The dark soliton density isFor $U=0$,a stationary notch reaching zero at its centre. For $U=1/2$,a shallower notch with minimum $1/2$. Since $\xi =x-Ut$, the first profile remains fixed and the second translates rigidly to the right at speed $1/2$.

The travelling-wave substitution $\xi =x-Ut$ givesWriteIts real part satisfiesFor $U>0$, the imaginary part also givesWith $\alpha =\sqrt{1-2U^2}$ and a translation chosen so that the notch is centred at zero,Its limiting modulus is $\sqrt{{\alpha }^2+2U^2}=1=d$, as required. The same expression at $U=0$ is obtained directly from the second-order equation.

For a stationary gas with constant phase, $\psi =a(x)$ is real and time independent. The field equation reduces toA positive constant solution obeys $d(1-d^2)=0$, so

Write $\psi =a{e}^{iS}$ with $a=\sqrt{\rho }$ and $v=S_x$. ThenwhileEquating imaginary parts givesand equating real parts and differentiating in $x$ givesThusThis is the hydrodynamic form of the defocusing nonlinear Schrödinger equation.

For the modified system, $F=x^2+y^2$ satisfiesWith $F(0)=1$, the logistic solution isConsequently,

The radially symmetric planar dynamical system formulas giveWith $r(0)=1$ and $\theta (0)=0$,again on the maximal interval for which the denominator is positive.

The only equilibrium is $(0,0)$. Its linearization iswhose eigenvalues are $\pm i$, so the linearized system has a centre.

The exact radial identity from part (i) givesThus the origin is asymptotically stable for $k<0$, neutrally stable for $k=0$, and unstable for $k>0$. Moreover,so trajectories rotate clockwise.

TakeThenWith $F(0)=1$, separation givesand henceon its maximal interval of existence.

The floor-half generating function isSumming $u_n-2u_{n-1}=\lfloor n/2\rfloor$ for $n\ge 1$ givesso

Sincea particular solution isAdding the homogeneous term $C{2}^n$ and imposing $u_0=1$ gives $C=5/3$. Therefore

The homogeneous characteristic equation isA linear particular solution $u_n=an+b$ givesso $a=-1/4$ and $b=-11/16$. ThusThe initial data yield $A=1$ and $B=-5/16$, hence

For the ordinary generating function of a recurrence, multiply by $x^n$ and sum for $n\ge 2$. Using $u_0=0,u_1=2$ givesEquivalently,Expanding each geometric series gives exactly the displayed formula for $u_n$, verifying consistency.

Rewrite the equation asOn $[2,6]$, one has $q(x)\le 1/3$. If a nontrivial solution had two zeros, choose two consecutive ones $\alpha <\beta$ in that interval. Then $\beta -\alpha \le 4<\pi \sqrt{3}$.

The solutionof ${\phi }^{\prime \prime }+\frac{1}{3}\phi =0$ is strictly positive on $[\alpha ,\beta ]$. Since $q\le 1/3$ and the coefficients are not identical on this interval, the Sturm comparison theorem requires $\phi$ to have a zero between $\alpha$ and $\beta$, a contradiction. Hence every nontrivial solution has

TakeIts consecutive zeros are $-1$ and $\pi -1$. For the stated equation,and $q_2$ is not identically equal to $1$ on the interval. The Sturm comparison theorem therefore shows that every nontrivial solution has at least one zero in

Suppose ${\phi }_2$ has no zero in $(x_1,x_2)$. It has a constant sign there, so replace it by its negative if needed and apply part (i). The conclusion is $q_1\equiv q_2$ on that interval. Taking the contrapositive proves the Sturm comparison theorem: unless the coefficients agree identically, ${\phi }_2$ has a zero between the consecutive zeros of ${\phi }_1$.

After multiplying either solution by $-1$ if necessary, suppose both are positive on $(x_1,x_2)$. SetThe equations giveSince ${\phi }_1$ is positive between consecutive zeros,ThereforeBut $W$ is nondecreasing, so both endpoint values coincide at zero and $W^{\prime }\equiv 0$. Since both solutions are strictly positive inside,This is the equality case in the wronskian proof of Sturm comparison.

LetThey are jointly normal because they are linear combinations of a Gaussian vector. When ${\sigma }_1={\sigma }_2=\sigma$,Hence uncorrelated jointly normal variables are independent. Their means and variances areandThereforeindependently.

Completing the square in $x_1$ in the joint density shows that the conditional distribution of a bivariate normal variable isThus its conditional density is

The bivariate normal distribution has density

The Markov inequality states that for a nonnegative random variable $Y$ and $a>0$,For $t>0$, the event $X\ge x$ is the same as $e^{tX}\ge e^{tx}$. Applying Markov's inequality to the nonnegative variable $e^{tX}$ gives the exponential Markov bound