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Linearizing at extinction gives $n_t=(r-\delta )n+D{n}_{xx}$. The principal eigenfunction for Neumann data at $0$ and Dirichlet data at $L$ is $\cos (\pi x/(2L))$, with growth rateA population is viable exactly when this is positive, requiring $r>\delta$ and $L>(\pi /2)\sqrt{D/(r-\delta )}$. Thus intrinsic growth must beat predation and the refuge must exceed the diffusion-dependent critical size.

The terms are logistic birth and crowding, $rn-r{n}^2/K$, linear predation mortality $-\delta n$, and diffusion $D{n}_{xx}$. The coast is reflecting, so $n_x(0)=0$; lethal habitat beyond $L$ gives the absorbing condition $n(L)=0$.

A word containing neither $01$ nor $10$ cannot change symbol. The language is therefore $0^{\ast }\cup 1^{\ast }$, hence is regular.

This language is not regular. Intersecting with the regular language $0^{\ast }110^{\ast }$ isolates strings whose two zero blocks must have equal length; the pumping lemma then gives a contradiction.

The pumping lemma for regular languages says that sufficiently long accepted words split as $xyz$ with $|xy|\le P$, $|y|>0$, and $x{y}^{k}z$ accepted for all $k\ge 0$. Applying it to $0^P{1}^{P}01$ pumps only zeros, so language (i) is not regular.

After multiplying the five-point equations by $-1$ if necessary, their matrix $A$ is symmetric positive definite. For any ordering write $A=D+L+L^T$, where $D=4I$. Gauss-Seidel method uses $M=D+L$ and has iteration matrix $-M^{-1}{L}^T$. In part (a) take $B=L^T$; then $A-B=M$ andwhich is positive definite. Therefore the iteration matrix has spectral radius below one for every ordering, proving convergence.

Let $Mx=(A-B)x$ and suppose $Hx=\lambda x$, so $-Bx=\lambda Mx$. Put $a=x^{\ast }Ax>0$ and $m=x^{\ast }Mx$. Since $A=M-(-B)$, $a=(1-\lambda )m$. AlsoPositive definiteness makes this positive, hence $|\lambda |<1$ for every eigenvalue and $\rho (H)<1$.

Tangential velocity is unchanged, while the downstream normal velocity is $U{\rho }_1/{\rho }_2$. Thus the incident and outgoing angles to the shock have tangents $1$ and ${\rho }_1/{\rho }_2$. Using the tangent-difference identity, the ratio from part (b), and $M^2=2U^2/c_1^2$ gives

Insert $h=\gamma p/[\rho (\gamma -1)]$, eliminate $p_2$ and the downstream normal speed using mass conservation, and solve for the density ratio. This gives

In the shock frame, conservation of mass, normal momentum, and energy givesThese are the Rankine-Hugoniot relations.

Integrating the viscous traction $2\rho \nu e_{r\theta }$ over the side and end surfaces givesIt is always negative. In steady state the cylinder stores no angular momentum, so the divergence theorem for angular-momentum balance requires viscous torque to cancel net advective influx.

Adding inward side flux and outward end-cap flux gives the net inward angular-momentum advectionIt is positive because $e^{s_0}>1+s_0$, and tends to the stated value.

The only vorticity component is $\omega =(1/r)(rv)\prime$. Its steady equation is $-\alpha r\omega \prime =2\alpha \omega +\nu (\omega \prime \prime +r^{-1}\omega \prime )$, solved by $\omega ={\omega }_0{e}^{-\alpha r^2/(2\nu )}$. Regularity at zero givesIts maximum satisfies $e^s=1+2s$, $s=\alpha r^{\ast 2}/(2\nu )$, so $r^{\ast }=O(\sqrt{\nu /\alpha })$.

The radial strain draws fluid inward, axial strain expels it away from the midplane, and $v(r)$ supplies swirl. In cylindrical coordinates $r^{-1}{\partial }_r(r(-\alpha r))+{\partial }_z(2\alpha z)=-2\alpha +2\alpha =0$, so the flow is incompressible.

The far field of a dipole is $[3(p\cdot \hat{x})\hat{x}-p]/(4\pi {\epsilon }_0{r}^3)$. Comparison yields the induced electric dipole moment