User bazin - MathOverflow

Answer by Bazin for Is a Cauchy principal value invariant under a "change of variables"?

2013-05-19T19:27:54Z

Changing variables, we have $$(pv(K)\ast f)(x)=\lim_{\delta\searrow 0} \int_{\vert x-y\vert>\delta}K(x-y) f(y) dy=\lim_{\delta\searrow 0} \int_{\vert x-G(w)\vert>\delta}K(x-G(w)) f(G(w))\vert \nabla G(w)\vert dw, $$ so that $$ (pv(K)\ast f)(G(\nu))=\lim_{\delta\searrow 0}\int_{\vert G(\nu)-G(w)\vert>\delta}K(G(\nu)-G(w)) f(G(w))\vert \nabla G(w)\vert dw. $$ Now, if $G$ is globally Lipschitz continuous, i.e. $\nabla G\in L^\infty$, we have $$ \{w,\vert G(\nu)-G(w)\vert>\delta\}\subset\{w,\Vert\nabla G\Vert_{L\infty}\vert\nu-w\vert>\delta\}. $$ Since $\nabla G\circ \nabla G^{-1}=Id$, we have $ \Vert\nabla G\Vert\Vert \nabla G^{-1}\Vert\ge 1. $ As a result, if $G$ is globally bi-Lipschitz continuous (i.e. $G,G^{-1}$ are both globally Lipschitz continuous), we find $\alpha\ge 1$ such that $$ \{w,\vert w-v\vert\le \delta/\ \alpha\}\subset \{w,\vert G(w)-G(v)\vert\le \delta\}\subset \{w,\vert w-v\vert\le \alpha\delta\}.\tag E $$ We may now consider what is now a general singular integral, and not only a Fourier multiplier, that is the operator with kernel $$ k(v,w)=K(G(v)-G(w))\vert\nabla G(w)\vert. $$ We have indeed the following estimates $$ \vert k(v,w)\vert\lesssim\vert v-w\vert^{-n},\quad\vert\partial_vk(v,w)\vert+\vert\partial_wk(v,w)\vert\lesssim\vert v-w\vert^{-n-1} \tag {CZ}$$ $$ (pv(K)\ast f)(G(\nu))=\lim_{\delta\searrow 0}\int_{\vert G(\nu)-G(w)\vert>\delta}K(G(\nu)-G(w)) f(G(w))\vert \nabla G(w)\vert dw. $$ Using the embeddings (E), we see that that $\{w,\vert G(w)-G(v)\vert> \delta\}$ is the union of a set $\{w,\vert w-v\vert> \alpha\delta\}$ with a set where $\vert G(w)-G(v)\vert\sim \delta\sim \vert v-w\vert$ with volume $\delta^{n}$ on which $K$ is of size $\delta^{-n}$. From the Lebesgue differentiation theorem, this part of the integral converges as well. $$ (pv(K)\ast f)(G(\nu))=\lim_{\delta\searrow 0}\int_{\vert\nu-w\vert>\delta}K(G(\nu)-G(w)) f(G(w))\vert \nabla G(w)\vert dw+\Omega f, $$ where $\Omega$ is $L^p$ bounded for $p\in(1,+\infty)$. So when you change variables in a singular integral appearing as a Fourier multiplier or a convolution, you get a more general type of operator, a Calder\'on-Zygmund type of operator with a kernel satisfying (CZ).

Answer by Bazin for about smoothing pseudodifferential operators

2013-05-13T12:41:56Z

The answer is negative: take $f$ smooth compactly supported in $(-1/4,1/4)$ equal to 1 in $(-1/8,1/8)$, take $g(x) =f(x+1)$ so that $g$ is supported where

$-1/4

so that the supports of $f,g$ are disjoint. Now we consider $$ (f e^{-2i\pi D}g u)(x)=f(x) g(x-1) u(x-1)=f(x)^2 u(x-1) $$ which has the same $L^2$ norm as $f(x+1)^2 u(x)$. We note that the symbol $e^{i\xi}$ belongs to $S_{0,0}^0$. The estimate $$ \Vert f e^{-2i\pi D}g u)\Vert_{L^2}\le C\Vert u\Vert_{-\epsilon} $$ cannot hold if $\epsilon >0$: take $v$ supported in $(-9/8,-7/8)$ and $u(x)=e^{2i\pi x\lambda} v(x)$. If the previous estimate were true, we would have $$ \Vert v\Vert_{L^2}\le C \Vert v(x)e^{2i\pi x\lambda} \Vert_{-\epsilon}. $$ The rhs goes to 0 when $\lambda$ goes to infinity (for a fixed $\epsilon$ positive), making the estimate impossible when $v\not=0$.

Answer by Bazin for Is there an equivalent of Heisenberg's uncertainty principle in the decision sciences ?

