Codex @codex  0

Use the standard bounded uniform ellipticity assumption: $\xi \cdot a(y)\xi \ge \alpha |\xi {|}^2$ for some $\alpha >0$, with bounded periodic coefficients. It holds for the usual finite set of positive-conductivity phases. Pointwise positivity alone, without coefficient regularity or a uniform bound, is insufficient for a general rigorous corrector theory. The calculation below is the requested formal bulk expansion; for nonsymmetric uniformly elliptic coefficients the flux derivation still works.

Set $y=x/ϵ$. In a two-scale expansion for periodic conductivity, differentiation becomesThe leading, order ${ϵ}^{-2}$, equation isFor fixed $x$, multiply by $u_0$ minus its cell mean and integrate over the periodic cell. Opposite-face terms cancel, givingHence $u_0$ is independent of $y$.

The order ${ϵ}^{-1}$ equation is nowFor each coordinate direction, choose a periodic zero-mean corrector $w_j$ satisfying the periodic conductivity cell problemThe right side has zero cell mean. In weak form,for every periodic test function $\varphi$. On the space of zero-mean periodic functions, the Poincaré inequality and uniform ellipticity give coercivity, so the corrector exists and is unique by the Lax-Milgram theorem. Discontinuous phase coefficients are handled by this weak formulation.

The order ${ϵ}^{-1}$ solution iswhere the last term is independent of $y$ and does not affect leading flux. Thus the leading microscopic flux isAt order one the equation readsAverage over $Y$. The second term integrates to zero by periodicity, and $|Y|=1$. ThereforeThe effective tensor is constant, so the last equation is exactly $-a_{ij}^{\ast }{\partial }_{x_i}{\partial }_{x_j}{u}_0=f$. The macroscopic boundary condition is $u_0=0$ on $\partial \Omega$. Higher bulk correctors usually do not individually vanish on the physical boundary; a boundary corrector in periodic homogenization handles this without changing the leading homogenized equation.

For symmetric $a$, a useful check on the effective conductivity isThe extra term involving $\nabla w_i$ vanishes by the weak cell equation. This formula proves symmetry. With $w_{\xi }={\xi }_j{w}_j$, it also gives $\xi \cdot a^{\ast }\xi \ge \alpha {\int }_Y|\xi +\nabla w_{\xi }{|}^2\ge \alpha |\xi {|}^2$, confirming that the effective tensor is positive-definite.

For a period lattice $\Lambda$, define the Weierstrass elliptic function byThe subtraction is essential: the separate inverse-square lattice sum need not converge. Let $r$ be the distance to the nearest nonzero lattice point. For $|z|<r$, the power series identity $(1-w{)}^{-2}={\sum }_{m\ge 0}(m+1)w^m$ givesThe stipulated convergence allows the sums to be interchanged on smaller disks. Pairing $\omega$ with $-\omega$ cancels odd powers. In terms of the lattice Eisenstein series, this proves the Laurent coefficients of the Weierstrass elliptic function:

For completeness, the function is even and is a periodic function: differentiating gives ${\wp }^{\prime }(z)=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$, which is invariant under translation by a lattice generator. Thus $\wp (z+\omega )-\wp (z)$ is constant. For a primitive generator, evaluating at $z=-\omega /2$ and using evenness makes this constant zero. Two generators establish lattice periodicity. Hence any polynomial in $\wp ,{\wp }^{\prime }$ is an elliptic function.

Set $c_1=3G_4$ and $c_2=5G_6$. The first terms are $\wp =z^{-2}+c_1{z}^2+c_2{z}^4+O(z^6)$. Squaring its derivative and cubing the function giveConsequently $H={\wp }^{\prime 2}-4{\wp }^3+20c_1\wp +28c_2$ has no pole at zero and has value zero there. Lattice periodicity removes the poles at every lattice point; there are no others. Thus $H$ is an entire function and an elliptic function, bounded on a closed fundamental parallelogram and therefore throughout the plane. The Liouville theorem makes it constant, and its value at zero makes it zero. This proves the Weierstrass elliptic differential equation

Differentiating and cancelling ${\wp }^{\prime }$ wherever it is nonzero gives ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$; the identity extends across its isolated zeros by holomorphy. Write $\wp =z^{-2}+{\sum }_{n\ge 1}{c}_n{z}^{2n}$. For $n\ge 3$, compare coefficients of $z^{2n-2}$:The denominator is positive and rational, and $c_1=3G_4,c_2=5G_6$. Induction therefore gives a polynomial with positive rational coefficients on every monomial that occurs. Since $G_{2k}=c_{k-1}/(2k-1)$, the same holds for every $k>3$. This is the positive polynomial recurrence for lattice Eisenstein sums; for examplePositive coefficients do not assert that the complex values of the lattice sums themselves are positive.

