An invariant box and the entire graph

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

A radial solution through the origin produced a solution of the radial system near \(r=0\). This lesson shows that in the dimension \(m=9\), that is for graphs in ten variables, the solution exists for all \(r>0\) with \(F,P>0\), and proves Theorem A of The affine maximal equation, following OpenAI [OpenAI-AM, Section 4]. The danger is that \(P\) reaches \(0\) at a finite radius, as it does when \(m=1\) (Exercise 3.4 of A radial solution through the origin). It is excluded by following three logarithmic derivatives, which satisfy an autonomous system and stay in a box whose upper corner comes from the singular profile \(r^{9/2}\).

Section 1 collects the facts on ordinary differential equations that are used, with proofs. Results of the core courses Real Analysis I and II are cited from the text of Lebl [Lebl] by number.

1. Solutions of ordinary differential equations

Let \(U\subseteq\mathbb R\times\mathbb R^N\) be open and \(G:U\to\mathbb R^N\) continuously differentiable. A solution on an interval \(I\) is a continuously differentiable \(Y:I\to\mathbb R^N\) with \((r,Y(r))\in U\) and \(Y'(r)=G(r,Y(r))\) for \(r\in I\); at an endpoint of \(I\) that belongs to \(I\) the derivative is one-sided. Integrals of vector-valued functions are taken componentwise, and \(|\cdot|\) is the Euclidean norm.

Lemma 1.1 (local existence and uniqueness). For \((r_0,Y_0)\in U\) there are \(h>0\) and a solution on \([r_0-h,r_0+h]\) with \(Y(r_0)=Y_0\). Two solutions on an interval \(I\) that agree at one point agree on \(I\).

Proof. Choose \(a,b>0\) with \(C=[r_0-a,r_0+a]\times\bar B(Y_0,b)\subseteq U\). On the compact set \(C\), \(|G|\le M_G\) and the derivative of \(G\) in \(Y\) has norm at most some \(L\). As \(\bar B(Y_0,b)\) is convex, with \(Z_\theta=Z+\theta(Y-Z)\), \[ \begin{aligned} &G(r,Y)-G(r,Z)\\ &\quad=\int_0^1D_YG(r,Z_\theta)(Y-Z)\,d\theta , \end{aligned} \] which gives \(|G(r,Y)-G(r,Z)|\le L|Y-Z|\) on \(C\). Let \(h=\min\bigl(a,\,b/(M_G+1),\,1/(2L+1)\bigr)\) and let \(E\) be the set of continuous \(Y:[r_0-h,r_0+h]\to\bar B(Y_0,b)\) with the supremum norm, a complete metric space [Lebl, Propositions 7.4.5 and 7.4.6]. The map \[ T(Y)(r)=Y_0+\int_{r_0}^rG(\rho,Y(\rho))\,d\rho \] sends \(E\) to itself, since \(|T(Y)(r)-Y_0|\le hM_G\le b\), and \(\|T(Y)-T(Z)\|\le hL\|Y-Z\|\), which is at most \(\frac12\|Y-Z\|\). Its fixed point [Lebl, Theorem 7.6.2] is a solution, by the fundamental theorem of calculus [Lebl, Theorem 5.3.3].

For uniqueness, let \(Y,\hat Y\) be solutions on \(I\) and \(S=\{r\in I:Y(r)=\hat Y(r)\}\), which is nonempty and closed in \(I\). If \(r_1\in S\), apply the construction at \((r_1,Y(r_1))\), with constants \(h_1,b_1\). By continuity there is \(h'\in(0,h_1]\) such that on \(J=I\cap[r_1-h',r_1+h']\) both \(Y\) and \(\hat Y\) stay within \(b_1\) of \(Y(r_1)\). Both satisfy \(Y(r)=Y(r_1)+\int_{r_1}^rG(\rho,Y(\rho))\,d\rho\) on \(J\) [Lebl, Theorem 5.3.1], so both are fixed points of the corresponding contraction on the continuous maps \(J\to\bar B(Y(r_1),b_1)\), and they agree on \(J\). So \(S\) is open in \(I\), and \(S=I\) because \(I\) is connected. \(\square\)

Lemma 1.2 (maximal solutions). For \((r_0,Y_0)\in U\) there is a solution \(Y\) on an open interval \((\alpha,\omega)\ni r_0\) with \(Y(r_0)=Y_0\) of which every solution through \((r_0,Y_0)\) is a restriction. If \(\omega<\infty\) and \(\mathcal C\subseteq U\) is compact, there is no \(r_1<\omega\) with \((r,Y(r))\in\mathcal C\) for all \(r\in[r_1,\omega)\).

