TITLE: Negation of uniform convergence
QUESTION [8 upvotes]: Suppose $f_{n}$ is a sequence of functions which does not convergence uniformly to $f$. Does this mean that there exists an $\varepsilon_{0} > 0$, an $x_{0}$, and a sequence of integers $n_{k} \rightarrow \infty$ such that $|f_{n_{k}}(x_{0}) - f(x_{0})| \geq \varepsilon_{0}$?

REPLY [25 votes]: If you know how to negate logical formulas with quantifiers, you can do this more or less mechanically.
Definition of uniform convergence can be written like this:
$$(\forall \varepsilon>0) (\exists n_0) (\forall x\in S) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon) $$
Negation of uniform convergence
$$(\exists \varepsilon>0) (\forall n_0) (\exists x\in S) (\exists n>n_0) (|f_n(x)-f(x)|\ge\varepsilon) $$
Pointwise convergence is defined as follows
$$(\forall \varepsilon>0) (\forall x\in S) (\exists n_0) (\forall n>n_0) (|f_n(x)-f(x)|<\varepsilon) $$
negation
$$(\exists \varepsilon>0) (\exists x\in S) (\forall n_0) (\exists n>n_0) (|f_n(x)-f(x)|\ge\varepsilon)$$
If you look closely at the negation of pointwise convergence, it is equivalent to condition from your question. Indeed, we have existence of $\varepsilon>0$ and existence of a point, which we many denote $x_0$, such that $|f_n(x_0)-f(x_0)|\ge\varepsilon$ happens for infinitely many $n$'s.
So any function which converges pointwise but not uniformly is a counterexample to the claim in your post.


TITLE: Numerically Solve PDE
QUESTION [0 upvotes]: I am looking for some help finding a numerical solution to a pde of the form:
$$C_t=f(x)C_x+\alpha C_{xx}$$
with initial condition for $C(x,t)$:
$$C(x,0)=\delta(x)$$
and boundary condition
$$C(\pm\infty,0)=0$$
I am using Mathematica's NDSolve function, but it doesn't seem to be able to find a solution for some chosen $f(x)$ and $\alpha$. I think this may be due to the condition's at infinity? Another idea I have is to turn it into a ODE via separation of variables. Any advice on resources or methods would be helpful.

REPLY [1 votes]: You can use a standard method, like the Crank-Nicolson method, on a truncated spatial domain. The conditions on this artificial boundary can be approximated by  setting the solution to be zero, or some function of $(x,t)$, if some adicional information on the asymptotics is available.


