existence and regularity of solutions to the minimal surface … · 2013-02-11 · 2 existence of...
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Existence and Regularity of Solutions to the MinimalSurface Equation
T. Begley, H. Jackson, J. Mathews, P. Rockstroh
February 11, 2013
1 Introduction
Minimal surfaces date back to the introduction of calculus of variations and were studied by bothEuler and Lagrange; indeed we will shortly present a derivation of the minimal surface equationas the Euler-Lagrange equations of the area functional on graphs. In this report we will beconcerned with establishing existence and regularity properies of solutions to the minimal surfaceequation. We proceed by constructing first a Lipschitz continuous solution, before showing thatthis solution necessarily satisfies additional regularity properties. There are many approaches tothe latter problem, the one we present very PDE theoretic and is based, in part, on the celebratedtheory of DeGiorgi, Nash and Moser. Finally we will drop the Lipschitz condition on our boundarydata and generalise to arbitrary continuous boundary conditions. As promised though, we firstderive the minimal surface equation by way of motivation.
1.1 Derivation of the Minimal Surface Equation
Suppose that Ω ⊂ Rn is a bounded domain (that is, Ω is open and connected). Fix φ : ∂Ω→ R,and introduce
L(Ω;φ) := u ∈ C0,1(Ω); u|∂Ω = φ, (1.1)
the set of Lipschitz functions on Ω whose restriction to ∂Ω is φ. We consider the problem ofminimising the area functional
A(u) :=
∫Ω
√1 + |Du|2 (1.2)
over functions u ∈ L(Ω;φ). Note that since Lipschitz functions are almost everywhere differ-entiable, we can make sense of this functional for all u ∈ L(Ω;φ). Suppose u ∈ L(Ω;φ) is aminimiser and that v ∈ C∞c (Ω). Clearly for all t ∈ R, u+ tv ∈ L(Ω;φ). Therefore the function
f : t 7→ A(u+ tv)
has a local minimum at t = 0, and so f ′(0) = 0. Differentiating f and evaluating at t = 0 leads to
f ′(0) =
∫Ω
Du ·Dv√1 + |Du|2
= 0, (1.3)
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which is precisely what it means for u to be a weak solution of the equation
div
(Du√
1 + |Du|2
)= 0.
Using the summation convention we may rewrite this in the following equivalent form
Mu := Di
(Diu√
1 + |Du|2
)= 0 in Ω. (1.4)
This is known as the minimal surface equation in divergence form and will be the object of ourstudy for the remainder of this report.
Remark. Returning to (1.3), we note that if we assume u ∈ C2(Ω), then integrating by partsyields
0 =
∫Ω
Du ·Dv√1 + |Du|2
= −∫
Ω
div
(Du√
1 + |Du|2
)v,
with the boundary terms vanishing since v is compactly supported in Ω. Since v ∈ C∞c (Ω) wasarbitrary, we conclude that u satisfies (1.4) in the classical sense as we would expect.
We now observe that equation (1.4) may be expanded as follows:
Mu = Di
(Diu√
1 + |Du|2
)=
[Diiu√
1 + |Du|2− DiuDijuDju
(1 + |Du|2)3/2
].
which may be equivalently expressed asn∑
i,j=1
(δij −
DiuDju
1 + |Du|2
)Diju = 0. (1.5)
From equation (1.5) we immediately recognize that this is a second order quasilinear PDE, sincethe coefficients of the highest order terms depend only on lower order terms. A quick calculationnow establishes that the minimal surface equation is elliptic. First we let
aij(x,Du) = δij −DiuDju
1 + |Du|2.
Recall that the PDE (1.5) is elliptic if for each x and u there exists a λ > 0 such that aij(x,Du)ξiξj ≥λ|ξ|2 for all ξ ∈ Rn. We observe
aijξiξj =
(δij −
DiuDju
1 + |Du|2
)ξiξj
= |ξ|2 − DiuξiDjuξj1 + |Du|2
Now by the Cauchy-Schwarz inequality we have (Du · ξ)2 ≤ |Du|2|ξ|2 which implies
aijξiξj ≥ |ξ|2(
1− |Du|2
1 + |Du|2
). (1.6)
Since |Du|2/(1 + |Du|2) < 1, (1.6) implies aijξiξj ≥ λξiξj which establishes that the PDE in (1.5)is elliptic.
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2 Existence of Solutions
In this section we prove the existence of Lipschitz continuous solutions u to the minimal surfaceequation (1.3), with given boundary data φ. We assume throughout that our domain Ω ⊂ Rn isbounded, with C2 boundary. For now we assume that φ ∈ C(∂Ω), later on we will need to makeadditional assumptions on the regularity of φ to deduce existence. Most of the material in thissection is based on [3] and [5], with many details expanded upon.
2.1 The functional A and its properties
We first prove a few basic properties of the area functional A,as defined in (1.2). Note that wecan make sense of A(u) for any u ∈ W 1,1(Ω). The following property will be useful in variousproofs throughout this section:
Theorem 2.1. For any function u ∈ W 1,1(Ω),∫Ω
√1 + |Du|2dx = sup
∫Ω
(gn+1 + u divg)dx; g ∈ C1c (Ω;Rn+1), ‖g‖∞ ≤ 1
. (2.1)
Proof. We first show that∫Ω
|Dv|dx = sup
∫ ′Ω
(v divg)dx; g ∈ C1c (Ω′;Rn), ‖g‖∞ ≤ 1
(2.2)
for all v ∈ W 1,1(Ω′) where Ω′ ⊂ Rn+1. To see this we use Cauchy Schwartz and integration byparts to get ∫ ′
Ω
(v divg)dx =
∫ ′Ω
(Dv) · (−g)dx ≤∫ ′
Ω
|Dv||g|dx ≤∫ ′
Ω
|Dv|dx,
and the other direction follows on choosing g to be approximations to −Dv/|Dv| in C1c . To
establish (2.1) let u ∈ C1(Ω) and introduce v := −xn+1 + u. Integrating by parts we have∫Ω
(gn+1 + u divg)dx =
∫Ω
(gn+1 −n+1∑i=1
∂u
∂xigi)dx =
∫Ω
(gn+1 −n∑i=1
∂u
∂xigi)dx =
∫Ω
(v divg)dx,
Noting that |Dv| =√
1 + |Du|2 and the result follows from (2.2).
Corollary 2.2 (Weak Lower Semicontinuity). Let ujj∈N ⊂ W 1,1(Ω) be a sequence of functionswhich converge in L1(Ω) to a function u ∈ W 1,1(Ω). Then
A(u) ≤ lim infj→∞
A(uj).
Proof. Let g ∈ C1c (Ω;Rn+1), with ‖g‖∞ ≤ 1. Then we have∫Ω
(gn+1 + u divg)dx = limj→∞
∫Ω
(gn+1 + uj divg)dx ≤ lim infj→∞
A(uj)
The first equality holds since g is compactly supported and continuously differentiable and sodivg is bounded on Ω. Weak lower semi-continuity now follows on taking the supremum over allg.
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Corollary 2.3 (Strict Convexity). The area functional A is strictly convex in the following sense:if u, v ∈ W 1,1(Ω) are such that Du 6= Dv then for all t ∈ (0, 1)
A(tu+ (1− t)v) < tA(u) + (1− t)A(v)
Proof. This follows immediately from the strict convexity of the function
x 7→√
1 + |x|2,
which follows from the fact that the Hessian is positive definite at every x.
We can make sense of the area functional for any u ∈ W 1,1(Ω), and in fact, any u ∈ BV(Ω). Forour purposes however it will be enough to restrict our attention to Lipschitz continuous functionson Ω, which are necessarily W 1,1(Ω) and so the area functional remains well defined on this classof functions.
Theorem 2.4 (Characterisation of W 1,∞). Let Ω be open and bounded, with ∂Ω of class C2 Thenu : Ω→ R is Lipschitz continuous if and only if u ∈ W 1,∞(Ω).
2.2 Lipschitz Functions and Minimisers of AThe next step is to choose a suitable space over which to minimise A. We want to choose asuitably large space in order that we may appeal to certain compactness properties in order toshow existence, while at the same time not discarding so much regularity that proving additionalregularity later becomes too hard. It turns out that the space of Lipschitz continuous functionsstrikes the right balance. We first present some definitions and useful facts. For a function uwhich is Lipschitz continuous on Ω, we denote its Lipschitz constant as
[u]Ω = sup
|u(x)− u(y)||x− y|
; x, y ∈ Ω, x 6= y
.
Definition. Let k > 0, and set Lk(Ω) to be the space of Lipschitz continuous functions withLipschitz constant less than or equal to k, i.e.
Lk(Ω) = u ∈ C0,1(Ω) : [u]Ω ≤ k.
For boundary data φ ∈ C(∂Ω), let L(Ω;φ) be the set of Lipschitz continuous functions whichagree with φ on ∂Ω, as defined in (1.1). Finally, let Lk(Ω;φ) be the set of Lipschitz continuousfunctions with Lipschitz constant less than or equal to k, and which agree with φ on ∂Ω, i.e.
Lk(Ω;φ) = Lk(Ω) ∩ L(Ω;φ).
Now we present two results about sequences of functions in these spaces. Combining them withthe Arzela-Ascoli theorem will allow us to conclude compactness of the space Lk(Ω;φ).
Lemma 2.5. Any sequence of functions uj ⊂ Lk(Ω) is equicontinuous in Ω.
Proof. Pick ε > 0, and let δ < εk. Then for any x, y ∈ Ω with |x− y| < δ, we have
|uj(x)− uj(y)| ≤ k|x− y| < kδ < ε,
for any j ∈ N. Hence the sequence is equicontinuous.
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Lemma 2.6. Any sequence of functions uj ⊂ Lk(Ω;φ) is uniformly bounded in Ω.
Proof. Let dmax := sup|x− y| : x, y ∈ Ω be the greatest distance between any two points in Ω.Since we are assuming that Ω is bounded, we know that dmax is finite. By compactness of ∂Ω andcontinuity of φ, we also know that ‖φ‖L∞(∂Ω) is finite. Therefore, using the uniform bound of kon the Lipschitz constants, we can see that
|uj(x)| ≤ ‖φ‖L∞(∂Ω) + kdmax <∞
for any x ∈ Ω and j ∈ N. Hence the sequence is uniformly bounded.
Using the above results, we can now very easily deduce the existence of a minimiser in the classLk(Ω, φ). From here it will take some extra work, as well as some additional assumptions on thedomain Ω, to show the existence of a minimiser in L(Ω, φ).
Theorem 2.7. Let φ ∈ C(∂Ω), and suppose that Lk(Ω, φ) is non-empty. Then there exists afunction u[k] ∈ Lk(Ω, φ) such that
A(u[k]) = infA(u) : u ∈ Lk(Ω, φ).
Proof. Let ujj∈N be a minimising sequence in Lk(Ω, φ). By Lemmas 2.5 and 2.6 the sequenceis equicontinuous and uniformly bounded. We can therefore apply the Arzela-Ascoli theorem tofind a subsequence which converges uniformly to a function u[k] ∈ L(Ω, φ). Since convergenceis uniform and Ω is bounded it follows that uj converge to u[k] in L1. Hence by weak lowersemicontinuity of A we deduce that
A(u[k]) ≤ lim infj→∞
A(uj) = infA(u) : u ∈ Lk(Ω, φ) ≤ A(u[k]),
which completes the proof.
Proposition 2.8. Suppose u[k] ∈ Lk(Ω;φ) is the minimum defined above for some k, that is
A(u[k]) = infA(v); v ∈ Lk(Ω, φ).
If in addition we have the strict inequality [u[k]]Ω < k, then u[k] is the minimum of A in L(Ω;φ),so
A(u[k]) = infA(v); v ∈ L(Ω, φ).
