Proof. Direct differentiation, interpreted as an identity of quadratic forms, gives \[ D^2u=\frac Fu H+\frac{F^2}{u^3} \left(dt-\frac tF\,dF\right)^2. \tag{4}\] The first term is positive on the \(x\) directions, and the second is positive on the remaining direction. Thus \(D^2u>0\). Put \(\zeta=t/F\) and use the pointwise basis \[X_i=e_i+\zeta F_i e_t\quad(1\leq i\leq m),\qquad e_t.\] Its change-of-basis matrix has determinant one. In this basis, (4) has diagonal blocks \((F/u)H\) and \(F^2/u^3\), so \[ \det D^2u=F^{m+2}u^{-(m+3)}\det H. \tag{5}\] The first equation in (3) is equivalent to \(P=(F/\det H)^{(m+2)/(m+3)}\). Therefore \[ w=(\det D^2u)^{-(m+2)/(m+3)}=P h(\zeta), \qquad h(\zeta)=(1+\zeta^2)^{k/2}. \tag{6}\]

We evaluate the ordinary Hessian of \(w\) in the same pointwise basis. Since \(d\zeta(X_i)=0\), twice differentiating \(t=F\zeta\) gives \[D^2\zeta(X_i,X_j)=-\frac\zeta F F_{ij}.\] Consequently \[D^2w(X_i,X_j)=hP_{ij}-\frac PF\zeta h'F_{ij}, \qquad w_{tt}=\frac P{F^2}h''.\] Taking the trace against the inverse blocks of \(D^2u\) yields \[ \frac Fu\mathop{\mathrm{tr}}\bigl((D^2u)^{-1}D^2w\bigr) =h\mathop{\mathrm{tr}}(H^{-1}D_x^2P) +\frac PF\bigl((1+\zeta^2)h''-m\zeta h'\bigr). \tag{7}\] Differentiation of \(h\) gives \[(1+\zeta^2)h''-m\zeta h' =k(1+\zeta^2)^{k/2-1} \bigl(1+(k-1-m)\zeta^2\bigr)=kh,\] because \(k=m+2\). The second equation in (3) makes (7) zero. Multiplication by the positive determinant in (5) gives (1). ◻

Proof. Seek \(F(r)=f(s)\) and \(P(r)=p(s)\), where \(s=r^2/2\). The integrated equations suggest the normalized derivative \(q=y/r\) and flux \(j=-y^{m-1}P'/(kr^m)\) for \(r>0\). We will construct their regular extensions as functions of \(s\) using the following averages. For continuous \(f,p\) on \([-\delta,\delta]\) taking values in \([1/2,3/2]\), define \[\begin{align*} q(s)&=\left(m\int_0^1 v^{m-1} f(v^2s)p(v^2s)^{-\gamma}\,dv\right)^{1/m}, \tag{10}\\ j(s)&=\int_0^1 v^{m-1}p(v^2s)^{1-\gamma}\,dv. \tag{11}\end{align*}\] On the closed subset \[\mathcal K=\{(f,p)\in C([-\delta,\delta])^2: \|f-1\|_\infty,\|p-1\|_\infty\leq1/2\},\] consider \[ \mathcal T(f,p)(s)= \left(1+\int_0^s q(a)\,da, \ 1-k\int_0^s j(a)q(a)^{1-m}\,da\right). \tag{12}\] The functions \(q,j\) have uniform positive upper and lower bounds. The maps \((f,p)\mapsto q\) and \((f,p)\mapsto jq^{1-m}\) are Lipschitz in the product supremum norm, with constants independent of \(\delta\): all scalar powers are evaluated on fixed positive compact intervals. Hence there are constants \(M,L\) independent of \(\delta\) such that \[\|\mathcal T(f,p)-(1,1)\|_\infty\leq M\delta, \qquad \|\mathcal T(f,p)-\mathcal T(\widetilde f,\widetilde p)\|_\infty \leq L\delta\|(f,p)-(\widetilde f,\widetilde p)\|_\infty.\] Choose \(M\delta\leq1/2\) and \(L\delta<1\). The contraction mapping theorem gives a fixed point in \(\mathcal K\).

The fixed point is smooth on \((-\delta,\delta)\). Indeed, continuity of \(q,j\) first gives \(f,p\in C^1\). If \(f,p\) are \(C^\ell\), differentiation under the integrals in (10)–(11) shows that \(q,j\) are \(C^\ell\), and (12) gives one further derivative. This proves the assertion by induction. The definition on both sides of \(s=0\) ensures that \(f(|x|^2/2)\) and \(p(|x|^2/2)\) are smooth in \(x\) at the origin.

