Euler equations and the order of singularities

Reconstructed by GPT-6 Astra (OpenAI), Ultra reasoning effort, 4 October 2026. Public domain (CC0).

The Euler equation records scaling. Repeating its operator produces powers of the logarithm, but the resulting chains do not have larger distributional order. The distinction that matters is whether a solution remains nonzero away from the origin. We prove the complete classification, the exceptional maps at degree zero, and the exact orders, including two tensor examples where adding the separate orders gives an unnecessarily large bound.

Our proof inputs are Order, positivity and distributional limits, Section 1 and its finite- extension; Cauchy kernels and boundary limits, Corollary 2.2; Finite parts of singular powers; Complex powers at a boundary; and Homogeneous extensions and angular moments, with its angular foundations, A1–A5. Those lessons supply the scalar power families, exact point jets, polar integration and all their underlying calculus. The zero-derivative theorem, polynomial factorization and the particular tensor constructions needed here are proved below.

Two elementary ways to solve an equation

We use complex-linear distributions. On the real line write

Zero-derivative fact. On any open interval , a distribution with is constant. Indeed, every compact test of integral zero has a compact primitive inside , obtained by integrating from a point to the left of its support. Hence annihilates such tests. Choose a compact test in with integral one. Applying this observation to gives , as claimed. The primitive is smooth by the fundamental theorem and vanishes on both sides of the support because the integral is zero.

Lemma 1.1 (a primitive and division by the coordinate). Every distribution on is a derivative, and every distribution is for some distribution . The two ambiguities are, respectively, constants and multiples of .

Proof. Fix with integral one, and define The integrand has integral zero. Thus is supported in the interval containing the supports of and . On any fixed test-support space, that interval is fixed. Its zeroth norm is bounded by its length times the integrand's sup norm, and its higher derivatives are derivatives of the integrand. This proves every required seminorm bound. Consequently is a distribution. Since and , its derivative is . The zero-derivative fact proves the ambiguity.

For division choose equal to one near zero and put Near zero the quotient is ; its -th derivative is , by bounded difference quotients. Away from zero the ordinary product rule bounds it by finitely many test derivatives. Its support lies in the fixed union of the original support and . Hence is a distribution, and proves . If , the identity gives . Conversely . This proves the second ambiguity directly.

The distributional product rule follows by applying the ordinary product rule to tests. It yields Thus differentiation takes to , and multiplication by takes it to .

Homogeneous solutions and logarithmic chains

Let and be the locally integrable indicators of the positive and negative half-lines. Write , and use the proved families Here is meromorphic and has no assigned value at a pole; is the symmetric finite part of U016; and are the entire distribution families of U017.

Theorem 2.1 (the two homogeneous solutions). For every complex , has dimension two. Off zero its members have the form If , a basis is . At , , a basis is , and the coefficients obey . The nonzero solutions supported at zero exist exactly at these negative integers and are multiples of the indicated jet.

For , At zero, , its derivative image is , and is the space of constant distributions.

Proof. U018, Lemma 1.1, proves Euler/scaling equivalence. On either half-line the product rule gives , since . The zero-derivative fact therefore proves (2.2).

At nonexceptional , the two meromorphic values have the independent pairs of half-line coefficients and . Subtracting their matching combination leaves a point-supported distribution. U018, foundation A3, proves that it is a finite sum of independent jets and that No such jet has eigenvalue , so the difference is zero.

At , U018, Theorem 4.1, gives the necessary and sufficient angular moment condition on the two-point sphere: . The symmetric finite part has coefficients . After subtracting , (2.4) leaves exactly a multiple of . These two terms are independent because only the second is supported at zero.

On the source of (2.3), ; on its target, . These identities prove both inverse compositions. Finally, the endpoint fundamental theorem gives and . U016's symmetric formula gives , while . The claimed critical images follow.

Near a fixed , choose two holomorphic families . Use at a nonexceptional , and at . In the latter case U017 proves so their values are independent. In every case . Put

Theorem 2.2 (complete generalized Euler spaces). The distributions , , , form a basis of , and On each half-line every member is a function Point-supported solutions remain exactly the multiples of at ; there are no longer chains consisting entirely of point-supported distributions.

Proof. Distributional parameter derivatives exist with the finite test bounds proved in U016 and U017. Applying a fixed differential operator commutes with them because it acts by a continuous map on tests. Differentiating times and dividing by proves the first identity in (2.7).

In a zero linear combination of the proposed basis, apply . The independence of kills the two top coefficients. Repeat with descending powers to kill every coefficient. For spanning, induct on . Expand in the shorter chains, raise each chain index by one to obtain with , and use Theorem 2.1 on . This proves the basis and dimension.

