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IOA UBOSTAD NIKOLAI u t p p u ∞ → u p u = + ) 1. ∆ div( = ncnrs,the contrast, In . Introduction := p 1 p | u Du . p p p 1 − p PRBLCEUTO IN EQUATION -PARABOLIC | | 2 Du u | 1 − 2 Du , p ∆ | normalised 2 | D p − p u − p 2 ssuid ernsmethod Perron’s studied. is ∆ 2 ⊂ u Du | p Du Du R u. ) n , | × , p | ( Lpaeoeao is operator -Laplace Du Du p −∞ prblcequation -parabolic | , ∞ , .Hr ∆ Here ). p i- u 2 NIKOLAIUBOSTAD to the non-linear equation (1.1) in general domains, not necessarily space-time cylinders. Equation (1.1) was studied by Does in connection with image pro- cessing, see [Doe09]. The existence of viscosity solutions on cylinders QT = Q (0, T ) was established by Perron’s method. Integral× to the potential theory is the regularity of boundary points. A point ζ0 ∂Ω is called regular if, for every continuous function f : ∂Ω R,∈ we have → lim u(η)= f(ζ0), η Ω. η→ζ0 ∈ The boundary values are prescribed as for an elliptic problem. For example, the points on Q t = T are not regular boundary points ×{ } of the cylinder QT . We characterize the regularity of boundary points in terms of barri- ers. We say that w is a barrier at ζ if w is a positive supersolution of (1.1) defined on the entire domain, such that w(ζ) = 0 and w(η) > 0 for ζ = η ∂Ω. To6 keep∈ the presentation within reasonable limits, we investigate only the case where the boundary data f is bounded. This is not a serious restriction. The related equation (1.2) u = Du 2−p∆ u, t | | p without the factor 1/p present, was investigated in [BG15]. For p 2, it was proved that a boundary point ζ is regular if, and only if, there≥ exists a barrier at ζ. The authors also showed that in the case of space - time cylinder QT , (x, t) ∂Q (0, T ], is a regular boundary point if and only if x ∂Q is a∈ a regular× boundary point for the elliptic p-Laplacian. ∈ The regularity of a point is a very delicate issue. Using the Petrovsky criterion, one can construct a domain where the origin is regular for 1 the equation ut = ∆u, while it is irregular for ut = 2 ∆u, cf [Wat12]. Therefore it is quite remarkable that as p , the domain in our Petrovsky criterion converges precisely to the→ ∞ domain in the Petro- vsky criterion for the normalised -parabolic equation (1.6) derived in [Ubo17], namely that the origin∞ is an irregular point for the domain enclosed by the hypersurfaces (x, t) Rn (0, ): x 2 = 4t log log t (1.3) { ∈ × ∞ | | − | | ||} and t = c , { − } for 0 We consider (1.1) instead of (1.2) because of the convergence prop- erties as p . Results similar to ours could easily be established for (1.2) by→ the ∞ same methods. Indeed, this was recently proven in [BBP17] . The regular, non-normalised p-parabolic equation (1.4) ut = ∆pu, 1 (1.5) ut = ∆∞u, and the related normalised -parabolic equation ∞ (1.6) u = Du −2∆ u := ∆N u, t | | ∞ ∞ where the -Laplace operator is given by ∞ 2 ∆∞u = D u Du,Du . Both of these have an interesting theory in their own right, see [JK06] regarding the normalised case and [CW03] for (1.5). As p , (1.1) converges to (1.6) and not to (1.5). Our first→∞ result is a characterization of regular boundary points via exterior spheres: Theorem 1.1 (Exterior Sphere). Let ζ0 =(t0, x0) ∂Ω, and suppose that there exists a closed ball (x, t): x x′ 2 ∈+ (t t′)2 R2 { | − | − ≤ 0} intersecting Ω precisely at ζ0. Then ζ0 is regular, if the intersection ′ ′ point is not the south pole, that is (x0, t0) =(x , t R0). If the point of intersection is the north pole, we must restrict6 the− radius of the sphere. We use a barrier function to prove the following Petrovsky criterion: 4 NIKOLAIUBOSTAD Theorem 1.2. The origin (0, 0) is a regular point for the domain en- closed by the hypersurfaces (x, t) Rn (0, ): x 2 = βt log log t and t = c , { ∈ × ∞ | | − | | ||} { − } for 0 1.1. Notation. In what follows Ω is an arbitrary domain in Rn ( , ). Q is a space-time cylinder: Q = Q (0, T ), ∂Ω is the× −∞ ∞ T T × Euclidean boundary of Ω and ∂pQT is the parabolic boundary of QT , i.e. (Q 0 ) (∂Q (0, T ]). (the ”bottom” and the sides of the cylinder.×{ The} top∪ is excluded.)