Book 6C

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Course 6Book 6C: Nonlinear PDE: Variational and Geometric MethodsChapter 3

Symmetry, Noether’s Theorem and Moving Planes

Invariance gives conservation laws, and maximum principles give symmetry.

39 min read · Updated Oct 23, 2026

Read with Evans, Partial Differential Equations, section 8.6 (invariance and Noether's theorem; 2nd edition) and section 9.5 (geometric properties of solutions: star-shaped level sets, radial symmetry by moving planes). Gidas, Ni and Nirenberg, "Symmetry and related properties via the maximum principle" (Comm. Math. Phys. 68, 1979), and Berestycki and Nirenberg, "On the method of moving planes and the sliding method" (Bol. Soc. Brasil. Mat. 22, 1991), are the original sources. This chapter proves Noether's theorem for translations and internal symmetries, the maximum principles for narrow and small domains, and the Gidas–Ni–Nirenberg symmetry theorem.

In this chapter · 8 sections
  1. 3.1Round stars
  2. 3.2Translations: the energy–momentum tensor
  3. 3.3Internal symmetries
  4. 3.4Maximum principles for thin and small domains
  5. 3.5Moving planes
  6. 3.6Bubbles
  7. 3.7History
  8. 3.8Exercises
  9. ↗Free resources

Symmetry enters PDE in two opposite directions. If an equation is symmetric, invariant under translations, rotations, scalings or some internal transformation of the unknown, then its solutions satisfy conservation laws: integrals that do not change in time, or divergence identities that hold for every solution. That is Emmy Noether's theorem (1918), the reason energy is conserved because the laws of physics do not change in time, and momentum because they do not change from place to place. Pohozaev's identity (6C.2 Critical Points and the Mountain Pass) is a Noether identity for scaling.

The other direction is subtler: if the problem is symmetric (the domain is a ball, the equation is rotation invariant), are its solutions symmetric too? Not always, since symmetric problems can have asymmetric solutions (a buckled beam chooses a side). But for positive solutions of −Δu=f(u)-\Delta u = f(u) in a ball with zero boundary values, the answer is yes, and the proof, the method of moving planes of Basilis Gidas, Wei-Ming Ni and Louis Nirenberg (1979), is a beautiful use of the maximum principle: reflect the solution in a plane, compare it with itself, and move the plane until it reaches the centre. The same method classifies the "bubbles" that appear when compactness fails at the critical exponent, and symmetry results of this kind, proved by maximum principles, are a recurring theme in the classification of solitons and singularity models of geometric flows.

By the end of this chapter you will be able to:

  • derive the energy–momentum tensor of a variational problem, and prove that it is divergence-free when the Lagrangian does not depend on xx;
  • prove Noether's theorem for internal symmetries, and derive energy, momentum and charge conservation for wave and Schrödinger equations;
  • see Pohozaev's identity as the conservation law of the scaling symmetry;
  • prove the maximum principle for narrow domains and, with the Alexandrov–Bakelman–Pucci estimate, for domains of small measure;
  • prove that positive solutions of −Δu=f(u)-\Delta u = f(u) in a ball vanishing on the boundary are radially symmetric and decreasing, and identify the critical-exponent bubbles.

Round stars

In the world Model Polytropes and the Lane–Emden equation

A star in equilibrium balances gravity against pressure. In the simplest models the pressure depends on the density as a power, P=Kρ1+1/NP = K\rho^{1 + 1/N} (a polytrope of index NN), and combining hydrostatic equilibrium with Newton's law of gravitation gives, after scaling, the Lane–Emden equation

−Δθ=θN,-\Delta\theta = \theta^N,

with θ\theta proportional to ρ1/N\rho^{1/N} and Δ\Delta the Laplacian in three dimensions. Jonathan Homer Lane derived it in 1870, Robert Emden studied it in depth in his 1907 book Gaskugeln ("gas spheres"), and Arthur Eddington used polytropes to build the first quantitative models of stellar interiors in the 1920s. The index N=1.5N = 1.5 models fully convective stars and white dwarfs at low density; N=3N = 3 is Eddington's standard model; N=5N = 5 is the critical exponent n+2n−2\frac{n+2}{n-2} for n=3n = 3, at which the star has infinite radius but finite mass.

Astronomers always assume the star is spherical and solve an ODE in the radius. Is that justified? A non-rotating fluid star should be round, and for the Lane–Emden equation on a ball with θ=0\theta = 0 on the surface, the Gidas–Ni–Nirenberg theorem of this chapter proves it: every positive solution is radially symmetric and decreasing outward. (Rotation, magnetic fields and companions break the symmetry, and real stars are slightly oblate; the theorem is about the idealised problem.)

Translations: the energy–momentum tensor

Consider a variational problem with Lagrangian L(p,z)L(p, z) that does not depend on xx, and its Euler–Lagrange equation (6A.11 Calculus of Variations and Gradient Flows)

∑i∂i(Lpi(∇u,u))=Lz(∇u,u).\sum_i\partial_i\big(L_{p_i}(\nabla u, u)\big) = L_z(\nabla u, u).

The independence of xx means the problem is invariant under translations u(x)↦u(x+τek)u(x) \mapsto u(x + \tau e_k). Noether's theorem says each such symmetry gives a divergence-free vector field.

Theorem 3.1 The energy–momentum tensor is divergence-free

If L=L(p,z)L = L(p, z) and u∈C2u \in C^2 solves the Euler–Lagrange equation, then for each k=1,…,nk = 1, \ldots, n,

∑i∂iTki=0,Tki=Lpi(∇u,u) ∂ku−L(∇u,u) δik.\sum_i\partial_i T_k^i = 0, \qquad T_k^i = L_{p_i}(\nabla u, u)\,\partial_ku - L(\nabla u, u)\,\delta_{ik}.

