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For a differentiable vector field in Cartesian coordinates, the divergence and curl arewith summation over repeated indices. In components the curl is $({\partial }_y{F}_z-{\partial }_z{F}_y,{\partial }_z{F}_x-{\partial }_x{F}_z,{\partial }_x{F}_y-{\partial }_y{F}_x)$. The contraction of two Levi-Civita symbols and the product rule giveRecognizing the directional derivatives gives the curl of a cross product identity

In Cartesian coordinates, use the Levi-Civita symbol and the Einstein summation convention to expandThe contraction of the two symbols uses the Kronecker delta, and the final step is the product rule. Thus the convective acceleration identity isPut $\mathbf{\omega }=\nabla \times \mathbf{u}$. The curl of a gradient vanishes, and the curl of a cross product identity givesSince the divergence of a curl is zero, taking the curl of the first identity yieldsTwice continuous differentiability suffices for the mixed-derivative identities used here.

For a twice continuously differentiable vector field in Cartesian coordinates, the curl and divergence arePut $\mathbf{\omega }=\nabla \times \mathbf{v}$. Using the Levi-Civita symbol, the Kronecker delta contraction and the product rule,The last term vanishes because the divergence of a curl is zero: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the curl of a cross product identity becomes

For orthogonal coordinates $q_i$ with position vector $\mathbf{r}(q_1,q_2,q_3)$, the scale factors of orthogonal coordinates and local unit vectors areOrthogonality gives $d{s}^2={\sum }_i{h}_i^{2}d{q}_i^2$, and right-handedness fixes ${\mathbf{e}}_1\times {\mathbf{e}}_2={\mathbf{e}}_3$. In cylindrical coordinates, $\mathbf{r}=(\rho \cos \varphi ,\rho \sin \varphi ,z)$, soThe radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\varphi$; ${\mathbf{e}}_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.

For the axial velocity profile, the cylindrical coordinates formula givesThe associated cross product isThis field has only a radial component depending on $\rho$, so its curl is zero. Also $\nabla \cdot \mathbf{u}={\partial }_z{u}_z=0$. The identity above therefore yieldsIn fact both directional derivatives vanish: the first differentiates $u_z{\mathbf{e}}_z$ only in the azimuthal direction, and the second differentiates $\mathbf{\omega }$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho =0$, despite the coordinate-basis singularity.

The Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. A holomorphic function is smooth, so equality of mixed partial derivatives givesThus its real and imaginary parts are harmonic functions.

An entire function is holomorphic at every finite point of the complex plane. The globally convergent power series for the exponential implies that $e^z$ and $e^{-z}$ are entire. Their difference therefore givesat every complex $z$, proving that the hyperbolic sine is entire. For powers, the base case $f_1(z)=z$ has derivative one everywhere. If $z^n$ is entire with derivative $n{z}^{n-1}$, the product rule gives $(z^{n+1}{)}^{\prime }=z^n+zn{z}^{n-1}=(n+1)z^n$ everywhere. Induction proves that every positive integer power is entire.

A function has a pole of order $n$ at $z_0$ when it is holomorphic in a punctured neighbourhood and $g(z)=(z-z_0{)}^{n}f(z)$ extends holomorphically to $z_0$ with $g(z_0)\ne 0$. An isolated essential singularity is neither a removable singularity nor a pole; equivalently its Laurent series has infinitely many nonzero negative-power terms. In particularhas an essential singularity at zero.

For $f_N$, the Laurent series about infinity isTo classify the point correctly, use $w=1/z$. Then $f_N(1/w)=w^{-N}{e}^w$, and $e^w$ is holomorphic with value one at zero. ThereforeThe pole at infinity is compatible with an essential singularity at the finite point zero. Negative powers of $z$ are regular powers of the local coordinate $w$ near infinity, so they do not make that point essential.

The $k_0^{2}uv$ terms cancel in $u{L}_{H}v-v{L}_{H}u$. For a ball $B_R$, the product rule and divergence theorem give Green's second identity in the formThus the whole-space integral vanishes when this boundary term tends to zero. For example, uniform bounds $u,v=O(R^{-1-ϵ})$ and $\nabla u,\nabla v=O(R^{-1-ϵ})$, with $ϵ>0$, make it $O(R^{-2ϵ})$.

