2025 Spring Qualifying Exam in Applied Mathematics (AI-generated)

Proof.

Let and seek . Then

At order ,

so write

At order ,

Put . Since ,

and

The resonant forcing terms are the coefficients of and . To avoid secular growth in , impose

Thus is constant and

The initial data give and , hence and . Therefore

The unperturbed equation with the same data has solution . Thus the perturbation mainly changes the frequency from to at leading multiple-scales order, while keeping the amplitude equal to .

Proof.

(a) The matrix exponential is

This series converges for every and satisfies

Therefore satisfies and . Uniqueness for linear ODEs gives the desired formula.

(b) Since is hyperbolic, no eigenvalue has real part zero. Let be the direct sum of generalized eigenspaces with , and let be the corresponding sum for . Then

Write . Then

On , the exponential decays as , so as . On , the exponential decays backward in time, so as . This proves the decomposition and describes the long-time dynamics.

(c) Let and be the spectral projections onto and . A bounded solution, if it exists, must be

The first integral converges because decays when , and the second converges because decays when . Differentiating under the integral sign gives

Boundedness follows from the exponential decay estimates and boundedness of .

For uniqueness, suppose are bounded solutions. Their difference solves . If has a nonzero unstable component, then is unbounded as ; if it has a nonzero stable component, then is unbounded as . Since is bounded on all of , both components are zero, so . Thus the bounded solution is unique.

Proof.

The linearization at is

At the eigenvalues are and , so there is a one-dimensional center direction and a one-dimensional stable direction.

For a center manifold with parameter , write

The coefficient of is because the center eigenspace at satisfies . The invariance equation is

Using in the equation gives, to second order,

Thus the reduced flow is

This is the normal form of a transcritical-type exchange of equilibria. The reduced equilibria are

The derivative of the reduced vector field at is . Since the transverse eigenvalue remains near , the origin is locally asymptotically stable for and unstable for . At , the reduced equation is , so the origin is semistable, not locally asymptotically stable in a two-sided neighborhood.

Proof.

Taylor expansion gives

Thus

which agrees with the exact Taylor expansion through . Therefore the local truncation error is and the global error is .

For stability on a finite time interval, one assumes is Lipschitz in and sufficiently smooth. Then the one-step map is Lipschitz with constant , so the global error satisfies a discrete Gronwall inequality.

For absolute stability, apply the method to , . Then

so

Hence

This is the absolute stability region of Heun's method.

Proof.

A vector solves the least squares problem if and only if the residual is orthogonal to the column space of :

These are the normal equations

The solution is unique if and only if , equivalently has full column rank. The residual is always unique because is the orthogonal projection of onto , and orthogonal projection onto a subspace is unique.

For the given matrix, solve :

There are infinitely many exact least squares solutions. Write

Minimize

The derivative is , so . Therefore

This is the minimum-norm least squares solution.

Proof.

(a) Suppose with . Then , because otherwise . Moreover

Thus is an eigenvalue of . The converse is identical, replacing and . Algebraic multiplicities agree from

when dimensions differ, with the natural interpretation.

(b) Since , the matrix splits as a block diagonal matrix

where has size and has size . Each block is an unreduced symmetric tridiagonal matrix, so every eigenvalue of each block is simple. If every eigenvalue of had multiplicity at least two, then every eigenvalue of would also have to be an eigenvalue of , and conversely except for counting. But has one more eigenvalue than . Since all eigenvalues inside each block are simple, at least one eigenvalue of the larger block cannot be shared. That eigenvalue appears with multiplicity one in .

(c) Let and . The first column of is , which has norm . A QR factorization can be obtained with

Then

One unshifted QR step gives

Thus

Proof.

The Lagrangian is independent of , so the Beltrami identity gives

Thus

Integrating gives the catenary family

The constants and are determined by

The problem statement says the constant need not be found explicitly. Along such an extremal, and

The strengthened Legendre condition holds, and the catenary extremal has no conjugate point for the minimizing branch connecting the endpoints. Hence the corresponding catenary is a weak local minimum.

Proof.

Let be a minimizing broken extremal, smooth on and and continuous at . Vary the curve by , where but is arbitrary. The first variation is

Integrating by parts on each side gives

The integral terms vanish because each smooth piece satisfies the Euler--Lagrange equation. Since is arbitrary and for a minimizer,

This is exactly the first Erdmann corner condition.

Proof.

We have

Thus the Euler--Lagrange equation is

Let

Solving gives

The Hamiltonian is

Hence

Since is conserved, one may set constant in magnitude along energy levels and integrate the resulting first order system.

The Hamilton--Jacobi equation is

Seek . Then

Thus and

For ,

Therefore a separated solution is