2024 Spring Qualifying Exam in Applied Mathematics (AI-generated)

Proof.

Let and write . At order , . At order ,

Put . Then and

Also

The resonant terms vanish if

Hence is constant and

With , this gives

Thus

The amplitude slowly increases from to , while the leading phase remains .

Proof.

A function is a Lyapunov function if has a fixed sign along trajectories; it is strong if the sign is strict away from equilibria. A periodic orbit would force to return to its initial value, contradicting strict monotonicity. Therefore a strong Lyapunov function excludes periodic orbits.

Here

If , then , with equality only at the origin, so there are no periodic orbits.

If , the origin is repelling because the linearization has trace and positive determinant. On the ellipse , the negative quartic terms dominate and one checks . Thus trajectories starting in a suitable annulus are trapped: the inner boundary repels and the outer ellipse points inward. By the Poincare--Bendixson theorem, the annulus contains a periodic orbit.

Proof.

The linearization is

At the eigenvalues are and , so the origin is nonhyperbolic and has a one-dimensional center manifold. Put . Then

and the equation has the form

On the center manifold . Substitution gives

Thus the origin changes stability at ; the reduced dynamics has the pitchfork-type cubic saturation shown above. Since the transverse eigenvalue stays near , the origin is locally asymptotically stable for , unstable for , and nonhyperbolic at .

Proof.

The method is the trapezoidal quadrature rule applied to , so its local truncation error is and the global order is . It is convergent under the standard one-step assumptions: consistency, Lipschitz continuity in , and solvability of the implicit equation for small .

For , , the method gives

so

A method is A-stable if whenever . Since

the trapezoidal method is A-stable.

Proof.

Symmetry is clear. If , then full column rank gives , so

Thus is positive definite. The least squares solution satisfies

For the given matrix,

Therefore

Proof.

The power method iterates and estimates the dominant eigenvalue by the Rayleigh quotient. The leading block has eigenvalues and , so the eigenvalues are . The convergence factor is about , close to , hence slow.

Apply the power method to . Choosing gives transformed eigenvalues . The eigenvalue corresponding to is dominant, and the convergence factor improves to .

Proof.

The Euler--Lagrange equation is

For , , so , a constant. The endpoints are also satisfied.

For any admissible , . The first variation vanishes, and the exact difference is

Thus is a global strong minimum, hence also a weak minimum.

Proof.

The momentum is

Solving for gives . Therefore

The Hamilton--Jacobi equation is

With ,

Thus a separated solution is

Hamilton's system follows from , .

Proof.

The flaw is that bounded sequences in infinite-dimensional spaces need not have strongly convergent subsequences. For example, is bounded in but has no strongly convergent subsequence.

The direct method fixes this by using weak compactness in a reflexive space, weak closedness of the admissible set, coercivity, and weak lower semicontinuity of the functional. These hypotheses allow a minimizing sequence to converge weakly to an admissible limit and allow the inequality

to pass the minimum to the limit.