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Differentiability at $x\in {\mathbb{R}}^2$ means that there is a linear map $Df{|}_x:{\mathbb{R}}^2\to \mathbb{R}$ such thatThe partial derivatives are $D_{i}f(x)=Df{|}_x(e_i)$ when the derivative exists; independently, they may be defined by the corresponding one-variable difference quotients.

Suppose the partial derivatives exist near $x=(x_1,x_2)$ and are continuous at $x$. Apply the one-dimensional mean value theorem along the two coordinate segments from $x$ to $x+h$. It gives intermediate points ${\xi }_h,{\eta }_h\to x$ such thatContinuity of the partials makes the remainder after subtractingequal to $o(\Vert h\Vert )$. Hence $f$ is differentiable and this is its derivative.

For the given function, writing $r=\sqrt{x^2+y^2}$, at $r>0$ we haveAt the origin both partial derivatives are zero, since $f(h,0)/h$ and $f(0,h)/h$ tend to zero. Neither partial is continuous there: along its corresponding coordinate axis the cosine term oscillates without a limit. Neverthelessso $f$ is differentiable at the origin with derivative zero. This illustrates that existence of partial derivatives does not imply their continuity.

The final assertion is false. DefineAgain $|g|\le r^2$, so $g$ is differentiable at the origin and it is smooth elsewhere. But along the $x$-axis,is unbounded near zero, and similarly $D_{2}g(0,y)$ is unbounded. Thus both partial derivatives can be unbounded in every neighbourhood of a point even when the function is differentiable everywhere.

Fourier transformation givesFor $x>0$, close the inverse-transform contour in the upper half-plane and take the residue at $k=im$; for $x<0$, close it in the lower half-plane. This gives the one-dimensional modified Helmholtz Green functionThe same formula works for complex $m$ with $\mathrm{Re}m>0$: it decays at both ends, is continuous at zero, and its derivative has the jump $G^{\prime }(0+)-G^{\prime }(0-)=-1$ required by the delta source.

Convolution therefore yields

Taking the vertical component of the curl of the momentum equation givesThe continuity equation gives ${\eta }_t+h_0\nabla \cdot \mathbf{u}=0$, and henceSince initially $\mathbf{u}=0$ and $\eta ={\eta }_0$,Taking the divergence of the momentum equation and usinggivesDifferentiate continuity in time and substitute the last two identities to obtain

Let the Rossby deformation radius beThe even, decaying steady solution ofwith continuous value and derivative at $x=\pm a$ isThe steady momentum balance is a geostrophic balance:Thus the flow is parallel to the two edges of the raised strip, in opposite $y$-directions on the two sides; for $f>0$, it points toward $+y$ on the left and $-y$ on the right. Its magnitude is concentrated within a few deformation radii of the edges.

The time-independent Schrödinger equation isIntegrating it over $(a-\epsilon ,a+\epsilon )$ and letting $\epsilon \to 0$ shows that ${\psi }^{\prime }$ has no jump because $V$ is finite. A jump in $\psi$ would create a delta term in ${\psi }^{\prime }$ and hence a derivative of a delta in ${\psi }^{\prime \prime }$, so $\psi$ is also continuous.

For a bound state $-V_0<E<0$, defineThen the proposed even wavefunction solves the equation away from the interfaces, andContinuity at $x=a$ gives $B\cos (ka)=A{e}^{-\eta a}$, while continuity of the derivative gives $-Bk\sin (ka)=-A\eta e^{-\eta a}$. Dividing yields the second required relationThus the lowest even state of the finite square well is the first-quadrant intersection, with $0<ka<\pi /2$, of the circle $k^2+{\eta }^2=2m{V}_0/{\hslash }^2$ and the curve $\eta =k\tan (ka)$.

The free energy is dimensionless in units with $k_{B}T=1$, so the integrand has momentum dimension $d$. Since a derivative has dimension one, the kinetic term givesThe mass term then givesThese are engineering dimensions.

The space of test functions isA sequence ${\phi }_j$ converges to $\phi$ in $\mathcal{D}$ when all supports lie eventually in one compact set $K$ andfor every multi-index $\alpha$. A distribution is a linear functional $u:\mathcal{D}\to \mathbb{C}$ such that, for every compact $K$, there are $C_K>0$ and $m_K\in {\mathbb{N}}_0$ withwhenever $\mathrm{supp}\phi \subseteq K$. Convergence in ${\mathcal{D}}^{\prime }$ is pointwise convergence on test functions.

Continuity plainly implies sequential continuity. Conversely, suppose the linear form is sequentially continuous but the displayed estimate fails for some compact $K$. For every $j$, choose ${\phi }_j$ supported in $K$ such thatThen ${\phi }_j\to 0$ in $\mathcal{D}$, while $u({\phi }_j)\nrightarrow 0$, a contradiction. Hence the seminorm estimate holds on every compact set and $u\in {\mathcal{D}}^{\prime }$.

For a translation vector $h$ and a multi-index $\alpha$, defineThese definitions extend ordinary translation and differentiation to distributions.

