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On the viscosity solutions to a degenerate parabolic differential equation

In this work, we study some properties of the viscosity solutions to a degenerate parabolic equation involving the non-homogeneous infinity-Laplacian.


Introduction
In this work, we study some properties of the viscosity solutions to a degenerate parabolic equation involving the non-homogeneous infinity-Laplacian. This may be viewed as related to the work done in [5], where we studied an eigenvalue problem for the infinity-Laplacian.
To make matters more precise, let Ω ⊂ IR n , n ≥ 2, be a bounded domain and ∂Ω be its boundary. Define Ω T = Ω × (0, T ), T > 0, and its parabolic boundary P T = (Ω × {0}) ∪ (∂Ω × [0, T )). Our goal is to study existence and properties of positive solutions φ(x, t) to (1.1) where f (x) and g(x, t) are continuous. The operator ∆ ∞ is the non-homogeneous infinity-Laplacian and is defined as The infinity-Laplacian is nonlinear and is a degenerate elliptic operator. As a result, solutions are to be understood in the viscosity sense. Questions involving the infinity-Laplacian have been attracting considerable attention recently. In particular, existence, uniqueness and local regularity have been topics of great interest. For greater motivation and context, we direct the reader to the works [3,4,8,9,14,15,17]. Moreover, degenerate parabolic equations, in general, are of great interest, see for instance, [13].
An additional reason for studying (1.1) is the connection that it has with the following problem for the infinity-Laplacian, namely, where −∞ < λ < ∞, is a parameter, and h ∈ C(Ω). It is shown in [5] that the there is a λ Ω > 0 such that the problem (1.2) has a solution u ∈ C(Ω) for 0 ≤ λ < λ Ω . We refer to λ Ω as the first eigenvalue of ∆ ∞ on Ω. The case λ < 0 has been discussed in [7]. Moreover, when λ = λ Ω and h = 0, then there is a positive eigenfunction u that solves (1.2). If we write φ(x, t) = v(x)w(t) then (1.1) leads to (1.3) ∆ ∞ v = kv 3 , in Ω, and 3w (t) = kw(t), 0 < t < ∞, where k is a constant. If v is a solution then φ(x, t) = v(x)e kt/3 solves the partial differential equation in (1.1). Thus, the problem in (1.2) has close connections to (1.1).
We will provide full justification of this computation in Section 2. A similar change of variables for positive solutions v to (1.3) (set w = log v) leads to (1.5) ∆ ∞ w + |Dw| 4 = k, in Ω.
We feel that these questions may be of some interest in the context of the infinity-Laplacian.
As a matter of fact the related equation ∆ ∞ η = η t has been studied in [1], [2] and [12]. In [1], the authors study the existence, uniqueness and regularity by employing a regularizing scheme. They show that solutions are Lipschitz continuous. In [2], the authors study asymptotic behavior of solutions to the same equation. The work in [12] presents a characterization of sub-solutions. We direct the reader to these works for a more in-depth discussion. All these works provide us the motivation to study (1.1).
Our goals in this work is to prove a comparison principle, existence of positive solutions and some results about the asymptotic behavior of positive solutions. We do not employ any regularizations and work in the setting of viscosity solutions. For a more detailed discussion about viscosity solutions, see [11]. Our work shows that positive solutions are continuous but provides no results on any additional regularity. It would be quite interesting to know if these are locally Holder continuous or even smoother. In this connection, it would be also interesting to know if a Harnack inequality holds for positive solutions.
We now describe the general lay out of the paper. In Section 2, we introduce definitions, notations, state some previously proven results relevant to our work. We prove also some preliminary results including the change of variables relating ( 1.1) and (1.3) to (1.4) and (1.5). We point out that the equation in (1.4) plays an important role in our discussions. In Section 3, we prove a maximum principle that holds for any solution φ to (1.1) regardless of its sign. Theorem 3.3 contains a comparison principle valid for positive solutions to (1.1).
