# Analytic coefficients and one-sided uniqueness *Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).* A solution that is zero on one side of a surface need not be zero on the other. A characteristic surface can carry a wave front or even a distribution concentrated on the surface. Holmgren's theorem gives a precise alternative: analytic coefficients and a noncharacteristic normal force a local zero neighborhood. The solution itself may be an arbitrary distribution. We first turn every finite differential order into a finite first-order system. We then use an analytic paraboloid touching the support to handle a surface that is only continuously differentiable. This separates the analytic argument from the geometry. Read [Fundamental solutions in shrinking cones][S], Sections 7.1–7.2, for the complete first-order distributional uniqueness proof, and [Cauchy bounds, root counts and analytic extensions][C] for its analytic estimates. Basic references are these preceding chapters, Crétier's notes [R], and Hörmander's account [H]. The [complete proof](analytic-coefficients-and-one-sided-uniqueness-formal.md) establishes the result for all finite scalar orders and for square systems with invertible highest normal coefficient. It requires no paid proof. ## 1. What a noncharacteristic normal means We use ordinary derivatives. For an order-\(m\) scalar operator \[ P=\sum_{|\beta|\le m}a_\beta(x)\partial_x^\beta, \qquad p_m(x,\xi)=\sum_{|\beta|=m}a_\beta(x)\xi^\beta, \tag{24} \] a surface \(\rho=0\) is noncharacteristic at \(x_0\) if \(p_m(x_0,d\rho(x_0))\ne0\). For an \(r\)-component square system, replace nonzero by invertible. Complex coefficients are allowed. Switching to \(D=-i\partial\) multiplies the principal symbol by \((-i)^m\), so it does not change this condition. **Holmgren's theorem.** Suppose the coefficients are real analytic near \(x_0\), the surface is \(C^1\), and its normal is noncharacteristic. A distribution solving \(Pu=0\) and vanishing on one side vanishes near \(x_0\). There is no assumption that \(u\) is a function, has boundary traces, or satisfies a bound at infinity. The conclusion is local. It gives uniqueness when a solution exists; it does not supply arbitrary-data existence. ## 2. Why the order does not restrict the theorem Use coordinates \((y,t)\), with the normal coefficient invertible. Collect all derivatives of the solution through total order \(m-1\): \[ W_{\alpha j}=\partial_y^\alpha\partial_t^j u, \quad |\alpha|+j\le m-1,\qquad \#W=r\binom{d+m}{m-1},\quad d=\dim y. \tag{25} \] Below the top derivative layer, time differentiation moves to the next component. At the top layer, a tangential derivative converts the desired time derivative into an already listed component. Only the last pure-time row needs the original equation. For example, if \(|\alpha|+j=m-1\) and \(\alpha_\ell>0\), then \[ \partial_tW_{\alpha j} =\partial_{y_\ell}W_{\alpha-e_\ell,j+1}. \tag{26} \] For the pure-time row, solve \(Pu=0\) for \(\partial_t^m u\). Every other term of order \(m\) has a tangential derivative; remove one of those derivatives and apply it to a component of \(W\). We have obtained \[ \partial_tW=A(y,t,\partial_y)W, \tag{27} \] where \(A\) is first order in tangential variables and has analytic matrix coefficients. Derivatives of a distribution vanish on every open set where that distribution vanishes. Thus the whole vector has the same open zero side, and the exact first-order theorem from [S] applies. ![All derivatives of total order at most two form a triangular six-component system.](figures/third-order-jet-system.png) *Figure 1. For one tangential coordinate and scalar order three, the vertices are \(W_{\alpha j}=\partial_x^\alpha\partial_t^j u\), with \(\alpha+j\le2\). A solid arrow from \(W_{\alpha j}\) to \(W_{\alpha,j+1}\) means \(\partial_tW_{\alpha j}=W_{\alpha,j+1}\). A dashed arrow from \((2,0)\) to \((1,1)\), or from \((1,1)\) to \((0,2)\), means that the target is differentiated once in \(x\). The circled pure-time vertex uses \(Pu=0\) for its final row. This is a derivative identity diagram; its arrows are not particle trajectories. See Example 1 and complete-proof Lemma 2.1.