Proof. By the product rule and the Euler–Lagrange equation,

∑i∂i(Lpi∂ku)=∑i∂i(Lpi) ∂ku+∑iLpi∂i∂ku=Lz∂ku+∑iLpi∂k∂iu.\sum_i\partial_i\big(L_{p_i}\partial_ku\big) = \sum_i\partial_i(L_{p_i})\,\partial_ku + \sum_iL_{p_i}\partial_i\partial_ku = L_z\partial_ku + \sum_iL_{p_i}\partial_k\partial_iu.

On the other hand, since LL depends on xx only through ∇u\nabla u and uu, the chain rule gives ∂k(L(∇u,u))=∑iLpi∂k∂iu+Lz∂ku\partial_k\big(L(\nabla u, u)\big) = \sum_iL_{p_i}\partial_k\partial_iu + L_z\partial_ku. The two expressions agree, so ∑i∂i(Lpi∂ku)−∂kL=0\sum_i\partial_i(L_{p_i}\partial_ku) - \partial_kL = 0.

The matrix TT is the energy–momentum tensor (or stress tensor) of the problem. For time-dependent problems, treat time as one of the coordinates.

Example 3.2 Energy and momentum for the wave equation

The wave equation utt=Δuu_{tt} = \Delta u is the Euler–Lagrange equation of ∬12(ut2−∣∇u∣2) dx dt\iint\frac12(u_t^2 - |\nabla u|^2)\,dx\,dt, with L=12(p02−∣p′∣2)L = \frac12(p_0^2 - |p'|^2) in the variables (t,x)(t, x), p0=utp_0 = u_t, p′=∇up' = \nabla u. Invariance under time translation (k=0k = 0) gives

∂t(12ut2+12∣∇u∣2)−div⁡(ut∇u)=0\partial_t\Big(\frac12u_t^2 + \frac12|\nabla u|^2\Big) - \operatorname{div}\big(u_t\nabla u\big) = 0

(Exercise 3.9): the energy density 12(ut2+∣∇u∣2)\frac12(u_t^2 + |\nabla u|^2) changes only by the flux −ut∇u-u_t\nabla u, so the total energy is conserved for solutions decaying at infinity (6B.3 The Wave Equation). Invariance under the space translation in xjx_j gives conservation of the momentum ∫ut ∂ju dx\int u_t\,\partial_ju\,dx. These are exactly the conserved quantities of a vibrating string or membrane.

Integrating a divergence identity. Noether identities are usually used integrated against a vector field or over a region. Integrating div⁡Tk=0\operatorname{div}T_k = 0 over a domain gives boundary relations; multiplying by xkx_k and summing gives the virial or Pohozaev-type identities. For L=12∣p∣2−F(z)L = \frac12|p|^2 - F(z), the scaling vector field x⋅∇x\cdot\nabla is not a symmetry (except at the critical exponent), but the identity ∑i,k∂i(xkTki)=∑kTkk\sum_{i,k}\partial_i(x_kT_k^i) = \sum_kT_k^k holds for every solution (using div⁡Tk=0\operatorname{div}T_k = 0), and ∑kTkk=∣∇u∣2−n(12∣∇u∣2−F(u))\sum_kT^k_k = |\nabla u|^2 - n\big(\frac12|\nabla u|^2 - F(u)\big). Integrated over UU with u=0u = 0 on ∂U\partial U, this is exactly Pohozaev's identity (Exercise 3.11). Scaling would be a symmetry if ∑kTkk\sum_kT_k^k were a divergence, which happens at the critical exponent, where the identity has no interior term left: that is why the critical exponent is special.

Internal symmetries

A symmetry can also act on the values of uu rather than on xx: rotating a complex-valued function by a phase, u↦eiτuu \mapsto e^{i\tau}u, or rotating a vector-valued one. Let u:U→Rmu : U \to \mathbb{R}^m, L=L(p,z)L = L(p, z) with p∈Rm×np \in \mathbb{R}^{m\times n}, and let RτR_\tau be a one-parameter group of linear maps of Rm\mathbb{R}^m with R0=IR_0 = I and generator A=ddτRτ∣τ=0A = \frac{d}{d\tau}R_\tau\big|_{\tau=0}.

Theorem 3.3 Noether's theorem for internal symmetries

Suppose L(Rτp,Rτz)=L(p,z)L(R_\tau p, R_\tau z) = L(p, z) for all τ\tau, pp, zz, where RτR_\tau acts on each column of pp. If u∈C2u \in C^2 solves the Euler–Lagrange equations ∑i∂i(Lpia)=Lza\sum_i\partial_i(L_{p_i^a}) = L_{z^a} (a=1,…,ma = 1, \ldots, m), then the current

ji=∑aLpia(∇u,u) (Au)aj^i = \sum_aL_{p_i^a}(\nabla u, u)\,(Au)^a

is divergence-free: ∑i∂iji=0\sum_i\partial_ij^i = 0.

Proof. Differentiate the invariance L(Rτ∇u,Rτu)=L(∇u,u)L(R_\tau\nabla u, R_\tau u) = L(\nabla u, u) in τ\tau at τ=0\tau = 0:

∑a,iLpia (A∂iu)a+∑aLza(Au)a=0.\sum_{a,i}L_{p_i^a}\,(A\partial_iu)^a + \sum_aL_{z^a}(Au)^a = 0.