Proof. Let 0 ≤ t ≤ 1 and v ∈ L(Ω;φ), and set vt = u[k] + t(v − u[k]). Then note that vt|∂Ω = φand if t is sufficiently small then [vt]Ω < k. To see this, find ε > 0 such that [u[k]]Ω = k − ε andchoose N sucht that [v − u[k]]Ω ≤ N (which is possible, since both v and u[k] Lipschitz). Thentake t = ε/2N and use the triangle inequality.Since u[k] is the minimiser of A in Lk(Ω, φ), we must then have A(uk) ≤ A(vt). Thus by theconvexity of the area functional we have
A(uk) ≤ A(vt) ≤ (1− t)A(uk) + tA(v),
which, upon rearrangement, completes the proof.
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To finish the proof of existence it remains to show two things. First, we must show that for klarge enough, Lk(Ω;φ) is non-empty. This is trivial since we have assumed a C2 domain withcontinuous boundary data. Hence we can produce a Lipschitz function agreeing with φ on theboundary simply by solving the Dirichlet problem for the Laplacian. According to Proposition 2.8,to show the existence of a minimiser in L(Ω, φ) it is sufficient to obtain estimates on the Lipschitzconstant of the minimisers u[k] in Lk(Ω, φ). We do this by formulating a weak maximum principlewhich will allow us to restrict our attention to boundary estimates on the Lipschitz constant, andthen obtain the desired estimates using the notion of a barrier function.
2.3 Maximum principle
Definition (Supersolutions and Subsolutions). function w ∈ Lk(Ω) is called a supersolution(subsolution) for the area functional A if for all v ∈ Lk(Ω;w) with v ≥ w (v ≤ w) on Ω we haveA(v) ≥ A(w).
Proposition 2.9 (Weak maximum principle). Let w and z be a supersolution and a subsolutionof A in Lk(Ω) respectively. If w ≥ z on the boundary ∂Ω then w ≥ z in Ω.
Proof. Seeking a contradiction suppose that the maximum principle doesn’t hold, then the setK := x ∈ Ω : w(x) < z(x) is non empty. Let v := maxz, w. Then it is clear that v = w onthe boundary, v ∈ Lk(Ω) and v ≥ w in Ω. Thus, since w is a supersolution, we have A(v) ≥ A(w).On the set K, v = z and therefore the area functional on K satisfies A(z;K) ≥ A(w;K). Byconsidering minz, w we similarly see that A(z;K) ≤ A(w;K), so we deduce
A(z;K) = A(w;K).
Since z = w on the boundary ∂K (as z and w are continuous), and z > w in K, there exists a setof non zero measure where Dw 6= Dz. By the strict convexity of the area functional we thereforeget
A(w + z
2;K
)<
1
2A(w;K) +
1
2A(z;K) = A(w;K).
But since w is a supersolution in Lk(Ω) and (w + z)/2 ≥ w on K we also have
A(w + z
2;K
)≥ A(w;K),
giving the desired contradiction.
Proposition 2.10. Let w and z be a supersolution and a subsolution of A in Lk(Ω) respectively.Then we have
supx∈Ω
[z(x)− w(x)] ≤ supy∈∂Ω
[z(y)− w(y)]. (2.3)
Proof. First, we note that if w is a supersolution then so is w + α for any α ∈ R. To see thissuppose v ∈ Lk(Ω;w + α) and v ≥ w + α. Then v − α ∈ Lk(Ω;w) and v − α ≥ w so weget A(v − α) ≥ A(w). w + α is then a supersolution since A(v − α) = A(v). Now chooseα = supy∈∂Ω[z(y)− w(y)], and note that for any x ∈ ∂Ω,
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z(x) ≤ w(x) + supy∈∂Ω
[z(y)− w(y)].
After applying the maximum principle we get that the above equation holds for all x ∈ Ω, andtaking the supremum over all x gives (2.3).
Corollary 2.11. If u and v both minimise the area functional A in Lk(Ω) then
supx∈Ω|u(x)− v(x)| ≤ sup
x∈∂Ω|u(x)− v(x)|.
Proof. This comes immediately from Proposition 2.10, since any area minimising function is botha supersolution and a subsolution.
Proposition 2.12 (Reduction to boundary estimates). Suppose u minimises the area functionalin Lk(Ω), then we have
[u]Ω ≤ supx∈Ω,y∈∂Ω
|u(x)− u(y)||x− y|
.
Proof. Let x1, x2 be two distinct points in Ω and set r = x2−x1. If u minimises area in Lk(Ω) thenur(x) := u(x+ r) minimises area in Lk(Ωr), where Ωr := x+ r : x ∈ Ω. Note that x2 ∈ Ω ∩ Ωr
so the intersection is non empty, and that both u and ur minimise area in Lk(Ω ∩ Ωr). We canapply Corollary 2.11, which gives us that there is some z ∈ ∂(Ω ∩ Ωr) such that
|u(x1)− u(x2)| = |ur(x2)− u(x2)| ≤ |ur(z)− u(z)| = |u(z + r)− u(z)|.
Then dividing by |r| = |x1 − x2| we get
|u(x1)− u(x2)||x1 − x2|
≤ |u(z)− u(z + r)||r|
.
Since z ∈ ∂(Ω ∩ Ωr) then either z ∈ ∂Ω, or z ∈ ∂Ωr in which case z + r ∈ ∂Ω. Thus one of z orz + r lie on the boundary ∂Ω, so
|u(x1)− u(x2)||x1 − x2|
≤ |u(z)− u(z + r)||r|
≤ supx∈Ω,y∈∂Ω
|u(x)− u(y)||x− y|
,
and the result follows by taking the supremum on the left hand side.
2.4 Barriers and completion of proof of existence
To complete the proof, we introduce the notion of a barrier, for which it is necessary to assumethat φ is Lipschitz. We show that the existence of barriers is sufficient for existence of a solutionand then construct barriers under certain assumptions on Ω to prove that a solution exists.
Definition (Upper and lower barriers). Let x ∈ Ω, and let d(x) = dist(x, ∂Ω). For c > 0 define
Σc := x ∈ Ω : d(x) < c and Γc := x ∈ Ω : d(x) = c.
Let φ ∈ C0,1(∂Ω). An upper barrier v+ relative to φ is a Lipschitz function defined on Σco forsome co, which satisfies
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(i) v+|∂Ω = φ (v+ agrees with φ on the outer edge),
(ii) v+ ≥ sup∂Ω φ on Γc0 (v+ lies above φ on the inner edge),
(iii) v+ is a supersolution on Σco .
A lower barrier v− is a Lipschitz function if it satisfies (1), v− ≤ inf∂Ω φ on Γc0 and is a subsolutionon Σco .
Theorem 2.13. Let φ be a Lipschitz function on ∂Ω, and suppose upper and lower barriers v+
and v− (relative to φ) exist. Then the area functional A achieves its minimum on L(Ω;φ).
Proof. Let Q := max([v+]Σc0 [v−]Σc0
), and choose k > Q large enough that Lk(Ω, φ) is non-empty.
Following earlier arguments, there exists an area minimising function u ∈ Lk(Ω, φ). We claim thatin fact [u]Ω < k and thus via Proposition 2.8 that u is the desired minimiser of A on L(Ω, φ). Webegin by observing that u also minimises area in Lk(Σco). Moreover it follows that
inf∂Ωφ ≤ u(x) ≤ sup
∂Ωφ ∀x ∈ Ω
To see this suppose the set K := x ∈ Ω : sup∂Ω φ < u(x) is non-empty. Since u is continuous,it follows that K is open and furthermore since u ≤ sup∂Ω φ on ∂Ω, we have u = sup∂Ω φ on ∂K.Now u minimises A on K, which in turn implies that u ≡ sup∂Ω φ on K, which is a contradiction.In particular we have v−(x) ≤ u(x) ≤ v+(x) for x ∈ Γc0 , and thus by the weak maximum principle,we deduce
v−(x) ≤ u(x) ≤ v+(x)
for x ∈ Σc0 . Since v− = u = v+ on ∂Ω, it now follows that if x ∈ Σc0 and y ∈ ∂Ω, thenu(x) − u(y) ≤ v+(x) − v+(y) if u(x) ≥ u(y) and u(y) − u(x) ≤ v−(y) − v−(x) if u(x) ≤ u(y).Therefore by the definition of Q
|u(x)− u(y)| ≤ Q|x− y|
for x ∈ Σc0 and y ∈ ∂Ω.
We now extend this to arbitrary x ∈ Ω. Suppose d(x) > c0, since inf∂Ω φ ≤ u(x) ≤ sup∂Ω φwe have the following inequality for y ∈ ∂Ω
|u(x)− u(y)| ≤ max
sup∂Ω
φ− u(y), u(y)− inf∂Ωφ
≤ max
v+(z)− v+(y), v−(y)− v−(z)
for any z ∈ Γc0 . Choosing z with |z − y| = c0 we conclude that
|u(x)− u(y)| ≤ max
[v+]Σc0c0, [v−]Σc0c0
≤ Q|x− y|.
In other words, by Proposition 2.12, [u]Ω ≤ Q < k and therefore Proposition 2.8 implies that u isa minimiser in L(Ω, φ).
To complete the proof of existence, we will construct an upper barrier. This is sufficient since ifv is an upper barrier relative to −φ then −v is a lower barrier relative to φ. In order to do so weneed to make additional assumptions about the domain Ω, and in particular we must introducethe notion of mean curvature.
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Definition. Assume that ∂Ω is C2. For y ∈ ∂Ω we denote by ν(y) and T (y) respectively the innernormal to ∂Ω at y and the tangent hyperplane to ∂Ω at y. The curvatures at y are determinedas follows. By a rotation of the coordinates we may assume that the xn coordinate axis lies inthe direction ν(y). Then in some neighbourhood N = N (y) of y, ∂Ω is then given by xn = ψ(x′)where x′ = (x1, . . . , xn−1), ψ ∈ C2(T (y)∩N ) and Dψ(y′) = 0. The eigenvalues κ1, . . . , κn−1 of theHessian matrix (D2ψ(y′)) are called the principal curvatures of ∂Ω at y, and the mean curvatureof ∂Ω at y is given by
H(y) =1
n− 1
n−1∑i=1
κi
We use the following facts about the distance function d = dist(x, ∂Ω), the proofs of which canbe found in [2].
(i) For sufficiently small c0, d ∈ C2(Σc0).
(ii) If d(x) = c ≤ c0 then −∆d(x) = nH(x) where H(x) is the mean curvature of Γc at x.
(iii) If the mean curvature H of ∂Ω is non-negative at every point, then d is superharmonic inΣc0 (so ∆d ≤ 0).
The assumption of non-negative mean curvature, also referred to as mean convexity, combinedwith these facts will allow us to construct an upper barrier, under the additional assumption thatφ ∈ C2(∂Ω). Later, in Section 5 we will show that the regularity assumptions on φ can be relaxed.
Theorem 2.14 (Existence of a Lipschitz solution to the MSE). Let Ω be a bounded C2 domainwhose boundary has non negative mean curvature, and φ ∈ C2(∂Ω). Then there exists u ∈W 1,∞(Ω) which minimises the area functional A, i.e. u solves∫
Ω
Du ·Dη√1 + |Du|2
= 0
for all η ∈ C1c (Ω) and the boundary condition u|∂Ω = φ.
Proof. We will construct a barrier of the form
v(x) = φ(x) + ψ(d(x)),
for some ψ ∈ C2[0, R] to be determined, where R < c0, and c0 is as above. We choose ψ such thatψ(0) = 0 and
ψ(R) ≥ 2 supΩ
|φ| =: L
so that v satisfies the first and second properties of an upper barrier. We also specify that ψ′(t) ≥ 1and ψ′′(t) < 0 which we will make use of shortly in our calculations. We just need to show thatv is a supersolution, and it is sufficient to show that v satisfies
Mv =n∑i=1
Di
(Div√
1 + |Dv|2
)≤ 0
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for all x ∈ Ω, which is equivalent to
ζ(v) := (1 + |Dv|2)3/2Mv = (1 + |Dv|2)∆v −n∑
i,j=1
DivDjvDijv ≤ 0.