* ## 3. The surface does not have to be analytic An analytic graph can be flattened by an analytic change of coordinates. A merely \(C^1\) graph cannot be used this way: it can destroy the coefficient regularity and need not even give a smooth change of variables for distributions. Instead choose affine coordinates so the graph is \(t=g(y)\), with \(g(0)=0\), \(\nabla g(0)=0\), and the zero side below it. On a small ball let \(|\nabla g|\le\varepsilon\). Then \(|g(y)|\le\varepsilon|y|\). Minimize on the local closed support \[ F(y,t)=t+\frac{2\varepsilon}{a}|y|^2, \qquad |y|\le a,\quad |t|\le2\varepsilon a. \tag{28} \] If the origin were in the support, the minimum would be at most zero. The bottom contains no support; the side and top have strictly positive values. The minimum is therefore attained at an interior support point. The level through that point is an analytic paraboloid. Its normal differs from the vertical normal by at most \(4\varepsilon\). Continuity keeps it noncharacteristic when \(\varepsilon\) is small. The analytic-graph theorem forces a zero neighborhood at the support point, which is impossible. This proves the \(C^1\) version. The argument does not assume a supporting quadratic graph at the original point. It finds a suitable supporting level at an interior minimum. Complete-proof Lemma 3.1 and Theorem 3.2 give every graph and compactness step. ## 4. Four examples ### Example 1: a third-order equation with complex coefficients. Consider \[ \begin{split} P={}&\partial_t^3+(1+x^2+it)\partial_x\partial_t^2 +(2-it)\partial_x^2\partial_t+i\partial_x^3\\ &+(1+t)\partial_t+x\partial_x+1. \end{split} \tag{29} \] The coefficient of \(\partial_t^3\) is one. Hence \(t=0\) is noncharacteristic everywhere. A distributional homogeneous solution that is zero for \(t<0\) near a point of that plane is zero near the point. In the order \[ W=(u,u_x,u_t,u_{xx},u_{xt},u_{tt}), \tag{30} \] the first five rows are \[ W_{0,t}=W_2,\quad W_{1,t}=W_4,\quad W_{2,t}=W_5, \quad W_{3,t}=\partial_xW_4,\quad W_{4,t}=\partial_xW_5. \tag{31} \] The final row is \[ \begin{split} W_{5,t}={}&-(1+x^2+it)\partial_xW_5 -(2-it)\partial_xW_4-i\partial_xW_3\\ &-(1+t)W_2-xW_1-W_0. \end{split} \tag{32} \] Nothing was diagonalized. At \((x,t)=(0,0)\) the time polynomial at tangential frequency one is \(\tau^3+\tau^2+2\tau+i\). It cannot have only real roots: a monic polynomial with real roots has real coefficients. The uniqueness argument does not require hyperbolicity. ### Example 2: a characteristic surface really can carry a solution. For \(P=\partial_t^2-\partial_x^2\), the normal to \(t=cx\) gives \(p_2(-c,1)=1-c^2\). The two exceptional slopes are \(c=\pm1\). On \(t=x\), \[ u(x,t)=H(t-x) \tag{33} \] is a nonzero distributional solution. Both \(\partial_t^2u\) and \(\partial_x^2u\) equal \(\delta'(t-x)\), so their difference is zero. The solution vanishes below the surface and is nonzero above it. This is permitted because that surface is characteristic. The distribution \(\delta(t-x)\) gives a solution concentrated on the same surface. ### Example 3: an analytic paraboloid touches a nonanalytic graph. Let the candidate closed set be \(K=\{t\ge-|x|^{3/2}\}\). This is a geometric example of a possible support set, not an asserted homogeneous PDE solution. Choose \[ \varepsilon=\frac18,\quad a=\frac1{144},\quad \kappa=\frac{2\varepsilon}{a}=36,\quad F=t+36x^2. \tag{34} \] For \(|x|\le a\), \(|g'(x)|=(3/2)\sqrt{|x|}\le1/8\). The minimum of \(F\) on \(K\) in the cylinder from (28) is achieved at \[ x_\pm=\pm\frac1{2304},\quad t_*=-\frac1{110592},\quad \mu=-\frac1{442368}. \tag{35} \] The supporting parabola is \(t=\mu-36x^2\). Its two contact normals are \[ dt\pm\frac1{32}\,dx. \tag{36} \] They are close to the vertical normal although the graph is not analytic at zero. Solution 5 