Since AA is constant, (A∂iu)a=∂i(Au)a(A\partial_iu)^a = \partial_i(Au)^a. Now compute the divergence of jj, using the Euler–Lagrange equations:

∑i∂iji=∑a,i∂i(Lpia)(Au)a+∑a,iLpia∂i(Au)a=∑aLza(Au)a+∑a,iLpia∂i(Au)a=0.\sum_i\partial_ij^i = \sum_{a,i}\partial_i(L_{p_i^a})(Au)^a + \sum_{a,i}L_{p_i^a}\partial_i(Au)^a = \sum_aL_{z^a}(Au)^a + \sum_{a,i}L_{p_i^a}\partial_i(Au)^a = 0.

For the Schrödinger equation iut=−Δu+V(x)uiu_t = -\Delta u + V(x)u with complex uu, the phase symmetry gives conservation of ∫∣u∣2\int|u|^2, the total probability (or charge), with current Im⁡(uˉ∇u)\operatorname{Im}(\bar u\nabla u) (Exercise 3.10). For harmonic maps into a sphere (6C.1 Regularity of Minimisers and Constraints), rotations of the target give the conservation laws div⁡(ua∇ub−ub∇ua)=0\operatorname{div}(u^a\nabla u^b - u^b\nabla u^a) = 0, which Frédéric Hélein used in 1991 to prove that weakly harmonic maps from surfaces are smooth, a case where De Giorgi–Nash–Moser fails (6C.7 Harmonic Maps and Their Heat Flow).

The general theorem (Evans, section 8.6.2) treats transformations that move both xx and uu, and it gives a conserved current for every one-parameter group of symmetries. Emmy Noether's 1918 paper also contained a "second theorem", for symmetries depending on arbitrary functions (gauge symmetries, diffeomorphisms): such symmetries give identities satisfied by the Euler–Lagrange expressions themselves, whether or not uu is a solution. The contracted Bianchi identity of Riemannian geometry is one (Exercise 3.15).

Maximum principles for thin and small domains

The second half of the chapter proves symmetry. Its key tool is a maximum principle for operators Δ+c(x)\Delta + c(x) whose zero-order coefficient cc has no sign, which is what comparing two solutions of a nonlinear equation produces. Such operators do not satisfy the maximum principle in general (Δu+λ1u=0\Delta u + \lambda_1u = 0 has the positive solution ϕ1\phi_1 with zero boundary values, and so does −ϕ1-\phi_1), but they do on domains that are thin or small, because there the Laplacian dominates.

Lemma 3.4 Maximum principle in narrow domains

Let Ω\Omega be bounded and contained in a slab {a<x1<a+d}\{a < x_1 < a + d\}, and let ∣c∣≤M|c| \leq M in Ω\Omega. Suppose w∈C2(Ω)∩C(Ω‾)w \in C^2(\Omega)\cap C(\overline\Omega) satisfies Δw+c(x)w≤0\Delta w + c(x)w \leq 0 in Ω\Omega and w≥0w \geq 0 on ∂Ω\partial\Omega. If d<π2Md < \frac{\pi}{2\sqrt M}, then w≥0w \geq 0 in Ω\Omega.

Proof. Let g(x)=cos⁡(π2d(x1−a−d2))g(x) = \cos\big(\frac{\pi}{2d}(x_1 - a - \frac d2)\big). On the slab the argument of the cosine lies in [−π4,π4][-\frac\pi4, \frac\pi4], so g≥12>0g \geq \frac{1}{\sqrt2} > 0, and −Δg=κg-\Delta g = \kappa g with κ=(π2d)2>M\kappa = \big(\frac{\pi}{2d}\big)^2 > M. Write w=gvw = gv. Then

Δw+cw=gΔv+2∇g⋅∇v+v(Δg+cg)=gΔv+2∇g⋅∇v−(κ−c)gv.\Delta w + cw = g\Delta v + 2\nabla g\cdot\nabla v + v(\Delta g + cg) = g\Delta v + 2\nabla g\cdot\nabla v - (\kappa - c)gv.

If ww, hence vv, were negative somewhere, vv would have a negative minimum at an interior point (it is ≥0\geq 0 on ∂Ω\partial\Omega). There ∇v=0\nabla v = 0 and Δv≥0\Delta v \geq 0, so Δw+cw≥−(κ−c)gv>0\Delta w + cw \geq -(\kappa - c)gv > 0, since κ−c>0\kappa - c > 0, g>0g > 0 and v<0v < 0. That contradicts Δw+cw≤0\Delta w + cw \leq 0.

Thinness in one direction is not always available. Smallness of volume is enough, thanks to the Alexandrov–Bakelman–Pucci estimate of 6B.8 Lᵖ Theory and Strong Solutions.

Lemma 3.5 Maximum principle in domains of small measure

Let Ω⊆BR\Omega \subseteq B_R and ∣c∣≤M|c| \leq M. There is δ=δ(n,R,M)>0\delta = \delta(n, R, M) > 0 such that if ∣Ω∣<δ|\Omega| < \delta, then Δw+cw≤0\Delta w + cw \leq 0 in Ω\Omega and w≥0w \geq 0 on ∂Ω\partial\Omega imply w≥0w \geq 0 in Ω\Omega.

Proof. Let v=−wv = -w, so Δv≥−cv\Delta v \geq -cv in Ω\Omega and v≤0v \leq 0 on ∂Ω\partial\Omega, and suppose m=sup⁡Ωv>0m = \sup_\Omega v > 0. Work on the open set Ω′={x∈Ω:v(x)>0}\Omega' = \{x \in \Omega : v(x) > 0\}. On its boundary v=0v = 0 (a boundary point inside Ω\Omega has v=0v = 0 by continuity; one on ∂Ω\partial\Omega has v≤0v \leq 0 and, as a limit of points with v>0v > 0, also v≥0v \geq 0). In Ω′\Omega', Δv≥−cv≥−Mv≥−Mm\Delta v \geq -cv \geq -Mv \geq -Mm. Apply ABP (6B.8 Lᵖ Theory and Strong Solutions) on Ω′\Omega', with aij=δija^{ij} = \delta_{ij}, D∗=1D^* = 1, f=−Mmf = -Mm and diameter at most 2R2R:

m=sup⁡Ω′v≤sup⁡∂Ω′v+2Rnα(n)1/n∥Mm∥Ln(Ω′)≤2RMnα(n)1/n ∣Ω∣1/n m.m = \sup_{\Omega'}v \leq \sup_{\partial\Omega'}v + \frac{2R}{n\alpha(n)^{1/n}}\big\|Mm\big\|_{L^n(\Omega')} \leq \frac{2RM}{n\alpha(n)^{1/n}}\,|\Omega|^{1/n}\,m.