If ${\tau }_{t{e}_i}u=u$ for every $t$, differentiating its pairing at $t=0$ gives ${\partial }_{i}u=0$. Conversely, if ${\partial }_{i}u=0$, then for every test functionThe pairing is constant in $t$, hence ${\tau }_{t{e}_i}u=u$. This proves both directions of translation invariance and vanishing distributional derivative.

Finally, the distributional differentiation obeys the linear chain rule under the linear coordinates $s=x-y$ and $t=x+y$. Thusand likewiseas distributions. Adding the two identities givesthe one-dimensional wave equation; this is the low-regularity form of the travelling waves in the D'Alembert formula.

Setso $0<x<\Delta$. Apart from material constants, the heat flux in part ii isThe derivative of its logarithm vanishes whenwhich gives the maximizing interface temperatureSubstitution into the flux law giveswhereThus the proportionality coefficient depends on the thermal conductivity, nonlinear thermal-expansion coefficient $\alpha$, kinematic viscosity, thermal diffusivity, gravity, and critical Rayleigh number. This is the maximum-flux law for penetrative convection in an ice-covered lake.

Chandrasekhar dynamical friction is the drag exerted by the overdense gravitational wake that a massive body creates in a background of lighter particles. It transfers orbital energy and angular momentum to the host, causing satellites, star clusters, and massive black holes to spiral inward and promoting galaxy mergers. In a homogeneous isotropic Maxwellian background its force iswhere $\log \Lambda$ is the Coulomb logarithm in stellar dynamics. Its scaling can be reconstructed from the strong-deflection impact parameter $b_{\mathrm{cr}}\sim GM/v^2$: the encountered mass rate is $\sim \rho v\pi b_{\mathrm{cr}}^2$, and multiplying by momentum change $\sim v$ gives $F_{\mathrm{df}}\sim G^2{M}^2\rho /v^2$.

For a spherical host with a flat galaxy rotation curve,This is a singular isothermal sphere with one-dimensional dispersion $\sigma =v_c/\sqrt{2}$, so $X=1$. DefineFor a constant-mass satellite on a circular orbit, the drag magnitude and its torque areIt follows thatThe quadratic radius dependence and inverse mass dependence explain why massive nearby satellites merge much faster than light or distant ones.

Let the satellite also have a flat internal rotation curve of speed $v_s$. Equating its edge density to the host density gives the tidal radiusBecause $M_s(<R_t)=v_s^2{R}_t/G$, the bound mass decreases linearly:Under the question's literal closure that the derivative of the remaining satellite's total orbital angular momentum equals the frictional torque,and henceIf stripped material is explicitly assigned the satellite's instantaneous specific orbital angular momentum, the balance for the bound remnant is instead $M_{s}d(v_{c}r)/dt=-r{F}_{\mathrm{df}}$; that convention gives $t_{\mathrm{df},\mathrm{strip}}=2t_{\mathrm{df},0}$. Both treatments show the robust point: tidal stripping weakens the drag as the orbit shrinks and substantially delays coalescence. In less idealized profiles the mass can fall faster than linearly, producing dynamical-friction stalling by tidal stripping.

For a cylinder of radius $r$ and length $L$, the mean curvature is $H=1/(2r)$ and the area is $2\pi rL$. Its Helfrich membrane energy per unit length is thereforeThe first term favors a narrow tube and the second penalizes its curvature. Setting the derivative with respect to $r$ to zero givesThe positive second derivative confirms a minimum.

A covering map is locally a disjoint union of homeomorphisms onto the base. A space is simply connected when it is path-connected and every loop is null-homotopic. By the uniformization theorem, the simply connected Riemann surfaces areup to analytic isomorphism.

A lattice is a discrete subgroupwith ${\omega }_1/{\omega }_2\notin \mathbb{R}$. The Weierstrass elliptic functionconverges normally on compact subsets of $\mathbb{C}\setminus L$, defines a meromorphic map $\mathbb{C}\to {\mathbb{C}}_{\infty }$, and is nonconstant because it has a double pole at every lattice point. Reindexing the normally convergent derivative series shows that ${\wp }_L^{\prime }$ is $L$-periodic; evenness fixes the integration constants, so ${\wp }_L(z+\omega )={\wp }_L(z)$.

The invariance makeswell-defined and unique on $\mathbb{C}/L$. It is analytic because the quotient projection is locally biholomorphic.

Neither map is a covering. For any nonzero $\omega \in L$, periodicity and oddness of ${\wp }_L^{\prime }$ giveso ${\wp }_L^{\prime }(\omega /2)=0$. The map is not a local homeomorphism there; since the quotient projection is locally biholomorphic, the descended map fails for the same reason.

There is no covering $\mathbb{C}/L\to {\mathbb{C}}_{\infty }$. The torus is connected, while every connected covering of the simply connected sphere is a homeomorphism. A torus is not homeomorphic to a sphere, for example because their fundamental groups are ${\mathbb{Z}}^2$ and $0$.

The $n$ roots of $T_n$ are distinct and lie in $(-1,1)$. Repeated use of Rolle's theorem shows that every root of $T_n^{(k)}$, for $0\le k<n$, also lies in $(-1,1)$. Its leading coefficient is positive, soTaking $k=r+1$ shows thatfor $0\le r\le n-1$; when $r=n-1$, the derivative is the positive constant $T_n^{(n)}$. This is the positivity of Chebyshev derivatives beyond the unit interval.

