TITLE: Help solving the PDE $u_x^2 + u_y^2 = u$
QUESTION [3 upvotes]: I'm trying to solve this Cauchy problem:
$$ u_x^2 + u_y^2 = u \\
u(x,0) = x^2+1$$
Here's what I've tried so far:
Letting $p(r,s) = u_x$, $q(r,s) = u_y$ and $z(r,s) = u$, we have:
$$ F = p^2 + q^2 - z = 0 $$
and thus:
$$ \begin{align} \frac{dx}{ds} &= F_p = 2p \\
\frac{dy}{ds} &= F_q = 2q \\
\frac{dz}{ds} &= pF_p + qF_q = 2p^2 + 2q^2 = 2z\\
\frac{dp}{ds} &= -F_x -pF_z = -(0) -p(-1) = p \\
\frac{dq}{ds} &= -F_y -qF_z = -(0) -q(-1) = q \\ \end{align}$$
Also, $\Gamma(r,0) = r^2 +1 $, and
$$ \begin{align} (r^2+1)' &= \phi_1 \cdot (r)' + \phi_2 \cdot (0)' \\
\implies 2r &= \phi_1 \end{align}$$
so that:
$$ \begin{align} (2r)^2 + \phi_2^2 &= r^2+1 \\
\implies 4r^2 + \phi_2^2 &= r^2+1 \\
\implies \phi_2^2 &= -3r^2 + 1 \\
\implies \phi_2 &= \pm \sqrt{ -3r^2 +1 } \end{align} $$
Which leads to:
$$ \begin{align} p &= e^s 2r \\
q &= e^s \sqrt{ -3r^2 +1 } \\
\implies x &= 4e^s r - 3r \\
\text{and } y & = 2e^s \cdot ( \pm \sqrt{ -3r^2 +1 } ) \mp \sqrt{ -3r^2 +1 }\end{align} $$
And I'm pretty sure by now it's gone way off the rails, because I can't solve for for $s$ and $r$ in terms of $x$ and $y$, but I can't figure out where I went wrong.
Could anybody please shed some light?

REPLY [1 votes]: You lost your way at the integration of $y$, all before looks good. You should have got
\begin{align}
x&=(4e^s-3)r&&=4(e^s-1)r+r\\
y&=\pm 2(e^s-1)\sqrt{1-3r^2}
\end{align}
Then
$$
\pm2yr=(x-r)\sqrt{1-3r^2}\\
4y^2r^2=(x-r)^2(1-3r^2)
$$
and at this point you have to solve this degree 4 polynomial in $r$. After selecting the correct branch ($x>r$ is one condition) you obtain the value of $z$ as
$$
z=e^{2s}(1+r^2)=\frac{(x+3r)^2}{16}(1+r^2).
$$


TITLE: If $x=1$, then its powers end with $1$
QUESTION [1 upvotes]: How to "show", that if $x=1$, then its powers will end with $1$?
Or, if $x=5$, how do we show it's powers end with $5$?

REPLY [6 votes]: If you know modular arithmetic: $\rm\; e\equiv 1,5\;\Rightarrow\; e^2\equiv e \;\Rightarrow\; e^k\equiv e\pmod{10}\;$ for all $\rm\; k>0$
If not, use induction: suppose $\rm\;\; e^2 = e + 10j\;$ and  $\rm\; (e + 10 m)^k = e + 10n$
then $\rm\; (e + 10 m)^{k+1} = (e + 10n) (e + 10m) = e + 10(j+em+en+10mn)$
REMARK Note how the use of modular arithmetic has allowed us to rephrase the problem in a completely arithmetical form that makes the induction step utterly trivial - namely that $\rm 1^k \equiv 1$ or, more generally that $\rm\; e^2 \equiv e\;\Rightarrow e^k \equiv e\;$ for $\rm\: k>0$. This is but one very simple illustration of how one may often reduce problems about integers to simpler problems about integers modulo $\rm\:m\:$. Because the integers modulo $\rm\:m\:$ satisfy many of  the same arithmetical laws as the integers (associative, commutative, distributive, etc) - i.e. they form a ring - we can reuse our strong intuition about integers to help discover proofs in this simpler structure integers modulo $\rm\:m\:$. And, if we're lucky, we can sometimes lift these simple solutions from the integers modulo $\rm\:m\:$ up to the integers. If you study abstract algebra you'll see many more examples of problems solved by passing to such simpler structures (technically they are known as quotient objects - structures modulo a congruence, i.e. a structure-compatible equivalence relation). Said in an informal way, such reduction to quotient objects is a method to divide and conquer mathematically.


TITLE: Prove that $\int_{0}^{1} |p(x)| dx \leq \frac{\pi}{2} $
QUESTION [7 upvotes]: Let the polynomial $p(x)= a_0 + a_1 x + . . . + a_n x^n$ have coefficients satisfying the relation $$\sum_{i=1}^{n} a_i^{2} = 1$$
Prove that $\int_{0}^{1} |p(x)| dx \leq \frac{\pi}{2} $.
I don't have any idea to prove this inequality, is there any reference to study about integrating polynomial ?

