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March 4, 2011

Harmonic Analysis Lecture Notes 16

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:54 pm

Method of Stationary Phase

Proposition (non-stationary phase) Let {a \in C^\infty_c} and {\phi \in C^\infty} (and real-valued) with {\phi \geq 1} on {supp(a)}, then

\displaystyle  	\bigg| \int e^{\pm i \lambda \phi(x)} a(x) \; dx \bigg| \lesssim_N \lambda^{-N}

for every integer {N \geq 0} and {\lambda > 0}.

Proof: Let us take the {e^{+ i \lambda \phi}} case,

\displaystyle  	I(\lambda) := \int e^{i\lambda \phi(x)} a(x) \; dx.

In the spirit of integration by parts,

\displaystyle  \begin{array}{rcl}  	\displaystyle I(\lambda) &=& \displaystyle \int \left[ \frac{1}{i \lambda \phi'(x)} \frac{d}{dx} e^{i\lambda \phi(x)}\right] a(x) \; dx \\ \\ 		&=& \displaystyle \int e^{i \lambda \phi(x)} \frac{d}{dx} \left( \frac{i}{\lambda \phi'(x)} a(x) \right) \; dx. 	\end{array}

Thus, by the quotient rule,

\displaystyle  	|I(\lambda)| \leq \frac{1}{\lambda} \left\{ \left\| \frac{a'}{\phi'}\right\|_{L^1} + \left\| \frac{a\phi''}{(\phi')^2}\right\|_{L^1} \right\}

More generally

\displaystyle  	e^{i\lambda \phi} = \left( \frac{1}{i \lambda \phi'(x)} \frac{d}{dx} \right)^N e^{i\lambda \phi}

and so

\displaystyle  	I(\lambda) = \int e^{i\lambda \phi} \left( \frac{d}{dx} \frac{i}{\lambda \phi'(x)} \right)^N a(x) \; dx.

Thus

\displaystyle  	| I(\lambda) | \lesssim \lambda^{-N} \sum_{k=0}^N \;\;\; \sum_{\tiny \begin{matrix} \alpha_1 + \cdots + \alpha_k + \beta = N \\ \alpha_j \geq 1 , \beta \geq 0 \end{matrix} } 	\left\| \frac{(\partial^\beta a)(\partial^{\alpha_1} \phi') \cdots (\partial^{\alpha_k} \phi')}{(\phi')^{N+k}} \right\|_{L^1}

\Box

The ideas we use work for sums as well. Traditionally, dealing with sums are much harder.

Theorem (van der Corput lemma) Suppose {\phi \in C^\infty} and {\phi^{(k)}(x) \geq 1} throughout the interval {[a,b]} for some {k \geq 1}. If {k =1}, then we additionally assume that {\phi'} is monotone. Then

\displaystyle  	\bigg| \int_a^b e^{\pm i \lambda \phi(x) } \; dx \bigg| \lesssim_k \lambda^{-\frac{1}{k}}

independent of {a,b,} and {\phi}.

Remark

  1. If {k = 1} then the phase is non-stationary; however, we cannot do better because of the end-point contributions from integration by parts (alternatively, the “amplitude function” is {\chi_{[a,b]}}, which is not smooth). If {\phi(x) = x}, we can compute exactly and see that we cannot do better.
  2. To get a bound which is independent of the length of {[a,b]}, we need to assume {\phi'} is monotone (or at least some additional hypothesis). Recall

    \displaystyle  		\text{Im}\; \int_a^b e^{i \phi(x)} \; dx = \int_a^b \sin(\phi(x)) \; dx.

    We define {\phi} heuristically, slow down when {\sin(x)} is positive and speed up when {\sin(x)} is negative. Doing this appropriately, we may make the right side arbitrarily large.

