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A remark on the existence of viscosity solutions for quasilinear elliptic equations
Nonlinear Analysis 65 (2006) 230 – 241 www.elsevier.com/locate/na
A remark on the existence of viscosity solutions for quasilinear elliptic equations Aris Tersenov∗ Department of Computer Science and Technology, University of Peloponnese, 22100 Tripoli, Greece Received 23 March 2005; accepted 13 September 2005
Abstract In the present paper we study the existence of viscosity solutions of the Dirichlet problem for quasilinear elliptic equations of the form −
n
aij (x)uxi xj + b(x, u, Du) = 0,
i,j =1
where a(x), b(x, u, p) are continuous functions and the function b(x, u, p) has an arbitrary growth with respect to p. Under some structure restrictions on b(x, u, p) suitable subsolution and supersolution satisfying boundary conditions are constructed. The existence and uniqueness of viscosity solutions is obtained by Perron’s method under assumption that the strong comparison result holds. 䉷 2005 Elsevier Ltd. All rights reserved. Keywords: Quasilinear equations; Viscosity solution; Dirichlet problem
1. Introduction Consider the equation Lu ≡ −
n
aij (x)uxi xj + b(x, u, Du) = 0
i,j =1
∗ Tel.: +30 2710 372204; fax: +30 2710 372160.
E-mail address: [email protected]. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.09.023
(1.1)
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
in a bounded open domain ⊂ R N coupled with the Dirichlet boundary condition u = 0, j
231
(1.2)
where x = (x1 , x2 , . . . , xn ), Du = (ux1 , . . . , uxn ) and j is the Lipschitz boundary of . Write Eq. (1.1) in the form F (x, u, Du, D 2 u) = −
n
aij (x)uxi xj + b(x, u, Du) = 0,
(1.3)
i,j =1
where F : R n × R × R n × S(n) → R and S(n) is the set of symmetric matrices. Assume that F is proper in the sense of [3] what means F (x, r, p, X)F (x, s, p, X) for r s,
(1.4)
F (x, r, p, X)F (x, r, p, Y ) for Y X,
(1.5)
where r, s ∈ R, x, p ∈ R n , X, Y ∈ S(n) and S(n) is equipped with its usual order. From (1.4) and (1.5) we have that F is nondecreasing in r and nonincreasing in X. In our case F (x, r, p, X) = −Tr(A(x)X) + b(x, r, p), where A(x) = {aij (x)} is a symmetric matrix, and the properness, in particular, means that A 0 and b(x, r, p) is nonincreasing in r. Assume that F is continuous and uniformly elliptic, that means I A(x)I for some positive and . Let us recall the precise meaning of the notion of continuous viscosity solution of (1.1), (1.2). The boundary condition is interpreted in the strict sense. Regarding the equation continuous viscosity solution means what viscosity means in [6]. That is, u is a continuous viscosity solution of F 0 (equivalently, a continuous viscosity subsolution of F = 0) if u is upper semicontinuous function on and for every ∈ C 2 () and local maximum point xˆ ∈ of u − , one has ˆ F (x, ˆ u(x), ˆ D(x), ˆ D 2 (x))0. The notion of a continuous viscosity solution of F 0 (equivalently, a continuous viscosity supersolution of F = 0) arises replacing upper semicontinuous by lower semicontinuous, max by min and reversing the inequality to ˆ F (x, ˆ u(x), ˆ D(x), ˆ D 2 (x))0. Continuous viscosity solution of F =0 is a function which is simultaneously a continuous viscosity subsolution and a continuous viscosity supersolution. We assume that the strong comparison result holds which means that a subsolution u∗ and a supersolution u∗ of F = 0 in satisfying u∗ ,(−u∗ ) are upper semicontinuous functions and u∗ u∗ on j, satisfy u∗ u∗ in . To assert that comparison holds via theorems of [3,6] one must impose structure on the continuity of F (x, r, p, X) in x. According to [3] if there exist > 0 such that (r − s)F (x, r, p, X) − F (x, s, p, X) = b(x, r, p) − b(x, s, p), s r, (x, p, X) ∈ × R n × S(n)
(1.6)
232
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
and there is a function : [0, ∞] → [0, ∞] that satisfies (0+) = 0 such that F (y, r, (x − y), Y ) − F (x, r, (x − y), X)(|x − y|2 + |x − y|)
(1.7)
whenever x, y ∈ , and
X 0
r ∈ R,
0 −Y
I −I
X, Y ∈ S(n)
−I I
,
where I stands for identity matrix and > 0 is a parameter that satisfies lim→∞ |x − y|2 = 0, then comparison holds. Our main purpose is to construct subsolutions and supersolutions for problem (1.1), (1.2) in the case when the function b(x, u, Du) has an arbitrary growth with respect to Du, and invoking Perron’s method to prove the existence of continuous viscosity solution for problem (1.1), (1.2). There are many papers devoted to the construction of subsolutions and supersolutions for (1.1), (1.2) and its parabolic analogue (see, for example, [4,7,13,14]), but as far as we know the results presented therein do not cover the case when b has an arbitrary growth with respect to Du and the boundary j is only Lipschitz. It is worth mentioning that the technique we present here handles the case, when b(x, 0, 0) has a variable sign. That means that u ≡ 0 is nor subsolution neither supersolution of (1.1), (1.2). We seek for a subsolution u∗ in the form u∗ = g(z(x)), where z(x) is a solution of the auxiliary problem ˜ ≡ Lz
n
aij (x)zxi xj − C2 |Dz| = ,
i,j =1
z
j
= 0,
> 0, C2 > 0 are constants and g(z) is a twice differentiable function which solves the certain boundary value problem for the second-order ordinary differential equation. Next we show that u∗ =−u∗ is the supersolution for the same problem. Having these sub and supersolutions one may invoke Perron’s method to obtain the existence and uniqueness of continuous viscosity solution of (1.1), (1.2). 2. Construction of subsolutions for the auxiliary problem We turn to the construction of subsolutions and supersolutions of (1.1), (1.2). In this section we will construct a subsolution u∗ and in the next one we will show that u∗ = −u∗ is a supersolution for the same problem. We seek for a subsolution u∗ of (1.1), (1.2) in the form u∗ (x) = g(z(x)),
(2.1)
where g(z) is a twice differentiable function in z with g (z) > 0, g (z) > 0 and assume that u∗ 0. From (1.4) it follows that Lu∗ −
n i,j =1
aij (x)u∗xi xj + b(x, 0, Du∗ ).
