Intrinsic Harnack's inequality for a general nonlinear parabolic equation in non-divergence form

We prove the intrinsic Harnack's inequality for a general form of a parabolic equation that generalizes both the standard parabolic $p$-Laplace equation and the normalized version arising from stochastic game theory. We prove each result for the optimal range of exponents and ensure that we get stable constants.


Introduction
We prove the intrinsic Harnack's inequality for the following general non-divergence form version of the nonlinear parabolic equation for the optimal range of exponents.The theorem states that a non-negative viscosity solution satisfies the following local a priori estimate u(•, t 0 − θr q ) ≤ u(x 0 , t 0 ) ≤ γ inf Br(x 0 ) u(•, t 0 + θr q ) (1.2) for a scaling constant θ which depends on the value of u.This intrinsic waiting time is the origin of the name and is required apart from the singular range of exponents where the elliptic Harnack's inequality holds [KPS23].We also establish stable constants at the vicinity of q = 2.
When q = p, the equation (1.1) is the standard p-parabolic equation for which the intrinsic Harnack's inequality was proven by DiBenedetto [DiB88] and Kwong [DK92], see also [DiB93].These results were generalized for equations with growth of order p by DiBenedetto, Gianazza, and Vespri [DGV08] and by Kuusi [Kuu08].When q ̸ = p, the equation (1.1) is in non-divergence form.For non-divergence form equations parabolic Harnack's inequalities and related Hölder regularity results were first studied by Cordes [Cor56] and Landis [Lan71].Parabolic Harnack's inequality for a non-divergence form equation with bounded and measurable coefficients was proven by Krylov and Safonov [KS81].Further regularity results for general fully nonlinear equations were proven by Wang [Wan90], see also [IS13].To the best of our knowledge, our proof is partly new even in the special case of the p-parabolic equation since it does not rely on the divergence structure.
The idea of the proof of the right inequality in (1.2) is to first locate a local supremum and establish a positive lower bound in some small ball around this point.Then we use specific subsolutions as comparison functions to expand the set of positivity over the unit ball for a specific time slice using the comparison principle.Our proof uses the connection of equation (1.1) and the p-parabolic equation established by Parviainen and Vázquez in [PV20] to construct suitable comparison functions.Heuristically, radial solutions to the non-divergence form problem can be interpreted as solutions to divergence form p-parabolic equation in a fictitious dimension d, which does not need to be an integer.The proof of the left inequality is based on estimating the values of a function in the specific time slice by using the other inequality with suitable radii and scaling of constants.
Our proofs often are split into three different cases because the behavior of solutions to (1.1) depends on the value of q.For the degenerate case q > 2, the right-side inequality is proven in [PV20] and we prove the singular case q < 2 as well as the case of q near 2.This is done separately to obtain stable constants as q → 2. For the left-side inequality, the singular case was proven in [KPS23] and we prove the remaining cases.
DiBenedetto's proof uses the theory of weak solutions but since the equation (1.1) is in nondivergence form, unless q = p, we use the theory of viscosity solutions instead.Because of this, we cannot directly use energy estimates as in [DGV08] or in [DiB93].Even defining solutions is non-trivial for this type of equations.A suitable definition taking singularities of the problem into account was established by Ohnuma and Sato [OS97].When q = 2, we get the normalized p-parabolic equation arising from game theory which was first examined in the parabolic setting in [MPR10].This problem has had recent interest for example in [JS17], [HL19], [DPZZ20] and [AS22].We also point out that normalized equations have been studied in connection to image processing [Doe11], economics [NP14] and machine learning [Cal19].The general form of (1.1) has been examined for example in [IJS19] and [PV20] in addition to [KPS23].
1.1.Results.We work with the exponent range which is optimal for the intrinsic Harnack's inequality as we prove in Section 7.For the elliptic version of the inequality where we get both estimates without waiting time, the optimal range is to assume (1.3) and q < 2, as we proved in [KPS23].The notation used for space-time cylinders is defined in the next section.
We prove this theorem in Sections 4 and 5 after first introducing prerequisites and proving auxiliary results in Sections 2 and 3.The theorem is proven by first establishing the right inequality, from now on called the forward Harnack's inequality, and then using this result to prove the left inequality, henceforth called backward Harnack's inequality.These names are standard in the literature.We prove the forward inequality by first locating the local supremum of our function and establishing a positive lower bound in some small ball around the supremum point.This differs from the integral Harnack's inequality used by DiBenedetto for weak solutions at this step [DiB93,Chapter VII].The proof of this integral inequality uses the divergence form structure of the p-parabolic equation and thus is not available to us without a new proof.Next, we expand the positivity set around the obtained supremum point by using suitably constructed viscosity subsolutions and the comparison principle.In the singular case, we first expand the set in the time direction using one comparison function and then expand it sidewise for a specific time slice using another one.In the degenerate case, a single Barenblatt-type function is enough to get a similar result.Yet we need a different comparison function to handle exponents near q = 2 if we wish to have stable constants as q → 2. We construct these viscosity subsolutions in Section 3.
For the backward Harnack's inequality, the singular case is proven as [KPS23, Theorem 5.2], and we prove the remaining cases in Section 5.The case q = 2 is a direct consequence of the forward inequality as we do not have to deal with intrinsic scaling.The proof of the degenerate case follows the proof of the similar result for the p-parabolic equation [DGV12, Section 5.2] and uses the forward inequality and proceeds by contradiction that the backward inequality has to hold.In Section 6 we prove covering arguments that take the intrinsic scaling into account.We do this by repeatedly iterating Harnack's inequality and choosing points and radii taking the intrinsic scaling into account.In the last Section 7, we prove that if q does not satisfy the range condition (1.3), it must vanish in finite time and thus cannot satisfy the intrinsic Harnack's inequality.Thus the range condition is optimal.