Proof: (by induction in {k}). For {k = 1} we argue as before

\displaystyle  \begin{array}{rcl}  	I(\lambda) &=& \displaystyle \int_a^b e^{i\lambda\phi(x)} \; dx \\ \\ 		&=& \displaystyle \int_a^{b} \frac{1}{i \lambda \phi'(x)} \left( \frac{d}{dx} e^{i\lambda \phi(x)} \right) \; dx \\ \\ 		&=& \displaystyle \left[ \frac{e^{i\lambda \phi(x)}}{i \lambda \phi'(x) } \right]^b_a + \int_a^b e^{i\lambda \phi(x)} \left( \frac{d}{dx} \frac{i}{\lambda \phi'(x)} \right) \; dx. 	\end{array}

Thus

\displaystyle  \begin{array}{rcl}  	I(\lambda) &\leq& \displaystyle \frac{2}{\lambda} + \frac{1}{\lambda} \bigg| \int_a^b \underbrace{\frac{d}{dx} \frac{1}{\phi'(x)}}_{\tiny \begin{matrix}\text{sign definite as} \\ \phi' \text{ is monotone} \end{matrix}} \; dx \bigg| \\ \\ 		&\leq& \displaystyle \frac{2}{\lambda} + \frac{1}{\lambda} \bigg| \frac{1}{\phi'(b)} - \frac{1}{\phi'(a)}\bigg| \leq \frac{3}{\lambda}. 	\end{array}

Now, let us do the induction step (assume the result is true for {k}, now show for {k+1}). By hypothesis, {\phi^{(k+1)} \geq 1}. Let us look at {\phi^{(k)}}. This vanishes at most at one point {c \in [a,b]}; if it does not vanish on {[a,b]}, then choose {c} to be the end point where {|\phi^{(k)}|} is least. Consider the partition of {[a,b]} into interval {(c- \delta, c+ \delta) \cap [a,b]} and its complement, which we will write as {J}. On {J}, we have {|\phi^{(k)}| \geq \delta} and {\phi^{(k)}} is monotone. Using the inductive hypothesis,

\displaystyle  	|I(\lambda)| \lesssim_k \delta + (\lambda \delta)^{-\frac{1}{k}}.

Choosing {\delta \approx \lambda^{-\frac{1}{k+1}},} we get {| I(\lambda)| \lesssim_k \lambda^{-\frac{1}{k+1}}}, as desired. \Box

Remark A computation using {\phi(x) = x^k} shows that these bounds are the best that we may expect.

Corollary Let {\phi} obey the hypothesis from the van der Corput lemma and let {\psi} be of bounded variation (or at least {C^1}). Consider

\displaystyle  	I(\lambda) = \int_a^b e^{\pm i \lambda \phi(x)} \psi(x) \; dx

then

\displaystyle  	| I(\lambda) | \lesssim_k \lambda^{-\frac{1}{k}} \big( \|\psi\|_{L^\infty} + \|\psi'\|_{L^1}\big)

Proof:

\displaystyle  \begin{array}{rcl}  	I(\lambda) &=& \displaystyle \int_a^b \psi(x) \frac{d}{dx} \int_a^x e^{i\lambda \phi(y)} \; dy \; dx \\ \\ 		&=& \displaystyle \Big[ \psi(x) \int_a^x e^{i \lambda \phi(y)} \; dy \Big]_a^b - \int_a^b \psi'(x) \int_a^x e^{i \lambda \phi(y)} \; dy \; dx. 	\end{array}

Using the van der Corput lemma twice,

\displaystyle  	| I(\lambda) | \lesssim_k |\psi(b)| \lambda^{-\frac{1}{k}} + \|\psi'\|_{L^1} \sup_{a \leq x \leq b} \Big| \int_a^x e^{i\lambda\phi(y)} \; dy \Big| \lesssim_k \lambda^{-\frac{1}{k}} \big( \|\psi\|_{L^\infty} + \|\psi'\|_{L^1}\big)

