2023 Fall Applied Mathematics

Part A

Choose any TWO of the following problems.

Problem A1.

Consider the planar ODE

(a) Find the linearization of at the fixed points and .

(b) Show that the triangular region bounded by the lines , , and , is negatively invariant under the flow of .

(c) Show that admits a heteroclinic orbit between the fixed points and which approaches along a center manifold.

Proof.

(a) Write

The fixed points satisfy and

Hence the fixed points are

The Jacobian matrix is

Since

we get

and

Thus the linearization at is

and the linearization at is

where and near .

(b) The triangular region is

To prove that is negatively invariant, it is enough to show that the vector field points out of along the boundary for forward time. Equivalently, it points into for the reversed flow.

On the side , where , we have

and

Thus the vector field points below the side , hence outward for forward time.

On the side , where , we have

Thus the vector field points to the right of the side , hence outward for forward time.

On the side , define

The region lies in . Along , we have

Since on this side,

for . Hence the vector field points toward , again outward for forward time.

Therefore the vector field points outward along each side of for forward time. Hence it points inward along each side for backward time. Therefore is negatively invariant.

(c) At , the linearization is

Its eigenvalues are

Thus has a one-dimensional center direction and a one-dimensional unstable direction.

Let

Then near , the system becomes

A center manifold has the form

The invariance equation is

then comparing powers of gives

Therefore

On this center manifold,

Hence for small , we have , so as . Equivalently, the center manifold branch inside approaches as .

Now choose a point on this center manifold branch inside . Since is negatively invariant, the backward orbit through this point remains in the compact set . Therefore its alpha-limit set is nonempty, compact, connected, and invariant.

Also,

By the Bendixson criterion, there are no periodic orbits in the interior of . Hence the alpha-limit set must contain an equilibrium. The only equilibria in are and .

The chosen orbit approaches as , and it is not the constant orbit at . Hence its alpha-limit cannot also be . Therefore the alpha-limit is .

Thus there exists an orbit such that

This is a heteroclinic orbit from to , and it approaches along a center manifold.

Problem A2.

Consider the boundary value problem

where for . Using WKB theory, show that a leading order asymptotic expansion for the solution is given by

where

Proof.

We look for WKB solutions of the form

Substituting into

and collecting the leading order terms gives

Thus

These two roots give a slow mode and a fast boundary-layer mode.

First consider the slow mode . Then to leading order. Substituting into the equation and keeping the leading order terms gives

Therefore

and hence

So the slow mode is

Now consider the fast mode

Write

Substituting this into the differential equation gives

At leading order,

Thus

Therefore

The fast mode is therefore

Combining the two leading order modes gives

Since on , we have

Therefore the fast mode is exponentially small at . Hence the boundary condition at gives

Thus

At , the exponential factors are , so

Therefore

This gives the desired leading order asymptotic expansion.

Problem A3.

Consider the initial value problem

where is locally Lipschitz. That is, there exist such that

for all .

(a) Consider the map defined by

and take where

Show that is a contraction on the space

where denotes the space of continuous functions with the supremum norm

What does this imply about solutions to the initial value problem ? You may use the contraction mapping theorem without proof.

(b) Now suppose and that the maximal interval of existence of the solution to the initial value problem is where . Show that is unbounded as .

Proof.

(a) First we show that maps into itself. Let . Then for all , the values lie in . Hence

Therefore

Since

we get

Thus .

Now let . Then

Using the Lipschitz condition,

Taking the supremum over gives

Since

we have . Therefore is a contraction.

By the contraction mapping theorem, has a unique fixed point in . A fixed point satisfies

Differentiating gives

Therefore the initial value problem has a unique local solution on .

Strictly speaking, the contraction mapping theorem applies directly to a closed complete ball. One may replace the strict inequality in the definition of by ; the above estimates still show that maps the closed ball into the strict ball.

(b) Suppose, toward a contradiction, that is bounded as . Then there exists such that

for all sufficiently close to .

Since , it is continuous. Hence is bounded on the compact ball

So there exists such that

for all .

For sufficiently close to , we have

Thus is Cauchy as . Therefore there exists such that

By the local existence theorem applied at the initial condition , there is a solution on some interval for some . Gluing this solution to the old solution on extends beyond , contradicting the maximality of .