For \(r>0\), the fixed point equations give \(y=rq(r^2/2)>0\). Changing variables \(\rho=rv\) in (10)–(11) then yields \[ y^m=m\int_0^r \rho^{m-1}F(\rho)P(\rho)^{-\gamma}\,d\rho, \qquad y^{m-1}P'=-k\int_0^r\rho^{m-1}P(\rho)^{1-\gamma}\,d\rho. \tag{13}\] Differentiation proves (8); the first equation implies \(y'>0\), and the second integral gives \(P'<0\). Finally \(q(0)=1\) and \(j(0)=1/m\), so \[f_s(0)=1,\qquad p_s(0)=-k/m=-11/9.\] These are exactly the claimed Hessians of the radial extensions. Shrinking \(\varepsilon\) if necessary completes the proof. ◻

Proof. Positivity of \(F,y,y'\) gives \(A,B>0\). Using (8), one may write \[B=\frac{r^mFP^{-\gamma}}{y^m}.\] Logarithmic differentiation of \(A\) and \(B\), followed by the second equation in (8), gives \[\begin{align*} \dot A&=A(1+B-A),\\ \dot B&=B(m+A+\gamma C-mB),\\ \dot C&=C(1+C-(m-1)B)+kAB. \end{align*}\] For the last identity, use \(\dot C=C+C^2-r^2P''/P\) and \(AB=r^{m+1}P^{-\gamma}/y^{m-1}\). Thus \[\dot D=(k-D)A-(m-1)D+BD\bigl(1+(1-\gamma)D\bigr).\] Substituting (9) proves (16).

Lemma 3 gives \(F=1+O(r^2)\), \(y=r+O(r^3)\), \(y'=1+O(r^2)\), and \(P'=-(11/9)r+O(r^3)\). Hence \(A\to0\), \(B\to1\), and \(D\to0\). The positivity of \(A,B\) and the local inequality \(P'<0\) also give \(D>0\) for small positive \(r\). ◻

Proof. Write \(V=(V_1,V_2,V_3)\) for the vector field in (16). The upper corner \(a_*=(9/2,7/2,33/7)\) satisfies \(V(a_*)=0\) by direct substitution. Each \(V_i\) is nondecreasing in the other two coordinates on \(\mathcal B\). The nonzero cross derivatives are \[\partial_BV_1=A,\quad \partial_AV_2=B,\quad \partial_DV_2=(12/11)B^2,\quad \partial_AV_3=11-D,\quad \partial_BV_3=D(1-D/11),\] all nonnegative because \(0\leq D\leq33/7<11\). On each upper face, comparison with \(a_*\) therefore gives the corresponding \(V_i\leq0\). On the lower faces \(A=0\), \(B=0\), and \(D=0\), the respective components are \(0\), \(0\), and \(11A\), hence nonnegative.

These weak boundary inequalities suffice for invariance. Let \(\Pi z\) be the coordinatewise projection of \(z\in\mathbb R^3\) onto \(\mathcal B\). The face inequalities imply \[(z-\Pi z)\cdot V(\Pi z)\leq0.\] On any bounded neighborhood of the box, \(V\) has a Lipschitz constant \(L\). Along a trajectory \(z\) in that neighborhood, the squared distance \(d^2=|z-\Pi z|^2\) consequently satisfies \[\frac{d}{d\log r}d^2 =2(z-\Pi z)\cdot V(z)\leq2L d^2.\] The squared distance to a closed box is continuously differentiable, so this identity also holds when the nearest face changes. Gronwall’s inequality shows that a trajectory starting in the box cannot leave it.

Finally, Lemma 4 gives strictly positive coordinates tending to \((0,1,0)\) as \(r\downarrow0\). They therefore lie below all three upper bounds for sufficiently small \(r>0\). Forward invariance proves the last assertion. ◻

Proof. Suppose its maximal outer endpoint were \(R<\infty\), and fix \(r_0\in(0,R)\). Lemma 5 gives \[0\leq A\leq9/2,\qquad 0\leq B\leq7/2, \qquad 0\leq C=BD\leq33/2.\] Since the derivatives of \(\log F,\log y,\log P\) with respect to \(\log r\) are \(A,B,-C\), respectively, integration gives, for \(r_0\leq r<R\), \[\begin{align*} F(r_0)&\leq F(r)\leq F(r_0)(R/r_0)^{9/2},\\ y(r_0)&\leq y(r)\leq y(r_0)(R/r_0)^{7/2},\\ P(r_0)(r_0/R)^{33/2}&\leq P(r)\leq P(r_0). \end{align*}\] Also \(|P'(r)|=C(r)P(r)/r\leq(33/2)P(r_0)/r_0\). Thus \((r,F,P,y,Z)\) remains in a compact subset of the domain \(r,F,P,y>0\) of the smooth system (14). Its right-hand side is bounded there, so the state has a limit at \(R\) in that domain. Local existence and uniqueness at this limit extend the solution past \(R\), a contradiction.

Smoothness for \(r>0\) follows from (14), and Lemma 3 supplies smoothness in \(x\) at the origin. The Hessian eigenvalues for \(r>0\) are \(y'>0\) and \(y/r>0\); at the origin the Hessian is \(I\). ◻

Proof of Theorem 1. Take the global radial functions of Proposition 6. By (8), they satisfy (3) for \(x\ne0\). The functions are smooth, \(F,P>0\), and \(D_x^2F>0\) everywhere, so both identities extend to \(x=0\) by continuity. Proposition 2 now shows that \[u(x,t)=\sqrt{F(|x|)^2+t^2},\qquad (x,t)\in\mathbb R^9\times\mathbb R,\] is smooth, has positive definite Hessian everywhere, and solves the classical affine maximal equation in dimension ten. Finally, \(u(0,t)=\sqrt{1+t^2}\) because \(F(0)=1\), so \(u\) is not a quadratic polynomial. ◻