derives all constants. ![An analytic supporting parabola touches a continuously differentiable cusp at two interior points.](figures/analytic-contact-at-a-C1-graph.png) *Figure 2. The displayed coordinates are \(X=2304x\) and \(Y=110592t\). The graph is \(Y=-|X|^{3/2}\); the supporting parabola is \(Y=-1/4-3X^2/4\). They touch at \((X,Y)=(\pm1,-1)\), which are the exact physical points in (35). The shaded candidate set lies above the graph. The normals labelled in physical coordinates are \(dt\pm dx/32\). This is a zoom of the interior contact region; the full cylinder side is at \(|X|=16\). No differential equation or solution is assigned to this candidate set. See Example 3, Solution 5 and complete-proof Theorem 3.2.* ### Example 4: a square system requires the whole vector. Take \[ P=\begin{pmatrix}\partial_t&-\partial_x\\ \partial_x&\partial_t\end{pmatrix}, \qquad p_1(\xi,\tau)= \begin{pmatrix}\tau&-\xi\\ \xi&\tau\end{pmatrix}. \tag{37} \] Its determinant is \(\tau^2+\xi^2\), so every real nonzero normal is noncharacteristic. Holmgren uniqueness applies to its two-component distribution solutions. The smooth vector \[ U=(\cos x\,\sinh t,\ \sin x\,\cosh t) \tag{38} \] satisfies \(PU=0\), but only its first component vanishes at \(t=0\). The vector does not have zero Cauchy data. If it is extended by zero to \(t<0\), the resulting vector satisfies \[ P(H(t)U)=(0,\sin x)\delta(t), \tag{39} \] so it is not a homogeneous one-sided solution. The theorem requires the vector to vanish on the open zero side; one zero component on the boundary does not suffice. ## 5. What support continuation now gives A noncharacteristic \(C^1\) exterior normal cannot occur at a support point of a homogeneous solution. Consequently a nonzero scalar constant-coefficient operator has no nonzero compactly supported homogeneous distribution on the whole space. There is also full-space causal uniqueness: if its principal symbol is nonzero at a normal \(N\), a homogeneous solution supported in \(N\cdot x\ge0\) is zero, even without a growth bound. Complete-proof Corollaries 4.1–4.3 prove these statements with explicit extrema. They give the local uniqueness input for support-exhaustion arguments; they do not replace the separate constructions of solutions. Nor does this theorem make all solutions analytic. For instance \(\delta(x)\), independent of \(t\), solves \(\partial_tu=0\) and remains singular in \(x\). What matters is the zero side and the direction of the surface. ## 6. Exercises 1. **Basic, 10 points.** For two tangential coordinates, scalar order four and two unknown components, count the jet components. Write the row for \(\alpha=(1,1)\), \(j=1\), choosing the first tangential coordinate. 2. **Intermediate, 12 points.** Flatten \(t=x^2\) for \(P=\partial_t^2-\partial_x^2\). Compute every transformed second- and first-order term. Locate the characteristic points of the graph. 3. **Intermediate, 12 points.** Verify both characteristic distributions in Example 2, including the signs of the two \(x\)-derivatives. Explain why setting the first component of Example 4 to zero at \(t=0\) does not give zero vector Cauchy data. 4. **Advanced, 14 points.** Prove the interior-minimum assertions of the \(C^1\) argument for an arbitrary closed candidate support in \(d\) tangential dimensions. Include the case where the minimum is zero. 5. **Intermediate, 12 points.** Derive every value in (35) and (36), and the two rescaled curves in Figure 2. Verify that the contact points lie in the interior of the full cylinder. 6. **Advanced, 14 points.** Prove causal uniqueness without a growth hypothesis by the minimum in complete-proof Corollary 4.3. Treat separately a chosen support point with \(t_0=0\). 7. **Intermediate, 12 points.** For \(u(x,t)=\delta(x)\), verify \(\partial_tu=0\), show that \(u\) is not smooth near any point of \(x=0\), and explain why its existence does not contradict uniqueness at the noncharacteristic surface \(t=0\). 8. **Advanced, 14 points.