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MATH2070 Chap.5 Unconstrained Nonlinear Optimisation

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Chapter 5 of 11 · MATH2070

Unconstrained Nonlinear Optimisation

Unconstrained nonlinear optimisation separates candidate generation from classification. An interior differentiable optimum must have zero gradient, but stationary points may be minima, maxima or saddles. The Hessian supplies local curvature through definiteness or, in two variables, a determinant test. A zero determinant makes the standard test inconclusive and requires direct or higher-order analysis.

Convexity strengthens a stationary local result to a global guarantee, while boundaries require a constrained method or explicit comparison. Candidate equations must be solved completely and each Hessian evaluated at the candidate rather than treated as a universal matrix. Positive diagonal entries alone do not establish a minimum when cross-curvature makes the matrix indefinite.

Degenerate stationary points can be resolved by inspecting directional behaviour or higher-order terms. The domain also controls the claim: an interior condition does not automatically cover boundary points, and local classification does not become global without convexity, concavity or a complete comparison argument.

In this chapter

What this chapter covers

  • 01

    Gradient conditions

  • 02

    Stationary systems

  • 03

    Hessian matrices

  • 04

    Definiteness

  • 05

    Saddle points

  • 06

    Convexity and global minima

Worked example · free

Classify a coupled quadratic

Q [4 marks]. Solve an original gradient system, test the Hessian and compute the exact minimum value. The marks shown are not a University mark allocation.
Hessian curvature classificationThree contour panels show a bowl, a hill and a saddle around stationary points.positive definitelocal minimumnegative definitelocal maximumindefinitesaddle point
Contour curvature around a stationary point determines the classification: bowls rise in every direction, hills fall in every direction, and saddles mix both behaviours.
  • 1Set every gradient component to zero.
  • 1Solve the stationary system.
  • 1Test Hessian definiteness.
  • 1State local or global and compute value.
Positive-definite curvature makes the quadratic strictly convex, so its unique stationary point is the unique global minimum.
Sia tip — A stationary point is a candidate until curvature or structure classifies it.
Glossary

Key terms

Stationary point
A point where every component of the gradient equals zero.
Hessian matrix
The symmetric matrix of second partial derivatives for a twice differentiable function.
Definite matrix
A matrix whose quadratic form keeps one strict sign for every non-zero direction.
Saddle point
A stationary point with improving and worsening directions nearby.
Directional test
A comparison of function behaviour along selected paths through a stationary point. It is useful when the Hessian is inconclusive, particularly for exposing a saddle through rises in one direction and falls in another.
Strict convexity
Curvature making each interior weighted average lie strictly below the matching weighted function values for distinct points. On a convex domain, it ensures that any attained minimiser is unique. A positive definite Hessian throughout the domain is a common sufficient test for twice-differentiable functions.
FAQ

Unconstrained Nonlinear Optimisation FAQ

Why is zero gradient not enough to prove a minimum?

The first-order condition identifies stationary candidates only. A point with zero gradient can be a local maximum, local minimum or saddle, so curvature or another comparison is needed.

What should be written when the Hessian determinant is zero?

State that the second-derivative test is inconclusive. Examine higher-order terms, factor the function, compare nearby paths or use convexity rather than assigning a classification from the diagonal entries.

When does a local stationary result become global?

For a differentiable convex function, every stationary point is a global minimum. Strict convexity additionally gives uniqueness. Without such structure, a positive-definite Hessian at one point proves only a local minimum.

Study strategy

Exam move

Organise every solution around claims rather than derivatives. First state the domain and identify whether interior stationary points are the only candidates; a restricted or non-open domain may also require boundary analysis. Compute the full gradient, solve all simultaneous equations and keep every candidate until classification.

Substitute each point back into the original derivatives to catch solutions introduced by algebraic manipulation. Build Hessian fluency in layers. For two variables, calculate both second derivatives and the mixed derivative, then evaluate the matrix at the candidate before applying determinant and leading-entry tests.

For larger or structured problems, practise recognising positive definiteness through eigenvalues, quadratic forms or leading principal minors when appropriate. Always distinguish a positive semidefinite Hessian from a positive definite one, because the weaker condition may not give strict local behaviour. Include deliberately inconclusive cases.

At a point where the Hessian test degenerates, inspect the function along different directions, compare nearby values or use the first non-zero higher-order term. Mix these with saddle examples whose positive diagonal entries tempt an incorrect minimum claim.

After local classification, ask separately whether convexity or concavity on the whole domain upgrades the result to a global conclusion; local calculus alone does not supply that statement. Keep a four-line audit beneath each problem: candidate set, Hessian at each candidate, classification evidence and global justification.

Rework one problem by completing the square or using geometry when possible, then compare that reasoning with the derivative method. Under time pressure, write the logical sequence before filling arithmetic so missing steps remain visible. Track errors by differentiation, equation solving, definiteness and scope of conclusion.

Revision is complete only when you can explain why a stationary point is necessary in the stated setting and why the chosen classification test is decisive or explicitly inconclusive.

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