About de Almeida, Marcelo Fernandes

Just another mathematician

ArXiv, comments, and “quality control”

Those of you who browsed the arXiv recently may have seen a link to a user survey on top of the page (as of now, apparently no longer online) (update: still available here, until April 26). I ignored it a few times, until a friend brought this particular bit to my attention. Sure enough, I took […]

via ArXiv, comments, and “quality control” — The Accidental Mathematician


Ramanujan film, the man who knew infinity

Turing Machine

ramanujan.jpghi all. ramanujan is one of the great/ inspiring/ legendary characters out of the pantheon of math heroes and they just released a major hollywood movie on his life starring Dev Patel and Jeremy Irons. its been a few great years lately for geeks of all stripes such as with the facebook movie, google movie, and the Turing movie (just a few that immediately come to mind). am really enjoying this moment in the spotlight or sun. if you are curious about such things, the etymology of geek vs nerd and the relation of “semantic drift” is now documented on dictionary.com, but the short story is that what was once a stigma is now an accolade/ badge of honor.[b15] and lets face it, mathematicians are long close to the stereotypical ultimate geeks. which reminds me of old joke:

Q. how can you tell if a mathematician is extroverted?


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Turning transport into diffusion

A remark very interesting where transport equations is linked to heat equations.


Consider the simple linear transport equation

$latex displaystyle partial_t f + vpartial_x f = 0,quad f(x,0) = phi(x) &fg=000000$

with a velocity $latex {v}&fg=000000$. Of course the solution is

$latex displaystyle f(x,t) = phi(x-tv), &fg=000000$

i.e. the initial datum is just transported in direction of $latex {v}&fg=000000$, as the name of the equation suggests. We may also view the solution $latex {f}&fg=000000$ as not only depending on space $latex {x}&fg=000000$ and time $latex {t}&fg=000000$ but also dependent on the velocity $latex {v}&fg=000000$, i.e. we write $latex {f(x,t,v) =phi(x-tv)}&fg=000000$.

Now consider that the velocity is not really known but somehow uncertain (while the initial datum $latex {phi}&fg=000000$ is still known exactly). Hence, it does not make too much sense to look at the exact solution $latex {f}&fg=000000$, because the effect of a wrong velocity will get linearly amplified in time. It seems more sensible to assume a distribution $latex {rho}&fg=000000$ of…

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Project 2.4: Boundary condition, linear equation and visualization of N-S equations. Full code

It appears that the experiments shows us a possibility of existence of a certain critical Reynolds number, in sense that there is a limier where the fluid become turbulent.


Hello even if am spending my weekend in Warsaw, I could not wait to show you final solution to Navier-Stokes and send you last part of numerical analysis (notebook). Let me start with animation and personal reflection:

To be honest I love big cities! Especially long journey in public transport through Warsaw can be really inspirational. City with population of 2 millions with such a history acts like a living organism itself with information carriers, call humans trying to minimize its own effort. It is not the end of this structure. Those carriers seams to be in the middle of some type of sophisticated computer, calculating infinite many extremely hard problems, like Navier – Stokes equations for leafs, cars, oceans or even planes within the blink of an eye. Yesterday I attended popular science lecture by one of leading polish scientists prof. Ryszard Horodecki and believe me or not, but…

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Mittag-Leffler family for an integro-differential equation

In a few months ago I post a note about a family of an integro-partial differential equations (2), there the main objective was to expose, without details, a first work to treat questions such as existence and symmetries of solutions in certain critical space, namely, homogeneous Morrey Spaces. On that work the Mittag-Leffler family below was central on the estimates,

\displaystyle \widehat{u}(t,\xi)=\mathbb{E}_{\alpha}(-t^{\alpha}\vert\xi\vert^2)\widehat{\varphi}(\xi) \ \ \ \ \ (1)

which is an integral form of the integro-partial differential equation

\displaystyle u_t(t,x)=\int_{0}^{t}r_{\alpha-1}(t-s) \Delta_{x}u(s,x)ds,\; u(0,x)=\varphi(x) \ \ \ \ \ (2)

where {1\leq\alpha<2}, {\Delta_{x}=\Sigma_{j=1}^{n}(\partial/\partial x_j)^2}, {u(t,x)=(u_{1}(t,x),\cdots,u_{n}(t,x))} for {t\in [0,\infty)} and {x\in\mathbb{R}^{n}} and {r_{\alpha}(s)=s^{\alpha-1}/\Gamma(\alpha)}, here {\Gamma(\alpha)} stands for Gamma function. In this post we will prove (1). Before, we need some properties of the Laplace transform. Given a “well” real-valued function {u_j(\cdot,x)} on {[0,\infty]}, we define the Laplace transform as

