Use integration by parts to exploit cancellation

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Use integration by parts to exploit cancellation

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[QUICK DESCRIPTION]

When faced with the task of bounding an integral such as
[math] \int_a^b f(x) g(x)\ dx,[/math]
where $f$ is "oscillatory" (e.g. $f(x) = e^{i \phi(x)}$ for some rapidly oscillating $\phi$) and $g$ is "smooth" (so that $g'$ is "small"), then it is often advantageous to use [[integration by parts]] to integrate $f$ and differentiate $g$.  In some cases one may wish to transfer a term from the smooth factor $g$ to the oscillatory factor $f$ to make $f$ easier to integrate (e.g. borrow a factor of $i \phi'(x)$ from $g$ to turn $f$ into $i \phi'(x) e^{i\phi(x)}$, which has an antiderivative of $e^{i \phi(x)}$).

[PREREQUISITES]

Undergraduate calculus

[EXAMPLE]

Fourier transform of bump functions or Schwartz functions is rapidly decreasing. Consider a Schwartz function $f\in\mathcal{S}(R)$, and denote by $\hat f$ its [[Fourier transforms front page|Fourier transform]]. We will use integration by parts and the fact that $f$ together with its derivatives of any order decay faster than any polynomial function at infinity to exhibit the rapid decay of $\hat f$ at infinity. Using the definition of $\hat f$ and 'borrowing' a factor of  $(-2\pi i \xi)$  from the function $f$ we write

[maths] \begin{align}\hat f (\xi) &= \int_\R f(x)e^{-2\pi i \xi x} dx=\int_\R \frac{f(x)}{(-2\pi i \xi)}\frac{d}{dx}\big( e^{-2\pi i\xi x}\big) dx =\\&= \frac{1}{2\pi i \xi} \int_\R e^{-2\pi i \xi x} f'(x) dx.\end{align}[/maths]

Using the same integration by parts trick on the higher order derivatives of $f$ it is not difficult to see that

[math] (2\pi i \xi)^k\hat f (\xi) = \int_\R e^{-2\pi i \xi x} f^{(k)}(x) dx ,[/math]

for any $k\in \N$. Now that the integration by parts has exploited the cancellation due to the oscillating factor $e^{-2\pi i\xi x}$ we can put absolute values everywhere and estimate

[math] |\hat f (\xi)| \leq  \frac{C}{|\xi|^k} \|f^{(k)}\|_{L^1(\R)} ,\quad k\in\N.[/math]

Of course this estimate makes sense since $f$ is a Schwartz function and thus $f^{(k)}\in L^1(\R)$ for any $k\in\N$.

A slightly more general statement is the following. Let $P:\R \to \C$ be a polynomial and write $D:\mathcal{S}(\R) \to \mathcal{S}(\R)$ for the differential operator $Df(x)=f'(x)$. We consider the differential operator $P(D):\mathcal{S}(\R) \to \mathcal{S}(\R)$, defined in the obvious way. Using the special case $P(t)=t$ that we consider before and the linearity of the Fourier transform we can now write the identity

[math] P(2\pi i \xi)\hat f (\xi) = \int_\R e^{-2\pi i \xi x} P(D)f (x) dx ,[/math]

for any $f\in\mathcal{S}(\R)$, which in turn implies the estimate

[math] |\hat f (\xi)| \leq  \frac{C}{|P(\xi)|} \|P(D)f\|_{L^1(\R)}.[/math]

[EXAMPLE vdc]

'''Deducing [[the van der Corput lemma for oscillatory integrals]] in the simplest case.''' Suppose that the phase function $\phi:\R \to \R$ obeys the bound $|\phi '(x)|\geq \lambda$ for all $x \in [a,b]$. Assume also that $\phi '$ is monotone. We want to prove the bound