TITLE: Linearization of the Klein-Gordon equation and decoupling of ''spinors''!
QUESTION [1 upvotes]: We know that the K-G equation is deduced from the Einstein relation:
$E^{2}=m^{2} +\vec{p}^{2} \;\;\;\;$    (with $c=1$)
It is known that :$E^{2}=\frac{m^{2}}{1-\beta^{2}}=\left(\frac{m}{1-\beta}\right)
 \left(\frac{m}{1+\beta}\right) \;\;\;\;,\beta=\frac{v}{c} $
we assume that $$ E^{2}=E_{1}E_{2}=\left(\frac{m}{1-\beta}\right)\left(\frac{m}{1+\beta}\right)$$
and$$\vec{p}^{\;2}=\vec{p}_{1}.\vec{p}_{2}=\left(\frac{m\vec{v}}{1-\beta}\right). \left(\frac{m\vec{v}}{1+\beta}\right)$$
the Einstein equation is written $$E_{1}E_{2}-(\vec{\sigma}.\vec{p}_{1})\,(\vec{\sigma}.\vec{p}_{2})=m^{2}$$
we can deduce the previous equation from :
$$ \left(E_{1}-\vec{\sigma}. \vec{p}_{1}\right) \left(E_{2}+ \vec{\sigma}.\vec{p}_{2}\right)=m^{2}$$
because:  $\;\;\vec{\sigma}.\left( E_{1} p_{2}-E_{2} p_{1} \right)\vec{n}=0$ and $\vec{\sigma}^{\,2}=1$
by the correspondence equations $E=i\hbar\frac{\partial }{\partial t}\;,\;\vec{p}=-i\hbar\vec{\nabla}$, we have  ;
$$ (i\hbar)^{2}\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}\right)\left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}\right)\psi_{1}\psi_{2}=m^{2}\psi_{1}\psi_{2}\;\;\;\;\;\;\;(1)$$
$$\psi(\vec{r},t)=\psi_{1}(\vec{r}_{1},t_{1})\psi_{2}(\vec{r}_{2},t_{2})$$ and $$\partial t^{2}=\frac{\partial\tau ^{2}}{1-\beta^{2}}=\left(\frac{\partial\tau }{1-\beta}\right)\left(\frac{\partial\tau }{1+\beta}\right)=\partial t_{1}\partial t_{2}$$
and $$\vec{\nabla}_{1}.\vec{\nabla}_{2}=\vec{\nabla}^{\;2}$$
for a massless particle, we have $$\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}\right)\left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}\right)\psi_{1}\psi_{2}=0$$
i.e:$$\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}\right)\psi_{1}=0$$ or $$\left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}\right)\psi_{2}=0$$
the two equations are similar to the the Weyl equations but the two spinors are different.
equation (1) can be written:
$$\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}\right)\left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}\right)\psi_{1}\psi_{2}=\frac{m^{2}}{(i\hbar)^{2}}\psi_{1}\psi_{2}$$
we see that it is the product of two equations :
$$\begin{cases}\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}\right)\psi_{1}=\frac{m}{i\hbar}\psi_{1} \\ \left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}\right)\psi_{2}=\frac{m}{i\hbar}\psi_{2}\end{cases}$$
$$\begin{cases}\left(\partial_{t_{1}}+\vec{\sigma}.\vec{\nabla}_{1}+i\frac{m}{\hbar}\right)\psi_{1}=0 \\ \left(\partial_{t_{2}}-\vec{\sigma}.\vec{\nabla}_{2}+i\frac{m}{\hbar}\right)\psi_{2}=0\end{cases}$$
we have two decoupled ''spinors'' and their product is the wave function that verifies the K-G equation.
Is there something that I have missed?

REPLY [1 votes]: Dirac originally arrived at spin-1/2 field equations by taking "square root" of the wave operator:
$$\square=\not{\partial}\cdot\not{\partial}$$where $\not{\partial}=A\partial_x+B\partial_y+C\partial_z+iD\partial_t$ (see this Wikipedia article). After taking square and equating with the wave operator, we observe that the coefficients satisfy the same commutation relation as the Pauli matrices.
A Dirac field is a positive frequency solution for the equation $(i\not{\partial}-m)\psi=0$. From this expression it is not clear how a spin-1/2 wave function $\psi$ is related to a scalar field $\phi$ which satisfies free K.G. field equation. OP claims that it is possible to write a scalar field as product of two spinor like quantities, which requires some explanation. This is not possible in general, for (intuitive)reasons that I have mentioned in the comments. However, if $\psi$ satisfies the Dirac's equation, then it is also a solution for K.G. field equation:
$$(i\not{\partial}-m)\psi=0\\
(i\not{\partial}+m)(i\not{\partial}-m)\psi=0\\
(\square+m^2)\psi=0$$which is expected since the primary motivation for solving Dirac equation was to arrive at restrictive class of solution for free K.G. field which could allow for positive definite current density. There is a more general result that any free massless spin-n/2 field in flat space-time will also satisfy the massless K.G. equation (see section 6.8 of Spinors and Space-Time Volume-II). It uses the idea of 2-spinor formalism (see Spinors and Space-Time Volume-I; An Introduction to Twistor Theory - S.A.Huggett & K.P.Tod). In flat space-time, if we are working with Minkowski null tetrad, then the Infeld-van der Waerden symbols are the Pauli matrices. It means that the 2$\times$2 operator:
$$\sigma_a\nabla^a={\sigma_a}^{AA'}\nabla^a:\iff\nabla^{AA'}$$We can extend this construction for a curved space-time. It is more intuitive to arrive at tensorial equation starting from the 2-spinor algebra (to get the motivation see my answer in an earlier post here). In the language of 2-spinors, a free z.r.m. spin-n/2 field equation is given by $\nabla^{AA'}\phi_{\underbrace{AB\cdots L}_n}=0$. On contracting with $\nabla_{MA'}$ operator we get (after some manipulation):
$$(\square+(n+2)R/12)\phi_{AB\cdots L}=2(n-1){\Psi_{(AB}}^{MN}\phi_{C\cdots L)MN}+2ie{\varphi_A}^M\phi_{MB\cdots L}$$You can find this expression in equation 6.8.35 of Spinors and space-time volume II. For flat space-time and if the spinor field is chargeless, the RHS is zero and the Ricci scalar is zero, hence we have $\square\phi_{AB\cdots L}=0$. Note that this expression is exact for $n>0$ only. However, it is possible to extend this definition to scalar field (n=0) by defining the limit where $n\to 0$ . This limit is well defined provided that $\Psi_{ABCD}=0$ (space-time is patch wise conformally flat) and $e\varphi_{AB}=0$ (no electromagnetic interaction), then we essentially end up with the conformally coupled scalar field equation:
$$\left(\square+\frac{R}{6}\right)\phi=0$$