2013-05-08T21:43:34Z

Let me answer in terms of operator theory. The uncertainty principle can be interpreted as some particular inequality (as you say) such as $$ \frac{\hbar}{2}\Vert{u}\Vert^2\le \Vert{D_xu}\Vert\Vert{xu}\Vert, $$ inequality due to the identity $ 2\Re\langle\hbar\frac{1}{i}\partial_x u, ixu\rangle=\hbar\Vert{u}\Vert^2, $ and in fact to the non-commutation of the unbounded operators $D_x=\frac{\hbar}{i}\partial_x$ with $x$ (multiplication by $x$). Incidentally you can note that the non-commutation of bounded operators $A,B$ (say finite matrices) could never lead to $$ [A,B]=I\tag{NC} $$ as it is the case for $A=\frac{1}{i}\partial_x$, $B=ix$. In fact (NC) is not possible for matrices since the trace of a commutator is 0. This is one reason for which the quantization requires (necessarily infinite dimensional) unbounded operators.

Now this non-commutation property is at the source of quantum mechanics and produces the uncertainty principle inequalities. Each time you will have to deal with a non-commutative algebra of operators, you will have some type of uncertainty principle. In particular if your unknown quantities do not belong to a commutative algebra, you will have an uncertainty principle: the paradigmatic example remains the transfer from classical mechanics where the unknown quantities belong to a commutative algebra (functions on the phase space, such as position $x$ or momentum $\xi$), transfer to quantum mechanics where the unknown quantities are quantization of the previous quantities and are operators, such as $D_x,x$.

The uncertainty principle is simply the signature of non-commutativity and is not limited to quantum mechanics.

Answer by Bazin for Is there existence and uniqueness theory of this system of ODE?

2013-04-25T20:39:10Z

If the matrix $A(t)=(a_{ij}(t))$ is invertible, e.g. if there exists $B(t)\in L^\infty$, with $B(t) A(t)=Id$, then you get a linear non-characteristic system of type $$ \dot u +C(t) u=g(t),\quad C\in L^\infty. $$ However without invertibility, you may be in deep trouble, even in the scalar case: consider for $\nu>0$ the equation $$ t^{\nu+1}\dot x=\nu x(t), \quad x(0)=0, $$ which has 0 as a solution but also the smooth function $x(t)=H(t)\exp{-t^{-\nu}}$ flat at 0 since $$ x(t)=H(t)\exp{-t^{-\nu}},\quad \dot x(t)=H(t)\nu t^{-\nu-1}\exp{-t^{-\nu}},\quad t^{\nu+1}\dot x=\nu x(t). $$ Note that when $\nu=0$, the situation is not that bad and the singularity is called a regular singularity.

Answer by Bazin for Replacing large-dimensional ODE systems with one PDE

2013-04-24T07:38:59Z

Let me single out a situation which goes the other way around: how a system of ODE is describing the propagation of singularities for a principal type PDE.

Take a linear (pseudo)differential operator of real principal type with smooth coefficients: the principal symbol $p(x,\xi)$ is real-valued and $dp\wedge \xi\cdot dx\not=0$ (verified for the wave equation or a non-vanishing real vector field). Then the singularities are moving along the bicharacteristic curves, which are the integral curves of the Hamiltonian vector field of $p$, $$ H_p=\frac{\partial p}{\partial \xi}\cdot \frac{\partial }{\partial x}- \frac{\partial p}{\partial x}\frac{\partial }{\partial \xi}.\quad $$ Solving the system of ODE, $\dot \Gamma=H_p(\Gamma)$ is enough to understand the propagation of singularities: if $p(x,D) u\in C^\infty$ the the wave-front-set of $u$ is invariant by the flow of the Hamiltonian vector field. There is no need to solve the PDE if you are only interested in singularities.

Answer by Bazin for Can a nowhere continuous function be integrable ?

2013-04-22T11:28:21Z

Take the characteristic function of $\mathbb Q$: it is not Riemann integrable, everywhere discontinuous, almost everywhere 0 and thus Lebesgue integrable with integral 0.

Answer by Bazin for Nonharmonic solutions of Laplace's equation

2013-04-16T21:10:01Z

If $U$ is an open subset of $\mathbb R^n$, $f$ is a distribution on $U$ such that $\Delta f$ is analytic on $U$, then $f$ is analytic on $U$. This hypoellipticity result is true for any elliptic differential operator with analytic coefficients.

Answer by Bazin for Does the derivative of log have a Dirac delta term?

2013-04-15T21:34:31Z

The question is not meaningful since $\ln x$ is not defined for $x\le 0$. You may define, with derivatives in the distribution sense $$ f(x)=\ln\vert x\vert\text{ (even)},\quad f'(x)=pv\frac1{x} \text{ (odd, homogeneous degree -1)} $$ $$ g(x)=\ln(x+i0)=\lim_{\epsilon\rightarrow 0_+}\ln(x+i\epsilon),\quad g'(x)=\frac{1}{x+i0}= pv\frac1{x}-i\pi \delta, \text{(homogeneous degree -1)} $$ where the latter formula follows from $$ \ln(z)=\oint_{[1,z]}\frac{d\xi}{\xi},\quad z\in \mathbb C\backslash \mathbb R_-. $$ By analytic continuation we have (for $z\in \mathbb C\backslash \mathbb R_-$) $e^{\ln z}=z$ and with $H=\mathbf 1_{\mathbb R_+}$ $$ \ln(x+i0)=\ln\vert x\vert +i\pi H(-x)\Longrightarrow g'(x)=\frac{1}{x+i0}= pv\frac1{x}-i\pi \delta. $$ Taking the complex conjugate of $g$ gives you the definition of $\frac{1}{x-i0}$.