[math] \bigg|\int_a^b e^{i \phi(x)} \ dx\bigg| \leq C / \lambda,[/math]

for $\lambda>0$. Here $ f(x)=e^{i\phi(x)}$ and $g(x)=1$ but more general functions $g$ can be considered as in [ref Example #vdcbump] below. 'Borrowing' a factor of $i\phi'(x)$ from the function $g(x)=1$ we can write $e^{i\phi(x)}=(e^{i\phi(x)})'\frac{1}{i\phi'(x)}$. Now we use integration by parts to get

[math] \int_a^b e^{i \phi(x)}dx = \frac{e^{i\phi(b)}}{i\phi'(b)} - \frac{e^{i\phi(a)}}{i\phi'(a)}- \int_a ^b e^{i\phi(x)} \frac{d}{dx}\bigg(\frac{1}{i\phi'(x)}\bigg) dx.[/math]

Now we have exploited the cancellation that comes from the oscillating factor $e^{i\phi(x)}$ so we can estimate more or less by brute force

[math] \bigg|\int_a^b e^{i \phi(x)}dx \bigg|\leq  \frac{1}{|\phi'(b)|} + \frac{1}{|\phi'(a)|}+ \int_a ^b \bigg| \frac{d}{dx}\bigg(\frac{1}{\phi'(x)}\bigg) \bigg|\ dx.[/math]

In order to complete the estimate we note that $\frac{d}{dx}(\frac{1}{\phi'(x)})$ has constant sign throughout $(a,b)$ since $\phi'$ is monotone and thus

[math] \bigg|\int_a^b e^{i \phi(x)}dx \bigg|\leq  2 \max \bigg\{\frac{1}{|\phi'(b)|},\frac{1}{|\phi'(a)|}\bigg\}\leq \frac{2}{\lambda}.[/math]

[EXAMPLE vdcbump]

'''Deducing van der Corput for bump functions from [[the van der Corput lemma for oscillatory integrals]].''' We now consider the slightly more general case than the one of [ref Example  #vdc] where $f(x)=e^{i\phi(x)}$ and $g(x)=\psi(x)$ is a smooth bump function on $[a,b]$. We will deduce van der Corput's lemma for bump functions, that is the estimate

[math] \bigg|\int_a^b e^{i \phi(x)}\psi(x) \ dx\bigg| \leq C_k / \lambda^{1/k} \big(\ |\psi(b)|+\int_a ^b |\psi'(x)|dx \ \big),[/math]

where $\phi:\R \to R$ obeys the bound $|\phi'(x)|\geq \lambda$ for $x\in [a,b]$, assuming we know [[the van der Corput lemma for oscillatory integrals]]. Using integration by parts we write

[maths]\begin{align} \int_a^b e^{i \phi(x)}\psi(x) \ dx &= \int_a ^b \frac{d}{dx} \int_a ^x e^{i\phi(t)}dt \ \psi(x) \ dx=\\&= \int_a^b e^{i \phi(x)} dx \ \psi(b)-\int_a^b \int_a ^x e^{i\phi(t)}dt\  \psi'(x)\ dx.\end{align}[/maths]

The basic point here is that van der Corput's estimate is ''uniform'' over choices of intervals $[a,b]$ where the hypothesis $|\phi'(x)|\geq \lambda$ holds. Thus it makes sense to estimate

[math] \bigg|\int_a^b e^{i \phi(x)}\psi(x) dx \bigg|\leq \sup_{a\leq s \leq b}\bigg|  \int_a ^s e^{i\phi(t)}dt \bigg| \ \big\{ |\psi(b)|+\int_a^b |\psi'(x)| dx \big\}.[/math]

Now the term $\sup_{a\leq s \leq b}\big|  \int_a ^s e^{i\phi(t)}dt \big|$ is controlled by [[the van der Corput lemma for oscillatory integrals]] by $C_k/\lambda^\frac{1}{k}$ so we get the desired estimate.

[EXAMPLE]

Here we give an example of the localization principle in [[the method of stationary phase]] (also known as the "principle of non-stationary phase"): if $a: \R \to \C$ is a bump function and $\phi: \R \to \R$ is a smooth phase which is never stationary on the support of $a$ (thus $\phi'$ is bounded away from zero on this support), then $\int_\R a(x) e^{i \lambda \phi(x)}\ dx$ is rapidly decreasing in $\lambda$.

