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A tensor is isotropic tensor if its components are unchanged under every proper orthogonal change of basis. The general isotropic rank-four tensor iswhere $a,b,c$ are scalars.

A tensor is totally antisymmetric tensor if interchanging any two indices changes its sign. Consequently every component with repeated indices vanishes. In ${\mathbb{R}}^3$:

These are all possibilities.

A tensor of rank $n$ is a multilinear map of $n$ vector or covector arguments, equivalently an element of an $n$-fold tensor product. In components it obeys one change-of-basis factor for each of its $n$ indices.

The first field isso its integral around $C$ is zero and it is conservative with potential $e^{x^2}\cos (yz)$.

Along $C$, $dr=(-\sin s,\cos s,0)ds$ andsoThus $B_2$ is not conservative and cannot be a gradient on the stated punctured region.

Finally,Its integral around $C$ is zero, and it is conservative with the displayed potential. Therefore the three integrals are respectivelyand exactly $B_1,B_3$ are conservative gradient fields.

The Stokes theorem states that for an oriented smooth surface $S$ with positively oriented boundary $\partial S$,A field is conservative vector field if its line integral is path-independent, equivalently if every closed line integral vanishes. If $F=\nabla \varphi$, the fundamental theorem for line integrals givesThis is zero for every closed curve, so every globally defined gradient field is conservative.

Similarly,since the last derivatives are symmetric in $i,j$. Thus

In components,because the Levi-Civita symbol is antisymmetric in $j,k$ while equality of mixed partial derivatives makes ${\partial }_j{\partial }_k\varphi$ symmetric. Hence

Close the surface with the unit disk $D$ in the plane $z=0$. Its outward normal is $-e_z$, but $F\cdot (-e_z)=-z^4=0$ there. AlsoThe $x$ and $y$ terms integrate to zero by symmetry. The divergence theorem therefore gives

Parametrize the upward-oriented paraboloid byThenOn the surface,The cubic trigonometric terms integrate to zero over $0\le \theta \le 2\pi$, so the surface integral is

Conditioning on $N$ and using the moment-generating function of a unit exponential variable,where the geometric series converges precisely when $\theta <p$. This is the moment-generating function of an exponential variable of rate $p$, so the geometric sum of exponential variables satisfies

This is the unit-rate Erlang distribution. For completeness, use convolution. The result is clear for $n=1$. Ifthen independence and the unit exponential density give, for $x>0$,Induction therefore proves

Setting $t=0$ in part (i) givesAlsosoPart (i) now factors aswhich proves that $K$ and $T$ are independent.

Put $Q={\sum }_{j=1}^n{q}_j$. Since the variables have continuous distributions, the minimum is unique almost surely. Integrating over its possible value givesThis is the joint tail calculation for competing exponential clocks.

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.