Hence the assumption was false. Therefore

along some sequence as . Equivalently, is unbounded as .

Part B

You must complete problem 1, and then choose ONE of problems 2 or 3.

Problem B1.

Numerical ODE problem.

Consider a linear multistep scheme of the form

for solving the ODE

Here, is the time step, is the total number of time steps, , and is the numerical approximation to at .

(a) Find the equation for the error , that describes how the error propagates in time, and describe the meaning of each of the terms in the equation.

(b) What is the highest order of accuracy this method can attain? Determine the coefficients that make the scheme reach this order of accuracy, assuming is a sufficiently accurate approximation of . Hint: The scheme is th order accurate if the test functions , and satisfy the scheme exactly with .

(c) How do you determine if the method is stable? Hint: Do not forget that this is a multistep method.

(d) Define A-stability for a numerical scheme for first order ODEs. Is the optimally accurate scheme you found A-stable?

Proof.

(a) Let

Define the local truncation residual by

Subtracting the numerical scheme from the exact equation with residual gives

The terms describe propagation of past errors. The two terms involving describe how nonlinear dependence on amplifies or damps error. The term is the local truncation error generated by inserting the exact solution into the numerical method.

(b) We use the order conditions with . The method is

For , we get

Using , this gives

For , we get

Equating the constant terms after using gives

For , we get

This yields

Solving through gives

There are four free coefficients, and these conditions give exactness through degree . The next condition, for , is not satisfied by these coefficients. Therefore the highest possible order is

The third-order method is

(c) Since this is a multistep method, stability is determined by a root condition.

For zero-stability, we consider the first characteristic polynomial

The method is zero-stable if every root of satisfies

and every root with is simple.

For the optimally accurate method,

The roots are

The root is simple, and the other root has modulus less than . Hence the optimally accurate method is zero-stable.

For absolute stability, one applies the method to the test equation

and sets

Then the recurrence becomes

The method is absolutely stable for a given if all roots of

satisfy , with unit-modulus roots simple.

(d) A numerical method is A-stable if, for the test equation

its region of absolute stability contains the whole closed left half-plane

Equivalently, whenever , the numerical solution is stable for every step size .

For the optimally accurate method, the stability polynomial is

Multiplying by gives

Take

Then , but the polynomial becomes

Its roots are

The root

has modulus greater than . Therefore the method is not absolutely stable at .

Hence the optimally accurate scheme is not A-stable.

Problem B2.

Least squares problem.

(a) Let be a real matrix. State the singular value decomposition (SVD) of . Briefly present an algorithm to solve the overdetermined system using the SVD.

(b) What are the advantages and disadvantages of using the SVD to solve overdetermined systems?

(c) Perform the SVD on the matrix

(d) Compute the pseudo-inverse of , the Moore-Penrose pseudo-inverse. Leave in factored form.

(e) Find the minimal length, least squares solution of the problem

Proof.

(a) The singular value decomposition of a real matrix is

where is an orthogonal matrix, is an orthogonal matrix, and is an diagonal matrix whose nonnegative diagonal entries

are the singular values of .

To solve an overdetermined system

in the least squares sense, compute

Then the minimal length least squares solution is

where

Here is obtained by replacing each nonzero singular value by and transposing the rectangular diagonal matrix.

(b) Advantages of the SVD method:

It is numerically stable, especially for ill-conditioned matrices. It reveals the numerical rank of . It works even when is rank deficient. It gives the minimal length least squares solution automatically.

Disadvantages:

It is more computationally expensive than methods such as QR factorization or normal equations. For very large problems, computing a full SVD can be too costly in both time and memory.

(c) Let

This matrix has rank . We can write

where

Thus the only nonzero singular value is

A possible choice of orthogonal matrices is

and

Then

(d) The Moore-Penrose pseudo-inverse is

where

Equivalently,

Since

we get

(e) Let

The minimal length least squares solution is

Thus

Therefore the minimal length least squares solution is

Problem B3.

Eigenvalue problem.

(a) Let be an matrix. Prove Gershgorin's theorem, which states: For , let

Every eigenvalue of falls within at least one of the closed discs in the complex plane with center and radius . Hint: Let

and assume that the largest component of in absolute value is . Consider the th equation of .