A tedious calculation which the reader is encouraged to check yields (where we have used sum-mation convention)
ζ(v) =(1 + |Dφ|2)∆φ−DiφDjφDijφ
+ψ′[2DiφDjd∆φ+ (1 + |Dφ|2)∆d−DidDjφDijφ−DiφDjφDijd]
+ψ′2[∆φ+ 2DiφDjd∆d−DidDjdDijφ] + ψ′3∆d
+ψ′′[1 + |Dφ|2 − (DiφDjd)2]. (2.4)
This follows on using the property that |Dd| = 1, so DidDijd = 0, and also noting that Diψ(d) =ψ′Did and Dijψ(d) = ψ′′DidDjd+ψ′Dijd. We now use the fact that ψ′ ≥ 1, so we have ψ′2 ≥ ψ′.We also note that φ and d are C2 in ΣR and hence the second order derivatives are bounded.Thus we can bound the first three terms in (2.4) (up to the term involving ψ′3) by Cψ′2 for someC, dependent on the region and φ. We now turn our attention to the term involving ψ′′, notingthat |Dφ|2− (DiφDjd)2 ≥ |Dφ|2−|Dφ|2|Dd|2 ≥ 0 by Cauchy-Schwartz, and since ψ′′ < 0 we thenget then final term is no greater than ψ′′. As a result, we reduce (2.4) to the inequality
ζ(v) ≤ ψ′′ + Cψ′2 + ψ′3∆d,
which further reduces toζ(v) ≤ ψ′′ + Cψ′2 (2.5)
since ∆d ≤ 0 in ΣR. It remains to choose ψ such that the right hand side of (2.5) is less than zero,and that ψ satisfies the technical conditions above. We can choose ψ by solving the differentialequation ψ′′ + Cψ′2 = 0 which gives
ψ(t) =1
Clog(1 + βt) + γ
We set γ = 0 so that ψ(0) = 0 and it is clear that ψ′′ < 0, so it remains to choose β and R suchthat v is an upper barrier, i.e such that ψ′ ≥ 1 and ψ(R) ≥ L. We calculate that
ψ′(t) =1
C
β
1 + βt≥ 1
C
β
1 + βR
and
ψ(R) =1
Clog(1 + βR),
thus we need β/(1+βR) ≥ C and βR ≥ eLC−1. One way of doing this is setting R = β−12 , which
gives us the inequalities β/(1 + β12 ) ≥ C and β
12 ≥ eLC − 1. By taking β sufficiently large (and
hence R sufficiently small) we construct an upper barrier, which completes the proof of existenceof a Lipschitz solution by Theorem 2.13.
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Note that the only part in the proof of existence that required ∂Ω to be C2 was in constructing anupper barrier. It’s a natural question to ask whether we can still be guaranteed a solution underweaker assumptions on the boundary. As we will show in the next section this is not the case. Ifonly a single point on the boundary has negative mean curvature then it is impossible to constructboundary data for which there is no minimiser. The situation for φ is better, and we will showlater how we can reduce the regularity assumptions on φ while still maintaining existence of aminimiser.
3 Necessity of mean convexity
In this section we show that mean convexity is essential in the proof of the existence theorem inSection 2. We will show that if there is a point where the mean curvature H is negative then wecan find some φ ∈ C2(∂Ω) such that there is no solution to the minimal surface equation withthese boundary data. To do this, we first present a version of the maximum principle.
Lemma 3.1. Let Ω be a bounded C2 domain and Γ a non-empty, closed subset of ∂Ω. Letu, v ∈ C2(Ω) ∩ C(Ω ∪ Γ) satisfy
Mv ≤Mu in Ω and v ≥ u on Γ.
Let Ωt := x ∈ Ω : dist(x, ∂Ω) ≥ t and let νt be the outward normal of Ωt. If for every open setV ⊃ Γ we have
limt→0+
∫∂Ωt\V
(1− Dv(y) · νt(y)√
1 + |Dv(y)|2
)dHn−1 = 0, (3.1)
then we havev ≥ u in Ω ∪ Γ.
Proof. Let ϕ = max0, u − v and ϕk = mink, ϕ, we want to show that ϕ ≡ 0 in Ω. SinceMv ≤Mu in Ω the same is true in Ωt for all t, since ϕk is positive we have∫
Ωt
(Mv −Mu)ϕk ≤ 0. (3.2)
DefineG(p) =
p√1 + |p|2
and Gi(p) =pi√
1 + |p|2.
Note that Mw = DiGi(Dw), so using integration by parts on (3.2) we get∫
∂Ωt
(G(Dv)−G(Du)) · νtϕk −∫
Ωt
(G(Dv)−G(Du)) ·Dϕk ≤ 0
Assume first v > u on Γ, and set V = x ∈ Ω : u(x) < v(x). Notice we have ϕk = Dϕk ≡ 0 onV and that V is open (as a subset of Ω) and contains Γ. Thus we have∫
Ωt\V(G(Du)−G(Dv)) ·Dϕk ≤
∫∂Ωt\V
(G(Du)−G(Dv)) · νtϕk. (3.3)
11
Now using the fundamental theorem of the calculus and the chain rule we see
Gi(Du)−Gi(Dv) =
∫ 1
0
d
dtGi(tDu+ (1− t)Dv)dt
=
∫ 1
0
∂Gi
dpj(tDu+ (1− t)Dv)
d
dt(tDju+ (1− t)Djv)dt
= Dj(u− v)
∫ 1
0
∂Gi
dpj(tDu+ (1− t)Dv)dt = Dj(u− v)bij (3.4)
where we defined
bij :=
∫ 1
0
∂Gi
dpj(tDu+ (1− t)Dv)dt.
Observe bij is elliptic. This follows from Section 1 since aij is elliptic due to Du and Dv beinguniformly bounded. Thus we have bijξiξj ≥ λt|ξ|2 for some λt > 0 for x ∈ Ωt. Proceeding, wehave∫
Ωt
λt|Dϕk|2 ≤∫
Ωt
bijDjϕkDiϕk by ellipticity
=
∫Ωt
bijDj(u− v)Diϕk since Dϕk = D(u− v) if 0 < u− v < k and 0 otherwise
=
∫Ωt\V
bijDj(u− v)Diϕk by definition of V
≤∫∂Ωt\V
(G(Du)−G(Dv)) · νtϕk by (3.3) and (3.4)
=
∫∂Ωt\V
(1−G(Dv) · νt)ϕk −∫∂Ωt\V
(1−G(Du) · νt)ϕk. (3.5)
Since |G(Du) ·νt| ≤ 1 by the Cauchy-Schwartz inequality,∫∂Ωt\V (1−G(Du) · νt)ϕk ≥ 0 and thus
(3.5) becomes∫Ωt
λt|Dϕk|2 ≤∫∂Ωt\V
(1−G(Dv) · νt)ϕk ≤ k
∫∂Ωt\V
(1−G(Dv) · νt) , (3.6)
where the second inequality follows from the definition of ϕk. We now let t 0. By hypothesisthe limit of the right hand side is zero so limt0
∫Ωtλt|Dϕk|2 = 0, from which we can conclude
|Dϕk| = 0. Since (3.6) holds for all values of k, we can take k →∞ and conclude that Dϕ ≡ 0 inΩ. This means that either u ≤ v in Ω, or u− v is constant in Ω If it is the second case then sinceu ≤ v on Γ we must have by continuity that u ≤ v in Ω. If we only have v ≥ u on Γ, we replacev by v + ε in the definition of V and then let ε→ 0 to get the result.
We now restate what we are trying to show, the non-existence of a solution.
Theorem 3.2. Let Ω be a bounded C2 domain and suppose that there is some x0 in ∂Ω with meancurvature H(x0) < 0. Then there exists φ ∈ C2(∂Ω) such that the Dirichlet problem
Mu = 0 in Ω
u|∂Ω = φ
has no solution u ∈ C2(Ω).
12
Using the version of the maximum principle we have just proved, we now show the theoremis true. We prove an inequality related to the values that u can take on the boundary and thenshow we can easily construct boundary data for which this does not hold.
Proof. We assume that u ∈ C2(Ω) is a solution to the Minimal Surface Equation in Ω. SinceH(x0) < 0 then by properties of the distance function we stated in §2.4 we have ∆d(x0) ≥ 0where d(x) = dist(x, ∂Ω) as before. Since d ∈ C2(Σc) for sufficently small c then there exists someR > 0 such that ∆d(x) > ε > 0 for x ∈ Ω ∩BR(x0). We claim that
sup∂Ω∩BR(x0)
u < sup∂Ω\BR(x0)
u+ C,
where C depends only on the domain Ω. To prove this we apply Lemma 3.1 to two differentfunctions. Let v(x) := α− βd(x)1/2 for x ∈ Ω ∩BR(x0) and let w(x) := λ− µ dist(x, ∂BR(x0))1/2
for x ∈ Ω\BR(x0), with constants to be chosen later. We firstly deal with v(x), we can calculatethat
Mv = − β
2√d(x) + β2/4
(∆d(x)− 1
2(d(x) + β2/4)
),
by checking that Div = −12βDidd
−1/2 and Dijv = −12βDijdd
−1/2 + 14βDiDjdd
−3/2 and recallingthat |Dd| = 1 and DidDijd = 0.
Since Mu = 0 to apply the lemma we need that Mv ≤ 0 in Ω ∩ BR(x0), and having β2 > 2ε
is sufficent since
∆d(x)− 1
2(d(x) + β2/4)> ε− 2
β2> 0.
Figure 1: Diagram of Ω and BR(x0)
We want to use the Lemma in Ω1 := Ω ∩BR(x0) and Γ1 = Ω ∩ ∂BR(x0). We check that (3.1)holds, to see this note that we have νt(y) = −Dd(y) and so the integrand reduces to
1− Dv(y) · νt(y)√1 + |Dv(y)|2
= 1 +βDd(y) · νt(y)
2√d(y) + β2/4
= 1− β
2√d(y) + β2/4
.
13
which is zero on the boundary of Ω. To apply the Lemma we just need to choose α > 0 such thatv(x) ≥ u(x) on Γ1, and since α ≥ v(x) ≥ α− β
√diam(Ω) we can choose
α = supΓ1
u+ β√
diam(Ω).
Applying the Lemma we then get for x ∈ Ω1 ∪ Γ1 that
u(x) ≤ v(x) ≤ supΓ1
u+ β√
diam(Ω).
Thus we get upon taking a supremum in Ω1 that
supΩ1
u ≤ supΩ∩∂BR(x0)
u+ β√
diam(Ω). (3.7)
Finally we can use the maximum principle (see [6]) which gives us supΩ1u = sup∂Ω1
u and thussince ∂Ω ∩BR(x0) ⊂ ∂Ω1 from (3.7) we have
sup∂Ω∩BR(x0)
u ≤ supΩ∩∂BR(x0)
u+ β√
diam(Ω). (3.8)
We now carry out a similiar procedure for w(x) with x ∈ Ω2 := Ω\BR(x0). We can see thatw(x) = λ − µ(|x − x0| − R)1/2. We again need that Mv ≤ 0 in Ω2 and a similiar calculation asbefore (with d(x) = |x− x0| −R) gives
Mv = − µ
2√|x− x0| −R + µ2/4
(n− 1
|x− x0|− 1
2(|x− x0| −R + µ2/4)
).
Sincen− 1
|x− x0|− 1
2(|x− x0| −R + µ2/4)≥ n− 1
|x− x0|− 2
µ2,
we can ensure the above expression is positive and hence Mv ≤ 0 by taking
µ >
√2 diam(Ω)
n− 1.