(2.2)
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
233
Using representation (2.1) we can express the derivatives of u∗ in terms of g and z Du∗ = g (z)Dz,
u∗xi xj = g (z)zxi zxj + g (z)zxi xj .
Hence (2.2) takes the form n
Lu∗ − g (z)
aij (x)zxi xj − g (z)
i,j =1
n
aij (x)zxi zxj + b(x, 0, g (z)Dz).
(2.3)
i,j =1
Multiplying (2.3) by (−1) and using the inequality I A(x)I we obtain −Lu∗ g (z) g (z)
n i,j =1 n
aij (x)zxi xj + g (z)
n
aij (x)zxi zxj − b(x, 0, g (z)Dz)
i,j =1
aij (x)zxi xj + g (z)|Dz|2 − b(x, 0, g (z)Dz).
(2.4)
i,j =1
So we have that if g (z)
n
aij (x)zxi xj + g (z)|Dz|2 − b(x, 0, g (z)Dz) 0,
(2.5)
i,j =1
then u∗ = g(z(x)) is a subsolution of (1.1). Write the function b(x, 0, v) in the following way: b(x, 0, v) = c1 (x) + c2 (x)v + b1 (x, v).
(2.6)
Put C1 = maxx∈ |c1 (x)| and C2 = maxx∈ |c2 (x)|. Obviously |b(x, 0, v)|C1 + C2 |v| + |b1 (x, v)|. Using the last inequality we rewrite (2.5) in the following way: ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 + g (z)|Dz|2 − |b1 (x, g (z)Dz)| 0. g (z) ⎣
(2.7)
i,j =1
We will split inequality (2.7) into two parts in order to obtain two independent problems with respect to g(z) and z(x). So instead of (2.7) we consider two inequalities ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 0 (2.8) g (z) ⎣ i,j =1
and g (z)|Dz|2 − |b1 (x, g (z)Dz)| 0. It is clear that from (2.8) and (2.9) follows (2.7).
(2.9)
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A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
Let z(x) satisfies the problem (auxiliary problem) ˜ ≡ Lz
n
aij (x)zxi xj − C2 |Dz| = > 0,
i,j =1
z
j
= 0.
(2.10)
We suppose that j is Lipschitz and as a consequence satisfies a uniform exterior cone condition. Using the results of Miller [8,9] concerning the construction of barriers on exterior cones for elliptic equations and results of Crandall et al. [4] we will prove the existence of viscosity solution for problem (2.10). The main goal of the proof is to obtain an estimate for z. To assert that comparison hold for (2.10) one can suppose that aij (x), i, j = 1, . . . , n are uniformly continuous in x and so for F1 (x, p, X) = −Tr(A(x)X) + C2 |p| + we have F1 (x, p, X) − F1 (y, p, X)(|x − y|)(1 + |p| + X)
(2.11)
for some continuous : [0, ∞) → [0, ∞) satisfying (0) = 0. Consider Pucci minimizing and maximizing operators P − (X) = Trace(X + ) − Trace(X − ), P + (X) = Trace(X + ) − Trace(X − ), where X+ , X− denote the positive and the negative part of X ∈ S(n) and Trace(X + ) is the sum of the positive eigenvalues of X. It is obvious that F1 satisfies the following condition: P − (X − Y ) − C2 |p − q|F1 (x, p, X) − F1 (y, q, Y ) P + (X − Y ) + C2 |p − q|.
(2.12)
For n2 and ∈ (0, ) let T = {x ∈ R N : xn (cos )|x|} be the closed right circular cone of aperture . Let n
L0 u ≡
aij (x)uxi xj .
i,j =1
Consider barriers of the form r f ( ), where r = |x| and = arccos(xn /|x|). From results of [8, Theorems 2, 3] it follows that for every ∈ (0, ) there exist ∈ (0, 1) and f ∈ C 2 ([0, )) depending only on , , n, , such that f (0) = 0 and f > 1 on [0, ] so that L0 (r f ( )) − r −2 .
(2.13)
Put w(x) = −r f ( ). Then, from (2.13) it follows that L0 (w(x)) r −2 .
(2.14)
Deriving a polar representation for |Dw| =
|Dw| =
wr2 + w 2
wx21 + · · · + wx2n it is easy to show that
1 −1 = r 2 f 2 + f 2 . r2
(2.15)
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
Using (2.14), (2.15) one can easily see that −2 −1 ˜ −r 2 f 2 + f 2 . Lw r
235
(2.16)
Due to the fact that f and f are bounded it immediately follows that for every > 0 there exist r, small enough, such that ˜ > Lw
(2.17)
holds. We have that is a bounded domain in R n with Lipschitz boundary, so it satisfies a uniform exterior cone condition. It means that there exist ∈ (0, ) and r0 > 0 so that for every ∈ j there is a rotation = () such that ∩ Br0 () ⊂ + T .
(2.18)
Here Br0 () denotes the open ball in R n of radius r0 centered at . Setting w (x)=w(−1 (x −)) we have w (x) − |x − |
w () = 0,
on ∩ Br0 ().
(2.19)
In particular, we have
w (x) − r0
on {x ∈ : |x − | = r0 }.
(2.20)
Taking r0 sufficiently small that (2.17) holds we arrive at ˜ > Lw
for x ∈ ∩ Br0 ().
(2.21)
We need to extend these local barriers to global ones. Choose any point y = and 2r1 < d ≡ distance(y, j), where we assume that r1 > 1. For > 0 put 1 1 . (2.22) − G(x) = |x − y| r1 Obviously 2 − 1 1 . − G(x) − r1 (2r1 )
(2.23)
A standard computation [5] shows that ˜ LG(x)
(( + 1) − (n − 1) − C2 |x − y|) > 0 |x − y|+2
(2.24)
on for large depending only on n, , , j and diam(). Note that due to (2.23) we can choose = 0 such that (2.24) and the inequality 1 1 G(x) > − 0 > − r0 2 r1
for x ∈
(2.25)
hold. Let satisfies the inequality
0 (d + diam())0 +2
((0 + 1) − (n − 1) − C2 (d + diam())).
