For $f=u+iv$ holomorphic in $z=x+iy$, the Cauchy-Riemann equations are $u_x=v_y$ and $u_y=-v_x$. Therefore the normal vectors to the two level curves obeyAt a point where $f^{\prime }\ne 0$, both gradients are nonzero, so their perpendicular tangent lines are also perpendicular. At critical points this argument does not assert that regular level curves exist.

TakeIts real part is $u=x^2-y^2$, so the specified first family is $u=-\alpha$. Its imaginary part is $v=2xy$, so the second family is $v=2\beta$. Since $\beta \ne 0$, their intersections exclude the origin, where $f^{\prime }(z)=2z$ would vanish. Thus these hyperbolas meet orthogonally as required.

For the chosen Fourier transform convention, Parseval's identity is $\int |f{|}^{2}dx=(2\pi {)}^{-1}\int |\tilde{f}{|}^{2}dk$. For smooth rapidly decaying functions its bilinear form follows by inserting the inverse transform of $g$ into $\int f\overline{g}dx$ and interchanging the integrals:Take $g=f$. Approximation by such functions extends the identity to L2 space, which includes the compact triangular function here.

For that triangle, $\int |f{|}^{2}dx=2{\int }_0^1(1-x{)}^{2}dx=2/3$, while $\tilde{f}(k)=[\sin (k/2)/(k/2){]}^2$. ThusgivingThe continuous limiting integrand at zero is one. This is the sinc fourth-power integral.

The factor satisfying the left condition is $x$, while $x^{-1}$ satisfies decay at infinity. Matching the value and unit derivative jump givesFor the specified compactly supported forcing, integrating over $0<\xi <1$ givesFor example, on $0<x<1$, the two pieces contribute $-x^2/6$ and $-x(1-x)/2$. Both $y_1$ and $y_1^{\prime }$ match at one, and the function tends to zero at both ends. A homogeneous difference $ax+b/x$ with those two limits must vanish, establishing uniqueness.

Each spatial eigenfunction acquires the temporal factor $e^{-\kappa {\lambda }_{mn}t}$ under the heat equation. Expanding the initial constant using the preceding coefficients gives the cooling of an initially uniform square:For $t>0$ the exponential factors justify differentiating the series and enforce the zero boundary temperature. As $t\downarrow 0$, it tends to $T_0$ at every interior point and in L2 space; the instantaneous boundary cooling is incompatible with continuity at the initial boundary corners, which is expected for this idealization.

The slowest decaying nonzero mode is $(1,1)$, givingat each fixed interior point when $T_0\ne 0$. The next contributing modes have $(m,n)=(1,3),(3,1)$ and decay as $e^{-10\kappa {\pi }^{2}t/{\ell }^2}$, making the leading-mode approximation exponentially accurate.

Multiply by the $(m,n)$ sine product and integrate over the square. Fourier orthogonality gives squared norm ${\ell }^2/4$, and thusThereforeThis double Fourier sine series equals one at interior points. At boundary points every summand vanishes, consistent with the endpoint value of the sine extension.

Take a nonzero separated solution $\phi =X(x)Y(y)$. Dividing by $XY$ separates the Laplacian eigenfunction equation into $X^{\prime \prime }=-\mu X$ and $Y^{\prime \prime }=-(\lambda -\mu )Y$. The zero-endpoint Dirichlet boundary conditions admit no nonzero linear or hyperbolic factors, so both separation constants must be positive squares. HenceAll separated nonzero solutions have this form. At a repeated eigenvalue, linear combinations of the corresponding modes are also solutions, though they need not remain separable.

The outward area element is $d\mathbf{S}=a^2\sin \theta d\theta d\varphi {\mathbf{e}}_r$. ThereforeSplit the sphere into the northern cap $0\le \theta \le \alpha$ and the southern cap $\alpha \le \theta \le \pi$. Their induced boundary orientations are $+C$ and $-C$, respectively. The Stokes theorem applies to ${\mathbf{A}}^{(+)}$ on the north and ${\mathbf{A}}^{(-)}$ on the south. Adding their fluxes gives $2\pi (1-\cos \alpha )+2\pi (1+\cos \alpha )=4\pi$, agreeing with direct integration.