By similiar logic to before, we have w(x) ≥ λ− µ√
diam(Ω) so choosing
λ = supΓ2
u+ µ√
diam(Ω)
ensures that v(x) ≥ u(x) in Γ2 := ∂Ω\BR(x0). We can check that the integral in (3.1) is zero asbefore by showing the integrand is zero on the boundary, so we can apply the Lemma to get that
u(x) ≤ w(x) ≤ supΓ1
u+ µ√
diam(Ω)
Finally taking the supremum as before and applying the maximum principle in Ω2 we concludethat
supΩ∩∂BR(x0)
u ≤ sup∂Ω\BR(x0)
u+ β√
diam(Ω) (3.9)
since Ω ∩ ∂BR(x0) ⊂ ∂Ω2. Combining (3.8) and (3.9) gives us the result we wanted,
sup∂Ω∩BR(x0)
u < sup∂Ω\BR(x0)
u+ C. (3.10)
14
In fact we have just proved that we can findφ ∈ c∞(Ω) for which there is no solution to theDirichlet problem. For example choose φ ∼ e−x
2
with peak at 2C in ∂Ω ∩ BR(x0) and less than Cin ∂Ω\BR(x0) .
4 Regularity
Having established the existence of a Lipschitz continuous solution, which we will henceforthdenote u, we now turn to the issue of regularity. Our goal is to show that the solution of theminimal surface equation that we have constructed is in fact smooth. This takes a lot of work,but the idea can be summarised roughly as follows. The first step is to prove that the solution uis in fact in W 2,2
loc . Having done this we can rewrite the minimal surface equation to obtain a linearequation for the derivatives of u. We apply the theory of DeGiorgi-Nash-Moser to conclude thatthe derivatives are Holder continuous. Since the coefficients in the minimal surface equation aregiven in terms of the derivatives of u, we may conclude that the coefficients are Holder continuous.This will then allow us to show that u is C2,α and hence a classical solution of the minimal surfaceequation. We finally conclude by appealing to Schauder estimates and a bootstrapping argumentto show that u is C∞.
4.1 Additional Regularity of the Solution
We will use following results to show that the solution is in W 2,2loc .
Definition. Assume that u ∈ L1loc(Ω), and Ω′ ⊂⊂ Ω. Then the i-th difference quotient of size h,
where 0 < |h| < dist(Ω′, ∂Ω), is defined as
δhi u(x) =u(x+ hei)− u(x)
h,
for i = 1, . . . , n and x ∈ Ω′. We also define δh := (δh1 , . . . , δhn).
The first two theorems we will only state, proofs may be found in Evans [1].
Theorem 4.1. (i) Suppose that 1 ≤ p <∞ and u ∈ W 1,p(Ω). Then for each Ω′ ⊂⊂ Ω
‖δhu‖Lp(Ω′) ≤ C‖Du‖Lp(Ω)
for some constant C and for all 0 < |h| < 12dist(Ω′, ∂Ω).
(ii) Assume that 1 < p <∞, u ∈ Lp(Ω′), and that there exists a constant C such that
‖δhu‖Lp(Ω′) ≤ C,
for all 0 < |h| < 12dist(Ω′, ∂Ω). Then
u ∈ W 1,p(Ω′), with ‖Du‖Lp(Ω′) ≤ C.
15
Theorem 4.2. Let u ∈ W 1,2(Ω) ∩W 1,∞(Ω) satisfy∫Ω
Fi(Du)Diϕ = 0, (4.1)
for all φ ∈ W 1,20 (Ω) with φ ≥ 0, where F is a smooth function satisfying the following two
conditions:
(i) DiFj = DjFi for all i, j,
(ii) There exist c, C > 0 such that c|ξ|2 ≤∑n
i,j=1DjFiξiξj ≤ C|ξ|2 for all ξ ∈ Rn, for all x ∈ Ω.
These conditions can be thought of as local uniform ellipticity conditions on F , and under theseassumption we have u ∈ W 2,2
loc .
Proof. We begin by constructing a suitable test function ϕ ∈ W 1,20 (Ω). In order to do so, let
η ∈ C∞c (Ω) and |h| < 12
dist(∂Ω, supp η) and define ϕ := δ−hk (η2δhku), where k ∈ 1, . . . , n.Also, note that δhk commutes with Dj and furthermore δhi commutes with D when δhi is appliedcomponentwise. A quick calculation shows that by substituting our choice of test function into(4.1) we get ∫
Ω
Fi(Du)Diδ−hk (η2δhku)dx = −
∫Ω
δhkFi(Du)Di(η2δhku)dx,
since η is compactly supported in Ω and 0 < |h| < 12
dist(∂Ω, supp η). By (4.1), the LHS of theabove expression is 0, so we have ∫
Ω
δhkFi(Du)Di(η2δhku)dx = 0. (4.2)
Now note that
Fi(Du(x+ hek))− Fi(Du(x)) = Fi(Du(x) + hδhkDu(x))− Fi(Du(x)).
Hence, by the fundamental theorem of calculus,
δhkFi(Du(x)) =1
h
∫ 1
0
d
dtFi(Du(x) + thδhkDu(x))dt
= Djδhku(x)
∫ 1
0
DjFi(Du(x) + thδhkDu(x))dt.
Now set
θij :=
∫ 1
0
DjFi(Du+ thδhkDu)dt.
Now if we set v := δhku, then we may rewrite (4.2) as∫Ω
θijDjvDi(η2v)dx = 0. (4.3)
16
Suppose we are given Ω1 ⊂⊂ Ω, then we can find sets Ω2 and Ω3 such that Ω1 ⊂⊂ Ω2 ⊂⊂ Ω3 ⊂⊂Ω. In addition, choose |h| < mink dist(∂Ωk,Ωk+1). Now choose η ∈ C∞c (Ω2) such that η = 1 onΩ1. By the ellipticity assumption, we find that there exist constants c and C such that
c|ξ|2 ≤ θijξiξj ≤ C|ξ|2 (4.4)
in Ω2. Now since η = 1 on Ω1, we have∫Ω1
|Dv|2 ≤∫
Ω2
|D(ηv)|2.
Using the first inequality in (4.4), we also have∫Ω2
|D(ηv)|2 ≤ 1
c
∫Ω2
θijDj(vη)Di(vη),
and moreover by the first ellipticity condition, we have θij = θji which implies
1
c
∫Ω2
θijDj(vη)Di(vη) =1
c
∫Ω2
θij(v2DjηDiη + 2ηvDjvDiη + η2DjvDiv).
Applying the product rule and the fact that∫
ΩθijDjvDi(η
2v)dx = 0 and (4.3) we find
1
c
∫Ω2
θij(v2DjηDiη + 2ηvDjvDiη + η2DjvDiv) =
1
c
∫Ω2
θijv2DjηDiη +
1
c
∫Ω
θijDjvDi(η2v)
=1
c
∫Ω
θijv2DjηDiη.
Finally using the second inequality in (4.4)
1
c
∫Ω
θijv2DjηDiη ≤
C
c
∫Ω2
v2|Dη|2 ≤ C
∫Ω2
v2,
and therefore ∫Ω1
|Dv|2 ≤ C
∫Ω2
v2. (4.5)
Now since v = δhku and u ∈ W 1,2loc (Ω) we have that v ∈ L2(Ω2) and
‖v‖L2(Ω2) ≤ C‖Du‖L2(Ω3) =: K.
By (4.5) and the above result, we have ∫Ω1
|Dv|2 ≤ CK. (4.6)
Hence, ‖δhi Du‖L2(Ω1) ≤ K which implies ‖D2u‖L2(Ω1) ≤ K and therefore u ∈ W 2,2loc (Ω).
Applying the above results to our specific situation, we first note that u is Lipschitz and thus isW 1,∞(Ω). Hence since Ω is bounded, we also have that u ∈ W 1,2(Ω). We can therefore applytheorem 4.2 with F (p) := p/
√1 + |p|2 which satisfies the ellipticity conditions to conclude that
u ∈ W 2,2loc .
17
4.2 DeGiorgi-Nash-Moser
4.2.1 Overview
The proof proceeds in multiple stages, many of which are very technical, so in the interest ofclarity we present a brief informal overview of the strategy. We have just shown that u ∈ W 2,2
loc .With this result in hand we perform the following trick. Fix some k = 1, . . . , n, let w := Dku andchoose a test function of the form Dkφ for φ ∈ C∞c (Ω). Then substituting this into (1.3) we have
0 =
∫Ω
DiuDi(Dkφ)√1 + |Du|2
= −∫
Ω
aij(x)DjwDiφ,
where
aij(x) =δij√
1 + |Du|2− DiuDju
(1 + |Du|2)32
.
In other words, w is a weak solution of the equation
Di(aij(x)Dju) = 0. (4.7)
Moreover, this equation is linear! The coefficients depend only on x since the solution u is fixed.In addition to this, we also have ellipticity and boundedness of the coefficients aij. Observe
|aij(x)| =
∣∣∣∣∣ δij√1 + |Du|2
− DiuDju
(1 + |Du|2)32
∣∣∣∣∣ ≤ 1√1 + |Du|2
(1 +
|Du|2
1 + |Du|2
)≤ 2,
and
aijξiξj =1√
1 + |Du|2
(|ξ|2 − DiuξiDjuξj
1 + |Du|2
)≥ |ξ|2√
1 + |Du|2
(1− |Du|2
1 + |Du|2
)≥ λ|ξ|2,
where
λ = infΩ
1
(1 + |Du|2)32
> 0.
This last inequality follows since u ∈ W 1,∞(Ω). We now study the function w as a weak solutionof the uniformly elliptic linear equation with bounded coefficients (4.7), and show that w is Holdercontinuous, which of course implies that u ∈ C1,α. This is done by appealing to the DeGiorgi-Nash-Moser theory. We first prove a local boundedness result using a powerful technique calledMoser iteration. Specifically we show that for any ball BR ⊂⊂ Ω and a subsolution u ∈ W 1,2(BR)we have that u+ ∈ L∞loc(BR) together with an estimate of the form
supBr
u+ ≤ C1
(R− r)np
‖u+‖Lp(BR).
This will then allow us to prove a weak Harnack inequality of the form
supBR
u ≤ C infBR/2
u.
Using this we can then show that the oscillation of u (which we will define in due course) decaysand this will allow us to deduce Holder continuity. Finally it is an easy matter to apply thistheory to the function w and conclude the desired result.
18
4.2.2 Local Boundedness
From here on we consider a uniformly elliptic second order partial differential equation withbounded coefficients written in divergence form which takes the following form
Lu = Di(aij(x)Dju) = 0. (4.8)
This is of course precisely the form of the equation (4.7), however, we do not need to appeal to thespecific structure of the minimal surface equation to prove the results which follow. We introducethe notions of weak subsolutions and supersolutions of (4.8).
Definition. We say that u ∈ W 1,2(Ω) is a subsolution (respectively supersolution) of (4.8) if forany φ ∈ W 1,2
0 (Ω) with φ ≥ 0 we have∫Ω
aijDiuDjφ ≤ 0 (resp. ≥ 0).
Theorem 4.3 (Local Boundedness). Suppose that aij ∈ L∞(B1) with ‖aij‖L∞ ≤ Λ and
aij(x)ξiξj ≥ λ|ξ|2 for all x ∈ B1 ξ ∈ Rn,
for some Λ and λ positive. Suppose that u ∈ W 1,2(B1) is a subsolution. Then u+ ∈ L∞loc(B1).Moreover, for each θ ∈ (0, 1) and p > 0 we have
supBθ
u+ ≤ C
(1− θ)np
‖u+‖Lp(B1),
where C = C(n, λ,Λ, p) is a positive constant.
Remark. As mentioned in the overview, we will use a powerful technique known as Moser it-eration to prove this. The idea is essentially as follows. By a clever choice of test function weestimate the Lp1 norm of u+ on a ball Br1 by the Lp2 norm of u+ on a larger ball Br2 wherep1 > p2, that is,
‖u+‖Lp1 (Br1 ) ≤ C‖u+‖Lp2 (Br2 ).