(b) Consider the matrix

Show that is positive definite. Hint: Use Gershgorin's theorem to show that is positive semi-definite. Then consider the equation , where the first component of equals .

(c) Let

Find the eigenvalues of and verify that Gershgorin's theorem holds.

Proof.

(a) Let be an eigenvalue of , and let be a corresponding eigenvector. Choose such that

Then .

The th component of

Thus

Taking absolute values gives

Since for every , we get

Because , we may divide by and obtain

Therefore lies in the Gershgorin disc centered at with radius . Hence every eigenvalue lies in at least one Gershgorin disc.

(b) The matrix is real symmetric, so all of its eigenvalues are real.

The Gershgorin discs are centered at . For the first and last rows, the radius is , and for the middle three rows, the radius is . Hence all eigenvalues lie in

Since the eigenvalues are real, all eigenvalues are nonnegative. Therefore is positive semi-definite.

It remains to show that is not an eigenvalue. Suppose

If , then the first equation gives

so . Then the second equation gives , and continuing gives .

Thus any nonzero solution of would have . Scaling, assume

The first equation gives

The second equation gives

Continuing in this way gives

But the last equation is

Substituting and gives

a contradiction. Therefore has no nonzero solution, so is not an eigenvalue.

Hence all eigenvalues are strictly positive. Therefore is positive definite.

(c) For

the characteristic polynomial is

Expanding gives

Thus

Therefore the eigenvalues are

Now compute the Gershgorin discs. The first and third rows have center and radius , so they give

The second row has center and radius , so it gives

Thus Gershgorin's theorem says that all eigenvalues lie in

Indeed,

Therefore Gershgorin's theorem is verified.

Part C

Choose any TWO of the following problems.

Problem C1.

Let be given by

(a) Show that is convex, continuously differentiable, and satisfies

(b) Deduce that any arc admissible for the problem

subject to

satisfies

(c) Show that the functions

satisfy

(d) Conclude that the problem defined in (b) admits no solution.

Proof.

(a) Since

we have

For ,

At , the left derivative is and the right derivative is also . Hence .

Also, for and for , and is nondecreasing across . Therefore is convex.

Now we prove

If , then

Also . Hence

If , then

Also

Thus . Hence

(b) Let be admissible. Since for all , we have

for almost every . Therefore

Because , the fundamental theorem of calculus gives

Thus

Since and is continuous, cannot be identically zero. Hence

Therefore

(c) For

we have

Since , we have

Thus

Let

Then runs from to . Hence

Therefore

Taking , we get

(d) From (b), every admissible arc satisfies

From (c), there exists an admissible sequence such that

Therefore

but no admissible satisfies . Hence the problem admits no solution.

Problem C2.

Assume .

(a) Prove the dual variational principle

(b) Write out Euler-Lagrange equations for both formulations.

Proof.

(a) Let

Let satisfy

in the sense of distributions. Since , integration by parts gives

Therefore

Complete the square:

Hence

Thus, for every admissible and every admissible ,

Taking the minimum over and the maximum over gives the weak duality inequality

Now let be the minimizer of . Its Euler-Lagrange equation is

in the weak sense. Define

Then

Thus is admissible for the dual problem. Also,

Therefore equality holds in the completed-square inequality, and

Hence

(b) For the primal problem, the Euler-Lagrange equation is

for every . Equivalently,

For the dual problem, introduce a Lagrange multiplier for the constraint

The Lagrangian is

Taking variations in gives

Taking variations in gives the constraint

Thus the Euler-Lagrange system for the dual formulation is

Equivalently,

Problem C3.

For each of the given Lagrangians, find the Hamiltonian and solve the Hamiltonian system.

(a)

(b)

Proof.

(a) The momentum is

Thus

The Hamiltonian is

Since

we get

Therefore

Hamilton's equations are

Solving the second equation gives

Then

Multiplying by the integrating factor ,

Since ,

Hence

Thus the Hamiltonian solutions are

where are constants.

(b) The momentum is

Thus

Solving for gives

where the sign agrees with the sign of .

The Hamiltonian is

Since

we have

Also

Therefore

Thus

Hamilton's equations are

Since is independent of , it is constant along solutions. Let

Then

The equation for becomes

Therefore

Integrating,

Thus

Since

we get

Hence

on any interval where . Therefore the Hamiltonian solutions are

where and are constants.