Prerequisites
When ∇u ̸ = 0, we denote where p > 1 and q > 1 are real parameters and the normalized or game theoretic infinity Laplace operator is given by Thus the equation (1.1) can be written as Let Ω ⊂ R n be a domain and denote Ω T = Ω × (0, T ) the space-time cylinder and its parabolic boundary.We will mainly work with the following type of cylinders where θ is a positive parameter that determines the time-wise length of the cylinder relative to r q .We denote the union of these cylinders as and when not located at the origin, we denote Apart from the case p = q, the equation (1.1) is in non-divergence form and thus the standard theory of weak solutions is not available, and we will use the concept of viscosity solutions instead.Moreover, the equation is singular for 2 > q > 1, and thus we need to restrict the class of test function in the definition to retain good a priori control on the behavior of solutions near the singularities.We use the definition first introduced in [IS95] for a different class of equations and in [OS97] for our setting.This is the standard definition in this context and it naturally lines up with the p-parabolic equation (p = q), where notions of weak and viscosity solutions are equivalent for all p ∈ (1, ∞)[JLM01, PV20, Sil21].See also [JJ11].

Denote
and also require that for g(x) := f (|x|), it holds that lim This set F(F ) is never empty because it is easy to see that f (r) = r β ∈ F(F ) for any β > max(q/(q − 1), 2).Note also that if f ∈ F(F ), then λf ∈ F(F ) for all λ > 0.
Define also the set and σ(r) > 0 for all r > 0}.
We use F(F ) and Σ to define an admissible set of test functions for viscosity solutions.

Note that by definition a function φ is automatically admissible in Ω
p u in Ω T if the following three conditions hold.
(1) u is lower semicontinuous, (2) u is finite in a dense subset of Ω T , (3) whenever an admissible φ ∈ C 2 (Ω T ) touches u at (x, t) ∈ Ω T from below, we have The existence and uniqueness for viscosity solutions of (1.1) is proven in [OS97, Theorem 4.8].
In our proof of the forward Harnack's inequality for the singular range we need a version of the corollary proven in [PV20, Corollary 7.2] for the case q < 2. The lemma remains largely the same except we change the signs of the exponents.We present the proof here for the convenience of the reader.
Lemma 2.3.Let u be a viscosity solution to (1.1) in Q 4r (1) and let the range condition (1.3) hold and assume q < 2. For any δ ∈ (0, 1), there exists C := C(n, p, q, δ) > 1 such that the following holds.Suppose that ω 0 > 1 is such that for a 0 := ω q−2 0 < 1 we have osc Qr(a 0 ) u ≤ ω 0 , and define the sequences holds as long as in the time-direction we have which holds if we choose C to satisfy C q δ 2−q ≥ 1.To prove the second claim we will use induction.
The case i = 0 holds by assumption.Suppose that the claim holds for some i = k meaning By induction assumption sup Q 1 (1) u k ≤ 1.By change of variables, we can rewrite Next, we will use the Hölder estimates proved in [IJS19] to estimate the oscillation.By [IJS19, Lemma 3.1], there exists a constant and by using [IJS19, Lemma 2.3] for y = x 0 and t = t 0 , there exists a constant By our induction assumption and the definition of ≤ 1 and thus C 1 and C 2 can be chosen independent of the solution.Now (2.1) can be estimated with (2.2) and (2.3) in the following way: Denote A := Q C −1 (δ q−2 C −q ) and let (x, t) ∈ A be the point where sup A u k is obtained and (ȳ, s) ∈ A be the point where inf A u k is obtained.Now for where the last inequality holds if we choose Thus by combining (2.1) and (2.4), we get osc Qr k+1 (a k+1 ) u ≤ δω k = ω k+1 as desired.□ A standard argument (see [PV20, Corollary 7.2]) together with the assumption C ≥ δ −1 now yields the following oscillation estimate.
Corollary 2.4.Let u be a viscosity solution to (1.1) in Q 4r (1) and let the range condition (1.3) hold.For any given ω 0 > 1 such that a 0 := ω q−2 0 satisfies osc Qr(a 0 ) u ≤ ω 0 , there exist constants Ĉ = Ĉ(n, p, q) > 1 and ν = ν(n, p, q) ∈ (0, 1) such that for any 0 < ρ < r it hold The proof of this well-known result is a direct calculation.This lemma directly generalizes for functions with time dependence and also for functions u where u(x − x) is radial for some vector x ∈ R n .When dealing with functions v : R × R + → R, we denote the spacial derivative by v ′ (r, t) and the time derivative by ∂ t v(r, t).
Our proofs use the following comparison principle, which is Theorem 3.1 in [OS97].
Theorem 2.5.Let Ω ⊂ R n be a bounded domain.Suppose that u is a viscosity supersolution and v is a viscosity subsolution to

Comparison functions
Comparison functions are used in the standard proof for the intrinsic Harnack's inequality for the divergence form equation to expand the positivity set around the supremum point using the comparison principle.In the degenerate case, a single Barenblatt-type solution is enough to get the estimate but in the singular case, we need two separate subsolutions.The Barenblatt solutions do not have compact support in the singular range and thus we need to find another type of comparison function.Because of the connection of the equation (1.1) and the usual p-parabolic equation examined in [PV20], we can use similar comparison functions as DiBenedetto in his proof for the singular range.We will need three different subsolutions to handle the singular case and values of q near q = 2.We denote throughout this section which is the time-scaling constant connecting (1.1) to the usual q-parabolic equation in the radial case.
Assume q < 2. We will use the following subsolution which is a time-rescaled version of the solution used in the p-parabolic case by DiBenedetto where R(t) := ηκ q−2 t + ρ q and κ and ρ are positive parameters and ξ > 1 is chosen independent of κ and ρ.By (•) + we denote the positive part of the function inside the bracket.
By construction supp Φ(•, 0) = B ρ (0) and for t ≥ 0, we get the expanding balls and the estimate We examine Φ in the domains We have Φ ∈ C ∞ (P κ,ξ ) ∩ C(P κ,ξ ) and as we see in the following lemma, we can choose the constant ξ to make Φ a viscosity subsolution to (1.1) in this set.
