Lecture 16

$p$-periodic distributions¶

Let $p>0$, a distribution $T$ is $p$-periodic if $\tau_p T = T$. Similar to the 1-periodic case, by choosing a partition of unity $\varphi_n$ subordinate to the intervals $[-np/2,np/2]$ we get

This time $\sum_{n\in \mathbb{Z}}\mathcal Ff(x+np) = (\omega_p(y),\mathcal Ff(y+x)) = (\mathcal F \tau_x\omega_p(y),f) =( e^{-2\pi i xy}\frac{1}{p}\omega_{\frac{1}{p}}(y),f(y)) = \frac{1}{p}\sum_{n\in \mathbb{Z}}f(\frac{n}{p})e^{-\frac{2\pi}{p}inx}$. Of course, one can also derive this by applying the Poisson summation formula directly.

This implies $\mathcal FT = \sum_{n\in \mathbb{Z}}(T(y),\varphi_0e^{-\frac{2\pi }{p}iny})\frac{1}{p}\delta_{\frac{n}{p}}$. If we write $\hat{T}(n) = \frac{1}{p}(T(y),\varphi_0e^{-\frac{2\pi}{p}iny})$, then by taking inverse transform (recall that $\mathcal F^{-1}\delta_{\frac{n}{p}} = e^{\frac{2\pi}{p}inx}$), we can write $T = \sum_{n\in \mathbb{Z}}\hat{T}(n)e^{\frac{2\pi }{p}inx}$, this is the Fourier expansion of $T$, which is parallel with the Fourier series theory.

Fundamental solution of $D+s$¶

Fundamental solution of a linear differential operator¶

The reason why fundamental solution is useful is because the following lemma and the fact that δ is the unit of convolution.

Lemma¶

$\frac{d}{dx}(f*T) = f*T'$. Consequently, $P(f*T) = f*PT$.

Proof. $((f*T)',g) = -(f*T,g') = -(T,f^-*g') = -(T,(f^-*g)')=(T',f^-*g) = (f*T',g)$. $\square$

It follows that if we can solve $PE = \delta$, then $P(g*E) = (g*PE) = g*\delta = g$, in other words, $g*E$ is solution of $Pf = g$ in sense of distribution!

The case of $D+s$¶

Let $D:=-i\frac{d}{dx}$ be the Dirac operator and $s$ be a complex number.

In problem set 3 we gave solution of the equation $(D+s)f = g$ by explicitedly calculating the inverse of $D+s$ using Fourier series and showed that the solution is given by $f = g*G$, where $G$ is given by Fourier series $\sum_{n\in \mathbb{Z}}\frac{1}{n+s}e^{inx}$.

But this method may not work for general differential operators. In this note we’ll revisit the problem using common method of solving differential equations from distribution theory. From the derivative theorem we see that $(D+s) G$ has Fourier coefficients 1, so as a $2\pi$-periodic distribution, $(D+s)G$ is equal to $2\pi\omega_{2\pi}$. From the above point of view, we can interprate $G$ as a fundamental solution of $2\pi$-periodic functions since $\omega_{2\pi}$ is the unit of convolution for $2\pi$-periodic functions. (Recall that convolving with $\omega_{2\pi}$ = periodization to period $2\pi$, but if it’s already $2\pi$-periodic then periodization does nothing). You may wonder why $(D+s)=2\pi\omega_{2\pi}$ instead of $\omega_{2\pi}$, this is because when we defined convolution at that time, we made a normalization by deviding $2\pi$.

Use Fourier transform to solve $D+s$¶

One of the applications of Fourier transform is that it can help us to find solution of partial differential equations in a direct way. As an example we consider the simple case $D+s$. Later we’ll use the Fourier transform to solve the wave equation and heat equation.

TBA.

The Cauchy kernel: A complex analysis interlude¶

Let $D$ be an open set with $\mathscr{C}^1$ bundary on $\mathbb{C}$. We adapt the identification $\mathbb{C}$ with $\mathbb{R}^2$ via $z$ with $x+iy$, where $i = \sqrt{-1}$. Then for $f(z)$ which is $\mathscr{C}^1$ on a neighbourhood of $D$, we have the following Cauchy integral formula

The Cauchy integral formula¶

$f(\zeta) = \frac{1}{2\pi i}(\int_{D}\frac{\partial_{\bar{z}}f(z)}{z-\zeta}dz\wedge d\bar{z}+\int_{\partial D}\frac{f(z)}{z-\zeta}dz)$.

In the real notation it is $f(\zeta) = \frac{1}{2\pi i }\int_{\partial D}\frac{f(z)}{z-\zeta}dz -\frac{1}{\pi}\int_{D}\frac{\partial_{\bar{z}}f(z)}{z-\zeta} d\lambda(z)$, where $d\lambda(z) = dxdy$ stands for the Lebesgue measure on $\mathbb{C}=\mathbb{R}^2$.