We iterate this process for a careful choice of sequences ri and pi with ri r > 0 and pi → ∞to get the inequality we desire. The details now follow, but we will first state a technical lemmathat will be needed towards the end of the proof of Theorem 4.3.
Lemma 4.4. Suppose that f(t) is bounded in [τ0, τ1], with τ0 ≥ 0. Suppose that for all τ0 ≤ t <s ≤ τ1 we have
f(t) ≤ θf(s) +A
(s− t)α(4.9)
for some fixed θ ∈ [0, 1). Then for all τ0 ≤ t < s ≤ τ1 we have
f(t) ≤ c(α, θ)A
(s− t)α.
19
Proof. Fix τ0 ≤ t < s ≤ τ1. Let τ ∈ (0, 1) and consider the sequence (tn)∞n=0 defined as follows:
t0 = t tn+1 = tn + (1− τ)τn(s− t).
Notice that limn→∞ tn = s and that (tn) is an increasing sequence. Applying (4.9) repeatedly weget that
f(t) = f(t0) ≤ θnf(tn) +A(s− t)−α
(1− τ)α
n−1∑k=0
θkτ−kα.
Now choose τ such that θτ−α < 1, then letting n→∞ we obtain
f(t) ≤ c(α, θ)A
(1− τ)α(s− t)−α.
Proof of Theorem 4.3. Step 1: We will first prove the special case θ = 12
and p = 2. Let k,m > 0and define the functions u := u+ + k and
um :=
u if u < mk +m if u ≥ m
.
Note Dum = 0 if u < 0 or u ≥ m and furthermore um ≤ u. Let β ≥ 0, η ∈ C1c (B1) and introduce
the test functionφ := η2( uβm u− kβ+1) ∈ W 1,2
0 (B1).
Since u ≥ um ≥ k it is clear that φ ≥ 0. Computing Dφ we see
Dφ = βη2 uβ−1m uD um + η2 uβmD u+ 2ηDη( uβm u− kβ+1)
= η2 uβm(βD um +D u) + 2ηDη( uβm u− kβ+1),
where we used the fact that u = um if u < m and D um = 0 if u ≥ m. Note that φ = 0 andDφ = 0 if u ≤ 0. Moreover u is a subsolution since u+ is and Di u = Diu
+. Therefore we have
0 ≥∫aijDi uDjφ =
∫aijDi u(βDj um +Dj u)η2 uβm + 2
∫aijDi u( uβm u− kβ+1)ηDjη. (4.10)
We have omitted the domain of integration for now, since these estimates hold when integratingover any ball Br(y) ⊂ B1. Using uniform ellipticity we also have the estimates
β
∫η2 uβmaijDi uDj um ≥ λβ
∫η2 uβm|D um|2, and (4.11)∫
η2 uβmaijDi uDj u ≥ λ
∫η2 uβm|D u|2. (4.12)
since either Di u = Di um or Dj um = 0 and aijDi umDj um ≥ λ|Dum|2. Furthermore, we may useHolder’s inequality to obtain the following estimate for the second integral in (4.10), after firstusing the Cauchy-Schwarz inequality.
2
∫aijDi u( uβm u− kβ+1)ηDjη ≥ −2Λ
∫η uβm u|D u||Dη|
≥ −2Λ
(∫|D u|2η2 uβm
) 12(∫|Dη|2 u2 uβm
) 12
. (4.13)
20
Combining the last three inequalities with (4.10), we get
0 ≥ λβ
∫η2 uβm|D um|2 + λ
(∫η2 uβm|D u|2 − 2Λ
λ
(∫|D u|2η2 uβm
) 12(∫|Dη|2 u2 uβm
) 12
)
≥ λβ
∫η2 uβm|D um|2 +
λ
2
∫η2 uβm|D u|2 − 2Λ2
λ
∫|Dη|2 u2 uβm,
where we have used the inequality ab ≤ a2
2+ b2
2with a =
(∫|D u|2η2 uβm
) 12 and b = 2Λ
λ
(∫|Dη|2 u2 uβm
) 12 .
We deduce
β
∫η2uβm|Dum|2+
∫η2uβm|Du|2 ≤ 2
(β
∫η2 uβm|D um|2 +
1
2
∫η2 uβm|D u|2
)≤ C
∫|Dη|2u2uβm.
(4.14)
Set w := uβ2m u. Then note that
|Dw|2 =
∣∣∣∣β2 uβ2−1
m uD um + uβ2mD u
∣∣∣∣2≤(β
2| um|
β2 |D um|+ | um|
β2 |D u|
)2
= | um|β|D u|2 + β| um|β|D um|2 +β2
4| um|β|D um|2
≤ (1 + β)(β uβm|D um|2 + uβm|D u|2).
The second line follows the fact that if u 6= um then D um = 0. Then from (4.14) we have∫|Dw|2η2 ≤ C(1 + β)
∫|Dη|2 u2 uβm = C(1 + β)
∫|Dη|2w2,
which implies by the product rule that∫|D(wη)|2 ≤ C(1 + β)
∫|Dη|2w2.
We apply the Gagliardo-Nirenberg-Sobolev inequality with χ := nn−2
> 1 if n > 2 or χ > 2 ifn = 2, and obtain (∫
|wη|2χ) 1
χ
≤∫|D(wη)|2 ≤ C(1 + β)
∫|Dη|2w2.
We will now choose a particular cut-off function η. Let 0 < r < R ≤ 1 and choose η ∈ C1c (BR)
with
η ≡ 1 in Br and |Dη| ≤ 2
R− r.
Then (∫Br
w2χ
) 1χ
≤ C1 + β
(R− r)2
∫BR
w2.
21
Recalling how w was defined, this is the same as(∫Br
u2χ uβχm
) 1χ
≤ C1 + β
(R− r)2
∫BR
u2 uβm.
Set γ := β + 2 ≥ 2 and recall that um ≤ u so that(∫Br
uγχm
) 1χ
≤ Cγ − 1
(R− r)2
∫BR
uγ.
Let m→∞ and conclude
‖ u‖Lγχ(Br) ≤(C
γ − 1
(R− r)2
) 1γ
‖ u‖Lγ(BR),
where C = C(n, λ,Λ) is independent of γ.Step 2: Define for i = 0, 1, . . . the sequences (γi) and (ri) by
γi := 2χi, ri :=1
2+
1
2i+1.
Then we have ri − ri−1 = 1/2i+1 and γi = χγi−1, so
‖ u‖Lγi (Bri ) ≤(C
(2χi−1 − 1)
( 12i+1 )2
) 1
2χi−1
‖ u‖Lγi−1 (Bri−1 )
≤ (32C)1
2χi−1 (4χ)i−1
2χi−1 ‖ u‖Lγi−1 (Bri−1 ).
Importantly, C depends only on n, λ and Λ! Therefore we can iterate this l times to obtain
‖ u‖Lγl (Brl ) ≤ (16C)∑li=1
1
2χi−1 (4χ)∑li=1
i−1
2χi−1 ‖ u‖Lγ0 (Br0 ) ≤ C‖ u‖L2(B1),
where we have used the fact that the series∑∞
i=11
2χi−1 and∑∞
i=1i−1
2χi−1 are both convergent sinceχ > 1. In particular
‖ u‖Lγl (B1/2) ≤ C‖ u‖L2(B1)
for all l. Now let l→∞. We have γl →∞ also and therefore
supB1/2
u ≤ C‖ u‖L2(B1).
Letting k 0 in the definition of u we arrive at the desired result
supB1/2
u+ ≤ C‖u+‖L2(B1).
Step 3: We now seek to generalise for p > 0 and θ ∈ (0, 1). We proceed by applying a dilationargument. Let R ≤ 1 and define u(y) := u(Ry) for y ∈ B1. Then it is trivial to verify that∫
B1
aijDiuDjφ ≤ 0 ∀φ ∈ W 1,20 (B1), with φ ≥ 0,
22
where aij(y) = aij(Ry). Notice that
‖aij‖L∞(B1) = ‖aij‖L∞(BR) ≤ Λ,
andaij(y)ξiξj ≥ λ|ξ|2 ∀ξ ∈ Rn, ∀y ∈ B1.
Hence we may apply the first part of the proof to u, and we write the resulting estimate in termsof u. This yields, for p ≥ 2 the following estimate
supBR/2
u+ ≤ C
Rnp
‖u+‖Lp(BR),
where C = C(n, λ,Λ, p) > 0. For θ ∈ (0, 1), we apply the above result on B(1−θ)R(y) for eachy ∈ BθR. This leads to
u+(y) ≤ supB((1−θ)R)/2(y)
u+ ≤ C
((1− θ)R)np
‖u+‖Lp(B(1−θ)R(y)) ≤C
((1− θ)R)np
‖u+‖Lp(BR),
which implies
supBRθ
u+ ≤ C
((1− θ)R)np
‖u+‖Lp(BR).
For R = 1 this is precisely the stated result for p ≥ 2.Step 4: Our final goal is to extend this result to p ∈ (0, 2) also. Notice first that∫
BR
(u+)2 ≤ ‖u+‖2−pL∞(BR)
∫BR
(u+)p,
and so
‖u+‖L∞(BθR) ≤C
((1− θ)R)n2
‖u+‖1− p2
L∞(BR)
(∫BR
(u+)p) 1
2
.
We apply Young’s inequality
ab ≤ ε
qaq +
1
εrq rbr for all a, b ≥ 0
1
q+
1
r= 1 ε > 0,
where we choose
a = ‖u+‖1− p2
L∞(BR), b =C
((1− θ)R)n2
(∫BR
(u+)p) 1
2
, q =2
2− p, r =
2
p, ε =
q
2.
This yields
‖u+‖L∞(BθR) ≤1
2‖u+‖L∞(BR) +
C
((1− θ)R)np
‖u+‖Lp(BR).
Setting f(t) := ‖u+‖L∞(Bt) for each t ∈ (0, 1] we have shown that for any 0 < r < R ≤ 1
f(r) ≤ 1
2f(R) +
C
(R− r)np
‖u+‖Lp(B1).
23
Therefore we may invoke Lemma 4.4 to deduce
f(r) ≤ C
(R− r)np
‖u+‖Lp(B1).
Let R 1, then for all θ < 1 we have
‖u+‖L∞(Bθ) ≤C
(1− θ)np
‖u+‖Lp(B1)
as desired, thus we are done.
An immediate consequence of Theorem 4.3 is the following which we obtain by a scalingargument.
Theorem 4.5. Let Ω be a bounded domain in Rn and
Lu = Di(aij(x)Dju),
where ‖aij‖L∞(Ω) ≤ Λ and aij(x)ξiξj ≥ λ|ξ|2 for all x ∈ Ω, ξ ∈ Rn. Suppose that u ∈ W 1,2(Ω) isa subsolution of L in Ω, that is ∫
Ω
aijDiuDjφ ≤ 0
for all φ ∈ W 1,20 (Ω) with φ ≥ 0 in Ω. Then for any ball BR(x) ⊂ Ω, r < R and p > 0 we have
supBr
u+ ≤ C
(R− r)np
‖u+‖Lp(BR), (4.15)
where C = C(n, λ,Λ, p) is positive.
4.2.3 Weak Harnack Inequality
Theorem 4.6 (John-Nirenberg Lemma). Suppose u ∈ L1(Ω) satisfies∫Br(x)
|u− uBr(x)| ≤Mrn ∀Br(x) ⊂ Ω,
where uBr(x) = 1|Br(x)|
∫Br(x)
u(y)dy. Then for any Br(x) ⊂ Ω∫Br(x)
ep0|u−uBr(x)|/M ≤ Crn,
for some positive p0 and C depending only on n.
For the proof see [4].