Proof. By Lemma 1.1 all solutions through \((r_0,Y_0)\) agree where their intervals overlap, so together they define a solution on the union \(I^*\) of their intervals. If \(I^*\) contained its supremum \(\omega\), Lemma 1.1 at \((\omega,Y(\omega))\) and uniqueness would give a solution on a longer interval; so \(I^*\) is open on the right, and likewise on the left.

Suppose \(\omega<\infty\) and \((r,Y(r))\in\mathcal C\) for \(r\in[r_1,\omega)\). Then \(|Y'|\le\max_{\mathcal C}|G|\) there, so \(Y\) is Lipschitz on \([r_1,\omega)\) and \(Y(r)\) has a limit \(Y_\omega\) as \(r\to\omega\), with \((\omega,Y_\omega)\in\mathcal C\subseteq U\). Setting \(Y(\omega)=Y_\omega\), the identity \(Y(r)=Y(r_1)+\int_{r_1}^rG(\rho,Y(\rho))\,d\rho\) extends to \(r=\omega\), so \(Y\) is a solution on \((\alpha,\omega]\). Lemma 1.1 at \((\omega,Y_\omega)\) gives a solution on \([\omega-h,\omega+h]\), which agrees with \(Y\) on \([\omega-h,\omega]\); joined with \(Y\) it is a solution on \((\alpha,\omega+h)\), contradicting the first part. \(\square\)

Lemma 1.3. If \(G\) is smooth, every solution is smooth.

Proof. If \(Y\in C^\ell\), then \(r\mapsto G(r,Y(r))\) is of class \(C^\ell\), so \(Y'\in C^\ell\) and \(Y\in C^{\ell+1}\). \(\square\)

2. The radial system as a regular system

From here on \[ m=9,\qquad k=11,\qquad\gamma=\tfrac{12}{11} . \] On the open set \(U\subseteq\mathbb R\times\mathbb R^4\) of the points \((r,F,P,y,Z)\) with \(r,F,P,y>0\) define the smooth map \[ \begin{gathered} G(r,F,P,y,Z)=(y,\ Z,\ g,\ G_4),\\ g=\frac{r^8FP^{-12/11}}{y^8},\\ G_4=-8\frac gyZ-\frac{11r^8P^{-1/11}}{y^8} . \end{gathered} \]

Lemma 2.1. Let \(F,P\) be profiles of class \(C^2\) on an interval \((a,b)\subseteq(0,\infty)\) with \(F,P,y>0\), where \(y=F'\). They solve the radial system of Graphs over a convex profile if and only if \(Y=(F,P,y,P')\) solves \(Y'=G(r,Y)\).

Proof. The first radial equation, \(y^8y'=r^8FP^{-12/11}\), says \(y'=g\). Expanding the second, \(8y^7y'P'+y^8P''=-11r^8P^{-1/11}\); given \(y'=g\), it says \(P''=-8\frac gyP'-11r^8P^{-1/11}/y^8\). \(\square\)

Let \(F,P\) be the solution of Proposition 2.1 of A radial solution through the origin, on \(0<r<\varepsilon\). Fix \(r_1\in(0,\varepsilon)\) and let \(Y=(F,P,y,Z)\) be the maximal solution of \(Y'=G(r,Y)\) through \((r_1,(F,P,F',P')(r_1))\), on \((\alpha,\omega)\) (Lemma 1.2). By Lemma 2.1 the local solution is a solution through the same point, so it is a restriction of \(Y\); as solutions stay in \(U\), where \(r>0\), \(\alpha=0\). Thus the local solution extends to \((0,\omega)\), with \(F,P,y>0\), \(Z=P'\), and \(y'=g>0\); it is smooth by Lemma 1.3. It remains to show \(\omega=\infty\).

3. Logarithmic derivatives

On \((0,\omega)\) define \[ \begin{gathered} A=\frac{ry}F,\qquad B=\frac{ry'}y,\\ C=-\frac{rP'}P,\qquad D=\frac CB , \end{gathered} \] which is possible because \(B>0\). A dot denotes \(r\,\frac d{dr}\), the derivative with respect to \(\log r\).

Lemma 3.1 (OpenAI). On \((0,\omega)\), \(A>0\), \(B>0\), and \[ \begin{aligned} \dot A&=A(1+B-A),\\ \dot B&=B\bigl(9+A+\tfrac{12}{11}BD-9B\bigr),\\ \dot D&=(11-D)A-8D+BD\bigl(1-\tfrac D{11}\bigr). \end{aligned} \] Moreover \((A,B,D)\to(0,1,0)\) as \(r\to0\), and \(A,B,D>0\) for all sufficiently small \(r>0\).