The section ends with a partial result about the strong minimum principle. In Section 4, we discuss some partial results about large time asymptotic behavior of φ and provide some results on unbounded domains. We do not address any existence issues in this context.
Section 5 contains a proof of the existence of a positive continuous solution φ to (1.1), see Theorem 5. 6. This leads to the existence of solutions to (1.4). Our approach is an adaptation of the ideas in [10,11] and uses the fact that one can find a sub-solution and a super-solution of (1.1) that attain initial and boundary data. The construction of such sub-solutions and super-solutions is carried out in Appendix I (Section 6). This have been done since the work is quite computational in nature and, we feel, is best done in a separate section. Appendix II (Section 7) contains some calculations that have been used in Section 5.

Notations and preliminary results
In this section we provide notations, definitions, some preliminary results and recall some useful lemmas proven previously in other works.
In what follows, Ω ⊂ IR n , n ≥ 2, will be a bounded domain, unless otherwise stated, and ∂Ω its boundary. Let Ω denote the closure of Ω and Ω c = IR n \ Ω denote the complement of Ω. The letters x, y, z will often denote points in IR n . We reserve the letter o for the origin in IR n . There will be occasion to write a point x = (x 1 , x 2 , · · · , x n ). Moreover, for each i = 1, · · · , n, e i will denote the unit vector along the positive x i axis. The letters s, t will denote points in IR + ∪ {0}. For T > 0, we define (2.1) Ω T = Ω × (0, T ) = {(x, t) ∈ IR n × IR : x ∈ Ω, 0 < t < T }.
By P T we will denote the parabolic boundary of Ω T and this is the set For future use, we set S T to be the side boundary, i.e., S T = ∂Ω × [0, T ) and I Ω = Ω × {0}, implying P T = I Ω ∪ S T . In this work, we use Ω T ∪ P T and Ω × [0, T ) interchangeably. Next, we describe open cylinders as follows. Let B r (x) ⊂ IR n be the ball of radius r that is centered at x. For r > 0 and τ > 0, we define the following cylinders (2.3) D r,τ (x, t) = B r (x) × (t − τ /2, t + τ /2), and F r,τ (x, t) = B r (x) × (t, t + τ ).
For a function u : Ω T ∪ P T :→ IR, we recall the definitions of semi-jets P + Ω T u(y, s) and P − Ω T u(y, s) where (y, s) ∈ Ω T , see [10,11] for a detailed discussion. Let S(n) denote the set of symmetric n × n matrices. Then (a, p, X) ∈ IR × IR n × S(n) is in P + Ω T u(y, s) if and (x m , t m ) → (y, s), u(x m , t m ) → u(y, s) and (a m , p m , X m ) → (a, p, X). The set P − Ω T u(y, s) may now be analogously defined.
We discuss the notion of a viscosity solution of (1.1) and (1.4). Set We say that a function u is a sub-solution of Π(φ) = 0 in Ω T , if u is upper semi-continuous in Ω T and ∈ Ω T and ∀(a, p, X) ∈ P + Ω T u(x, t).
We will write this as Π(u) ≥ 0 in Ω T . Similarly, a function v is a super-solution of Π(φ) = 0, in Ω T , if v is lower semi-continuous in Ω T and Formally, we write Π(v) ≤ 0. A function u is a solution to Π(u) = 0, if u is continuous in Ω T and is both a sub-solution and a super-solution in Ω T . Analogous definitions hold for the in Ω T and In this case, we write Γ(v) ≥ 0.
We say that u is a sub-solution of ( 1.1) Similarly, u is a super-solution of ( 1.1) If both (2.9) and (2.10) hold then u solves ( 1.1), that is, u ∈ C(Ω × [0, T )), Π(u) = 0 in Ω T and u attains the initial data and the side condition.
We now show that studying positive solutions to Π(φ) = 0 in Ω T is the same as studying solutions to Γ(η) = 0 in Ω T , see (1.1) and (1.4). This equivalence will be used repeatedly in the work. Before stating this result, we make an elementary observation.
