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Use suffix notation, with repeated indices summed from one to three. The cross product, divergence, and curl have components $(\mathbf{A}\times \mathbf{B}{)}_i={ϵ}_{ijk}{A}_j{B}_k$, $\nabla \cdot \mathbf{A}={\partial }_i{A}_i$, and $(\nabla \times \mathbf{A}{)}_i={ϵ}_{ijk}{\partial }_j{A}_k$. Here $ϵ$ is the Levi-Civita symbol.

The product rule givesUsing the cyclic identity ${ϵ}_{ijk}={ϵ}_{kij}$ in the first term and ${ϵ}_{ijk}=-{ϵ}_{jik}$ in the second identifies these terms as $B_k(\nabla \times \mathbf{A}{)}_k$ and $-A_j(\nabla \times \mathbf{B}{)}_j$. ThereforeFor the second identity, contract two Levi-Civita symbols, using ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$:The four component expressions translate intoBoth derivations require only continuously differentiable vector fields; no additional divergence-free assumptions are used.

A conservative vector field is the gradient of a single-valued differentiable scalar potential. Its line integral is path independent and equals the potential difference. If $\psi \mathbf{A}=\nabla F$ with a twice continuously differentiable potential, equality of mixed partial derivatives givesThe product rule then gives ${\psi }_y{A}_1+\psi A_{1,y}={\psi }_x{A}_2+\psi A_{2,x}$. Rearranging proves the integrating-factor equation for a planar vector field:This argument does not require division by $\psi$, so it remains valid even at its zeros. A nonvanishing integrating factor is normally sought when one wants to recover useful potential curves for the original field.

The Batchelor entrainment hypothesis makes the inward entrainment velocity proportional to the local plume speed:where $\alpha >0$ is the dimensionless entrainment coefficient. For an upward plume $w>0$, this is $u_e=\alpha w$. The hypothesis represents turbulent ingestion of ambient fluid; it is a closure rather than a consequence of mass conservation.

Let $A=b^2$, suppressing the common factor $\pi$ from cross-sectional integrals. In the Boussinesq approximation, use a constant reference density ${\rho }_r$ in the leading-order mass conservation equation. Dividing that equation by ${\rho }_r$ gives volume conservation:The factor two is the ratio of the perimeter $2\pi b$ to the suppressed area factor $\pi$. Density differences are retained in buoyancy, even though they are neglected in the leading inertial and volume terms.

To derive the density-deficit balance, first leave the small density differences explicit in the supplied mass equation. Because the ambient density ${\rho }_0(z)$ is time independent, the product rule givesThus the ambient mass entrained through the perimeter cancels exactly against its contribution to the volume budget, yielding the required buoyancy balance. It is understood to the accuracy of the underlying Boussinesq approximation; the volume equation is not an additional exact constraint on a fully variable-density fluid.

For later use define reduced gravity $g^{\prime }=g({\rho }_0-\rho )/{\rho }_r$ and the squared buoyancy frequency $N^2=-g{\rho }_{0z}/{\rho }_r$. With $B=A{g}^{\prime }$ and $Q=Aw$, the balance isA stably stratified ambient has $N^2>0$, so upward motion consumes the plume's positive density deficit; an unstable ambient reverses this sign.

By differentiability implies continuity, $f(x+h)\to f(x)$. Add and subtract $f(x+h)g(x)$ in the difference quotient:The two quotient limits exist by the given differentiability. Taking the limit and using continuity yields the product rule:Thus the product is differentiable at the specified point; differentiability elsewhere is not needed.

At zero, $g(0)=0$ andby continuity of $f$ at zero. Thus $g$ is always differentiable there, with $g^{\prime }(0)=0$, regardless of whether $f^{\prime }(0)$ exists.

At $x\ne 0$, if $f$ is differentiable, the product rule applied to $t^{2}f(t)$ proves that $g$ is differentiable. Conversely, if $g$ is differentiable at $x$, the function $t^{-2}$ is differentiable on a neighborhood of $x$. Apply the product rule to $f(t)=t^{-2}g(t)$ on that neighborhood. This proves the differentiability restored by a quadratic zero equivalence:At a nonzero differentiability point, $g^{\prime }(x)=2xf(x)+x^2{f}^{\prime }(x)$.

If $f$ is differentiable at $x$, the product rule applied to $f\cdot f$ givesThis proves sufficiency of the first alternative. No additional growth condition is required; if $f(x)=0$, the derivative of the square is zero.