REPLY [13 votes]: For $x\in [0,1)$, by Cauchy-Schwarz inequality,
$$|p(x)|^2\leq \sum_{i=0}^{n} a_i^{2}\cdot \sum_{i=0}^{n} x^{2i}\leq \frac{1}{1-x^2}.$$
Hence
$$\int_{0}^{1} |p(x)| dx \leq \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} dx=\frac{\pi}{2}.$$
P.S. I am assuming that $\sum_{i=0}^{n} a_i^{2} \leq 1$. If we have the weaker condition $\sum_{i=1}^{n} a_i^{2} = 1$ the inequality does not hold. Take for example $p(x)=2$.


TITLE: How to do a proof of sequences?
QUESTION [1 upvotes]: Prove that for all $k > 100$, the quantity $\frac {1}{k^2}$ is in the interval $(0, 0.0001)$.

REPLY [0 votes]: An exmple of how to write it. Assume that any reader will not require you to prove that $a>b>0 \implies 0<a^{-1}<b^{-1}.$
$\forall k\in \Bbb R$
$(k>10^2\to k>10^2>0\to k^2>10^4>0\to 0<(k^2)^{-1}<(10^2)^{-1}\to 0<k^{-2}<10^{-4}).$
I used $\to$ for $\implies$ to fit it into one line. Can't fit the $\forall$ part into the same line.


TITLE: Collapsing a wave function without hitting the detector
QUESTION [0 upvotes]: If a free particle is placed at the origin (in 1D) with a wave function that consists of a superposition of the particle moving in both the +/- X direction, and a single detector is placed on the positive X axis, what would happen?
I was thinking that the wave function would collapse at the time that the +X state would impinge on the detector, and there is a 50% chance of the detector being hit. If the particle does not hit the detector, the wave function collapses into the -X state. So, the wave function is collapsed even though the particle seems to have never interacted with the detector. Is this accurate? Would the -X state then be 'fixed', or can it evolve over time, become a superposition again, and eventually hit the detector at a later time?

REPLY [1 votes]: So, the wave function is collapsed even though the particle seems to have never interacted with the detector. Is this accurate?

Totally. There is even a whole class of measurements called quantum nondemolition measurements, where you can detect whether or not a single photon could set off a trigger without actually setting off the trigger.

Would the -X state then be 'fixed', or can it evolve over time, become a superposition again, and eventually hit the detector at a later time?

That statement seems confused. If your wave separated and half ran away and half ran over to your detector then half the time your detector fires and now the running away branch is not an issue. Or else the other half of the time, the detector doesn't go off and the running away branch becomes the entire wave.
And as a wave it will then evolve according to the Schrödinger equation, like always.


TITLE: Physics of K meson
QUESTION [1 upvotes]: $K^{0}$ meson consists of a $d$ quark and an $\bar{s}$ antiquark.
Its antiparticle $\bar{K^{0}}$ consists of an $s$ quark and a $d$ antidown quark.
$$|K^{0}\rangle=|d\bar{s}\rangle$$
$$|\bar{K}^{0}\rangle=|\bar{d}s\rangle$$
There are two versions of $K^{0}$ meson. They have the same mass but different decay lifetimes.
K-zero-short:
$$K_{S}^{0}=\frac{1}{\sqrt{2}}\left[|K^{0}\rangle+|\bar{K}^{0}\rangle\right]$$
with the observed lifetime $\tau_{S}=90$ ps.
K-zero-long:
$$K_{L}^{0}=\frac{1}{\sqrt{2}}\left[|K^{0}\rangle-|\bar{K}^{0}\rangle\right]$$
with the observed lifetime $\tau_{L}=50$ ns.
What is the physical interpretation of the minus sign? I know that $\pm 1$ are the eigenvalues of $CP$ operator and $K_{S}^{0}$ and $K_{L}^{0}$ are its eigenstates. But what is the physical difference between $K_{S}^{0}$ and $K_{L}^{0}$ besides that they decay the way they do.

REPLY [2 votes]: Strictly speaking it is not true, as you say, that the mass eigenstates are the CP eigenstates in the neutral kaon system. This is exactly where CP violation was observed in first place by Cronin and Fitch.
But lets assume CP symmetry holds (which is a good assumptions since the CP violation is tiny in the neutral kaon system). What is the difference between the two Kaons? It is not the particle content. We define 'particles' by their well defined lifetime and mass. It may look weird that these states are not the same but on the other hand why should they? In the lagrangian to your qft you have terms like $\psi^\dagger M \psi$ where $M$ could be a non-diagonal Matrix. Diagonalizing it creates mixed flavour states but diagonal mass eigenstates. Every oscillation of particles means that the flavour eigenstates are not matching with the mass eigenstates, like in B Meson or neutrino oscillations.
If you now observe two Kaons being produced in the same process but with different lifetimes you will conclude that they have to be different particles. Your minus sign then determines which Kaon will be the long-living and which the short-living one, since CP eigenstates can only decay into final states with the same eigenvalue (assuming CP holds). Neutral Kaons can decay into 2 or 3 pions. Now you could calculate what the CP eigenvalues of a 2 and a 3 pion system and conclude which Kaon will decay into which pion system. This also allows you to see why one of the Kaons will decay faster by looking at the phase space dimension.
I hope this helps a bit.