Positive solutions and their potentials

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

This lesson reduces the theorem of Spherical means and Newton potentials to a bound, uniform over all solutions, for a localized energy (Proposition 5.1). It follows [OpenAI-LE, Section 2]. Fix an integer \(n\ge2\), exponents \(p,q>0\) and weights \(A,B\in\mathbb R\). A solution is a pair of functions \(u,v\) that are continuous and positive on \(\mathbb R^n\), twice continuously differentiable on \(\mathbb R^n\setminus\{0\}\), and satisfy \[ \begin{gathered} -\Delta u=|x|^Av^p,\\ -\Delta v=|x|^Bu^q\\ \text{on }\mathbb R^n\setminus\{0\}. \end{gathered}\tag{0.1} \] Results of Spherical means and Newton potentials are cited as "Lemma 2.1 of the first lesson" and so on, and its notation is kept. A constant is universal if it depends only on \(n,p,q,A,B\) and on auxiliary parameters named explicitly, but not on the solution.

1. Reductions at the origin

Lemma 1.1 (OpenAI). If a solution exists, then \(n\ge3\), \(A>-2\) and \(B>-2\).

Proof. By Lemma 2.1 of the first lesson, the flux \(F_u(r)=r^{n-1}\overline u\,'(r)\) satisfies \[ \begin{aligned} &F_u'(r)\\ &\quad =-r^{n-1+A}\,\overline{v^p}(r)<0\\ &(r>0), \end{aligned}\tag{1.1} \] since \(|x|^A\) is constant on spheres; so \(F_u\) is strictly decreasing, and its limit \(L\in(-\infty,\infty]\) as \(r\downarrow0\) exists. If \(L\neq0\), then for small \(r\) the flux has a fixed sign and \(|F_u|\ge\kappa>0\); integrating \(\overline u\,'(r)=F_u(r)r^{1-n}\) against \(\int_0r^{1-n}\,dr=\infty\) shows that \(\overline u(r)\) is unbounded as \(r\downarrow0\), contradicting \(\overline u(r)\to u(0)\). So \(L=0\), and \(F_u(r)<0\) for every \(r>0\). The same holds for \(v\).

If \(n=2\), then \(r\overline u\,'(r)\le F_u(1)<0\) for \(r\ge1\), so \(\overline u(r)\le\overline u(1)+F_u(1)\log r\), which becomes negative; this contradicts \(u>0\).

Let \(n\ge3\). Since \(v\) is continuous and \(v(0)>0\), there are \(c>0\) and \(r_0>0\) with \(v^p\ge c\) on \(B_{r_0}\). For \(r<r_0\), integrating (1.1) over \([r/2,r]\) and using \(F_u(r/2)<0\), \[ \begin{aligned} &-\overline u\,'(r)=-r^{1-n}F_u(r)\\ &\quad \ge c\,r^{1-n}\int_{r/2}^rs^{n-1+A}\,ds=c_A\,r^{1+A} \end{aligned} \] with a constant \(c_A>0\) (the integral is \(r^{n+A}\) times a positive constant, which is \(\log2\) when \(n+A=0\)). If \(A\le-2\), then \(\int_0r^{1+A}\,dr=\infty\), and \(\overline u(r)=\overline u(r_0)+\int_r^{r_0}(-\overline u\,')\to\infty\) as \(r\downarrow0\), contradicting the continuity of \(u\) at \(0\). So \(A>-2\), and in the same way \(B>-2\). \(\square\)

From now on \(n\ge3\) and \(A,B>-2\). Define the sources \[ \begin{gathered} f(x)=|x|^Bu(x)^q,\\ g(x)=|x|^Av(x)^p\quad(x\neq0),\\ f(0)=g(0)=0, \end{gathered} \] so that \(-\Delta u=g\) and \(-\Delta v=f\) off the origin. Near \(0\), \(f\le C|x|^B\) and \(g\le C|x|^A\) with a constant depending on the solution, and \(n+A,n+B>0\); so \(f\) and \(g\) are locally integrable.

2. Potentials

Proposition 2.1. For every solution, \(-\Delta u=g\) and \(-\Delta v=f\) in distributions on \(\mathbb R^n\), and \[ K*g\le u,\qquad K*f\le v\qquad\text{on }\mathbb R^n . \] There are constants \(c_u,c_v\ge0\) with \(u=c_u+K*g\) and \(v=c_v+K*f\) on \(\mathbb R^n\setminus\{0\}\). The potentials are continuously differentiable off the origin, with \[ \begin{gathered} \nabla u(x)=\int\nabla K(x-y)g(y)\,dy,\\ \nabla v(x)=\int\nabla K(x-y)f(y)\,dy,\\ \nabla K(z)=-(n-2)\frac{z}{|z|^2}K(z), \end{gathered} \] both integrals converging absolutely for \(x\neq0\).