Since f (0) = g(0) = 0, the inequalities stated in the lemma hold. One can prove analogously the inequalities for e c .
We now prove a result about change of variables.
Lemma 2.2. Let T > 0 and suppose that φ : Ω T → IR + and φ ≥ m, for some m > 0. Then in Ω T , and analogously, Proof: We prove (i). The proof of (ii) is similar. We recall (2.6). It is clear that φ is upper semi-continuous in Ω T if and only if η is upper semi-continuous in Ω T .
Remark 2. 3. The proof in Lemma 2.2 can be easily adapted to show equivalence for super- We now list some results that have been proven in previous works. We start with the following versions of the comparison principle for the ∆ ∞ . See Lemmas 2.2 and 2.3 in [7]. Lemma 2.4. Suppose that f ∈ C(Ω) and f > 0, or The next two results have been proven in [5]. For a definition of the first eigenvalue λ Ω , see (3.2) and Theorem 4.1 in [5].
If h(x) = 0 and inf ∂Ω b > 0 then the above has a unique positive solution u.
We now recall a result about the radial version of Theorem 2.6, on a ball, that will be used in this work, see Theorem 6.1 in [5]. Let Ω = B R (o), R > 0. We set λ B = λ B R (o) and r = |x|.
(iii) Also, u defined by is a radial solution to (2.15) and we have, Remark 2. 10. From part (iii) of Theorem 2.9 with δ ≥ 0, Fix R in what follows. Clearly, δ/m is a decreasing continuous function of λ. Fixing δ and considering m = m(λ), we obtain Thus, m is an increasing C 1 function of λ. By Lemma 2.7, m(λ) → ∞ as λ ↑ λ B . Next, we fix m and consider δ as a function of λ. Thus, δ is continuous and decreasing, and Lemma 2.7 implies that δ → 0 as λ increases to λ B .
We consider the problem in (2.15) with λ < 0. Existence and uniqueness of a solution follow from Theorem 5.5 and Corollary 7.5 in [7].
Lemma 2. 11. Let u > 0 be continuous and defined in [0, R], and δ > 0. Fix λ > 0; if we define u(r) by Proof: Set r = |x|. It is clear from (2.17) that u(o) = δ, u is continuous on [0, R] and C ∞ , in r > 0. Also, u is an increasing function, and in r > 0, This leads to (2.18) du dr Since u(o) = δ and u(x) ≥ δ, it follows quite easily that Dψ(o) = 0. Next, we estimate, using Using ( 2.17), we find that Using this on the left side of (2.19), Since u(r) is continuous at r = 0, we can make u(r) ≤ 2δ, by taking r small. Using s = 2δ in the cubic polynomial in (2.20), for small r,   Then u(R) = sup B R (o) u, and, if we fix R, then u(R) is a C 1 increasing function of λ and grows unboundedly as λ increases. Similarly, if we fix λ then u(R) grows unboundedly as R increases. Take R > 0, large, so that u(R) ≥ 2δ. To get a better estimate for the latter, we write (2.23) In the second integral on the right side of (2.23), we use the inequality , for large R.
Next, we verify that if u(x) is as in Theorem 2.9 or as in Lemma 2.11 then φ(x, t) = u(x)e kt , for an appropriate k, solves ∆ ∞ φ = 3φ 2 φ t in the sense of viscosity.
One can now argue in a similar fashion that ∆ ∞ φ ≤ 3φ 2 φ t . The lemma holds.

Maximum and comparison principles
Our main goal in this section is to prove a maximum principle and a comparison principle for (1.1). We have included both here since our proof of the comparison principle applies only to solutions that are positive or have one sign in Ω T . It is not clear to us if a more general version holds. The maximum principle, on the other hand, applies to solutions that may change sign. We also provide a partial result regarding the strong minimum principle.
We start by proving the weak maximum principle for sub-solutions to (1.1). We note that φ is a sub-solution if and only if −φ is a super-solution. This will imply the weak minimum principle. We introduce the following notation. For any x ∈ IR n , write x = (x 1 , x 2 , · · · , x n ) and define the projections Proof: Let 0 < τ < T with τ close to T . We show that the weak maximum principle holds in Ω τ for any τ < T . Fix τ < T . Note that Ω τ is compactly contained in Ω T ∪ P T .