Proof. The computation is the same for every \(m\), with \(k=m+2\) and \(\gamma=\frac{m+3}{m+2}\). Positivity of \(F,y,y'\) gives \(A,B>0\). By the first radial equation, \[ B=\frac{r^mFP^{-\gamma}}{y^m},\qquad AB=\frac{r^{m+1}P^{-\gamma}}{y^{m-1}} . \] Taking \(r\frac d{dr}\log\) of \(A\) and of \(B\), \[ \begin{aligned} \dot A&=A\Bigl(1+\frac{ry'}y-\frac{rF'}F\Bigr)\\ &=A(1+B-A),\\ \dot B&=B\Bigl(m+\frac{rF'}F-\gamma\frac{rP'}P-m\frac{ry'}y\Bigr)\\ &=B(m+A+\gamma C-mB). \end{aligned} \] Next \(\dot C=C+C^2-r^2P''/P\), and the second radial equation gives \[ \begin{aligned} P''={}&-(m-1)\frac{y'}yP'\\ &-\frac{kr^{m-1}P^{1-\gamma}}{y^{m-1}}, \end{aligned} \] so \[ \begin{gathered} -\frac{r^2P''}P=-(m-1)BC+kAB,\\ \dot C=C\bigl(1+C-(m-1)B\bigr)+kAB . \end{gathered} \] With \(C=BD\), \[ \begin{aligned} \dot D&=\frac{\dot C}B-D\frac{\dot B}B\\ &=(k-D)A-(m-1)D\\ &\quad+BD\bigl(1+(1-\gamma)D\bigr), \end{aligned} \] and \(m=9\), \(k=11\), \(\gamma=\frac{12}{11}\) give the stated system.

Near \(r=0\), Proposition 1.3 of A radial solution through the origin gives, with \(s=r^2/2\), \(F=f(s)=1+O(r^2)\), \(y=rq(s)=r+O(r^3)\), \(y'=q(s)+r^2q'(s)=1+O(r^2)\), \(P=1+O(r^2)\) and \(P'=rp'(s)=-\frac{11}9r+O(r^3)\). Hence \(A=r^2+O(r^4)\), \(B=1+O(r^2)\) and \(C=\frac{11}9r^2+O(r^4)\), so \((A,B,D)\to(0,1,0)\). Since \(P'<0\) for small \(r>0\), \(C\) and \(D\) are positive there. \(\square\)

4. The invariant box

Write \(V=(V_1,V_2,V_3)\) for the right side of the system in Lemma 3.1, a polynomial vector field on \(\mathbb R^3\), and let \[ \mathcal B=[0,\tfrac92]\times[0,\tfrac72]\times[0,\tfrac{33}7]. \]

Lemma 4.1 (OpenAI). (a) \(V(\frac92,\frac72,\frac{33}7)=0\).

(b) On the faces of \(\mathcal B\): \[ \begin{aligned} V_1(\tfrac92,B,D)&=\tfrac92\bigl(B-\tfrac72\bigr)\le0,\\ V_2(A,\tfrac72,D)&=\tfrac72\bigl(A+\tfrac{42}{11}D-\tfrac{45}2\bigr)\le0,\\ V_3(A,B,\tfrac{33}7)&=\tfrac{44}7A-\tfrac{264}7+\tfrac{132}{49}B\le0, \end{aligned} \] and \(V_1=0\) where \(A=0\), \(V_2=0\) where \(B=0\), \(V_3=11A\ge0\) where \(D=0\), for \((A,B,D)\in\mathcal B\).

(c) If \(z:[s_0,s_1)\to\mathbb R^3\), with \(s_1\le\infty\), is continuously differentiable with \(\dot z=V(z)\) and \(z(s_0)\in\mathcal B\), then \(z(s)\in\mathcal B\) for all \(s\in[s_0,s_1)\).

Proof. (a) and the face formulas are direct substitutions; for instance \(V_2(\frac92,\frac72,\frac{33}7)=\frac72(\frac92+18-\frac{45}2)=0\). On the box, \(B\le\frac72\) gives the first inequality, \(A\le\frac92\) and \(D\le\frac{33}7\) give \(A+\frac{42}{11}D\le\frac92+18=\frac{45}2\), and \(A\le\frac92\), \(B\le\frac72\) give \(\frac{44}7A+\frac{132}{49}B\le\frac{198}7+\frac{66}7=\frac{264}7\).