\displaystyle \mathcal{L}(u_j(\cdot,x))(s)=\int_0^{\infty}e^{-st}u_j(t,x)dt. \ \ \ \ \ (3)

Given {\lambda\in (0,\infty)}, let  {\partial_tu, u,v:[0,\lambda]\times\mathbb{R}^{n}\rightarrow\mathbb{R}^n} be certain  {\mathbb{R}^n}-valued functions for each {t\in(0,\lambda)} fixed, then

\displaystyle \mathcal{L}(\partial_tu)(s)=-u(0,x)+s\mathcal{L}(u)(s) \ \ \ \ \ (4)


\displaystyle \mathcal{L}(u\ast^{t}v)(s)=\mathcal{L}(u)(s)\mathcal{L}(v)(s) \ \ \ \ \ (5)

where {\ast^{t}} stands for convolution operator on variable {t}, that is, {u\ast^{t}v(t)=\int_0^t u(t-s,x)v(s,x)ds}. Hence, applying the Laplace transform in (2), we obtain

\displaystyle s\mathcal{L}(u)(s)-\varphi(x)=\mathcal{L}(r_{\alpha-1})(s)\mathcal{L}(\Delta_{x} u)(s). \ \ \ \ \ (6)

In view of

\displaystyle \begin{array}{rcl} \mathcal{L}(t^{d})(s)=\int_0^{\infty}e^{-ts}t^{d}dt=\frac{\Gamma(d+1)}{s^{d+1}}, \;\; (d>0) \end{array}


\displaystyle \begin{array}{rcl} \mathcal{L}(\frac{\partial^2}{\partial x_i^2}u)(s)=\frac{\partial^2}{\partial x_i^2}\mathcal{L}(u)(s) \end{array}

we have that the equation (6) can be written as

\displaystyle s\mathcal{L}(u)(s)-\varphi(x)=s^{1-\alpha}\Delta_{x}\mathcal{L}(u)(s). \ \ \ \ \ (7)

For {j=1,\cdots n}, let {u_j\in \mathcal{S}(\mathbb{R}^{n})} (Schwartz space) and let {\widehat{u_j}} be the Fourier transform of {u_j} given by

\displaystyle \widehat{u_j}(t,\xi)=\int_{\mathbb{R}^{n}}e^{-ix\cdot \xi}u_j(t,x)dx. \ \ \ \ \ (8)

If {u_j\in C^{\infty}(\mathbb{R}^{n})} with compact support, easily seems that

\displaystyle \widehat{\frac{\partial^{2} u_j}{\partial x_i^{2}}}(t,\xi)=-\xi_j^{2}\,\widehat{u_j}(t,\xi)\Rightarrow \widehat{\Delta_{x} u}(t,\xi)=-\vert \xi\vert^2\,\widehat{u}(t,\xi). \ \ \ \ \ (9)

As {\widehat{\mathcal{L}(u)(s)}(\xi)=\mathcal{L}(\widehat{u}(\cdot,\xi))(s)} it follows from (7) and (9) that

\displaystyle s\mathcal{L}(\widehat{u}(\cdot,\xi))(s)=\widehat{\varphi}(\xi)-s^{1-\alpha}\vert \xi\vert^2 \mathcal{L}(\widehat{u}(\cdot,\xi))(s)\nonumber \ \ \ \ \ (10)

which yields

\displaystyle \mathcal{L}(\widehat{u}(\cdot,\xi))(s)=\frac{\widehat{\varphi}(\xi)}{s+s^{1-\alpha}\vert\xi\vert^2}=\frac{s^{\alpha-1}}{s^{\alpha}+\vert\xi\vert^2} \widehat{\varphi}(\xi). \ \ \ \ \ (11)

This motivates to define Mittag-Leffler function as a complex integral on certain curve. Indeed, firstly notice that the complex valued function {f(z)=\frac{e^zz^{\alpha-1}}{z^{\alpha}+\vert\xi\vert^2}} has singularities in