Calculating and using (3.1), we get We obtain a contradiction. For the possibility φ(y, s) = M , where (y, s) ∈ Ω τ , we apply the above argument with s k = s.

The above implies that sup
This is a contradiction and the lemma holds.
Remark 3.2. Lemmas 3.1 and 2.2 implies that the maximum principle holds for sub- Our next goal in this section is to prove a comparison principle for ( 1.1). Unlike the maximum principle proven in Lemma 3.1, we will require that the sub-solutions and super-solutions have one sign. There are simple situations where this is not needed. For instance, if u is a subsolution and v, a super-solution, and inf P T v ≥ sup P T u then by the maximum principle, Proof: We employ the ideas in [11] after making the change of variables described in Lemma 2.2, also see Remark 2. 3. We note that since u ≤ v on P T , by Lemma 3.1, u is bounded in Define η(x, t) = log u(x, t) and ζ(x, t) = log v(x, t). Then η is bounded and ζ is bounded, in particular, from below. By Lemma 2.2, Our goal is to prove the comparison principle for (3.3). We record the following calculation for later use. For ε > 0 (to be determined later) if w is a sub-solution we have (3.5) sup Fix ε > 0 and recall from (3.4) that η ε is a sub-solution of (3.3) and By ( 3.5) and (3.6), η ε − ζ has a positive maximum in Ω T . Set M = sup Ω T (η ε − ζ) and for Then for each α > 0, M α ≥ M ≥ δ/2. Since both η ε and ζ are bounded, there exists an It is known that as and hence, we may assume that the sequences x α , z α lie in a set compactly contained in Ω.
We now show that there are 0 < t 0 < T 0 < T and α 0 , small enough, such that Suppose that the first inequality is false. There is a sequence α m → 0 such that t αm → 0. Since The last inequality follows from (3.6) and the hypothesis that η ≤ ζ on P T . This is a contradiction and the assertion in (3.8) Recalling (3.7), there are numbers a α , b α ∈ IR and X α , Y α ∈ S(n) such that (see (2.4)) Moreover, Using (3.3), (3.4), (3.8) and with (2.8), we have Using (3.9), This is a contradiction and η ≤ ζ in Ω T . To complete the proof, we recall that u = e η and v = e ζ . The theorem holds.
Remark 3. 4. The comparison principle in Theorem 3.3 continues to hold for equations of the type where −∞ < σ < ∞ is a constant.
Corollary 3. 5. Suppose that η is upper semi-continuous and ζ is lower semi-continuous in Moreover, if η and ζ are solutions then η, ζ ∈ C(Ω T ∪ P T ) and In particular, if η = ζ on P T then η = ζ in Ω T .
Proof: Let k = sup P T (η − ζ) and ζ k = ζ + k. Then ζ k continues to be a super-solution and ζ k ≥ η on P T . By Theorem 3.3 and Remark 3.4, sup Ω T (η − ζ k ) ≤ sup Pt (η − ζ k ) = 0, and the claim holds. The second claim now follows quite easily.
Remark 3. 6. Let u and v be as in Theorem 3.
and u, v be positive in Ω T ∪ P T . Then This follows from Corollary 3.5 by writing η = log u and ζ = log v. For solutions, we obtain This leads to uniqueness for positive solutions to (1.1).
The above result is analogous to the result in Lemma 2.3 in [5], where we showed a comparison principle for positive solutions of problems related to the eigenvalue problem for the infinity-Laplacian.
