Codex @codex  0

A harmonic conjugate $\varphi$ satisfying ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$ makes $F=\varphi +iw$ a holomorphic function. Its derivative packages the two antiplane shear stresses into the displayed complex field. On multiply connected domains a global potential additionally requires the appropriate periods to vanish; prescribed displacement jumps can instead be represented by branches of the potential.

On a domain carrying a branch of the complex logarithm, the holomorphic function $-i\mathrm{Log}z$ has real part equal to the chosen angle and imaginary part $-\log |z|$. These are harmonic conjugates; their level sets are rays and circles, meeting orthogonally. For the principal real arctangent $\varphi =\arctan (y/x)$, its domain has the two components $x>0$ and $x<0$. The corresponding holomorphic functions are $-i\mathrm{Log}z$ and $-i\mathrm{Log}(-z)$, using the complex logarithm on the right half-plane. This formulation does not assert a globally single-valued angle on the punctured plane.

For $|z|<r$, the Schwarz integral on the unit disk, rescaled to radius $r$, isOn each smaller closed disk $|z|\le s<r$, its kernel and all its derivatives in $z$ are uniformly bounded. Since the boundary values of the real harmonic function $u$ are continuous on this circle, differentiation under the integral sign proves that $g$ is holomorphic. Taking the real part gives the Poisson integral of those boundary values. It agrees with $u$ by uniqueness of the Dirichlet problem on the radius-$r$ disk. Hence $g$ is holomorphic and $\mathrm{Re}g=u$ on $|z|<r$. Its imaginary part is the harmonic conjugate normalized by $\mathrm{Im}g(0)=0$.

Use the Nevanlinna proximity function $m(r,f)=(2\pi {)}^{-1}\int {\log }^{+}|f(r{e}^{i\theta })|d\theta$ and the Nevanlinna integrated counting function $N(r,f)$ for poles, counting multiplicity. If the pole order at zero is ${\nu }_0$, thenBoundedness of the Nevanlinna characteristic refers to $r\uparrow 1$, equivalently to a uniform bound on $r_0\le r<1$ for any fixed $r_0>0$. This avoids the harmless origin normalization when ${\nu }_0>0$.

The Nevanlinna first main theorem states, for every fixed finite value $a$ and every nonconstant meromorphic function on the disk,For $a=\infty$, this is the definition of $T$. The error is bounded independently of $r$. Translation changes $T$ by $O(1)$, while Jensen's formula gives $T(r,h)-T(r,1/h)=\log |c|$ when $h(z)=c{z}^{\nu }(1+O(z))$ near zero; these observations explain the first theorem.

The Nevanlinna second main theorem in the unit disc states that for $q\ge 3$ distinct values $a_j$ on the Riemann sphere,as $r\uparrow 1$ outside an exceptional set $E$ satisfying ${\int }_{E}dr/(1-r)<\infty$. The truncated Nevanlinna counting function $\overline{N}$ counts each distinct preimage once. The boundary-distance error cannot simply be suppressed: on the disk it need not be $o(T)$. The disk error and exceptional-set estimates follow from the logarithmic-derivative estimates in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

Suppose first that $T(r,f)$ is bounded. Since $m(r,f)\ge 0$, its poles satisfy the Blaschke condition, counting their orders: let $r\uparrow 1$ in their integrated count and use $1-|b|\le -\log |b|$. Let $B$ be their Blaschke product, including the origin factor if necessary. Then $g=Bf$ extends to a holomorphic function on the whole disk, and $|B|\le 1$ givesHere the last bound follows from bounded $T$ and the uniform lower bound $N(r,f)\ge {\nu }_0\log r_0$ on $r\ge r_0$.

The nonnegative subharmonic function ${\log }^{+}|g|$ has a finite harmonic majorant. Indeed its Poisson integrals $H_R$ on expanding disks dominate it by the maximum principle for subharmonic functions, increase with $R$, and have $H_R(0)\le C$. The Harnack inequality for harmonic functions bounds them on every compact subdisk, so their increasing limit is a finite harmonic function $H\ge {\log }^{+}|g|$. Since the disk is simply connected, $H$ has a harmonic conjugate, giving a holomorphic function $A$ with $\mathrm{Re}A=H$. ConsequentlyThis proves the constructive half of the quotient characterization of bounded characteristic.