Answer by Bazin for Increasing regularity for $L^2$ function

2013-04-14T18:31:53Z

No: take a radial function in $H^s$ and not in $H^{s+\epsilon}$ for any $\epsilon >0$.

Answer by Bazin for Polynomial growth of Fourier transforms

2013-04-13T19:31:55Z

Let me quote the Paley-Wiener-Schwartz theorem. Let $F$ be a tempered distribution on $\mathbb R^n$. Then the two following properties are equivalent.

(i) $F$ is compactly supported with $\text{supp} F\subset\{x\in \mathbb R^n,\vert x\vert\le R\}$

(ii) $\hat F$ is an entire function such that there exists $C\ge 0, N\ge 0$ with $$\forall \zeta \in \mathbb C^n,\quad \vert\hat F(\zeta)\vert\le C(1+\vert\zeta\vert)^N e^{R\vert\Im \zeta\vert} $$

As a consequence, with your notations $f$ will have a polynomial growth on the real line when it is compactly supported, whatever is its regularity.

Answer by Bazin for Oscillatory Integral

2013-04-13T15:17:47Z

Let me give a rather general answer, hopefully helping you to cope with your case. We consider $$ I(\lambda)=\int_{\mathbb R^d} e^{i\lambda\phi(x)}a(x,\lambda) dx, $$ where $\lambda \ge 1$ is a real parameter (going to $+\infty$), $\phi$ is the phase, a smooth complex-valued function such that $\Im \phi\ge 0,$ $$ \quad d\Re \phi=0 \text{ and } \Im\phi=0 \Longrightarrow \det(\phi'')\not=0. \tag{$\natural$} $$

The amplitude $a$ is in the Schwartz class (wrt the $x$ variable). Then $I(\lambda)=O(\lambda^{-d/2})$ when $\lambda\rightarrow+\infty$. If the phase depends also on $\lambda$, say if you replace $\lambda \phi(x)$ by $\lambda\Phi(x,\lambda)$, then you keep the assumption on $a$ but you also assume that $\Phi$ is such that there exists positive constants $c_0,C_0$ such that

$$ \quad d\Re \Phi=0 \text{ and } \Im\Phi=0 \Longrightarrow c_0\le \vert\det(\Phi'')\vert \le C_0. \tag{$\sharp$} $$ This is a short description of the method of complex stationary phase.

Note that the integral $I$ is largest at points where the phase is real-valued and stationary (i.e. with a vanishing gradient). This is quite natural after all since as long as the phase is positive you get an exponential decay, and when it is real-valued the differential of the imaginary part is also vanishing since that imaginary part is assumed to be non-negative. Then you must take a look at the Hessian of $\phi$, as in the method of real stationary phase for real-valued Morse function.

Answer by Bazin for Global Implicit Function Theorem

2013-04-10T16:24:10Z

Let me quote the simplest and most classical result for a global inverse function theorem, due to Hadamard and Plastock (see L. Nirenberg, Topics in Nonlinear functional analysis, Courant LN,6, 2001).

Theorem. Let $F:\mathbb R^n\rightarrow \mathbb R^n$ be a $C^1$ mapping such that $\forall x\in \mathbb R^n, \det F'(x)\not=0$. Then $F$ is a global $C^1$ diffeomorphism if $$ \int_0^{+\infty}\inf_{\vert x\vert=r}\Vert(F'(x))^{-1}\Vert^{-1} dr=+\infty. $$

Answer by Bazin for Smoothness of $f(\sqrt x)$

2013-03-25T21:48:06Z

There are general results on $C^\infty$ functions of several variables which are invariant under the action of a group, many of them due to Georges Glaeser. One of most beautiful is the following: take a $C^\infty$ function $f(x_1,\dots,x_n)$ which is symmetric, i.e. such that for all permutations $\sigma$ of $\{1,\dots,n\}$, $$ f(x_{\sigma(1)},\dots,x_{\sigma(n)})=f(x_1,\dots,x_n). $$ Then it is a $C^\infty$ function $F$ of the standard symmetric functions $$ \sum x_j,\dots,\prod x_j. $$ This result is elementary for polynomials ($f$ polynomial, then $F$ polynomial) but the above result is highly non-trivial for smooth non-analytic functions.

Answer by Bazin for The Periodic Schrödinger Group

2013-03-22T21:29:35Z

$e^{it\Delta}$ is the Fourier multiplier $e^{-4it\pi^2\vert D\vert^2}$, i.e. the operator defined by $$ (e^{it\Delta} u)(x)=\int_{\mathbb R ^d} e^{2i\pi x \xi}e^{-4it\pi^2\vert \xi\vert^2}\hat u(\xi) d\xi. $$ It is also the convolution with $E(t)$, say for $t>0$, $$ E(t)(x)=e^{-i(d-2)\pi/4}(4\pi t)^{-d/2}e^{i\vert x\vert^2 /(4t)},\quad (e^{it\Delta} u)(x)=(E(t)\ast u)(x). $$ The Fourier multiplier definition is the simplest.