× ζ, η Rn R are points in space-time, that is ζ =(x, t). ∈ × We denote by Du the gradient of u(x, t) taken with respect to the spatial coordinates x, and D2u is the spatial Hessian matrix of u. a, b is the Euclidean inner product of the vectors a, b Rn. and xh yi ∈ ⊗ denotes the tensor product of the vectors x, y, that is (x y)i,j = xiyj. The space of lower semi-continuous functions from Ω to⊗ R is denoted by LSC(Ω), while USC(Ω) contains the upper semicontinuou∪ {∞} s ones. ON THE NORMALISED p-PARABOLICEQUATION 5 2. Several solutions We derive several explicit solutions to (1.1), and identify the funda- mental solution. 2.1. Uniform propagation. Assume u(x, t)= w( a, x bt), a Rn, b R. We then get h i− ∈ ∈ u = bw′, t − ′ uxi = aiw , ′′ uxixj = aiajw , and hence ∆u = ∆N = a 2w′′. ∞ | | Inserting this into (1.1), we get that w must satisfy p 1 w′ + w′′ a 2 − =0, | | bp with solution − ζ w(ζ)= A + B e m , or − 1 (ha,xi−bt) u(x, t)= A + B e m , with m = a 2 p−1 . and ζ = a, x bt. | | bp h i − 2.2. Separable solution. Assume u(x, t) = f(r)+ g(t), r = x . We get | | ′ ut = g , n 1 ∆u = f ′′ + − f ′, r N ′′ ∆∞u = f . Thus g′(t)= c, and n 1 f ′′(r)+ − f ′(r)= c (p 1)r − So p n 1 cp 2 p 1 − f(r)= r + c − r p−1 + c , 2 n + p 1 p n 2 − Setting c1 = c2 = 0, we get the solution p n p 2 p 1 − u(x, t)= c x +2 − x p−1 +2t . n + p| | p n| | − 6 NIKOLAIUBOSTAD 2.3. Heat Equation transformation. We search for solutions on the form u(x, t)= v(rν, t), where r = x and ν is a critical exponent to be determined. This gives | | ∂2 n 1 ∂ ∆u = v(rν, t)+ − v(rν, t), ∂r2 r ∂r ∂2 ∆N u = v(rν, t). ∞ ∂r2 Calculating, we get ∂ v(rν, t)= νrν−1v′, ∂r ∂2 v(rν, t)= ν(ν 1)rν−2v′ + ν2r2ν−2v′′, ∂r2 − where ∂v v′ = , ρ = rν ∂ρ Hence we get ∆u = ν(ν 1)rν−2v′ + ν2r2ν−2v′′ + νrν−2v′, − ∆N u = ν(ν 1)rν−2v′ + ν2r2ν−2v′′. ∞ − Inserting this into (1.1) and collecting terms we get p 1 ν (2.1) v = ν2 − r2ν−2v′′ + rν−2[(p 2)(ν 1) + n + ν]v′. t p p − − We want to eliminate the first order terms in (2.1), so we demand (p 2)(ν 1) + n + ν =0, − − or p n ν = − . p 1 − Then (1.1) reads 2 (p n) 2−n v = − ρ p−n v , t p(p 1) ρρ − where p−n ρ = x p−1 . | | ON THE NORMALISED p-PARABOLICEQUATION 7 2.4. Similarity I. We make the ansatz x 2 u(x, t)= F (ζ), ζ = | | , t We calculate F ′(ζ)ζ u = , t − t 2F ′(ζ)x Du = , t 2F ′(ζ) 4F ′′(ζ) D2u = I + (x x). t t2 ⊗ Hence 2F ′(ζ) 4F ′′(ζ) tr(D2u)= n + x 2, t t2 | | 2F ′(ζ) 4F ′′(ζ) ∆N u = + x 2. ∞ t t2 | | So, if u is a solution then F ′(ζ)ζ F ′(ζ) F ′′(ζ)ζ α β =0, − t − t − t with α, β as in (2.3). If t = 0; 6 F ′′(ζ) d α 1 = log F ′(ζ)= . F ′(ζ) dζ −βζ − β Integrating the above gives ζ α log F ′(ζ)= log ζ, −β − β or ζ − α − s F (ζ)= C s β e β ds. ˆ0 This leads to the solution |x|2/t − α − s (2.2) u(x, t)= C s β e β ds. ˆ0 This solution is not differentiable where x = 0. However, (2.2) is a solution outside the line 0 (0, ) Rn (0, ), and a subsolution or supersolution depending{ } on × the∞ sign⊂ of C×in all∞ of Rn (0, ). × ∞ 8 NIKOLAIUBOSTAD 2.5. Similarity II. We note that if u(x, t) is a solution of (1.1), then so is v(x, t)= u(Ax, A2t). We search for solutions on the form x 2 u(x, t)= g(t)f(ζ), ζ = | | . t Inserting this into (1.1), we get g(t)f ′(ζ)ζ u = g′(t)f(ζ) , t − t 2g(t)f ′(ζ) Du = x, t and 2g(t)f ′(ζ) 4g(t)f ′′(ζ) D2u = I + (x x). t t2 ⊗ Hence we see 2g(t)f ′(ζ) 4g(t)f ′′(ζ) tr(D2u)= n + x 2, t t2 | | and 2g(t)f ′(ζ) 4g(t)f ′′(ζ) ∆N u = + x 2. ∞ t t2 | | Therefore, u is a solution to (1.1) if g(t)f ′(ζ)ζ p + n 2 g(t)f ′(ζ) p 1 g(t)f ′′(ζ)ζ g′(t)f(ζ) =2 − +4 − . − t p t p t p+n−2 p−1 with α =2 p and β =4 p , tg′(t)f(ζ) αg(t)f ′(ζ)= g(t)ζ(f ′(ζ)+ βf ′′(ζ)). − − ζ for t > 0. The right hand side of this is zero if f(ζ) = e β . Inserting this back in, we see that α f(ζ) tg′(t)+ g(t) =0, β with solution − α g(t)= t β . Together, this gives 2 − α − |x| u(x, t)= t β e βt , (2.3) p + n 2 p 1 α =2 − , β =4 − . p p ON THE NORMALISED p-PARABOLICEQUATION 9 This is a solution for t> 0, and if we replace t by t we get a solution for negative t, as well.1 − Remark 2.1. As p , α 2 and β 4. This gives that (2.3) converges to the fundamental→ ∞ solution→ → 2 1 − |x| W (x, t)= e 4t . √t of the normalised -parabolic equation found in [JK06]. Compare also (2.3) with the fundamental∞ solution to the heat equation, 2 1 − |x| H(x, t)= e 4t . (4πt)n/2 3. Comparison and Perron solutions In this Section we present several basic facts regarding the existence of solutions to (1.1), and present Perron’s method. We start with the definition of viscosity solutions. If Du = 0, we replace the operator p with its lower or upper semicontinuous envelope: A Definition 3.1. A lower semicontinuous function u L∞(Ω) is a viscosity supersolution of (1.1) provided that, if u φ has∈ a minimum at ζ Ω for φ C2(Ω), then − 0 ∈ ∈ φ (ζ ) φ(ζ ) 0, if Dφ(ζ ) =0, t 0 −Ap 0 ≥ 0 6 φ (ζ ) 1 tr(D2φ(ζ )) p−2 λ(D2φ(ζ )) 0, if Dφ(ζ )=0. ( t 0 − p 0 − p 0 ≥ 0 An upper semicontinuous function u L∞(Ω) is a viscosity subsolu- ∈ tion of (1.1) provided that, if u φ has a maximum at ζ0 Ω for φ C2(Ω), then − ∈ ∈ φ (ζ ) φ(ζ ) 0, if Dφ(ζ ) =0, t 0 −Ap 0 ≤ 0 6 φ (ζ ) 1 tr(D2φ(ζ )) p−2 Λ(D2φ(ζ )) 0, if Dφ(ζ )=0. ( t 0 − p 0 − p 0 ≤ 0 A function that is both a viscosity sub- and supersolution is called a viscosity solution. 2 2 Here λ(D φ(ζ0)), λ(D φ(ζ0)) denotes the smallest and largest eigen- 2 2 values of the Hessian matrix D φ(ζ0), and tr(D φ(ζ0)) is its trace. It turns out that the second condition in Definition 3.1 can be re- laxed. This is the Lemma 2 in [MPR10]. 1It has recently come to the author’s attention that this solution also was found in [BG13]. 10 NIKOLAIUBOSTAD Lemma 3.2. An upper semicontinuous function u L∞(Ω), is a vis- cosity subsolution of (1.1) provided that, if u φ has∈ a maximum at ζ Ω for φ C2(Ω), then either − 0 ∈ ∈ φ (ζ ) φ(ζ ) 0, if Dφ(ζ ) =0, t 0 −Ap 0 ≤ 0 6 or φ (ζ ) 0, if Dφ(ζ )=0,D2φ(ζ )=0. t 0 ≤ 0 0 A similar result holds for viscosity supersolutions. We define p-parabolic functions in Ω as follows: Definition 3.3. A function u LSC(Ω) L∞(Ω) is a supersolution to (1.1) if it satisfies the following∈ comparison∩ principle: On each set of the form Q = Q (t , t ) with closure in Ω, and t1,t2 × 1 2 for each solution h to (1.1) continuous up to the closure of Qt1,t2 , and h u on ∂ Q , then ≤ p t1,t2 h u in Q . ≤ t1,t2 In Banerjee–Garofalo [BG15], they use the name generalized su- per/subsolution instead of super/subparabolic function. They prove that these are the same as the viscosity super/subsolutions in a given domain. Hence we can use the term parabolic interchangeably with viscosity solution. The assumption that supersolutions are bounded is not needed. See Theorem 2.6 in [BBP17]. We shall improve this to include comparison on general domains Σ compactly contained in Ω, see Lemma 4.3. Using the classical comparison principle for cylindrical domains QT = Q (0, T ) and a covering argument, we can prove Theorem 3.10 from [BG15].× This is the comparison principle, essential for Perron’s method to work. Theorem 3.4. Suppose u is a supersolution bounded from above and v is a subsolution bounded from below of (1.1) in a bounded open set Ω Rn+1. If at each point ζ ∂Ω we have ⊂ 0 ∈ lim sup v(ζ) lim inf u(ζ), ζ→ζ0 ≤ ζ→ζ0 then v u in Ω. ≤ ON THE NORMALISED p-PARABOLIC EQUATION 11 3.1. The Perron Method. We start with a definition. Let f : ∂Ω R be a continuous function. → Definition 3.5. A function u belongs to the upper class f if u is a viscosity supersolution in Ω and U lim inf u(η) f(ζ) η→ζ ≥ for ζ ∂Ω. ∈ Likewise, a function v belongs to the lower class f if v is a viscosity subsolution in Ω and L lim sup v(η) f(ζ) η→ζ ≤ for all ζ ∂Ω. We define∈ the upper solution H (ζ) = inf u(ζ): u , f { ∈ Uf } and the lower solution H (ζ) = sup v(ζ): v . f { ∈Lf } Note that at each point the inf and sup are taken over the functions. Remark 3.6. The Comparison Principle, Theorem 3.4, gives immedi- ately that v u in Ω, for v and u , and hence ≤ ∈Lf ∈ Uf H H . f ≤ f Whether Hf = Hf holds in general, is a more subtle question. We need that the Upper and Lower Perron solutions, indeed, are viscosity solutions to (1.1). Theorem (Theorem 3.12 in [BG15]). The upper Perron solution Hf and the lower Perron solution Hf are solutions to (1.1) in Ω. An integral part of