(2.26)
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A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
Hence, we have ˜ LG(x) > . Put
W (x) =
(2.27)
G(x) max(G(x), w (x))
for x ∈ , |x − | > r0 , for x ∈ , |x − | r0 .
(2.28)
From (2.19), (2.23) it follows that W (x) agrees with w (x) in a neighborhood of ∈ j relative to . From (2.20), (2.25) it follows that W (x) agrees with G(x) on \Br2 () for some 0 < r2 < r0 . At last in view of (2.21) and (2.27) we conclude that ˜ (x) > LW
on .
(2.29)
Thus we have constructed the global barrier W (x) for every ∈ j that has the following properties: W (x) ∈ C(),
W () = 0,
on \{}
W (x) > 0
(2.30)
and ˜ (x) > LW
on .
(2.31)
Finally we put W (x) = sup W (x).
(2.32)
∈j
(x) ≡ 0 will From (2.30) to (2.32) it follows that W (x) is a subsolution of (2.10). Obviously W be a supersolution of (2.10). Having the comparison holds we may invoke the Perron’s method to obtain the existence and uniqueness of the viscosity solution of (2.10). Due to the fact that W (x) is a subsolution and W˜ ≡ 0 is a supersolution of (2.10), we can estimate the function z(x) in the following way: (2.33)
W (x)z(x)0. Obviously from (2.20) and (2.25) we obtain 1 − 0 < z(x)0 r1
for x ∈ .
(2.34)
3. Solvability of the problem for the function g(z) Consider inequality (2.9). In order to construct twice differentiable function g(z) which will satisfy inequality (2.9) we need the following estimate |Dz| M, where M is some constant, to hold in . It is obvious that F1 (x, Dz, D 2 z)= ni,j =1 aij zxi xj −C2 |Dz|− satisfies the following inequality: |F1 (x, Dz, D 2 z) − F1 (x, D z¯ , D 2 z¯ )| = C2 ||D z¯ − |Dz|| C2 |D z¯ − Dz| n C2 |¯zxi − zx1 | i
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
237
for all zxi xj , zxi , z¯ xi , x and is nonincreasing in z. If we suppose that aij ∈ C (), i, j =1, . . . , n then from results of Safonov [10] it follows that there exists a classical solution z ∈ C 2+ ()∩C 0,1 () of problem (2.10) that coincides with the viscosity solution we have built above. Thus we have |Dz|M in . In the case when aij , i, j = 1, . . . , n are only uniformly continuous functions in x one can consider the Pucci minimizing operator P − (D 2 z∗ ) = Trace(D 2+ z∗ ) − Trace(D 2− z∗ ), where D 2+ z∗ , D 2− z∗ denote the positive and the negative part of D 2 z∗ ∈ S(n) and Trace(D 2+ z∗ ) is the sum of the positive eigenvalues of D 2 z∗ . It is obvious that P − (D 2 z∗ ) − C2 |Dz∗ |
n
aij (x)z∗xi xj − C2 |Dz∗ |.
(3.1)
i,j =1
Consider now the Dirichlet problem for Bellman equation P − (D 2 z∗ ) − C2 |Dz∗ | = , z∗ = 0.
(3.2)
In [10] Safonov proved that (3.2) has a classical solution z∗ ∈ C 2+ () ∩ C 0,1 () and from (3.1) it follows that this solution is a subsolution for z. Obviously z∗ ≡ 0 will be a supersolution for z. So we have constructed the subsolution and the supersolution which are Lipschitz continuous in and satisfy boundary condition. One can easily check that the existence of mentioned above sub and supersolutions guarantee that problem (2.10) has a viscosity solution z which is Lipschitz continuous in some neighborhood of the boundary j (see also [1]). It follows from [2] (see also 1, [4,11,12]) that (2.11) and (2.12) are enough to guarantee that the solution of (2.10) is Cloc () for all 0 < < ¯ = ¯(, , n) ∈ (0, 1). So, for the gradient of the viscosity solution of (2.10) we have that |Dz|M in . Suppose that the function b1 (x, g (z)Dz) satisfies the following structure condition: b1 (x, g (z)Dz) b˜1 (x, g (z)Dz)|Dz|2
(3.3)
for some continuous function b˜1 . Obviously one may have the estimate |b˜1 (x, g (z)Dz)| (g (z))
(3.4)
for some continuous positive function . From (3.3), (3.4) it follows that (2.9) holds if g (z) − (g (z))0
(3.5)
takes place. Consider the equation g (z) − (g (z)) = 0.
(3.6)
Firstly we will specify boundary conditions for (3.6) in order to construct g(z) such that inequality (2.7) will take place. Recall that we seek for a subsolution of (1.1), (1.2) in the form u∗ = g(z(x)), hence from (2.10) we obtain u∗ = g(z(x)) = g(0) = 0. (3.7) j
j
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A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
If z(x) is the solution of problem (2.10) and g(z) is the solution of (3.6), then (2.7) holds if g (z)
C1
(3.8)
for all x ∈ . From (3.6) we have that g (z) is a nondecreasing function of z. From (2.33) it follows that z varies on the interval [−K, 0], where K < 1/r10 depends on , j, , and diam(). Hence if we put g (z)(−K) =
C1 ,
(3.9)
then for all z ∈ [−K, 0] the function g(z) will satisfy inequality (3.8). From (3.6), (3.7) and (3.9) it follows that if g(z) satisfies the problem g (z) − (g (z)) = 0,
g(0) = 0,
g (z)(−K) =
C1 ,
(3.10)
then (2.7) holds. The main problem that arises here is the existence of a solution of (3.10).An obvious substitution g (z) = reduces (3.10) to the problem (z) − () = 0,
(−K) =
C1 .