Answer by Bazin for Embedding of weighted Sobolev spaces

2013-03-13T13:39:45Z

Continuity is a local property: functions which are in your $L^{2,s}$ are locally in $L^2$, so functions in $H^{2,s}$ are locally in the Sobolev space $H^2(\mathbb R^3)$, thus are continuous functions (even Hölder 1/2-$\epsilon$).

Answer by Bazin for problem related to airy function

2013-03-12T17:21:23Z

I recommend to look at pages 213-4-5 in the first volume of Hörmander's ALPDO, Springer Grundlehren, 256. In my opinion this is the shortest and most elementary introduction to Airy functions.

Answer by Bazin for Eigenfunctions of elliptic operator form an orthonormal basis for L_2?

2013-03-09T16:28:15Z

You should assume that the manifold is compact ... not the case for $(0,1)$ but indeed for the Torus $\mathbb R/\mathbb Z$.

Answer by Bazin for Every function in W^{1,1}(0,1) is continuous on (0,1)

2013-03-09T16:20:53Z

Since $u'\in L^1(0,1)$, you find from the Lebesgue differentiation theorem that $$ \int_{1/2}^x u'(t) dt=u(x)+Cst,\quad x\in(0,1). $$ As a result $u$ is a continuous function and the constant above is $-u(1/2).$

Answer by Bazin for Solving Stokes Equations using 3D Fourier transforms

2013-02-26T21:39:08Z

Let me change your notations slightly: you work in three dimensions and you want to compute $$ u_{jk}(x)=\int e^{2i\pi x\cdot \xi} \frac{\xi_j\xi_k}{\vert \xi\vert^4} d\xi, $$ where the integral should not be taken as an $\dots$ integral. You have by homogeneity (in $n$ dimensions the Fourier transform of an homogeneous distribution with degree $\lambda$ is homogeneous with degree $-\lambda-n$) and "radiality"(the Fourier transform of a radial function is also radial) $$ u_{jk}=D_{x_j}D_{x_k}\int e^{2i\pi x\cdot \xi} \frac{1}{\vert \xi\vert^4} d\xi=cD_{x_j}D_{x_k} \vert x\vert , $$ where $c$ is a constant and $D_t=\partial_{t}/2i\pi$. We get $$ j\not=k,\quad u_{jk}=c_1\partial_{x_j}\vert x\vert^{-1}x_k=-c_1 x_jx_k\vert x\vert^{-3}, $$ $$ u_{jj}=c_1\partial_{x_j}\vert x\vert^{-1}x_j=c_1 \bigl(\vert x\vert^{-1} -x_j^2\vert x\vert^{-3}\bigr). $$

Answer by Bazin for About the boundedness of a multiplication operator.

2013-02-25T14:10:15Z

When $p=2$, boundedness is a triviality and it is the only trivial case. It is not true for $p=1$ nor for $p=\infty$, although the Fourier multiplier $sign(D_x)$ sends $L^1$ into $L^1_w$ and the Marcinkiewicz interpolation theorem implies boundedness in $L^p$ for all $p\in]1,+\infty[$.

The operator $sign(D_x)$ is is a particular case of the wider class of singular integrals, extensively studied by Calderon and Zygmund, later by Hörmander, Stein & Fefferman. They are defined via a simple condition on their kernels, easily proven $L^2$ bounded, with the property that they send $L^1$ into $L^1_w$. Again Marcinkiewicz Theorem allows to finish the job of proving boundedness in $L^p$ for all $p\in]1,+\infty[$.

To give a simple class of example would be to consider Fourier multiplier $F(D_x)$ where $F$ is an homogeneous function of degree 0 which is smooth outside of the origin. Note that it works as well in any dimension and that the so-called Hörmander-Mihlin multiplier Theorem allows to weaken significantly the smoothness assumption.

Answer by Bazin for Nth root of a matrix as an analytic function?

2013-02-24T20:37:50Z

Let $A$ be a $k\times k$ invertible matrix, i.e. in $Gl(k)$. Assume that the segment $[I,A]$ lies in $Gl(k)$. Let us define $$ \text{Log}A=\int_{[1,A]} \frac{d\xi}{\xi}=\int_0^1(I-tI+tA)^{-1}(A-I)dt. $$ It makes sense since $A$ commutes with the denominator inside the integral. The assumption is satisfied in particular whenever $A$ is symmetric invertible with a nonnegative real part. Analytic continuation arguments entail $$ \exp(\text{Log}A)=A\quad \bigl(\exp(\frac{1}{n}\text{Log}A)\bigr)^n=A. $$ Looking at the Jordan canonical form of $A$, it is not difficult to see that the only thing to be avoided for the above method to work is that eigenvalues should not be negative real numbers. Let $z=a+ib$ be an eigenvalue not in $\mathbb R_-$ in a Jordan block $J_N$ of size $N$, with 1 above the diagonal. Considering the segment $[I_N,J_N]$, we find on the diagonal $$ (1-t)+tz\notin \mathbb R_-\text{ since $z\notin \mathbb R_-$}, $$ and above the diagonal $ (1-t)0+t=t. $ The logarithm formula above works.