the theory of Perron solutions are the boundary regularity and the barrier functions. Definition 3.7. We say that ζ ∂Ω is a regular boundary point if 0 ∈ lim Hf (ζ)= f(ζ0). ζ→ζ0 for every continuous function f : ∂Ω R. → Note that we instead could have used Hf in the above, since Hf = H . − −f 12 NIKOLAIUBOSTAD Remark 3.8. The Petrovsky condition in Theorem 1.3 shows that a point can be regular for ut = pu but not for ut = qu, p < q. Hence it would be more accurate to useA the term p-regular,A but we use regular where no confusion will arise. Definition 3.9. A function w is a barrier at ζ ∂Ω if 0 ∈ (1) w> 0 and w is p-superparabolic in Ω, (2) lim inf w(ζ) > 0 for ζ = η ∂Ω, ζ→η 0 6 ∈ (3) limζ→ζ0 w(ζ)=0. Using barrier functions, we can prove the following classical result, which is Theorem 4.2 in [BG15]: Theorem 3.10. A boundary point ζ0 is regular if and only if there exists a barrier at ζ0. The existence of a barrier is a local property in the following sense: Let Ω˜ be another domain such that B Ω=˜ B Ω ∩ ∩ for an open ball B centered at ζ0. Suppose there is a barrier w at ζ0, and let m = inf w(ζ): ζ ∂B Ω˜ . { ∈ ∪ } It now follows that the function min(w, m) in B Ω˜, v = ∪ m in Ω B ( \ is a barrier in Ω. Indeed, since w is assumed to be a barrier in Ω,˜ w ˜ > 0 by the definition, and therefore m > 0 and v > 0. Since |Ω Ω˜ B = Ω B we see that lim inf v(η) = lim inf w(η) > 0 on ∩ ∩ η→ζ η→ζ ∂Ω˜ B and lim inf v(η)= m> 0 elsewhere. At last, lim v(ζ)= ∩ η→ζ ζ→ζ0 limζ→ζ0 w(ζ) = 0. From this we get the following useful corollary. Corollary 3.11. Let Ω˜ Ω, and let ζ be a common boundary point. ⊂ 0 If ζ0 is not a regular point for Ω˜, then it is not a regular point for Ω. Proof. Let Ω˜ Ω, and let ζ be an irregular boundary point for Ω.˜ ⊂ 0 Assume that ζ0 is regular for Ω. Then Theorem 3.10 gives that there exists a barrier, w, in Ω. The above implies the existence of a barrier in Ω,˜ contradicting the irregularity of ζ0. A classical application of the theory of viscosity solutions is the fol- lowing convergence lemma. ON THE NORMALISED p-PARABOLIC EQUATION 13 Lemma 3.12. Assume that up p is a sequence of viscosity solutions of { } u u =0. t −Ap Assume further that u contains a subsequence u that converges { p}p { pj }j uniformly to a function u∞ in Ω. Then, as j , the upj converge to u, the viscosity solution of the normalised →∞-parabolic equation (1.6), that is ∞ u ∆N u =0. t − ∞ Proof. We show that viscosity subsolutions of (1.1) converge to viscos- ity subsolutions of (1.5). The proof for supersolutions is similar. We say that u USC(Ω) is a viscosity subsolution to (1.6) if, for every function φ ∈C2(Ω) such that u ψ has a maximum at ζ , we have ∈ − 0 φ (ζ ) ∆N ψ(ζ ) 0, for Dφ(ζ ) =0 t ∞ − ∞ ∞ ≤ ∞ 6 φ (ζ ) Λ(D2ψ(ζ )) 0, for Dφ(ζ )=0. ( t ∞ − ∞ ≤ ∞ Assume that u φ has a maximum at ζ for φ C2(Q ). ∞ − ∞ ∈ T 1. Assume first that Dφ(ζpj ) = 0 for j greater than some number N. By definition of viscosity subsolution,6 we then have 1 pj 2 N (3.1) φt(ζpj ) ∆φ(ζpj ) − ∆∞φ(ζpj ) 0. − pj − pj ≤ Since u u uniformly, standard arguments, cf [?] gives that the pj → ∞ maximum points ζpj converge to a maximum point ζ∞ of u∞ φ. Hence, letting j in (3.1) we see that − →∞ φ (ζ ) ∆N φ(ζ ) 0. t ∞ − ∞ ∞ ≤ 2. If Dφ(ζpj )=0 for j > N, we have 1 p 2 φ (ζ ) tr(D2φ(ζ )) − Λ(D2φ(ζ )) 0 t pj − p pj − p pj ≤ and arguing as in the first case, we get as p ; j →∞ φ (ζ ) Λ(D2φ(ζ )) 0. t ∞ − ∞ ≤ This shows that u∞ is indeed a viscosity subsolution of the normalised -parabolic equation. ∞ Remark 3.13. The existence of a uniformly convergent subsequence of up is not known to exist in general. Does finds such an example for the initial-boundary value problem with smooth boundary data in [Doe09]. 14 NIKOLAIUBOSTAD 4. Exterior Sphere Condition We use the barrier characterization to prove Theorem 1.1. We repeat the result here for completeness. Theorem (Exterior sphere). Let ζ0 =(t0, x0) ∂Ω, and suppose that there exists a closed ball (x, t): x x′ 2 +(t ∈t′)2 R2 intersecting { | − | − ≤ 0} Ω precisely at ζ0. Then ζ0 is regular, if the intersection point is not the south pole, that is (x , t ) =(x′, t′ R ). 