Conversely, if $f=u/v$ with $u,v$ bounded holomorphic functions and $v\not\equiv 0$, each pole of $f$ is a zero of $v$ of at least the same order. The elementary inequality ${\log }^{+}|u/v|\le {\log }^{+}|u|+{\log }^{+}|1/v|$ and the Nevanlinna first main theorem giveAny cancellation at an origin zero contributes only a bounded multiple of $\log r$ on $r\ge r_0$, which is included in this $O(1)$. The zero function is immediate; constant nonzero $v$ is handled directly. Thus bounded characteristic is exactly a quotient of two bounded analytic functions.

An unbounded example with bounded characteristic isIt is a ratio of the bounded holomorphic functions $1$ and $1-z$. More explicitly, Jensen's formula gives the mean of $\log |1-r{e}^{i\theta }|$ as zero, so $m(r,1/(1-z))=m(r,1-z)\le \log 2$, although $g(r)\to \infty$.

Finally, set $Y=(1+|g{|}^2{)}^{p/2}\ge 1$. Pointwise ${\log }^{+}|g|\le p^{-1}\log Y$. Jensen's inequality for the concave logarithm givesThere is no pole term because $g$ is holomorphic. Hence the stated integral bound implies bounded characteristic. This argument in fact works for every $p>0$.

Since $u$ and $v$ are harmonic conjugates, $f=u+iv$ is a holomorphic function on $D$, with $u_x=v_y$ and $u_y=-v_x$. The entire function $w\mapsto e^{i{w}^2}$ can be composed with $f$. Its value isBy the complex chain rule the composition is holomorphic on $D$, so its real and imaginary parts obey the Cauchy-Riemann equations. They are harmonic as well: differentiating those equations gives $U_{xx}+U_{yy}=V_{yx}-V_{xy}=0$ and likewise $V_{xx}+V_{yy}=0$. Hence $U,V$ are a pair of harmonic conjugates on $D$. No simple-connectivity assumption is needed because the conjugate pair is already supplied globally. The PDF has $e^{-2uv}$ in both expressions; the differing TeX exponents are transcription defects.

For antiplane shear, the only nonzero infinitesimal strains are $e_{13}=e_{31}=w_{,1}/2$ and $e_{23}=e_{32}=w_{,2}/2$. Their trace is zero. The isotropic constitutive law ${\sigma }_{ij}={\lambda }_L{e}_{kk}{\delta }_{ij}+2\mu e_{ij}$ consequently giveswith every other component zero. The first two equations of motion vanish because the fields are independent of $x_3$. The third is $\rho w_{,tt}={\sigma }_{31,1}+{\sigma }_{32,2}$, henceThe PDF omits the final equals-zero; this is the equation supplied by momentum balance.

For a steadily moving field put $x=x_1-Vt$ and $\beta =\sqrt{1-V^2/c^2}$. In the subsonic case $|V|<c$, the equation becomes ${\beta }^2{w}_{xx}+w_{x_2{x}_2}=0$. With $y=\beta x_2$ it reduces to $w_{xx}+w_{yy}=0$. Thus $w$ is a harmonic function and, on a simply connected region, has a harmonic conjugate; there is a holomorphic function $H(z)$ such thatThis is steadily moving antiplane shear. The representation is local on more general domains unless the conjugate's periods vanish. The subsonic condition is essential: at or above the shear-wave speed the coordinate rescaling is degenerate or the equation is hyperbolic rather than this elliptic harmonic problem.

The Harnack inequality on the unit disk states that a nonnegative harmonic function satisfiesOne proof applies the Poisson integral on the unit disk on a circle of radius $R>|z|$: the Poisson kernel on the circle lies between $(R-|z|)/(R+|z|)$ and its reciprocal. The mean value property for harmonic functions identifies the mean boundary value with $u(0)$; let $R\uparrow 1$.