By taking absolute values everywhere (or by the [[base times height]] bound) we see that this integral is already bounded in $\lambda$.  To get one order of decay, one can rewrite the integral (using the [[adding and subtracting|multiplying and dividing]] trick), taking advantage of the non-vanishing of $\phi'$, as

[math] \int_\R \frac{a(x)}{i \lambda \phi'(x)} i \lambda \phi'(x) e^{i \lambda \phi(x)}\ dx.[/math]

The point is that the second factor $i \lambda \phi'(x) e^{i \lambda \phi(x)}$ is the derivative of $e^{i \lambda \phi(x)}$.  Integrating by parts, we can express this as

[math] -\int_\R \frac{d}{dx}(\frac{a(x)}{i \lambda \phi'(x)}) e^{i \lambda \phi(x)}\ dx.[/math]

Applying the [[base times height]] bound now gives a bound of $O(1/\lambda)$, giving one order of decay. Repeating this process, one can show arbitrary amounts of decay in $\lambda$ (note that while many derivatives of $\phi$ will begin appearing, only the first derivative $\phi'$ will appear in the ''denominator'' of the integrand).

One can also deduce this example from [ref Example #vdc] above by the trick "[[linearize the phase]]".

[EXAMPLE]

This technique works well for establishing ''energy identities'' or ''energy inequalities'' in PDE.  We illustrate this with the Korteweg-de Vries equation

[math kdv] u_t + u_{xxx} = 6 u u_x[/math]

where $u: \R \times \R \to \R$ is a function, which for simplicity we will assume to be smooth, with all derivatives rapidly decreasing as $x \to \infty$, in order to ignore issues about justifying integration by parts, [[differentiation under the integral sign]], etc..  We claim that the mass

[math] M(t) := \int_\R u(t,x)^2\ dx[/math]

is conserved.  To see this, it of course suffices to show that $\partial_t M(t)$ vanishes.  Differentiating under the integral sign, we obtain

[math] \partial_t M(t) = \int_\R 2 u u_t\ dx,[/math]

where we have suppressed the independent variables $t,x$ for brevity.  Using [eqref kdv], this becomes

[math] \partial_t M(t) = \int_\R - 2 u u_{xxx} + 12 u^2 u_x\ dx.[/math]

To simplify this using integration by parts, we first observe that $3 u^2 u_x$ is a total derivative $(u^3)_x$, and so by integration by parts (or the fundamental theorem of calculus) the second term $\int_R 12 u^2 u_x\ dx$ vanishes.

What about the first term $-2 \int_\R u u_{xxx}\ dx$?  The problem here is that one factor, $u_{xxx}$, is absorbing too many of the derivatives.  But we can rebalance the derivatives by integration by parts, writing

[math] -2\int_\R u u_{xxx}\ dx = 2 \int_\R u_x u_{xx}\ dx.[/math]

(Intuitively, $u_{xxx}$ has a lot of cancellation in it, while $u$ has a lot of regularity (at least relative to $u_{xxx})$, and integration by parts is the perfect tool to rebalance the two.)  But now we observe that $2 u_x u_{xx}$ is also a total derivative $(u_x^2)_x$, so $\partial_t M(t)$ vanishes as required.

[GENERAL DISCUSSION]

The technique also works for improper integrals such as $\int_\R f(x) g(x)\ dx$ provided there is enough decay at infinity.  In fact, the method tends to work even better in this case, as one does not pick up boundary terms.

In some cases, the method can be iterated, integrating an oscillation repeatedly and differentiating all other factors.  This often leads to bounds which are ''rapidly decreasing'' in some key parameter that controls the amount of oscillation.

The trick also works in higher dimensions; one picks a direction (or vector field) to integrate one factor in, and differentiates the other (using Stokes' theorem if necessary).

See also [[partial summation]] for a discrete version of this trick.

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