Lemma 3.1.Let the range condition (1.3) hold and q < 2. There exists a constant ξ := ξ(n, p, q) so that Φ is a viscosity subsolution to (1.1) in R n × 0, κ 2−q ρ q ηξ .Proof.The function Φ ≡ 0 outside P κ,ξ , so it is enough for us to check that Φ is a viscosity subsolution on the boundary and inside this set.Let us first look at the points where ∇Φ ̸ = 0, because here we can use the radiality of Φ in spacial coordinates and a simple calculation to simplify our statement to the form where We use the following notation during the calculation By direct calculation inside P ′ κ,ξ , we have By the calculations above, we have Introduce the two sets where δ > 0 is a constant to be chosen.Now inside E 1 we can estimate what is inside the last brackets from above if we choose δ small enough that the last inequality holds.Notice also that both F ∈ [0, 1] and ρ qξ R(t) ξ ∈ [0, 1] and thus Here we can choose δ to be small enough to guarantee that the right side of the equation is negative and this can be done without dependence on ξ.
Let us next focus on E 2 .By the range of t, we have and thus for δ we chose above, we have We can now choose ξ to be large enough to guarantee that the right side is negative.
Thus combining the estimates (3.5) and (3.6), we have that L(ϕ) ≤ 0 in the entire P k,ξ and thus by just multiplying the positive scaling factor in the definition of L away, we have that ϕ is a classical subsolution.
We still need to check the points where ∇Φ = 0, because there the simplification we did earlier in (3.3) does not hold.Also because of the singular nature of the equation (1.1), the concept of a classical solution does not really make sense at these points and we need to use the definition of viscosity solutions.By similar calculation to the radial case using the same notation, we have and thus the gradient vanishes at the origin as 1 q−1 > 0 and in the set ∂B as there F = 0.This latter set happens to be the lateral boundary of the support of Φ.By our previous calculation (3.4), the time derivative of Φ is which clearly satisfies ∂ t Φ ≤ 0 at the critical points as the first term is negative and the second is zero if either z = 0 or F = 0. Let φ ∈ C 2 be an admissible test function touching Φ at a critical point (x, t) from above.For any such function ∂ t φ(x, t) = ∂ t Φ(x, t) ≤ 0 and thus Φ is a viscosity subsolution in P κ,ξ .The zero function is also a viscosity subsolution so Φ is a viscosity subsolution in the entire R n × 0, κ 2−q ρ q ηξ as we already verified the boundary.□ The comparison function Φ defined in (3.1) does not give us stable constants as q → 2 because the radius we use it for blows up.We can extend the proof of the degenerate case slightly below q = 2 with a different comparison function and use this to get stable constants in our inequality for the whole range (1.3).Let ρ and κ be positive parameters and define the function Here ν > 1 is a constant and (3.8) The function (3.7) is a time-rescaled version of the comparison function introduced by DiBenedetto in [DiB93,VII 3(i)].We also introduce a number This number q(ν) will define the size of the interval around q = 2, where G is a viscosity subsolution.
Lemma 3.2.Let q ∈ (4 − q(ν), 7/3).There exists a ν := ν(n, p) > 1 independent of q such that G is a viscosity subsolution to Proof.We prove this statement by first showing that G is a classical subsolution in the support of this function apart from the points where ∇G = 0 and dealing with the boundary and rest of the space afterward.The function G is radial with respect to space and thus we can perform our calculations in radial coordinates.Define Again whenever g ′ ̸ = 0, we can use the radiality and a quick calculation to simplify our statement to the form Inside S ′ , we have and thus For the spatial derivatives, we have so finally Therefore by the calculation above, we have where the exponent of Σ(t) is zero because of (3.8).We introduce two sets and note that as λ is decreasing with respect to q, we have Inside E 1 , the first term can be small depending on the data but the lower bound we chose for z q q−1 ensures that the rest of the terms are negative on their own without dependence on ν.Using (3.10) and estimate (3.11), it follows that in E 1 it holds where â = max{a, a q−1 }.The last inequality holds because we assumed that q < 7 3 .Notice that this estimate holds for all ν but only for q ∈ (4 − q(ν), 7 3 ) depending on the ν we pick.In E 2 , we have and we can ensure that L(g) is negative by choosing a suitably large ν.We again estimate using (3.10) and (3.11) that inside E 2 , it holds L(g) ≤ −νηF q q−1 + aλ(ν)ηF so that for this ν, we have L(g) ≤ 0 in E 2 by (3.13).Notice that this choice of ν depends on n and p but not q and that 4 − q(ν) > 8 5 for all ν ≥ 1.Thus for this choise of ν, we get that by (3.12) and (3.13), we have L(g) ≤ 0 in the classical sense in S ′ for all q ∈ (4 − q(ν), 7 3 ).We still have to check the points where ∇G = 0.The gradient for the original function (3.7) is which exists and vanishes at the origin as 1 q−1 > 0 and also vanishes when F = 0, that is when x ∈ ∂R.Using the time derivative we calculated in (3.9), we have that for x ∈ ∂R it holds ∂ t G = 0 and for x = 0, we have Let φ ∈ C 2 be an admissible test function touching Φ at a critical point (x, t) from above.For any such function ∂ t φ(x, t) = ∂ t G(x, t) ≤ 0 and thus Φ is a viscosity subsolution in S. In R n × R + \ S, any admissible test function touching G from above must have zero time-derivative and thus G is a viscosity subsolution in the entire R n × R + .□ We need one more comparison function to handle expanding the sidewise positivity set in our proof of the singular forward Harnack's inequality in Theorem 4.1.This differs from the degenerate case where only one Barenblatt type comparison function is used [PV20, Theorem 7.3].
Let k and ν be positive parameters and consider cylindrical domains with annular cross-section For these parameters and a constant ζ, we define set the size of the hole in the annular cross-section of our cylinder to be of suitable radius.In our proof of the forward inequality these are picked in (4.5) and (4.6).We present the proof in detail for the ease of the reader and to fix some typos in the literature.