For an explicit half-line interpretation, transfer a distribution on by The test map and its inverse preserve compact supports in the respective domains and bound all derivatives by the chain rule. Transposing the Euler operator shows that transfers to : indeed . Thus . The zero-derivative fact, followed successively by subtraction of ordinary polynomial primitives, proves that is a polynomial of degree below . Scalar substitution in its pairing recovers (2.8). On the negative half-line use ; its absolute Jacobian is again , giving the other polynomial. Finally, (2.4) diagonalizes on the finite independent jet expansion. Only its zero eigenvalue can survive, regardless of .

With , the exact scaling formula on is To prove it, differentiate times in the parameter, divide by , and use . It follows on every basis vector, hence on the space.

What fails at the parameter zero

Theorem 3.1 (maps between generalized spaces). If , differentiation maps isomorphically to . With on the target, its inverse is At zero the images instead are Furthermore, with codimension one at each inclusion. The critical differentiation kernel is the constants, and the critical multiplication kernel is .

Proof. Multiplying the finite series in (3.1) by gives , since . The corresponding series also inverts on . The identities and give injectivity and surjectivity. The intertwining in (1.4) places the displayed inverse in the correct space.

The two chains at zero show that applying removes exactly their last levels and has image ; at degree the same argument applies to . If for , then . Conversely, if this last condition holds, Lemma 1.1 supplies a primitive , and . This proves the third image formula.

If with , then . Conversely, divide a given with that derivative property by , using Lemma 1.1. The resulting satisfies , hence . This proves the fourth image formula.

The first inclusion in (3.3) follows from (1.4). If , then , and the primitive argument puts in . The zero-derivative fact makes the kernel of differentiation on each one-dimensional. Therefore the three spaces in (3.3) have dimensions . To justify this dimension subtraction directly, extend a basis of the kernel to a basis of the finite-dimensional domain; the images of the added vectors span the image and are independent by subtraction of a kernel vector. The multiplication kernel follows from Lemma 1.1, and belongs to every .

Polynomial facts used below. Every nonconstant complex polynomial factors into linear factors, and coprime polynomial factors have a Bézout identity. Here are the needed proofs.

If a nonconstant polynomial had no zero, would be holomorphic everywhere by the scalar quotient rule. Writing shows for sufficiently large : divide the finite lower-term sum by and let increase. On the remaining compact disk, continuity and the absence of zeros give a positive minimum. Thus is bounded by one fixed constant on the whole plane. U013, Corollary 2.2, gives on every centered circle of radius . Letting tend to infinity makes this derivative zero. The fundamental theorem along line segments makes constant, a contradiction. Hence has a root . The finite identity divides by . Repeating on the quotient proves factorization by induction on degree, including multiplicities.

Polynomial division is obtained by subtracting the leading monomial multiple of the divisor; each subtraction lowers the remainder degree and the process terminates. Repeated division, exchanging divisor and nonzero remainder, strictly lowers the latter's degree. The last nonzero remainder divides both original polynomials by back-substitution through the divisions; every common divisor divides each remainder and hence . Back-substitution also expresses as a polynomial combination of the originals. Iterating this procedure gives the same statement for any finite list. If the list has no common nonconstant divisor, rescale the final constant to one to obtain its Bézout identity. By factorization, having no common root implies precisely this condition.

Corollary 3.2 (polynomials in the Euler operator). For a nonzero complex polynomial of degree , If its distinct roots have multiplicities , then

Proof. Divide out the nonzero leading coefficient and put , . The have no common root: at any root , . For a single root , which has the same property. The proved polynomial facts give . Thus for , The -th term is killed by . Conversely each such kernel is contained in . On , the operators vanish for , because contains . Since their sum is the identity, the remaining operator is the identity there. Applying these operators to a zero sum proves directness. Theorem 2.2 gives dimension . A nonzero constant has zero kernel and .

Scaling forces a sharp order bound

Order at most means that on each fixed compact test support , . Exact order is the smallest such nonnegative integer.

Lemma 4.1 (a strict bound from disjoint annuli). If has order at most and its nonzero restriction off zero is homogeneous of degree , then

Proof. Some annular test has ; every compact subset of the punctured space fits inside an annulus. For , put . These tests have supports in one ball and uniformly bounded derivatives through order . The punctured scaling law gives A negative real exponent contradicts the order estimate. If its real part is zero, choose a geometric sequence whose closed supporting annuli are pairwise disjoint. For example, if the original radii are , take successive ratios smaller than . Multiply each test by a scalar of modulus one making its pairing equal to . Every finite sum is a smooth compact test vanishing near zero. At each point at most one summand or its derivatives is nonzero, so the sums have one uniform bound. Their pairings are , a contradiction. This excludes equality as well.