The normalized one-particle spatial density gives the local number densityThe radial probability density is proportional toIts logarithmic derivative vanishes whenso the most likely radius is

Take the divergence of the field equation. Commuting derivatives and antisymmetry of $F^{\mu \nu }$ giveCurrent conservation then yieldsFor $m>0$,This is the Lorenz constraint in Proca theory; it follows dynamically rather than being a gauge choice.

For a circular orbit, the derivative ofvanishes. ThusA real finite angular momentum therefore requires

Fix $w\in \mathbb{C}$ and apply the stated clockwise argument principle to $h=f-w$. Its logarithmic derivative is periodic, so the integrals over opposite sides of a fundamental cell cancel. Hence $P-Z=0$. The poles of $f-w$ are precisely the poles of $f$, with unchanged multiplicities, while its zeros are the solutions of $f(z)=w$. Thus the value multiplicity of an elliptic function is the same fixed pole count for every finite $w$, counting multiplicity.

The Hartogs theorem says that for every set $X$ there is an ordinal that does not inject into $X$. Indeed, the well-orderings of subsets of $X$ form a set, since each is a relation in $\mathcal{P}(X\times X)$. By replacement, their order types form a set of ordinals. Its supremum plus one cannot be the order type of a well-ordered subset of $X$, and hence cannot inject into $X$. The least such ordinal is the Hartogs ordinal $h(X)$.

Ordinal exponentiation is defined byfor nonzero limit $\lambda$. Fix $\alpha ,\beta$ and induct on $\gamma$. The zero case is immediate. If the identity holds at $\gamma$, thenAt a limit $\lambda$, continuity of exponentiation and right continuity of ordinal multiplication give

Put $\rho ={\omega }^{\alpha }$. The setis nonempty and bounded, so let $\beta =\sup B$. Continuity of multiplication in its right argument gives $\rho \beta \le \gamma$. There is a unique tail $\delta$ with $\gamma =\rho \beta +\delta$. If $\delta \ge \rho$, then
$\rho (\beta +1)=\rho \beta +\rho \le \gamma$, contrary to maximality. Hence $\delta <\rho$. If
$\rho \beta +\delta =\rho {\beta }^{\prime }+{\delta }^{\prime }$ with both remainders below $\rho$ and $\beta <{\beta }^{\prime }$, thena contradiction. This proves existence and uniqueness in the division by an additively indecomposable ordinal.

Now let $X$ be nonempty and well ordered. Its least element is not a limit point, so $X^{\prime }\ne X$. Every derivative is itself a well-ordered subset. If $X^{(\alpha )}$ were nonempty for every $\alpha <h(X)$, then each differencewould be nonempty. Choosing its least member would inject $h(X)$ into $X$, since the nested successive differences are disjoint. This contradicts Hartogs' lemma. Thus some derivative is empty, as asserted by the derived-set iteration of a well-order.

For an ordinal $\xi$, a point $\gamma <\xi$ is a limit point exactly when it is a nonzero limit ordinal. Division by $\omega$ writes $\gamma =\omega \beta +n$ with $n<\omega$; this is a nonzero limit exactly when $n=0$ and $\beta >0$. HenceWithin this derived set, $\omega \beta$ is a limit point exactly when $\beta$ is a nonzero limit ordinal, equivalently $\beta =\omega \eta$ for some $\eta >0$. Therefore the derived sets of an ordinal satisfyFinally ${\omega }^{\prime }=\varnothing$, so the index of $\omega$ is $1$, while
$({\omega }^2{)}^{\prime }=\{\omega ,2\omega ,3\omega ,\dots \}$ and
$({\omega }^2{)}^{\prime \prime }=\varnothing$, so the index of ${\omega }^2$ is $2$.

A transcritical bifurcation has normal formIts equilibrium branches are $x=0$ and $x=\mu$. Linearization gives derivative $\mu$ on the first and $-\mu$ on the second, so the branches cross at $(\mu ,x)=(0,0)$ and exchange stability: for $\mu <0$, $x=0$ is stable and $x=\mu$ unstable; for $\mu >0$ the labels reverse.

A small constant perturbationhas equilibria $x=(\mu \pm \sqrt{{\mu }^2+4\epsilon })/2$. For $\epsilon >0$ the two branches avoid the crossing; for $\epsilon <0$ there is a parameter interval with no equilibrium, bounded by two saddle-node points. Thus arbitrarily small perturbations change the bifurcation diagram, so the crossing is not structurally stable.

A subset $X\subset {\mathbb{R}}^N$ is a smooth $k$-manifold if every $x\in X$ has a neighborhood parametrized by a smooth map $\varphi :U\subset {\mathbb{R}}^k\to X$ that is a homeomorphism onto its image and has derivative of rank $k$. DefineIf $\psi$ is another parametrization, the transition maps satisfyThe transition derivative is invertible, so the two images coincide. Thus the tangent space is independent of the parametrization.