Lemma 3.3.Let the range condition (1.3) hold and q < 2. There exist constants ζ := ζ(n, p, q) and Θ := Θ(n, p, q) such that for every 0 < ν < 1 and k > 0, (3.15) is a viscosity subsolution to the equation Proof.The function Ψ is radial and thus we will again do our calculations in radial coordinates.Define so that ψ = vw.Whenever ψ ′ ̸ = 0, we can simplify our statement to the form We have (3.17) Here d = (n−1)(q−1) + 1 and A consists of four latter terms.Next, we will prove that A is negative for suitably large z and we prove the technical part of this as a separate lemma after finishing this proof.Let Z(p, q, N ) be the positive constant given by the Lemma 3.4 proven below and note that to use this lemma, we will need to restrict θ to make sure that z ≥ Z for all (r, t) ∈ C(θ).
The correct choice turns out to be as plugging this into the definition of z, we get Thus by Lemma 3.4 and thus combining this with (3.17), we get Next, we estimate and by direct calculation and thus we get and plug in our estimates (3.19) and (3.20) to get By our definition of w and z Using these we can further estimate (3.21) to get Now finally if we assume we have q F = q 1+z < λ 2 and can choose the ζ that satisfies For this ζ, the estimate (3.22) becomes L(ψ) ≤ 0 and we have that ψ is a classical subsolution.
To ensure that only z satisfying both z ≥ Z and (3.23) are in our annulus C ′ (θ), we need to further restrict θ we picked in (3.18) to make sure that t < k 2−q ζ q−1 r q λ 2q q−1 η −1 in the set.By the definition of C ′ (θ), we have r ≥ ν so it suffices to choose ν q k 2−q =: Θ(n, p, q)ν q k 2−q all estimates hold and ψ is a classical subsolution in C ′ (θ).We still have to check the points where ∇Ψ = 0.By direct calculation, denoting and it is easy to see that ∇Ψ(x) = 0 if and only |x| = 1 or x = 0.The origin is outside our domain so let (y, s) be an arbitrary point such that |y| = 1 and s ∈ (0, θ) and let φ ∈ C 2 be an admissible test function touching Ψ from above at (y, s).At such point and same trivially holds when touching a point in (R n \ B 1 (0))×(0, θ) and thus Ψ is a viscosity subsolution in (R n \ B ν (0)) × (0, θ).This finishes the proof of Lemma 3.3.□ Next, we will prove Lemma 3.4 that we used in the above proof to show that A was negative.
Lemma 3.4.There exists a constant Z = Z(p, q, n) such that for all z ≥ Z and all r ∈ (0, 1) we have We begin by showing that K < 1 using the range condition (1.3).
We first consider the case where p < n+1 2 and q > 2(n−p) n−1 .Since the latter inequality implies we obtain using the upper bound on p.Similarly, we estimate In the case p > n+1 2 , we have directly and so hence we have K < 1 for all exponents satisfying (1.3).Now observe that this implies Denote Z = K 1−K so that by above we have z z + 1 ≥ K for all z ≥ Z. (3.24) Now, we estimate E(r) separately in the cases r ≥ 2 3 and r < 2 3 .If we first assume r ≥ 2 3 , this implies r 2 1−r 2 ≥ 4 5 =: a so using (3.24), we can estimate where the last identity follows from the definition of K 1 .If r ≤ 2 3 , we discard the second term with r and estimate again using (3.24) to get where we used the definition of K 2 .□

Forward intrinsic Harnack's inequality
In their paper [PV20], Parviainen and Vázquez prove the forward Harnack's inequality for viscosity solutions of (1.1) in the degenerate case q > 2. In this section, we prove the remaining singular case q < 2 and the case of q near 2. For the proof of the same results for the standard singular p-parabolic equation see [DiB93, VII.9].In the proof we first rescale the equation into a simpler form, locate the local supremum of the function in some specific cylinder and, use oscillation estimates to show that there exists some small ball on a time slice where the function is strictly larger than the value depending on the singularity of the equation (1.1).Barenblatt-type solutions have an infinite speed of propagation for q < 2 and hence do not work as comparison functions similarly to the degenerate case.In the strictly singular case, we next use a comparison function constructed in Lemma 3.1 to expand the set of positivity in the time direction to get a similar lower bound extended from one time slice to a space-time cylinder.Finally, we use a second comparison function constructed in Lemma 3.3 to widen the set of positivity in the spacial direction to fill the entire ball we are interested in and get the final estimate.At the end of this section, we prove the inequality for values of q near 2.This case is similar to the degenerate case and only requires one comparison function but here we use one constructed in Lemma 3.2 instead of the Barenblatt solution used in the degenerate case.This method gives us stable constants as q → 2 from either side.
Theorem 4.1.Let u ≥ 0 be a viscosity solution to (1.1) in Q − 1 (1) and let the range condition Then there exist µ = µ(n, p, q) and c = c(n, p, q) such that u(x 0 , t 0 ) ≤ µ inf where θ = cu(x 0 , t 0 ) 2−q , whenever (x 0 , t 0 ) + Q 4r (θ) ⊂ Q − 1 (1).Remark 4.2.The constants µ and c can be picked to be stable as q → 2 from either side as we show in the proof.As q approaches the lower bound in (1.3), µ tends to infinity, and c tends to zero.As q → ∞, both µ and c tend to infinity.
Proof of Theorem 4.1.The proof for the degenerate case q > 2 is given as [PV20, Theorem 7.3] and thus we can focus on the singular case q < 2. Consider the rescaled equation v(x, t) = 1 u(x 0 , t 0 ) u(x 0 + rx, t 0 + u(x 0 , t 0 ) 2−q tr q ) which solves where Q = B 4 (0) × (−4 q , 4 q ).Now it is enough to show that there exists positive constants c 0 and µ 0 such that inf because then by the definition of v, we have For the first part of the proof, we make the extra assumption q < 2 and deal with values near q = 2 afterward.Proof for q = 2 is easy but we need to deal with values near it separately to ensure that we get stable constants as q → 2 from either side.We will prove (4.1) in the following steps.
Step 1: Locating the supremum.First, we will need to locate the supremum of v in Q and establish a positive lower bound for v in some small ball around the supremum point.
