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

Part A

Problem A1.

Let be a flow, . (a) For , define the omega limit set . Show that if the forward orbit is bounded, then is nonempty. (b) Suppose has nonempty omega limit set . Show that is invariant; that is, if , then for all . (c) Suppose satisfies . Show that must be a fixed point. (d) Suppose and there exist satisfying for all and . Show that either is a fixed point or lies on a periodic orbit.

Proof.

(a) For , the omega limit set is

Equivalently,

If the forward orbit is bounded, then each closed tail

is a nonempty compact set, and the family is nested: when . By compactness, the intersection of these nested compact sets is nonempty. Hence .

(b) Let , so for some . For fixed , for all large we have , and the flow property and continuity give

Since , this proves . Thus is invariant.

(c) Suppose as . For any fixed ,

Thus is fixed by the whole flow, and in particular it is an equilibrium.

(d) Let and . Passing to a subsequence, set

Then

If , then lies on a periodic orbit, with period dividing unless is an equilibrium. If , then

because the vector field is continuous and . The numerator tends to along the chosen sequence, so . Hence is a fixed point. Therefore either is fixed or lies on a periodic orbit.

Problem A2.

(a) By converting to polar coordinates, show that

has at least one periodic orbit. Clearly state any theorems you use. (b) By considering the surface or otherwise, show that

admits at least one periodic orbit.

Proof.

(a) Put , . A direct computation gives

It is convenient to use as the independent variable and set . Since ,

and therefore

This linear equation has a unique -periodic solution,

which is positive for every , since its oscillatory amplitude is . Hence gives a closed nonconstant trajectory. Thus the planar system has at least one periodic orbit.

(b) Let . Along solutions of the three-dimensional system,

On the surface , this equals

Therefore is invariant. On , where , the first two equations reduce exactly to the planar system in part (a). The periodic orbit found in part (a) therefore lifts to a periodic orbit of the three-dimensional system.

Problem A3.

Using WKB theory, find a leading order asymptotic expansion for the solution of

where is smooth and on . Then describe how to approach the case where and for some .

Proof.

Let

For , the standard WKB ansatz gives two leading order solutions

To satisfy , take their odd combination relative to :

The boundary condition gives

Thus the leading order approximation is

Equivalently, away from this is dominated by the exponentially growing branch normalized at .

If and , then is a simple turning point. The WKB form fails there because the amplitude becomes singular. One introduces a stretched Airy variable of size near , solves the local Airy equation, and matches the Airy asymptotics to WKB solutions on the two sides of the turning point.

Part B

Problem B1.

There is a one-parameter family of two-stage, second-order Runge--Kutta methods for :

(a) Verify that these schemes, for all values of , are second order accurate. (b) Show that these schemes, for all values of , have exactly the same absolute stability domain. (c) Verify that the absolute stability domain, along the negative real axis, extends exactly over .

Proof.

(a) Expanding about gives

where all derivatives are evaluated at . Hence

The exact solution satisfies

Thus the local truncation error is and the method is second order accurate for every .

(b) Apply the method to , . Then

Therefore

The stability function is independent of , so all schemes in the family have exactly the same absolute stability domain.

(c) The absolute stability domain is

On the negative real axis write , . Then

The condition is

The right inequality gives , hence . The left inequality is automatically true for real because . Hence the interval is exactly

Problem B2.

Consider the least squares minimization problem of finding a minimum-norm solution of

where is a real matrix and is given. (a) Express the minimum-norm least squares solution as a solution of a system . Determine when the solution is unique, when multiple solutions exist, and whether no solution is possible. (b) Under the conditions that there is a unique solution, express the minimum-norm least squares solution as a function of the eigenvalues of the matrix from part (a).

Proof.

Let the least squares problem be , with and .

(a) The normal equations are

Thus and . A least squares solution always exists because the normal equations are always consistent: lies in . The least squares solution is unique exactly when has full column rank, equivalently is positive definite. If is rank deficient, there are multiple least squares solutions, but among them there is a unique one of smallest Euclidean norm, namely .