If ∣Ω∣<δ=(nα(n)1/n2RM)n|\Omega| < \delta = \big(\frac{n\alpha(n)^{1/n}}{2RM}\big)^n, the factor in front of mm is less than 11, so m≤0m \leq 0, a contradiction.

Both lemmas give only w≥0w \geq 0. To upgrade this to w>0w > 0, and to get a sign for the normal derivative at the boundary, we need the strong maximum principle and Hopf's lemma of 6A.5 Maximum Principles in the presence of a zero-order term. For non-negative supersolutions they carry over.

Lemma 3.6 Strong maximum principle with a zero-order term

Let Ω\Omega be connected, ∣c∣≤M|c| \leq M, and w∈C2(Ω)w \in C^2(\Omega) with w≥0w \geq 0 and Δw+cw≤0\Delta w + cw \leq 0 in Ω\Omega. Then either w≡0w \equiv 0 or w>0w > 0 in Ω\Omega. If w>0w > 0 in Ω\Omega, and at a point x0∈∂Ωx_0 \in \partial\Omega where Ω\Omega has an interior ball, ww is C1C^1 up to x0x_0 with w(x0)=0w(x_0) = 0, then ∂w∂ν(x0)<0\frac{\partial w}{\partial\nu}(x_0) < 0 for the outward normal ν\nu.

Proof. Since w≥0w \geq 0, cw≥−Mwcw \geq -Mw, so Δw−Mw≤0\Delta w - Mw \leq 0. Put u=−w≤0u = -w \leq 0, so Δu−Mu≥0\Delta u - Mu \geq 0. First a weak maximum principle: a function zz with Δz−Mz>0\Delta z - Mz > 0 cannot have a non-negative interior maximum, since there Δz≤0\Delta z \leq 0 and −Mz≤0-Mz \leq 0. Now repeat the proof of Hopf's lemma in 6A.5 Maximum Principles (the lemma there with aij=δija^{ij} = \delta_{ij}, b=0b = 0): on the ball BrB_r touching ∂Ω\partial\Omega at x0x_0, with u<0u < 0 inside, the barrier v=e−λ∣x∣2−e−λr2v = e^{-\lambda|x|^2} - e^{-\lambda r^2} satisfies Δv−Mv≥(4λ2∣x∣2−2nλ−M)e−λ∣x∣2>0\Delta v - Mv \geq \big(4\lambda^2|x|^2 - 2n\lambda - M\big)e^{-\lambda|x|^2} > 0 on the annulus r2<∣x∣<r\frac r2 < |x| < r once λ\lambda is large. With z=u+εvz = u + \varepsilon v, Δz−Mz>0\Delta z - Mz > 0 in the annulus and z≤0z \leq 0 on its boundary for small ε\varepsilon, so z≤0z \leq 0 in the annulus by the weak principle just proved, and z(x0)=0z(x_0) = 0. Hence ∂νu(x0)≥−ε∂νv(x0)>0\partial_\nu u(x_0) \geq -\varepsilon\partial_\nu v(x_0) > 0, i.e. ∂νw(x0)<0\partial_\nu w(x_0) < 0. The first statement follows exactly as the strong maximum principle followed from Hopf's lemma in 6A.5 Maximum Principles: if {w=0}\{w = 0\} and {w>0}\{w > 0\} were both non-empty, a largest ball in {w>0}\{w > 0\} touching {w=0}\{w = 0\} at an interior point x1x_1 would give ∂νw(x1)<0\partial_\nu w(x_1) < 0, while x1x_1 is an interior minimum of ww, so ∇w(x1)=0\nabla w(x_1) = 0.

Moving planes

Theorem 3.7 Gidas–Ni–Nirenberg

Let BB be the unit ball in Rn\mathbb{R}^n, ff locally Lipschitz, and u∈C2(B‾)u \in C^2(\overline B) a solution of

−Δu=f(u),u>0  in B,u=0  on ∂B.-\Delta u = f(u), \quad u > 0 \ \text{ in } B, \qquad u = 0 \ \text{ on } \partial B.

Then uu is radially symmetric, u(x)=v(∣x∣)u(x) = v(|x|), and v′(r)<0v'(r) < 0 for 0<r<10 < r < 1.

Proof. Set-up. It suffices to prove symmetry in the x1x_1 direction (then apply it to every direction, by rotation invariance of the problem). For 0<λ<10 < \lambda < 1 let Tλ={x1=λ}T_\lambda = \{x_1 = \lambda\}, the cap Σλ={x∈B:x1>λ}\Sigma_\lambda = \{x \in B : x_1 > \lambda\}, and xλ=(2λ−x1,x2,…,xn)x^\lambda = (2\lambda - x_1, x_2, \ldots, x_n) the reflection of xx in TλT_\lambda. The reflected cap lies inside BB (reflecting a point of the cap towards the centre moves it closer to the centre; Exercise 3.12). On Σλ\Sigma_\lambda compare uu with its reflection:

wλ(x)=u(xλ)−u(x).w_\lambda(x) = u(x^\lambda) - u(x).

