Mathematics 174 – Applied Mathematics 2 Final Problem Set April 27, 2000

Ground Rules

This is an open book, open notes, open library, open Maple, open SW 335, unlimited time exam. If you find something of use in a library book, please footnote it (this is not necessary for topics covered in our course textbook, Haberman). You must work independently, and you may not consult other faculty members in working on these problems. When you have completed the problem set, please write a brief statement saying that you have abided by these ground rules, sign it, and turn it in with your solutions.

Due: No later than 5:00 pm, Friday, May 5.

Directions

Do problem I below, and any two of problems II, III, IV. (If you submit solutions for all four problems (or parts of them) I will give Extra Credit points.)

I. (40) In this problem, you will see another example of how Fourier Transforms can be applied to solve PDE. Consider the following problem for the linearized Korteweg-deVries equation

(1)  \{ \begin{matrix}u_{t} = u_{xxx} \\ u(x, 0) = f(x) \end{matrix}

(This is a simplified version of an equation used to model the motion of water waves in long, shallow channels like canals – u(x, t) represents the vertical displacement at position x along the canal, at time t. For a long canal, and a wave that starts due to some disturbance of the water surface in the middle, there are essentially no boundary conditions, so we take −∞ < x < ∞.)

A) Take the Fourier Transform (with respect to x) of both sides of the PDE to find an ODE (with independent variable = t) satisfied by U(ω, t) = F[u(x, t)] and solve it to find U(ω, t). Don’t forget to include a “constant of integration” here, which will depend on ω. Take the inverse transform to give one way find a solution of (1).

B) Use the convolution theorem for the Fourier Transform to show that the solution u(x, t) can be written as

u(x, t) = f(x) ∗ G(x, t) = ∫_{−∞}^{∞} f(s)G(x − s, t) ds

where the function G is given by

G(x, t) = \frac{1}{2π} ∫_{−∞}^{∞} cos(ω^{3}t − ωx) dω = \frac{1}{π} ∫_{0}^{∞} cos(ω^{3}t − ωx) dω.

This is called the fundamental solution – it is the solution corresponding to u(x, 0) = δ(x) (Dirac delta).

By a happy chance, there is a built-in function called AiryAi in Maple (with an efficient method for computing values) that is very closely related to this function:

G(x, t) = \begin{matrix}AiryAi & ( \frac{−x}{\sqrt[3]{3t}}) \\ ^{3} & \sqrt{3t} \end{matrix}

This is good because the integral representation above is very inefficient for evaluating G(x, t) – it’s similar to a very slowly-converging alternating series.

C) Use Maple to generate plots of the solution of (1) if

f(x) = \{ \begin{matrix}sin(πx) & if x ∈ [0, 1] \\ 0 & otherwise \end{matrix}

and describe what the solution u(x, t) appears to be doing as t increases. Note: there are several different ways to think about computing the convolution formula from part B (either integrating directly, or by using the FFT in a suitable way to approximate). Explain how you are doing that step.

II. (30) Consider a Sturm-Liouville form differential equation:

(2)  \frac{d}{dx} (p(x)\frac{dy}{dx}) + q(x)y = 0,

and assume p(x) > 0 for all x ∈ [a, b] (as in a regular Sturm-Liouville eigenvalue problem). In this problem, we will study the zeroes of solutions of (2), and prove a result called the Sturm Separation Theorem – part B below. Let y_{1}, y_{2} be any two solutions of (2). The Wronskian of y_{1}, y_{2} is defined to be:

W [y_{1}, y_{2}](x) = det ( \begin{matrix}y_{1}(x) & y_{2}(x) \\ \frac{dy_{1}}{dx} (x) & \frac{dy_{2}}{dx} (x) \end{matrix} )

A) Show that W [y_{1}, y_{2}](x) = C/p(x) for some constant C. (Hint: What is the derivative \frac{d}{dx}W [y_{1}, y_{2}](x)?) Use this to show that if the Wronskian of y_{1} and y_{2} is zero at any c ∈ [a, b], then y_{1} and y_{2} are linearly dependent. You will need to use the fact that for all c ∈ [a, b], and all A, B ∈ R, there is a unique solution of (1), together with the initial conditions y(c) = A, \frac{dy}{dx}(c) = B.

B) Let y_{1}, y_{2} be two linearly independent solutions of (2). Show that between any two consecutive zeroes of y_{1} in [a, b], there is precisely one zero of y_{2} as follows: Assume x_{1} < x_{2} are the consecutive zeroes of y_{1}, but that y_{2} is not zero on the interval [x_{1}, x_{2}]. Consider the function \frac{y_{1}}{y_{2}} and apply the “most valuable theorem” (MVT) of calculus.

C) What does this result say about the zeroes of the Bessel functions J_{n}(z) and Y_{n}(z)?