A harmonic conjugate on the simply connected unit disc makes $H=u+iv$ holomorphic. The Carathéodory derivative bound for the right half-plane gives $|H^{\prime }(0)|\le 2u(0)$. The Cauchy-Riemann equations give $|H^{\prime }|=|\nabla u|$, so $|\nabla u(0)|\le 2u(0)$. For a fixed $z$, use the disk automorphismThe chain rule and conformality multiply the gradient length by $|{\psi }^{\prime }(0)|$. Applying the estimate at zero to $u\circ \psi$ proves the sharp gradient bound for positive harmonic functions:Sharpness follows from $u(\zeta )=\mathrm{Re}[(1+\zeta )/(1-\zeta )]$, which has equality at zero, and then composing with ${\psi }^{-1}$. These extremizers are positive multiples of the Poisson kernel on the circle with its boundary pole fixed at one point.

For antiplane shear, the infinitesimal displacement is $(0,0,w(x_1,x_2))$. The isotropic linear elasticity constitutive law gives ${\sigma }_{13}=\mu w_{,1}$ and ${\sigma }_{23}=\mu w_{,2}$, together with the symmetric entries ${\sigma }_{31},{\sigma }_{32}$. All other stress tensor components vanish. Static force balance without body force becomes $\mu (w_{,11}+w_{,22})=0$, so $w$ satisfies the Laplace equation.

Locally choose a harmonic conjugate $\varphi$ by ${\varphi }_{,1}=w_{,2}$ and ${\varphi }_{,2}=-w_{,1}$. Compatibility follows from the Laplace equation. These are the Cauchy-Riemann equations for $F=\varphi +iw$, and henceThus the complex stress potential for antiplane shear supplies both stresses and, by integration, the displacement. A global potential must have the displacement periods appropriate to the cracked body; these periods also help fix homogeneous solutions.

For the additional crack-induced field, set $G=\mu F^{\prime }$ and $p(t)={\sigma }_{23}^A(t,0)$. Only $p$ loads the prospective faces, since ${\sigma }_{13}^A$ is tangential to their normal. Denote upper and lower boundary values by $G^{+},G^{-}$. The imposed cancellation of face tractions is $\mathrm{Re}{G}^{\pm }=-p$. The traction data and geometry are invariant under the displacement reflection $w(x_1,x_2)\mapsto -w(x_1,-x_2)$. For the induced correction with no independently added loading, uniqueness up to a rigid translation allows the odd displacement choice. Then $w_{,2}$ is even, $w_{,1}$ is odd, and $G^{-}=\overline{G^{+}}$ on the real axis. Therefore the Hilbert problem for an antiplane crack isThe second condition expresses continuation through uncracked material. There $G$ is real and the additional ${\sigma }_{13}$ vanishes. This scalar Riemann-Hilbert problem is supplemented by at most inverse-square-root stress singularities at the crack tips and the usual correction with no additional remote resultant or imposed dislocation. Without a far-field or period normalization, the face conditions alone permit homogeneous terms. The formulas below select this normalized induced field and explicitly identify the excluded terms.

Take $u(z)=\log |z|$. In polar coordinates its Laplacian is $u_{rr}+r^{-1}{u}_r=-r^{-2}+r^{-2}=0$, so it is harmonic on the annulus. If $u=\mathrm{Re}f$ for a single-valued holomorphic function, the Cauchy-Riemann equations would give $f^{\prime }=u_x-i{u}_y=1/z$. But for $1<\rho <2$,The first equality follows by evaluating the primitive $f$ around a closed curve. This contradiction verifies the absence of a global harmonic conjugate.

For the final function, the quotient of squared distances is $|z+1{|}^2/|z-1{|}^2$. On $|z|>1$, both $1+1/z$ and $1-1/z$ lie in the open right half-plane, so their principal logarithms are single-valued and holomorphic. DefineThe series converges locally uniformly outside the unit circle, andThus this is the required single-valued holomorphic function, even on the larger domain $|z|>1$. Its derivative is $2/(z+1)-2/(z-1)$, whose two periods around the holes cancel. The logarithmic cancellation on an exterior annulus explains why this example succeeds while $\log |z|$ fails; using independent multivalued logarithms of $z+1$ and $z-1$ would conceal that cancellation.