(3.11)
It is well-known that if is a continuous function then there exists a solution of (3.11) at least in some neighborhood of the point z = −K, i.e. on the interval [−K − , −K + ] for some > 0 (further we will consider the existence of a solution of problem (3.11) only on [−K, −K + ]). It is clear that the interval of existence of a solution of (3.11) for fixed depends on and C1 . It is also clear that in general when C1 / → ∞ one may have → 0. Hence for arbitrary C1 / (3.10) fails to have a twice differentiable solution on [−K, 0]. Thus, in order to have existence for (3.10) one has to put some restriction on the smallness of C1 (for fixed ). We will formulate this restriction in the following form Lemma 1. There always exist , K, C1 such that there exist a twice differentiable solution of (3.10). Proof. Consider the problem (z) − () = 0,
(−K) = 0.
(3.12)
There exists a solution of (3.12) at least on [−K, −K + 0 ] for some 0 , which depends only on . Choose r1 > 1 and sufficiently large 0 to have 1 0 − + 0 > . r 0 2
(3.13)
Then from (3.13) it follows that −K + 0 >
0 . 2
(3.14)
Without loss of generality we can assume that is a Lipschitz function. Using the theorem on continuous dependence of solutions of (3.12) on the initial data we conclude that there
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
239
exists = () such that if C1 ,
(3.15)
then problem (3.11) has a solution on [−K, 0]. So we have shown that there exists = () such that if (3.15) holds, then problem (3.10) has a twice differentiable solution on [−K, 0]. We summarize the results of this construction in the following Lemma 2. Let z(x) be a solution of problem (2.10) and g(z) be a solution of (3.10). If satisfies (2.26) and C1 satisfies (3.15), where depends only on , then u∗ = g(z(x)) will be a subsolution of problem (1.1), (1.2). The construction of a supersolution we carry out exactly in the same way as in the case of subsolution. We seek for a supersolution u∗ 0 of (1.1), (1.2) in the form u∗ (x) = h(z(x)),
(3.16)
where h(z) is a twice differentiable function in z with h (z) < 0, h (z) < 0. Hence Lu∗ − h (z)
n
aij (x)zxi xj − h (z)
i,j =1
n
aij (x)zxi zxj + b(x, 0, h (z)Dz)
i,j =1
0.
(3.17)
Using the inequality I A(x)I we obtain Lu∗ − h (z)
n
aij (x)zxi xj − h (z)|Dz|2 − |b(x, 0, h (z)Dz)| 0.
(3.18)
i,j =1
Using representation (2.6) for the function b(x, 0, v) we conclude that |b(x, 0, v)|C1 + C2 |v| + |b1 (x, v)|. So we can rewrite (3.18) in the following way: ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 − h (z)|Dz|2 Lu∗ − h (z) ⎣ i,j =1
− |b1 (x, h (z)Dz)|0.
(3.19)
We split inequality (3.19) into two parts in order to obtain two independent problems with respect to h(z) and z(x). So instead of (3.19) we consider two inequalities ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 0 (3.20) −h (z) ⎣ i,j =1
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A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
and −h (z)|Dz|2 − |b1 (x, h (z)Dz)| 0.
(3.21)
One can easily see that if z(x) satisfies problem (2.10) and h(z) is the solution of h (z) + (|h (z)|) = 0,
h(0) = 0,
h (z)(−K) = −
C1 ,
(3.22)
where is the same as in (3.4), then u∗ (x) is the supersolution of (1.1), (1.2). Thus we obtain that if u∗ = g(z(x)) is a subsolution of (1.1), (1.2) then u∗ = −u∗ is a supersolution of the same problem. Having the comparison holds we may invoke the Perron’s method to obtain the existence and uniqueness of the viscosity solution of (1.1), (1.2). Hence the following theorem holds Theorem 1. Let aij (x), b(x, u, Du) be continuous functions. Suppose that F (x, r, p, X) = −Tr(A(x)X) + b(x, r, p), where A(x) = {aij (x)}, is uniformly elliptic and satisfies (1.3)–(1.7). Suppose that b(x, 0, Du) satisfies (2.6), (3.3) and b(x, 0, 0) = c1 (x) is sufficiently small (in the sense of Lemmas 1, 2), then there exists a unique continuous viscosity solution of (1.1), (1.2). Remark 1. It is worth to mention that if |b(x, 0, ∇u)| c1 (x) + c2 (x)|∇u| (i.e. b1 (x, v) = 0 in (2.6)), then problem (1.1), (1.2) has a unique continuous viscosity solution without any smallness restriction on b(x, 0, 0) = c1 (x). In that case the subsolution u∗ = g(z(x)) will be a linear function z of the form u∗ = (c1 /)z. Remark 2. If we suppose, for example, that |b(x, 0, ∇u)| c1 (x) + c2 (x)|∇u| + c3 (x)|∇u|2 , where |c3 (x)|C3 and for z which is the solution of (2.10) we have |∇z| C4 then g(z) will be the solution of the problem g (z) − Mg = 0, 2
g(0) = 0,
g (−K) =
C1 ,
M = C3 C42 .
So the subsolution u∗ in that case will be − M M M u∗ = − K − z − ln − K ln , M C1 C1 where the smallness of b(x, 0, 0) = c1 (x) means that c1 (x)
. MK
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[5] D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, Springer, Berlin, Heidelberg, New York, 1983. [6] H. Ishii, P.L. Lions, Viscosity solutions of fully nonlinear second-order elliptic partial differential equations, J. Differential Equations 83 (1990) 26–78. [7] J. Li, J. Liu, Q. Yu, Uniqueness and existence of viscosity solutions of the Dirichlet boundary problem for nonlinear elliptic second order equations, J. Math. Study 35 (1) (2002) 26–31. [8] K. Miller, Barriers on cones for uniformly elliptic operators, Ann. Mat. Pura Appl. 76 (1967) 93–105. [9] K. Miller, Extremal barriers on cones with Phragmen–Lindelof theorems and other applications, Ann. Mat. Pura Appl. 90 (1971) 297–329. [10] M.V. Safonov, On the classical solution of nonlinear elliptic equations of second order, Izv. Akad. Nauk SSSR Ser. Mat. 52 (1988) 1272–1287 (in Russian) (translation in Math. USSR-Izv. 33 (1989) 597–612). [11] A. Swiech, W 1,p -interior estimates for solutions of fully nonlinear, uniformly elliptic equations, Adv. Differential Equations 2 (1997) 1005–1027. [12] N.S. Trudinger, Holder gradient estimates for fully nonlinear elliptic equations, Proc. R. Soc. Edinburgh 108 A (1988) 57–65. [13] Q. Yi, J.N. Zhao, Regularity of viscosity solutions of the Dirichlet boundary problem for a class of quasilinear degenerate parabolic equations, Chinese J. Contemp. Math. 23 (1) (2002) 63–74. [14] R.J. Zhu, Existence, uniqueness and regularity of viscosity solutions of fully nonlinear parabolic equations under natural structure conditions, Beijing Shifan Daxue Xuebao 30 (3) (1994) 290–300.