\displaystyle a_{\alpha}(\xi)=|\xi|^{\frac{2}{\alpha}}e^{\frac{i\pi}{\alpha}} ,\;\;\;b_{\alpha}(\xi)=|\xi|^{\frac{2}{\alpha}}e^{-\frac{i\pi}{\alpha}},\text{ for }\xi\in\mathbb{R}^{n}. \ \ \ \ \ (12)

Let {\gamma} be the standard Hankel’s curve in {\mathbb{C}}, positive oriented, such that {a_{\alpha}(\xi)\in Int(\gamma)} and {b_{\alpha}(\xi)\in Int(\gamma)}, that is, let {\gamma=r_1+r_2+C_r} be a parametrized curve given by {r_1(t)=te^{i\theta}}, {r_2(t)=te^{-i\theta}} for {t\in (r,\infty)} and {C_r(t)=r e^{it}} for {t\in (-\theta,\theta)}. As we want to chose {\gamma} such that {a_{\alpha}(\xi)\in Int(\gamma)} and {b_{\alpha}(\xi)\in Int(\gamma)}, we shall suppose that {r^{\alpha}>\vert \xi\vert^2>\epsilon^{\alpha}>0} and one defines Mittag-Leffler function as

\displaystyle \begin{array}{rcl} \mathbb{E}_{\alpha,\beta}(-\vert\xi\vert^2)=\frac{1}{2\pi i}\int_{\gamma}\frac{e^zz^{\alpha-\beta}}{z^{\alpha}+\vert\xi\vert^2}dz,\; (\alpha>0,\beta>0). \end{array}

Using residue theorem, a characterization very important of this definition was obtained in [Fujita] (see also [Hirata-Miao][Fujita2]).

Proposition 1 If {1<\alpha<2} and {\beta=1}, we have

\displaystyle \begin{array}{rcl} L^{1}(\mathbb{R}^{n})\ni\mathbb{E}_{\alpha}(-|\xi|^{2})=\frac{1}{\alpha}(\exp(a_{\alpha}(\xi))+\exp(b_{\alpha}(\xi)))+l_{\alpha}(\xi) \end{array}


\displaystyle l_{\alpha}(\xi)= \begin{cases} \frac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\frac{|\xi|^{2}s^{\alpha-1}e^{-s} }{s^{2\alpha}+2|\xi|^{2}s^{\alpha}\cos(\alpha\pi)+|\xi|^{4}}ds & \text{ if }\xi\neq0\\ 1-\frac{2}{\alpha}, & \text{ if }\xi=0. \end{cases} \ \ \ \ \ (13)

Some corollaries can be obtained by Lemma above. Indeed, taking {t=\vert\xi\vert^{\frac{2}{\alpha}}s^{\frac{1}{\alpha}}} one has

\displaystyle \begin{array}{rcl} \vert \mathbb{E}_{\alpha}(-|\xi|^{2})\vert&\leq& \frac{2}{\alpha}+\vert l_{\alpha}(\xi)\vert\\ &\leq&\frac{2}{\alpha}+\frac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\frac{e^{-\vert\xi\vert^{\frac{2}{\alpha}}s^{\frac{1}{\alpha}}}}{s^{2}+2s\cos(\alpha\pi)+1}ds\\ &\leq&\frac{2}{\alpha}+\frac{\sin(\alpha\pi)}{\pi}\int_{0}^{\infty}\frac{1}{s^{2}+2s\cos(\alpha\pi)+1}ds\\ &=&\frac{2}{\alpha} +(1-\frac{2}{\alpha})=1, \end{array}

more general (see de Almeida, Ferrreira, L.F.C).

Proposition 2 Let {1\leq\alpha<2} and {0\leq\delta<2.} There is {C>0} such that

\displaystyle \left\vert \frac{\partial^{k}}{\partial\xi^{k}}\left[ \left\vert \xi\right\vert ^{\delta}\mathbb{E}_{\alpha}(-|\xi|^{2})\right] \right\vert \leq C\left\vert \xi\right\vert ^{-\left\vert k\right\vert },\text{ } \ \ \ \ \ (14)

for all {k\in(\mathbb{N}\cup\{0\})^{n}} with {\left\vert k\right\vert \leq\lbrack n/2]+1} and for all {\xi\neq0.}