0 0 6 − 0 Proof. We use the exterior sphere to construct a suitable barrier func- tion at ζ0. Define 2 2 w(x, t) = e−aR0 e−aR , R2 = x x′ 2 +(t t′)2, − | − | − for a constant a> 0 to be determined. Clearly w(x0, t0) = 0, and close to (x0, t0) we have (4.1) δ < x x′ , 2R < t t′. | − | − 0 − We prove that w is a viscosity supersolution. Calculating the deriva- tives, we get 2 Dw(x, t)=2ae−aR (x x′), − 2 (4.2) w (x, t)=2ae−aR (t t′), t − 2 D2w(x, t)=2ae−aR (I 2a(x x′) (x x′)). n − − ⊗ − This shows that Dw = 0 precisely when x = x′. According to Definition 3.1, we need to check the cases x = x′ and x = x′ separately. 1. Assume that x = x′. Then the point of6 contact is not the north pole. It suffices to show6 that w is a classical supersolution. Inserting the derivatives into (1.1) we get 1 p 2 w ∆w + − ∆N w t − p p ∞ 2 1 2 n 1 =2a(t t′) e−aR e−aR (p 1)(2a 4a2r2)+2ar − − −p − − r 2 p 1 p + n 2 =2a e−aR (t t′)+2a − x x′ 2 − . − p | − | − p In light of (4.1), we have 2 p 1 p + n 2 w w > 2a e−aR 2R +2a − δ2 − t −Ap − 0 p − p For the right hand side of this to be positive, we must have p 1 p + n 2 R + a − δ2 > − , − 0 p 2p ON THE NORMALISED p-PARABOLIC EQUATION 15 and choosing a big enough to ensure this, shows that w is super- parabolic. ′ ′ 2. If the point of intersection is the north pole, i.e (x0, t0)=(x , t + ′ R0), we can find points arbitrarily close to the line x = x such that 2 p + n 2 w w =2a e−aR R − + ǫ, t −Ap 0 − p for any ǫ> 0. We see that we must demand that the radius R0 satisfies p + n 2 R α/2, α =2 − 0 ≥ p for w to be a barrier in this case. Assume now that x = x′. We need to verify that for every φ C2(Ω) touching w from below at (x′, t) we have ∈ 1 p 2 (4.3) φ (x′, t) tr(D2φ(x′, t)) + − λ(D2φ(x′, t)). t ≥ p p Assume to the contrary that there is a φ such that w φ has a minimum at (x′, t), but that − 1 p 2 φ (x′, t) < tr(D2φ(x′, t)) + − λ(D2φ(x′, t)). t p p Since w φ has a minimum, we must have − φ (x′, t)= u (x′, t), Dφ(x′, t)= Du(x′, t),D2u(x′, t) D2φ(x′, t). t t ≥ This implies, for any z Rn ∈ D2w z,z D2φ z,z h i≥h i and, since D2w is a scalar multiple of the identity matrix, tr(D2w) z 2 tr(D2φ) z 2 | | ≥ | | at (x′, t). Hence 1 p 2 tr(D2w) z 2 + − D2w z,z p | | p h i 1 p 2 tr(D2φ) z 2 + − D2φ z,z ≥ p | | p h i 1 p 2 tr(D2φ) z 2 + − λ(D2φ) z 2 ≥ p | | p | | >φ z 2 = w z 2. t| | t| | Inserting x = x′ in (4.2) and dividing by z 2 this is | | 1 2 p 2 2 2 2an e−aR + − 2a e−aR > 2a e−aR (t t′), p p − 16 NIKOLAIUBOSTAD or n + p 2 − > (t t′). p − This is a contradiction because of our restriction on the radius, and hence (4.3) must hold, and w is a supersolution even in this case. The condition R0 α/2 restricts the set of exterior spheres usable in a positive way. The≥ author does not know if this restriction can be circumvented. The exclusion of the south pole (x , t )=(x′, t′ R ) in the above 0 0 − 0 is strictly necessary, since then for (x, t) close to (x0, t0) we could have (t t′) < 0 and x x′ = x′ x′ =0, − | − | | − | and so 2 p + n 2 w w =2a e−aR (t t′) − < 0 t −Ap − − p for any positive a, since p + n 2. ≥ Another way to see that it is necessary to exclude the south pole is to consider the Dirichlet problem on the cylinder Q = Q (0, T ). T × Example 4.1. Suppose that f : ∂QT R is continuous. Theorem 3.4 and Theorem 2.6 in [BG15] gives the→ existence of a unique viscosity solution h in QT . Now construct the upper and lower Perron solutions Hf and Hf . Since both are p-parabolic in QT , uniqueness gives that Hf = Hf = h, regardless of what values we choose at that part of the boundary where t = T . Indeed, h itself need not be in either the upper or lower class, because we may not have that either h > f or h < f on the plane t = T . However, if we define ǫ h˜ = h(x, t)+ , T t − we see that ǫ h˜ h˜ =0+ , t −Ap (T t)2 − so h˜ is in f for ǫ> 0, and in f for ǫ< 0. Therefore, it is possible for every pointU on the top of the cylinderL to be irregular. we can say that f is resolutive in this case. We provide another example of an irregular boundary point. ON THE NORMALISED p-PARABOLIC EQUATION 17 Example 4.2 (Latest moment on heat