Since u(xλ)u(x^\lambda) also solves the equation,

Δwλ+cλ(x)wλ=0,cλ(x)=f(u(xλ))−f(u(x))u(xλ)−u(x)\Delta w_\lambda + c_\lambda(x)w_\lambda = 0, \qquad c_\lambda(x) = \frac{f(u(x^\lambda)) - f(u(x))}{u(x^\lambda) - u(x)}

(and cλ=0c_\lambda = 0 where the denominator vanishes). Because ff is Lipschitz on [0,max⁡u][0, \max u], ∣cλ∣≤M|c_\lambda| \leq M for a constant MM independent of λ\lambda. On the boundary of the cap: on the flat part Tλ∩BT_\lambda\cap B, wλ=0w_\lambda = 0; on the curved part ∂B∩{x1>λ}\partial B\cap\{x_1 > \lambda\}, u(x)=0u(x) = 0 while xλx^\lambda is an interior point of BB, so wλ=u(xλ)>0w_\lambda = u(x^\lambda) > 0.

Claim: wλ>0w_\lambda > 0 in Σλ\Sigma_\lambda for every λ∈(0,1)\lambda \in (0, 1).

Step 1: start near the edge. For λ\lambda close to 11, the cap lies in the slab {λ<x1<1}\{\lambda < x_1 < 1\} of width 1−λ<π2M1 - \lambda < \frac{\pi}{2\sqrt M}, so Lemma 3.4 gives wλ≥0w_\lambda \geq 0. And wλ≢0w_\lambda \not\equiv 0 (it is positive on the curved boundary). Lemma 3.6 gives wλ>0w_\lambda > 0 in the (connected) cap.

Step 2: move the plane. Let λ0=inf⁡{λ∈(0,1):wμ>0 in Σμ for all μ∈[λ,1)}\lambda_0 = \inf\{\lambda \in (0, 1) : w_\mu > 0 \text{ in } \Sigma_\mu \text{ for all } \mu \in [\lambda, 1)\}, and suppose λ0>0\lambda_0 > 0. By continuity in λ\lambda, wλ0≥0w_{\lambda_0} \geq 0 in Σλ0\Sigma_{\lambda_0}, and as in Step 1 it is not identically zero, so wλ0>0w_{\lambda_0} > 0 there. Let δ\delta be the constant of Lemma 3.5 for R=1R = 1 and this MM, and choose a compact K⊂Σλ0K \subset \Sigma_{\lambda_0} with ∣Σλ0∖K∣<δ2|\Sigma_{\lambda_0}\setminus K| < \frac\delta2. On KK, wλ0≥η>0w_{\lambda_0} \geq \eta > 0, so by uniform continuity wλ>0w_\lambda > 0 on KK for all λ\lambda in some interval (λ0−ε,λ0](\lambda_0 - \varepsilon, \lambda_0], and also, shrinking ε\varepsilon, ∣Σλ∖K∣<δ|\Sigma_\lambda\setminus K| < \delta. On the open set Σλ∖K\Sigma_\lambda\setminus K we have Δwλ+cλwλ=0\Delta w_\lambda + c_\lambda w_\lambda = 0, and wλ≥0w_\lambda \geq 0 on its boundary (on ∂K\partial K by the previous sentence, on TλT_\lambda and the curved boundary as in the set-up). By Lemma 3.5, wλ≥0w_\lambda \geq 0 on Σλ∖K\Sigma_\lambda\setminus K, hence on all of Σλ\Sigma_\lambda, and by Lemma 3.6 wλ>0w_\lambda > 0. This holds for every λ∈(λ0−ε,λ0]\lambda \in (\lambda_0 - \varepsilon, \lambda_0], contradicting the definition of λ0\lambda_0. So λ0=0\lambda_0 = 0.

Step 3: symmetry. Letting λ→0\lambda \to 0, u(−x1,x′)≥u(x1,x′)u(-x_1, x') \geq u(x_1, x') for x1>0x_1 > 0. Applying the same argument in the direction −e1-e_1 gives the reverse inequality, so uu is symmetric under x1↦−x1x_1 \mapsto -x_1. Every direction is equivalent, so uu is symmetric under every reflection through a hyperplane through the origin, hence radial.

Step 4: monotonicity. For λ∈(0,1)\lambda \in (0, 1), wλ>0w_\lambda > 0 in Σλ\Sigma_\lambda and wλ=0w_\lambda = 0 on TλT_\lambda. By Lemma 3.6 (the cap has an interior ball at each point of Tλ∩BT_\lambda\cap B), the normal derivative of wλw_\lambda into the cap is positive: ∂1wλ>0\partial_1w_\lambda > 0 on TλT_\lambda. There, ∂1wλ=−∂1u(xλ)−∂1u(x)=−2∂1u(x)\partial_1w_\lambda = -\partial_1u(x^\lambda) - \partial_1u(x) = -2\partial_1u(x) (as xλ=xx^\lambda = x on the plane). So ∂1u<0\partial_1u < 0 on {x1=λ}\{x_1 = \lambda\} for every λ∈(0,1)\lambda \in (0, 1), which for the radial function u=v(∣x∣)u = v(|x|) means v′(r)<0v'(r) < 0 for 0<r<10 < r < 1.

The proof never solved the equation. It compared the solution with its own reflection, used the maximum principle to keep the comparison one-signed, and slid the plane until it could go no further. That only the sign of uu and the boundary condition entered, together with the Lipschitz bound on ff, is why the method is so robust: it works for fully nonlinear equations and on unbounded domains with suitable decay.

