A radial solution through the origin
Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).
Graphs over a convex profile reduced the affine maximal equation for \(\sqrt{F(x)^2+t^2}\) to the radial system \[ \begin{aligned} y^{m-1}y'&=r^{m-1}FP^{-\gamma},\\ (y^{m-1}P')'&=-k\,r^{m-1}P^{1-\gamma},\qquad F'=y, \end{aligned} \] for profiles \(F,P\) of radial functions, with \(k=m+2\) and \(\gamma=\frac{m+3}{m+2}\). The system divides by powers of \(y\), and \(y=F'\) vanishes at \(r=0\) for a smooth radial \(F\). This lesson constructs a solution with \(F(0)=P(0)=1\) that is smooth as a function of \(x\) near the origin, following OpenAI [OpenAI-AM, Section 3]. The idea is to integrate the system once from \(r=0\), to write it in the variable \(s=r^2/2\), and to solve the resulting integral equations with the contraction mapping principle.
Throughout, \(m\ge1\). Results of the core courses Real Analysis I and II are cited from the text of Lebl [Lebl] by number.
1. The fixed-point problem
Fix \(\delta\in(0,1]\). Let \(X=C([-\delta,\delta])^2\), the pairs \((f,p)\) of continuous functions, with the norm \(\|(f,p)\|=\max(\|f\|_\infty,\|p\|_\infty)\). It is complete: a Cauchy sequence has Cauchy components, which converge uniformly [Lebl, Proposition 7.4.5]. The set \[ \begin{aligned} \mathcal K=\bigl\{(f,p)\in X:{}&\|f-1\|_\infty\le\tfrac12,\\ &\|p-1\|_\infty\le\tfrac12\bigr\} \end{aligned} \] is closed in \(X\), hence complete [Lebl, Proposition 7.4.6]. For \((f,p)\in\mathcal K\) and \(s\in[-\delta,\delta]\) put \[ \begin{aligned} Q(s)&=m\int_0^1v^{m-1}f(v^2s)\,p(v^2s)^{-\gamma}\,dv,\\ j(s)&=\int_0^1v^{m-1}p(v^2s)^{1-\gamma}\,dv , \end{aligned} \] \(q=Q^{1/m}\), and define \(T(f,p)=(\tilde f,\tilde p)\) by \[ \begin{aligned} \tilde f(s)&=1+\int_0^sq(a)\,da,\\ \tilde p(s)&=1-k\int_0^sj(a)\,q(a)^{1-m}\,da . \end{aligned} \] For \(v\in[0,1]\) the point \(v^2s\) lies in \([-\delta,\delta]\), so the integrands are defined.
Lemma 1.1 (bounds). For \((f,p)\in\mathcal K\), the functions \(q\) and \(j\) are continuous on \([-\delta,\delta]\), and \[ \begin{gathered} \lambda\le q\le\Lambda,\\ \tfrac1m(\tfrac32)^{1-\gamma}\le j\le\tfrac1m2^{\gamma-1},\\ \lambda=\bigl(\tfrac12(\tfrac32)^{-\gamma}\bigr)^{1/m},\qquad\Lambda=\bigl(\tfrac32\,2^\gamma\bigr)^{1/m}. \end{gathered} \]
Proof. The integrands are continuous on \([0,1]\times[-\delta,\delta]\), so \(Q\) and \(j\) are continuous [Lebl, Proposition 7.5.12]. Since \(f,p\) take values in \([\frac12,\frac32]\) and \(a\mapsto a^{-\gamma}\) is decreasing, \(\frac12(\frac32)^{-\gamma}\le fp^{-\gamma}\le\frac32\,2^\gamma\); as \(m\int_0^1v^{m-1}dv=1\), \(Q\) lies between the same bounds, and \(q=Q^{1/m}\) between \(\lambda\) and \(\Lambda\). Since \(1-\gamma<0\), \(p^{1-\gamma}\) lies between \((\frac32)^{1-\gamma}\) and \(2^{\gamma-1}\), and \(\int_0^1v^{m-1}dv=\frac1m\). \(\square\)
Lemma 1.2 (contraction). There are constants \(M\) and \(L\), depending only on \(m\), such that for every \(\delta\in(0,1]\) and all \((f,p),(\hat f,\hat p)\in\mathcal K\), \[ \begin{aligned} &\|T(f,p)-(1,1)\|\le M\delta,\\ &\|T(f,p)-T(\hat f,\hat p)\|\\ &\quad\le L\delta\,\|(f,p)-(\hat f,\hat p)\| . \end{aligned} \]
Proof. By Lemma 1.1, \(|\tilde f(s)-1|\le\Lambda|s|\) and \(|\tilde p(s)-1|\le\frac km2^{\gamma-1}\lambda^{1-m}|s|\), which gives the first bound with \(M=\max(\Lambda,\frac km2^{\gamma-1}\lambda^{1-m})\).
