Ahmed Mohammed. Harnack Inequality for Non-divergence Structure Semi-linear Elliptic Equations

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1 Harnack Inequality for Non-divergence Structure Semi-linear Elliptic Equations International Conference on PDE, Complex Analysis, and Related Topics Miami, Florida January 4-7, 2016

2 An Outline 1 The Krylov-Safonov Harnack Inequality 2 Some Technical Results 3 A global L -Estimate on Solutions 4 Harnack Inequality for Semilinear Equations

3 The Krylov-Safonov Harnack Inequality Harnack inequality is an important tool in the derivation of many qualitative properties of solutions to second order elliptic as well as parabolic PDEs. The most recent and remarkable application was demonstrated in Perelman s use of a version of Harnack inequality for Ricci flow to settle the Poincaré conjecture. If M is a closed 3-manifold with trivial fundamental group, then M is diffeomorphic to S 3. Let Ω R n be a bounded domain. We will be concerned with a non-divergence structure elliptic operator Lu := n i, j=1 a ij (x) 2 u x i x j + n i=1 b i (x) u x i + c(x)u.

4 The coefficients are only assumed to be measurable on Ω and we suppose that L is uniformly elliptic on Ω. That is, there are constants 0 < λ Λ such that λ ξ 2 n a ij (x)ξ i ξ j Λ ξ 2 (x, ξ) Ω R n. i, j=1 We also suppose that the lower order coefficients satisfy We use the notation: b(x) ;Ω Θ, c(x) ;Ω Θ. L 0 u := n a ij (x) 2 u. x i x j i, j=1

5 One of the celebrated results in the area of second order elliptic equations in non-divergence forms is the Harnack inequality of Krylov and Safonov for non-negative solutions of Lu = f in Ω which we recall below. Theorem (Krylov and Safonov) Let u W 2,n (B(z, 2R)) and f L n (B(z, 2R)) satisfy u 0 in B(z, 2R) and Lu = f in B(z, 2R). Then ( ) sup u C B(z,R) inf B(z,R) u + R f L n (B(z,2R)) where C := C ( n, Λ/λ, ΘR, M(z) R 2) and M(z) := sup B(z,2R) c.

6 Some Technical Results We are interested in establishing a Harnack inequality for non-negative solutions of equations of the form Lu = f (u) in Ω. We need the following conditions on f. (f-1) f is continuous, non-decreasing on [0, ), and f (t) > 0 for t > 0. (f-2) There is a constant θ > 1 such that lim inf t f (θt) θf (t) > 1.

7 Suppose f satisfies (f-1) and (f-2). It is possible to show that there are constants C > 0, t 0 > 0 and q > 1 such that f (t) Ct q t t 0. As a consequence, the following two conclusions hold for any f that satisfies (f-1) and (f-2). 1 f satisfies the so called Keller-Osserman condition: ds t <, where F (t) = f (s) ds. 1 F (s) 0 2 f is super linear at infinity. That is, given a constant µ > 0 there is t µ > 0 such that f (t) µ t t t µ.

8 Suppose f satisfies the Keller-Osserman condition. We introduce: Ψ(t) := t ds F (s) F (t), t > 0. It is well-known that Ψ is a decreasing and continuous function such that lim Ψ(t) = 0. t We let Φ denote the inverse of Ψ so that Φ(t) ds F (s) F (Φ(t)) = t, 0 < t < Ψ(0+). Let us note that Φ(0+) =. As an example, if f (t) = t q for some q > 1 then Φ(t) = ct 2/(q 1) for some c = c(q).

9 The following lemma provides the crucial result for obtaining the Harnack inequality. Lemma Suppose f satisfies (f-1) and (f-2). Then lim sup t 0 + t 2 f (Φ(t)) Φ(t) := α <. We note the following. Remark Suppose f satisfies (f-1) and (f-2). Then for any ɛ > 0 we have t 2+ɛ f (Φ(t)) t 2 ɛ f (Φ(t)) lim = 0. If α > 0, lim =. t 0 + Φ(t) t 0 + Φ(t)

10 Suppose f satisfies (f-1) and the Keller-Osserman condition. Given a ball B = B(z, R) in R n, and a constant κ > 0 it is well-known that the problem { w = κf (w) in B w = on B admits a C 2 radial solution w(x) = ϕ( x z ). Moreover ϕ satisfies { (r n 1 ϕ ) = r n 1 κf (ϕ), 0 r < R, ϕ(0) > 0, ϕ (0) = 0, ϕ(r) =.