We do this by using Hölder continuity results.For all τ ∈ [0, 1) and σ ∈ (0, 1) to be chosen later define nested expanding cylinders and the numbers Notice that M 0 = 1 = N 0 and lim as v is bounded.Therefore by continuity, the equation M τ = N τ must have a largest root τ 0 ∈ [0, 1), which satisfies Especially for τ := 1+τ 0 2 we have By continuity of v, it achieves the value M τ 0 at some point (x, t) ∈ Q τ 0 (1) and for the radius Let a 0 = ω q−2 0 and note that a 0 R q = ω q−2 0 R q ≤ R q and thus by (4.2), we have osc Thus we can use Corollary 2.4 to find Ĉ := Ĉ(n, p, q) > 1 and ν := ν(n, p, q) ∈ (0, 1) such that osc Bρ(x 0 ) v(•, t) ≤ Ĉω 0 ρ R ν for any 0 < ρ < R. Pick ρ = δR for some δ := δ(n, p, q) small enough to satisfy 1 − Ĉδ ν 2 Step 2: Time expansion of positivity.We have managed to prove the positivity of v in a small ball for time t and now we intend to improve this estimate to get positivity in a time cylinder.Consider the translated comparison function Φ(x − x, t − t) introduced in (3.1) for choices κ and ρ introduced above.By Lemma 3.1, we have thus that Φ is a viscosity subsolution to (1.1) in (x, t) + P κ,ξ := B ( R(t− t)) 1 q (x) × t, t + κ 2−q ρ q ηξ for time dependent radius R(t) := ηκ q−2 t + ρ q .We choose where σ is the constant we did not yet choose in the definition of our cylinders Q ρ .Now by (4.3) and (3.2), it holds and by positivity v ≥ Φ on the spatial boundary.Thus by the comparison principle Theorem (4.4) We do not have a way to know the exact location of t inside Q τ 0 but we know that t ∈ (−1, 0) and σ ∈ (0, 1).Thus as (σ, 2σ) ⊂ t∈(−1,0) we have estimate (4.4) for all (x, t) ∈ B ρ (x) × (σ, 2σ).As q ↗ 2, we have σ ↘ 0 and hence the set converges towards an empty set.To get the estimate for values of q near 2, we repeat a similar argument but with a different comparison function.
Step 3: Sidewise expansion of positivity.We will next expand the positivity set of v over B 1 (x) for a specific time slice using yet another comparison function to finally get the estimate (4.1).Choose we got from (4.4), Illustration of the time expansion.We use the comparison principle over the light gray cylinder and get the final estimate over the gray cylinder.and let θ be given by (3.16) for this k and ν.We have (4.4) for all (x, t) ∈ B ρ (x) × (σ, 2σ) so we have the same estimate with a smaller constant σ = min{θ, σ}.We want to use a translated and scaled version of the comparison function where we introduced Ψ in (3.15).This rescaled Ψ is a viscosity subsolution to (1.1) in Ĉ by Lemma 3.3.Notice that this Ψ vanishes for x ∈ ∂B 3 (x) or t = σ and that everywhere in Ĉ. Combining this estimate with (4.4) we have for all (x, t) ∈ {|x − x| = ρ}×(σ, 2σ) by continuity of v. Thus we can again use the comparison principle to get Ψ ≤ v in the entire set Ĉ.

Thus by taking infimum over
so we have proved estimate (4.1) for c 0 := 2σ.
Case of q near 2. The case of q near 2 is quite similar to the proof we presented above.Let ε > 0 be a small number to be fixed later and assume that q ∈ (2 − ε, 2 + ε).This time we define a family of nested expanding cylinders by so they no longer depend on the constant σ.We define for β > 0 to be chosen later similar to the proof of the degenerate case.Again let τ 0 ∈ [0, 1) be the largest root of equation M τ = N τ to ensure . By continuity of v, it achieves the value M τ 0 at some point (x, t) ∈ Q τ 0 (1) and for radii R = 1−τ 0 2 , we have sup For q < 2, we can repeat the same steps as we used to obtain (4.3) to find ρ := δR for some δ := δ(n, p, q) small enough so that For q ≥ 2, we repeat the same steps but use [PV20, Corollary 7.2] instead of Corollary 2.4 and we get the same estimate.
This is where we need the special subsolution constructed in Lemma 3.2.Let κ and ρ be the constants we set above and Consider the translated version G(x − x, t − t) which is a viscosity subsolution to (1.1) in R n × R + as long as our exponent q is close enough to 2. Let ν be the constant given by Lemma 3.2 and pick ε = min 4(1+2ν)  1+4ν − 2, 1 3 , 1 ν .The first two numbers ensure that G is a viscosity solution by Lemma 3.2 and the restriction ε ≤ 1 ν is here to ensure that λ(ν) ≥ 0 for all q in our range.

At time level
where Here we used R = 1−τ 0 2 and defined and c 0 = 3 and since |x| < 1 and t ∈ (−1, 0], these choices ensure and thus B 2 (0) ⊂ supp G(x − x, c 0 − t).In the set supp G(x − x, t − t) = B ρ (x), we have and similarly G ≤ v on the rest of ∂ p B 2 (x) × [ t, c 0 ] because we assumed v to be positive.Hence by the comparison principle Theorem 2.5 inf 1 − 1 3 q q−1 q q−1 + =: µ 0 (n, p, q), so we have proven (4.1).Notice that all constants used here are stable as q → 2 from either side.□

Backward intrinsic Harnack's inequality
In this section, we will prove the backward intrinsic Harnack's inequality for the optimal range of exponents (1.3).We proved the singular case as Theorem 5.2 in [KPS23] but the degenerate case has not been proven before to the best of our knowledge for equation (1.1).The degenerate case is proven for the standard p-parabolic equation [DGV12, Section 5.3].
All proofs are based on using the forward inequality in a specific way taking into account the intrinsic scaling.In the degenerate case, we move backward in time centered at x 0 seeking for a time where the function obtains a value larger than µu(x 0 , t 0 ).We handle the case of such time existing and not existing separately and show that in both cases we get the backward inequality using the forward inequality.The main difference to the singular case is that when q ≥ 2, we have to assume that u(x 0 , t 0 ) > 0 or the inequality will not hold.The case q = 2 follows directly from forward Harnack's inequality as we do not have to worry about the intrinsic scaling.In the singular case, the amount of space needed around our space-time cylinder depends on n, p, and q but we improve this result using covering arguments in the next section.