Remark 3. 7. The comparison principle in Theorem 3.3 can be extended to include the case when inf P T v = 0. We assume that both u and v are positive in Ω T . Let The right hand side increases as δ decreases to zero. Since u > 0 is upper semi-continuous We claim that for all 0 Let (y, s) ∈ P T be a limit point of (y k , s k ), then s ≤T . If (y, s) ∈ N , working with a subsequence of (y k , s k ), we obtain If (y, s) ∈ P T \ N then u/v is upper semi-continuous at (y, s) and Both cases lead to contradiction and our claim holds. Since ε is arbitrary, Since U δ ⊂ ΩT , we have u ≤ v in ΩT . Suppose that, for some (y, s) ∈ Ω T , we have (u/v)(y, s) > 1. Clearly, s < T , selecting s <T < T , (3.11) leads to a contradiction.
The conclusion holds.
We end this section by proving a weaker version of the strong minimum principle, see, for instance, [16]. We show that if a super-solution attains its minimum, on Ω T , at some point y ∈ Ω and at some instant s, then, at y, it has the same value at all instants prior to s. This will be shown by using the comparison principle in Lemma 3. 3. Recall (2.3) for the definition This lemma uses the minimum on the parabolic boundary P T . One could easily state the same with the minimum being taken on Ω T .
Remark 3. 9. An analogous result holds for sub-solutions. If η is upper semi-continuous in Ω T ∪ P T and solves and η M ≥ 0. The claim follows from Lemma 3.8.
Remark 3. 10. Using Lemma 2.2, the conclusions of Lemma 3.8 and Remark 3.9 apply to sub-solutions and super-solutions of ∆ ∞ φ = 3φ 2 φ t . 4. Asymptotic behaviour and some results on IR n × (0, T ) In this section, we prove some results when Ω = IR n , and also some asymptotic results on a bounded domain Ω. We do not address the question of existence of solutions in the two situations we discuss below.
where λ ≥ 0 and δ ≥ 0. Also, the solution ψ λ can be extended to all of IR n . Then the See Remark 2.12 and Lemmas 2.11 and 2. 13. Also shown in Remark 2.12 is that this solution super-solution, and φ − = M e −kt is a sub-solution. Both are bounded in Ω T .
We begin by showing that if the initial data is bounded then all bounded non-negative solutions satisfy a maximum principle. A slight generalization will be proven later. Let y ∈ Ω; for r > 0 and T > 0, call B r,T (y) = B r (y) × (0, T ) and let P r,T (y) denote the parabolic boundary of B r,T (y). We use Lemma 2.11 and Remark 2.12 for our proof.
In the next lemma, we replace the restriction of boundedness by an exponential type growth and obtain the maximum principle.
where f : Ω → IR + is continuous and bounded. Suppose that h : Proof: We assume that h is not bounded. Set ν = inf Ω f , µ = sup Ω f and assume that ν > 0. Let y ∈ Ω. For any r > 0 and T > 0, call B r,T (y) = B r (y) × (0, T ) and P r,T (y) its parabolic boundary.

Lemma 4.4.
Let Ω ⊂ IR n be a bounded domain and f : Ω → IR + be continuous. Suppose that φ ∈ C(Ω ∞ ∪ P ∞ ), φ > 0, solves the following problem: Assume that m and M , as in (4.6), Proof: For every T > 0, working in Ω T , the maximum principle in Lemma 3.1 implies , and for any Hence, µ t is decreasing in t.

Existence of positive solutions
We now address the question of existence of positive solutions to (1.1). Restated, let Ω ⊂ IR n be bounded and T > 0, we seek a solution φ ∈ C(P T ∪ Ω T ), φ > 0, that solves In this section, we assume 0 < m ≤ M < ∞. Moreover, we will take m < M , for otherwise, φ(x, t) = m, ∀(x, t) ∈ Ω T is the unique positive solution. These notations introduced above will be followed in the rest of this section, in Appendix I and Appendix II, see Sections 6 and 7.
We adapt the ideas described in [10] and [11] in what follows. Let u, v : P T ∪ Ω T → IR + .
Suppose that u is upper semi-continuous in P T ∪Ω T and v is lower semi-continuous in P T ∪Ω T .