24
Theorem 4.7 (Weak Harnack Inequality). Suppose that aij ∈ L∞(Ω) with λ|ξ|2 ≤ aijξiξj ≤ Λ|ξ|2for some 0 < λ ≤ Λ < ∞ and suppose that u ∈ W 1,2(Ω) is a non-negative supersolution of theequation Dj(aijDiu) = 0 in Ω, that is∫
Ω
aijDiuDjφ ≥ 0 ∀φ ∈ W 1,20 (Ω) φ ≥ 0. (4.16)
Then for all BR ⊂ Ω, 0 < p < n/(n− 2) and 0 < θ < τ < 1 we have
infBθR
u ≥ C
(1
Rn
∫BτR
up) 1
p
, (4.17)
where C = C(n, p, λ,Λ, θ, τ).
Proof. We will prove the case R = 1. As with Local Boundness (Theorem 4.3) the general resultwith follow from a scaling argument. The proof proceeds in two steps.Step 1: We show first that the statement holds for some p0 > 0. Set u := u+ k for some k > 0so that u > 0, and define v := u−1. We start by deriving an equation for v. Let φ ∈ W 1,2
0 (B1)with φ ≥ 0 and use u−2φ as the test function in (4.16). We have∫
B1
aij u−2DiuDjφ− 2
∫B1
aijφ u−3DiuDj u ≥ 0.
We note that D u = Du and Dv = − u−2D u and therefore since aij ≥ 0∫B1
aijDivDjφ ≤ 0.
Applying Theorem 4.5 we getsupBθ
v ≤ C‖v‖Lp(Bτ )
where C = C(θ, τ, p, n, λ,Λ) > 0. In other words
infBθ
u ≥ C
(∫Bτ
u−p)− 1
p
= C
(∫Bτ
u−p∫Bτ
up)− 1
p(∫
Bτ
up) 1
p
.
If we can show that there exists some p0 > 0 such that∫Bτ
u−p0∫Bτ
up0 ≤ C(n, λ,Λ, τ) (4.18)
then (4.17) holds in the case p = p0. We will show that there is some p0 > 0 such that for allτ < 1 ∫
Bτ
ep0|w| ≤ C(n, λ,Λ, τ), (4.19)
where w := log u− |Bτ |−1∫Bτ
log u. From this (4.18) will follow since∫Bτ
up0∫Bτ
u−p0 =
∫Bτ
ep0 log u
∫Bτ
e−p0 log u =
∫Bτ
ep0(w+|Bτ |−1∫Bτ
log u)∫Bτ
e−p0(w+|Bτ |−1∫Bτ
log u)
≤ Ce|Bτ |−1
∫Bτ
log u−|Bτ |−1∫Bτ
log u ≤ C(n, λ,Λ, τ).
25
We first derive an equation for w. Choose the test function u−1φ in (4.16) where φ ∈ L∞(B1) ∩W 1,2
0 (B1). Then since Dw = u−1D u we have∫B1
aijφDiwDjw ≤∫B1
aijDiwDjφ ∀φ ∈ L∞(B1) ∩W 1,20 (B1) φ ≥ 0.
On replacing φ with φ2, and using ellipticity, Cauchy-Schwarz inequality and Holder’s inequalitywe get ∫
B1
|Dw|2φ2 ≤ 1
λ
∫B1
aijφ2DiwDjw ≤
2
λ
∫B1
aijφDiwDjφ
≤ C
∫B1
|Dw||φ||Dφ| ≤ C
(∫B1
|Dw|2ζ2
) 12(∫
B1
|Dφ|2) 1
2
.
and hence, in particular, ∫B1
|Dw|2ζ2 ≤ C
∫B1
|Dφ|2 ∀ζ ∈ C1c (B1).
For B2r(y) ⊂ B1 let ζ be such that suppζ ⊂ B2r(y), ζ ≡ 1 in Br(y) and |Dζ| ≤ 2/r, then∫Br(y)
|Dw|2 ≤ Crn−2.
Therefore by Holder’s inequality and Poicare’s inequality
1
rn
∫Br(y)
|w − wBr(y)| ≤1
rn/2
(∫Br(y)
|w − wBr(y)|2) 1
2
≤ 1
rn/2
(r2
∫Br(y)
|Dw|2) 1
2
≤ C.
That is, w ∈ BMO and so by the John-Nirenberg lemma∫Bτ
ep0|w| ≤ C. (4.20)
Step 2: We next show that the result holds for all p < n/(n − 2). Note that for 0 < p ≤ p0
we have ep|w| ≤ ep0|w| and so we already have that the result holds for p < p0. Extending thisresult requires the use of the Moser iteration technique and the ideas are similar to those in theproof of Theorem 4.3. We give an outline of the argument here. We wish to show that for all0 < r1 < r2 < 1 and 0 < p2 < p1 < n/(n− 2) that(∫
Br1
up1
) 1p1
≤ C
(∫Br2
up2
) 1p2
(4.21)
for C = C(n, λ,Λ, r1, r2, p1, p2) > 0. Choose the test function φ = u−βη2 where β ∈ (0, 1) andη ∈ C1
c (B1). Then plugging this into (4.16) we have
0 ≤∫B1
aijDiuDj( u−βη2) = −β
∫B1
aijDiuη2 + 2
∫B1
aijDiu u−βηDjη
26
implying by ellipticity, Cauchy-Schwarz inequality and Holder’s inequality that
β
∫B1
|D u|2 u−β−1η2 ≤ C
∫B1
aijDi uDj u u−β−1η2 ≤ C
∫B1
aijDi u u−βηDjη
≤ C
∫B1
|D u| u−βη|Dη| ≤(∫
B1
|Du|2 u−β−1η2
) 12(∫
B1
|Dη|2 u1−β) 1
2
,
and hence ∫B1
|D u|2 u−β−1η2 ≤ C
β2
∫B1
|Dη|2 u1−β.
Let γ := 1− β ∈ (0, 1) and w := uγ/2, then Dw = γ2uγ/2−1 and so∫
B1
|Dw|2η2 ≤ C
(1− γ)2
∫B1
|Dη|2w2
and thus ∫B1
|D(wη)|2 ≤ C
(1− γ)2
∫B1
w2|Dη|2.
As in the proof of Theorem 4.3, Sobolev embedding theorem and an appropriate choice of cut-offfunction imply, with χ = n/(n− 2) gives us that for 0 < r < R < 1(∫
Br
uγχ) 1
γχ
≤(
C
(1− γ)2
1
(R− r)2
) 1γ(∫
BR
uγ) 1
γ
,
for any γ ∈ (0, 1). Hence we can obtain (4.17) for R = 1 after finitely many iterations.
The following result is now an immediate corollary of Theorems 4.5 and 4.7.
Theorem 4.8 (Moser’s Harnack Inequality). Let u ∈ W 1,2(Ω) be a non-negative solution of (4.8)in Ω, that is ∫
Ω
aijDiuDjφ = 0 ∀φ ∈ W 1,20 (Ω).
Then for any ball BR ⊂ ΩsupBR
u ≤ C infBR/2
u, (4.22)
where C = C(n, λ,Λ) > 0.
4.2.4 Holder Continuity of the Derivative
In this section, we will use the Harnack inequalities proved in the last section to show Holdercontinuity of solutions. We first state a simple technical lemma.
Lemma 4.9. Let τ < 1, γ > 0 and suppose ω is non-decreasing on (0, R] and satisfies
ω(τr) ≤ γω(r)
for all r ≤ R. Then for all r ≤ R
ω(r) ≤ C( rR
)αω(R)
for constants C = C(γ, τ) and α = α(γ, τ). In fact we have α = log γlog τ
.
27
Proof. Let r ≤ R and choose k such that τ kR < r ≤ τ k−1R. Then
ω(r) ≤ ω(τ k−1R) ≤ γk−1ω(R).
Moreover we have
γk = (τ k)log γlog τ ≤
( rR
) log γlog τ
.
Hence
ω(r) ≤ 1
γ
( rR
) log γlog τ
ω(R).
Theorem 4.10. Let u ∈ W 1,2(Ω) be a weak solution in Ω of (4.8), that is∫Ω
aijDiuDjφ = 0
for all φ ∈ W 1,20 (Ω). Then u ∈ C0,α(Ω) for α ∈ (0, 1) depending only on n, λ and Λ, and moreover,
for any BR ⊂ Ω,
|u(x)− u(y)| ≤ C
(|x− y|R
)α(1
Rn
∫BR
u2
) 12
,
for all x, y ∈ BR2
.
Proof. We prove the result for R = 1. Fix some ball B1 ⊂ Ω and let r ∈ (0, 1]. Define M(r) :=supBr u and m(r) := infBr u. Then by Theorem 4.5, we know that M(r) < ∞ and m(r) > −∞.It suffices to show that
ω(r) := M(r)−m(r) ≤ Crα(∫
B1
u2
) 12
,
for any r ≤ 12. To do so we observe that M(r)− u is a non-negative solution of (4.8) in Br, and
so Moser’s Harnack inequality (Theorem 4.8) implies
supB r
2
(M(r)− u) ≤ supBr
(M(r)− u) ≤ C infB r
2
(M(r)− u).
Thus,
M(r)−m(r
2
)≤ C
(M(r)−M
(r2
)).
Similarly, we can apply Moser’s Harnack inequality to u−m(r) on Br to obtain
M(r
2
)−m(r) ≤ C
(m(r
2
)−m(r)
).
Adding these two inequalities we obtain
ω(r) + ω(r
2
)≤ C
(ω(r)− ω
(r2
)),
which implies
ω(r
2
)≤ γω(r),
28
where γ = C−1C+1
< 1. We apply Lemma 4.9 to obtain.
ω(ρ) ≤ Cραω
(1
2
),
for all ρ ∈ (0, 1/2] where α = log γ
log 12
. Finally by Theorem 4.3 we have
ω
(1
2
)≤ C
(∫B1
u2
) 12
,
from which we deduce the result.
4.3 Holder Continuity of the Second Derivatives
In this section we will show that the solution u is in C2,α, and hence is a solution of the minimalsurface equation in the classical sense. The results of this section may be found in [7].
Lemma 4.11. Let σ(r) be a non-negative monotonically increasing function defined on [0, R0]satisfying
σ(r) ≤ γ(( r
R
)µ+ δ)σ(R) + κRν
for all 0 < r ≤ R ≤ R0, where γ, ν < µ and δ > 0 are constants. Then there exists δ0(γ, µ, ν) > 0such that, provided δ ≤ δ0, we have
σ(r) ≤ γ1
( rR
)νσ(R) + κ1r
ν
for all 0 < r ≤ R ≤ R0 and where γ1 = γ1(γ, µ, ν) and κ1 = κ1(γ, µ, ν, κ) with κ1 = 0 if κ = 0.
Proof. Let 0 < τ < 1, and R < R0. Then by assumption
σ(τR) ≤ γτµ(1 + δτ−µ)σ(R) + κRν .
We choose τ such that 2γτµ = τλ for some ν < λ < µ, (note that we may assume without loss ofgenerality that 2γ > 1). Assume that δ0τ
−µ ≤ 1. It follows that if δ ≤ δ0
σ(τR) ≤ τλσ(R) + κRν ,
and thus, by iterating, that
σ(τ kR) ≤ τλσ(τ k−1R) + κτ (k−1)νRν
≤ τ kλσ(R) + κτ (k−1)νRν
k−1∑j=0
τ j(λ−ν)
≤ τ kν(σ(R) +
κτ (k−1)νRν
1− τλ−ν
).
29
Now let r ≤ R and choose k such that τ kR < r ≤ τ k−1R as in the proof of Lemma 4.9 , then
σ(r) ≤ σ(τ k−1R) ≤ τ (k−1)λσ(R) +κRντ (k−1)ν
1− τ (λ−ν)
≤ Cτ ν(k−1)(σ(R) + κRν)
≤ γ1
( rR
)νσ(R) + κ1r
ν ,
which completes the proof.
We now a state two theorems whose proofs are found in [7], and are needed to show the C2,α
regularity. In these theorems n,is as usual, the dimension of the space.