For a harmonic function $u$, put $g=u_x-i{u}_y$. Its Wirtinger derivative isso the Cauchy-Riemann equations make $g$ holomorphic. In the simply connected domain it has a single-valued primitiveThe integral is path independent: local holomorphic primitives make its value invariant under path homotopies, and being simply connected makes any two paths with the given endpoints homotopic, as in Question 3.

Since $F^{\prime }=g$, the real part satisfies $(\mathrm{Re}F{)}_x=u_x$ and $(\mathrm{Re}F{)}_y=\mathrm{Re}(ig)=u_y$. Hence $\mathrm{Re}F-u$ is constant on the connected domain. Adding a real constant corrects it, producing a single-valued holomorphic $f$ with $\mathrm{Re}f=u$. The imaginary part is a harmonic conjugate and is unique up to an additive real constant. This is the global harmonic conjugate criterion by vanishing periods in the simply connected case.

A real harmonic function on a planar open set is a twice continuously differentiable function satisfying $f_{xx}+f_{yy}=0$. The ordered pair $(f,g)$ consists of harmonic conjugates when $f+ig$ is holomorphic; the corresponding Cauchy-Riemann equations are $f_x=g_y$ and $f_y=-g_x$.

To prove composition of harmonic conjugate pairs, denote $P=p(u,v)$ and $Q=q(u,v)$, and assume the inner pair's image lies in the outer pair's domain. Their equations are $u_x=v_y$, $u_y=-v_x$, $p_u=q_v$ and $p_v=-q_u$. The chain rule givesThus the composite satisfies the Cauchy-Riemann equations. Its components are $C^2$, and differentiating gives $P_{xx}+P_{yy}=Q_{yx}-Q_{xy}=0$ and similarly $Q_{xx}+Q_{yy}=0$. The composite pair is harmonic conjugate, including when the inner map has a critical point.

The principal real arctangent is defined here on $x\ne 0$, so work on either of the two open half-planes. Put $r=\sqrt{x^2+y^2}$. Direct differentiation givesHence ${\varphi }_{xx}+{\varphi }_{yy}=0$: $\varphi$ is harmonic on its domain. The Cauchy-Riemann equations for $f=\varphi +i\psi$ requireIntegrating these partial derivatives gives the harmonic conjugateThe constants may be chosen independently on the two components. The harmonic angle and logarithmic radius representation supplies the holomorphic functionwhere $\mathrm{Log}$ is the branch of the complex logarithm on the right half-plane. On the left half-plane it is the angle of $-z$, rather than the principal angle of $z$, that equals $\arctan (y/x)$. This distinguishes the literal arctangent from a globally defined argument; no single-valued complex logarithm exists on the whole punctured plane.

With $C_0=0$, the level sets $\varphi =C$, for $-\pi /2<C<\pi /2$, are the two opposite rays of $y=x\tan C$ with the origin removed. On each half-plane only one ray remains. The level sets $\psi =K$ are circles $r=e^{-K}$, with their two points on $x=0$ excluded from the literal domain. The rays and circles meet at right angles, as expected for harmonic conjugates.

A hyperbolic Riemann surface in potential theory is a noncompact Riemann surface admitting a positive Green function on a Riemann surface. With pole $p$, this is a positive function harmonic off $p$, having local form $-\log |\zeta |$ plus a harmonic function in a coordinate $\zeta (p)=0$, and obtained as the minimal such positive function by exhaustion. This is a noncircular definition for the present proof. For a simply connected surface it is equivalent to disk conformal type; for arbitrary surfaces potential-theoretic hyperbolicity and disk universal-cover type should be distinguished.

The unit disk is an example. At pole zero its function is $G(z,0)=-\log |z|$: it is positive in the punctured disk, harmonic there, has the correct logarithmic singularity, and tends to zero at the boundary. If another positive function has the same singularity and is harmonic elsewhere, subtract $G$ and apply the maximum principle for harmonic functions on disks of radius $r<1$; the boundary lower bound is $\log r$. Letting $r\uparrow 1$ shows that the other function dominates $G$. Thus $G$ is indeed the positive Green function on a Riemann surface. At a general pole it is

Here is a Green-function exhaustion proof of disk uniformization. The main analytic idea is to build conformal maps on bordered disks from their Green functions and pass to an extremal limit. The positive global Green function prevents that limit from becoming constant.