\Box

An application of this lemma

Definition A sequence {\{x_k\}_{k=1}^\infty \subseteq {\mathbb R}^z / {\mathbb Z}^n} is equi-distributed if

\displaystyle  	\frac{1}{K} \sum_{k=1}^K f(x_k) \stackrel{K\rightarrow \infty}{\longrightarrow} \int_{{\mathbb R}^n / {\mathbb Z}^n} f(y) \; dy

for all continuous functions {f}. Analogously, a curve {x : [0,\infty) \rightarrow {\mathbb R}^n / {\mathbb Z}^n} is equi-distributed if

\displaystyle  	\frac{1}{T} \int_0^T f(x(t)) \; dt \stackrel{T\rightarrow\infty}{\longrightarrow} \int_{{\mathbb R}^n / {\mathbb Z}^n} f(y)

for all continuous functions {f}.

Remark

  1. The notion is unchanged if we replace {f \in C({\mathbb R}^n/{\mathbb Z}^n)} by {\chi_E} or {\chi_{\mathcal{O}}} for arbitrary closed/open sets (Urysohn’s lemma).
  2. This replacement does not work for {F_\sigma}‘s or {G_\delta}‘s (heuristically because they mimic the sequences too well).
  3. That {e^{2\pi i t^2}} is not uniformly continuous is implied by the fact that the curve {(t^2, (t+h)^2)} is equi-distributed in {{\mathbb R}^2 / {\mathbb Z}^2} when {h \neq 0}, because

    Lemma If {x(t)} is equi-distributed then it is dense.

    Proof: Choose {f} to be a bump in the open set that {x(t)} misses. \Box

Theorem (Weyl’s criterion) The curve {x(t)} is equi-distributed in {{\mathbb R}^n/{\mathbb Z}^n} if and only if

\displaystyle  	\frac{1}{T}\int_0^T e^{2\pi i \vec m \cdot \vec x(t)} \; dt \rightarrow 0

for all {\vec m \in {\mathbb Z}^n\setminus\{0\}}.

Proof: The forward implication is a fortiori clear. For the reverse implication, trigonometric polynomials are dense. \Box

Theorem Let {p_1(t), \ldots, p_n(t)} be rationally independent “of constants”: that is, if {\vec m \cdot \vec p(t) := \sum m_j p_j(t) \equiv const} with {\vec m \in {\mathbb Z}^n}, then {\vec m \equiv 0}. Then the curve {\vec p (t)} is equi-distributed in {{\mathbb R}^n / {\mathbb Z}^n}.

Remark Conversely, if {\vec m \cdot \vec p \equiv c} where {\vec m \in {\mathbb Z}^n \setminus\{ 0\}}, then the curve is no equi-distributed: it lives in a co-dimension one sub-manifold.

Proof: By Weyl’s criterion, we need to show that {\frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt \stackrel{T \rightarrow \infty}{\longrightarrow} 0} for all {\vec m \in {\mathbb Z}^n \setminus\{0\}}. For fixed {\vec m}, we know that {\vec m \cdot \vec p(t)} is a non-constant polynomial of degree {k \geq 1}, so {\vec m \cdot \vec p(t) = a_k t^k + a_{k-1}t^{k-1} + \cdots + a_0} with {a_k \neq 0}. Changing variables {t = uT} gives

\displaystyle  	\frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt = \int_0^1 e^{2\pi iT^k \phi_T(u)} \; du

where

\displaystyle  	\phi_T(u) = T^{-k} \vec m \cdot \vec p(Tu) = a_ku^k + a_{k-1} T^{-1} u^{k-1} + \cdots + a_0T^{-k}.