Case 1 (q<2): The case q < 2 is Theorem 5.2. in [KPS23] where we get the theorem for constant σ = 6 α where α = (2µ) q−2 q < 1. Apart from the non-emptyness of U α this proof extends directly to the case q = 2 but this can be done easier as we do not have intrinsic time scaling in this case.
Case 2 (q=2): Let t = t 0 − cr 2 and y ∈ B r (x 0 ).Now by Theorem 4.1 at the point (y, t), we get an estimate (5.1) This holds for any y ∈ B r (x 0 ) and thus by taking supremum over all of them we get sup as desired.The use of Harnack's inequality in (5.1) is justified because in the space direction B 4r (y) ⊂ B 5r (x 0 ) ⊂ B 6 α r (x 0 ) and in the time direction we have for any σ > 5.
Inequality (5.2) combined with Theorem 4.1 proves the inequality in the case q = 2.
Case 3 (q>2): Finally, let q > 2 where we again have to deal with the time-scaling.Let ρ be a radius such that (x 0 , t 0 ) + Q 6ρ (θ) ⊂ Q − 1 (1) for θ = cu(x 0 , t 0 ) 2−q and define the set Now T is either empty or non-empty.If it happens that T ̸ = ∅, there exists a largest τ ∈ T by continuity of u.For a time like this, it must hold that t 0 − τ > cu(x 0 , τ ) q−2 ρ q = c(2µu(x 0 , t 0 )) 2−q ρ q , (5.3) because otherwise we can choose β ∈ (0, 1) such that τ + cu(x 0 , τ ) q−2 ( βρ) q = t 0 and use Theorem 4.1 on the point (x 0 , τ ) for radius βρ to get 2µu(x 0 , t 0 ) = u(x 0 , τ ) ≤ µ inf This is a contradiction assuming that we have suitable space to use the forward Harnack's inequality.This is automatically satisfied in space as we are centered at x 0 and in time we have where the last inequality holds for all µ > 1 as 1 + (2µ) 2−q 4 q < 2 • 4 q < 6 q for q ≥ 2.Here we used the fact that τ > t 0 − θ(4ρ) q and u(x 0 , τ ) = 2µu(x 0 , t 0 ) by the definition of T .Set and notice that by (5.3), it holds s ∈ (τ, t 0 ) and Assume thriving for a contradiction that there exists y ∈ B ρ (x 0 ) such that u(y, s) = 2µu(x 0 , t 0 ), ( and note that s + cu(y, s) 2−q ρ q = t 0 .
Therefore assuming there is enough room to use Theorem 4.1, we get We have enough room in space as B 4ρ (y) ⊂ B 5ρ (x 0 ) and by (5.4) and (5.5) in time, it holds where the last inequality holds for all µ > 1 because for q > 2 we have 2µ + 4 Therefore such y ∈ B ρ (x 0 ) cannot exist and we have u(y, s) < 2µu(x 0 , t 0 ) for all y ∈ B ρ (x 0 ) and thus by definition of s sup 2−q q ρ ≤ ρ and rewrite (5.6) as This combined with Theorem 4.1 for radius r and taking 2µ gives which is what we wanted.If it happens that T = ∅, we have u(x 0 , t) < 2µu(x 0 , t 0 ) for all t ∈ (t 0 − θ(4ρ) q , t 0 ) (5.8) by continuity of u.Assume thriving for a contradiction that sup which implies by continuity that there exists a point x * ∈ B r (x 0 ) such that u(x * , t 0 − θr q ) = 2µ 2 u(x 0 , t 0 ). (5.10) Assuming we have enough room to use Theorem 4.1 around the point (x * , t 0 − θr q ), we have The required space here is (x 0 , t 0 ) + Q 5r ⊂ Q − 1 because we need to make sure that B 4r (x * ) ⊂ B 1 .In time we do not need more room because t 0 − θr q − cu(x * , t 0 − θr q ) 2−q (4r) q = t 0 − c u(x 0 , t 0 ) 2−q + 4 q 2µ 2 u(x 0 , t 0 ) where the last inequality holds assuming q ≥ 2 and 1 + 4 q (2µ 2 ) 2−q < 6 q which is true for any µ ≥ 1 as (2µ 2 ) 2−q ≤ 1.We can estimate the time level by using equation (5.10) to get where the last inequality follows from q > 2 and taking µ > 1. Therefore because x 0 ∈ B r (x * ), combining equations (5.11) and (5.8) we get a contradiction 2µ 2 u(x 0 , t 0 ) = u(x * , t 0 − θr q ) ≤ µu(x 0 , •, t 0 − θr q + cu(x * , t 0 − θr q ) 2−q r q ) < 2µ 2 u(x 0 , t 0 ).Thus inequality (5.9) cannot hold and we have sup Dividing both sides by 2µ 2 and combining this with Theorem 4.1 gives us u(•, t 0 +θr q ).(5.12) as desired.Because T has to be either empty or non-empty, combining (5.7) and (5.12) gives us Harnack's inequality for constant γ = 2µ 2 .

The covering argument
The intrinsic Harnack requires a lot of room around the target cylinder if µ from Theorem 4.1 happens to be large.The amount of needed room can be reduced by using a covering argument but details about this are hard to find in the literature.In the time-independent case, this can be done easily by covering by small balls but in our case, the intrinsic scaling in the time direction can cause problems with the sets.We apply the covering argument in two steps: We first prove, in Lemma 6.1 below, that we can reduce the needed room as much as we want in the time variable by relaxing our constant in the space variable.Then by a second covering argument, we prove that we can gain back what we lost in the space variable without relaxing the time direction.
We only consider the forward Harnack's inequality as the presented proof can be directly modified for the backward version.We point out, however, that since the argument iteratively applies Harnack's inequality, it yields different constants c and µ for the backward and forward versions.
