Put $s=z-1$ and seek the leading Frobenius series term $y\sim s^{\rho }$. The coefficients of $y^{\prime \prime }$ and $y^{\prime }$ become $-2s(1+s)$ and $2+s$, respectively. The lowest power, $s^{\rho -1}$, therefore has coefficient $-2\rho (\rho -1)+2\rho$. The indicial equation isThe term $-y$ contributes only at the next order and does not enter this equation.

Take the spatial Fourier transform with the same convention. The partial differential equation becomes $i{\partial }_t\hat{f}-k^2\hat{f}=0$, so $\hat{f}(k,t)=e^{-i{k}^{2}t}{\hat{f}}_0(k)$. Inverting and substituting the transform of the initial function yields the convolutionComplete the square, $iks-it{k}^2=i{s}^2/(4t)-it[k-s/(2t){]}^2$, and put $\ell =\sqrt{t}[k-s/(2t)]$. This givesThe oscillatory Gaussian is a Fresnel integral, understood as the limit with a positive Gaussian damping factor. That prescription also justifies the transform manipulations, first for rapidly decreasing smooth data and then by L2 space extension. The resulting kernel is the free-particle propagator in units $\hslash =1$, $m=1/2$; its phase and normalization agree with the multiplier $e^{-i{k}^{2}t}$.

Extend $f$ evenly to $F(x)=f(|x|)$ on the whole real line, and use the Fourier transform convention $\hat{F}(k)={\int }_{\mathbb{R}}F(x)e^{-ikx}dx$. Its odd sine contribution vanishes, giving $\hat{F}(k)=2{\int }_0^{\infty }f(x)\cos (kx)dx={\hat{f}}_c(k)$ for $k>0$; the transform is also even. The Fourier inversion theorem therefore gives, for $x>0$,This is the inverse Fourier cosine transform with the factor-of-two forward convention used here.

The initial function is odd, so all $a_n$, including $a_0$, vanish. Set $k_n=n\pi /L$. For $k_n\ne 1$, product-to-sum and integration by parts giveConsequently the coefficients of the periodic sine with an endpoint-correcting linear term areInsert them into the sine terms of the preceding solution. If $L=m\pi$ for a positive integer $m$, the original initial profile is simply $\sin x$, with $k_m=1$. The apparent $0/0$ in the displayed coefficient has removable limit $b_m=1$, while $b_n=0$ for $n\ne m$. In that case

Separation of variables gives $X^{\prime \prime }=-k^{2}X$ and $T^{\prime }=-\kappa k^{2}T$. The periodic matching of both the function and its first derivative selects $k=n\pi /L$, with the constant mode at $n=0$. Thus the periodic diffusion on a circle solution iswhere the Fourier coefficients of the initial function areThe formula for $a_n$ also gives $a_0$ at $n=0$. The constant mode preserves the initial spatial mean; every nonconstant Fourier mode decays. For initial data in L2 space, the series approaches those data in L2 space as $t\downarrow 0$ and is smooth at positive times.

Use the right-handed orthonormal basis $({\mathbf{e}}_r,{\mathbf{e}}_{\theta },{\mathbf{e}}_{\varphi })$. The cross product with ${\mathbf{e}}_r$ rotates tangential components by a right angle; taking its cross product again in the opposite order projects a gradient onto the tangent plane. The radial curl of angular gradient fields then depends only on the rotated-gradient part. All coordinate calculations are made away from the polar coordinate singularities and extend wherever the original fields are smooth.

Away from the origin, $\hat{\mathbf{r}}=\nabla r$. Hence the curl of a gradient gives $\nabla \times \hat{\mathbf{r}}=0$. Its divergence is the Laplacian of the radial coordinate, which in spherical coordinates isEquivalently, differentiating $x_i/r$ in Cartesian coordinates gives $3/r-r^2/r^3=2/r$.