Positivity is needed. Sign-changing solutions in a ball need not be radial: for −Δu=λu-\Delta u = \lambda u the second eigenfunctions of the disc, J1(j1,1r)cos⁡θJ_1(j_{1,1}r)\cos\theta, are odd in x1x_1. And the ball is needed: in an annulus, positive solutions need not be radial: Charles Coffman showed in 1984 that in thin planar annuli −Δu=up-\Delta u = u^p has positive non-radial solutions, more and more of them as the annulus gets thinner.

Bubbles

On all of Rn\mathbb{R}^n, a moving-plane argument "from infinity" classifies the positive solutions at the critical exponent.

Theorem 3.8 Caffarelli–Gidas–Spruck (statement)

Every positive C2C^2 solution of −Δu=un+2n−2-\Delta u = u^{\frac{n+2}{n-2}} on Rn\mathbb{R}^n, n≥3n \geq 3, is of the form

u(x)=(n(n−2))n−24(λλ2+∣x−x0∣2)n−22u(x) = \big(n(n - 2)\big)^{\frac{n-2}{4}}\Big(\frac{\lambda}{\lambda^2 + |x - x_0|^2}\Big)^{\frac{n-2}{2}}

for some λ>0\lambda > 0 and x0∈Rnx_0 \in \mathbb{R}^n.

These are the bubbles of 4A.11 Sobolev Embeddings and Critical Exponents: they all have the same energy ∫∣∇u∣2\int|\nabla u|^2, they are the extremals of the sharp Sobolev inequality (Aubin and Talenti, 1976), and they are the profiles that concentrate when Palais–Smale fails at the critical exponent (6C.2 Critical Points and the Mountain Pass). Luis Caffarelli, Basilis Gidas and Joel Spruck proved the theorem in 1989 (without assuming any decay at infinity, which is the hard part; Gidas, Ni and Nirenberg had proved it under a decay assumption). Geometrically, the bubbles are the round metric of the sphere, pulled back to Rn\mathbb{R}^n by stereographic projection and written as a conformal factor (Exercise 3.14).

Where this goes Symmetry and conservation in the rest of the guide

Perelman's functionals are invariant under diffeomorphisms, and Noether's second theorem is why their first variations satisfy identities (such as the contracted Bianchi identity) that make the Ricci flow a gradient flow only up to diffeomorphism (12A.2 Ricci Flow as a Gradient Flow, Exercise 3.15). Maximum-principle symmetry theorems recur in the classification of singularity models: Simon Brendle proved in 2013 that every three-dimensional steady gradient Ricci soliton that is non-collapsed is rotationally symmetric, hence is the Bryant soliton, settling a question Perelman raised; his argument uses maximum principles for the linearised soliton equation in the same spirit as moving planes (11B.1 Ricci Solitons, 12B.1 κ-Solutions). In mean curvature flow, Brendle and Choi (2019) proved by related methods that every non-compact, strictly convex, non-collapsed ancient solution in R3\mathbb{R}^3 is, up to symmetries, the bowl soliton.

History

Emmy Noether proved her theorems in 1918, in a paper motivated by the problem of energy conservation in Einstein's general relativity, at the invitation of Hilbert and Klein. Jonathan Lane (1870) and Robert Emden (1907) studied polytropic stars. The method of moving planes was invented by Aleksandr Alexandrov in 1956 to prove that the only compact embedded hypersurfaces of constant mean curvature are round spheres ("Alexandrov's soap bubble theorem"); James Serrin used it in 1971 for an overdetermined problem in fluid mechanics, and Gidas, Ni and Nirenberg adapted it to semilinear equations in 1979. Henri Berestycki and Louis Nirenberg simplified it with the small-domain maximum principle in 1991. Caffarelli, Gidas and Spruck's classification dates from 1989.

Recall Where we stand

For a Lagrangian independent of xx, the energy–momentum tensor Tki=Lpi∂ku−LδikT_k^i = L_{p_i}\partial_ku - L\delta_{ik} is divergence-free on solutions: energy and momentum conservation for waves. An internal symmetry with generator AA gives the divergence-free current Lpi⋅AuL_{p_i}\cdot Au: charge conservation for Schrödinger, rotation currents for harmonic maps into spheres. Contracting TT with the scaling field gives Pohozaev's identity. Operators Δ+c\Delta + c with bounded cc of any sign satisfy the maximum principle on narrow domains (by a cosine weight) and on domains of small measure (by ABP). The method of moving planes reflects a positive solution of −Δu=f(u)-\Delta u = f(u) in a ball, compares it with itself on caps, and slides the plane to the centre: the solution is radial and decreasing (Gidas–Ni–Nirenberg). On Rn\mathbb{R}^n, positive solutions at the critical exponent are exactly the bubbles. 6C.4 The Yamabe Problem and Critical Exponents puts the bubbles to work: they are the round sphere in disguise, and their energy is the threshold below which the Yamabe problem, the model variational problem at the critical exponent, regains compactness.

Exercises

Exercise 3.9 Energy for the wave equation from Noether

With L=12(p02−∣p′∣2)L = \frac12(p_0^2 - |p'|^2) in the variables (x0,x′)=(t,x)(x_0, x') = (t, x), compute T00T_0^0 and T0iT_0^i for i≥1i \geq 1 from Theorem 3.1 and check that ∂0T00+∑i∂iT0i=0\partial_0T_0^0 + \sum_i\partial_iT_0^i = 0 is the energy identity of Example 3.2.

Solution

Lp0=p0=utL_{p_0} = p_0 = u_t and Lpi=−pi=−∂iuL_{p_i} = -p_i = -\partial_iu. So T00=Lp0ut−L=ut2−12(ut2−∣∇u∣2)=12(ut2+∣∇u∣2)T_0^0 = L_{p_0}u_t - L = u_t^2 - \frac12(u_t^2 - |\nabla u|^2) = \frac12(u_t^2 + |\nabla u|^2), and T0i=Lpiut=−ut∂iuT_0^i = L_{p_i}u_t = -u_t\partial_iu. The identity is ∂t12(ut2+∣∇u∣2)−div⁡(ut∇u)=0\partial_t\frac12(u_t^2 + |\nabla u|^2) - \operatorname{div}(u_t\nabla u) = 0.