The product rule givesApply the divergence theorem, with outward unit normal $\mathbf{n}$, and use the vanishing Laplacian of the harmonic function $g$:The boundary value of $f$ is one. Another application of the divergence theorem therefore givesThis is the gradient pairing with a boundary-constant function and a harmonic function, obtained directly from Green's first identity. The harmonic function must be regular throughout the volume; an interior singularity can supply an additional boundary flux.

Use $D=(d/dt{)}_S$, $D^{\prime }=(d/dt{)}_{S^{\prime }}$, and ${\mathbf{v}}^{\prime }=D^{\prime }\mathbf{r}$. By the rotating-frame derivative formula, $D\mathbf{r}={\mathbf{v}}^{\prime }+\mathbf{\omega }\times \mathbf{r}$. Apply the product rule for the cross product and the same rotating-frame derivative formula again:We have $D{\mathbf{v}}^{\prime }=D^{\prime }{\mathbf{v}}^{\prime }+\mathbf{\omega }\times {\mathbf{v}}^{\prime }$ and $D\mathbf{\omega }=D^{\prime }\mathbf{\omega }$, since $\mathbf{\omega }\times \mathbf{\omega }=0$. ThereforeRearranging this identity into an equation of motion in a rotating frame produces the apparent Coriolis acceleration, Euler acceleration and centrifugal acceleration, with signs opposite to their corresponding terms on the right of this inertial acceleration identity. No translational term is needed because the two origins coincide.

Take complex-valued functions. Pointwise multiplication is associative and commutative, with identity the constant function $1$, whose norm is one. The product rule givesTo prove completeness, let $(f_n)$ be a Cauchy sequence in this norm. The functions and their derivatives converge uniformly to continuous functions $f,g$. Passing to the limit in $f_n(t)-f_n(0)={\int }_0^t{f}_n^{\prime }(s)ds$ gives $f(t)-f(0)={\int }_0^{t}g(s)ds$. The fundamental theorem of calculus yields $f^{\prime }=g$, including the one-sided endpoint derivatives. Thus $f\in C^1[0,1]$ and $f_n\to f$ in the stated norm. Consequently $A$ is a unital commutative Banach algebra.

Let $u(t)=t$. If $\lambda \notin [0,1]$, then $(u-\lambda 1{)}^{-1}(t)=1/(t-\lambda )$ belongs to $C^1[0,1]$. If $\lambda \in [0,1]$, the function $u-\lambda 1$ vanishes somewhere and cannot have a pointwise inverse. Thus ${\sigma }_A(u)=[0,1]$. Every character of an algebra satisfies $\varphi (u)=t_0$ for some $t_0\in [0,1]$, by part (d), and therefore $\varphi (p(u))=p(t_0)$ for every polynomial $p$.

Polynomials are dense in this stronger norm, not just the supremum norm. By the Weierstrass approximation theorem, approximate the real and imaginary parts of $f^{\prime }$ uniformly by polynomials, combining them into $q_n\to f^{\prime }$. Put $p_n(t)=f(0)+{\int }_0^t{q}_n(s)ds$. Then $p_n$ is a polynomial andContinuity of the algebra character now gives $\varphi (f)=\lim p_n(t_0)=f(t_0)$. Conversely each point evaluation is a continuous algebra character. Hence the character space of C1 on a compact interval consists exactly of the evaluation functionals, and part (c) givesDistinct points give distinct maximal ideals, since the coordinate function separates them.

Away from zero and the zeros of $\cos (\pi /x)$, the sign function is locally constant and the absolute value function is differentiable at the value of the cosine. The product rule and chain rule giveThus the triangle inequality and the bounds on sine and cosine imply $|f^{\prime }(x)|\le 2|x|+\pi$. Part (c) rules out differentiability at every nonzero cosine zero, and part (b) gives $f^{\prime }(0)=0$, so this estimate covers every point where the derivative exists.

For a finite interval $I$, choose $L<\infty$ such that $|x|\le L$ throughout $I$. ThenThe apparent $x^{-2}$ factor in the derivative of the oscillation cancels against the multiplying $x^2$, which is what makes a uniform bound possible even near zero.

The first-order chain rule, with $x,y$ regarded as functions of $u,v$, isApply the product rule when differentiating the first identity in $v$, and apply the chain rule to $f_x,f_y$. Equality of mixed partial derivatives for a $C^2$ function givesThus $\boxed{H=x_{uv},K=y_{uv}}$. These terms record the second derivatives of the coordinate transformation, not of $f$.