We use right-angled paths connecting two points to deal with space and time variables separately.Given (x, t), (y, s) ∈ R n+1 , we denote by γ That is, γ (y,s) (x,t) first moves from (x, t) to (y, t) in space, and then from (y, t) to (y, s) in time.For a ∈ R, we denote by ⌈a⌉ ∈ Z the number a rounded up to the nearest integer.Lemma 6.1.Let u ≥ 0 be a viscosity solution to (1.1) in Q − 1 (1) and let the range condition Then for any σ t > 1 there exist µ = µ(n, p, q, σ t ), α = α(n, p, q, σ t ), c = c(n, p, q, σ t ) and σ x = σ x (n, p, q, σ t ) such that , then we move along the path γ If no such point exists, then by continuity we must have and the claim of the lemma follows.
If the iteration does not end prematurely, we continue until t * i ≥ t.When that happens, we apply Harnack's inequality one more time with a radius ρ ≤ ρ i−1 so that we obtain an estimate at the exact time level t.We define i t as the smallest natural number such that t * i t+1 ≥ t.We have i t ≤ ⌈κ⌉ as otherwise cμ (q−2)i u(x 0 , t 0 ) 2−q ρ q i = t 0 + cu(x 0 , t 0 ) 2−q ⌈κ⌉ i=0 κ −1 r q = t 0 + cu(x 0 , t 0 ) 2−q (⌈κ⌉ + 1)κ −1 = t, which would be against the definition of i t.Consequently, the procedure yields the estimate u(x 0 , t 0 ) ≤ μ⌈κ⌉+1 u(x, t).
We still need to verify that there is enough room to apply Harnack's inequality throughout the iteration.To this end, notice that the biggest jump in the time direction that we can do is J = cu(x 0 , t 0 ) 2−q κ −1 r q .Therefore we always have room in time direction, since in the worst case the jump starts from t − J, and then we have (using that c ≤ c) t − J + J σq = t 0 + cu(x 0 , t 0 ) 2−q r q + (σ q − 1)cu(x 0 , t 0 ) 2−q κ −1 r q ≤ t 0 + cu(x 0 , t 0 ) 2−q r q + c(σ q t − 1)u(x 0 , t 0 ) 2−q r q ≤ t 0 + cu(x 0 , t 0 ) 2−q (σ t r) q < 1.
Illustration of the Harnack chain in the proof of Theorem 6.2 when q < 2. If q > 2, the paraboloids get steeper instead.
If no such point exists, then by continuity we must have u(x 0 , t 0 ) ≤ μu(x, t), and we end the iteration.
Set t * i := t i−1 + cu(x i−1 , t i−1 ) 2−q ρ q .By Harnack's inequality in (6.2) we have where x * i is the point in B ρ (x i−1 ) that is closest to x.Now, if u(x i−1 , t i−1 ) < μu(x * i , t * i ), then we move along the path γ If no such point exists, then by continuity we must have and we end the iteration.
If the iteration does not end prematurely, we continue until t * i ≥ t.When that happens, we apply the Harnack's inequality (6.2) one more time with a radius smaller or equal to ρ so that we hit t.We define i t as the smallest natural number such that t * i t+1 ≥ t.That is, t î is the time from which it remains to apply the Harnack's inequality (6.2) one more time to reach the target time t.Next, we show that our selections of the constants ensure the finiteness of i t and that x i t = x.For finiteness, we observe that which would be against the definition of i t.Next we estimate the smallest i x ∈ {1, . ..} such that x i x = x (observe that the construction ensures that x i = x for all i ≥ i x).At each iteration step, unless we have already reached x, we move at least αρ closer towards x.Since |x 0 − x| ≤ r, we thus have We want to show that i x ≤ i t, as this implies that x i t = x.For this end, we may assume that the Harnack chain does not skip in time direction using the paths γ (x, t) (x i ,t i ) , as otherwise the chain automatically reaches x.Using this we conclude that i x ≤ i t must hold since otherwise which is against the definition of i t.Thus the procedure reaches x before we apply Harnack's inequality one last time.This yields the estimate u(x 0 , t 0 ) ≤ μ⌈(αϱ) −1 ⌉+1 u(x, t).
We still need to check that we have room to use Harnack's inequality.The room in space is clear from the definition of ρ since For the room in time, observe that we always end the Harnack at most the time level t.Therefore, the worst-case scenario would be if our biggest possible jump in Harnack's inequality ended up at t. Since the sequence u(x i , t i ) is decreasing, the biggest possible jump is J := cu(x 0 , t 0 ) 2−q ρ q , if q < 2, cu(x 0 , t ⌈(αϱ) −1 ⌉ ) 2−q ρ q , if q > 2.
To land on t, the jump would have to start from t − J. Thus it suffices to ensure that ( t − J) + Jσ q = t + (σ q − 1)J < 1.This holds, since if q < 2, we have t + (σ q − 1)J = t + (σ q − 1)cu(x 0 , t 0 ) 2−q ρ q = t 0 + cu(x 0 , t 0 ) 2−q r q + (σ q − 1)cu(x 0 , t 0 ) 2−q ( cϱ q c ) ≤ t 0 + cu(x 0 , t 0 ) 2−q (σρ) q < 1, and if q > 2, we have t + (σ q − 1)J = t + cu(x 0 , t k ) 2−q ρ q (σ q − 1) = t 0 + cu(x 0 , t 0 ) 2−q r q + cu(x 0 , t 0 ) 2−q r q (σ q − 1) Optimality of the range of exponents Intrinsic Harnack's inequality may fail outside of the range condition (1.3) as for such exponents, viscosity solutions of (1.1) vanish in finite time as we will prove in this section.The solutions of the standard p-parabolic equation in the corresponding subcritical exponent range behave in a similar way.Idea, behind the proof is to use the equivalence result proven by Parviainen and Vázquez [PV20] to transfer the problem onto a one-dimensional divergence form equation and then to prove that a solution to this equation vanishes.We use the weak formulation for a time-mollified solution with a suitable test function after first proving that this formulation holds for all weak solutions as the separate lemma.Next, we simplify both sides of the formulation, estimate using Sobolev's inequality and ultimately get a vanishing upper bound for the norm of the solution.We do this first in bounded domains and then prove the global result using convergence and stability results.This global result Proposition 7.5 gives us a counterexample to the intrinsic Harnack's inequality 1.1 and thus proves that range (1.3) is optimal.