** Formally flatten \(t=|x|^{3/2}\) in the transport operator \(\partial_t-\partial_x\). Identify the coefficient that loses analyticity, and explain the further obstruction for pulling back arbitrary distributions by this change. Show how the actual \(C^1\) proof avoids both difficulties. ## 7. Complete solutions ### Solution 1 Here \(d=2\), \(m=4\), \(r=2\), so (25) gives \(2\binom{6}{3}=40\) scalar components. Each component of the original unknown contributes twenty derivatives of total order at most three. The multi-index \((1,1)\) and \(j=1\) have total order three and are at the top layer. Commutation of distributional derivatives gives \[ \partial_tW_{(1,1),1} =\partial_{y_1}W_{(0,1),2}. \tag{40} \] The target has total order \(0+1+2=3\), so it is in the collected vector. The equation holds componentwise for the two original unknowns. These forty components satisfy relations because they are derivatives of the same unknown; the reduction only needs a system they satisfy, not forty independent data. ### Solution 2 Put \(s=t-x^2\). In the new coordinates \(\partial_t=\partial_s\) and \(\partial_x^{\rm old}=Q=\partial_x-2x\partial_s\). Apply \(Q\) twice to a test function. The derivative of its coefficient contributes \(-2\partial_s\): \[ Q^2=\partial_x^2-4x\partial_x\partial_s +4x^2\partial_s^2-2\partial_s. \tag{41} \] Therefore the transformed wave operator is \[ (1-4x^2)\partial_s^2+4x\partial_x\partial_s -\partial_x^2+2\partial_s. \tag{42} \] The normal of the original graph is \((-2x,1)\), and the original principal symbol at it is \(1-4x^2\). This agrees with the coefficient of \(\partial_s^2\). The graph is characteristic at exactly \(x=\pm1/2\). Around zero the coefficient is nonzero, so one-sided uniqueness holds there. The term \(2\partial_s\) must be retained; squaring a variable-coefficient vector field is not the same as squaring its symbol. ### Solution 3 Let \(s=t-x\). Then \(\partial_tH(s)=\delta(s)\), \(\partial_xH(s)=-\delta(s)\). Differentiating again gives \(\partial_t^2H(s)=\delta'(s)\) and \(\partial_x^2H(s)=\delta'(s)\), since the second \(x\)-derivative introduces a second minus sign. Thus their difference is zero. For \(\delta(s)\), the same calculation gives \(\delta''(s)\) for both second derivatives. Neither distribution is zero near the characteristic line. In Example 4, at \(t=0\) the vector is \((0,\sin x)\). A vector Cauchy value is zero only if both entries are zero. Product differentiation, and the fact that the normal coefficient of \(P\) is the identity matrix, give \(P(HU)=H(PU)+\delta(t)U(x,0)\). This is (39), with its nonzero boundary source. The first component's zero trace cannot cancel that second-component source. ### Solution 4 The candidate closed set lies in \(t\ge g(y)\), where \(g(0)=0\) and \(|g(y)|\le\varepsilon|y|\) on \(|y|\le a\). Intersect it with the closed cylinder (18). If it contains the origin, that intersection is nonempty and compact, so the continuous function (19) attains a minimum no larger than \(F(0,0)=0\). At the bottom, \(t=-2\varepsilon a<-\varepsilon a\le g(y)\), so no candidate point occurs. At the lateral boundary, \(F\ge-\varepsilon a+2\varepsilon a=\varepsilon a>0\). At the top, \(F\ge2\varepsilon a>0\). Thus every minimizing point is interior. These strict boundary inequalities hold also when the minimum is zero. On a neighborhood of a minimizer, candidate points satisfy \(F\ge\mu\). The level is an analytic graph because \(\partial_tF=1\). Its normal's tangential part has length \(2\kappa|y|\le4\varepsilon\). Noncharacteristic normals form an open set: the determinant is continuous and nonzero at the original normal. Choosing the displayed normal neighborhood inside that open set permits the analytic-graph uniqueness theorem. At a support point this is a contradiction; no regularity of the closed support itself was used. ### Solution 5 On the lower boundary of \(K\), minimize \(h(x)=-|x|^{3/2}+36x^2\). For \(x>0\), \[ h'(x)=-\frac32\sqrt{x}+72x. \tag{43} \] Apart from