Theorem 4.12 (Campanato’s Theorem). If for all 0 < r ≤ R0 and all x ∈ Ω we have∫Br(x)
|u− uBr(x)|p ≤ γrn+pα
with constants γ, p and 0 < α < 1, then u ∈ C0,αloc (Ω).
Theorem 4.13 (Campanato Estimates). Let (Aij) be a matrix with |Aij| ≤ Λ for all i, j andAijξiξj ≥ λ|ξ|2 for all ξ ∈ Rn where 0 < λ ≤ Λ < ∞ are constants. Suppose u ∈ W 1,2(Ω) is aweak solution of the equation
Dj(AijDiu) = 0. (4.23)
Then for all x0 ∈ Ω and 0 < r < R < dist(x0, ∂Ω) we have∫Br(x0)
|u|2 ≤ c1
( rR
)n ∫BR(x0)
|u|2∫Br(x0)
|u− uBr(x0)|2 ≤ c2
( rR
)n+2∫BR(x0)
|u− uBR(x0)|2.(4.24)
Theorem 4.14. Suppose that aij(x) are Cα(Ω) for i, j = 1, . . . , n and some α ∈ (0, 1) and satisfyaij(x)ξiξj ≥ λ|ξ|2 for all ξ ∈ Rn and x ∈ Ω, and that |aij| ≤ Λ on Ω, where 0 < λ ≤ Λ < ∞ areconstant. Then any weak solution u ∈ W 1,2(Ω) of the equation
Dj(aijDiu) = 0 (4.25)
is C1,α′(Ω) for all α′ ∈ (0, α).
Proof. Step 1: Let x0 ∈ Ω. We write
aij(x) = aij(x0) + (aij(x)− aij(x0)) (4.26)
and define Aij := aij(x0). Then for v satisfying (4.25)
Dj(AijDiu) = Dj((aij(x0)− aij(x))Diu) = Djfj(x) (4.27)
where we have introduced the function
f j(x) := (aij(x0)− aij(x))Diu(x).
30
In other words, u satisfies∫Ω
AijDivDjφ =
∫Ω
f jDjφ ∀φ ∈ W 1,20 (Ω). (4.28)
Let BR(x0) ⊂ Ω be a ball, and let w ∈ W 1,2(BR(x0)) be a weak solution of the Dirichlet problem
Dj(AijDiw) = 0 in BR(x0)
w = u on ∂BR(x0).(4.29)
Then w satisfies ∫BR(x0)
AijDiwDjφ = 0 ∀φ ∈ W 1,20 (BR(x0)). (4.30)
Such a w can be shown to exist by the Lax Milgram Theorem, since it guarantees the existenceof z ∈ W 1,2
0 (BR(x0)) with
B(φ, z) =
∫BR(x0)
AijDizDjφ = −∫BR(x0)
AijDiuDjφ ∀φ ∈ W 1,20 (BR(x0)), (4.31)
and then w given by w = u+z is easily seen to be a solution of (4.29). According to Theorem 4.2,we may deduce that w is in W 2,2
loc (BR(x0)), and so since the equation for w is linear with constantcoefficients, it follows that Dkw satisfies the same equation on BR(x0) for k = 1, . . . , n althoughnaturally the boundary conditions will be different). Hence, by Campanato estimates (4.24) weget ∫
Br(x0)
|Dw|2 ≤ C( rR
)n ∫BR(x0)
|Dw|2. (4.32)
Step 2: Next we observe that since u = w on ∂BR(x0), u−w is an admissible test function, andhence ∫
BR(x0)
AijDiwDjw =
∫BR(x0)
AijDiwDju
Combining this, the Cauchy-Schwarz inequality and ellipticity we obtain the estimate
λ
∫BR(x0)
|Dw|2 ≤∫BR(x0)
AijDiwDjw =
∫BR(x0)
AijDiwDju
≤∫BR(x0)
|ADw||Du| ≤(∫
BR(x0)
‖A‖2|Dw|2) 1
2(∫
BR(x0)
|Du|2) 1
2
≤ nΛ
(∫BR(x0)
|Dw|2) 1
2(∫
BR(x0)
|Du|2) 1
2
,
which implies ∫BR(x0)
|Dw|2 ≤(nΛ
λ
)2 ∫BR(x0)
|Du|2. (4.33)
Moreover, (4.28) and (4.30) together imply∫BR(x0)
AijDi(u− w)Djφ =
∫BR(x0)
f jDjφ ∀φ ∈ W 1,20 (BR(x0)).
31
Letting φ = u− w we obtain∫BR(x0)
|D(u− w)|2 ≤ 1
λ
∫BR(x0)
AijDi(u− w)Dj(u− w) =1
λ
∫BR(x0)
f jDj(u− w)
≤ 1
λ
(∫BR(x0)
|D(u− w)|2) 1
2
(∫BR(x0)
∑j
|f j|2) 1
2
,
implying ∫BR(x0)
|D(u− w)|2 ≤ 1
λ2
∫BR(x0)
∑j
|f j|2. (4.34)
Step 3: We now start to combine all of these estimates. First we use (4.32) and (4.33) to show∫Br(x0)
|Du|2 ≤ 2
∫Br(x0)
|Dw|2 + 2
∫Br(x0)
|D(u− w)|2
≤ C( rR
)n ∫BR(x0)
|Dw|2 + 2
∫Br(x0)
|D(u− w)|2
≤ C( rR
)n ∫BR(x0)
|Du|2 + 2
∫Br(x0)
|D(u− w)|2.
Next observe that∫Br(x0)
|D(u− w)|2 ≤∫BR(x0)
|D(u− w)|2 ≤ 1
λ2
∫BR(x0)
∑j
|f j|2
≤ 1
λ2supi,jx∈Ω
|aij(x0)− aij(x)|2∫BR(x0)
|Du|2 ≤ CR2α
∫BR(x0)
|Du|2,(4.35)
the last inequality following since each aij is in C0,α by assumption. Finally we arrive at theestimate ∫
Br(x0)
|Du|2 ≤ γ(( r
R
)n+R2α
)∫BR(x0)
|Du|2.
We apply Lemma 4.11 with σ(r) :=∫Br(x0)
|Du|2, µ = n, and ν = n − ε for some ε > 0. Then
there exists some δ0(γ, n, ε) > 0 such that provided R2α0 ≤ δ0, then for all 0 < r ≤ R ≤ R0∫
Br(x0)
|Du|2 ≤ C( rR
)n−ε ∫BR(x0)
|Du|2, (4.36)
where the constant C depends on ε, n and γ.Step 4: The second part of the proof is devoted to deriving an analogous inequality from thesecond Campanato estimate (4.24). Using the same arguments as were used to establish (4.32),we can show that∫
Br(x0)
|Dw − (Dw)Br(x0)|2 ≤ C( rR
)n+2∫BR(x0)
|Dw − (Dw)BR(x0)|2. (4.37)
32
Next we note that ∫BR(x0)
|Dw − (Dw)BR(x0)|2 ≤∫BR(x0)
|Dw − (Du)BR(x0)|2,
since for any real-valued g ∈ L2(BR(x0))∫BR(x0)
|g − (g)BR(x0)|2 = infκ∈R
∫BR(x0)
(g − κ)2.
To see this note that the function F (κ) :=∫BR(x0)
(g − κ)2 is convex and differentiable and has a
critical point at κ = (g)BR(x0), and so F attains its minimum here. Furthermore, we also have thefollowing estimate by ellipticity∫
BR(x0)
|Dw − (Du)BR(x0)|2 ≤1
λ
∫BR(x0)
Aij(Diw − (Diu)BR(x0))(Djw − (Dju)BR(x0))
=1
λ
∫BR(x0)
Aij(Diw − (Diu)BR(x0))(Dju− (Dju)BR(x0))
+1
λ
∫BR(x0)
Aij(Diu)BR(x0)(Dju−Djw).
The second integral vanishes since Aij(Diu)BR(x0) is constant and u − w ∈ W 1,20 (BR). Applying
the Cauchy-Schwarz inequality as we did to deduce (4.33), we have∫BR(x0)
|Dw − (Dw)BR(x0)|2 ≤Λ2n2
λ2
∫BR(x0)
|Du− (Du)BR(x0)|2. (4.38)
Step 5: Finally∫Br(x0)
|Du− (Du)Br(x0)|2 ≤ 3
∫Br(x0)
|Du−Dw|2 + 3
∫Br(x0)
|Dw − (Dw)Br(x0)|2
+ 3
∫Br(x0)
|(Du)Br(x0) − (Dw)Br(x0)|2,
and∫Br(x0)
|(Du)Br(x0) − (Dw)Br(x0)|2 = |Br(x0)|(
1
|Br(x0)|
∫Br(x0)
|Du−Dw|)2
≤ |Br(x0)|
(1
|Br(x0)||Br(x0)|1/2
(∫Br(x0)
|Du−Dw|2)1/2
)2
=
∫Br(x0)
|Du−Dw|2.
Combining the last two inequalities and using (4.35) gives∫Br(x0)
|Du− (Du)Br(x0)|2 ≤ 3
∫Br(x0)
|Dw − (Dw)Br(x0)|2 + 6
∫Br(x0)
|Du−Dw|2
≤ 3
∫Br(x0)
|Dw − (Dw)Br(x0)|2 + CR2α
∫BR(x0)
|Du|2,
33
Applying the second Campanato estimate (4.24) to the first integral and then applying (4.38), wehave∫
Br(x0)
|Du− (Du)Br(x0)|2 ≤ C( rR
)n+2∫BR(x0)
|Du− (Du)BR(x0)|2 + CR2α
∫BR(x0)
|Du|2
≤ C( rR
)n+2∫BR(x0)
|Du− (Du)BR(x0)|2 + CRn+2α−ε
where we have applied (4.36) with r = R and R = R0 to deduce the second inequality. Finallyapplying Lemma 4.11 we get∫
Br(x0)
|Dv − (Dv)Br(x0)|2 ≤
(C1
(1
R
)n+2α−ε ∫BR(x0)
|Du− (Du)BR(x0)|2 + C2
)rn+2α−ε.
This holds for all 0 < r ≤ R and so the claim now follows from Campanato’s Theorem.
We apply the above theorem to the partial derivatives of u and the equation (4.7) to deducethat u is C2,α.
4.4 Schauder Theory
Having shown that the solution u is a classical solution, we now appeal to the interior Schauderestimates to show that u is smooth. Roughly speaking, Schauder estimates may be used to showthat the second derivatives of a classical solution, have the same regularity as the coefficients inthe equation, so in our case in particular we can show that aij ∈ Ck,α(Ω) implies u ∈ Ck+2,α(Ω).Since the coefficients aij are written in terms of the first derivatives of u, a bootstrapping argumentwill allow us to deduce the result we are after. The first two results we shall state without proof,they may be found in, for example, [2].
Theorem 4.15 (Dirichlet Problem). Let B ⊂ Rn be a ball, and let
L = aijDij + biDi + c
be a uniformly elliptic operator with coefficients aij, bi and c ∈ C0,α(B), and c ≤ 0 on B. Thenthe Dirichlet problem
Lu = f in B
u = g on ∂B,
for f ∈ C0,α(B) and g ∈ C(∂B) has a unique solution u ∈ C2,α(B) ∩ C(Ω).
Theorem 4.16 (Interior Schauder Estimates). Let α ∈ (0, 1). Suppose Ω is a domain andu ∈ C2,α(Ω) solves the uniformly elliptic equation
Lu = aijDiju+ biDiu+ cu = f in Ω,
where aij, bi, c and f are in C0,α(Ω). Then for any Ω′ ⊂⊂ Ω
‖u‖C2,α(Ω′) ≤ C(‖u‖C(Ω) + ‖f‖C0,α(Ω))
where C = C(n, α, L,Ω′,Ω) ∈ (0,∞).