Proof. We give a proof of this using a distribution version of divergence theorem (or Green formula in 2 dimensional case), see Problem set 8 problem 3. The result is

We’ll use the complex version of this result. Choose a parametrization $\gamma(t) = (x(t),y(t))$ of $\partial D$, such that the tangent is given by $(x'(t),y'(t))$ (or $\gamma'(t) = x'(t)+iy'(t)$) and invards normal is given by $(-y'(t),x'(t))$ (or $i\gamma'(t) = i(x'(t)+iy'(t)) = -y'(t)+ix'(t)$). Then $ds =|\gamma'(t)|dt$, $n = \frac{1}{|\gamma'(t)|}(-y'(t),x'(t))$ so $n_1ds = -y'(t)dt$ and $n_2 ds = x'(t)dt$. Then $(-\partial_x[\chi_D],f) =\int_{D}\frac{\partial f}{\partial x}dxdy = -(n_1ds,f) = \int f(x(t),y(t))y'(t)dt = \int_{\partial D}fdy$. Similarly $(-\partial_y[\chi_D],f) = -\int_{\partial D}fdx$. Then we get

or $\partial_{\bar z}[\chi_D] = \frac{1}{2i}dz|_{\partial D}$. One can easily translate the proof using green formula or Stokes theorem. Let $\chi_D$ be the characteristic function of $D$, $\chi_{B_\epsilon}$ be the characteristic function of a small ball of radius ε centered at ζ.

Apply the identity (2) to $D\setminus B_{\epsilon}$ and acts to the function $\frac{f(z)}{z-\zeta}$ we get $\int_{\partial D}\frac{f(z)}{z-\zeta}dz - \int_{\partial B_\epsilon}\frac{f(z)}{z-\zeta}dz = 2i(\int_D \frac{\partial_{\bar{z}}f}{z-\zeta}dxdy - \int_{B_\epsilon}\frac{\partial_{\bar{z}}f}{z-\zeta}dxdy)$. Note that by parametrizing $\partial B_{\epsilon}$ with $\gamma(t) = \epsilon e^{it}+\zeta$, we get $\int_{\partial B_\epsilon}\frac{f(z)}{z-\zeta}dz = i\int_{0}^{2\pi}f(\zeta+\epsilon e^{i\theta})d\theta$ which goes to $2\pi f(\zeta)$ when $\epsilon \to 0$, by dominated convergence theorem.

Since $\int_{B_\epsilon}|\frac{\partial_{\bar z}f}{z-\zeta}|dxdy$ $\leq \sup_{B_{\epsilon}}|\partial_z f|\cdot \int_{B_{\epsilon}} \frac{1}{r}rdrd\theta = 2\pi\epsilon \sup_{B_\epsilon}|\partial_{\bar z}f|\to 0$ when $\epsilon \to 0$, we get $2\pi i f(\zeta) = \int_{\partial D}\frac{f(z)}{z-\zeta}dz -2i\int_{D}\frac{\partial_{\bar{z}}f}{z-\zeta}dxdy$. From the notation $dz\wedge d\bar{z} = (dx+idy)\wedge (dx-idy)=-2idx\wedge dy$, we get $f(\zeta) = \frac{1}{2\pi i }(\int_{\partial D}\frac{f(z)}{z-\zeta}dz + \int_D \frac{\partial_{\bar{z}}f(z)}{z-\zeta}dz\wedge d\bar{z})$.

Consequences of the Cauchy integral formula¶

Holomorphic case¶

If $\partial_{\bar{z}}f = 0$ on $D$, then $f(\zeta) = \frac{1}{2\pi i }\int_{\partial D}\frac{f(z)}{z-\zeta}dz$.

This is the usual Cauchy integral formula. Note that this holds for general simple closed curves than $\mathscr{C}^1$ case, see Ahlfors’ book.

Cauchy kernel¶

$\frac{1}{\pi}\frac{1}{z-\zeta}$ is the fundamental solution of $\partial_{\bar{z}}$.

Proof. Let φ be a smooth function of compact support on $\mathbb{C}$. Let $D$ be a domain with $\mathscr{C}^1$ boundary which contains the support of φ. Then by Cauchy integral formula $\varphi(\zeta) = \frac{1}{2\pi i }\int_{\partial D}\frac{\varphi(z)}{z-\zeta}dz - \frac{1}{\pi}\int_{D}\frac{\partial_{\bar z}\varphi(z)}{z-\zeta}d\lambda(z)$ $= 0 -\frac{1}{\pi}\int_D \frac{\partial_{\bar z}\varphi}{z-\zeta}\;d\lambda(z)$ because the support of φ is inside $D$ and hence $\varphi\equiv 0$ on $\partial D$.

It follows that $(\partial_{\bar z} [\frac{1}{\pi}\frac{1}{z-\zeta}],\varphi) = -\frac{1}{\pi}\int_D \frac{\partial_{\bar z}\varphi}{z-\zeta}\;d\lambda(z) = \varphi(\zeta) = (\delta_\zeta,\varphi)$.