Suppose $x,y$ have order three but $x\ne y,y^2$. Their cyclic subgroups have trivial intersection: any common nonidentity element would generate both subgroups and force $x=y$ or $y^2$. The nine products $x^i{y}^j$, $0\le i,j<3$, are distinct. Indeed $x^i{y}^j=x^k{y}^{\ell }$ implies $x^{i-k}=y^{\ell -j}$ in that intersection, so $i=k$ and $j=\ell$ modulo three. Nine distinct elements cannot fit in a group of order six, proving the claim without assuming commutativity.

In a group of order six, element orders divide six. If there were no element of order three or six, every nonidentity element would have order two, making the group abelian by the preceding argument. Choose $a\ne e$ and $b\notin \{e,a\}$. Then $\{e,a,b,ab\}$ is a subgroup of order four: its elements are distinct, it is closed because $a,b$ commute and square to $e$, and each element is its own inverse. This contradicts Lagrange's theorem, since four does not divide six. Hence an element of order three or six exists.

For proper rotations the Levi-Civita symbol is invariant, so contracting the two rank-one tensors ${\partial }_j$ and $v_k$ gives $c_i={ϵ}_{ijk}{\partial }_j{v}_k$ with transformation $c_i^{\prime }=R_{i\ell }{c}_{\ell }$. Thus the curl is a vector under proper rotations. Under an orientation-reversing orthogonal change it acquires an additional determinant sign: it is an axial vector when $v$ is a polar vector.

Commuting smooth partial derivatives and using epsilon contraction gives the curl of the curl identity:Now take the curl of Faraday's law and use the vacuum Ampère-Maxwell equation: $\nabla \times \nabla \times E=-{\partial }_t(\nabla \times B)=-c^{-2}{E}_{tt}$. Since $\nabla \cdot E=0$, the identity gives $\Delta E-c^{-2}{E}_{tt}=0$. Taking the curl of the Ampère-Maxwell equation gives the same result for $B$, using $\nabla \cdot B=0$. Hence both satisfy the electromagnetic wave equation

The unique symmetric and antisymmetric decomposition is $S=(T+T^T)/2$, $A=(T-T^T)/2$. Both pieces are Cartesian tensors: transposing $T^{\prime }=RT{R}^T$ gives $(T^{\prime }{)}^T=R{T}^T{R}^T$, so $S^{\prime }=RS{R}^T$ and $A^{\prime }=RA{R}^T$. Hence this is a decomposition into a symmetric tensor and an antisymmetric second-rank tensor, rather than just a decomposition of numerical arrays.

For the axial vector of an antisymmetric second-rank tensor, contract the two Levi-Civita symbols:ThusWith this convention $Ax=x\times a$, so the displayed matrix is the negative of the cross-product matrix for $a\times x$.

For a Cartesian tensor under a proper orthogonal change $R$, the rank-$n$ law is $T_{i_1\cdots i_n}^{\prime }=R_{i_1{j}_1}\cdots R_{i_n{j}_n}{T}_{j_1\cdots j_n}$, with repeated indices summed. An isotropic tensor has $T^{\prime }=T$ for every such rotation.

For rank two, $T=RT{R}^T$. A half-turn about a coordinate axis, obtained by applying a quarter-turn twice, reverses each off-diagonal component with exactly one index on that axis. Invariance forces every off-diagonal entry to vanish. A quarter-turn about the third axis exchanges $T_{11}$ and $T_{22}$; quarter-turns about the other axes equate them with $T_{33}$. Thus the isotropic second-rank tensor isConversely $R(\lambda I)R^T=\lambda I$, proving sufficiency.

A holomorphic function is complex differentiable throughout an open region; equivalently it is locally represented by a convergent complex power series. For $f=u+iv$, the Cauchy-Riemann equations are $u_x=v_y$, $u_y=-v_x$. With continuous first partial derivatives, these equations imply holomorphicity and $f^{\prime }=u_x+i{v}_x$.