zero, the critical point satisfies \(\sqrt{x}=1/48\), hence \(x=1/2304\). By evenness there is a second point at its negative. The derivative is negative before the positive point and positive after it, so these are the two minima. At a contact, \[ t=-\frac1{48^3}=-\frac1{110592}, \qquad 36x^2=\frac1{147456}, \qquad \mu=-\frac1{442368}. \tag{44} \] The normal of \(F\) is \(dt+72x\,dx\), giving \(dt\pm dx/32\). Also \(2\varepsilon a=1/576\); the contact has \(|x|0\), at a minimum \(|2\delta(y-y_0)|\le2\sqrt{\delta t_0}\). Choose \(\delta\) small enough to make all these normals noncharacteristic. The support is on one side of the analytic minimum level and the homogeneous equation holds near the point. Holmgren uniqueness contradicts membership in the support. For \(t_0=0\), the only point in the sublevel is \((y_0,0)\), the minimum is zero, and \(dG=dt\) there. This normal is noncharacteristic by hypothesis, giving the same contradiction without a smallness estimate. Compactness was needed only for one sublevel, not for the whole support. No transform or growth bound of the solution was used. ### Solution 7 For a compactly supported smooth test \(\phi\), \[ \langle\partial_tu,\phi\rangle =-\int_{\mathbb R}\partial_t\phi(0,t)\,dt=0. \tag{47} \] Thus \(u\) solves the equation. Its support is \(x=0\), and it is nonzero on every neighborhood meeting that line. If it were a smooth function on such a neighborhood, that function would vanish on the dense open set \(x\ne0\), and continuity would make it zero everywhere. This contradicts its nonzero action on a nonnegative test meeting the line. The surface \(t=0\) has noncharacteristic normal for \(\partial_t\), but \(u\) is not zero on its lower side. A test supported near \(x=0\) and at a small negative time has nonzero pairing. The premise of the one-sided theorem therefore fails. Holmgren uniqueness does not imply smoothness of solutions in unrelated directions. ### Solution 8 For \(s=t-g(x)\), a formal first-order chain rule on smooth functions gives \[ \partial_t-\partial_x =(1+g'(x))\partial_s-\partial_x. \tag{48} \] Here \(g'(x)=(3/2)\operatorname{sgn}(x)\sqrt{|x|}\), with value zero at the origin. This is continuous but is not analytic at zero. The transformed coefficients no longer satisfy the analytic first-order theorem. There is a separate distributional issue: \(t=s+g(x)\) is only a \(C^1\) diffeomorphism. Composing an arbitrary smooth test with it need not give a smooth test, so the scalar distributional pullback formula (11) is not available for arbitrary distributions. The formal calculation alone does not repair this. The actual proof uses only affine coordinates to describe the \(C^1\) graph. Affine changes preserve analytic coefficients and act smoothly on distributions. It then minimizes \(t+\kappa x^2\) on a compact local support set. The minimum level is an analytic graph with an explicit analytic inverse coordinate change. Its normal is kept in the original noncharacteristic neighborhood. The analytic theorem is applied only at that contact, so neither loss of analyticity nor a nonsmooth distributional pullback occurs. ## References [S]: ../../AN02-L113.html#7-the-analytic-uniqueness-argument-used-by-the-barrier [C]: ../../AN02-L045.html - **[S]** *Fundamental solutions in shrinking cones*, Sections 7.1–7.2. The full first-order analytic test construction and one-sided distributional theorem. - **[C]** *Cauchy bounds, root counts and analytic extensions*, Lemmas 1.1 and 3.1. - **[R]** Romain Crétier, *Cauchy-Kovalevska Theorem, Characteristics and Holmgren Theorem*, Sections 2, 3 and 5. [Author-hosted notes](https://www.imo.universite-paris-saclay.fr/media/filer_public/2f/54/2f545d60-a80b-41b4-ae17-58f05160651c/notes_ck.pdf). Freely readable background. - **[H]** Lars Hörmander, “Remarks on Holmgren's uniqueness theorem,” *Annales de l'Institut Fourier* **43** (1993), no. 5, 1223–1251. [Original article](https://www.numdam.org/item/10.5802/aif.1371.pdf), DOI 10.5802/aif.1371.