Exercise 3.10 Charge conservation

For iut=−Δu+V(x)uiu_t = -\Delta u + V(x)u with complex uu and real VV, show directly that ∂t∣u∣2+div⁡(2Im⁡(uˉ∇u))=0\partial_t|u|^2 + \operatorname{div}\big(2\operatorname{Im}(\bar u\nabla u)\big) = 0, and deduce that ∫∣u∣2\int|u|^2 is constant for decaying solutions. Which symmetry of the equation is responsible?

Solution

∂t∣u∣2=2Re⁡(uˉut)=2Re⁡(uˉ⋅i(Δu−Vu))=−2Im⁡(uˉΔu)\partial_t|u|^2 = 2\operatorname{Re}(\bar uu_t) = 2\operatorname{Re}\big(\bar u\cdot i(\Delta u - Vu)\big) = -2\operatorname{Im}(\bar u\Delta u), since Re⁡(iuˉVu)=0\operatorname{Re}(i\bar uVu) = 0. And div⁡(uˉ∇u)=∣∇u∣2+uˉΔu\operatorname{div}(\bar u\nabla u) = |\nabla u|^2 + \bar u\Delta u, whose imaginary part is Im⁡(uˉΔu)\operatorname{Im}(\bar u\Delta u). So ∂t∣u∣2+2div⁡Im⁡(uˉ∇u)=0\partial_t|u|^2 + 2\operatorname{div}\operatorname{Im}(\bar u\nabla u) = 0. The phase symmetry u↦eiτuu \mapsto e^{i\tau}u, generated by A=iA = i.

Exercise 3.11 Pohozaev from the energy–momentum tensor

For L=12∣p∣2−F(z)L = \frac12|p|^2 - F(z), so Tki=∂iu ∂ku−(12∣∇u∣2−F(u))δikT_k^i = \partial_iu\,\partial_ku - \big(\frac12|\nabla u|^2 - F(u)\big)\delta_{ik}: (a) show ∑i,k∂i(xkTki)=∑kTkk=(1−n2)∣∇u∣2+nF(u)\sum_{i,k}\partial_i(x_kT_k^i) = \sum_kT_k^k = \big(1 - \frac n2\big)|\nabla u|^2 + nF(u) for solutions; (b) integrate over UU, with u=0u = 0 on ∂U\partial U, and recover Pohozaev's identity of 6C.2 Critical Points and the Mountain Pass.

Solution

(a) ∂i(xkTki)=δikTki+xk∂iTki\partial_i(x_kT_k^i) = \delta_{ik}T_k^i + x_k\partial_iT_k^i, and summing, the second term vanishes by Theorem 3.1. ∑kTkk=∣∇u∣2−n(12∣∇u∣2−F)\sum_kT^k_k = |\nabla u|^2 - n\big(\frac12|\nabla u|^2 - F\big). (b) The divergence theorem gives ∫∂U∑i,kxkTkiνi=∫U((1−n2)∣∇u∣2+nF)\int_{\partial U}\sum_{i,k}x_kT_k^i\nu_i = \int_U\big((1 - \frac n2)|\nabla u|^2 + nF\big). On ∂U\partial U, ∇u=(∂νu)ν\nabla u = (\partial_\nu u)\nu and F(u)=F(0)=0F(u) = F(0) = 0, so ∑i,kxkTkiνi=(∂νu)2(x⋅ν)−12(∂νu)2(x⋅ν)=12(∂νu)2(x⋅ν)\sum_{i,k}x_kT^i_k\nu_i = (\partial_\nu u)^2(x\cdot\nu) - \frac12(\partial_\nu u)^2(x\cdot\nu) = \frac12(\partial_\nu u)^2(x\cdot\nu). Rearranging gives n−22∫∣∇u∣2−n∫F=−12∫∂U(∂νu)2(x⋅ν)\frac{n-2}{2}\int|\nabla u|^2 - n\int F = -\frac12\int_{\partial U}(\partial_\nu u)^2(x\cdot\nu).

Exercise 3.12 The reflected cap stays in the ball

For 0<λ<10 < \lambda < 1 and x∈Bx \in B with x1>λx_1 > \lambda, show that ∣xλ∣<1|x^\lambda| < 1, where xλ=(2λ−x1,x′)x^\lambda = (2\lambda - x_1, x'). (Compare (2λ−x1)2(2\lambda - x_1)^2 with x12x_1^2.)

Solution

∣xλ∣2−∣x∣2=(2λ−x1)2−x12=4λ(λ−x1)<0|x^\lambda|^2 - |x|^2 = (2\lambda - x_1)^2 - x_1^2 = 4\lambda(\lambda - x_1) < 0 since λ>0\lambda > 0 and x1>λx_1 > \lambda. So ∣xλ∣<∣x∣<1|x^\lambda| < |x| < 1.

Exercise 3.13 The narrow-domain constant

On the slab {0<x1<d}\{0 < x_1 < d\} in one dimension, the function w=sin⁡πx1dw = \sin\frac{\pi x_1}{d} satisfies w′′+π2d2w=0w'' + \frac{\pi^2}{d^2}w = 0, w=0w = 0 at the ends, and −w-w is a negative solution. Explain why this shows that some smallness condition on dd relative to MM is necessary in Lemma 3.4, and compare the threshold πM\frac{\pi}{\sqrt M} obtained from this example with the lemma's π2M\frac{\pi}{2\sqrt M}.