(c) Let \(\Pi z\) be the point of \(\mathcal B\) obtained by clamping each coordinate of \(z\) to its interval. For every \(z\in\mathbb R^3\), \[ (z-\Pi z)\cdot V(\Pi z)\le0 . \] Indeed, the coordinates in which \(z\) and \(\Pi z\) differ are those where \(z\) lies above or below the interval; for such a coordinate \(\Pi z\) lies on the corresponding upper or lower face, where by (b) the component of \(V\) is \(\le0\) on an upper face and \(\ge0\) on a lower face, opposite in sign to \(z_i-(\Pi z)_i\). The function \(\Delta(z)=|z-\Pi z|^2\) is continuously differentiable with gradient \(2(z-\Pi z)\), since in each coordinate it is the squared distance to an interval.

Let \(\mathcal N\) be the set of points within distance \(1\) of \(\mathcal B\), a compact convex set; on it \(|V(z)-V(z')|\le L_V|z-z'|\) for some \(L_V\), by the mean value inequality for the polynomial \(V\). Suppose \(z\) leaves \(\mathcal B\), and let \(s_*\) be the supremum of the \(s\) with \(z([s_0,s])\subseteq\mathcal B\); then \(s_*<s_1\) and \(z(s_*)\in\mathcal B\). Choose \(s_2\in(s_*,s_1)\) with \(z([s_*,s_2])\subseteq\mathcal N\). For \(s\) in that interval, with \(z=z(s)\), \[ \begin{aligned} \frac d{ds}\Delta(z)&=2(z-\Pi z)\cdot V(\Pi z)\\ &\quad+2(z-\Pi z)\cdot\bigl(V(z)-V(\Pi z)\bigr)\\ &\le2L_V\Delta(z), \end{aligned} \] so \(e^{-2L_V(s-s_*)}\Delta(z(s))\) is nonincreasing and vanishes at \(s_*\). Hence \(z(s)\in\mathcal B\) on \([s_*,s_2]\), contradicting the choice of \(s_*\). \(\square\)

5. The entire solution and the proof of Theorem A

Proposition 5.1 (OpenAI). \(\omega=\infty\): the radial solution exists for all \(r>0\), with \(F,P,y,y'>0\).

Proof. By Lemma 3.1 there is \(r_0\in(0,\varepsilon)\) with \((A,B,D)(r_0)\in\mathcal B\). The curve \(s\mapsto(A,B,D)(e^s)\) satisfies \(\dot z=V(z)\), so by Lemma 4.1 it stays in \(\mathcal B\) for \(r_0\le r<\omega\). Suppose \(\omega<\infty\). For \(r_0\le r<\omega\), \[ \begin{gathered} 0\le A\le\tfrac92,\qquad0\le B\le\tfrac72,\\ 0\le C=BD\le\tfrac{33}2 . \end{gathered} \] Since \(r\frac d{dr}\log F=A\), \(r\frac d{dr}\log y=B\) and \(r\frac d{dr}\log P=-C\), integration from \(r_0\) to \(r\) gives \[ \begin{gathered} F(r_0)\le F(r)\le F(r_0)(\omega/r_0)^{9/2},\\ y(r_0)\le y(r)\le y(r_0)(\omega/r_0)^{7/2},\\ P(r_0)(r_0/\omega)^{33/2}\le P(r)\le P(r_0), \end{gathered} \] and \(|P'(r)|=CP/r\le\frac{33}2P(r_0)/r_0\). So \((r,F,P,y,P')\) stays in a compact box contained in \(U\), as all its lower bounds for \(r,F,P,y\) are positive. This contradicts Lemma 1.2. \(\square\)

Proof of Theorem A. Let \(F,P\) be the radial functions on \(\mathbb R^9\) with the profiles of Proposition 5.1. On the ball \(|x|<\varepsilon\) they coincide with the smooth functions of Proposition 2.1 of A radial solution through the origin, and on \(\mathbb R^9\setminus\{0\}\) they are compositions of smooth maps; so they are smooth and positive on \(\mathbb R^9\). Their Hessian \(D^2F\) is positive definite: for \(x\ne0\) its eigenvalues are \(y'>0\) and \(y/r>0\) (Lemma 3.1 of Graphs over a convex profile), and \(D^2F(0)=I\). By Lemma 2.1 and Corollary 3.2 of Graphs over a convex profile, the two equations \[ \begin{gathered} \det D^2F=FP^{-12/11},\\ \operatorname{tr}\bigl((D^2F)^{-1}D^2P\bigr)+11\frac PF=0 \end{gathered} \] hold on \(\mathbb R^9\setminus\{0\}\); both sides are continuous on \(\mathbb R^9\), so they hold at \(0\) as well. Proposition 2.1 of Graphs over a convex profile shows that \[ \begin{gathered} u(x,t)=\sqrt{F(x)^2+t^2},\\ (x,t)\in\mathbb R^9\times\mathbb R , \end{gathered} \] is smooth, has positive definite Hessian everywhere, and solves the affine maximal equation on \(\mathbb R^{10}\). Finally \(u(0,t)=\sqrt{1+t^2}\), whose second derivative \((1+t^2)^{-3/2}\) is not constant, while the restriction of a polynomial of degree at most two to the line \(x=0\) has constant second derivative. So \(u\) is not a polynomial of degree at most two. \(\square\)