(b) In the unique case, is symmetric positive definite. Write its spectral decomposition as

Then

If is only positive semidefinite, the minimum norm least squares solution is obtained by replacing by on the zero eigenspaces, that is,

Problem B3.

Let be an matrix. (a) State the power method for computing the largest eigenvalue of . (b) Let

Explain why the rate of convergence of the power method applied to is slow. (c) Choose a suitable shift so that the shifted power method converges to the largest eigenvalue of . Can this improve the rate of convergence? If so, why?

Proof.

(a) The power method chooses a nonzero initial vector and repeats

An eigenvalue estimate can be taken as the Rayleigh quotient

If has a dominant eigenvalue and has a nonzero component in the dominant eigenvector direction, then converges in direction to the dominant eigenvector and .

(b) The matrix

is triangular, so its eigenvalues are . The largest eigenvalue is , but the next largest magnitude is . The asymptotic convergence factor of the power method is approximately

which is close to . Hence convergence is slow.

(c) For the shifted power method one applies the power method to . The transformed eigenvalues are

To converge to the eigenvector associated with the original eigenvalue , choose so that is the largest transformed magnitude. A useful choice is

for which

The convergence factor becomes approximately , which is better than . Thus the shift can improve convergence, although the improvement here is modest because the eigenvalue constrains how far one can separate the transformed dominant eigenvalue from the others.

Part C

Problem C1.

Consider the variational problem

with

Show that the Euler--Lagrange equation admits a first integral, solve the resulting first-order ODE, verify that the minimizer is , and conclude that this function yields a weak minimum.

Proof.

The functional is

The Lagrangian is independent of , so the Hamiltonian

is conserved. Since

we get

where is constant. Thus

The solutions are catenaries

The boundary conditions are satisfied by

Indeed and .

To verify that this extremal gives a weak minimum, note that along this curve and

Thus the strengthened Legendre condition holds. Since the catenary found above has no conjugate point on this short interval and satisfies the Euler--Lagrange equation with fixed endpoints, the standard sufficient condition for a weak local minimum applies. Hence yields a weak minimum.

Problem C2.

Dirichlet's principle was historically stated as follows: if is bounded below, then there exists a minimizer such that . Riemann's proposed proof chooses a minimizing sequence , claims that since it is bounded it has a convergent subsequence , and concludes that is the desired minimizer. Find the flaw in this proof and present sufficient conditions for the existence of a minimizer.

Proof.

The flaw is the assertion that a bounded minimizing sequence has a convergent subsequence. In finite-dimensional Euclidean space this is true by Bolzano--Weierstrass, but in infinite-dimensional function spaces bounded sets are generally not compact in the norm topology. Even if a subsequence converges weakly, one must also know that the admissible class is weakly closed and that the functional is weakly lower semicontinuous.

A standard direct-method existence theorem uses the following sufficient conditions: (i) The admissible set is nonempty and weakly sequentially closed in a reflexive Banach space. (ii) The functional is coercive on , so every minimizing sequence is bounded. (iii) The functional is weakly sequentially lower semicontinuous.

Then a minimizing sequence has a weakly convergent subsequence , the closedness of gives , and lower semicontinuity gives

Therefore is a minimizer.

Problem C3.

Consider the Lagrangian

(a) Write out the Euler--Lagrange equation for fixed endpoint curves. (b) Derive the Hamiltonian and write out the Hamiltonian system. (c) Derive the Hamilton--Jacobi equation and solve it.

Proof.

Here

(a) We compute

The Euler--Lagrange equation is

Since

we obtain

(b) The momentum is

Therefore

The Hamiltonian system is

Since is conserved, is constant up to sign. Then

Thus

(c) The Hamilton--Jacobi equation is

A separated complete integral is obtained by taking

where is an arbitrary constant. Indeed,

and hence

The sign of covers both separated branches.