balls). Recall the self-similar solution derived in Section 2.5. We define the fundamental solution to (1.1) as α |x|2 − β − βt Hp(x, t)= t e . Analogous to the heat equation and the p-parabolic equation, we define the normalised p-parabolic balls by the level sets (4.4) H (x x, t t) >c p 0 − 0 − We want to prove that the latest moment, or ”centre” (x0, t0) of (4.4) is not a regular point. Fix c> 0. We can assume that (x0, t0)=(0, 0), so that (4.4) reads x 2 − α − | | ( t) β e β(−t) >c − for t< 0. But this is equivalent to log c x 2 < t( + α log t ), | | β | | and this inequality defines a domain containing the one in the Petrovsky criterion 1.2. Hence the origin must be irregular. We prove that it suffices to consider arbitrary domains in the equiv- alent definition of p-parabolic functions. The proof follows the same idea as in [KL96]. Lemma 4.3. A function u LSC(Ω) L∞(Ω) is p-superparabolic if and only if for each domain Σ∈ with compact∩ closure in Ω, and for each solution h C(Σ) to (1.1), the condition h u on ∂Σ implies h u in Σ. ∈ ≤ ≤ Proof. Assume first that Definition 3.3 holds. If Σ is a box or finite union of boxes, the result is clearly true. The case where Σ is arbitrary follows by covering the set h u + ǫ with finitely many boxes. { ≥ } For the other direction, let Qt1,t2 be a box with closure in Ω and let h C(Qt1,t2 ) be p-parabolic, and so that h u on ∂pQt1,t2 . Assume that∈ ≤ Q =(a , b ) (a , b ). 1 1 ×···× n n Let δ > 0 be so that δ < t2 t1, and choose a hyperplane Pδ such that the points (x, t δ) with −x = a and (y, t ) with y = b belong to 2 − 1 1 2 1 1 Pδ. let Σ be the subset of Qt1,t2 that contains all the points below the hyperplane, that is all (x, t) with t 5. The Petrovsky Criterion We provide the proof of the Petrovsky Criterion, repeated here for completeness. Theorem. The origin (x, t)=(0, 0) is a regular point for (1.1) in the domain Ω enclosed by the hypersurfaces (5.1) (x, t) Rn ( , 0) : x 2 = βt log log t and t = c , { ∈ × −∞ | | − | | ||} { − } for a small constant 0 2 2 − |x| x (5.2) w (x, t)= e βt f ′(t)+ | | f(t) + g′(t), t βt2 2 2f(t) − |x| (5.3) Dw(x, t)= x e βt , − βt ON THE NORMALISED p-PARABOLIC EQUATION 19 and 2 2 2f(t) − |x| 4f(t) − |x| D2w(x, t)= e βt I + e βt x x − βt n t2β2 ⊗ (5.4) 2 2f(t) − |x| 2 = e βt I + x x . βt − n βt ⊗ From (5.4) we see that 2 2 2f(t) − |x| 2 x (5.5) tr(D2w(x, t)) = e βt n + | | . βt − βt From (5.3) and (5.4), (or observing that w(x, t) = G(r, t), and so N ∆∞w = Grr), we get 2 2 Dw Dw 2f(t) − |x| 2 x (5.6) D2w , = e βt 1+ | | . Dw Dw βt − βt | | | | From (5.6) and (5.5), we calculate 1 p 2 Dw Dw w = tr(D2w)+ − D2w , , Ap p p Dw Dw | | | | 2 2 2 2 1 2f(t) − |x| 2 x p 2 2f(t) − |x| 2 x = e βt n + | | + − e βt 1+ | | p · βt − βt p · βt − βt 2 2 2f(t) − |x| n p 2 2 x 1 p 2 = e βt − + | | + − βt − p − p βt p p 2 2 f(t) − |x| n + p 2 x p 1 = e βt 2 − + | | 4 − βt − p βt p 2 2 f(t) − |x| x = e βt α + | | , βt − t where n + p 2 α =2 − . p This, together with (5.2), gives (5.7) w w t −Ap 2 2 2 2 − |x| x f(t) − |x| x =e βt f ′(t)+ | | f(t) + g′(t) e βt α + | | βt2 − βt − t 2 2 2 − |x| x f(t) αf(t) x f(t) =e βt f ′(t)+ | | + | | + g′(t) βt2 βt − βt2 2 2 − |x| αf(t) |x| =e βt f ′(t)+ + g′(t)e βt . βt 20 NIKOLAIUBOSTAD Choose 1 1 f(t)= c , g(t)= , − log t δ+1 log t δ | | || | | || for constants 0 1 |x|2 c(δ + 1) α w w> e βt − c + δ t −Ap t log t δ+1 log t − β | | || | | || For this to be positive, the expression inside the parentheses must be negative. Choosing α (5.9) δ = c β ensures this, and with this choice w is superparabolic in this case. Assume that x = 0 so that Dw = 0. From (5.4) and (5.5) we deduce 2n tr(D2w(0, t)) = c , βt log t δ+1 | | || and 2 λ(D2w(0, t)) = c . βt log t δ+1 | | || ON THE NORMALISED p-PARABOLIC EQUATION 21 Since 1 c(δ + 1) w (0, t)= f ′(t)+ g′(t)= − + δ , t t log t δ+1 log t | | || | | || Definition 3.1 demands that we verify 1 c(δ + 1) − + δ t log t δ+1 log t | | || | | || 1 2n p 2 2 c + − c ≥p · βt log t δ+1 p · βt log t δ+1 | | || | | || for t< 0. This is equivalent to c(δ + 1) 2cn 2c(p 2) p + n 2 α − + δ + − =2c − = c . log t ≤ pβ pβ pβ β | | || Because of our choice