For the second, write \(\Delta=\|(f,p)-(\hat f,\hat p)\|\). On \([\frac12,\frac32]\) the derivatives of \(a\mapsto a^{-\gamma}\) and \(a\mapsto a^{1-\gamma}\) are bounded in absolute value by \(c_1=\gamma2^{\gamma+1}\) and \(c_2=(\gamma-1)2^\gamma\), so by the mean value theorem [Lebl, Theorem 4.2.4] these functions are Lipschitz there with those constants. Hence, at every point, \[ \begin{aligned} &|fp^{-\gamma}-\hat f\hat p^{-\gamma}|\\ &\quad\le|f-\hat f|\,p^{-\gamma}+\hat f\,|p^{-\gamma}-\hat p^{-\gamma}|\\ &\quad\le\bigl(2^\gamma+\tfrac32c_1\bigr)\Delta , \end{aligned} \] so \(|Q-\hat Q|\le(2^\gamma+\frac32c_1)\Delta\), and \(|j-\hat j|\le\frac{c_2}m\Delta\). The functions \(b\mapsto b^{1/m}\) on \([\lambda^m,\Lambda^m]\) and \(b\mapsto b^{1-m}\) on \([\lambda,\Lambda]\) are Lipschitz, again by the mean value theorem, and products of bounded Lipschitz functions are Lipschitz. So \(|q-\hat q|\le c_3\Delta\) and \(|jq^{1-m}-\hat j\hat q^{1-m}|\le c_4\Delta\) with constants depending only on \(m\). Integrating over an interval of length \(|s|\le\delta\) gives the second bound with \(L=\max(c_3,kc_4)\). \(\square\)
Proposition 1.3. Choose \(\delta\in(0,1]\) with \(M\delta\le\frac12\) and \(L\delta\le\frac12\). Then \(T\) maps \(\mathcal K\) into itself and has exactly one fixed point \((f,p)\in\mathcal K\). The functions \(f\) and \(p\) are smooth on \((-\delta,\delta)\), with \[ \begin{gathered} f'=q,\qquad p'=-k\,j\,q^{1-m},\\ f(0)=p(0)=1,\\ f'(0)=1,\qquad p'(0)=-\frac km . \end{gathered} \]
Proof. The functions \(\tilde f,\tilde p\) are continuous, and the first bound of Lemma 1.2 gives \(T(\mathcal K)\subseteq\mathcal K\). The second makes \(T\) a contraction with constant \(\frac12\), so the contraction mapping principle [Lebl, Theorem 7.6.2] gives a unique fixed point.
At the fixed point, \(f=1+\int_0^sq\) and \(p=1-k\int_0^sjq^{1-m}\) with continuous integrands, so by the fundamental theorem of calculus [Lebl, Theorem 5.3.3] \(f\) and \(p\) are continuously differentiable with the stated derivatives.