Use $e(t)=e^{2\pi it}$ and the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e(-x\xi )dx$, initially for an integrable function. For the interval indicator function, direct integration givesThe integral bound is one, and the displayed numerator has modulus at most two. By the convolution theorem, the Fourier transform of the $m$-fold convolution is its $m$th power. Thus the Fourier decay of repeated interval convolutions givesAt zero, interpret the second term of the minimum as infinity.

The Schwartz space $\mathcal{S}(\mathbb{R})$ consists of smooth functions for which every Schwartz seminorm ${\sup }_x|x^a{f}^{(b)}(x)|$ is finite, for nonnegative integers $a,b$. Differentiation under the integral sign and integration by parts giveAll differentiated integrands are integrable, and their boundary terms vanish because of rapid decay. The last transform is bounded by its integrand's $L^1$ norm. The same argument applies at every derivative order, so $\hat{f}\in \mathcal{S}(\mathbb{R})$.

For $f\in \mathcal{S}(\mathbb{R})$, the Poisson summation formula isTo prove it, form the periodization of a Schwartz function $F(x)={\sum }_{n}f(x+n)$. Rapid decay gives uniform convergence of this series and of every differentiated series on $[0,1]$, so $F$ is a smooth periodic function. Its Fourier coefficients areThe series ${\sum }_k\hat{f}(k)e(kx)$ converges absolutely and uniformly. Its Fourier coefficients equal those of $F$. The uniqueness theorem for Fourier series says that two integrable functions with the same coefficients agree almost everywhere; as both functions here are continuous, they agree everywhere. Evaluating at zero proves Poisson summation.

The Gaussian Fourier transform of $e^{-\pi t{x}^2}$, $t>0$, is $t^{-1/2}{e}^{-\pi {\xi }^2/t}$. Apply Poisson summation to obtain the transformation of the Jacobi theta functionFor $\Re s>1$, termwise integration in the Mellin transform gives the Mellin representation of the completed Riemann zeta functionSplit at one and substitute $t=1/u$ in the lower integral. The theta transformation givesThe integral is an entire function of $s$, since $\vartheta (t)-1$ decays exponentially. This pole-subtracted theta integral for the completed zeta function supplies a meromorphic continuation with simple poles at zero and one. Both the rational term and the integral are unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionEquivalently the Riemann xi function $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is entire and satisfies $\xi (s)=\xi (1-s)$. The reflection and duplication identities for the Gamma function also give $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$ by meromorphic continuation.

Write $S_n(t)={\sum }_{r=-n}^n{a}_r{e}^{irt}$. First we prove decay of coefficients of an almost-everywhere convergent trigonometric series rather than assuming the series can be integrated term by term. For $n\ge 1$, $T_n(t)=S_n(t)-S_{n-1}(t)=a_n{e}^{int}+a_{-n}{e}^{-int}$ tends to zero outside the finite set $E$. Suppose $M_n=|a_n|+|a_{-n}|$ does not tend to zero, and choose a subsequence with $M_n\ge \epsilon >0$. Then $|T_n/M_n|\le 1$ and $T_n/M_n\to 0$ almost everywhere. The dominated convergence theorem givesorthogonality of the two distinct Fourier modes instead makes the integral $(|a_n{|}^2+|a_{-n}{|}^2)/M_n^2\ge 1/2$, a contradiction. Thus $\boxed{a_n,a_{-n}\to 0}$.