First exhaust the noncompact simply connected surface $R$ by relatively compact domains $R_n$ containing $p$, each with a smooth Jordan boundary. This topological step does not use conformal uniformization: take regular neighborhoods of finitely many coordinate disks covering successive compact sets and fill bounded complementary components. Simple connectedness and the Jordan curve theorem make the filled neighborhoods topological disks; enlarge them to be nested.

Each $R_n$ has a zero-boundary Green function on a Riemann surface $G_n(\cdot ,p)$. One analytic construction, independent of conformal uniformization, is as follows. Equip the compact bordered domain with a smooth conformal metric. Multiply $-\log |\zeta |$ by a smooth cutoff function equal to one near $p$ and zero near the boundary; call the result $s$. Distributionallywhere $k$ is smooth and supported away from $p$. Solve $\Delta v=k$ with zero boundary data. This auxiliary Dirichlet problem can be obtained by minimizing $\frac{1}{2}\int |\nabla v{|}^2+\int kv$ over functions with zero boundary trace: the Poincaré inequality gives coercivity, minimization in a Hilbert space gives a weak solution, and interior and smooth-boundary elliptic regularity give a smooth solution. Then $G_n=s+v$ is harmonic away from $p$, has the logarithmic pole, and vanishes at the boundary. It is positive by the maximum principle for harmonic functions, applied after removing a sufficiently small circle around its positive pole. This construction uses ordinary linear elliptic existence, not the desired uniformization theorem. Equivalently, the same bordered-domain Green functions can be constructed by the Perron method for the Dirichlet problem using local logarithmic barriers.

On $R_n\setminus \{p\}$ choose local harmonic conjugates $G_n^{\ast }$. Their period around the pole is $-2\pi$, and every closed loop is homologous to an integer multiple of that loop. Thereforeis a single-valued holomorphic function. It extends across $p$ with one simple zero and has no others. Since $G_n>0$, $|f_n|<1$; since $G_n=0$ on the boundary, $|f_n|\to 1$ there. Thus $f_n:R_n\to \mathbb{D}$ is proper. For any $a\in \mathbb{D}$, choose a smooth interior contour sufficiently close to the boundary that $|f_n|>|a|$ everywhere on it. The contour surrounds $p$ and all possible preimages of $a$, because $|f_n|>|a|$ throughout a thin boundary collar. By Rouché's theorem, $f_n-a$ has the same number of zeros inside as $f_n$, namely one. Hence $f_n$ is onto and one-to-one, with nonzero derivative: it is a biholomorphism from $R_n$ to the disk.

Write near $p$Multiply $f_n$ by a unit constant so that, in the fixed coordinate, $f_n^{\prime }(p)=e^{-c_n}>0$. Domain monotonicity gives $G_n\le G_{n+1}$: their difference is harmonic even at $p$ and nonnegative on $\partial R_n$. The global positive Green function on a Riemann surface $G_R$ similarly majorizes every $G_n$. Comparing the regular parts shows that $c_n$ increases and is bounded above by the finite regular part of $G_R$ at $p$. Consequently

On each fixed compact subdomain, the $f_n$ are eventually bounded holomorphic functions. Montel theorem and a diagonal subsequence therefore give a locally uniform convergence limit $F:R\to \mathbb{D}$ with $F(p)=0$ and $F^{\prime }(p)=d>0$. The maximum modulus principle keeps its values strictly inside the disk. It is injective: for fixed $q$, the functions $f_n-f_n(q)$ have no zeros away from $q$; Hurwitz's theorem implies that their nonconstant limit has no other zeros. This is the locally uniform limit of univalent functions argument.

There is also an extremality property. For any holomorphic function $g:R\to \mathbb{D}$ with $g(p)=0$, the function $g\circ f_n^{-1}$ is a disk self-map fixing zero. The Schwarz lemma gives $|g^{\prime }(p)|\le f_n^{\prime }(p)$. Passing to the limit yields $|g^{\prime }(p)|\le d=F^{\prime }(p)$.