Pointwise vanishing alone, as stated in the source, is insufficient. Smooth fields agreeing with $u=\sin (k_{0}r)/r$ and $v=\cos (k_{0}r)/r$ outside a ball both tend to zero, but their distant boundary integral is $-4\pi k_0$. One may smooth them inside the ball without changing that flux. The requested zero identity therefore needs sufficient decay or boundary/radiation conditions that actually eliminate the boundary term.

For the generalized plane wave eigenfunctions, their orthogonality is distributional:Here $\int e^{iqx}dx=2\pi \delta (q)$ is the Fourier representation of the Dirac delta function. These eigenfunctions are not ordinary square-integrable functions or decaying test fields.

Expand $\Phi =\int A_k{\psi }_k{d}^{3}k$ and define $V_k=\int {\psi }_k^{\ast }(r^{\prime })V(r^{\prime })d^3{r}^{\prime }$. Since $L_H{\psi }_k=(k_0^2-k^2){\psi }_k$, projection of the source equation gives $(k_0^2-k^2)A_k=V_k$. Formally dividing and substituting back givesAt the shell $k=k_0$, a pole prescription is required; homogeneous solutions can otherwise be added. The formula is an inverse of the Helmholtz equation with that convention still to be selected.

Let $H$ denote the kernel in this source formula, so $L_{H}H=+{\delta }^{(3)}$. Put $l=|r-r^{\prime }|$ and align the Fourier polar axis with $r-r^{\prime }$. Angular integration givesConsequently the radial Fourier representation of the Helmholtz Green function givesThe last Green function written in the PDF has the opposite denominator. It is obtained by setting $G=-H$, the common convention $L_{H}G=-{\delta }^{(3)}$:Thus the two displayed source and Green formulas are consistent only when this minus sign is retained; $G$ cannot also be the positive-delta inverse kernel.

Finally the integral is not an ordinary improper integral across $k=k_0$. A Cauchy principal value yields $G_{\mathrm{PV}}(l)=\cos (k_{0}l)/(4\pi l)$. The outgoing prescription $k^2-k_0^2-i0$ yields the outgoing Green function for the three-dimensional Helmholtz equation, $G_{+}(l)=e^{i{k}_{0}l}/(4\pi l)$; the opposite sign gives the incoming wave. The Sokhotski–Plemelj formula accounts for their difference. The source does not specify which prescription is intended, so the boxed radial expression must be read with one chosen consistently.

Using the Levi-Civita symbol and the Einstein summation convention in Cartesian coordinates, the product rule givesThe cyclic permutation in the first term preserves the sign, whereas exchanging the first two indices in the second reverses it. Recognizing the two curls proves

Substitute $y=v{P}_n$ into the Legendre differential equation and expand using the product rule. The terms proportional to $v$ cancel because $P_n$ is itself a solution. What remains isMultiplying by $P_n$ identifies an exact derivative:The latter form is convenient for reduction of order on intervals where $P_n\ne 0$ and $x\ne \pm 1$.

Summing the geometric series givesThere is a particularly short verification using the product rule:Also $y(0)=1$ and $y^{\prime }(0)=0$. The radius of convergence of the series is one, set by the three complex cube roots of unity; the closed expression continues the solution beyond that disk wherever its denominator is nonzero. The real initial-value solution extends on the interval $(-\infty ,1)$.

Write the curl and cross product with the Levi-Civita symbol, then use its contraction identity:Expanding with the product rule givesRecognize the divergence in the first and third terms, and the directional derivatives in the other two. The curl of a cross product identity is thereforeThe derivatives act on both vector fields. No assumption of vanishing divergence or spatially constant fields has been made.

For a sufficiently smooth vector field $A$ on a volume with piecewise smooth closed boundary and outward unit normal vector, the divergence theorem statesExpanding the divergence of a cross product by the product rule gives $\nabla \cdot (F\times G)=G\cdot (\nabla \times F)-F\cdot (\nabla \times G)$. Apply the theorem to $A=F\times G$ to obtain

Work in a Cartesian coordinate system, write ${\partial }_j=\partial /\partial x_j$, and use the Einstein summation convention. For a twice continuously differentiable vector field, the contraction of two Levi-Civita symbols givesHere the Kronecker deltas select components, and equality of mixed partial derivatives gives the curl of the curl identityFor the second identity, the same contraction followed by the product rule yieldsThus the curl of a cross product isThe directional derivatives in the last two terms act on the vector field immediately to their right.