Solution

With c=π2d2c = \frac{\pi^2}{d^2} the function −w-w satisfies Δ(−w)+c(−w)=0≤0\Delta(-w) + c(-w) = 0 \leq 0, is 00 on the boundary, and is negative inside: the conclusion fails when d=πMd = \frac{\pi}{\sqrt M}. So the maximum principle can hold only for d<πMd < \frac{\pi}{\sqrt M} (the first Dirichlet eigenvalue π2d2\frac{\pi^2}{d^2} must exceed MM). The lemma's threshold π2M\frac{\pi}{2\sqrt M} is smaller by a factor 22, the price of using g=cos⁡g = \cos on a shifted interval so that gg stays bounded below; the sharp statement is that the maximum principle for Δ+c\Delta + c with c≤Mc \leq M holds exactly when the first eigenvalue of −Δ-\Delta on the domain exceeds MM.

Exercise 3.14 The bubble in three dimensions

For n=3n = 3 the critical exponent is 55. Show that u(x)=31/4(1+∣x∣2)−1/2u(x) = 3^{1/4}(1 + |x|^2)^{-1/2} solves −Δu=u5-\Delta u = u^5 on R3\mathbb{R}^3, using the radial Laplacian. Check that it is the case λ=1\lambda = 1, x0=0x_0 = 0 of Theorem 3.8, and that ∫R3u6 dx<∞\int_{\mathbb{R}^3}u^6\,dx < \infty.

Solution

With u=c(1+r2)−1/2u = c(1 + r^2)^{-1/2}: u′=−cr(1+r2)−3/2u' = -cr(1 + r^2)^{-3/2}, u′′=−c(1+r2)−3/2+3cr2(1+r2)−5/2u'' = -c(1 + r^2)^{-3/2} + 3cr^2(1 + r^2)^{-5/2}, and Δu=u′′+2ru′=c(1+r2)−5/2(−3(1+r2)+3r2)=−3c(1+r2)−5/2\Delta u = u'' + \frac2ru' = c(1 + r^2)^{-5/2}\big(-3(1 + r^2) + 3r^2\big) = -3c(1 + r^2)^{-5/2}. So −Δu=3c(1+r2)−5/2-\Delta u = 3c(1 + r^2)^{-5/2} and u5=c5(1+r2)−5/2u^5 = c^5(1 + r^2)^{-5/2}, equal when c4=3c^4 = 3. The formula gives (3⋅1)1/4(11+∣x∣2)1/2(3\cdot1)^{1/4}\big(\frac{1}{1 + |x|^2}\big)^{1/2}. And u6∼r−6u^6 \sim r^{-6} at infinity, integrable in three dimensions.

Exercise 3.15 Rehearsal: Noether's second theorem and the Bianchi identity

On a closed manifold, the total scalar curvature S(g)=∫MR dV\mathcal S(g) = \int_MR\,dV is invariant under diffeomorphisms: S(ϕ∗g)=S(g)\mathcal S(\phi^*g) = \mathcal S(g). Its first variation in a direction hh is ddsS(g+sh)=∫M⟨12Rg−Ric⁡,h⟩ dV\frac{d}{ds}\mathcal S(g + sh) = \int_M\big\langle\frac12Rg - \operatorname{Ric}, h\big\rangle\,dV (9A.5 Computing Curvature, 12A.2 Ricci Flow as a Gradient Flow). (a) An infinitesimal diffeomorphism generated by a vector field XX changes gg by h=LXgh = \mathcal L_Xg, with components ∇iXj+∇jXi\nabla_iX_j + \nabla_jX_i. Show that invariance gives ∫⟨12Rg−Ric⁡,LXg⟩ dV=0\int\big\langle\frac12Rg - \operatorname{Ric}, \mathcal L_Xg\big\rangle\,dV = 0 for every XX. (b) Integrate by parts (using the symmetry of 12Rg−Ric⁡\frac12Rg - \operatorname{Ric}) to get −2∫Xj∇i(12Rgij−Rij) dV=0-2\int X^j\nabla^i\big(\frac12Rg_{ij} - R_{ij}\big)\,dV = 0 for all XX, and conclude the contracted Bianchi identity ∇iRij=12∇jR\nabla^iR_{ij} = \frac12\nabla_jR. This is an identity, true for every metric, not only for critical points: exactly the content of Noether's second theorem for a symmetry group depending on arbitrary functions. Perelman's F\mathcal F-functional has the same diffeomorphism invariance, and its corresponding identity is what allows the Ricci flow to be written as its gradient flow modulo diffeomorphisms (12A.2 Ricci Flow as a Gradient Flow).

Solution

(a) The family gs=ϕs∗gg_s = \phi_s^*g, with ϕs\phi_s the flow of XX, has S(gs)=S(g)\mathcal S(g_s) = \mathcal S(g) and ddsgs∣s=0=LXg\frac{d}{ds}g_s|_{s=0} = \mathcal L_Xg, so the first variation in this direction vanishes. (b) With E=12Rg−Ric⁡E = \frac12Rg - \operatorname{Ric} symmetric, ⟨E,LXg⟩=Eij(∇iXj+∇jXi)=2Eij∇iXj\langle E, \mathcal L_Xg\rangle = E^{ij}(\nabla_iX_j + \nabla_jX_i) = 2E^{ij}\nabla_iX_j, and integrating by parts on the closed manifold, ∫2Eij∇iXj=−2∫Xj∇iEij\int2E^{ij}\nabla_iX_j = -2\int X_j\nabla_iE^{ij}. Since this vanishes for every XX, ∇iEij=0\nabla_iE^{ij} = 0, that is, 12∇jR−∇iRij=0\frac12\nabla^jR - \nabla_iR^{ij} = 0.

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