Proof. The sources are nonnegative, locally integrable, and continuous off the origin, so Lemma 3.2 and Proposition 3.3 of the first lesson apply to \((u,g)\) and to \((v,f)\). \(\square\)

For \(x,y\in B_R\) we have \(|x-y|<2R\), hence \(K(x-y)\ge c_n(2R)^{2-n}\). With Proposition 2.1 this gives the ball bounds \[ \begin{gathered} u(x)\ge c\,R^{2-n}\int_{B_R}g,\\ v(x)\ge c\,R^{2-n}\int_{B_R}f\\ (x\in B_R), \end{gathered}\tag{2.1} \] with \(c>0\) depending only on \(n\).

3. Small products and scaling

Lemma 3.1 (OpenAI). Every solution has \(pq>1\).

Proof. Let \(m_u(R)=\min_{\overline B_R}u>0\) and \(m_v(R)=\min_{\overline B_R}v>0\). Inserting \(g\ge|x|^Am_v(R)^p\) and \(f\ge|x|^Bm_u(R)^q\) on \(B_R\) into (2.1), and using \(\int_{B_R}|x|^A\,dx=\frac{|S|}{n+A}R^{n+A}\), \[ \begin{gathered} m_u(R)\ge c\,R^{2+A}m_v(R)^p,\\ m_v(R)\ge c\,R^{2+B}m_u(R)^q . \end{gathered} \] Combining them, \(m_u(R)^{1-pq}\ge c\,R^{2+A+p(2+B)}\), and the exponent on the right is positive. If \(pq<1\), the left side is at most \(u(0)^{1-pq}\); if \(pq=1\), it is \(1\). Either way \(R\to\infty\) gives a contradiction. \(\square\)

From now on also \(pq>1\). Define \[ \begin{gathered} \alpha=\frac{2+A+p(2+B)}{pq-1},\\ \beta=\frac{2+B+q(2+A)}{pq-1},\\ a=\frac1{q+1},\\ b=\frac1{p+1}. \end{gathered}\tag{3.1} \] Then \(\alpha,\beta>0\), and \[ \begin{gathered} p\beta=\alpha+2+A,\\ q\alpha=\beta+2+B . \end{gathered}\tag{3.2} \]

Lemma 3.2 (scaling). For every solution and every \(R>0\), the pair \(u_R(x)=R^\alpha u(Rx)\), \(v_R(x)=R^\beta v(Rx)\) is a solution, with sources \(f_R(x)=R^{\beta+2}f(Rx)\) and \(g_R(x)=R^{\alpha+2}g(Rx)\).

Proof. \(-\Delta u_R(x)=R^{\alpha+2}g(Rx)\), which is \(R^{\alpha+2+A}|x|^Av(Rx)^p\), which is \(R^{\alpha+2+A-p\beta}|x|^Av_R(x)^p=|x|^Av_R(x)^p\) by (3.2), and similarly for \(v_R\); also \(f_R(x)=|x|^Bu_R(x)^q\), which is \(R^{q\alpha-B}f(Rx)=R^{\beta+2}f(Rx)\). \(\square\)

The dilations are centered at the origin, where the weights are centered; translations do not preserve (0.1).

4. Centered estimates

Proposition 4.1 (OpenAI). There is a universal constant \(C\) such that every solution satisfies, for all \(R>0\), \[ \begin{gathered} \int_{B_R}f\le C\,R^{n-2-\beta},\\ \int_{B_R}g\le C\,R^{n-2-\alpha}. \end{gathered}\tag{4.1} \] The constants of Proposition 2.1 vanish: \(u=K*g\) and \(v=K*f\) on \(\mathbb R^n\setminus\{0\}\). Moreover, the bounds (4.2) hold: \(\int_{B_4}(f+g+u+v)\le C\), and, for \(x\in B_2\) and \(0<\ell\le1\), \[ \begin{aligned} &\int_{|x-y|\ge\ell}K(x-y)\bigl(f(y)+g(y)\bigr)\,dy\\ &\quad\le C\ell^{2-n}. \end{aligned} \]

Proof. Unit ball. Let \(M_f=\int_{B_1}f\) and \(M_g=\int_{B_1}g\), both positive. By (2.1) with \(R=1\), \(u\ge cM_g\) and \(v\ge cM_f\) on \(B_1\), so \(M_f\ge c_fM_g^q\) and \(M_g\ge c_gM_f^p\) with universal \(c_f,c_g>0\). Hence \(M_f\ge c_fc_g^qM_f^{pq}\), and since \(pq>1\) this bounds \(M_f\) by a universal constant; likewise \(M_g\). Applied to the scaled solution of Lemma 3.2, \(\int_{B_1}f_R=R^{\beta+2-n}\int_{B_R}f\le C\), which is (4.1).