Now, note that {\Big| \frac{d^k \phi_T}{du^k} \Big| = |a_k| > 0} uniformly in {T}. By Van der Corput’s lemma,

\displaystyle  	\Big| \frac{1}{T} \int_0^T e^{2\pi i \vec m \cdot \vec p(t)} \; dt \Big| \lesssim |a_k T^k|^{-1/k}

uniformly in {T}. This goes to zero as {T \rightarrow \infty}. \Box

Proposition (Multi-dimensional non-stationary phase) Fix {a \in C_c^\infty({\mathbb R}^d)} and {\phi \in C^\infty({\mathbb R}^d)} with {\nabla \phi \neq 0} on {\mathrm{supp}(a)}. Then

\displaystyle  	\Big| \int_{{\mathbb R}^d} e^{\pm i \lambda \phi(x)} a(x) \; dx \Big| \lesssim_{N,a,\phi} \lambda^{-N}

for any integer {N \geq 0}.

Proof: By introducing a partition of unity, we can assume that {| \frac{\partial \phi}{\partial x_j} | \geq \frac{1}{2d} \min_{\mathrm{supp}(a)} |\nabla \phi| > 0} on the support of {a} for some fixed {1 \leq j \leq d}. Now writing, with {x'=(x_1, \ldots, \hat{x_j}, \ldots, x_d)},

\displaystyle  	\int_{{\mathbb R}^d} e^{\pm i \lambda \phi(x)} a(x) \; dx = \int_{\text{compact set}} \left( \int e^{\pm i \lambda (x_j, x')} a(x_j, x') \; dx_j \right)\; dx'.

Now, we apply the one dimensional result to {\int e^{\pm i \lambda (x_j, x')} a(x_j, x') \; dx_j } to get {O(\lambda^{-N})} with a constant that depends continuously on {x'}. \Box

An alternative proof is to use {\frac{\nabla \phi}{i \lambda |\nabla \phi|^2} \cdot \nabla e^{i \lambda \phi} = e^{i\lambda \phi}} and integrate by parts sufficiently many times. For example

\displaystyle  	\int e^{i \lambda \phi(x)} a(x) \; dx = \int e^{i\lambda \phi} \nabla \cdot \left( \frac{ia\nabla\phi}{\lambda |\nabla \phi|^2}\right) \; dx.

Proposition (Morse Lemma) Let {\phi \in C^\infty({\mathbb R}^d)} have a non-degenerate critical point at {0 \in {\mathbb R}^d}; that is,

\displaystyle  	\phi(0) = 0 \hbox{\hskip 28pt but \hskip 28pt} \det \left( \frac{\partial^2 \phi}{\partial x_i \partial x_j}\right) \neq 0.

Then there is a local diffeomorphism {x \mapsto y} in a neighborhood of {x =0} with {0 \mapsto },

\displaystyle  	\frac{\partial y_j}{\partial x_i}(0) = \partial_{ij}

and

\displaystyle  	\phi(x) = \phi(0) + \frac{1}{2} \sum_{i,j} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0) y_i(x) y_j(x)

Proof: By Taylor’s formula (with integral remainder)

\displaystyle  \begin{array}{rcl}  	\phi(x) &=& \displaystyle \phi(0) + x \cdot \nabla \phi(0) + \int_0^1 (1-t) \left( \frac{d^2}{dt^2} \phi(t,x)\right) \; dt \\ \\ 		&=& \phi(0) + \int_0^1 (1-t) \sum_{i,j} x_ix_j \frac{\partial^2 \phi}{\partial x_i \partial x_j}(tx) \; dt 	\end{array}

Note that, for {|x|} sufficiently small, we can write

\displaystyle  	\int_0^1 (1-t) x_ix_j \frac{\partial^2 \phi}{\partial x_i \partial x_j}(tx) \; dt = \frac{1}{2}\big(T(x)^t \phi''(0) T(x) \big)_{i,j} = \frac{1}{2} \sum_{k,l} T_{kj}(x) \frac{\partial^2 \phi}{\partial x_k \partial x_l}(0) T_{li}(x)

with {T(x)} a smooth, non-singular matrix with {T(0) = \mathrm{Id}}. (Eg. first make a linear change of variables to reduce {\phi''(0)} to {\mathrm{diag}(\pm 1, \cdots, \pm 1)}, then explicitly do the completing the square argument).