For the given transformation,The mixed-derivative coefficient is zero, while $x_u{x}_v=x$, $y_u{y}_v=-y^2/x$, and $1/v^2=y/x$. The required partial differential equation is therefore exactly $g_{uv}=0$. Integrating first in $u$ and then in $v$ on a coordinate rectangle giveswhere $A,B$ are arbitrary twice continuously differentiable functions. The Jacobian determinant of the transformation is $-2u/v$, so this calculation uses patches with $u,v\ne 0$. On a positive-quadrant patch we can take $u=\sqrt{xy}$ and $v=\sqrt{x/y}$, obtainingChanging branches merely redefines the arbitrary functions.

For completeness, the original equation also makes sense in quadrants with $xy<0$, where that particular real square-root transformation is unavailable. Multiplying the equation by $x$ givesOn any quadrant with $xy\ne 0$, put $a=\log |x|$, $b=\log |y|$; then set $\xi =a+b$, $\eta =a-b$. The equation becomes $4f_{\xi \eta }=0$. Thus an equivalent general local answer on every such quadrant isHere $\Phi ,\Psi$ are arbitrary $C^2$ functions on the relevant intervals. If a solution is required across $y=0$, its quadrant expressions must extend with matching derivatives; on that line the equation itself imposes $x{f}_{xx}(x,0)+f_x(x,0)=0$, so its trace is $a_0\log |x|+b_0$ on each interval with $x\ne 0$. The displayed coordinate transformations do not assert invertibility at $y=0$.

Let $D=d/dx$ and write the constant-coefficient linear ordinary differential equation as $p(D)y=0$. Since $D^k{e}^{\lambda x}={\lambda }^k{e}^{\lambda x}$,The exponential never vanishes, so $\boxed{e^{\lambda x}\text{ solves the equation }⟺p(\lambda )=0}$.

For the second claim, the product rule givesfor $k\ge 1$, with the $k=0$ term treated separately. Summing the coefficients yieldsA root of multiplicity at least two has $p(\mu )=p^{\prime }(\mu )=0$, proving $x{e}^{\mu x}$ is also a solution. This directly establishes the repeated-root contribution without assuming a solution formula. If the displayed coefficient list has leading zeros, the exponential calculation still applies to its actual polynomial degree.

Retain $\Delta =\mathrm{div}\nabla$ and define the heat operator by $L={\partial }_t-{\Delta }_q$. A Riemannian heat kernel $H(p,q,t)$ satisfies $L_{q}H=0$ for $t>0$ and has the delta initial limitOn standard Euclidean space the kernel isOn the Riemannian manifold, the Gaussian function factor in the local construction uses $r(p,q)=\mathrm{dist}(p,q)$ in place of $|p-q|$.

Choose an open neighbourhood $\mathcal{U}$ of the diagonal such that every pair $(p,q)\in \mathcal{U}$ is joined by a unique short minimizing geodesic and $q={\exp }_{p}x$ with the inverse exponential map smooth in both variables. It can be obtained from normal coordinates tubes of radius smaller than the local injectivity radius, or from normal neighbourhoods with geodesic convexity. Then $r^2$ is smooth on $\mathcal{U}$. Write the Riemannian volume form in normal coordinates as $d{V}_q=J(p,x)dx$, where $J>0$ is smooth and $J(p,0)=1$.

For fixed $p$, radial divergence in normal coordinates givesFor $g=(4\pi t{)}^{-d/2}{e}^{-r^2/(4t)}$, use $|\nabla r|=1$ away from the diagonal and the product rule to obtainThe identities extend smoothly in the expressions involving $r{\partial }_r$, even though $r$ itself is not smooth on the diagonal. The singular powers cancel ifThe correct initial normalization is $w_0(p,p)=1$. These heat-kernel transport equations have the explicit solutionsand, recursively,To derive the integral, multiply the radial equation by $r^{i-1}{J}^{1/2}$ and integrate the derivative of $r^i{J}^{1/2}{w}_i$ from zero to $r$. The zero integration constant is forced by smoothness when $i\ge 1$. Changing variables from radial distance to $s\in [0,1]$ gives the displayed nonsingular expression. Its integrand is smooth in $(p,x,s)$, so differentiation under the integral proves inductively that every $w_i$ is smooth across the diagonal. In particular $w_i(p,p)=({\Delta }_q{w}_{i-1})(p,p)/i$.

Summing the product-rule identity makes adjacent terms telescope. For $S_k=g{\sum }_{i=0}^k{t}^i{w}_i$, the only term remaining isThis also fixes the sign convention: with the nonnegative Laplace-Beltrami operator $P=-\Delta$, the right-hand side would instead be $+g{t}^k{P}_q{w}_k$.