III. (30) In this problem, you will see a proof of the Uncertainty Principle, which implies (among other things) that signals cannot be simultaneously arbitrarily sharply time-limited and frequency-limited. The famous Heisenberg uncertainty principle from quantum mechanics is also derived from this fact about Fourier analysis. Throughout this problem, we will use the following notation (similar to our inner-products from Sturm-Liouville theory): Let f(x), g(x) be “nice” signals (i.e. piecewise smooth, and in L^{1}(R)). We write

(3)  ⟨f, g⟩ = ∫_{−∞}^{∞} f(x)\overline{g(x)} dx

where \overline{g(x)} is the complex conjugate of g(x) (i.e. if g(x) = u(x) + iv(x), then \overline{g(x)} = u(x) − iv(x)). We also write ∥f∥ = \sqrt{⟨f, f⟩} = \sqrt{∫_{−∞}^{∞} |f(x)|^{2} dx} (called the L^{2}-norm of f).

A) First, we need an important fact called the Plancherel identity. Show using the Fourier

Inversion Theorem that

⟨F[f], F[g]⟩ = \frac{1}{2π} ⟨f, g⟩,

where the left-hand side is the inner product from (3) above, but for functions of ω. From this deduce the Plancherel identity:

\frac{1}{2π} ∥f∥^{2} = ∥F[f]∥^{2}.

B) We assume all signals are “centered” at x = 0. The statement of the Uncertainty Principle will involve a measure of the “effective width” of a signal and its Fourier transform. For the effective widths we will use the following definitions. The effective width of f is

\begin{aligned}W_{f} & = \frac{∥xf(x)∥}{∥f(x)∥} \\ & = [\frac{1}{∥f∥^{2}} ∫_{−∞}^{∞} x^{2}|f(x)|^{2} dx]^{\frac{1}{2}} \end{aligned}

(this is related to a variance as defined in statistics; if that doesn’t mean anything to you, ignore it!!) Similarly, we define the effective width of the Fourier transform to be

\begin{aligned}W_{F} & = \frac{∥ωF[f](ω)∥}{∥F[f](ω)∥} \\ & = [\frac{2π}{∥f∥^{2}} ∫_{−∞}^{∞} ω^{2}|F[f](ω)|^{2} dω]^{\frac{1}{2}} \end{aligned}

Compute the effective width W_{f} for f(x) = e^{−ax^{2}} (scaled Gaussian), and W_{F} for the Fourier transform of the Gaussian. What is the product W_{f}W_{F} in this case?

C) The Uncertainty Principle is the statement:

W_{f}W_{F} ≥ \frac{1}{2}

for all real signals f. (Hence, in particular, W_{f} and W_{F} cannot both be small!) Prove this statement by 1) recognizing that W_{F} is equal to \frac{∥f′∥}{∥f∥}, 2) applying the Cauchy-Schwarz inequality:

∥xf(x)∥∥f′(x)∥ ≥ |∫_{−∞}^{∞} xf(x)f′(x) dx| ,

then 3) working with the integral on the right above. (You may assume x(f(x))^{2} → 0 as x → ±∞.)

IV. (30) In this problem, you will work through most of the steps of a proof of the Shannon-Nyquist Sampling Theorem, which is the following statement:

Sampling Theorem. Let f be piecewise smooth and in L^{1}(R). Assume that F(ω) = 0 for all |ω| ≥ L. Then

(4)  f(x) = ∑_{n=−∞}^{∞} f (\frac{nπ}{L} ) \frac{sin(Lx − nπ)}{Lx − nπ}.

The meaning of this is that if f is frequency limited (that is, F[f](ω) = 0 for all |ω| ≥ L), then we can reconstruct f exactly (in principle) from sampled values f (\frac{nπ}{L} ), provided that the sampling interval is \frac{π}{L} = \frac{2π}{2L}. Any sampling interval Δx < \frac{2π}{2L} would also work.

A) By the Fourier Inversion Theorem,

f(x) = ∫_{−∞}^{∞} F[f](ω)e^{−iωx} dω = ∫_{−L}^{L} F[f](ω)e^{−iωx} dω

Show that the Fourier series of F (ω) = F[f](ω) on [−L, L] is:

F (ω) ∼ \frac{1}{2L} ∑_{n=−∞}^{∞} f (\frac{nπ}{L} ) e^{\frac{inπω}{L}}

(see section 3.6 in Haberman, or the class notes, for the notation of the complex form of the Fourier series).

B) Let H_{L}(ω) be the periodic function defined by the Fourier series in part A. Explain why

F (ω) “=” H_{L}(ω)Π_{L}(ω),

so

F (ω) “=” \frac{1}{2L} ∑_{n=−∞}^{∞} f (\frac{nπ}{L} ) e^{\frac{inπω}{L}} Π_{L}(ω)

where Π_{L}(ω) = 1 if |ω| ≤ L and = 0 otherwise. Why are several equalities here written with quotation marks? (These are several technical details that are being overlooked for simplicity!)

C) Derive equation (4) by taking the Inverse Fourier Transform of both sides of the last equation.

Have a wonderful summer!