11 The following properties are easy consequences of the above initial value problem. ϕ (r) > 0 for 0 r < R. D 2 w(x) is positive definite on B. 0 ϕ (r) r κf (ϕ), 0 r < R. n w(x) R κ f (w(x)), x B. n

12 κ n f (ϕ(r)) ϕ (r) κ f (ϕ(r)), 0 r < R. 2κ n R Ψ(ϕ(0)) 2κ R, thus ( ) ( ) 2κ Φ 2κ R ϕ(0) Φ n R.

13 Let us now summarize the above discussion as follows. Summary Suppose f satisfies the Keller-Osserman condition. Given κ > 0, R > 0 and z R n there is a strict convex C 2 radial function w such that w = κ f (w) in B(z, R) and w = on B(z, R). Moreover, Φ( 2κ R) w(z) Φ ( ) 2κ n R. Furthermore, since ϕ is non-decreasing and w(x) = ϕ( x z ) we have w(x) Φ( 2κ R) x B(z, R).

14 A Global L - Estimate for solutions Let us recall the following fact from the theory of matrices. Given any m m symmetric real matrix A, and any m m positive semidefinite real matrix B λ min (A) tr (B) tr (AB) λ max (A) tr (B). We now state the following lemma. Lemma Suppose f satisfies (f-1) and (f-2). There is R 0 > such that for any given z Ω, and 0 < R < R 0 with B := B(z, R) Ω, then the equation Lw = f (w) in B and w = on B admits a super-solution. ( ) ( R R Moreover Φ 2Λ w(z) Φ ). 2nΛ

15 Proof: Let w be the convex solution of { u = κf (u) in B(z, R) u = on B(z, R) given in the summary where κ := (4Λ) 1. Recalling that f is super linear at infinity ( due to condition (f-2)) we choose t 0 > 0 such that f (t) 4Θt t t 0. By shrinking R 0 if necessary we assume that Φ(R 0 / 2Λ) t 0. Consequently w(x) t 0 in B(z, R) for 0 < R < R 0. Let A(x) denote the n n symmetric real matrix A(x) = (a ij (x)), x Ω.

16 Then Lw 1 2 f (w) = L 0w 1 4 f (w) + b w + c(x)w 1 4 f (w) = tr ( A(x)D 2 w(x) ) 1 4 f (w) + b w + c(x)w 1 4 f (w) λ max (A(x)) tr ( D 2 w(x) ) 1 4 f (w) + Θ w + Θw 1 4 f (w) Λ w(x) 1 4 ( ΘRκ f (w) + n f (w) + w Θ f (w) ) 4w = κλ f (w) 1 4 f (w) + ΘRκ ( }{{} n f (w) + w Θ f (w) ). 4w }{{} = 0, since κλ = 1/4 0, since w t 0

17 Thus Lw 1 ΘRκ f (w) 2 n f (w) in B(z, R), 0 < R < R 0. Recalling that κ = (4Λ) 1, we pick R 0 such that 0 < R 0 < 2nΛ Θ. One last condition needed on f is the following. (f-3) The functions t f (t) t is non-decreasing on (0, ). Given x Ω we will use the notation d(x) for dist (x, Ω).

18 The following estimate on solutions of Lu = f (u) on Ω is another crucial result used in the derivation of Harnack inequality. Theorem Suppose f satisfies (f-1), (f-2) and (f-3). There is a non-increasing function η : (0, ) (0, ) such that for any non-negative solution u of Lu = f (u) in Ω we have u(x) η(d(x)), x Ω. Proof: Let R 0 > 0 be as in the above theorem, and w be the convex super-solution of Lw = f (w) in Ω given above with t 0 chosen such that f (t) 4Θt for t t 0. Recall that w t 0 in B(z, R). Observe that c(x) f (w) w 0 x B(z, R).

19 Let Lv := n a ij (x)v xi x j + i j=1 n j=1 ( b j (x)v xj + c(x) f (w) ) v. w We note that Lw = Lw f (w) 0 in B(z, R). To construct a non-increasing function η : (0, ) (0, ) such that u η(d) in Ω for any solution u of Lu = f (u) in Ω, we look at two cases. Case 1. Suppose d(z) < R 0, and let 0 < R < d(z). We consider the ball B := B(z, R). Suppose Ω 0 := {x B : u(x) > w(x)} is non-empty. We note Ω 0 B. As a consequence of (f-3) we see that ( f (u) Lu := u u f (w) w ) 0 in Ω 0. Since u w on Ω 0, we conclude by the (Alexandroff-Bakelman-Pucci) maximum principle that u w on Ω 0, which is a contradiction.