No constants. If \(c_u>0\), then \(u\ge c_u\) off the origin, so \(\int_{B_R}f\ge c_u^q\frac{|S|}{n+B}R^{n+B}\). Since \((n+B)-(n-2-\beta)=\beta+2+B\), which is \(q\alpha>0\), this contradicts (4.1) for large \(R\). So \(c_u=0\), and likewise \(c_v=0\).

Far sources. Let \(x\in B_4\) and \(j\ge3\). On the shell \(2^j\le|y|<2^{j+1}\) we have \(|x-y|\ge2^{j-1}\), so its contribution to \(K*f(x)\) is at most \(c_n2^{(j-1)(2-n)}\int_{B_{2^{j+1}}}f\le C2^{-j\beta}\), and its contribution to \(K*g(x)\) is at most \(C2^{-j\alpha}\). Summing over \(j\), the potentials of \(f\mathbf 1_{\{|y|\ge8\}}\) and \(g\mathbf 1_{\{|y|\ge8\}}\) are bounded by a universal constant on \(B_4\).

Near sources. Since \(\int_{B_4}K(x-y)\,dx\le\int_{B_{12}}K\) for \(y\in B_8\), Tonelli's theorem and (4.1) bound \(\int_{B_4}K*(g\mathbf 1_{B_8})\) and \(\int_{B_4}K*(f\mathbf 1_{B_8})\). With \(u=K*g\) and \(v=K*f\), this proves the bound on \(\int_{B_4}(u+v)\), and (4.1) bounds \(\int_{B_4}(f+g)\).

Tails. For \(|x-y|\ge\ell\), \(K(x-y)\le c_n\ell^{2-n}\), so the sources in \(B_8\) contribute at most \(C\ell^{2-n}\); the sources outside \(B_8\) contribute at most \(C\le C\ell^{2-n}\), because \(\ell\le1\) and \(n\ge3\). \(\square\)

The estimates are centered at the origin; a ball \(B_\ell(x)\subseteq B_4\) has source mass at most that of \(B_4\). For each solution, the energy densities \(fu=|x|^Bu^{q+1}\) and \(gv=|x|^Av^{p+1}\) are locally integrable, but no bound on them uniform over solutions is known at this point.

5. The energy and the plan

Let \[ \begin{aligned} \gamma&=a(n+B)+b(n+A)-(n-2)\\ &=\frac{n+B}{q+1}+\frac{n+A}{p+1}-(n-2), \end{aligned} \] so that the hypothesis of the theorem is \(\gamma>0\). A computation with (3.1) gives the relations (5.1): \[ \begin{gathered} \gamma=(1-a-b)\,(\alpha+\beta+2-n),\\ 1-a-b=\frac{pq-1}{(p+1)(q+1)}>0, \end{gathered} \] so \(\gamma>0\) implies \(\alpha+\beta+2-n>0\). Changing variables in Lemma 3.2 gives the identity (5.2): \[ \begin{aligned} &\int_{B_1}\bigl(f_Ru_R+g_Rv_R\bigr)\\ &\quad=R^{\alpha+\beta+2-n}\int_{B_R}(fu+gv). \end{aligned} \]

Proposition 5.1 (uniform energy bound; OpenAI). Let \(n\ge3\), \(A,B>-2\), \(pq>1\) and \(\gamma>0\). There is a universal constant \(C\) such that every solution satisfies \[ \int_{B_1}\bigl(|x|^Bu^{q+1}+|x|^Av^{p+1}\bigr)\,dx\le C . \]

Proposition 5.1 is proved in Pressure and the energy bound, after A localized virial identity and Intervals and source layers.

Theorem 5.2 (OpenAI 2026). Let \(n\ge2\), \(p,q>0\), \(A,B\in\mathbb R\) and \(\gamma>0\). Then (0.1) has no solution.