Recall from (2.3) the definition of the cylinder D δ,δ (x, t) = B δ (x) × (t − δ/2, t + δ/2). We now introduce the following definitions, see [11]. For b : Ω T ∪ P T → IR, we define the upper semi-continuous envelope of b as Similarly, we define the lower semi-continuous envelope of b as We show that u us sub is a sub-solution and v s sup is a super-solution of (5.1). The proof is an adaptation of Proposition 4.3 in [11], we provide details for completeness, also see [10]. Lemma 5.2. Suppose that w is upper semi-continuous on Ω T ∪P T , (z, θ) ∈ Ω T and (a, p, X) ∈ P + Ω T u(z, θ). Assume that for each m = 1, 2, · · · , there is a w m : Ω T ∪ P T → IR, w m upper semi-continuous, such that (i) there exists a sequence (z m , θ m ) ∈ Ω T with (z m , θ m , w m (z m , θ m )) → (z, θ, w(z, θ)), (ii) if (x, t) ∈ Ω T and for j = 1, 2, · · · , (x j , t j ) ∈ Ω T is such that (x j , t j ) → (x, t) then Then there is a subsequence w m k , k = 1, 2, · · · , and sequences (y m k , s m k ) ∈ Ω T and , w(z, θ), a, p, X), as k → ∞.
The first follows from the hypothesis (i) of the lemma and the second from (5.13). Note that the sequence (y , s ) depends on k. Taking , m > max(M k , L k ), we select an element of the sequence {(z m , θ m )} and an element of the sequence {(y , s )} with m = . Call this index m k . Note that both (z m k , θ m k ) and (y m k , s m k ) are in D r,r (o, θ). From (5.9), for each k = 1, 2 · · · and ∀(x, t) ∈ D r,r (o, θ), we have Letting k → ∞, using (5.13), i.e., max(|s m k − θ|, |θ m k − θ|, |y m k | 2 ) ≤ 2/k and the hypothesis (i) of the lemma, we have Next, we write Using these in (5.14) we obtain Then (a + 4kδ(s m k − θ), p + 4δy m k + Xy m k , X + 4δI) ∈ P + Ω T w m k (y m k , s m k ). By our selection of s m k , k|s m k −θ| ≤ 2, see (5.13). Working with a further subsequence, if needed, (a+4kδ(s m k − θ), p+4δy m k +Xy m k , X +4δI) → (a+4cδ, p, X +4δ), for some appropriate c with −2 ≤ c ≤ 2.
Since the set of (b, q, Y ) ∈ IR × IR n × S(n) such that there is a sequence (x j , t j ) ∈ Ω T , θ), b, q, Y ) is closed, it contains (a + 4cδ, p, X + 4δI). . Remark 5. 3. An analogous version of Lemma 5.2 holds when w is lower semi-continuous.
We work with −w.
Recall the definitions of ℵ sub , ℵ sup , u sub and v sup from (5.4), (5.5) and Lemma 5. 1. Let u us sub be the upper semi-continuous envelope of u sub , see (5.6).
Lemma 5. 4. Let u us sub be defined by (5.4), (5.5) and (5.6). Then u us sub is a sub-solution of (5.1) and u us sub (x, t) = h(x, t), ∀(x, t) ∈ P T . Similarly, v ls sup is a super-solution of ( 5.1) and Moreover, u us sub ≤ v ls sup in Ω T .
The proof for v ls sup is similar. The rest of the lemma follows from Lemma 5.1 and Theorem 3. 3. Call (5.15) u = u us sub and v = v ls sup .
By Lemma 5.4, u ls ≤ v us in Ω T and u ls ( Our goal is to show that φ(x, t) = supφ ∈ℵ solφ (x, t) is a solution to (5.1) or (1.1). Before, we prove this, we need the following lemma (see Lemma 4.4 in [11]), also see the calculations pertaining to this in Appendix II.
Lemma 5. 5. Suppose that w : Ω T → IR + satisfies Assume that w ls is not a super-solution, i.e., ∆ ∞ w ls ≤ 3(w ls ) 2 (w ls ) t does not hold in Ω T .
