Weak equations and classical functions

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, September 2026. Self-checked by GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Public domain (CC0).

Source/proof self-check and prerequisite integration by GPT-6 Astra (OpenAI), Ultra, October 2026. Historical authorship and component terms are retained.

Distributional differentiation always exists, but its result need not be an ordinary function. A jump contributes a point mass. In the other direction, a differential equation with a continuous right side can force a distribution to become a classical function. The distinction depends on the equation and its coefficients.

We use Local data and compatible products for restriction, derivative signs and the product rule, and Tensor products and parameter-dependent distributions for separated variables and smooth parameter pairings. The smooth-kernel lesson supplies compact pairings, integration through a distribution and the complete integral Taylor proof. The supplied scalar foundation, §§12–13, proves calculus, cutoffs and differentiation of uniformly convergent series; the finite-dimensional algebra foundation, §10, and measure foundation, §§15.0–15.1, supply matrix algebra and integration. We preserve the radial matrix-transport route and prove its full ordered-series construction here, including its inverse and smooth parameter dependence.

A zero derivative determines a constant

Theorem 1.1 (distributional constants). If is a nonempty open interval and satisfies , there is a unique scalar such that

The interval may be unbounded. On a disconnected open set the constant can differ between its interval components.

Proof. A test with integral zero is the derivative of a test in the same interval. Extend it by zero and put

It is zero to the left of its support and, because the integral is zero, also zero to its right. The interval hull of the compact support lies compactly inside , so . Hence .

Choose with integral one. The test has integral zero, so . Take . Its uniqueness follows by testing on . Restriction proves the componentwise statement.

Corollary 1.2 (continuous right side). If and with , then is represented by a function, and its classical derivative is .

Proof. Fix . The function is , and integration by parts shows its distributional derivative is . Thus , so Theorem 1.1 gives .

This argument rules out hidden point-supported terms when the derivative is prescribed by a continuous function.

Independence of one coordinate

Theorem 2.1 (a constant distributional parameter). Let be open and be a nonempty open interval. If satisfies , there is a unique with

Equivalently , where is the distribution of integration on . The same formula equals .

Proof. Choose with integral one, and define . The map is continuous on every test support space, with compact image support. Thus is a distribution.

For , its integral is smooth and compactly supported in the projection of . Extend tests in by zero and set

The integrand has total integral zero for each . Therefore is zero before and after one compact interval in containing both -supports. Its -support lies in a compact subset of ; differentiation under the integral proves smoothness. Hence and . Pairing with gives .

Tests give uniqueness. The tensor theorem in the prerequisite identifies (2.1) with and gives its other iterated formula. The parameter pairing has one common compact -support and compact -support, so its ordinary integral is well-defined.

Corollary 2.2 (a continuous weak derivative is classical). Let be continuous functions on open , and suppose as distributions. Then the classical partial derivative exists at every point and equals .

If all first distributional partial derivatives of a continuous are continuous functions, then .

Proof. This is local, so work on a box where the distinguished coordinate is . Fix and put

The function is continuous and has continuous classical -derivative . Fubini and one-dimensional integration by parts against a compact test show distributionally; no derivative in is needed.

The continuous function has zero distributional -derivative. By Theorem 2.1 it is . The proof identifies with the continuous function

Thus and give the same distribution. Two continuous functions with that property agree pointwise, by the nonnegative-bump argument in the localization lesson. Hence , and its classical -derivative is at every point. Covering proves the assertion. If every partial derivative is continuous, telescope along a short coordinate polygon from to and use the one-dimensional fundamental theorem on its segments. After subtracting , the error is at most times the maximum variation of these partials within distance of . Continuity makes this , proving total differentiability and hence regularity.

For this single-coordinate assertion, continuity of is a necessary hypothesis. The distribution has zero -derivative but is not a continuous function.

All weak partials control regularity

Theorem 2.3 (all continuous weak partials). Let be open, , and . Suppose every first distributional partial is represented by a continuous complex function: Then exactly one represents , and its classical partials are the given . Neither continuity nor local integrability of is a hypothesis. The statement applies componentwise to finite vectors.

