2026 Spring Qualifying Exam in Applied Mathematics (AI-generated)

Proof.

If , then for each there is a unique maximal solution. The flow is in and ; if , the flow is correspondingly in initial data.

If and remained bounded as , then would stay in a compact set where is Lipschitz and bounded. Starting from times , the local existence theorem would extend the solution a uniform positive time beyond , hence beyond , contradiction. Thus is unbounded as .

If , then

For finite , Gronwall gives a finite bound on . Therefore no finite-time blowup is possible. The same argument backward in time gives existence for all .

Proof.

The linearization is

with eigenvalues and . The center eigenspace is tangent to . Let the center manifold be

The invariance equation is

To order , the left side has no term, while the right side is , so . To order , the right side gives , while the left side still has no term, so . Thus

The reduced flow is

If , the leading term has the same sign on both sides of the origin, so the origin is not locally asymptotically stable. If , then

The origin is locally asymptotically stable precisely when ; unstable when ; and higher-order analysis is needed when from the displayed terms, though the given leading nonlinearities give no attracting term. Hence, from the computed normal form, stability occurs for and .

Proof.

The reduced outer equation is

so

The reduced equation is singular at , and the right boundary gives

Thus

As , , so the left boundary condition requires a boundary layer. Balance near , giving the scale . The layer correction is , and . Therefore

This satisfies the left condition to leading order and the right condition up to an exponentially small layer contribution.

Proof.

Let . Subtracting the scheme from the exact identity gives

where

The local error and zero-stable two-step structure give global order . Starting values must be generated with at least second-order accuracy, for example using a second-order Runge--Kutta method for .

For , , the recurrence is

The absolute stability domain consists of those for which both roots satisfy the multistep root condition. On the imaginary axis the method is stable for with , and the stability region is the corresponding interval on the imaginary axis:

Proof.

The columns are orthogonal with norms and . Thus singular values are and . Take

This gives an SVD . The least squares solution satisfies

Here

so

For any matrix, by the SVD. If with , then . For nonsingular , applying this to gives , hence .

Proof.

Write

Then

If and , divide numerator and denominator by to get .

If but and , the same argument starts with the second term and gives .

If the first two relevant terms are present, then

for an explicit constant depending on and . Thus

Proof.

The momentum is , so . The Hamiltonian is

Hamilton's equations are

The Hamilton--Jacobi equation is

For ,

Thus

and

Proof.

The direct method takes a minimizing sequence, proves boundedness in , extracts a weakly convergent subsequence using reflexivity, proves the limit is admissible, and uses weak lower semicontinuity to pass to the limit.

If in and each has trace , continuity of the trace operator gives the trace of equal to . Also the functional is continuous linear, so . Hence the admissible set is weakly sequentially closed.

The gradient term is convex and weakly lower semicontinuous. If is convex, then is weakly lower semicontinuous under the stated growth assumptions. Coercivity follows from the -gradient term together with the trace and mean constraints and Poincare-type inequalities. Therefore attains its minimum.

Proof.

Introduce a Lagrange multiplier for the constraint. The first variation of

with constraint gives

Dot with :

Since , differentiating gives , and differentiating again gives

Thus

For ,

Substitution shows the normal components cancel, leaving