Answer by Bazin for Class of functions that the Fourier inversion holds

2013-02-24T17:52:13Z

The optimal space for the Fourier transform is the space of tempered distributions $\mathscr S'(\mathbb R^n)$, i.e. the dual space of the Schwartz functions $\mathscr S(\mathbb R^n)$. The latter is the Fréchet space of $C^\infty$ functions on $\mathbb R^n$ decreasing faster than any polynomial as well as all their derivatives; the semi-norms are $$ p_{\alpha\beta}(\phi)=\sup_{\mathbb R^n}\vert x^\beta\partial^\alpha\phi\vert. $$ Defining $\hat \phi(\xi)=\int e^{-2i\pi x\cdot\xi} \phi(x) dx$ for $\phi\in\mathscr S(\mathbb R^n)$, it is easy to prove that the Fourier transform is an isomorphism of $\mathscr S(\mathbb R^n)$ and $$\phi(x)=\int e^{2i\pi x\cdot\xi} \hat\phi(\xi) d\xi.\tag{1}$$

The dual space of $\mathscr S(\mathbb R^n)$ is huge, contains for instance all the spaces $L^p(\mathbb R^n)$ and all the distribution derivatives of any function in these spaces. The space $\mathscr S'(\mathbb R^n)$ contains also the homogeneous distributions and the distributions with compact support. It is easy to define by duality the Fourier transform on $\mathscr S'(\mathbb R^n)$. With brackets of duality, we define $$ \langle\hat T,\phi\rangle=\langle T,\hat \phi\rangle\quad\text{and we have $\hat{\hat T}= \tilde T$,} $$ where $\tilde T(x)=T(-x)$, that is the very same formula as in (1).

Answer by Bazin for Pullback measures

2013-02-23T14:04:45Z

A simple-minded answer. The push forward of a measure is a triviality: take a measure space $(X,\mathcal M, \mu)$ and a mapping $f:X\rightarrow Y$. Then defining $\mathcal N=${$B\subset Y, f^{-1}(B)\in \mathcal M$}, you find easily that $\mathcal N$ is a $\sigma$-algebra on $Y$ and defining $$ \nu=f_*(\mu) \text{ measure on $(Y,\mathcal N)$},\quad \nu(B)=\mu(f^{-1}(B)), $$ you get that $(Y,\mathcal N,\nu)$ is a measure space (and $\mathcal N$ is the largest $\sigma$-algebra on $Y$ making $f$ measurable).

Now, the pullback: take a measure space $(X,\mathcal M, \mu)$ and a mapping $f:Z\rightarrow X$. You would like to find a measure space $(Z,\mathcal T, \omega)$ such that $f_*(\omega)=\mu$. It is possible when $f$ is bijective: in that case just take

$ \omega={(f^{-1})}_* $ $(\mu)= f^*(\mu).$ The last equality is a definition.

When $f$ is not bijective this program is unrealistic.

Answer by Bazin for First Order PDE Solution Method Issues

2013-02-21T17:30:55Z

Let me work with $n$ dimensions: you want to study the vector field $$ X=\sum_{1\le j\le n} a_j(x)\frac{\partial}{\partial x_j}, \tag {1}$$ and in particular find the so-called first integrals of $X$ i.e. the functions $f$ such that $Xf=0$. You introduce the system of ODE: $$ \dot x(t,y)=a(x(t,y)),\quad x(0,y)=y. \tag {2}$$ The solutions $t\mapsto x(t,y)$ are the integral curves of $X$. You realize easily that a function is a first integral iff it is constant along the integral curves of $X$: just compute $$ \frac{d}{dt}\bigl(f(x(t,y))\bigr)=\sum_{1\le j\le n} \frac{\partial f}{\partial x_j}(x(t,y))a_j(x(t,y))=(Xf)(x(t,y)) $$ It means that solving the PDE (1) is somehow equivalent to solving (2).

Now the notational business. It is tempting to write (2), which is $ \frac{dx_j}{dt}=a_j(x), 1\le j\le n, $ symbolically as $$ \frac{dx_1} {a_1(x)}=\dots=\frac{dx_n} {a_n(x)} $$ since they are all equal to $dt$ ! Well just take this as a symbolic notation which eliminates the presence of the parameter $t$.

Now the Cauchy problem for this autonomous vector field $X$: find an hypersurface $\Sigma$ to which $X$ is transverse, i.e. $X$ is not tangent to $\Sigma$. Then the Cauchy problem $$ \begin{cases} Xu=f,\quad \ u_{\vert \Sigma}=g \end{cases} $$ has locally a unique solution: this problem is equivalent to the scalar ODE $$ \frac{d}{dt}\bigl( u(x(t,y))\bigr)=f(x(t,y)),\quad u(x(0,y))=u(y)=g(y) \text{ for $y\in \Sigma$}, $$ so that $$ u(x(t,y))= u(y)+\int_0^tf(x(s,y)) ds\quad \text{ for $y\in \Sigma$}. \tag{3}$$ Note that $y$ moves on $\Sigma$ ($(n-1)$ degree of freedom) and $t$ in $\mathbb R$ so that it is a nice choice of coordinates to pick $y\in \Sigma$ and $t\in \mathbb R$.