Proof. Fix a ball with compact closure in . Choose so small that the closed -neighborhood of is a compact subset of . This follows from compactness of and openness of . Fix a nonnegative smooth test , supported in the unit ball, with integral one; the exact smooth cutoff construction in Metric and topological foundations, Section 13.10 provides such a test by normalization. Put , , and define on the open -neighborhood of For fixed , kernels on each compact parameter neighborhood are tests supported in one compact subset of . The parameter-differentiation lemma in Tensor products and parameter-dependent distributions therefore proves smoothness and differentiation of this actual pairing. The two derivative signs give In the second equality the minus in the distribution derivative cancels . Uniform continuity on the fixed compact neighborhood gives

We first justify convergence of (2.3b) to the original distribution. For , extend by zero and define All these tests, for sufficiently small , have support in one compact subset of . For every multiindex , differentiating the ordinary integral and using uniform continuity of proves uniform convergence . Thus in a fixed test-support space. Moreover Here the interchange with is justified without a Fubini assertion about an arbitrary functional. On the fixed compact kernel support, choose the actual finite derivative seminorm controlling . Riemann sums for the -integral of converge in that seminorm, because all the required -derivatives are continuous on a compact product. Their scalar pairings converge to the ordinary integral on the left. This is the same finite-seminorm integral argument proved in the smooth-approximation lemma in When a kernel is smooth. No uniform estimate in is needed for this identity at each fixed .

Retain the value . The ordinary fundamental theorem on the segment , , in the convex ball gives By (2.3d), converges uniformly on to This formula defines a continuous function; we have not assumed that the continuous list already has a potential. For any coordinate segment inside , the classical derivatives in (2.3c) give Uniform convergence yields the exact limiting identity It includes negative with the oriented integral. Dividing by proves the classical partial .

For completeness, these continuous partials give total differentiability by a direct estimate. For sufficiently small, the coordinate polygon from to stays in . Telescope (2.3i) along it and subtract . The absolute remainder is at most Indeed and the suprema tend to zero. Hence .

It remains essential to recover the actual constant. Choose with integral one. Equations (2.3f)–(2.3h) give For each , pass to the limit in (2.3g), using (2.3f), to obtain Thus represents the original , with .

Continuous representatives are unique: if a continuous difference has a nonzero value at , choose a complex scalar making . This remains strictly positive on a smaller ball. Pairing with a nonnegative nonzero bump there contradicts that represents zero. Therefore the functions agree on all overlaps. They form a unique function on . To identify its global distribution, apply the finite test partition of the local uniqueness theorem in Local data and compatible products to , which is zero on each ball. Its derivatives are locally the , hence are so globally. This proves the theorem without connectedness or a prescribed normalization.

Corollary 2.4 (higher regularity from weak partials). If every in (2.3a) belongs to , , the representative is in : the first derivatives supplied by Theorem 2.3 are precisely the given functions, so all successive classical derivatives through order exist and are continuous. Their distributional identities follow by integration by parts locally, then restriction and uniqueness. Smooth yield a smooth representative.

More generally, fix an integer . Suppose every distributional derivative with is continuous. For each , all first weak partials of are among these continuous functions, because distributional partials commute by the differential rules in Local data and compatible products. Theorem 2.3 gives a unique representative for each such derivative. Descend one order at a time. If every derivative of order has a representative, all first partials of any derivative of order have that regularity. The preceding paragraph makes it . At this yields the unique representative of . There is no hypothesis on lower-order derivatives in advance, and uniqueness identifies their classical and weak versions at every step.

This assertion uses all multiindices of the fixed total order. It makes no claim that a list of only pure high-order derivatives, or one selected directional derivative, supplies the same conclusion.

First-order systems without commuting matrices

For a scalar smooth coefficient , the integrating factor satisfies . For matrices the same exponential expression generally fails. We need the order of multiplication fixed.

The following normalized radial transport statement supplies the matrix factor: on a star-shaped open containing zero, any smooth matrix with has a smooth invertible normalized solution

Proof of the radial transport statement. Use the matrix norm . Summing first in proves . Define, for , The second expression follows from the fundamental theorem and defines the value at . It is jointly smooth in . For every compact parameter set , its radial hull is the compact image of , lies inside , and controls all derivatives in this formula.

Let and recursively set Unrolling the recursion integrates the ordered product over . Induction by the outer integral gives the simplex volume . If bounds the coefficient and all its parameter derivatives through order on the compact set, each list of differentiations has assignments to the factors. Consequently, for , For each fixed derivative order, the ratio of consecutive scalar bounds tends to zero, so the bound is eventually dominated by a geometric series. The series of every such derivative converges uniformly. The proved uniform differentiation theorem in scalar §13.7 shows these are the actual parameter derivatives. Summing the integral recursions gives ; the fundamental theorem gives . Repeated differentiation of this equation supplies all -derivatives as well.