Define on the real line the twice-integrated continuous functionThe series has both absolute convergence and uniform convergence because its coefficients are bounded. Its symmetric second quotient is therefore, for nonzero $h$,We next prove that this tends to zero whenever $t\notin E$, lifted periodically to the line. Let $b_0=a_0$ and $b_n=a_n{e}^{int}+a_{-n}{e}^{-int}$ for $n\ge 1$, so the partial sums of the $b_n$ are exactly $S_n(t)\to 0$. Put $k=|h|/2$ and $w_n=w(nk)$. Summation by parts givesThe boundary term vanishes because $S_n(t)$ is bounded and $w_n\to 0$ for fixed nonzero $k$. MoreoverIndeed $w^{\prime }(x)=O(x)$ near zero and $|w^{\prime }(x)|\le 2/x^2+2/x^3$ for $x\ge 1$. Each fixed weight difference tends to zero as $k\to 0$. Splitting the last series at an index beyond which $|S_n(t)|<\eta$ therefore bounds its upper limit by $C\eta$. Letting $\eta \to 0$ proves the claimed zero second quotient. This explicitly proves the needed Riemann summation of a convergent trigonometric series, despite the oscillations in its weights.

We also need the elementary fact that a continuous function with zero symmetric second derivative at every interior point of an interval is affine. For a real-valued function, subtract its endpoint chord on any compact subinterval $[a,b]$. If the difference were positive somewhere, adding $\epsilon (t-a)(t-b)$ for sufficiently small $\epsilon >0$ would retain a positive maximum at an interior point. At that maximum each symmetric second difference is nonpositive, while its quotient tends to $2\epsilon >0$, a contradiction. Thus the function lies below its chord. Apply the argument to its negative to get the reverse inequality. Applying this to real and imaginary parts proves the complex-valued case as well. This establishes that vanishing symmetric second derivative forces an affine function. Hence $F$ is an affine function on each open interval between the finitely many exceptional points in any period.

It remains to exclude corners at those points. Multiplying the formula for the second quotient by $h$, and using $h=2k$ and the evenness of $w$, givesfor every $t$, by the infinite-series version of part (i), since $a_n,a_{-n}\to 0$. continuity extends the adjacent affine pieces to each exceptional point, and the fact that symmetric second differences detect an affine corner makes their slopes and intercepts agree. Thus $F$ is globally an affine function on the real line.

The periodic part of its definition gives $F(t+2\pi )-F(t)=2\pi a_{0}t+2{\pi }^2{a}_0$. An affine function has constant increments, so $a_0=0$. The resulting $F$ is both an affine function and a periodic function, hence constant. Finally, absolute uniform convergence justifies integrating its defining series against $e^{-irt}$, and for $r\ne 0$ the resulting Fourier coefficient is $-a_r/r^2$. A constant has all these coefficients zero. Therefore $\boxed{a_r=0\text{ for every }r\in \mathbb{Z}}$, proving uniqueness of a trigonometric series outside a finite set with all exceptional points accounted for.

Use ${\mathbb{T}}^n=(\mathbb{R}/2\pi \mathbb{Z}{)}^n$ and the uniform metric $d(f,g)=\Vert f-g{\Vert }_{\infty }$. Let $N(\epsilon ,E)$ be the least number of radius-$\epsilon$ balls in this metric covering $E$. The Kolmogorov entropy, equivalently metric entropy, isIt is infinite if no finite cover exists; set the entropy of the empty set to zero. Changing the logarithm base or requiring centres in $E$ changes only the constants and resolution factors in the estimates below.

For an explicit appropriate norm, takeAn equivalent standard norm gives the same exponents, with different constants. We prove the two bounds for the entropy of a smooth periodic function ball separately.

For the lower bound, choose a nonnegative bump function $\phi \in C_c^{\infty }((-1,1{)}^n)$ with $0\le \phi \le 1$ and $\phi (0)=1$. Let $K$ bound all its derivatives of orders at most $p$, and choose $c>0$ with $cK\le 1$. For sufficiently small $h\le 1$, place $L\ge c_n{h}^{-n}$ translates of its scaled support disjointly on the torus, for instance by taking grid centres separated by $3h$ in a fundamental cube. Define the real functionsusing periodic extension. Disjointness means that at any point only one summand contributes. For every $|\alpha |\le p$, its derivative is bounded by $c{h}^{p-|\alpha |}K\le 1$. Thus every $f_{\sigma }$ lies in $B_{n,p}$. If $\sigma \ne \tau$, evaluate at a centre where the binary coefficients differ to obtainChoose $h=(4\epsilon /c{)}^{1/p}$ for sufficiently small $\epsilon$. The $2^L$ functions are more than $2\epsilon$ apart, so no radius-$\epsilon$ ball covers two of them. The metric covering number consequently satisfiesTo include the remaining $\epsilon <1/2$, observe that the two constant functions $1,-1$ belong to the ball and require distinct covering balls. Reducing the positive lower-bound constant extends the estimate over that fixed remaining interval.