Now, set {y_i(x) = \sum_k T_{ik}(x) x_k}. Note that {y(0) = 0} and {\frac{\partial y_i}{\partial x_j}(0) = \sum_{k} T_{ik}(0) \delta_{kj} = \delta{ij}} \Box

Theorem Let {a \in C_c^\infty}, and {\phi \in C^\infty} with a non-degenerate critical point at {x =0} but otherwise, {\nabla \phi} does not vanish on {\mathrm{supp}(a)}. Then

\displaystyle  	\int_{{\mathbb R}^d} e^{i\lambda \phi(x)} a(x) \; dx = e^{i\lambda \phi(0)} a(0) \lambda^{-\frac{d}{2}} \det \left( \frac{1}{2\pi i} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0)\right)^{-\frac{1}{2}} + O_{a,\phi}(\lambda^{-\frac{d+2}{2}})

as {\lambda \rightarrow \infty}.

Note that the {\frac{1}{2\pi i} \frac{\partial^2 \phi}{\partial x_i \partial x_j}(0)} (wait… I forgot what I wanted to say here)

Remark One can use a partition of unity (and spatial translations) to generalize to multiple stationary phase points.

Proof: By the Morse lemma, we can find some small neighborhood of {x = 0} where we can change variables {x = \Phi(y)} to simplify {\phi}. Now use a smooth partition of unity to write {a(x) = a_1(x) + a_2(x)} where {a_1} is supported in the neighborhood of {x = 0} where the change of variables is defined, while {a_2(x} vanishes in a neighborhood of {x=0}. ({a_1}, {a_2 \in C_c^\infty} with {\mathrm{supp}(a_j) \subseteq \mathrm{supp}(a))}. By the non-stationary phase position (and {\nabla \phi \neq 0} on {\text{supp}(a_2)}),

\displaystyle  	\int e^{i \lambda \phi(x)} a_2(x) \; dx = O(\lambda^{-N})

for {N} as large as we wish. This leaves

\displaystyle  	\int e^{i\lambda \phi(x)} a_1(x) \;dx = \int e^{i \lambda ( \phi(0) + \frac{1}{2} y^t \phi''(0) y)} a_1 \circ \phi(y) \det(\Phi' x) |.

Lets call {a_1 \circ \Phi(y) \det(\Phi'(y)) = \tilde a(y)}; it is {C_c^\infty} and {\tilde a(0) = a_1 \Phi \circ \phi(0) = a(0)}.

Remembering that {\int fg = \int \hat f \check g}, we see that

\displaystyle  	\mathrm{above} = e^{i\lambda \phi(0)} \det \left( \lambda \frac{\phi''}{2\pi i}\right)^{-\frac{1}{2}} \iint e^{-2\pi^2 i\lambda^{-1} \xi^t (\phi'')^{-1} \xi} \check{\tilde{a}}(\xi) \; d\xi

which has the correct asymptotics. \Box

Application of the theorem

For the following discussion, we shall denote the Fourier transform (and its inverse) by

\displaystyle  \begin{array}{rcl}  	\hat f(x) &=& \displaystyle (2\pi)^{-d/2} \int e^{-ix \cdot \xi} f(x) \; dx \\ \\ 	f(x) &=& (2\pi)^{-d/2} \int e^{+ix \cdot \xi} f(x) \; dx 	\end{array}

noting that {\| \hat f\|_{L^2} = \|f\|_{L^2}}.

Example (Schrödinger’s equation) Consider the long-time behavior of the solutions to

\displaystyle  	i \partial_t \psi = -\frac{1}{2} \Delta \psi

for {\psi: {\mathbb R}_t \times {\mathbb R}_x^d \rightarrow {\mathbb C}} with {\psi(0,x) = \psi_0(x)}.