20 Therefore u w in B(z, R). In particular, we have independently of 0 < R < d(z), ( ) R u(z) w(z) Φ. 2nΛ Letting R d(z) we conclude that ( ) d(z) u(z) Φ z Ω with d(z) < R 0. 2nΛ Case 2: Now consider the set Ω 1 = {x Ω : R 0 ɛ < d(x)}. Let ( ) R0 ɛ v ɛ (x) := Φ x Ω 1. 2nΛ Clearly, by Case 1, u v ɛ on Ω 1.

21 We also recall that, since c(x) Θ on Ω and f (t) 4Θt for Φ(R 0 / 2Λ) t 0 ( ) ( ( )) R0 ɛ R0 ɛ c(x)φ f Φ. 2nΛ 2nΛ Therefore, arguing as in Case 1 we find that ( ) R0 ɛ u(x) v ɛ (x) = Φ 2nΛ on Ω 1. We then let ɛ 0 +. Putting the two cases together, we conclude that u(x) η(d(x)) in Ω where ( ) t Φ if 0 < t < R 0 2nΛ η(t) := ( ) R0 Φ if t R 0. 2nΛ

22 Harnack Inequality for Semi-linear Equations Given x Ω we use the notation δ j (x) = j d(x), j = 1, 2. 3 Theorem (Harnack Inequality) Let Ω R n be a bounded domain, and suppose that f satisfies (f-1), (f-2) and (f-3). There is a positive constant C, independent of any non-negative solution u of Lu = f (u) in Ω, and z Ω such that sup u C B(z,δ 1 (z)) inf u. B(z,δ 1 (z)) Proof: Let u 0 be any solution of Lu = f (u) in Ω. Fix ɛ > 0. Then L(u + ɛ) = ɛc(x) in Ω where Lw := a ij (x)w xi x j +b i (x)w xi +(c(x) V ɛ (x))w, and V ɛ (x) := f (u(x)) u(x) + ɛ.

23 We invoke the Harnack inequality of Krylov and Safonov to obtain ( ) sup (u + ɛ) C inf (u + ɛ) + B(z,δ 1 (z)) B(z,δ 1 (z)) ɛθ Ω 1/n diam Ω, (HI ɛ ) where the constant C depends on n, Λ ( ) λ, Θ diam Ω, and d 2 (z) max c ɛ(x), x B(z,δ 2 (z)) and c ɛ (x) := c(x) V ɛ (x). To show that this constant C in (HI ɛ ) is independent of z Ω, and ɛ > 0 it suffices to show that d 2 (z)m ɛ (z) is uniformly bounded, independently of ɛ, on Ω where M ɛ (z) := max V ɛ(x). x B(z,δ 2 (z))

24 To this end, let us first recall that u(x) η(d(x)) x Ω, where η : (0, ) (0, ) is the non-increasing function defined earlier. Thus by condition (f-3) - which implies t (t + ɛ) 1 f (t) is non-decreasing on (0, ) for any ɛ > 0 - we have V ɛ (x) = f (u(x)) u(x) + ɛ f (η(d(x))) η(d(x)) + ɛ f (η(d(x))) η(d(x)), x Ω. If x B(z, δ 2 (z)), then d(x) d(z)/3 = δ 1 (z). Therefore for any x B(z, δ 2 (z)), recalling that η is non-increasing and using (f-3) once again, we estimate V ɛ (x) f (η(d(x))) η(d(x)) f (η(δ 1(z))) η(δ 1 (z)).

25 Hence for any z Ω we have ( ) d 2 (z)m ɛ (z) d 2 (z) max V ɛ(x) x B(z,δ 2 (z)) d 2 (z)f (η(δ 1 (z))) η(δ 1 (z)) = 9 δ2 1 (z)f (η(δ 1(z))) η(δ 1 (z)) C. Therefore we have shown that the constant C in (HI ɛ ) is independent of z Ω, ɛ > 0 and the solution u. Now we let ɛ 0 in (HI ɛ ) to get the desired Harnack Inequality.