators. This public event preceded the 1859 book and is stronger evidence than an attempt to infer influence from a loose resemblance between two later texts.[ref]
On the Tendency of Species to Form Varieties by Charles Darwin and Alfred Russel Wallace. 1858, introductory communication dated 30 June and the materials read on 1 July; printed in the Linnean Society journal
There were also resources and disagreements beyond professed transformists. Geology supplied evidence of a long history; palaeontology documented succession and extinction; geographical distribution raised questions about affinities between inhabitants of neighbouring regions. Yet a commitment to extinction did not logically entail common descent, and an appeal to slowly acting geological causes did not oblige an author to accept changing species. The historical question is which inferences particular investigators drew from shared evidence, not whether a predecessor can be recruited to a modern camp merely because one premise resembles a later one.
What distinguished the 1859 synthesis was the scale and integration of the argument. Variation, heredity, the multiplication of organisms and limited opportunities for reproduction made unequal success consequential across generations. Branching descent could then connect adaptation, classification, geographical distribution and the fossil record. The theory did not require a breeder intentionally choosing the direction of change. At the same time, the first edition discussed other possible causes of variation and acknowledged serious difficulties. It should not be retrospectively equipped with a later genetic account of heredity.[ref]
On the Origin of Species by Charles Darwin. First edition, Introduction and chapters III–V, IX–XIII: the mechanism, acknowledged difficulties and the convergence of evidence
The appropriate judgment separates novelty of a general possibility, novelty of a mechanism and persuasiveness of a synthesis. Earlier and contemporary transformists defeat the first claim of solitary originality; the 1858 communication defeats the strongest version of the second. Neither fact erases Charles Darwin's achievement in constructing a sustained case. Equally, the premise of constant change is too crude: stable conditions, unequal rates and lineages persisting with little visible modification are compatible with a history of descent. Historical priority is best assessed by identifying the actual claim and evidence, rather than allocating one indivisible discovery called evolution.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 62 2 a Solution Created 2026-10-03 Updated 2026-10-14
Vary the scalar field and integrate by parts, taking the variation to vanish at the boundary:Thus the curved-spacetime Klein-Gordon equation is . In the flat FRW metric, and . A spatially homogeneous field therefore hasConsequentlyDots now denote cosmic time derivatives. The term is Hubble friction caused by expansion of the comoving volume.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 62 2 b Solution Created 2026-10-03 Updated 2026-10-14
For a variation of the covariant metric tensor, and . Writing , one obtainsThe specified metric-variation convention therefore givesFor , its invariant gradient square is . Hence , while and . Lowering one index makes , withPotential energy has negative pressure, whereas scalar kinetic energy contributes positive pressure. This sign distinction is what permits accelerated expansion.
Past exam of the mathematics course of the University of Cambridge 2013 ib Paper 4 16A Solution Created 2026-09-24 Updated 2026-10-14
For a variation preserving both endpoint values and slopes, at the endpoints. The first variation isIntegrate the last term twice and the middle term once. All boundary terms vanish, givingThe fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equationFor the present functional this becomes . Its general solution is . The left endpoint gives and . The right endpoint then givesThe coefficient determinant is , so , . The stationary function is .
For any admissible perturbation , exact quadratic expansion, followed by the stationarity equation and integration by parts, givesApply the supplied Poincaré inequality first to , then to , whose endpoint values also vanish:Thus for every admissible perturbation, proving a global maximum through the clamped second-variation inequality.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 16 1 a Solution Created 2026-10-03 Updated 2026-10-06
Let be the configuration smooth manifold, and let be a smooth Lagrangian on its tangent bundle. For a path with fixed endpoints, its action isThe principle of stationary action requires the first variation to vanish for every fixed-endpoint variation. In a coordinate chart take , with . Differentiation under the integral and integration by parts giveRepeated coordinate indices are summed. The endpoint term vanishes. Since compactly supported variations can be chosen independently in each coordinate, the fundamental lemma of the calculus of variations yields the Euler-Lagrange equationsConversely these equations make the displayed first variation zero for every fixed-endpoint variation. Variations localized in coordinate charts establish the same assertion for paths on . Thus the Euler-Lagrange equations are exactly the stationary-path condition, not necessarily a condition for an action minimum.
Past exam of the mathematics course of the University of Cambridge 2017 ib Paper 1 4D Solution Created 2026-09-24 Updated 2026-10-05
Take a smooth variation with on the boundary. Assume enough differentiability for the following operations, for example a smooth bounded open set, , and a twice continuously differentiable integrand. Differentiation under the integral givesBy integration by parts, the boundary term vanishes and the first variation isStationarity for every such variation, including every compactly supported smooth one, and the fundamental lemma of the calculus of variations give the Euler-Lagrange equationFor the quadratic integrand, , , and . Thuswith the prescribed Dirichlet boundary condition. For real , a stationary function is also the unique minimizer, if it exists: for any nonzero admissible , the linear term vanishes and . This last positivity uses the zero boundary values; it also holds at .
Past exam of the mathematics course of the University of Cambridge 2017 ii Paper 3 23I Solution Created 2026-09-24 Updated 2026-10-05
The length of a curve is . Its arc length parameter is . Regularity gives , so the inverse function theorem gives a smooth inverse and has .