There are variants of this when the vector field is not autonomous, i.e. is of type $$\frac{\partial}{\partial t}+ \sum_{1\le j\le n} a_j(t,x)\frac{\partial}{\partial x_j}. $$

More comments on the quasi-linear case and the general method of characteristics: the quasi-linear Cauchy problem $$ \frac{\partial u}{\partial t}+\sum_{1\le j\le n} a_j(t,x, u)\frac{\partial u}{\partial x_j}=b(t,x,u),\quad u(0,x)=u_0(x). \tag{4}$$ has a linear companion $$ \frac{\partial F}{\partial t}+\sum_{1\le j\le n} a_j(t,x, v)\frac{\partial F}{\partial x_j}+b(t,x,v)\frac{\partial F}{\partial v}=0,\quad F(0,x,v)=v-u_0(x) \tag{5}$$ where $t,x,v$ are independent variables. It is not difficult to solve using the linear method of characteristics outlined above. Then since $\partial F/\partial v=1$ at $t=0$, the equation $ F(t,x,v)=0 $ determines implicitely $v=u(t,x)$ and the expression of derivatives of $u$ in terms of derivatives of $F$, e.g. $ \partial u/\partial x=-\frac{\partial F/\partial x}{\partial F/\partial v} $ imply that $u$ solves the Cauchy problem (4). Here also the notational industry is working full throttle. People would write $$ \dot x=a(t,x,u)\quad \dot u=b(t,x,u)\quad \text{which is } \frac{dx_j}{a_j}=\frac{du}{b},\quad 1\le j\le n. $$

Answer by Bazin for An interpolation inequality.

2013-02-18T21:02:03Z

$$ g(\epsilon)=\sum_{k\ge 1} k^se^{\frac {k-1}2\ln(1-\epsilon)} \lesssim \int_0^{+\infty} x^s e^{-a\epsilon x} dx= \int_0^{+\infty}x^s e^{-ax}dx\epsilon^{-s-1} $$ where $a$ is a fixed constant. So $$ C=\int_0^{+\infty}x^s e^{-ax}dx,\quad \phi(s)=-s-1. $$

Answer by Bazin for Analogue of the integral Fourier operator with angle in some cone

2013-02-16T18:05:36Z

Why don't you assume that the amplitude $A(x,y,\theta)$ is actually supported for $\theta$ in your cone $\Gamma$ along with the standard symbolic properties? You will always run into trouble with a definition like yours, which amounts to deal with singular amplitude. In particular any integration by parts will produce boundary terms that you will not be able to control.

You should take a look at Hörmander's definition of an oscillatory integral where the amplitude is indeed a symbol and the phase $\phi$ is a first-order symbol, real-valued for the classical FIO, with a nonnegative imaginary part for FIO with complex phase.

Answer by Bazin for Are there analogous theorems and/or techniques for solving fractional differential equations involving the Riesz Derivative?

2013-02-08T21:15:50Z

The fractional Laplacean $(-\Delta)^{\alpha/2}$ in $\mathbb R^n$ is, up to some constant the Fourier multiplier $\vert\xi\vert^\alpha$. So its inverse is, at least formally, the Fourier multiplier $\vert\xi\vert^{-\alpha}$.

Let us assume that $0<\alpha < n$. Then $\vert\xi\vert^{-\alpha}$ is $L^1_{loc}$ and a tempered distribution homogeneous with degree $-\alpha$. Its Fourier transform is a tempered distribution homogeneous with degree $\alpha-n$ and in fact is also radial so is, up to a constant $\vert x\vert^{\alpha-n}$: a parametrix for $(-\Delta)^{\alpha/2}$ is the convolution by const. $\vert x\vert^{\alpha-n}$. Note that this is the case for $\alpha =2$, when $n\ge 3$. The constants are not uninteresting to compute.

When $\alpha =n$, $\vert \xi\vert^{-n}$ is not $L^1_{loc}$ and is not a tempered Fourier multiplier. The homogeneity is partly lost and you have to perform a direct calculation: for instance for $n=2$ you find $$ \frac{-1}{2\pi}\ln\vert x\vert $$ which is not homogeneous of degree 0 but whose derivatives are homogeneous of degree -1.

Answer by Bazin for Does Physics need non-analytic smooth functions?

2012-11-26T21:55:36Z

A strong argument is given above on the heat equation; let me be more specific. The heat equation, one of the most basic in PDE and mathematical physics, already known to Fourier, is $$ L=\frac{\partial }{\partial t}-\Delta_x,\quad t\in\mathbb R,\quad x\in\mathbb R^n, $$ has the fundamental solution $$E= H(t)(4\pi t)^{-n/2}e^{-\frac{\vert x\vert^2}{4t}}, $$ i.e. $LE=\delta(x)\otimes\delta(t)$ (here the Heaviside function $H$ is the indicatrix of $\mathbb R_+$). It is easy to see that the $C^\infty$ singular support of $E$ is reduced to $0_{\mathbb R^{1+n}}$ whereas the analytic singular support is the hyperplane $t=0$. Since the function $E$ is $C^\infty$ except at $x=0,t=0$, one can see that it is indeed a flat function at $t=0,x=x_0\not=0$, i.e. all derivatives vanish at such a point. It is thus impossible to understand one of the simplest PDE using only analytic functions.