TITLE: How do you project vectors?
QUESTION [1 upvotes]: So, I am relatively new to the concept of vectors. I'm not to sure on how to "project a vector", as shown below.
For $\textbf{v} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$ and $\textbf{w} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, compute $\text{proj}_{\textbf{w}} \textbf{v}$.
And... For $\textbf{v} = \begin{pmatrix} -10 \\ 6 \end{pmatrix}$ and $\textbf{w} = \begin{pmatrix} 15 \\ -9 \end{pmatrix}$, compute $\text{proj}_{\textbf{w}} \textbf{v}$.
These questions are virtually the same, but I don't know how to calculate the projections, thanks!

REPLY [1 votes]: Recall what is the application of the dot product between vectors in physics:
$$
\mathbf{v} \cdot \mathbf{w} = \text{scalar} = \left| {proj_\mathbf{w} \mathbf{v}} \right| \cdot \left| \mathbf{w} \right|
$$
then
$$
\left| {proj_\mathbf{w} \mathbf{v}} \right| = \frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{w} \right|}}
$$
and
$$
proj_\mathbf{w} \mathbf{v} = \left| {proj_\mathbf{w} \mathbf{v}} \right|\frac{\mathbf{w}}
{{\left| \mathbf{w} \right|}} = \left( {\frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\left| \mathbf{w} \right|^2 }}} \right)\mathbf{w} = \left( {\frac{{\mathbf{v} \cdot \mathbf{w}}}
{{\mathbf{w} \cdot \mathbf{w}}}} \right)\mathbf{w}
$$


TITLE: Solving Laplace Equation using Separation of Variables inside an annulus
QUESTION [0 upvotes]: Question: find the bounded function $u(x,y)$ that satisfies the following conditions

$$\nabla^2u(x,y)=0, \qquad 4<x^2+y^2<16$$
$$u(x,y)=x, \qquad x^2+y^2=4$$
$$u(x,y)=y, \qquad x^2+y^2=16$$

First of all I transform this problem to polar coordinates, obtaining

$$u_{rr}+\frac{1}{r}u_{r}+\frac{1}{r^2}u_{\theta\theta}=0, \qquad 3<r<4$$
$$u(3,\theta)=3\cos(\theta), \qquad -\pi\leq\theta\leq\pi$$
$$u(4,\theta)=4\sin(\theta), \qquad -\pi\leq\theta\leq\pi$$

I'll jump to the expression for the general solution to a Laplace Equation in polar coordinates, since where I'm getting stuck at is when evaluating the boundary conditions. A general expression would be the following:
$$u(r,\theta)=A+B\ln(r)+\sum_{n=1}^\infty (C_n\,r^n+D_n\,r^{-n})(E_n\sin n\theta+F_n\cos n\theta)$$
Since the boundary conditions are just a sine and a cosine, I thought it would be almost trivial to find expressions for the coefficients. But I've been stuck for quite some time now, how would one solve this system of equations to find said coefficients?

REPLY [1 votes]: There is no reason why $\sin n\theta$ and $\cos n\theta$ has to share the same function of $r$, which is probably why you are having troubles
Quicker way: We know the solution has to be of the form
$$
u(r,\theta)=f(r)\sin\theta+g(r)\cos\theta.
$$
So substituting, we want to solve
$$
f''(r)+\frac1rf'(r)-f(r)=0,\quad f(3)=0,\quad f(4)=4
$$
and similarly $g$.  That gives you a solution in terms of the modified Bessel functions $I_0,K_0$.