Reparametrize a length minimizer to constant speed on and use the energy functional . By Cauchy-Schwarz inequality, every competitor has , with equality for constant speed, so this parametrization also minimizes . For a fixed-endpoint variation with tangential vector field , differentiation and integration by parts giveThe boundary term is zero. Arbitrary smooth tangential of compact interior support are realizable by the stated variation fact. Taking to be a nonnegative cutoff times proves . Hence a length minimizer is a geodesic after constant-speed reparametrization. The original parameter need not be affine: arbitrary varying-speed parametrizations preserve length but have nonzero covariant acceleration. This qualification is necessary for the literal wording.
For distinct nonantipodal on the punctured unit sphere, there is a unique shorter great circle arc, of length . A minimizing curve exists exactly when that shorter arc avoids the removed north pole. If it avoids the pole it attains the spherical lower bound. If it passes through the pole, arbitrarily small smooth detours have lengths tending to , but equality would force the unique shorter arc, which is unavailable. Thus the infimum is not attained; the longer great circle arc is not a substitute minimizer.
For antipodal in the punctured sphere, one can choose a great circle semicircle avoiding the north pole; it attains length . Thus every admissible antipodal pair has a minimizer. If identical endpoints are included despite the earlier distinctness stipulation, the infimum is zero, but no smooth regular curve attains it; only a nonregular constant curve does.
Past exam of the mathematics course of the University of Cambridge 2019 ib Paper 4 16A a Solution Created 2026-09-24 Updated 2026-09-29
Write the admissible variation as , where at both ends. Differentiating the functional under the integral sign gives the first variationAfter integration by parts, with the boundary term vanishing,Thus the Euler-Lagrange equation isA second differentiation gives the second variation
Past exam of the mathematics course of the University of Cambridge 2021 ib Paper 2 13D Solution Created 2026-09-24 Updated 2026-09-25
Take a variation with . Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is
Linearizing the equation of motion about givesso the Jacobi equation isWhen has no zero on ,The total derivative integrates to zero because , hence
For the simple harmonic oscillator, the Jacobi equation is . Put and chooseIf , then throughout , so is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.
Past exam of the mathematics course of the University of Cambridge 2021 ib Paper 4 13D a Solution Created 2026-09-24 Updated 2026-09-25
Let , where the variation satisfies . Differentiating the functional at gives its first variationUsing integration by parts and the fixed endpoint conditions,If this vanishes for every admissible , the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation
Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 15B a Solution Created 2026-09-24 Updated 2026-09-25
Write the action density asUnder a compactly supported variation ,Integrating the first term by parts in time and the second in space, and discarding the boundary terms, givesThe principle of stationary action requires this to vanish for every . Sincedivision by yields the field Euler-Lagrange equationThis is the equation generated by the inflaton action in an expanding universe.
Past exam of the mathematics course of the University of Cambridge 2023 ib Paper 2 13C a Solution Created 2026-09-24 Updated 2026-09-25
Take a variation , where the differentiable function obeysbecause both and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on and make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation
Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2003 Paper 1 10 Solution Created 2026-10-18 Updated 2026-10-20
Choose homogeneous Dirichlet boundary conditions, , and nonzero admissible functions with finite and positive . Variations obey the same zero endpoint values. At a stationary Rayleigh quotient, set . ThenThe boundary term vanishes, and the fundamental lemma of the calculus of variations gives the Sturm-Liouville equation . Thus stationary functions are eigenfunctions. Conversely, multiply that equation by and integrate by parts to obtain ; its first variation then vanishes by the same calculation. This proves that stationary values are the corresponding eigenvalues. Homogeneous Neumann boundary conditions also make the displayed boundary terms vanish; general Robin boundary conditions can require a boundary contribution to the quadratic form.
For , , on , the polynomial trial function satisfies the endpoint conditions. Its Rayleigh quotient usesso the polynomial estimate is . The trigonometric trial function givesand hence . Both are upper bounds on the lowest eigenvalue, by the Rayleigh-Ritz method; neither trial is an exact eigenfunction, so the inequalities are strict.
Their closeness has a useful explanation beyond their similar single-arch shapes. Both trial functions are symmetric under , forcing . The reflection-symmetric trials for a linearly weighted Rayleigh quotient therefore reduce this problem to a constant multiple of the ordinary Dirichlet energy quotient. The sine is the best trial within this symmetric class, and the polynomial's value is only a factor larger. The unrestricted lowest eigenfunction can be asymmetric because the weight is asymmetric.
Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge 2003 Paper 1 9 Solution Created 2026-10-18 Updated 2026-10-20
Vary by with . The first variation of the functional isThe boundary term from integration by parts is zero. The fundamental lemma of the calculus of variations therefore gives . For , independently vary and with zero endpoint variations; the two Euler-Lagrange equations areFor the optical path, Fermat principle makes stationary. In cylindrical coordinates its integrand is , with . There is no explicit dependence. The Beltrami identity for several dependent variables givesThe conserved quantity is the axial component of the optical momentum of a ray. As , squaring and inverting yieldsNow suppose and seek a path of radius . The angle is a cyclic coordinate, so is conserved. At fixed radius this is strictly increasing in , forcing constant. The radial Euler-Lagrange equation reduces toConsequently the helical rays in a radially graded refractive index haveA real nonzero helix exists at such a radius when , and its helical pitch is . If , the solution is instead a straight axial light ray. Thus radial dependence permits constant-radius rays for suitable profiles; an arbitrary radial profile need not support a nonzero real helix. This qualification is necessary, for example when the refractive index increases with radius.