of δ in (5.9), the above inequality is satisfied for all t, and (5.8) satisfies the second condition in Definition (3.1). It remains to show that w is a viscosity supersolution. Let φ C2(Ω) touch w from below at (0, t). Since w φ has a minimum at (0∈, t), we have − 2 2 wt = φt, Dw = Dφ, D w > D φ 2 2f(t) I at this point. Since D w(0, t)= βt n, a scalar multiple of the identity matrix, this implies λ(D2w((0, t)) > Λ(D2φ(0, t)) >λ(D2φ(0, t)), where λ is the smallest eigenvalue and Λ is the greatest. Since n 2 2 tr(D φ(0, t)) = λi(D φ(0, t)), i=1 X we get φt(0, t)= wt(0, t) 1 p 2 tr(D2w(0, t)) − λ(D2w(0, t)) ≥ p − p 1 p 2 tr(D2φ(0, t)) − λ(D2φ(0, t)), ≥ p − p which implies that w is indeed a viscosity supersolution. 2 2. Since (5.1) implies |x| < log log t , we see − βt | | || 1 1 1 c w(x, t) > c elog | log |t|| + = − > 0, − log t δ+1 log t δ log t δ | | || | | || | | || for 0 3. w is continuous in Ω, so we only need to check that the restriction of w to ∂Ω is positive. We see 1 −βt log | log |t|| 1 1 c w(x, t) = c e βt + = − > 0. |∂Ω − log t δ+1 log t δ log t δ | | || | | || | | || 4. We see that lim f(t) = lim g(t)=0. t→0− t→0− Since x 2 < βt log log t 0, we see | | − | | || →2 − |x| e βt = ( log t ) O | | || as t 0−. Therefore → lim w(x, t)=0, (x,t)→(0,0−) and (4) in the Definition is satisfied. Together, these points show that (5.8) is indeed a barrier at (0, 0), and hence the origin is a regular point for the domain (5.1). Remark 5.2. Since β 4 as p , we see that (5.1) converges to the Petrovsky criterion→ for the →-parabolic ∞ equation (1.3). Note also that the result is completely independent∞ of the number of spatial variables n. We now turn to the proof that Theorem 1.2 is sharp; any constant greater than β in (5.1) will produce domain containting Ω where the origin is irregular. Theorem. The origin is not a regular point for the domain Ω enclosed by the hypersurfaces (x, t) Rn ( , 0) : x 2 = β(1 + ǫ)t log log t (5.10) { ∈ × −∞ | | − | | ||} and t = c , { − } for any ǫ> 0. Proof. The proof proceeds by constructing a domain Ω˜ contained in Ω, with the origin as common boundary point. We then show that (0, 0) is irregular for Ω,˜ and Lemma 3.11 then implies that (0, 0) regarded as a boundary point of Ω is irregular, too. We shall construct a smooth function w so that (1) w is subparabolic in Ω,˜ (2) w is continuous on Ω˜ (0, 0) , (3) The upper limit of w\{at interior} points converging to (0, 0) is greater than its upper limit for the points converging to (0, 0) along the boundary. ON THE NORMALISED p-PARABOLIC EQUATION 23 To see why the existence of such a w implies that the origin is irregular, consider the boundary data f : ∂Ω˜ R defined as follows. Let f = w near (0, 0), and set → f(0, 0) = lim v(x, t). ∂Ω˜∋(x,t)→(0,0) As we shall see, this limit exists. For the rest of the boundary, contin- uously extend f to a large constant b. If b is large enough, the comparison principle implies that every function u f which satisfies u f on ∂Ω˜ also satisfies u w in Ω˜ since w is a∈ subsolution U by (1). Taking≥ the infimum over all such≥ u, we see H w in Ω,˜ and hence by point (3) in the definition of w; f ≥ lim sup Hf (x, t) lim sup w(x, t) Ω˜ ∋(x,t)→(0,0) ≥ Ω˜∋(x,t)→(0,0) > lim sup w(y,s)= f(0, 0), ∂Ω˜ ∋(y,s)→(0,0) and so (0, 0) is not a regular point for Ω.˜ Our function w will be on the form 2 −|x| k (5.11) w(x, t)= f(t)e βt + g(t), 1 for suitable functions f and g. Here k ( 2 , 1) will be chosen later, and 1 2 2 2 −|x| k x (k k ) αk (5.12) w w = e βt f ′(t)+ f(t)| | − + f(t) +g′(t), t −Ap βt2 βt with α as in (2.3). Choose 1 1 f(t)= − and g(t)= , log t 1+ǫ1 log log t | | || | | || where ǫ1 is a positive constant. The case x = 0. We see that Dw = 0 precisely when x = 0. We show that (5.11)6 is a classical subsolution when x = 0. 