A remark on the existence of viscosity solutions for quasilinear elliptic equations Aris Tersenov∗ Department of Computer Science and Technology, University of Peloponnese, 22100 Tripoli, Greece Received 23 March 2005; accepted 13 September 2005
Abstract In the present paper we study the existence of viscosity solutions of the Dirichlet problem for quasilinear elliptic equations of the form −
n
aij (x)uxi xj + b(x, u, Du) = 0,
i,j =1
where a(x), b(x, u, p) are continuous functions and the function b(x, u, p) has an arbitrary growth with respect to p. Under some structure restrictions on b(x, u, p) suitable subsolution and supersolution satisfying boundary conditions are constructed. The existence and uniqueness of viscosity solutions is obtained by Perron’s method under assumption that the strong comparison result holds. 䉷 2005 Elsevier Ltd. All rights reserved. Keywords: Quasilinear equations; Viscosity solution; Dirichlet problem
1. Introduction Consider the equation Lu ≡ −
n
aij (x)uxi xj + b(x, u, Du) = 0
i,j =1
∗ Tel.: +30 2710 372204; fax: +30 2710 372160.
E-mail address: [email protected]. 0362-546X/$ - see front matter 䉷 2005 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2005.09.023
(1.1)
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
in a bounded open domain ⊂ R N coupled with the Dirichlet boundary condition u = 0, j
231
(1.2)
where x = (x1 , x2 , . . . , xn ), Du = (ux1 , . . . , uxn ) and j is the Lipschitz boundary of . Write Eq. (1.1) in the form F (x, u, Du, D 2 u) = −
n
aij (x)uxi xj + b(x, u, Du) = 0,
(1.3)
i,j =1
where F : R n × R × R n × S(n) → R and S(n) is the set of symmetric matrices. Assume that F is proper in the sense of [3] what means F (x, r, p, X)F (x, s, p, X) for r s,
(1.4)
F (x, r, p, X)F (x, r, p, Y ) for Y X,
(1.5)
where r, s ∈ R, x, p ∈ R n , X, Y ∈ S(n) and S(n) is equipped with its usual order. From (1.4) and (1.5) we have that F is nondecreasing in r and nonincreasing in X. In our case F (x, r, p, X) = −Tr(A(x)X) + b(x, r, p), where A(x) = {aij (x)} is a symmetric matrix, and the properness, in particular, means that A 0 and b(x, r, p) is nonincreasing in r. Assume that F is continuous and uniformly elliptic, that means I A(x)I for some positive and . Let us recall the precise meaning of the notion of continuous viscosity solution of (1.1), (1.2). The boundary condition is interpreted in the strict sense. Regarding the equation continuous viscosity solution means what viscosity means in [6]. That is, u is a continuous viscosity solution of F 0 (equivalently, a continuous viscosity subsolution of F = 0) if u is upper semicontinuous function on and for every ∈ C 2 () and local maximum point xˆ ∈ of u − , one has ˆ F (x, ˆ u(x), ˆ D(x), ˆ D 2 (x))0. The notion of a continuous viscosity solution of F 0 (equivalently, a continuous viscosity supersolution of F = 0) arises replacing upper semicontinuous by lower semicontinuous, max by min and reversing the inequality to ˆ F (x, ˆ u(x), ˆ D(x), ˆ D 2 (x))0. Continuous viscosity solution of F =0 is a function which is simultaneously a continuous viscosity subsolution and a continuous viscosity supersolution. We assume that the strong comparison result holds which means that a subsolution u∗ and a supersolution u∗ of F = 0 in satisfying u∗ ,(−u∗ ) are upper semicontinuous functions and u∗ u∗ on j, satisfy u∗ u∗ in . To assert that comparison holds via theorems of [3,6] one must impose structure on the continuity of F (x, r, p, X) in x. According to [3] if there exist > 0 such that (r − s)F (x, r, p, X) − F (x, s, p, X) = b(x, r, p) − b(x, s, p), s r, (x, p, X) ∈ × R n × S(n)
(1.6)
232
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
and there is a function : [0, ∞] → [0, ∞] that satisfies (0+) = 0 such that F (y, r, (x − y), Y ) − F (x, r, (x − y), X)(|x − y|2 + |x − y|)
(1.7)
whenever x, y ∈ , and
X 0
r ∈ R,
0 −Y
I −I
X, Y ∈ S(n)
−I I
,
where I stands for identity matrix and > 0 is a parameter that satisfies lim→∞ |x − y|2 = 0, then comparison holds. Our main purpose is to construct subsolutions and supersolutions for problem (1.1), (1.2) in the case when the function b(x, u, Du) has an arbitrary growth with respect to Du, and invoking Perron’s method to prove the existence of continuous viscosity solution for problem (1.1), (1.2). There are many papers devoted to the construction of subsolutions and supersolutions for (1.1), (1.2) and its parabolic analogue (see, for example, [4,7,13,14]), but as far as we know the results presented therein do not cover the case when b has an arbitrary growth with respect to Du and the boundary j is only Lipschitz. It is worth mentioning that the technique we present here handles the case, when b(x, 0, 0) has a variable sign. That means that u ≡ 0 is nor subsolution neither supersolution of (1.1), (1.2). We seek for a subsolution u∗ in the form u∗ = g(z(x)), where z(x) is a solution of the auxiliary problem ˜ ≡ Lz
n
aij (x)zxi xj − C2 |Dz| = ,
i,j =1
z
j
= 0,
> 0, C2 > 0 are constants and g(z) is a twice differentiable function which solves the certain boundary value problem for the second-order ordinary differential equation. Next we show that u∗ =−u∗ is the supersolution for the same problem. Having these sub and supersolutions one may invoke Perron’s method to obtain the existence and uniqueness of continuous viscosity solution of (1.1), (1.2). 2. Construction of subsolutions for the auxiliary problem We turn to the construction of subsolutions and supersolutions of (1.1), (1.2). In this section we will construct a subsolution u∗ and in the next one we will show that u∗ = −u∗ is a supersolution for the same problem. We seek for a subsolution u∗ of (1.1), (1.2) in the form u∗ (x) = g(z(x)),
(2.1)
where g(z) is a twice differentiable function in z with g (z) > 0, g (z) > 0 and assume that u∗ 0. From (1.4) it follows that Lu∗ −
n i,j =1
aij (x)u∗xi xj + b(x, 0, Du∗ ).