Theorem 4.2 (exact order for generalized Euler solutions). If is supported at zero, then for some and its exact order is . Otherwise its exact order is the least nonnegative integer such that In particular, increasing the logarithmic chain length does not increase this order.

Proof. First suppose is nonexceptional. On a small disk about , arrange and avoid all zeros of the following denominator. U016's meromorphic derivative identity gives For the denominator is the empty product one. The identity follows first by integration by parts where the endpoint terms vanish, and then by the meromorphic identity theorem already proved in U016. On any fixed compact test support, parameter derivatives of every finite order insert logarithmic powers and derivatives of the scalar denominator. They are bounded by , because is uniformly integrable at zero on a smaller disk; U016's gamma companion evaluates and bounds these integrals. Reflection has the same estimate. The basis of Theorem 2.2 therefore has order at most .

At , (4.3) gives . Set . The denominator in (4.4) has exactly one simple zero at , and its integral is For this is holomorphic with all local parameter derivatives bounded by a test norm. Write the resulting meromorphic family as , where is its known delta-jet residue. On a fixed small circle , both (4.4) and have uniform bounds. The removable-pole statement from U016 makes holomorphic inside. U013's scalar Cauchy coefficient formula, applied after each test pairing, bounds the -th coefficient of by . Thus every coefficient is a distribution of order at most .

The same holds after reflection. In the U017 identity the poles cancel. Multiplying by the analytic phase adds only finite scalar multiples of Laurent coefficients and the residue to each Taylor coefficient, all with bounds. Their parameter derivatives span the exceptional generalized space, so its order is at most .

For the lower bound, suppose and is nonzero off zero. Choose an annular with . From (2.10), The polynomial is nonzero because , and minimality of gives . If this real part is negative, the exponential growth in dominates any nonzero polynomial and the pairing is unbounded. This follows, for example, from the exponential series lower bound with larger than any specified polynomial degree, and the leading-term bound for . If the real part is zero and is nonconstant, its modulus also tends to infinity by its leading term. Either case contradicts an order- estimate. If is constant, phase-aligned disjoint annuli as in Lemma 4.1 give the contradiction instead. Therefore order is impossible. When , the upper bound already gives exact order zero.

A nonzero point-supported solution is with , by Theorem 2.2. It has order at most . If , choose a compact smooth with , for instance a cutoff times . The tests have uniformly bounded norm but jet values of size . This excludes smaller order. For the nonzero delta has exact order zero.

Thus has order one and has order two, as do their logarithmic parameter derivatives when nonzero off zero. At degree , has order one whereas has order two.

Tensor singularities can share one derivative

The tensor notation used in this section has a direct construction. U016 gives . For a planar test supported in , define The double difference equals ; its absolute value is at most . Thus the first integral is absolutely convergent after cancellation, independent of increasing , and bounded by . Applying the one-dimensional cancellation twice and Fubini gives precisely the iterated principal-value pairing, in either order. It also proves that removing strips , from the ordinary integral converges to as both cutoffs tend to zero, independently.

The trace in the second line is a compact smooth line test, and its first derivative is bounded by . Hence also has order at most two. These explicit formulas define They require no general tensor-product existence or uniqueness theorem. For tensor tests their pairings are the products of the individual factor pairings, by the same formulas.

Theorem 5.1 (orders of two planar tensors). The distribution is homogeneous of degree and has exact order one. The distribution is homogeneous of degree and has exact order two.

Proof. Substitution in (T1), together with the density in , gives . For , the derivative in its traced test adds and the one-dimensional principal-value substitution has degree zero as a pairing on dilated tests, so . Both are nonzero off the origin. Near , is the nonzero ordinary function ; near , tests with make . Lemma 4.1 forces order at least one for and at least two for . The bound above already proves the latter exact order.

To improve the bound for , write . Its angular distribution is with symmetric principal values at , interpreting the interval periodically at zero. We justify this formula and its estimates. Choose a fixed small radius around each singular angle. There the denominator is , an odd function with absolute value at least ; the latter bound follows by continuity of at zero and by . The constant term cancels on every symmetric punctured interval. The remaining quotient is bounded by , by the fundamental theorem. On the complementary arcs the denominator is bounded away from zero. Thus the principal value exists and For the last equality, reflection preserves every symmetric exclusion and changes the denominator's sign.