6 24 NIKOLAIUBOSTAD Inserting derivatives into (5.12), we get 2 2 2 −|x| (1 + ǫ1) x (k k ) βt k wt pw = e − | | − −A t log t 2+ǫ1 − βt2 log t 1+ǫ1 − (5.13) | 2 | || | | || αk |x| k 1 + e βt . βt log t 1+ǫ1 t log2 log t log t | | || · | | ||·| | || Multiplying by t log t 1+ǫ1 < 0, we see that the sign of (5.13) coincides with the sign of·| | || 2 2 ǫ1 1+ ǫ1 x αk |x| k log t + | | (k k2)+ e βt | | || . − log t βt − β − log2 log t | | | | || We can choose t small enough that | | 1+ ǫ αk 1 < , log t β | | and then 2 2 ǫ1 x 2αk |x| k log t (5.14) | | (k k2)+ e βt | | || < 0. βt − β − log2 log t | | || This inequality is satisfied if x is so small that | | 2 ǫ1 2αk |x| k log t (5.15) < e βt | | || . β log2 log t | | || or if x so large that | | x 2(k k2) 2αk (5.16) | | − > . β t β | | We argue that at least one of these inequalities must hold. Indeed, fix t so that | | ǫ α (5.17) 1 log log t > 4 . 2 | | || β 1. In the case (5.15), we take logarithms to get 2αk x 2 log < | | k + ǫ log log t 2 log log log t , β βt 1 | | || − | | || ON THE NORMALISED p-PARABOLIC EQUATION 25 or x 2 2αk | | k < ǫ log log t 2 log log log t log β t 1 | | || − | | || − β | | < ǫ1 log log t log k | | || − ǫ < ǫ log log t 1 log log t 1 | | || − 2 | | || ǫ = 1 log log t > 2 2 | | || for t small enough. Hence (5.14) is satisfied. 2.| |On the other hand, if (5.16) holds, we calculate x 2 2α α | | > > 4 . β t β(1 k) β | | − Since we chose t according to (5.17), we have that at least one of the inequalities (5.15)| | or (5.16) is satisfied for any x = 0, and w is a subsolution. 6 The case x = 0. Then Dw = 0, and according to Definition 3.1 we need to show that for every φ C2(Ω) so that w φ has a maximum at (0, t), we have ∈ − 1 p 2 (5.18) φ (0, t) tr(D2φ(0, t))+ − Λ(D2φ(0, t)). t ≤ p p We show that w itself satisfies this condition. An argument similar to the one in the proof of Theorem 1.2 then shows that w is a viscosity subsolution. Inserting the derivatives at (0, t), we see that (5.18) reads αf(t) f ′(t)+ g′(t) k, ≤ − βt or 1+ ǫ 1 αk 1 + . −t log t 2+ǫ1 log2 log t log t t ≤ βt log t 1+ǫ1 ·| | || | | ||·| | || · ·| | || This is the same as 1+ ǫ log t ǫ1 αk 1 | | || + 0, log t − log2 log t β ≤ | | || | | || but this inequality is the same as the one in (5.14), and because of our choices of t and k. This shows that the condition (5.18) holds, and w is a subsolution| | even in this case. 26 NIKOLAIUBOSTAD Now we consider the level set w(x, t)= m, m< 0, and calculate 2 1 − |x| k 1 w(x, t)= − e βt + = m log t ǫ1+1 log log t | | || 2 | | || 1 − |x| k 1 − e βt = m ⇐⇒ log t ǫ1+1 − log log t | 2| || | | || − |x| k 1 e βt = log t ǫ1+1 m ⇐⇒ | | || log log t − | | || x 2 1 | | k =(ǫ + 1) log log t + log m , ⇐⇒ − βt 1 | | || log log t − | | || or simply ǫ +1 1 1 (5.19) x2 = βt 1 log log t + log m . − k | | || k log log t − | | || Letting Ω˜ denote the domain enclosed by (5.19) and the hyperplane t = c< 0, we have that for m< 0, the function v (5.11) is negative in Ω,˜ and w(x, 0) = 0. This shows that the origin is an irregular boundary point for Ω.˜ The inclusion Ω˜ Ω requires that ⊂ ǫ +1 1 1 1 log log t + log m < (1 + ǫ) log log t k | | || k log log t − | | || | | || for small t . Fix k close to 1 and α close to 0 so that | | ǫ +1 ǫ 1 < 1+ . k 2 Thus we have to verify that 1 1 k ǫ + m log t 2 , log log t | | ≤| | || | | || but this obviously holds for small t since the left-hand side is bounded. | | Hence Ω˜ Ω for ǫ1 and c close to 0, and k close to 1, (0, 0) is an irregular boundary⊂ point for Ω as well. Acknowledgement The author would like to thank Jana Bj¨orn and Vesa Julin for discov- ering a flaw in the original proof of the Petrovsky criterion. ON THE NORMALISED p-PARABOLIC EQUATION 27 References [Bar52] Grigory I. Barenblatt. On self-similar motions of compressible fluid in a porous medium. Prikladnaya Matematika i Mekhanika (Applied Mathe- matics and Mechanics (PMM)), 1952. [BBP17] Anders Bj¨orn, Jana Bj¨orn, and Mikko Parviainen. The tusk condition and Petrovski criterion for the normalized p-parabolic equation. arXiv preprint arXiv:1712.06807, 2017. [BG13] Agnid Banerjee and Nicola Garofalo. 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Then Hθ is continuous up to ∂Σ, and we have u H ≥ θ in all of Σ since the inequality holds on ∂Σ. Hence ǫ u(x, t) h(x, t) ≥ − t + δ t 2 2 − in Σ, and letting ǫ, δ 0, we get → u h ≥ in the box Qt1,t2 .
On the Normalized $ P $-Parabolic Equation in Arbitrary Domains