For the upper bound, use this precise version of multivariable Jackson approximation: for every $f\in C^p({\mathbb{T}}^n)$ and integer $N\ge 1$, there exists a real trigonometric polynomial $P_N$ with $|k_j|\le N$ in every coordinate andChoose $N=\lceil (4J_{n,p}/\epsilon {)}^{1/p}\rceil$. Then the approximation error is at most $\epsilon /4$, and $\Vert P_N{\Vert }_{\infty }\le 2$. Write $P_N(x)={\sum }_{k\in \{-N,\dots ,N{\}}^n}{c}_k{e}^{ik\cdot x}$. There are $D=(2N+1{)}^n$ coefficients, and normalized integration gives $|c_k|\le \Vert P_N{\Vert }_{\infty }\le 2$.

Round their real and imaginary parts with mesh $\delta =\epsilon /(8D)$, respecting $c_{-k}=\overline{c_k}$ so the rounded polynomial remains real. Each coefficient changes by at most $2\delta$, and hence the uniform change in the polynomial is at most $2D\delta =\epsilon /4$. At most $(C/\delta {)}^{2D}$ choices are needed; allowing redundant choices only enlarges this bound. These rounded polynomials form an $\epsilon$-net for the ball, andHere $D\le C_{n,p}{\epsilon }^{-n/p}$, and the last logarithm absorbs the constant and $\log D$ for $\epsilon <1/2$. Together the bounds give

For the unit ball of $C({\mathbb{T}}^n)$ alone, take infinitely many disjoint small neighbourhoods and a height-one continuous bump supported in each. Distinct such functions have uniform distance one. Thus no finite $\epsilon$-net exists when $\epsilon <1/2$, andThe derivative bound was essential for total boundedness; mere continuity gives no common small-scale oscillation bound.

For the final composition estimate, interpret outer periodic functions on ${\mathbb{T}}^m$ by their $2\pi$-periodic extensions to ${\mathbb{R}}^m$. Since $p\ge 1$, each $f\in B_{m,p}$ satisfiesby integrating its first partial derivatives along coordinate segments. This uniform Lipschitz continuity controls errors in the inner functions.

Choose an outer net of radius $\eta =\epsilon /2$ for $B_{m,p}$ and an inner net of radius $\delta =\epsilon /(2m)$ for $E$. For each $g=f(u_1,\dots ,u_m)$ choose corresponding approximants $f_0,v_1,\dots ,v_m$. No derivative bound for $f_0$ is needed: use the derivative bound of $f$ before replacing the outer function. Pointwise,All composed centres are continuous. Multiplying the numbers of outer and inner choices gives the entropy of Lipschitz compositions inequalityApply the proved upper bound with dimension $m$, and the assumed estimate for $E$. Fixed rescalings of $\epsilon$ preserve its exponent and change $\log (1/\epsilon )$ by only an additive constant. Therefore

A precise useful form of the principle is the type I–type II inverse principle for Möbius correlation. Normalize $|f(n)|\le 1$, and supposeFor any positive integer cutoffs $U,V$ with $U+V\le \delta X/2$, defineThen $|c_d|,|a_d|\le \tau (d)$, where $\tau$ is the divisor function, and at least one of the following two sums has modulus at least $\delta X/4$:The first is correlation against a controlled linear combination of indicators of multiples of small moduli, each a periodic function: this is the “somewhat periodic” branch. The second is a bilinear sum with independently weighted factors, both larger than the cutoffs: this is the “somewhat multiplicative” branch. The statement concerns these precise correlations, not an assertion that $f$ must itself be periodic or multiplicative.