As the equation is space-translation invariant (it is also time-translation invariant, but we will not use that here), we take the spatial Fourier transform to find

\displaystyle  	i \partial_t \hat \psi (t,\xi) = \frac{1}{2} |\xi|^2 \hat \psi (t,\xi)

with {\hat \psi(0,\xi) = \hat \psi_0(\xi)}, which we take (for now) to be Schwartz {\mathcal{S}({\mathbb R}^d)}. Therefore {\hat \psi(t,\xi) = e^{-t|\xi|^2/2}\hat \psi_0(\xi)}, i.e. {\psi(t,x) = e^{-it|\nabla|^2/2} \psi_0 = e^{it\Delta/2}\psi_0}. Taking inverse Fourier transforms, we get

\displaystyle  	\psi(t,x) = (2\pi)^{-d/2} \int e^{i x \cdot \xi - \frac{1}{2} i t|\xi|^2} \hat \psi_0(\xi) \; d\xi

and so we may apply the stationary phase theorem. Looking for stationary phase points,

\displaystyle  	0 = \nabla \phi = x - t\xi \iff \xi = \frac{x}{t}.

Also,

\displaystyle  	\mathrm{Hessian}(\phi) = \partial_{\xi_i} \partial_{x_j} \left[ -\tfrac{1}{2} |\xi|^2 + \frac{x}{t} \cdot \xi \right] = -\delta_{ij}

So the theorem says

\displaystyle  	\psi(t,x) = [e^{it\Delta/2}\psi_0](x) = (2\pi)^{-d/2} \cdot e^{\frac{i}{2} |x|^2/t} \cdot \hat \psi_0\left(\tfrac{x}{t}\right) \cdot e^{-i\frac{d\pi}{4}} \cdot \left(\tfrac{1}{2\pi}\right)^{-d/2} t^{-d/2} + O(t^{-\frac{d+2}{2}})

with the big-{O} notation for {\frac{x}{t}} constant and {t \rightarrow +\infty} (i.e., this describes the decay if we travel along a world-line). Simplifying the above, we get

\displaystyle  	\psi(t,x) = e^{-i\frac{d\pi}{4}} t^{-d/2} e^{i|x|^2 / (2t)} \hat \psi_0\left(\tfrac{x}{t}\right) + \mathrm{``smaller"}

with the interpretation that {\xi \equiv \mathrm{momentum} \equiv \mathrm{velocity}} and so {\mathrm{mass} = 1}. Note that for {t \rightarrow -\infty}, the asymptotics pick up a “discrete” change

\displaystyle  	\psi(t,x) = e^{+i\frac{d\pi}{4}} t^{-d/2} e^{i|x|^2 / (2t)} \hat \psi_0\left(\tfrac{x}{t}\right) + \mathrm{``smaller"}.

Proposition We have, uniformly in {t} and {\psi_0}, the dispersive estimate,

\displaystyle  	\|e^{it\Delta/2}\psi_0 \|_{L_x^\infty} \lesssim |t|^{-d/2} \|\psi_0\|_{L_x^1}

and the “Fraunhofer formula”

\displaystyle  	\left\|e^{it\Delta/2} \psi_0(x) - e^{-i\frac{d\pi}{4} + i \frac{|x|^2}{2t}} t^{-d/2} \hat \psi_0\left(\tfrac{x}{t}\right) \right\|_{L_x^2} \stackrel{t\rightarrow \infty}{\longrightarrow} 0

for every {\psi_0 \in L^2({\mathbb R}^d)}.

Remark The motivation for the dispersive estimate comes from the combination of our stationary phase computation and the a priori estimate {\| \hat \psi_0\|_{L^\infty} \leq (2\pi)^{-d/2} \|\psi_0\|_{L^1}}.