A heat parametrix is an approximate kernel with the same delta initial limit and an error under $L$ that is sufficiently regular, or flat, at $t=0$ to be corrected by time convolution. On a closed compact manifold choose a smooth cutoff $\chi (p,q)$ equal to one near the diagonal and supported in $\mathcal{U}$. A finite-order explicit parametrix isNear the diagonal its error is $-\chi g{t}^N{\Delta }_q{w}_N$. Derivatives of $\chi$ produce errors supported a positive distance from the diagonal, where $e^{-r^2/(4t)}$ decays faster than every power of $t$. Taking $N$ large makes the whole error as regular at time zero as any specified finite number of derivatives requires. Its initial delta limit follows from $J(p,0)=w_0(p,p)=1$ and the Euclidean Gaussian function change of variables $x=\sqrt{t}y$.

One may obtain a smooth error flat at zero by summing all coefficients with time cutoffs. Let $\rho \in C^{\infty }([0,\infty ))$ equal one near zero and zero for arguments at least one, and choose ${ϵ}_i\downarrow 0$ sufficiently fast. SetFor each $t>0$ the sum is locally finite. Choosing ${ϵ}_i$ successively so that the $i$th term and its derivatives through order $\lfloor i/2\rfloor$ have bounds $2^{-i}$ on their cutoff transition regions gives the usual smooth asymptotic sum. For any fixed derivative order, its tail is then smaller than an arbitrarily high power of $t$; the transport identities cancel all earlier orders. Consequently $R=L_{q}P$ extends smoothly and flatly to $t=0$.

Here is the Volterra parametrix correction, with the order of the Volterra convolution of kernels factors specified. DefineSince $L$ acts in the second spatial variable and $P$ has the delta initial limit, $L_q(Q\ast P)=Q+Q\ast R$. ThusIndeed, $Q+R+Q\ast R=0$ by the geometric-series cancellation, whence $L_{q}H=0$. If $|R|\le C$ on a bounded time interval and $v=\mathrm{vol}(M)$, thenThis proves convergence; the smooth flat error permits the corresponding derivative estimates. The correction has zero initial limit, so $H$ retains the delta initial condition. Uniqueness of the heat equation on a closed compact manifold identifies this kernel with the global Riemannian heat kernel.

The local coefficient construction requires no compactness. The global uniformly bounded convolution argument just given uses a closed compact manifold, the usual spectral-geometry setting. For noncompact or incomplete Riemannian manifolds one must specify the heat realization and justify global convergence separately; a canonical choice is the minimal kernel obtained as the increasing limit of Dirichlet kernels on a smooth relatively compact exhaustion. This gives the minimal heat semigroup rather than asserting an unqualified uniqueness statement at infinity or at a missing boundary.

Away from zero, the product and composition of the displayed smooth functions are continuous and differentiable. At zero,so $f$ remains continuous. Hence the continuity set is $\boxed{\mathbb{R}}$.

The difference quotient at zero is $f(h)/h=\sin (1/h)$. For $h_n=(\pi /2+2\pi n{)}^{-1}$ it equals one, and for $k_n=(3\pi /2+2\pi n{)}^{-1}$ it equals minus one. Both sequences tend to zero, so no derivative exists there. For $x\ne 0$, the product rule and chain rule giveThus the differentiability set is $\boxed{\mathbb{R}\setminus \{0\}}$.

The fundamental theorem of calculus connects the Riemann integral with differentiation. Its first part says that for continuous $h:[a,b]\to \mathbb{R}$, the accumulation function $H(x)={\int }_a^{x}h(t)dt$ is continuous on $[a,b]$ and differentiable inside it, withTo prove this, boundedness of $h$ gives $|H(y)-H(x)|\le \Vert h{\Vert }_{\infty }|y-x|$, proving continuity. For an interior point and a sufficiently small nonzero increment $u$,Its absolute value is at most ${\sup }_{|t-x|\le |u|}|h(t)-h(x)|$, which tends to zero by continuity of $h$ at $x$. This proves the derivative formula. The same estimate gives the appropriate one-sided endpoint derivatives.

The second part says that if $Q$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $Q^{\prime }=h$ with $h$ continuous, thenIndeed $(Q-H{)}^{\prime }=0$, so the mean value theorem makes $Q-H$ constant, and evaluating at the endpoints proves the formula. A useful stronger version only requires $Q^{\prime }$ to be Riemann integrable. For a partition $a=x_0<\cdots <x_N=b$, the mean value theorem on each subinterval provides ${\xi }_j\in (x_{j-1},x_j)$ such thatAs the mesh tends to zero, these tagged Riemann sums converge to ${\int }_a^b{Q}^{\prime }$, proving that version too. Its integrability hypothesis must not be omitted.