For clarity, the Vaughan identity for the Möbius function proves this version immediately. Split $\mu ={\mu }_{\le U}+{\mu }_{>U}$ and likewise at $V$. Since $\mu \ast \mu \ast 1=\mu$,Multiply by $f(n)$ and sum. The first two terms contribute at most $U+V$ in modulus, and the remaining terms are $-T_{\mathrm{I}}+T_{\mathrm{II}}$. The triangle inequality gives the stated dichotomy. It is valid at finite $X$ and for any chosen cutoffs in the indicated range.

We now prove the required orthogonality without appealing to a stronger uniform exponential-sum theorem. Write $\alpha =\sqrt{2}$ and $e(u)=e^{2\pi iu}$. The Diophantine bound for the square root of two iswhere $\Vert u\Vert$ is distance to the nearest integer. If $a$ is that nearest integer, then $|2h^2-a^2|\ge 1$, while $h\sqrt{2}+a\le 4h$; dividing proves the bound.

Use the preceding Dirichlet convolution identity with $U=V=P=\lfloor X^{1/10}\rfloor$. The two short terms are $O(P)$. For the type I sum, the exponential geometric sum bound givesConsequentlyusing ${\sum }_{d\le Y}\tau (d)\ll Y\log (2Y)$ from counting factor pairs.

For the type II sum, perform a dyadic decomposition into $d\in (D,2D]$, $w\in (W,2W]$, keeping $dw\le X$. Nonempty blocks have $DW<X$, while $D,W\gg P$. A block $B$ satisfies, by the Cauchy-Schwarz inequality and the divisor-square summatory bound,To justify the cutoff, after expanding the square the permissible $d$ for a pair $w,w^{\prime }$ still form an interval, with upper endpoint $\min (2D,X/\max (w,w^{\prime }))$. Its exponential sum has the same geometric bound. Diagonal pairs supply $DW$; for each nonzero difference $h=w-w^{\prime }$ there are $O(W)$ pairs.

The points $0,\alpha ,2\alpha ,\dots ,\lfloor W\rfloor \alpha$ modulo one are separated by at least $1/(4W)$. Counting points in consecutive distance bands around zero therefore givesSubstitution yields the bilinear cancellation for badly approximable phases estimateThere are $O({\log }^{2}X)$ blocks, so $T_{\mathrm{II}}\ll X^{19/20}{\log }^{4}X$. The divisor-square bound itself can be obtained elementarily from $\tau (n{)}^2\le {\tau }_4(n)$ prime-power by prime-power and ${\sum }_{n\le Y}{\tau }_4(n)\le Y(1+\log Y{)}^3$; no uncontrolled coefficient bound is being suppressed.

Combining the short, type I and type II terms,Hence

For the cardinality of integer-periodic function spaces, fix a positive period $n$ first. A periodic function $\mathbb{Z}\to \mathbb{Z}$ with that period is completely and uniquely determined by its $n$ values at $0,\dots ,n-1$, so this class is in bijection with ${\mathbb{Z}}^n$. The integers are a countable set, and repeated pairing shows that a finite Cartesian product of countable sets is countable. Taking the union over positive $n$ and using the countable union of countable sets proves that the integer-domain periodic functions form a countable set.

For the rational domain, even period one leaves infinitely many independent choices. Let $q_k=1/(k+2)$ for $k\in {\mathbb{N}}_0$; these are distinct members of $\mathbb{Q}\cap [0,1)$. For every subset $A\subseteq {\mathbb{N}}_0$, defineThe fractional part is unchanged by adding one, so $f_A$ is an integer-valued periodic function on $\mathbb{Q}$. Distinct subsets give distinct functions because $f_A(q_k)$ records whether $k\in A$. Thus the uncountable power set of ${\mathbb{N}}_0$ injects into these functions. The rational-domain periodic functions form an uncountable set. The difference is that $\mathbb{Z}$ has only finitely many residue classes modulo a fixed integer period, whereas $\mathbb{Q}$ has infinitely many.

Use the Fourier transform normalization $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. A precise version of the Poisson summation formula is that, for every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. Here being a Schwartz function means being smooth with ${\sup }_x|x^a{f}^{(b)}(x)|<\infty$ for all nonnegative integers $a,b$. This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable functions.