For the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$, the product rule givesThe mixed dot products cancel. The divergence theorem therefore yields Green's second identity,The fields and boundary are assumed sufficiently regular. A singular Green function can be treated by excising a small ball around its source and taking the radius to zero, or using distributions.

Use the product rule for the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The mixed dot products cancel. Applying the divergence theorem with the outward unit normal gives Green's second identity,The fields and boundary are assumed sufficiently regular; a Green-function singularity is handled by excising a small ball or by using distributions.

Apply the product rule to the vector field $F=\varphi \nabla \psi -\psi \nabla \varphi$:The two mixed dot products cancel. The divergence theorem, using the outward unit normal, then gives Green's second identity:The usual assumptions are a sufficiently regular boundary and fields with the derivatives and integrability needed for the theorem; singular sources can instead be treated by excising small balls or using distributions.

Apply the product rule in Einstein notation, using the Levi-Civita symbol to express the cross product:The third line uses a cyclic permutation of indices in the first term and an interchange of two indices in the second. Rearranging proves the divergence of a cross product identity:

Use the Levi-Civita symbol and the contraction of two Levi-Civita symbols in Cartesian coordinates. With repeated indices summed,The last line is the product rule. Reading its four terms in vector notation proves the curl of a cross product identity:

Suppose ${\Psi }_1$ and ${\Psi }_2$ have the same boundary data, and put $u={\Psi }_1-{\Psi }_2$. Then $\Delta u=0$ in $V$ and $u=0$ on its boundary. For real $u$, the product rule for the divergence givesThe divergence theorem therefore yieldsThe nonnegative integrand must vanish, so $\nabla u=0$ throughout $V$. On each connected component $u$ is constant, and its zero boundary value makes that constant zero. Hence $\boxed{{\Psi }_1={\Psi }_2}$, proving uniqueness. For complex fields, apply the same argument to real and imaginary parts, or use $\nabla \cdot (u^{\ast }\nabla u)$ and its real part.

This is Dirichlet uniqueness for the Poisson equation applied to the difference of two harmonic functions. The proof assumes a bounded volume with a sufficiently regular boundary and solutions. In an exterior or otherwise unbounded domain one must additionally control the boundary contribution at infinity; prescribed values only on the finite boundary do not generally ensure uniqueness.

Let $G(k)=\tilde{g}(k)$. Since $g^{\prime }=-xg$, differentiation under the Fourier transform integral and integration by parts giveThus $G(k)=G(0)e^{-k^2/2}$. The Gaussian integral gives $G(0)=\sqrt{2\pi }$, proving the Fourier transform of a Gaussian formulaFor the higher derivatives, start with $r_0=1$. If $g^{(m)}=r_{m}g$, the product rule givesIf $r_m$ has degree $m$ and leading coefficient $(-1{)}^m$, the recurrence gives degree $m+1$ and leading coefficient $(-1{)}^{m+1}$. Induction proves the assertion for every $m$. These polynomials are $r_m=(-1{)}^m{\mathrm{He}}_m$, where ${\mathrm{He}}_m$ is the Probabilists' Hermite polynomial.

Apply the product rule to the vector field $\Phi \nabla \Psi -\Psi \nabla \Phi$:The gradient cross terms cancel. Integrating and applying the divergence theorem with outward unit normal $\mathbf{n}$ proves Green's second identity:Choose the Dirichlet Green function $G_D(\mathbf{r},{\mathbf{r}}^{\prime })$ with ${\Delta }_{\mathbf{r}}{G}_D=\delta (\mathbf{r}-{\mathbf{r}}^{\prime })$ and $G_D=0$ on $S$. Taking $\Psi =G_D$ and using the Dirac delta distribution, the left side becomes $\Phi ({\mathbf{r}}^{\prime })$ while the second boundary term vanishes. HenceThe plus sign is tied to the convention $\Delta G_D=+\delta$, whose free-space fundamental solution is negative in three dimensions. The singularity may equivalently be handled by excising a small sphere about ${\mathbf{r}}^{\prime }$ before applying the smooth identity.

Since $\mathbf{a}$ is constant, the product rule gives $\nabla \cdot (\mathbf{a}\psi )=\mathbf{a}\cdot \nabla \psi$. Applying the divergence theorem therefore givesThis holds for every constant vector $\mathbf{a}$, so each component of the difference vanishes. Consequently the gradient has the boundary representation