An integrable integrand need not give an everywhere differentiable accumulation function. For a counterexample, let $f(t)=0$ for $t<1/2$ and $f(t)=1$ for $t\ge 1/2$. This bounded step function is Riemann integrable, butIts left and right derivatives at $1/2$ are zero and one. Thus $F$ need not be differentiable at every interior point. The first part of the fundamental theorem of calculus does apply at every point where the integrand is continuous.

A derivative need not have a Riemann integral. Define on all of $\mathbb{R}$At zero $f(h)/h=h\sin (1/h^2)\to 0$, so $f^{\prime }(0)=0$; elsewhere the product rule and chain rule giveAt $x_n=(2\pi n{)}^{-1/2}$, this equals $-2/x_n\to -\infty$. Hence $f^{\prime }$ is unbounded on $[0,1]$, whereas a properly Riemann integrable function on a compact interval must be bounded. The requested Riemann integral therefore need not exist, even though $f$ is differentiable everywhere. This example is an unbounded derivative obstruction to Riemann integrability; it does not contradict the theorem's version that explicitly assumes integrability of the derivative.

Use suffix notation, with repeated indices summed. Contracting the two Levi-Civita symbols givesApplying the product rule to $h=A_i{B}_i$ and substituting the two evolution equations yieldsIn the second line we renamed dummy indices in $A_i{B}_j{\partial }_j{u}_i$ and then combined product derivatives. HenceNo zero-divergence assumption was used in this local identity.

When both fields are solenoidal vector fields, the product rule for divergence rewrites this as a conservation law:Because the volume is fixed, differentiation under its integral and the divergence theorem giveunder the two tangency conditions. Thus the volume integral is conserved. This is helicity conservation by tangent boundary conditions; when $\mathbf{B}=\nabla \times \mathbf{A}$, the density is the magnetic helicity density.

For the given axisymmetric field, the curl in spherical coordinates givesThere are no $\varphi$ derivatives and $A_r=0$. ConsequentlyThe volume form in spherical polar coordinates is $r^2\sin \theta drd\theta d\varphi$, soHere $B_r=0$ at $r=a$, as required for the magnetic-field tangency condition in the conservation argument; tangency of $\mathbf{u}$ is a separate hypothesis.

In Cartesian coordinates, independent coordinate variables satisfywhere ${\delta }_{ij}$ is the Kronecker delta. Applying the product rule and chain rule to $u_i=r^{\alpha }{x}_i$ gives the Jacobian matrixThis tensor is symmetric in $i,j$. Hence its contraction with the antisymmetric Levi-Civita symbol vanishes:Similarly, $v_i={ϵ}_{ijk}{k}_j{u}_k$ and the constancy of $\mathbf{k}$ giveThe trace of the Jacobian matrix iswhich vanishes for $\alpha =-3$. The curl of a cross product identity, with $\mathbf{k}$ constant, now givesFor $\alpha =-3$,soAll identities for this singular case hold on $r>0$. In particular, zero divergence here is a pointwise assertion away from the origin, not an assertion that the flux through a sphere enclosing the singularity is zero.

Since the common radius of convergence is infinite, the termwise differentiation of a power series theorem applies on the entire real line. For the even series, differentiation and the index change $m=n-1$ giveThe odd series similarly gives $g^{\prime }(x)={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n}/(2n)!=f(x)$. Thus $\boxed{f^{\prime }=-g,g^{\prime }=f}$ everywhere. By the product rule,The mean value theorem implies that a differentiable function with zero derivative is constant on the real line. At zero, the series give $f(0)=1$ and $g(0)=0$. Consequently $\boxed{f(x{)}^2+g(x{)}^2=1\text{ for every }x\in \mathbb{R}}$. This derives the identity from the power series without assuming a trigonometric identity in advance.

Use the Einstein summation convention and the contraction of two Levi-Civita symbols. For the first cross product term,The last term cancels $(F\cdot \nabla )G_i$. Interchanging $F,G$ gives the analogous cancellation with $(G\cdot \nabla )F_i$. The surviving terms are $F_j{\partial }_i{G}_j+G_j{\partial }_i{F}_j={\partial }_i(F_j{G}_j)$ by the product rule. This proves the dot-product gradient identityThe proof only requires continuously differentiable vector fields; neither field needs to be an irrotational vector field.