(2.2)
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
233
Using representation (2.1) we can express the derivatives of u∗ in terms of g and z Du∗ = g (z)Dz,
u∗xi xj = g (z)zxi zxj + g (z)zxi xj .
Hence (2.2) takes the form n
Lu∗ − g (z)
aij (x)zxi xj − g (z)
i,j =1
n
aij (x)zxi zxj + b(x, 0, g (z)Dz).
(2.3)
i,j =1
Multiplying (2.3) by (−1) and using the inequality I A(x)I we obtain −Lu∗ g (z) g (z)
n i,j =1 n
aij (x)zxi xj + g (z)
n
aij (x)zxi zxj − b(x, 0, g (z)Dz)
i,j =1
aij (x)zxi xj + g (z)|Dz|2 − b(x, 0, g (z)Dz).
(2.4)
i,j =1
So we have that if g (z)
n
aij (x)zxi xj + g (z)|Dz|2 − b(x, 0, g (z)Dz) 0,
(2.5)
i,j =1
then u∗ = g(z(x)) is a subsolution of (1.1). Write the function b(x, 0, v) in the following way: b(x, 0, v) = c1 (x) + c2 (x)v + b1 (x, v).
(2.6)
Put C1 = maxx∈ |c1 (x)| and C2 = maxx∈ |c2 (x)|. Obviously |b(x, 0, v)|C1 + C2 |v| + |b1 (x, v)|. Using the last inequality we rewrite (2.5) in the following way: ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 + g (z)|Dz|2 − |b1 (x, g (z)Dz)| 0. g (z) ⎣
(2.7)
i,j =1
We will split inequality (2.7) into two parts in order to obtain two independent problems with respect to g(z) and z(x). So instead of (2.7) we consider two inequalities ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 0 (2.8) g (z) ⎣ i,j =1
and g (z)|Dz|2 − |b1 (x, g (z)Dz)| 0. It is clear that from (2.8) and (2.9) follows (2.7).
(2.9)
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Let z(x) satisfies the problem (auxiliary problem) ˜ ≡ Lz
n
aij (x)zxi xj − C2 |Dz| = > 0,
i,j =1
z
j
= 0.
(2.10)
We suppose that j is Lipschitz and as a consequence satisfies a uniform exterior cone condition. Using the results of Miller [8,9] concerning the construction of barriers on exterior cones for elliptic equations and results of Crandall et al. [4] we will prove the existence of viscosity solution for problem (2.10). The main goal of the proof is to obtain an estimate for z. To assert that comparison hold for (2.10) one can suppose that aij (x), i, j = 1, . . . , n are uniformly continuous in x and so for F1 (x, p, X) = −Tr(A(x)X) + C2 |p| + we have F1 (x, p, X) − F1 (y, p, X)(|x − y|)(1 + |p| + X)
(2.11)
for some continuous : [0, ∞) → [0, ∞) satisfying (0) = 0. Consider Pucci minimizing and maximizing operators P − (X) = Trace(X + ) − Trace(X − ), P + (X) = Trace(X + ) − Trace(X − ), where X+ , X− denote the positive and the negative part of X ∈ S(n) and Trace(X + ) is the sum of the positive eigenvalues of X. It is obvious that F1 satisfies the following condition: P − (X − Y ) − C2 |p − q|F1 (x, p, X) − F1 (y, q, Y ) P + (X − Y ) + C2 |p − q|.
(2.12)
For n2 and ∈ (0, ) let T = {x ∈ R N : xn (cos )|x|} be the closed right circular cone of aperture . Let n
L0 u ≡
aij (x)uxi xj .
i,j =1
Consider barriers of the form r f ( ), where r = |x| and = arccos(xn /|x|). From results of [8, Theorems 2, 3] it follows that for every ∈ (0, ) there exist ∈ (0, 1) and f ∈ C 2 ([0, )) depending only on , , n, , such that f (0) = 0 and f > 1 on [0, ] so that L0 (r f ( )) − r −2 .
(2.13)
Put w(x) = −r f ( ). Then, from (2.13) it follows that L0 (w(x)) r −2 .
(2.14)
Deriving a polar representation for |Dw| =
|Dw| =
wr2 + w 2
wx21 + · · · + wx2n it is easy to show that
1 −1 = r 2 f 2 + f 2 . r2
(2.15)
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
Using (2.14), (2.15) one can easily see that −2 −1 ˜ −r 2 f 2 + f 2 . Lw r
235
(2.16)
Due to the fact that f and f are bounded it immediately follows that for every > 0 there exist r, small enough, such that ˜ > Lw
(2.17)
holds. We have that is a bounded domain in R n with Lipschitz boundary, so it satisfies a uniform exterior cone condition. It means that there exist ∈ (0, ) and r0 > 0 so that for every ∈ j there is a rotation = () such that ∩ Br0 () ⊂ + T .
(2.18)
Here Br0 () denotes the open ball in R n of radius r0 centered at . Setting w (x)=w(−1 (x −)) we have w (x) − |x − |
w () = 0,
on ∩ Br0 ().