As proven by Parviainen and Vázquez, radial viscosity solutions to (1.1) are equivalent to weak solutions of the one-dimensional equation Here, denoting by u ′ the radial derivative of u, is heuristically the usual radial q-Laplacian in a fictitious dimension If d happens to be an integer, then solutions to (7. ) and u ′ (0, t) = 0 is a continuous weak solution to (7.1) if we have We define time-mollification and prove a basic result for it in Lemma 7.2 below for the convenience of the reader.Let ε > 0 and η ε : R → [0, ∞) be the standard mollifier such that supp η ε ⊂ (−ε, ε).The time-mollification of u ∈ L 1 ((0, R) × (0, T )) is defined by Lemma 7.2.Let u be a continuous weak solution to (7.1) and let u ε denote the mollification (7.2).Then for all 0 Because η ε is even, we have by partial integration (the boundary terms vanish since φ(0, Thus by Fubini's theorem Similarly, for the space derivative we have By the last two displays, we obtain Since φ h := ξ h ϕ is Lipschitz, it satisfies (7.3) by the first part of the proof and a simple approximation argument.Since by continuity all t 1 , t 2 ∈ (0, T ) satisfy as h → 0, the claim of the lemma follows.□ Our proof of finite extinction uses the following Sobolev's inequality, which is heuristically speaking the Gagliardo-Nirenberg inequality for radial functions in the fictitious dimension d.The standard formulation of the Gagliardo-Nirenberg inequality requires q < n and hence does not work for our one-dimensional case.
Proof.Suppose first that q = 1.We denote Integrating by parts and using that g(r) = 0 for large r, we obtain This we can further estimate as , which implies the desired inequality.□ Now we have the needed tools to state and prove the finite extinction of solutions.We do this by first proving the result for solutions of a Dirichlet problem in simple cylinders and then expanding this result to the entire space by convergence results.The existence of global solutions with extinction in finite time is a counterexample for the intrinsic Harnack's inequality as we show at the end of this section.The proof uses the following notation for the weighted Lebesgue norm ||v|| L q (r d−1 ,(0,R)) := R 0 |v| q r d−1 dr 1 q .
We only consider radially symmetric initial data in what follows.The finite extinction holds in the general situation by comparison principle.
Proposition 7.4.Assume q does not satisfy the range condition (1.3) and let R > 0. Let u be a viscosity solution of on ∂B R for any t ∈ (0, T ).
Proof.The existence of a solution u ∈ C(B R × [0, T ]) to the Cauchy-Dirichlet problem (7.5) can be proven for example by modified Perron's method (see [PV20, Theorem 2.6]) and the comparison principle ensures that it is radial.Therefore, by the equivalence result [PV20, Theorem 4.2], u is a continuous weak solution to where u 0 ∈ L ∞ ((−R, R)), u(−R, t) = u(R, t) = 0 for any t ∈ (0, T ).
Let s = d(2 − q) q and notice that s > 1 because we assumed q < 2d d+1 .We define the test function φ := u s−1 ε,h − h s−1 , where u ε,h := u ε + h for ε, h > 0 and u ε denotes the time-mollification.We add this h to ensure that our function remains strictly positive as we have negative exponents during the calculation.Then φ is an admissible test function and by Lemma 7.2 we have for all 0 < t 1 < t 2 < T .We rewrite the first term on the left-hand side using integration by parts and Fubini's theorem Hence, since u s−1 ε,h − φ = h s−1 , the equation (7.6) becomes 1 s R 0 (u s ε,h (r, t 1 ) − u s ε,h (r, t 2 ))r d−1 dr − h s−1 R 0 (u ε (r, t 2 ) − u ε (r, t 1 ))r d−1 dr = A ε,h .
Since we eliminated the time derivative, we may let ε → 0 to obtain 1 s 0 (u s h (r, t 1 ) − u s h (r, t 2 ))r d−1 dr − h s−1 R 0 (u(r, t 2 ) − u(r, t 1 ))r d−1 dr = A h .(7.7) Next, we rewrite A h as follows where by Sobolev's inequality in Theorem 7.3 forcing vanishing boundary values Here C 1 is the constant in Sobolev's inequality.Since s = d(2−q) q , we have by the last two displays lim inf h→0 A h ≥ C 1 (p − 1)(s − 1) q − 1 q s + q − 2 Observe that this implies in particular that v is decreasing.Next, we derive a distributional inequality which implies that v must in fact vanish for large times.For this end, let κ := q − 2 + s and observe that for any 0 < a < b we have Let then φ ∈ C ∞ 0 (0, T ) be non-negative.Next, we apply the integration by parts formula for difference quotients and the fact that v is decreasing together with the above elementary inequality.This way, we obtain by dominated convergence theorem Here we can use the estimate (7.8) where the last two identities follow from continuity and the computation Hence we have established the distributional inequality T 0 −v 2−q (t)φ ′ (t) + (2 − q)C 2 φ(t) dt ≤ 0 for all non-negative φ ∈ C ∞ 0 (0, T ).
1) are equivalent to radial weak solutions of the q-parabolic equation in B R × (0, T ) ⊂ R d+1 by [PV20, Section 3].If d ̸ ∈ N, we still have an equivalence between radial viscosity solutions of (1.1) and continuous weak solutions of (7.1) as proven in [PV20, Theorem 4.2].A weak solution of (7.1) is in a weighted Sobolev space but we are only interested in continuous solutions and thus will assume this in the following definition.The description of the exact definition in the elliptic case is in [Sil21, Definition 2.2].The following definition is written in a slightly different form but is equivalent to the definition given by Parviainen and Vazquez [PV20, Definition 4.1].We use the notation dz := r d−1 dr dt for the natural parabolic measure for this problem and denote the distributional derivative of v by v ′ and define it by R 0 v ′ φ dr = − R 0 vφ ′ dr for all φ ∈ C ∞ 0 ((0, R)) so it coincides with standard derivative for differentiable functions.Definition 7.1.Let