Proof, given Proposition 5.1. Lemmas 1.1 and 3.1 exclude \(n=2\), weights \(A\le-2\) or \(B\le-2\), and \(pq\le1\). In the remaining range, apply Proposition 5.1 to every scaled solution of Lemma 3.2. By (5.2), \(R^{\alpha+\beta+2-n}\int_{B_R}(fu+gv)\le C\) for all \(R>0\). For \(R\ge1\) the integral is at least \(\int_{B_1}(fu+gv)>0\), and \(\alpha+\beta+2-n>0\) by (5.1), so the left side tends to infinity, a contradiction. \(\square\)

Corollary 5.3 (the Lane–Emden conjecture). For \(n\ge3\) and \(p,q>0\) with \(\frac1{p+1}+\frac1{q+1}>\frac{n-2}n\), the system \(-\Delta u=v^p\), \(-\Delta v=u^q\) has no positive twice continuously differentiable solution on \(\mathbb R^n\).

Proof. Take \(A=B=0\); the condition is \(\gamma>0\) divided by \(n\). \(\square\)

6. Exercises

Exercise 6.1 (easy). Verify (3.2) and (5.1).

Exercise 6.2 (easy). Show that \(n=5\), \(p=q=4\), \(A=B=3\) satisfies \(\gamma>0\), although \(\frac1{p+1}+\frac1{q+1}<\frac{n-2}n\). Why can the weighted theorem not be deduced from the unweighted one at the same exponents?

Exercise 6.3 (medium). Show that the strict inequality \(\gamma>0\) cannot be weakened to \(\gamma\ge0\): for \(A=B=0\), \(n\ge3\) and \(p=q=\frac{n+2}{n-2}\), check that \(\gamma=0\) and that \(u=v\), with \(u\) equal to \(\bigl(n(n-2)\bigr)^{(n-2)/4}(1+|x|^2)^{-(n-2)/2}\), is a positive solution.

Exercise 6.4 (medium). Let \(pq>1\). Show from the two inequalities in the proof of Lemma 3.1 that \(\min_{\overline B_R}u\le CR^{-\alpha}\) for all \(R>0\), with a universal \(C\), and explain why this agrees with the scaling of Lemma 3.2.

7. Solutions

6.1. \(p\beta-\alpha=\frac{p(2+B)+pq(2+A)-(2+A)-p(2+B)}{pq-1}\), which is \(2+A\), and symmetrically \(q\alpha-\beta=2+B\). Next \[ \begin{aligned} &(1-a-b)(\alpha+\beta)\\ &\quad=\frac{(q+1)(2+A)+(p+1)(2+B)}{(p+1)(q+1)}\\ &\quad=b(2+A)+a(2+B), \end{aligned} \] and \((1-a-b)(2-n)\) equals \((2-n)-a(2-n)-b(2-n)\); adding gives \(a(n+B)+b(n+A)-(n-2)=\gamma\).

6.2. \(\gamma=\frac85+\frac85-3=\frac15>0\), while \(\frac15+\frac15=\frac25<\frac35\). The unweighted theorem at \(p=q=4\) in dimension \(5\) is not available (those exponents are not subcritical without weights), and the weights change the equation, so neither statement implies the other at these exponents.

6.3. \(\gamma=\frac{2n}{p+1}-(n-2)\), which is \(\frac{2n(n-2)}{2n}-(n-2)=0\). For \(U=(1+r^2)^{-k}\) with \(k=\frac{n-2}2\), \[ \begin{aligned} &U''+\frac{n-1}rU'\\ &\quad=(1+r^2)^{-k-2}\bigl(-2kn(1+r^2)\\ &\qquad+4k(k+1)r^2\bigr)\\ &\quad=-n(n-2)(1+r^2)^{-(n+2)/2}, \end{aligned} \] since \(2kn=4k(k+1)=n(n-2)\). So \(-\Delta U=n(n-2)U^{(n+2)/(n-2)}\), and \(u=\lambda U\) with \(\lambda^{4/(n-2)}=n(n-2)\) satisfies \(-\Delta u=u^{(n+2)/(n-2)}\).

6.4. From \(m_u(R)\ge cR^{2+A}m_v(R)^p\) and \(m_v(R)\ge cR^{2+B}m_u(R)^q\) we get \(m_u(R)\ge c\,R^{2+A+p(2+B)}m_u(R)^{pq}\), that is, \(m_u(R)^{pq-1}\le CR^{-(2+A+p(2+B))}\), or \(m_u(R)\le CR^{-\alpha}\). The scaled solution \(u_R(x)=R^\alpha u(Rx)\) has \(\min_{\overline B_1}u_R=R^\alpha m_u(R)\), and the bound says exactly that this is at most \(C\), the same estimate for the solution \(u_R\) on the unit ball.

References