Smoothness. Suppose \(f,p\in C^\ell((-\delta,\delta))\) for some \(\ell\ge1\), and fix \(\delta'<\delta\). The integrand \(v^{m-1}f(v^2s)p(v^2s)^{-\gamma}\) has partial derivatives in \(s\) up to order \(\ell\), continuous on \([0,1]\times[-\delta',\delta']\): by the chain rule \(\partial_s^i[f(v^2s)]=v^{2i}f^{(i)}(v^2s)\), and \(p\ge\frac12\). Differentiating under the integral sign \(\ell\) times [Lebl, Theorem 9.1.1], \(Q\in C^\ell([-\delta',\delta'])\), and \(q=Q^{1/m}\in C^\ell\) because \(Q\ge\lambda^m>0\). In the same way \(j\) and \(q^{1-m}\) are of class \(C^\ell\). So \(f'\) and \(p'\) are of class \(C^\ell\) on \((-\delta',\delta')\), and \(f,p\in C^{\ell+1}\) there. As \(\delta'<\delta\) is arbitrary, induction gives \(f,p\in C^\infty((-\delta,\delta))\).
Values at \(0\). \(Q(0)=m\int_0^1v^{m-1}f(0)p(0)^{-\gamma}\,dv=1\), so \(f'(0)=q(0)=1\); and \(j(0)=\frac1m\), so \(p'(0)=-\frac km\). \(\square\)
2. The radial functions
Proposition 2.1 (OpenAI). Let \((f,p)\) be the fixed point of Proposition 1.3, \(\varepsilon=\sqrt{2\delta}\), and on the ball \(B=\{x\in\mathbb R^m:|x|<\varepsilon\}\) define \[ \begin{gathered} F(x)=f\bigl(\tfrac12|x|^2\bigr),\\ P(x)=p\bigl(\tfrac12|x|^2\bigr). \end{gathered} \] Then \(F,P\in C^\infty(B)\) are positive, \(F(0)=P(0)=1\), \[ D^2F(0)=I,\qquad D^2P(0)=-\frac kmI , \] and \(D^2F(x)\) is positive definite for every \(x\in B\). Their profiles satisfy, for \(0<r<\varepsilon\) and with \(y=F'\), \[ y>0,\qquad y'>0,\qquad P'<0, \] and the radial system.
Proof. The map \(x\mapsto\frac12|x|^2\) is smooth with values in \([0,\delta)\), so \(F\) and \(P\) are smooth, and they take values in \([\frac12,\frac32]\). With \(s=\frac12|x|^2\), \(\partial_iF=f'(s)x_i\) and \(\partial_{ij}F=f''(s)x_ix_j+f'(s)\delta_{ij}\), so \(D^2F(0)=f'(0)I=I\), and in the same way \(D^2P(0)=p'(0)I\).
The profile of \(F\) is \(F(r)=f(r^2/2)\), so \(y(r)=rf'(r^2/2)=rq(r^2/2)>0\) for \(0<r<\varepsilon\). Fix such \(r\) and put \(s=r^2/2\). Since \(f(v^2s)=F(rv)\) and \(p(v^2s)=P(rv)\), the substitution \(\rho=rv\) [Lebl, Theorem 5.3.5] gives \[ \begin{aligned} y(r)^m&=r^mQ(s)\\ &=m\int_0^r\rho^{m-1}F(\rho)P(\rho)^{-\gamma}\,d\rho,\\ r^mj(s)&=\int_0^r\rho^{m-1}P(\rho)^{1-\gamma}\,d\rho . \end{aligned} \] Moreover \(P'(r)=rp'(s)=-krj(s)q(s)^{1-m}\), and \(y^{m-1}=r^{m-1}q(s)^{m-1}\), so \[ \begin{aligned} y^{m-1}P'&=-kr^mj(s)\\ &=-k\int_0^r\rho^{m-1}P(\rho)^{1-\gamma}\,d\rho . \end{aligned} \] Differentiating both integral identities [Lebl, Theorem 5.3.3] gives \(my^{m-1}y'=mr^{m-1}FP^{-\gamma}\) and \((y^{m-1}P')'=-kr^{m-1}P^{1-\gamma}\), which is the radial system. The first identity shows \(y'>0\), and the last integral shows \(P'<0\).
