Stack Exchange Network

Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange
32

There are a few references for your interests: Rodino, Wong, Zhu (eds.) - Pseudo-Differential Operators: Analysis, Applications and Computations, Rodino, Schulze, Wong - Pseudo-Differential Operators: Partial Differential Equations and Time-Frequency Analysis, Rodino, Wong (eds.) - New Developments in Pseudo-differential Operators, Boggiatto, Rodino, Toft, ...


24

This is more of an extended comment than an answer, but with regards to typesetting in $\LaTeX$, let me point out that typing \mathrm{d} should not take longer than typing d, as you shouldn't be doing either throughout your paper! The semantically correct thing to do is to define a macro representing your desired differential operator, for example, \...


15

The short answer is yes, absolutely, and the theory of such operators is part of microlocal analysis. The basic ingredient is that differential operators can be written as integral operators (in an appropriate generalized sense) via the Fourier transform. E.g. $$ \frac{d}{dx} f(x) = \frac{d}{dx} \int e^{2\pi i kx} \hat{f}(k) dk = \int e^{2 \pi i k x} (2\pi ...


12

Many excellent journals and books use $d$ in the italics form, such as the Journal of the American Mathematical Society (e.g., recent article by Terence Tao), London Mathematical Society Proceedings (e.g., equations 74 and 75 of this recent paper) and Spivak's Calculus. Given that reference quality publications use $d$--and that it is faster and cleaner to ...


11

The axiom of choice is equivalent to the assertion that any vector space (no matter how large) has a basis. Given a choice of basis of an arbitrary vector space $V$, one can represent a linear transformation $T : V \to V$ using an infinite row-finite matrix (this means that only finitely many entries in each row are nonzero). However, in practice this is ...


9

Let $E\rightarrow M$ and $F\rightarrow M$ be vector bundles with spaces $\Gamma(E)$ and $\Gamma(F)$ of smooth sections. Consider a linear partial differential operator of order $k$, which is a map \begin{align} L=\sum_{|\alpha|\leq k}\ell_\alpha\partial^\alpha:\Gamma(E)&\rightarrow\Gamma(F)\\ S&\mapsto L(S)=\sum_{|\alpha|\leq k}\ell_\alpha\partial^\...


9

Quick answer: there is a standard to follow. Longer answer: while physicists write differential operators in upright fonts (because they follow the standards), mathematicians tend to typeset differential operators as variables (because we are lazy). I am joking, but it should be clear that $dx$ is not $d \cdot x$, and that $d$ is essentially an operator: ...


9

There is another way to show something is a linear partial differential operator. One can characterize partial differential operators in a coordinate-free manner using commutators: Let $E\to M$ and $F\to M$ be two vector bundles over $M$. Consider the set, $\mathcal{L}(\Gamma(E),\Gamma(F))$, of $\mathbf{R}$-linear operators between sections of $E$ and ...


8

It's helpful to translate this question into the notation of differential forms. If the vector field $\mathbf F$ corresponds to the $1$-form $F_1\,dx + F_2\,dy + F_3\,dz$, then $\text{curl}\,\mathbf F$ corresponds to the $2$-form $d\omega$ and the dot product $\mathbf F\cdot \text{curl}\,\mathbf F$ is the coefficient of the $3$-form $\omega\wedge d\omega$. ...


7

The operator $e^{t(x+D)}$ is an exponential generating function for the sequence $(x+D)^n$. The commutator of the two operators $x$ and $D$ is given by $[x,D] = xD - Dx = -1$ and $[[x,D],x] = [[x,D],D] = 0$ so by Baker-Campbell-Hausdorff we have $$e^{(x+D)t}=e^{xt}e^{tD}e^{\frac{t^2}{2}}$$ or $$\sum_{n=0}^\infty \frac{(x+D)^nt^n}{n!} = \sum_{i=0}^\infty \...


6

Certainly if $\alpha$ is a $(0,1)$-form which is $\overline{\partial}$-exact, it must be $\overline{\partial}$-closed as well, since $\overline{\partial}^2 = 0$. Thus if $\alpha$ is not $\overline{\partial}$-closed, it cannot be $\overline{\partial}$-exact. Assume then that $\alpha$ is $\overline{\partial}$-closed. In this case, my understanding is that ...


6

By the Mean Value Theorem, for every positive integer $n$, $$\frac{f(n+1)-f(n)}{1}=f'(c_n),$$ for some $c_n$ between $n$ and $n+1$. Since $f(n+1)-f(n)\to 0$, $f'(c_n)$ must have small absolute value when $n$ is large. So if the limit of $f'(x)$ exists, it must be $0$.


6

Hint: Denote by $$D:(C^{1}[0,1],\Vert\cdot\Vert_{\infty})\rightarrow(C[0,1],\Vert\cdot\Vert_{\infty})$$ the differential operator. Let $f_{n}(x)=\sin(nx)$. Note that $\Vert f_{n}\Vert_{\infty}=1$ (for $n\geq 2$). However, $(Df_{n})(x)=n\cos(nx)$, and hence $\Vert Df_{n}\Vert_\infty=n$. What does this mean?


6

The two question are very much related; I hope this goes some way to help you. I have been suitably vague at points to hopefully get the ideas across but happy to try and answer any further concerns (to the best of my ability!). 1) Given a vector field $X$, the question arises: does this give rise to a family of curves? That is, can we find a family of ...


5

The symbol of a nonlinear differential operator is defined as the symbol of its linearization.


