Linearize the phase

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Linearize the phase

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[QUICK DESCRIPTION]

When faced with the task of controlling an integral containing a phase such as $e^{i \lambda \phi(x)}$, consider using a change of variables (or a Taylor-type approximation) to replace $\phi$ with a simpler phase, such as a linear phase or a quadratic phase in normal form.  This potentially allows one to use the tools of [[Fourier transforms front page|Fourier analysis]] or contour integration.

[PREREQUISITES]

Harmonic analysis

[EXAMPLE]

Suppose one wants to show that a one-dimensional integral $\int_\R a(x) e^{i\lambda \phi(x)}\ dx$ is rapidly decreasing in $\lambda$, where $a$ is a bump function and $\phi$ is a smooth phase which has no stationary points on the support of $a$ (and in particular, is strictly monotone on this support).  Then one can perform a change of variables $y = \phi(x)$ to replace the integral with a Fourier integral $\int_\R \tilde a(y) e^{i \lambda t}$, where $\tilde a$ is another bump function (which can be written explicitly in terms of $a$, the change of variables function $\phi$, and the Jacobian factor $\phi'$).  Since the Fourier transform of a bump function is rapidly decreasing, the claim follows.

Note that one can also establish this claim using "[[use integration by parts to exploit cancellation]]" or [[the method of stationary phase]].

[EXAMPLE]

Now suppose one wants to understand the one-dimensional integral $\int_\R a(x) e^{i\lambda \phi(x)}\ dx$, where $\phi$ now has one stationary point on the support of $a$, say at the origin $x=0$.  Suppose also that $\phi$ is non-degenerate with $\phi' '(0) > 0$.  By Taylor expansion, $\phi(x) = \phi(0) + \frac{\phi' '(0)}{2} x^2 + O(x^3)$ near the origin.  If we eliminate the contribution away from the origin (which is rapidly decreasing in $\lambda$ by the previous example), and then perform a smooth change of variables $\phi(x) = \phi(0) + t^2$, then one is faced with an integral of the form $e^{i \lambda \phi(0)} \int_\R \tilde a(t) e^{i \lambda t^2}\ dt$, where $\tilde a$ is a bump function with $\tilde a(0) = a(0) / (\phi' '(0)/2)^{1/2}$.  To proceed further, we [[create an epsilon of room]] and use the explicit integral

[math] \int_\R e^{-\varepsilon t^2} e^{i \lambda t^2}\ dt = (\pi / (\varepsilon - i \lambda) )^{1/2},[/math]

(where we take a standard branch cut of the logarithm) which can be established by contour integration (or using the [[square and rearrange]] trick).  One can then use "[[adding and subtracting]]" to write $\tilde a(t) = \tilde a(0) e^{-\varepsilon t^2} + (\tilde a(t) - \tilde a(0) e^{-\varepsilon t^2})$.  The contribution of the main term is

[math] e^{i \lambda \phi(0)} \frac{a(0)}{(\phi''(0)/2)^{1/2}} (\frac{\pi}{ \varepsilon - i \lambda})^{1/2}.[/math]

The contribution of the error term is $O(1/\lambda)$.  This can be seen by noting that one can pull a factor of $t$ out of the amplitude $(\tilde a(t) - \tilde a(0) e^{-\varepsilon t^2})$, leaving a remainder which is still smooth; now we can "[[use integration by parts to exploit cancellation]]", taking advantage of the fact that we can antidifferentiate $t e^{i\lambda t^2}$.  putting everything together, and letting $\varepsilon \to 0$, we obtain

[math]\int_\R a(x) e^{i\lambda \phi(x)}\ dx = a(0) e^{i \lambda \phi(0)} (\frac{2\pi}{-i\lambda \phi' '(0)})^{1/2} + o(\lambda^{-1/2}).[/math]

In fact a full asymptotic expansion in powers of $\lambda^{-1/2}$ can be obtaind by refining this method, leading to the [[method of stationary phase]].

Note also that the asymptotics are consistent with [[base times height]] heuristics: the (signed) amplitude of the integrand at the stationary point $0$ is $a(0) e^{i \lambda \phi(0)}$, while the width of the interval where the phase is stationary (in the sense that $\phi(x)$ only differs from $\phi(0)$ by $O(1)$ is about $O( 1 / (\lambda \phi''(0))^{1/2} )$, as can be seen from Taylor expansion.

[EXAMPLE]

(Talk about how wave packet decompositions are used to analyse FIOs, or to analyse the restriction problem)

[GENERAL DISCUSSION]

In higher dimensions, placing a stationary phase in normal form may require [[w:Morse theory]].

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