Proof: As a Fourier multiplier, {e^{it\Delta/2}} is also just a convolution operator. To compute the convolution kernel, take a {\psi_0 \in \mathcal{S}({\mathbb R}^d)} and find

\displaystyle  \begin{array}{rcl}  	\psi(t,x) &=& \displaystyle (2\pi)^{-d/2} \int e^{i \xi \cdot x - i t|\xi|^2/2} \hat \psi_0(\xi) \; d\xi \\ \\ 		&=& \displaystyle \lim_{\epsilon \downarrow 0} \; (2\pi)^{-d/2} \int e^{i \xi \cdot x - (\epsilon+i t)|\xi|^2/2} \hat \psi_0(\xi) \; d\xi \\ \\ 		&=& \displaystyle \lim_{\epsilon \downarrow 0} \; (2\pi)^{-d/2} \int \int e^{i \xi \cdot (x-y) - (\epsilon+i t)|\xi|^2/2} \psi_0(y) \; dy \; d\xi \\ \\ 		&\vdots& \scriptsize\hbox{using Fubini and Gaussian integrals} \\ 		&=& e^{-i\pi d/4}(2\pi t)^{-d/2} \int e^{+ i|x-y|^2/(2t)} \psi_0(y) \; dy 	\end{array}

In particular, the above computation gives the dispersive estimate

\displaystyle  	\|e^{it\Delta/2} \psi_0\|_{L_x^\infty} \leq (2\pi t)^{-d/2} \| \psi_0\|_{L_x^1}.

Furthermore, we get

\displaystyle  	[e^{it\Delta} \psi_0] (x) = (2\pi t)^{-d/2} e^{-i\pi d/2 + i|x|^2/(2t)} \int e^{i x\cdot y/t} \underbrace{e^{i y^2/(2t)}}_{\scriptsize \begin{matrix} \text{the only factor making}\\ \text{Fraunhofer non-exact} \end{matrix}} \psi_0(y) \; dy.

Thus,

\displaystyle  \begin{array}{rcl}  	\text{LHS (Fraunhofer)} &=& \displaystyle \Big\| (2\pi t)^{-d/2} \int (1 - e^{iy^2/(2t)}) e^{i x\cdot y/t} \psi_0(y) \; dy \Big\|_{L_x^2} \\ \\ 	\tiny \begin{bmatrix} \text{Plancherel and changing}\\ \text{variables to } x' = \frac{x}{t} \end{bmatrix}	&=& \| (1-e^{i|y|^2/(2t)}) \psi_0(y)\|_{L_x^2}, 	\end{array}

which, by the dominated convergence theorem with dominating function {4|\psi_0(y)|^2}, goes to {0}. \Box

As an aside, we discuss the result in the setting of optics. If we shine a laser through a screen with an aperture, let us denote {\psi_0(y_1,y_2) = (\text{amplitude }) e^{i \text{phase}}} with {y_1, y_2} being the horizontal and vertical axes, respectively. At the far screen we have

\displaystyle  	\psi(x_1, x_2) = \int_{{\mathbb R}^2} e^{i \sqrt{L^2 + |x-y|^2}} \; \psi_0(y) \; dy

where {\sqrt{L^2 + |x-y|^2}} is the distance from the source point to the observation point. Because {\sqrt{L^2 + |x-y|^2} \approx \sqrt{L^2 + x^2} - \frac{x \cdot y}{\sqrt{L^2 + x^2}} + (\text{smaller})} and that {\sqrt{L^2 + x^2} \approx L + \frac{x^2}{2L} + \cdots}, the dominant term is essentially the Fourier transform.

Also

\displaystyle  \begin{array}{rcl}  	0 &=& (\partial_t^2 - \Delta_{x,y,z}) ( e^{i\omega (t-z)} \psi(x,y,z)) \\ 		&=& \big( 2i\omega \psi_z + \underbrace{\psi_{zz}}_{\text{``negligible''}} - \Delta_{x,y} \psi \big) e^{i\omega(t-z)} 	\end{array}

cf Paraxial approximations.

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