Theorem 5. 6. Let Ω ⊂ IR n be a bounded domain and T > 0. Let u and v be as in (5.15) and ℵ sol be as in (5.16). Then is C(Ω T ∪ P T ) and is a solution of (5.1) and (1.1).
Proof: The details are as in Theorem 4.1 in [11]. It is clear that Thus, φ(x, t) = h(x, t), ∀(x, t) ∈ P T . By Lemmas 5.2 and (5.4), φ us is a sub-solution of (5.1) and by Lemma 5.1 and Theorem 3.3, φ us ≤ v. Thus, φ us ∈ ℵ sol . By definition, φ us = φ and φ is a sub-solution. If φ ls is not a super-solution at some (z, θ) ∈ Ω T , then take φ ε as defined in Lemma 5.5. If ε > 0 is small enough, φ ε = h on P T and u ≤ φ ε . Since φ ε is a sub-solution, by Theorem 3.3, φ ε ≤ v and φ ε ∈ ℵ sol . Since φ ε > φ some where in Ω T , this contradicts the definition of φ and hence φ ls is a super-solution. By Theorem 3.3, φ ≤ φ ls . By (5.2), φ = φ us = φ ls . Thus, φ is continuous and solves (5.1) or (1.1). 6. Appendix I: Sub-solutions and super-solutions for the problem in (1.1) In this Appendix, we construct sub-solutions and super-solutions to (1.1). These examples of sub-solutions and super-solutions will be arbitrarily close to the boundary data, in a local sense and prove important for the Perron method.
These follow quite easily.
We set λ R to be the first eigenvalue of ∆ ∞ on B R (o). These notations will be followed in the rest of this section. Next, we make an elementary observation that will be used in what follows.

Part 1: Sub-solutions
We divide our work into two steps. First, we tackle the initial data and then consider the data along ∂Ω × [0, T ). Some of the basic ideas used in constructing the sub-solutions will also prove useful in constructing super-solutions.
In what follows, ε > 0 is small enough such that m − 2ε > 0.
We summarize as follows. Recalling (6.2), ( 6.3) and the calculations done above, we see that for every y ∈ Ω and ε > 0, small, The construction in (6.3) and (6.4) and its analogue will be used repeatedly in this section.
We will construct a sub-solution in a region R ⊂ D δ 0 ,τ 0 and then extend it to the rest of Ω T as a sub-solution. The region R will be the union of two opposing cones in IR n+1 along t, with their bases joined at t = s. See Figure 1. In what follows, we select 0 < τ ≤ τ 0 and 0 < δ ≤ δ 0 , constants k and c to ensure that we have a sub-solution. Also, to make our calculations easier, we will work with the inequality in Ω T and e η will be the desired sub-solution for (1.1), see Lemma 2.2. Assume that h(y, s) > m.
A simple calculation gives us Since, δ → 0 as τ → 0, we can choose appropriate values of τ and δ such that D δ,τ (y, s) ⊂ D δ 0 ,τ 0 (y, s). This ensures that (6.10) is satisfied. The region R will be in D δ,τ (y, s). We write R = R + ∪ R − , where R + denotes the region in the upper cone and R − that in the lower cone. We describe these next. Set r = |x − y|, and choose c, k, τ and δ as in (6.11). Define, in 0 ≤ r ≤ δ and s − τ ≤ t ≤ s + τ , The apex of R + is (y, s + τ ) and that of R − is (y, s − τ ). Also, at t = s, 0 ≤ r ≤ kτ /c = δ.
We show next thatη is a sub-solution in Ω T . To do this, we note that it is enough to check the definition on ∂R ∩ Ω T and the set common to both R + and R − . Recall that these lie in the set B δ (y) × [s − τ, s + τ ]. We describe these as follows.
The base S common to both the cones is described in (iii).

Part 2: Super-solutions
Next, we construct super-solutions to (6.1). Our approach will be similar to what was done in Part 1. Let M = sup (x,t)∈P T h(x, t). As done in Part 1, we divide our work into two steps.