(2.19)
In particular, we have
w (x) − r0
on {x ∈ : |x − | = r0 }.
(2.20)
Taking r0 sufficiently small that (2.17) holds we arrive at ˜ > Lw
for x ∈ ∩ Br0 ().
(2.21)
We need to extend these local barriers to global ones. Choose any point y = and 2r1 < d ≡ distance(y, j), where we assume that r1 > 1. For > 0 put 1 1 . (2.22) − G(x) = |x − y| r1 Obviously 2 − 1 1 . − G(x) − r1 (2r1 )
(2.23)
A standard computation [5] shows that ˜ LG(x)
(( + 1) − (n − 1) − C2 |x − y|) > 0 |x − y|+2
(2.24)
on for large depending only on n, , , j and diam(). Note that due to (2.23) we can choose = 0 such that (2.24) and the inequality 1 1 G(x) > − 0 > − r0 2 r1
for x ∈
(2.25)
hold. Let satisfies the inequality
0 (d + diam())0 +2
((0 + 1) − (n − 1) − C2 (d + diam())).
(2.26)
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A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
Hence, we have ˜ LG(x) > . Put
W (x) =
(2.27)
G(x) max(G(x), w (x))
for x ∈ , |x − | > r0 , for x ∈ , |x − | r0 .
(2.28)
From (2.19), (2.23) it follows that W (x) agrees with w (x) in a neighborhood of ∈ j relative to . From (2.20), (2.25) it follows that W (x) agrees with G(x) on \Br2 () for some 0 < r2 < r0 . At last in view of (2.21) and (2.27) we conclude that ˜ (x) > LW
on .
(2.29)
Thus we have constructed the global barrier W (x) for every ∈ j that has the following properties: W (x) ∈ C(),
W () = 0,
on \{}
W (x) > 0
(2.30)
and ˜ (x) > LW
on .
(2.31)
Finally we put W (x) = sup W (x).
(2.32)
∈j
(x) ≡ 0 will From (2.30) to (2.32) it follows that W (x) is a subsolution of (2.10). Obviously W be a supersolution of (2.10). Having the comparison holds we may invoke the Perron’s method to obtain the existence and uniqueness of the viscosity solution of (2.10). Due to the fact that W (x) is a subsolution and W˜ ≡ 0 is a supersolution of (2.10), we can estimate the function z(x) in the following way: (2.33)
W (x)z(x)0. Obviously from (2.20) and (2.25) we obtain 1 − 0 < z(x)0 r1
for x ∈ .
(2.34)
3. Solvability of the problem for the function g(z) Consider inequality (2.9). In order to construct twice differentiable function g(z) which will satisfy inequality (2.9) we need the following estimate |Dz| M, where M is some constant, to hold in . It is obvious that F1 (x, Dz, D 2 z)= ni,j =1 aij zxi xj −C2 |Dz|− satisfies the following inequality: |F1 (x, Dz, D 2 z) − F1 (x, D z¯ , D 2 z¯ )| = C2 ||D z¯ − |Dz|| C2 |D z¯ − Dz| n C2 |¯zxi − zx1 | i
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
237
for all zxi xj , zxi , z¯ xi , x and is nonincreasing in z. If we suppose that aij ∈ C (), i, j =1, . . . , n then from results of Safonov [10] it follows that there exists a classical solution z ∈ C 2+ ()∩C 0,1 () of problem (2.10) that coincides with the viscosity solution we have built above. Thus we have |Dz|M in . In the case when aij , i, j = 1, . . . , n are only uniformly continuous functions in x one can consider the Pucci minimizing operator P − (D 2 z∗ ) = Trace(D 2+ z∗ ) − Trace(D 2− z∗ ), where D 2+ z∗ , D 2− z∗ denote the positive and the negative part of D 2 z∗ ∈ S(n) and Trace(D 2+ z∗ ) is the sum of the positive eigenvalues of D 2 z∗ . It is obvious that P − (D 2 z∗ ) − C2 |Dz∗ |
n
aij (x)z∗xi xj − C2 |Dz∗ |.
(3.1)
i,j =1
Consider now the Dirichlet problem for Bellman equation P − (D 2 z∗ ) − C2 |Dz∗ | = , z∗ = 0.
(3.2)
In [10] Safonov proved that (3.2) has a classical solution z∗ ∈ C 2+ () ∩ C 0,1 () and from (3.1) it follows that this solution is a subsolution for z. Obviously z∗ ≡ 0 will be a supersolution for z. So we have constructed the subsolution and the supersolution which are Lipschitz continuous in and satisfy boundary condition. One can easily check that the existence of mentioned above sub and supersolutions guarantee that problem (2.10) has a viscosity solution z which is Lipschitz continuous in some neighborhood of the boundary j (see also [1]). It follows from [2] (see also 1, [4,11,12]) that (2.11) and (2.12) are enough to guarantee that the solution of (2.10) is Cloc () for all 0 < < ¯ = ¯(, , n) ∈ (0, 1). So, for the gradient of the viscosity solution of (2.10) we have that |Dz|M in . Suppose that the function b1 (x, g (z)Dz) satisfies the following structure condition: b1 (x, g (z)Dz) b˜1 (x, g (z)Dz)|Dz|2
(3.3)
for some continuous function b˜1 . Obviously one may have the estimate |b˜1 (x, g (z)Dz)| (g (z))
(3.4)
for some continuous positive function . From (3.3), (3.4) it follows that (2.9) holds if g (z) − (g (z))0
(3.5)
takes place. Consider the equation g (z) − (g (z)) = 0.