For $e^z=e^x\cos y+i{e}^x\sin y$, the equations hold and $u_x+i{v}_x=e^x(\cos y+i\sin y)=e^z$. For $\sin z=\sin x\cosh y+i\cos x\sinh y$, they also hold, and $u_x+i{v}_x=\cos x\cosh y-i\sin x\sinh y=\cos z$. Therefore

Substituting $a_n=2^n/n!$ into the $r=0$ recurrence, and $a_n=1/n!$ into the $r=1/2$ recurrence, satisfies every coefficient equation. They give $e^{2x}$ and $\sqrt{x}{e}^x$. Thus on a positive interval not crossing a singular point,Their Wronskian is $e^{3x}(2x-1)/(2\sqrt{x})$, nonzero for $x>0$, $x\ne 1/2$, proving independence on each ordinary interval. Both functions are analytic near $1/2$, so this is an apparent regular singular point: a pole in normalized coefficients need not force a singular solution. The Wronskian can vanish at that singular point without making the two functions linearly dependent on an ordinary interval.

The coefficients in the Taylor series areFor either series, the ratio test gives absolute term ratio $|x|/(n+1)$ or $2|x|/(n+1)$, tending to zero for each fixed real or complex $x$. Both have infinite radius of convergence.

Use the Frobenius method and write $y=x^r{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$, with coefficients at negative indices defined as zero. The lowest power gives the indicial equation $r(2r-1)=0$, so $r=0$ or $r=1/2$. Equating the coefficient of $x^{n+r-1}$ gives, for $n\ge 1$,For $r=0$ and $a_0=1$, the first coefficients are $1,2,2,4/3,\dots$. For $r=1/2$ they are $1,1,1/2,1/6,\dots$, multiplying $x^{1/2}$. The distinct nonintegrally separated indicial roots give two independent Frobenius series. The half-power solution is a generalized power series, rather than an ordinary Taylor series at zero.

Put $\beta =\Vert b\Vert >0$ and choose an oriented orthonormal basis with third vector $b/\beta$. The Levi-Civita symbol term has action ${ϵ}_{ijk}{b}_k{x}_j=x\times b$, the negative of $b\times x$. Thus this axisymmetric linear map has block formThe transverse eigenvalues $1\pm i\beta$ are not real. The only real eigenvalue and its eigenvectors are thereforeThe determinants are $\det B^{+}=(1+{\beta }^2{)}^2>0$ and $\det B^{-}=(1-{\beta }^2)(1+{\beta }^2)$, proving that $B^{+}$ is always invertible, and $B^{-}$ is invertible exactly when $\Vert b\Vert \ne 1$. The complex transverse pair is important for this determinant even though the question asks for real eigenvalues.

Decompose ${\mathbb{R}}^n=\mathrm{span}\{a\}\oplus a^{\perp }$. The rank-one matrix update $a{a}^T$ acts as multiplication by $\Vert a{\Vert }^2$ along $a$ and as zero on its orthogonal complement. ThusThe perpendicular eigenvalue has multiplicity $n-1$ and is absent when $n=1$. Consequently $A^{+}$ is always an invertible matrix, while

For the curl of a cross product, the Cartesian contraction isThe product rule then givesFor the fluid equation, replace the advective term by $\mathbf{w}\times \mathbf{u}+\nabla (|\mathbf{u}{|}^2/2)$ using the preceding identity. Take its curl. Both gradient terms have zero curl, while $\nabla \cdot \mathbf{w}=\nabla \cdot (\nabla \times \mathbf{u})=0$ and incompressibility gives $\nabla \cdot \mathbf{u}=0$. The vorticity equation is consequentlyThe left side is the material derivative of vorticity; the right is vortex stretching. It follows here because the given force per mass is a gradient, in addition to the flow being inviscid and incompressible.

Use $w_j={ϵ}_{j\ell m}{\partial }_{\ell }{u}_m$ and contract the Levi-Civita symbols:This proves the convective acceleration identity, with the sign set by the displayed order $\mathbf{w}\times \mathbf{u}$.