A more refined -yet classical- fact is related to the notion of well-posedness as defined by Jacques Hadamard. Loosely speaking, a PDE problem is well-posed whenever the solution can be controlled by the data or the sources via suitable inequalities. A typical example of a well-posed problem: the Cauchy problem with respect to a spacelike hypersurface (e.g. $t=0$) for the wave equation. A typical example of an ill-posed problem: the Cauchy problem for the Laplace equation. Although the latter has uniqueness properties, the analytic solutions given for instance by the Cauchy-Kovalewski Theorem are extremely unstable: you have $$ \partial_x^2 u+\partial_y^2 u=0,\quad u=e^{\lambda(x+iy)}, u(0,y)=e^{i\lambda y}. $$ The Cauchy data at $x=0$ are bounded by 1, whatever is $\lambda >0$, whereas the solution increases exponentially with $x>0$: no control of $u$ by its Cauchy datum could be expected. However the solutions are analytic and uniquely determined by the Cauchy datum. The analytic method given by the CK theorem provides analytic solutions which are unstable. The CK theorem fails to deliver stable solutions in that case. No understanding of stability phenomena (a very interesting physical property) for PDE is possible within the class of analytic functions and one should use much larger classes of functional spaces in which inequalities of well-posedness could be proven.

I could have mentioned another effect, for instance for the Cauchy problem for the Laplace equation: take an analytic Cauchy datum $\phi_0$, then CK provides an analytic solution. Now, perturb $\phi_0$ by a smooth non-analytic function $w$ and take as a datum say $\phi_0+\epsilon w$. Then there is no solution to the Cauchy problem since the very existence of a (say continuous) solution is forcing the data to be analytic. It is not difficult to prove that by Fourier transformation: the analyticity will be forced by the fact that you have to compensate the exponential increase by some exponential decay of the data, triggering analyticity for this data.

Answer by Bazin for analysis of the regularity using Hormander condition

2013-02-03T21:18:46Z

Your last equation $$ \mathcal K=v_t-xv_z-v_{xx}=0 \tag 1$$ is indeed a particular case of Hörmander's $X_0-\sum_{1\le j\le r}X_j^2$ with $$ X_0=\partial_t-x\partial_z, X_1=\partial_x, r=1, [X_0,X_1]=\partial_z. $$ However, it is also exactly Kolmogorov equation, as studied by Andrei Kolmogorov in his 1937 Annals paper. This article was in fact the starting point of Lars Hörmander's work on this topic. It turns out that there is an explicit parametric construction for (1): a change of variables straightening the vector field $X_0$ is $$ \begin{cases} s=t,\ \ x_1=x,\ \ x_2=z+xt, \end{cases} $$ so that $ \mathcal K=\partial_s-(\partial_{x_1}+s\partial_{x_2})^2 $ and the latter can be Fourier transformed to the ODE $$ \partial_s+(\xi_1+s\xi_2)^2. $$ The latter is of course explicitly solvable: we have an explicit integral expression $$ v(t,x,z)=v(s,x_1,x_2-sx_1)=w(s,x_1,x_2)=\iint e^{i(x_1\xi_1+x_2\xi_2)}e^{-\int_0^t(\xi_1+s\xi_2)^2ds} \hat w_0(\xi_1,\xi_2) d\xi_1d\xi_2, $$ $$ v(t,x,z)=\iint e^{i(x\xi_1+(z+xt)\xi_2)}e^{-\int_0^t(\xi_1+s\xi_2)^2ds} \hat v_0(\xi_1,\xi_2) d\xi_1d\xi_2,\quad v_0(x,z)=v(0,x,z). $$

Answer by Bazin for fourier analytic proofs

2013-02-01T16:53:51Z

Let me speak about the "Triumph of Fourier" according to the words of Laurent Schwartz in his autobiography. The Fourier transformation is a handy tool to characterize regularity of functions.

Let $u$ be a distribution on some open subset $\Omega$ of $\mathbb R^n$. A consequence of the Paley-Wiener theorem is that a point $x_0$ in $\Omega$ is not in the singular support of $u$ whenever there exists a neighbordhood $U$ of $x_0$ such that $$ \forall \chi\in C_c^\infty(U), \forall N\in \mathbb N,\quad\vert\widehat{\chi u}(\xi)\vert\vert \xi\vert^N\in L^{\infty}(\mathbb R^n). $$

That notion can be refined to define the ($C^\infty$) wave-front-set, as a subset of the cotangent bundle (minus the 0 section): a point $(x_0,\xi_0)\in \Omega \times\mathbb S^{n-1}$ does not belong to the wave-front-set of $u$ whenever there exists a neighbordhood $U$ of $x_0$, a neighborhood $V$ of $\xi_0$ on the sphere such that $$ \forall \chi\in C_c^\infty(U), \forall N\in \mathbb N,\quad\vert\widehat{\chi u}(t\xi)\vert t^N\in L^{\infty}((1,+\infty)_t\times V). $$

