2023 Spring Qualifying Exam in Applied Mathematics (AI-generated)

Proof.

The reduced outer equation is , so . It satisfies neither endpoint exactly, but because the highest derivative is multiplied by , endpoint layers are expected. The dominant layer balance is , so the layer scale is at and at .

At , put . The layer correction satisfies and must decay as , so . Since and , .

At , put . The decaying correction is . Since and , . Hence

Proof.

The linearization is

Thus the origin is hyperbolic for : stable for and unstable for .

At , the center variable is and the stable variable is . Seek . To leading order, the stable equation gives

On the center manifold,

This is a supercritical pitchfork normal form. The origin is locally asymptotically stable for , unstable for , and at the reduced equation is asymptotically stable along the center direction while the transverse direction is stable. Hence the origin is also locally asymptotically stable at , though not exponentially stable.

Proof.

Use polar coordinates. The radial equation is

Thus for sufficiently small and for sufficiently large. The angular equation satisfies . Hence there is a trapping annulus containing no equilibrium. By the Poincare--Bendixson theorem, a periodic orbit exists.

Because the vector field points outward on the inner boundary and inward on the outer boundary, the periodic orbit obtained by trapping is expected to be locally asymptotically stable.

Proof.

The method is

Taylor expansion shows agreement with the exact solution through terms of order , hence local error and global order .

Let . Subtracting the numerical update from the exact integral form gives

where is the consistency error. Lipschitz continuity gives stability by discrete Gronwall.

For , ,

Thus

Proof.

The SVD is , where are orthogonal and contains the singular values. Here

Thus the singular values are and . A convenient swaps the order if one wants decreasing singular values. In fact the columns of are orthogonal, so the right singular vectors are the coordinate axes and the left singular vectors are normalized columns:

The pseudoinverse is

For ,

Proof.

If , then

Thus is an eigenvalue of .

For the Gershgorin part, the hypotheses imply each Gershgorin disk is centered at with radius the same number, so every disk lies in the closed right half-plane. Since the matrix is symmetric, all eigenvalues are real, hence nonnegative. Also the zero row-sum condition gives , so is an eigenvalue. If the graph associated to nonzero off-diagonal entries is connected, the nullspace is exactly ; in general its dimension equals the number of connected components.

Proof.

For (a), , so and Euler--Lagrange gives , hence . Without endpoint conditions, extremals form a family. Since , any extremal satisfying the imposed admissible endpoint conditions is a weak minimum.

For (b), Euler--Lagrange gives

With , ,

Since and the second variation is

for nonzero admissible variations, this extremal is a strict weak minimum.

Proof.

The momentum is , so . Hence

Hamilton's equations are

Thus is constant, say , so and , .

The Hamilton--Jacobi equation is

or

A complete integral is

locally, with the usual interpretation that more general solutions follow from characteristics.

Proof.

Let . Taylor expansion and the Euler--Lagrange equation give

Since and ,

Thus is a minimizer.