There is no contradiction with the flux of a globally smooth curl through a closed surface: neither potential is regular on the entire sphere. On their overlap they differ by $2{\mathbf{e}}_{\varphi }/(r\sin \theta )=2\nabla \varphi$, a local gauge transformation. The longitude is not a globally single-valued scalar, so that difference can have nonzero circulation around $C$.

For an axisymmetric azimuthal vector field, the curl in spherical coordinates has componentsThe angular expressions are $\sin \theta A_{\varphi }^{(+)}=(1-\cos \theta )/r$ and $\sin \theta A_{\varphi }^{(-)}=-(1+\cos \theta )/r$, each with $\theta$ derivative $\sin \theta /r$. Also $r{A}_{\varphi }$ is independent of $r$, so the second component vanishes. Consequently, on each potential's regular patch,

Along the circle with increasing $\varphi$, the line element is $d\mathbf{r}=a\sin \alpha {\mathbf{e}}_{\varphi }d\varphi$. Thus the two circulations areThe half-angle identities simplify them toHere $0<\alpha <\pi$ so both potentials are defined on the whole curve.

On the sphere, constant $\theta$ gives a latitude circle of radius $a\sin \theta$ and height $a\cos \theta$. Constant $\varphi$ gives a meridian semicircle in a vertical plane through the $z$ axis. The orthonormal basis of spherical coordinates isHere ${\mathbf{e}}_r$ points outward, ${\mathbf{e}}_{\theta }$ along increasing polar angle, and ${\mathbf{e}}_{\varphi }$ along increasing azimuth. At either pole the radial unit vector is unambiguous, but the two tangential coordinate vectors depend on the undefined longitude. The figure shows their directions away from the poles.

This coordinate ambiguity does not make a vector field singular if its tangential coefficient vanishes suitably. In Cartesian coordinates,The first is smooth at the north pole and zero there; the second is smooth and zero at the south pole. Their actual singular axes are the opposite ones. These are the regular patches of a two-patch vector potential for a radial monopole.

Set $a_n=1-1/(2n)$, $b_n=1+1/(2n)$. The finite pulse has Laplace transform ${\bar{S}}_n(p)=n(e^{-p{a}_n}-e^{-p{b}_n})/p$. Since $f(0)=0$, transforming the equation givesThe partial fraction decomposition $1/[p(p-1)]=1/(p-1)-1/p$ has inverse $e^t-1$. By the Laplace transform time-shift rule,This finite-pulse response of a first-order linear equation gives the requested outer regions:For completeness, while the pulse is active the solution is $n(e^{t-a_n}-1)$. It is continuous at both switching times. After the forcing stops it grows as $e^t$, because the homogeneous equation is $f^{\prime }=f$. The apparent pole at $p=0$ in the full transform is removable; the Bromwich contour may be taken with $c>1$.

The unilateral Laplace transform is $\bar{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ in its region of absolute convergence of a Laplace transform. Under the usual inversion hypotheses, the Bromwich inversion formula isThe vertical Bromwich contour is directed upward, with $c$ inside the transform's convergence half-plane and to the right of all its singularities. At discontinuities the standard symmetric inversion gives the mean of the one-sided values.

For a piecewise continuously differentiable $f$ with an integrable transformed derivative, integration by parts gives the Laplace transform of a derivative:Here the upper endpoint vanishes by the given decay condition, and right continuity at zero supplies the lower endpoint.

The square-root factor gives an algebraic branch point at $z=0$. The logarithm is locally analytic there but has value an odd multiple of $i\pi$, which is nonzero, so it cannot cancel that branching. The zero and pole of $(z-1)/(z+1)$ give logarithmic branch points at $z=1$ and $z=-1$, respectively. Thus the finite singularities areInfinity is also a branch point. On the logarithm branch tending to zero there, $\log [(z-1)/(z+1)]=-2/z+O(z^{-3})$, so the product behaves as $-2z^{-1/2}$. Other logarithm branches add $2\pi ik\sqrt{z}$ and retain branching. A chosen branch cut joins or separates these branch points; its ordinary interior points are not additional branch points of the underlying multivalued function.