Using an exercise,

\displaystyle \int_{0}^{\infty}e^{-t}\mathbb{E}_{\alpha}(at^{\alpha})dt=\frac{1}{1-a},\;\; (r^{\alpha}>\vert a\vert>0)\text{ and } (1< \alpha< 2),\nonumber \ \ \ \ \ (15)

we have

\displaystyle \mathcal{L}(\mathbb{E}_{\alpha}(-t^{\alpha}\vert\xi\vert^2))(s)=\frac{s^{\alpha-1}}{s^{\alpha}+\vert\xi\vert^2}. \ \ \ \ \ (16)

Taking in (11) the inverse Laplace transform and using (16), we obtain {\widehat{u}(t,\xi)=\mathbb{E}_{\alpha}(-t^{\alpha}\vert\xi\vert^2)\widehat{\varphi}(\xi).}

Mikhlin-Hormander theorem for Morrey spaces


In the last post we given a proof of the following theorem.

Theorem 1 (Mikhlin-Hormander theorem) Let {k>n/2} and {\sigma\in C^{k}(\mathbb{R}^{n}\backslash\{0\})}. If for {\vert\gamma\vert\leq k} we have

\displaystyle \sup_{r>0}r^{\vert \gamma\vert}\left(\frac{1}{r^n}\int_{\frac{r}{2}<\vert\xi\vert<2r}\vert (D^{\gamma}_{\xi}\sigma)(\xi)\vert^{2}d\xi\right)^{\frac{1}{2}}\leq L \ \ \ \ \ (1)

then {T_{\sigma}:=\mathcal{F}^{-1}\sigma(\xi)\mathcal{F}f} is a Fourier multiplier on {L^p}, {1<p<\infty}. In particular, {T_{\sigma}} is a Fourier multiplier on {L^p} if {\sigma\in\Sigma_1^0(\mathbb{R}^n)}, that is, if

\displaystyle \vert D^{\gamma}\sigma(\xi)\vert \leq L \vert \xi\vert ^{-\vert\gamma\vert},\;\xi\neq 0. \ \ \ \ \ (2)

In this post, we would like to extend this theorem for a class of Morrey-type spaces. In other words

Theorem 2 Let {1<p<\infty} and {0<\mu<n}. If {\sigma \in\Sigma_{1}^{0}(\mathbb{R}^{n})}, the Fourier multiplier {T_{\sigma}} on {L^{p}} can be extended to {\mathcal{M}_{p,\mu}}, that is, there exists {C>0} such that

\displaystyle \Vert T_{\sigma}f\Vert_{\mathcal{M}_{p,\mu}}\leq C\,L\Vert f\Vert_{\mathcal{M}_{p,\mu}}, \ \ \ \ \ (3)

for all {f\in\mathcal{M}_{p,\mu}}.

 Proof. A proof can be found in [Taylor] (see also [Kozono1Kozono2]. For reader convenience we give some details. This key theorem was important in my recent article in join with Ferreira, Lucas C.F.

We have that {T_{\sigma}} is a convolution operator with kernel {k_{\sigma}(z)=(\sigma(\xi))^{\vee}(z)}. As {\sigma\in \Sigma_1^0} it follows from [p.26, Stein] that

\displaystyle \vert \partial_{z}^{\gamma}k_{\sigma}(z)\vert\leq C L |z|^{-n-\vert\gamma\vert},\; z\neq 0, \ \ \ \ \ (4)

Based in (4) and Minlin-Hormander theorem we obtain a proof of the Theorem 2. Indeed, firstly one splits {f\in\mathcal{M}_{p,\mu}(\mathbb{R}^{n})} as

\displaystyle f=f_{0}+\sum_{j=1}^{\infty}g_{j},


\displaystyle f_{0}=\chi_{B_{2r}(x_{0})}f,\text{ }\,g_{j}=f\chi_{A_{rj}}\text{ and } A_{rj}=\{x\,:\,2^{j}r\leq|x_{0}-x|\leq2^{j+1}r]\}.

Defining {k_{j}(x,y)=\chi_{B_{r}(x_{0})}(x)k(x-y)\chi_{A_{rj}}(y)} and {T_{\sigma,j}(f)(x)=\int_{\mathbb{R}^{n}}k_{j}(x,y)f(y)dy} easily gets by Hölder inequality and estimate (4) the following