5

The space $L(C^\infty[a,b])$ of continuous linear operators on $C^\infty[a,b]$ is - excepting the trivial case $a = b$ - not a Fréchet space (in the usual topologies). The usual topologies on $L(C^\infty[a,b])$ are the topology of uniform convergence on all bounded subsets, often (well, sometimes, at least ;) denoted $L_b(C^\infty[a,b])$, the topology of ...


5

Apply the operator to $f(x)$ $\hat Af=\frac {df}{dx}+xf$ $\hat A^2f=\hat A\left(\frac {df}{dx}+xf\right)=\frac {d^2f}{dx^2}+\left(f+x\frac {df}{dx}\right)+x\frac {df}{dx}+x^2f$


5

Assuming that $f(x)$ is twice differentiable, we have $$ \frac{d^2}{dx^2}[f(x)]=\frac{d}{dx}\left[\frac{d}{dx}[f(x)]\right]$$ $$=\frac{d}{dx}\left[\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}\right]$$ $$=\lim\limits_{h\to 0} \frac{\frac{d}{dx}[f(x+h)-f(x)]}{h}$$ $$ =\lim\limits_{h\to 0} \frac{\lim\limits_{k\to 0}\frac{f(x+h+k)-f(x+k)-f(x+h)+f(x)}{k}}{h} $$ $$ =...


5

The aim of this answer is to make the question self-contained, and propose a generalization of the answer by Sangchul Lee. The basic formula to be employed is (anticipating with $y$ instead of $x$ and $g$ instead of $f$): $$ e^{\lambda\frac{d}{dy}}\,g(y) = g(y+\lambda) $$ Which is easily proved with Taylor series expansions for differential operators and ...


4

Roughly speaking, it is the anti-symmetric part of the full derivative due to the "odd permutation = sign change" trait of the differential forms, for a 1-form in $\mathbb{R}^3$ you gave if we write it as $\displaystyle\omega = \sum^{3}_{j=1} f_i dx_i$, it is : $$ \begin{aligned} d\omega &= \sum^{3}_{i=1} \sum^{3}_{j=1} \frac{\partial f_i}{\partial x_j}...


4

[Some of the links in the comments are dead.] Fundamental solutions of the wave equations are truly singular, and before distributions it was almost impossible to describe. If you don't care about the constant in front, it can be described by repeated (distributional) derivatives of (regular) functions. For n = odd, let m = (n-1)/2 and $$ E = \square^m H(...


4

Higher dimensional cases will follow from Poisson's and Kirchhoff's formulas.


4

What you are calling $E$ here, sounds like the Green's function of the wave operator....look at the bottom of the wiki page D'Alembert operator for the explicit form.


4

Let's do this in $n$ variables, and over $\mathbb C$. The polynomials $P$ such that $\ker \partial P_0 \subset \ker \partial P$ form an ideal $J$ in ${\mathbb C}[z_1,\ldots,z_n]$ which contains $P_0$. The question is whether this ideal is generated by $P_0$. For $f(z_1,\ldots,z_n) = \exp(\sum_j a_j z_j)$, where $a_j \in \mathbb C$, we have $(\partial P_0) ...


4

If I understand the question correctly, the answer is no, because it is not even true for the square of an operator: solutions to $\frac{d^2}{dx^2}f(x)=0$ are not just sums of two of the constant solutions of $\frac{d}{dx}f(x)=0$.


4

It seems the answer is yes. We can write $F(x,y)=\sum_{d=0}^Nf_d(x,y)$ where $f_d$ is a homogeneous polynomial of degree $d$. Since for all $n$, $P_n(x,y):=(x+iy)^n+(x-iy)^n$ and $Q_n(x,y):=(x+iy)^n-(x-iy)^n$ are harmonic (because holomorphic), we should have $\partial_FP_n=\partial_FQ_n=0$. If $d>n$, we have $\partial_{f_d}P_n=\partial_{f_d}Q_n=0$. We ...


4

Let $G = (V, E)$ be a locally finite graph; this means that each vertex has finite degree. The Laplacian operator $\Delta$ acting on the space of functions $f : V \to \mathbb{R}$ is given by $$(\Delta f)(v) = \sum_{v \to w} (f(w) - f(v))$$ where $v \to w$ indicates an edge from the vertex $v$ to the vertex $w$. When $G = \mathbb{Z}$ with edges between ...


4

I obtained those identities by using $\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$ etc. (without worrying about convergence of operators, I admit). In general, you can use the Fourier-Transform of your operator as follows: $$\begin{array}{rl} D\left(z,\frac d{dz}\right)f(z) &= D\left(z,\frac d{dz}\right)\frac1{2\pi}\int_{-\infty}^\infty dk \int_{-\infty}^\...


4

Adding a little to @DanielFischer's answer: the space $C^\infty[a,b]$, being a nested intersection, is a projective limit of Banach spaces. Thus, any (continuous linear) map to it from a TVS is exactly induced (uniquely) from a collection of (continuous linear) maps to the limitands $C^k[a,b]$. In particular, this is true of self-maps, so every such just ...


4

Ultimately, I think the reason is that $x$ is a differentiable function of $t$ which particularly means that as $\Delta t\rightarrow 0$, $\Delta x\rightarrow 0$. That said I really don't like this approach to the chain rule. It is a bit cobbled together. I feel that this is the better way to do chain rule (which is much clearer from the get go, I think). ...


Only top voted, non community-wiki answers of a minimum length are eligible