For \(0<|x|<\varepsilon\), the eigenvalues of \(D^2F(x)\) are \(y'\) and \(y/r\) by Lemma 3.1 of Graphs over a convex profile, both positive; at \(x=0\), \(D^2F(0)=I\). \(\square\)
3. Exercises
3.1 (easy). Show that \(f''(0)=\dfrac{m+\gamma k}{m(m+2)}\). For \(m=9\) deduce \(f''(0)=\frac7{33}\) and \[ F(r)=1+\tfrac12r^2+\tfrac7{264}r^4+O(r^6). \]
3.2 (easy). Show that \(|a^{-\gamma}-b^{-\gamma}|\le\gamma2^{\gamma+1}|a-b|\) for \(a,b\in[\frac12,\frac32]\).
3.3 (medium). Let \(F,P\) be positive radial functions of class \(C^2\) on a ball around \(0\) in \(\mathbb R^m\), \(m\ge2\), whose profiles satisfy the radial system on \((0,\varepsilon)\) with \(y=F'>0\) there. Show that \(y(r)^m=m\int_0^r\rho^{m-1}FP^{-\gamma}\,d\rho\). (So the integral form used above is forced by the radial system.)
3.4 (medium). Let \(m=1\), so \(k=3\) and \(\gamma=\frac43\), and suppose the solution of Proposition 2.1 extends to a solution of the radial system on an interval \((0,R)\) with \(F,P,y>0\). Show that \(R\le\sqrt{2/3}\). So in two variables this construction does not produce an entire solution, in accordance with the theorem of Trudinger and Wang quoted in The affine maximal equation.
4. Solutions
3.1. Differentiating under the integral sign at \(s=0\), \[ \begin{aligned} Q'(0)&=m\int_0^1v^{m+1}\bigl(f'(0)-\gamma p'(0)\bigr)\,dv\\ &=\frac{m}{m+2}\Bigl(1+\frac{\gamma k}m\Bigr). \end{aligned} \] Since \(Q=q^m\) and \(q(0)=1\), \(q'(0)=Q'(0)/m=\frac{m+\gamma k}{m(m+2)}\), and \(f''=q'\). For \(m=9\), \(\gamma k=12\) and \(f''(0)=\frac{21}{99}=\frac7{33}\). By Taylor's theorem \(f(s)=1+s+\frac7{66}s^2+O(s^3)\), and \(s=r^2/2\) gives the expansion.
3.2. The derivative of \(a^{-\gamma}\) is \(-\gamma a^{-\gamma-1}\), of absolute value at most \(\gamma2^{\gamma+1}\) on \([\frac12,\frac32]\); apply the mean value theorem.
3.3. A radial function of class \(C^1\) has \(\nabla F(0)=0\) (its gradient at \(0\) is invariant under all rotations), so \(y(r)\to0\) as \(r\to0\). On \((0,\varepsilon)\), \((y^m)'=my^{m-1}y'=mr^{m-1}FP^{-\gamma}\). Integrate over \([r_0,r]\) and let \(r_0\to0\).
3.4. For \(m=1\) the factor \(y^{m-1}\) is \(1\), and the second equation reads \(P''=-3P^{-1/3}<0\). Near \(0\), \(P'(r)=rp'(r^2/2)\), so \(P'(0)=0\), and \(P'(r)=\int_0^rP''<0\) for \(0<r<R\). Hence \(P\le1\) on \([0,R)\), so \(P^{-1/3}\ge1\) and \(P''\le-3\). Integrating twice from \(0\) gives \(P'(r)\le-3r\) and \(P(r)\le1-\frac32r^2\) for \(0<r<R\). If \(R>\sqrt{2/3}\), this gives \(P(\sqrt{2/3})\le0\), contradicting \(P>0\).
References
- [OpenAI-AM] OpenAI, A Smooth Nonquadratic Entire Affine Maximal Graph in Dimension Ten, OpenAI Math Release preprint, 5 October 2026, Section 3. https://github.com/openai/math/tree/main/preprints/Smooth-Nonquadratic-Affine-Maximal-Graph-in-Dimension-Ten-October-5-2026
- [Lebl] J. Lebl, Basic Analysis I and II, version 6.3; the text of the core courses Real Analysis I and Real Analysis II. https://www.jirka.org/ra/