Put $r_0=|{\mathbf{x}}_0|$ and ${\mathbf{x}}_0^{\ast }=a^2{\mathbf{x}}_0/r_0^2$ for $0<r_0<a$. The Dirichlet Green function inside a sphere isOn $|\mathbf{x}|=a$, the distance identity $|\mathbf{x}-{\mathbf{x}}_0^{\ast }|=(a/r_0)|\mathbf{x}-{\mathbf{x}}_0|$ cancels the two terms. The inverse point lies outside the sphere, so it introduces no interior source. This weighted inversion is the Kelvin transform. For a central source, take the limit to get $G_B(\mathbf{x};0)=-(4\pi {)}^{-1}(1/|\mathbf{x}|-1/a)$.

Let ${\mathbf{x}}_0^{-}=(x_0,y_0,-z_0)$ be the reflection of the source across the plane. The method of images gives the Dirichlet Green functionThe two source distances agree on $z=0$, so the boundary value vanishes. The reflected source lies outside the upper half-space, leaving only the required Dirac delta distribution inside. Both terms vanish at infinity.

Multiplying by $y$ and integrating with the zero endpoint values first gives $\lambda {\int }_1^b{y}^{2}dx/x={\int }_1^{b}x(y^{\prime }{)}^{2}dx>0$ for a nonzero real eigenfunction; hence $\lambda >0$. Using the printed change of variables $\xi =\lambda \ln x$, write $y(x)=Y(\xi )$. Then $(x{y}^{\prime }{)}^{\prime }={\lambda }^2{Y}_{\xi \xi }/x$, so $Y_{\xi \xi }+Y/\lambda =0$. The first endpoint gives $Y=A\sin (\xi /\sqrt{\lambda })$, and the other gives $\sin (\sqrt{\lambda }\ln b)=0$. ThusThis is the logarithmic change of variable in a Sturm-Liouville problem. The measure $dx/x=d(\ln x)$ gives squared weighted norm $A_n^2\ln b/2$, so an orthonormal choice is $A_n=\sqrt{2/\ln b}$.

Multiply the equation for $y_n$ by $y_m$ and the equation for $y_m$ by $y_n$, then subtract. The common $h$ terms cancel, and integration by parts gives the Lagrange identity for a Sturm-Liouville operator:At each endpoint the two pairs $(y_n^{\prime },y_n)$ and $(y_m^{\prime },y_m)$ satisfy the same nontrivial homogeneous linear equation. They therefore lie on the same one-dimensional line, and their determinant vanishes. The boundary expression is zero. Since the eigenvalues are distinct, division gives the required orthogonality of Sturm-Liouville eigenfunctions:For complex-valued eigenfunctions of a real self-adjoint problem one may choose real eigenfunctions, or use the conjugated version as a weighted inner product.

The product rule gives ${\psi }^{\prime }=e^{-x^2/2}(y^{\prime }-xy)$ and ${\psi }^{\prime \prime }=e^{-x^2/2}[y^{\prime \prime }-2x{y}^{\prime }+(x^2-1)y]$. Substituting in the Schrödinger equation cancels the $x^2$ and constant terms, leaving the Hermite differential equationFor $y={\sum }_{n\ge 0}{a}_n{x}^n$, matching coefficients givesThe power series solutions of the Hermite differential equation split into independent even and odd series. With initial data $(y(0),y^{\prime }(0))=(1,0)$ and $(0,1)$, respectively, their first three terms areEach nonterminating parity series has infinite radius of convergence: the ratio of successive terms of that parity has magnitude asymptotic to $2|x{|}^2/n$. Their initial data prove linear independence. If a coefficient shown vanishes at a special parameter, the recurrence supplies the terminating series instead.

A nonzero polynomial solution of degree $N$ requires $\lambda =N$, by its highest-degree term, and the recurrence then terminates the matching parity. Thus all possible values are $\lambda =0,1,2,\dots$. Up to nonzero scalar multiples, the degree-at-most-three solutions are the Hermite polynomials