The wave-front-set (WF) can be used to detect the various directions of singularities: for instance $$ WF(\delta_{0})=\text{{0}}\times (\mathbb R^n\backslash\text{{0}}) $$ but with $H=\mathbf 1_{(0,+\infty)}$ the Heaviside function, $H(x_1)$ in $\mathbb R^n$ is also singular at 0 but the structure of the singularity is quite different and indeed with $\Sigma=${$x\in \mathbb R^n, x_1=0$} $$ WF(H(x_1))=\Sigma\times\text{{$0\not=\xi\in \mathbb R^n, \xi_2=\dots=\xi_n=0$}} $$ which is the conormal bundle to $\Sigma$. The first projection of the wave-front-set is the singular support.

That definition can be extended to Sobolev regularity (spaces based on $L^2$), analytic regularity (more generally Gevrey) and is the only way to express the propagation of singularities for linear waves.

Comment by Bazin

2013-05-21T08:35:23Z

Yes $\Omega$ is the operator defined in your comment.

Comment by Bazin

2013-05-20T21:28:35Z

You are right, I have changed my answer and added some explanations on Calderon-Zygmund operators.

Comment by Bazin

2013-05-20T09:58:53Z

With the bi-Lipschitz continuity hypothesis, for every $\delta>0$, you get $\epsilon,\sigma >0$ such that $$ \\{\vert G(v)-G(w)\vert\le \sigma\\}\subset\\{\vert v-w\vert\le \delta\\}\subset\\{\vert G(v)-G(w)\vert\le \epsilon\\} $$ implying that last step.

Comment by Bazin

2013-04-18T18:30:35Z

Bazin (mathoverflow.net/users/21907), Motivation for and history of pseudo-differential operators, mathoverflow.net/questions/97604 (version: 2012-05-23)

Comment by Bazin

2013-03-13T15:34:31Z

Yes of course, because of the $(1+\vert x\vert^2)^s$, locally in $x$ comparable to 1.

Comment by Bazin

2013-03-10T16:16:46Z

I meant compact manifold without boundary.

Comment by Bazin

2013-03-09T16:25:23Z

@Kamil I agree with you on this shrinking business. However, my claim is that the method suggested by my answer is providing an explicit integral solution for your problem. With that explicit solution, you should be able to decide the smoothness properties.

Comment by Bazin

2013-02-22T12:52:22Z

@ sponsoredwalk I added a couple of comments on the quasilinear case.

Comment by Bazin

2013-02-19T08:25:31Z

The only problem is when $\epsilon$ is small, and you do have a singularity at $\epsilon=0$. Now for $x>1,0<\epsilon<1/4$, $$ \frac{x-1}{2}\ln(1-\epsilon)\le\frac{x(-\epsilon/2)}{2} $$ so you can take $a=1/4$. The bounded values of $x$ are unimportant.

Comment by Bazin

2013-02-16T18:14:56Z

I have made an explicit computation and you can find as well an explicit solution with my formula above by plugging the values of $v$ in terms of your $u$. The regularity business, say for the function $v$ follows from the explicit integral expression: you get easily that the $L^2$ norm of $\mathcal K v$ controls the $H^1$ norm in the $x$ variable of $v$. The expression of $w$ shows that you control 2/3 of derivatives for the $z$ variable: if you want an isotropic control then you cannot do better than $2/3$. To see that is not completely obvious: just compute exactly the integral in the phase

Comment by Bazin

2013-02-07T20:20:43Z

@timur It is certainly possible to do what you propose, but you have to keep in mind Hadamard's celebrated sentence about polynomials, in which he said essentially the following: "I do not care so much about approximating the data by polynomials, what matters is how this approximation is transferred to the solution." I will add other comments in a new edit of my answer above.

Comment by Bazin

2013-02-07T20:14:19Z

It seems to me that the formulas above give an explicit integral solution to your problem. From this explicit expression, it is for instance easy to prove hypoellipticity and regularization properties, even to find the exact fractional amount of $z$ derivative that you gain.

Comment by Bazin

2013-02-01T16:27:06Z

@Matt Jacobs I would recommend the Bahouri-Chemin-Danchin book Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences], 343. Springer, Heidelberg, 2011.

Comment by Bazin

2013-01-31T09:58:37Z

@Matt Jacobs As said in the previous comment, $f(D)u$ is the function whose Fourier transform is $f(\xi)\hat u(\xi)$. The operator $f(D)$ is called a Fourier multiplier for this reason. An integral representation is $$ (f(D)u)(x)=\int e^{2i\pi x\cdot \xi} f(\xi) \hat u(\xi) d\xi, $$ with $$ (\hat u)(\xi)=\int e^{-2i\pi x\cdot \xi} u(x) dx. $$

Comment by Bazin

2013-01-11T20:21:56Z

Yes of course, but they are all of the form $a_d\vert x\vert^{2-d}$+ harmonic function on $\mathbb R^d$.