For the inverse, use , , and . The same bounds apply while retaining the reverse factor order. Thus , and gives . A square matrix with a left inverse is injective, hence surjective by finite-dimensional linear algebra; therefore too. Both matrices are smooth in all their parameters.

Uniqueness uses the same estimate. The difference of two solutions with the same initial value satisfies . Iterating this times bounds it by , which tends to zero. Hence .

Set . For , the identity and uniqueness give , and hence . Differentiating at from the left proves . At , the coefficient is zero, so ; the equation holds there too. The inverse is . Finally, any other smooth normalized radial solution restricted to a ray solves the same regular initial-value equation, so it equals . This proves (3.1) on the whole star-shaped domain, with no commutativity assumption.

To obtain the integrating factor we need on , take , , and

This is smooth, vanishes at zero, and is star-shaped. Equation (3.1) gives for , hence at zero too by continuity. Its ordinary transpose, with no complex conjugation, gives

The matrix and its inverse are smooth on all of . This is an exact specialization of the transport proof just supplied.

Theorem 3.1 (distributional systems become classical). Let , , and . If

then is represented by a vector function and satisfies (3.4) classically. With as in (3.3), every distributional solution has the form

Proof. Apply the distributional product rule entry by entry. Retaining the multiplication order,

The right side is continuous. Corollary 1.2 applied to each component gives , a vector. Multiplying by the smooth inverse proves (3.5) and the regularity claim. Conversely, classical differentiation of (3.5) gives the equation. The constant is its value at , because .

This also applies componentwise to open subsets of the line, with one independent constant vector on each interval component.

Corollary 3.2 (higher-order scalar equations). Let , let be smooth on open , and let . If satisfies

then and (3.6) holds classically.

Proof. Work on an interval component. Set . It satisfies , with , the first rows of having a single in their next column, and its last row . Theorem 3.1 makes every component of .

The distributional identities now agree with the classical derivatives of those components. Starting with , induction gives ; its -th derivative is , which is continuous. The last equation is then classical.

A smooth nonvanishing leading coefficient can be divided out, so the same conclusion holds locally wherever it is nonzero. If it vanishes, singular solutions may remain: , although is discontinuous.

Jumps produce concentrated derivatives

Theorem 4.1 (derivative across a jump). Let , with open, and let be on . Suppose its ordinary derivative there is integrable on a neighborhood of . Then the finite one-sided limits , exist. Any assignment of gives the same locally integrable distribution, whose derivative is

Proof. For fixed sufficiently close to ,

Absolute integrability makes the right side have a finite limit as . The same argument to the left gives the other limit. In particular is bounded near , hence locally integrable; its value at one point does not affect its integrals.

For a compact smooth test, integrate by parts separately to the left of and to the right of . The outer boundary terms vanish, and

Local boundedness of , integrability of , and the one-sided limits justify passage to . This gives , exactly (4.1). The argument is local near ; on the rest of the support ordinary integration by parts applies.

The coefficient is the right limit minus the left limit. The sign is determined by the distributional convention, not by a choice of a value at the discontinuity.

For a piecewise function whose derivatives through order have finite one-sided limits and whose ordinary -th derivative is locally integrable at , iteration gives

To justify the iteration, Theorem 4.1 applied first to gives its first jump term. Apply it to the ordinary derivative on each side at each subsequent step. Each previously obtained delta derivative is differentiated once, while the new ordinary derivative contributes its own jump. Induction produces precisely the indices in (4.2). The stated hypotheses ensure all those ordinary derivatives define locally integrable functions.

Exercises

1. No hidden singularity — foundation. Determine every distributional solution of on the line. Explain why the integrating-factor argument also excludes delta terms.
2. Two derivative levels — intermediate. Let for and for . Compute its first and second distributional derivatives. Check each concentrated coefficient from (4.2).
3. An unsmoothed transverse variable — intermediate. On , let . Prove , but is not a continuous function. Identify the missing hypothesis if one tries to apply Corollary 2.2.
4. A degenerate leading coefficient — foundation. Verify distributionally and explain why this does not contradict Corollary 3.2. Give another distributional solution obtained by adding a constant.
5. Matrix order — advanced. Put , , and . Let , . Compare the coefficient of in with that in . Show the difference is , so the naive exponential is not the integrating factor.