Finally suppose $F$ omits $a\in \mathbb{D}$. Necessarily $a\ne 0$. Apply the disk Möbius transformation ${\varphi }_a(w)=(w-a)/(1-\overline{a}w)$. The nowhere-zero function ${\varphi }_a\circ F$ has a global holomorphic logarithm because $R$ is simply connected, hence a holomorphic square root $h$. It maps into the disk and $|h(p)|=\sqrt{|a|}$. Normalize it by setting $g={\varphi }_{h(p)}\circ h$, so $g(p)=0$. Direct differentiation givesThe strict inequality follows from $(1-\sqrt{|a|}{)}^2>0$. This contradicts extremality. Therefore $F$ omits no point, and we conclude

For a holomorphic function, compute its complex derivative along the real and imaginary directions. The real direction gives $f^{\prime }=u_x+i{v}_x$, while division by an imaginary increment gives $f^{\prime }=v_y-i{u}_y$. Equating real and imaginary parts proves the Cauchy-Riemann equations:For the specified real part, $u_x=-e^{-x}\cos y$ and $u_y=-e^{-x}\sin y$. Integrating $v_y=u_x$ gives $v=-e^{-x}\sin y+C(x)$; the other relation in the Cauchy-Riemann equations then forces $C^{\prime }(x)=0$. Thus the harmonic conjugate and the holomorphic function are

For complex differentiability at a point, the derivative must have the same value along real and imaginary increments. Along real increments it equals $u_x+i{v}_x$, while along imaginary increments it equals $(u_y+i{v}_y)/i=v_y-i{u}_y$. Equating real and imaginary parts gives the Cauchy-Riemann equationsFor the given real part,Integrating the Cauchy-Riemann equations gives a harmonic conjugateThis open set is open and connected; $F$ is holomorphic there with derivative $-z^{-2}$. The origin must be excluded because the real part is undefined there. On this connected open set, any other harmonic conjugate differs by a real constant: the Cauchy-Riemann equations make both partial derivatives of the difference zero.

Liouville theorem states that every bounded entire function is constant.

Because $\mathbb{C}$ is simply connected, the harmonic function $u$ has a global harmonic conjugate $v$, sois entire. Positivity givesLiouville's theorem makes $e^{-F}$ constant. Differentiating this nonzero constant gives $F^{\prime }=0$, so $F$, and hence $u$, is constant.

Harmonic means twice continuously differentiable with $u_{xx}+u_{yy}=0$. On the simply connected plane, choose an entire harmonic conjugate so $F=u+iv$ is entire. If $u\ge 0$, then $e^{-F}$ is bounded; Liouville's theorem makes it, and hence $u$, constant.

Compute the complex derivative along horizontal and vertical increments. Along the real direction, $f^{\prime }=u_x+i{v}_x$; along the imaginary direction, $f^{\prime }=(u_y+i{v}_y)/i=v_y-i{u}_y$. Equality gives the Cauchy-Riemann equationsThe real and imaginary parts of a holomorphic function are smooth. Equality of their mixed partial derivatives then gives $u_{xx}+u_{yy}=v_{yx}-v_{xy}=0$ and $v_{xx}+v_{yy}=-u_{yx}+u_{xy}=0$, proving that both are harmonic functions.

For the first real part, $u_x=2x$ and $u_y=-2y$. The Cauchy-Riemann equations require $v_y=2x$ and $v_x=2y$, so its harmonic conjugate is $v=2xy+C$. ThereforeFor the second real part, $u=x/(x^2+y^2)$ is the real part of $1/z$. Its harmonic conjugate is $-y/(x^2+y^2)+C$, givingThese are the most general answers on a connected open domain: the difference of two candidates has zero real part, so the Cauchy-Riemann equations make its imaginary part constant. The second function cannot extend holomorphically through zero.

The Cauchy-Riemann equations give $v_x=2x$, $v_y=2-2y$. Integrating and matching both derivatives gives the harmonic conjugateThe real part of this polynomial is exactly the required $2x(1-y)$.

Here $v_x=-3x^2+3y^2+4y$ and $v_y=6xy+4x$. The harmonic conjugate and resulting polynomial areIndeed $\mathrm{Re}(-i{z}^3)=3x^{2}y-y^3$ and $\mathrm{Re}(2z^2)=2x^2-2y^2$.