(3.6)
Firstly we will specify boundary conditions for (3.6) in order to construct g(z) such that inequality (2.7) will take place. Recall that we seek for a subsolution of (1.1), (1.2) in the form u∗ = g(z(x)), hence from (2.10) we obtain u∗ = g(z(x)) = g(0) = 0. (3.7) j
j
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A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
If z(x) is the solution of problem (2.10) and g(z) is the solution of (3.6), then (2.7) holds if g (z)
C1
(3.8)
for all x ∈ . From (3.6) we have that g (z) is a nondecreasing function of z. From (2.33) it follows that z varies on the interval [−K, 0], where K < 1/r10 depends on , j, , and diam(). Hence if we put g (z)(−K) =
C1 ,
(3.9)
then for all z ∈ [−K, 0] the function g(z) will satisfy inequality (3.8). From (3.6), (3.7) and (3.9) it follows that if g(z) satisfies the problem g (z) − (g (z)) = 0,
g(0) = 0,
g (z)(−K) =
C1 ,
(3.10)
then (2.7) holds. The main problem that arises here is the existence of a solution of (3.10).An obvious substitution g (z) = reduces (3.10) to the problem (z) − () = 0,
(−K) =
C1 .
(3.11)
It is well-known that if is a continuous function then there exists a solution of (3.11) at least in some neighborhood of the point z = −K, i.e. on the interval [−K − , −K + ] for some > 0 (further we will consider the existence of a solution of problem (3.11) only on [−K, −K + ]). It is clear that the interval of existence of a solution of (3.11) for fixed depends on and C1 . It is also clear that in general when C1 / → ∞ one may have → 0. Hence for arbitrary C1 / (3.10) fails to have a twice differentiable solution on [−K, 0]. Thus, in order to have existence for (3.10) one has to put some restriction on the smallness of C1 (for fixed ). We will formulate this restriction in the following form Lemma 1. There always exist , K, C1 such that there exist a twice differentiable solution of (3.10). Proof. Consider the problem (z) − () = 0,
(−K) = 0.
(3.12)
There exists a solution of (3.12) at least on [−K, −K + 0 ] for some 0 , which depends only on . Choose r1 > 1 and sufficiently large 0 to have 1 0 − + 0 > . r 0 2
(3.13)
Then from (3.13) it follows that −K + 0 >
0 . 2
(3.14)
Without loss of generality we can assume that is a Lipschitz function. Using the theorem on continuous dependence of solutions of (3.12) on the initial data we conclude that there
A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
239
exists = () such that if C1 ,
(3.15)
then problem (3.11) has a solution on [−K, 0]. So we have shown that there exists = () such that if (3.15) holds, then problem (3.10) has a twice differentiable solution on [−K, 0]. We summarize the results of this construction in the following Lemma 2. Let z(x) be a solution of problem (2.10) and g(z) be a solution of (3.10). If satisfies (2.26) and C1 satisfies (3.15), where depends only on , then u∗ = g(z(x)) will be a subsolution of problem (1.1), (1.2). The construction of a supersolution we carry out exactly in the same way as in the case of subsolution. We seek for a supersolution u∗ 0 of (1.1), (1.2) in the form u∗ (x) = h(z(x)),
(3.16)
where h(z) is a twice differentiable function in z with h (z) < 0, h (z) < 0. Hence Lu∗ − h (z)
n
aij (x)zxi xj − h (z)
i,j =1
n
aij (x)zxi zxj + b(x, 0, h (z)Dz)
i,j =1
0.
(3.17)
Using the inequality I A(x)I we obtain Lu∗ − h (z)
n
aij (x)zxi xj − h (z)|Dz|2 − |b(x, 0, h (z)Dz)| 0.
(3.18)
i,j =1
Using representation (2.6) for the function b(x, 0, v) we conclude that |b(x, 0, v)|C1 + C2 |v| + |b1 (x, v)|. So we can rewrite (3.18) in the following way: ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 − h (z)|Dz|2 Lu∗ − h (z) ⎣ i,j =1
− |b1 (x, h (z)Dz)|0.
(3.19)
We split inequality (3.19) into two parts in order to obtain two independent problems with respect to h(z) and z(x). So instead of (3.19) we consider two inequalities ⎡ ⎤ n aij (x)zxi xj − C2 |Dz|⎦ − C1 0 (3.20) −h (z) ⎣ i,j =1
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A. Tersenov / Nonlinear Analysis 65 (2006) 230 – 241
and −h (z)|Dz|2 − |b1 (x, h (z)Dz)| 0.
(3.21)
One can easily see that if z(x) satisfies problem (2.10) and h(z) is the solution of h (z) + (|h (z)|) = 0,
h(0) = 0,
h (z)(−K) = −
C1 ,
(3.22)
where is the same as in (3.4), then u∗ (x) is the supersolution of (1.1), (1.2). Thus we obtain that if u∗ = g(z(x)) is a subsolution of (1.1), (1.2) then u∗ = −u∗ is a supersolution of the same problem. Having the comparison holds we may invoke the Perron’s method to obtain the existence and uniqueness of the viscosity solution of (1.1), (1.2). Hence the following theorem holds Theorem 1. Let aij (x), b(x, u, Du) be continuous functions. Suppose that F (x, r, p, X) = −Tr(A(x)X) + b(x, r, p), where A(x) = {aij (x)}, is uniformly elliptic and satisfies (1.3)–(1.7). Suppose that b(x, 0, Du) satisfies (2.6), (3.3) and b(x, 0, 0) = c1 (x) is sufficiently small (in the sense of Lemmas 1, 2), then there exists a unique continuous viscosity solution of (1.1), (1.2). Remark 1. It is worth to mention that if |b(x, 0, ∇u)| c1 (x) + c2 (x)|∇u| (i.e. b1 (x, v) = 0 in (2.6)), then problem (1.1), (1.2) has a unique continuous viscosity solution without any smallness restriction on b(x, 0, 0) = c1 (x). In that case the subsolution u∗ = g(z(x)) will be a linear function z of the form u∗ = (c1 /)z. Remark 2. If we suppose, for example, that |b(x, 0, ∇u)| c1 (x) + c2 (x)|∇u| + c3 (x)|∇u|2 , where |c3 (x)|C3 and for z which is the solution of (2.10) we have |∇z| C4 then g(z) will be the solution of the problem g (z) − Mg = 0, 2
g(0) = 0,
g (−K) =
C1 ,
M = C3 C42 .
So the subsolution u∗ in that case will be − M M M u∗ = − K − z − ln − K ln , M C1 C1 where the smallness of b(x, 0, 0) = c1 (x) means that c1 (x)
. MK
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