6. Exercises

6.1 (easy). Check that \(V(\frac92,\frac72,\frac{33}7)=0\) for all three components.

6.2 (medium). For the singular profiles \(F=r^{9/2}\), \(P=cr^{-33/2}\) of Exercise 4.2 of Graphs over a convex profile, show that \(A,B,C,D\) are the constants \(\frac92,\frac72,\frac{33}2,\frac{33}7\). So the upper corner of \(\mathcal B\) is the constant value of \((A,B,D)\) along the singular example \(u_0\) of The affine maximal equation.

6.3 (medium). Show that \((0,1,0)\) is a zero of \(V\), and that the matrix of derivatives of \(V\) there has eigenvalues \(2\), \(-9\) and \(-7\). Relate the eigenvalue \(2\) to the behaviour \(A=r^2+O(r^4)\) found in Lemma 3.1.

6.4 (medium). Show that for \(r\ge1\) \[ F(1)+y(1)(r-1)\le F(r)\le F(1)\,r^{9/2}, \] so the solution \(u\) of Theorem A grows at least linearly and at most like \(|x|^{9/2}\) in the \(x\) directions.

7. Solutions

6.1. At the corner, \[ \begin{aligned} V_1&=\tfrac92\bigl(1+\tfrac72-\tfrac92\bigr)=0,\\ V_2&=\tfrac72\bigl(9+\tfrac92+\tfrac{12}{11}\cdot\tfrac72\cdot\tfrac{33}7-\tfrac{63}2\bigr)\\ &=\tfrac72\bigl(\tfrac{27}2+18-\tfrac{63}2\bigr)=0,\\ V_3&=\tfrac{44}7\cdot\tfrac92-\tfrac{264}7+\tfrac72\cdot\tfrac{33}7\cdot\tfrac47\\ &=\tfrac{198-264+66}7=0 . \end{aligned} \]

6.2. \(y=\frac92r^{7/2}\) and \(y'=\frac{63}4r^{5/2}\) give \(A=ry/F=\frac92\) and \(B=ry'/y=\frac72\); \(P'=-\frac{33}2P/r\) gives \(C=\frac{33}2\), and \(D=C/B=\frac{33}7\).

6.3. At \(A=0\), \(B=1\), \(D=0\): \(V_1=0\), \(V_2=1\cdot(9-9)=0\), \(V_3=0\). The derivatives there are \(\partial_AV_1=1+B-2A=2\), \(\partial_BV_1=\partial_DV_1=0\); \(\partial_AV_2=B=1\), \(\partial_BV_2=9+A+\frac{24}{11}BD-18B=-9\), \(\partial_DV_2=\frac{12}{11}B^2=\frac{12}{11}\); \(\partial_AV_3=11-D=11\), \(\partial_BV_3=D(1-\frac D{11})=0\), \(\partial_DV_3=-A-8+B(1-\frac{2D}{11})=-7\). The matrix \[ \begin{pmatrix}2&0&0\\1&-9&\tfrac{12}{11}\\11&0&-7\end{pmatrix} \] has characteristic polynomial \((2-\lambda)(-9-\lambda)(-7-\lambda)\), by expanding along the first row. The solution leaves \((0,1,0)\) as \(\log r\) increases, along the direction of the eigenvalue \(2\): \(A\) grows like \(e^{2\log r}=r^2\).

6.4. Since \(B>0\), \(y\) is increasing, so \(F(r)=F(1)+\int_1^ry\), which is at least \(F(1)+y(1)(r-1)\). Since \(r\frac d{dr}\log F=A\le\frac92\) on \([1,\infty)\) (the solution stays in \(\mathcal B\) for \(r\ge r_0\), and \(r_0\) may be taken below \(1\)), \(\log F(r)-\log F(1)\le\frac92\log r\). The function \(u\) satisfies \(F(|x|)\le u(x,t)\le F(|x|)+|t|\).

References