\displaystyle \begin{array}{rcl} \vert T_{\sigma,j}(f)(x)\vert^{p}&\leq& \left(\int_{\mathbb{R}^{n}}\vert k_{j}(x,y)f(y)\vert dy\right)^{p}\\ \\ &\leq& \left(\int_{\mathbb{R}^{n}}\vert k_{j}(x,y)\vert dy\right)^{p/q}\int_{\mathbb{R}^{n}}\vert k_{j}(x,y)\vert\, \vert f(y)\vert^{p} dy\\ \\ &=&\left(\int_{A_{rj}(x_0)}\chi_{_{D_{r}(x_0)}}(x)\vert k_{\sigma}(x-y)\vert dy\right)^{p/q}\int_{\mathbb{R}^{n}}\vert k_{j}(x,y)\vert\, \vert f(y)\vert^{p} dy\\ \\ &\leq& \left(CL(2^{j}r)^{-n} vol(A_{rj}(x_0))\right)^{p/q}\int_{\mathbb{R}^{n}}\vert k_{j}(x,y)\vert\, \vert f(y)\vert^{p} dy\\ \\ &\leq& C L^{p/q}\int_{\mathbb{R}^{n}}\vert k_{j}(x,y)\vert\, \vert f(y)\vert^{p} dy, \end{array}

where {q} denotes the conjugate exponent of {p}. Using Tonelli’s theorem and once more estimate (4) we have

\displaystyle \begin{array}{rcl} \int_{\mathbb{R}^{n}}\vert T_{\sigma,j}(f)(x)\vert^{p}dx &\leq& CL^{p/q} \int_{\mathbb{R}^{n}}\vert k_{j}(x,y)\vert dx \int_{\mathbb{R}^{n}}\vert f(y)\vert^{p}dy\\ \\ &=& CL^{p/q} \int_{D_{r}(x_0)}\vert k_{\sigma}(x-y)\vert\chi_{_{A_{rj}}}(y) dx \int_{\mathbb{R}^{n}}\vert f(y)\vert^{p}dy\\ \\ &\leq& CL^{p/q+1} 2^{-jn} \int_{\mathbb{R}^{n}}\vert f(y)\vert^{p}dy. \end{array}

Now we ready to proof the theorem. By Mikhlin-Hormander theorem and the last estimate one has

\displaystyle \begin{array}{rcl} \left\Vert T_{\sigma}f\right\Vert _{L^{p}(B_{r}(x_{0}))} & \leq&\left\Vert T_{\sigma}f_{0}\right\Vert _{L^{p}(\mathbb{R}^{n})}+\sum_{j=1}^{\infty}\left\Vert T_{\sigma}g_{j}\right\Vert _{L^{p}(B_{r}(x_{0}))}\nonumber\\ \\ & \leq& CL\left\Vert f_{0}\right\Vert _{L^{p}(\mathbb{R}^{n})}+\sum_{j=1}^{\infty}\left\Vert T_{\sigma,j}(\chi_{A_{rj}}f)\right\Vert_{L^{p}(\mathbb{R}^{n})}\nonumber\\ \\ &\leq& CL\left\Vert f_{0}\right\Vert _{L^{p}(\mathbb{R}^{n})}+\sum_{j=1}^{\infty}\left(\int_{\mathbb{R}^n} \vert T_{\sigma,j}(\chi_{A_{rj}}f(x))\vert^{p}\right)^{\frac{1}{p}}\nonumber\\ \\ & \leq& CL\left\Vert f\right\Vert _{L^{p}(B_{2r}(x_{0}))}+\sum_{j=1}^{\infty}CL^{(p/q+1)/p}2^{-jn/p}\left\Vert \chi_{A_{rj}}f\right\Vert_{L^{p}(\mathbb{R}^{n})}\nonumber\\ \\ & \leq&2^{\frac{\mu}{p}}CL\left\Vert f\right\Vert _{p,\mu}r^{\frac{\mu}{p}}+CL^{1/q+1/p}\sum_{j=1}^{\infty}2^{-jn/p}\left\Vert f\right\Vert _{p,\mu}(2^{j}r)^{\frac{\mu}{p}}\nonumber\\ \\ & \leq& 2C(1+\sum_{j=1}^{\infty}2^{-j(\frac{n-\mu}{p})})L\left\Vert f\right\Vert _{p,\mu}r^{\frac{\mu}{p}}, \end{array}

which yields (3), because the series above is convergent.

Mat-1.C – Harmonic Analysis

I Woke Up In A Strange Place

For those of you who followed my Harmonic Analysis notes a couple of years back, here is an updated (and slightly more polished) version in pdf  form, from an advanced course I gave at Aalto university:


the notes can now be found here


and they’re slightly updated.

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