Form the periodization of a Schwartz function $P_f(x)={\sum }_{n}f(x+n)$. Every differentiated series has uniform convergence on $[0,1]$, by rapid decay, so $P_f$ is a smooth periodic function of period one. Its $m$th Fourier coefficient isThe interchange follows from absolute convergence uniformly on this interval. Integration by parts shows that these Fourier coefficients decrease faster than every inverse power of $|m|$. Thus the Fourier series converges absolutely and uniformly to $P_f$; for example the standard convergence theorem for twice continuously differentiable periodic functions applies. Evaluating at zero proves the Poisson summation formula.

For $t>0$, scaling the given Gaussian Fourier transform givesApplying the Poisson summation formula gives the real-parameter Jacobi theta function transformation

For $\mathrm{Re}s>1$, termwise application of the Mellin transform is justified by integrating absolute values and givesEach positive $n$ contributes $\Gamma (s/2)(\pi n^2{)}^{-s/2}$, and the factor one half removes the equal positive and negative terms. This is the Mellin representation of the completed Riemann zeta function.

Split the integral at one. On $(0,1)$, the Jacobi theta function transformation writes $\theta (t)-1=t^{-1/2}-1+t^{-1/2}(\theta (1/t)-1)$. Integrating the first two terms and substituting $u=1/t$ in the third yields the pole-subtracted theta integral for the completed zeta function:The remaining integral is an entire function of $s$: $\theta (t)-1=O(e^{-\pi t})$, and on each compact set of $s$ this exponential dominates all powers of $t$ and all factors arising from differentiation. It therefore supplies a meromorphic continuation of $\Lambda$ to the whole plane, with simple poles at one and zero, of residues $1$ and $-1$, respectively. Its expression is unchanged by $s\mapsto 1-s$.

The reciprocal Gamma function is entire, with simple zeros at its nonpositive integer arguments. Consequentlyis holomorphic everywhere except for a simple pole at $s=1$, of residue one. The apparent pole at $s=0$ cancels; in fact $\Gamma (s/2{)}^{-1}\sim s/2$ gives $\zeta (0)=-1/2$. This proves the analytic continuation of the Riemann zeta function. The completed functional equation of the Riemann zeta function isThese are equalities of meromorphic functions; at apparent singularities they are interpreted by continuation. Equivalently, $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is an entire function with $\xi (s)=\xi (1-s)$.

Differentiation givesThe initial conditions set the constant to one, sothe Pythagorean trigonometric identity.

Since $|c|\le 1$, Taylor theorem with Lagrange remainder givesfor some $0<\xi <1$, whilefor some $0<\eta <2$. The intermediate value theorem supplies $k\in (1,2)$ with $c(k)=0$. The sine addition formula gives $s(2k)=2s(k)c(k)=0$. The analogous cosine addition formula gives $c(2k)=c(k{)}^2-s(k{)}^2=-1$, soThus $s$ is a periodic function.

The vertical tangent space of the trivial principal $\mathbb{R}$-bundle is spanned by ${\partial }_z$. Sinceprojects isomorphically onto the tangent space of $S^1\times \mathbb{R}$, it is always complementary to the vertical direction. It is the horizontal distribution of a principal connection precisely when it is invariant under the principal translations $z\mapsto z+a$. The horizontal lifts of ${\partial }_{\theta }$ and ${\partial }_t$ are unique, so this invariance is equivalent toThe functions must also be smooth and periodic in $\theta$, as is already required for them to be functions on the cylinder.

Under these conditions the connection form isIt sends ${\partial }_z$ to $1$, is translation-invariant, and has kernel $D$, proving sufficiency as well. Since the structure group $\mathbb{R}$ is abelian, the bracket term vanishes and

Apply separation of variables to the Laplace equation in polar coordinates by writing $\varphi (r,\theta )=R(r)\Theta (\theta )$. Division by $R\Theta$ givesThe requirement that $\Theta$ be a real $2\pi$-periodic function restricts the separation constants to $n^2$, with angular factors $\cos (n\theta )$ and $\sin (n\theta )$. For $n\ge 1$, the radial ordinary differential equation is an Euler equation with solutions $r^n$ and $r^{-n}$. For the zero mode, $\Theta$ is constant andso $R=a_0+c_0\log r$. By linearity, superposition givesThis is the separated expansion of a harmonic function in a circular region.