34
Theorem 4.17 (Interior Regularity). Let k ≥ 0 be an integer and α ∈ (0, 1). Let Ω ⊂ Rn be openand suppose that u ∈ C2(Ω) satisfies
Lu = aijDij + biDiu+ cu = f in Ω
for some elliptic operator L with coefficients aij, bi, c ∈ Ck,α(Ω) and some f ∈ Ck,α(Ω). Thenu ∈ Ck+2,α(Ω).
Proof. We will prove the simpler case where the operator L takes the form L = aijDij. The moregeneral result is proved in a similar fashion with slightly more complicated calculations.Step 1: We first show that aij, f ∈ C0,α(Ω) and u ∈ C2(Ω) implies that u ∈ C2,α(Ω). This isa direct consequence of Theorem 4.15. Let B ⊂⊂ Ω be a ball and consider solutions v to theDirichlet problem
aijDijv = f in B,
v = u on ∂B.
Theorem 4.15 tells us that there exists a solution in C(B) ∩ C2,α(B), however we also know thatu is a C(B) ∩ C2(B) solution and that there exists at most one solution in C(B) ∩ C2(B). Thuswe conclude that u ∈ C2,α(B) and hence that u ∈ C2,α.Step 2: We next show that aij, f ∈ C1,α(Ω) and u ∈ C2,α(Ω) implies that u ∈ C3,α(Ω). This isdone via a difference quotient argument. Choose Ω′′′, Ω′′ and Ω′ such that Ω′′′ ⊂⊂ Ω′′ ⊂⊂ Ω′ ⊂⊂ Ωand h0 > 0 such that both dist(Ω′′′, ∂Ω′′) < h0 and dist(Ω′′, ∂Ω′) < h0. We then apply δhl for somel ∈ 1, . . . , n and 0 < h < h0 to both sides of the equation Lu = f and rearrange to get
L(δhl u) = aij(x)Dij(δhl u) = δhl f − δhl aij(x)Diju(x+ hel) on Ω′′.
We then apply Theorem 4.16 to arrive at the estimate since δhl ∈ C2,α(Ω1)
‖δhl u‖C2,α(Ω′′′) ≤ C(‖δhl u‖C(Ω′′) + ‖δhl f‖C0,α(Ω′′) + ‖δhl aij‖C0,α(Ω′′)‖Dij(·+ hel)‖C0,α(Ω′′). (4.39)
We next observe that
supx∈Ω′′|δhl aij(x)| = sup
x∈Ω′′
∣∣∣∣1h∫ 1
0
d
dtaij(x+ htel)dt
∣∣∣∣ = supx∈Ω′′
∣∣∣∣1h∫ h
0
Dlaij(x+ tel)dt
∣∣∣∣ ≤ supx∈Ω′|Dlaij(x)|,
and thus δhl aij and hence continuous, and also notice that for x, y ∈ Ω′′
|δhl aij(x)− δhl aij(y)| =∣∣∣∣1h∫ h
0
(Dlaij(x+ tel)−Dlaij(y + tel))dt
∣∣∣∣ ≤ [Dlaij]C0,α(Ω′)|x− y|α.
These together imply ‖δhl aij‖C0,α(Ω′′) ≤ ‖Dlaij‖C0,α(Ω′), and similar calculations yield an estimateof the same form for f . Hence from (4.39) we arrive at
‖δhl u‖C2,α(Ω′′′) ≤ C(‖Dlu‖C(Ω′) + ‖Dlf‖C0,α(Ω′) + ‖Dlaij‖C0,α(Ω′)‖Diju‖C0,α(Ω′)).
This bound is independent of h, and therefore it follows that the set δhl uh<h0 is bounded andequicontinuous. The same is true for the derivatives of δhl u. Therefore by the Arzela-Ascoli
35
theorem we can find a sequence (hj)j∈N with hj 0 such that δhjl u converges in C2(Ω′′′). Now
u is continuously differentiable on Ω and so δhjl u converges uniformly to Dlu on Ω′′′. Thus we
conclude that δhjl u converges to Dlu in C2(Ω′′′), which implies Dlu ∈ C2,α(Ω′′′). Since l and Ω′′′
were arbitrary, we conclude that u ∈ C3,α(Ω).Step 3: Let k ≥ 2. We now wish to show that aij, f ∈ Ck,α(Ω) and u ∈ Ck+1,α(Ω) impliesu ∈ Ck+2,α(Ω). Let β be a multiindex with |β| = k − 1. Then applying Dβ to both sides of theequation Lu = f , we obtain
L(Dβu) = Dβf −∑γ<β
β!
γ!(β − γ)!Dβ−γaijD
γDiju in Ω.
Since the right hand side is in C1,α(Ω) and Dβu ∈ C2,α, we may apply step 2 to deduce that Dβu ∈C3,α(Ω), and hence that u ∈ Ck+2,α. The proof of the theorem now follows by induction.
5 Gradient bound
In this section, we want to find a bound of the gradient of a solution u to the minimal surfaceequation. We will state the main theorem here, but we will have to do a bit of work before wecan prove it. This section is based on [6], and any proofs omitted are proved there or in [2].
Theorem 5.1 (Gradient Bound). Let Ω be a bounded C2 domain and suppose that u solves theminimal surface equation in Ω. Then for any y ∈ Ω, we have the following bound on gradient ofu:
|Du(y)| ≤ exp
(C
(1 + sup
x∈Bd(y)
[u(x)− u(y)
d
])),
where C = C(u) and d = dist(y, ∂Ω).
We will split the proof up into a series of propositions, but first of all we will define our notationand present a few useful results. We assume throughout this section, as we did in the statementof Theorem 5.1, that Ω ⊂ Rn is a bounded C2 domain and that u solves the minimal surfaceequation in Ω. We will write G(u) for the graph of u, namely
G(u) = (x, u(x)) : x ∈ Ω
and write a =√
1 + |Du|2 for the area element of u on its graph. The graph can equivalently beviewed as a level set of the function Φ : Ω × R → R defined by Φ(x, xn+1) = xn+1 − u(x). Theunit normal to G(u) is given by
ν =DΦ
|DΦ|. (5.1)
Lemma 5.2. If ν is the normal to G(u) as defined in (5.1), then
div(ν) := Diνi = −nH,
where H is the mean curvature of the graph G(u).
36
Lemma 5.3. Let u be a solution to the minimal surface equation, φ ∈ C1c (Ω) a compactly sup-
ported, differentiable test function, and ν the normal to G(u) as defined in equation (5.1). Then∫Ω
νiDiφ = 0.
Proof. Since u is a solution to the minimal surface equation, we know that it has mean curvature,H, equal to zero. Therefore, by Lemma 5.2, we know that
divν = Diνi = 0 in Ω.
We then multiply by φ ∈ C1c (Ω), integrate over Ω, and integrate by parts to finish the proof.
Definition (Tangential gradient). For a differentiable function g defined in some open set con-taining the graph G(u), we define the tangential gradient of g on G(u) to be
∇g = Dg − (Dg · ν)ν.
Note that for y ∈ G(u), the tangential gradient ∇g(y) is the projection of Dg(y) onto the tangentplane to G(u) at y.
Lemma 5.4. Let u be a solution to the MSE, and suppose that we have a function φ ∈ C1c (Ω×R).
Define the function φ : Ω→ R by setting
φ(x′) = φ(x′, u(x′)),
where x′ = (x1, . . . , xn) ∈ Ω. Then on G(u) we have
∇iνn+1Diφ = ∇iνn+1∇iφ = ∇iνn+1∇iφ.
Proposition 5.5. Let w = log√
1 + |Du|2. Then for any y ∈ Ω, and ρ > 0 such that Bnρ (y) ⊂ Ω,
we have the following inequality:
w(y) ≤ 1
ωnρn
∫G(u)∩Bn+1
ρ (y)
w dHn.
We split the proof of this proposition up into two lemmas.
Lemma 5.6. If u is a solution to the MSE, then the function w = log√
1 + |Du|2 is subharmonic,in the sense that
∇i∇iw ≥ 0.
37
Proof. From Lemma 5.3, we know that ∫Ω
νiDiφ = 0,
where ν is the tangent to G(u), for any φ ∈ C1c (Ω). Therefore, for any function φ such that
Dkφ ∈ C1c (Ω) for some k = 1, . . . , n+ 1, we can integrate by parts to see that∫
Ω
DkνiDiφ = 0.
Now let us choose φ so that it is non-negative in Ω, andDi(νkφ) ∈ C1c (Ω) for each k, i = 1, . . . , n+1.
Then we can use νkφ as our test functions in the equation above and sum over k to get∫Ω
DkνiDi(νkφ) = 0,
and therefore ∫Ω
(Dkνi)(Diνk)φ+
∫Ω
νk(Dkνi)(Diφ) = 0. (5.2)
It can be shown thatDkνiDiνk =
∑k
(κk)2 ≥ 0,
where κk are the principal curvatures, and that
νkDkνiDi = −a∇iνn+1.
Substituting these into (5.2), we can see that∫Ω
(∑k
(κk)2
)φ−
∫Ω
a∇iνn+1Diφ = 0.
Therefore, since we have chosen φ so that it is non-negative in Ω, we now know that∫Ω
a∇iνn+1Diφ ≥ 0. (5.3)
Now, if we find a function φ ∈ C1c (Ω × R) such that φ(x) = φ(x, u(x)), then by Lemma 5.4 we
have ∫Ω
a∇iνn+1Diφ =
∫G(u)
∇iνn+1∇iφ dHn.
Combining this with (5.3) gives ∫G(u)
∇iνn+1∇iφ dHn ≥ 0. (5.4)
38
To move on from here note that, by the definition of the normal vector ν and the area element a,we have that νn+1 = 1/a. If we also use a test function φ in (5.4) which has the from φ = aϕ, wecan re-write the equation as:
0 ≤∫G(u)
∇iνn+1∇iφ dHn
= −∫G(u)
1
a2∇ia∇i(aϕ) dHn
= −∫G(u)
1
a2∇ia∇iaϕ dHn −
∫G(u)
1
a∇ia∇iϕ dHn
= −[∫G(u)
|∇a|2
a2ϕ+∇a · ∇ϕ
adHn
].
We have seen before that w = log a, which means that∇w = 1a∇a. Therefore, the above inequality
can be re-written as ∫G(u)
|∇w|2ϕ+∇w · ∇ϕ dHn ≤ 0. (5.5)
We now use the integration by parts formula on the graph G(u),∫G(u)
∇f · ∇g dHn = −∫G(u)
g∆f dHn,
to see that ∫G(u)
(|∇w|2 −∆w
)ϕ dHn ≤ 0.
Therefore, since we have implicitly assumed that ϕ is positive on G(u), we can conclude that
∆w ≥ |∇w|2 ≥ 0,
i.e. the function w = log√
1 + |Du| is subharmonic on the graph of u.
Lemma 5.7 (Mean Value Inequality). Suppose that w ∈ C2(Ω× R) is a non-negative, subhar-monic function on G(u). Then for any y ∈ G(u) and any ρ > 0 such that Bn
ρ (y) ⊂ Ω, we havethat
w(y) ≤ 1
ωnρn
∫G(u)∩Bn+1
ρ (y)
w dHn,
where ωn is the measure of the n-dimensional unit ball.
Proof of Propositon 5.5. All of the hard work has now been done: we use Lemma 5.6 to see thatthe function w = log
√1 + |Du|2 is subharmonic, which means that we can apply Lemma 5.7 to
conclude that we have the inequality
w(y) ≤ 1
ωnρn
∫G(u)∩Bn+1
ρ (y)
w dHn,
for any y ∈ G(u) and ρ > 0 such that Bnρ (y) ⊂ Ω.
39
Our aim now is to bound the integral on the right hand side of the inequality in Proposition 5.5.
Proposition 5.8. For any y ∈ Ω, and ρ > 0 such that Bn+1ρ (y) ⊂ Ω× R, we have the following
inequality: ∫G(u)∩Bn+1
ρ (y)
w dHn ≤ ωnρn +
∫Bnρ (y)∩|u−u(y)|<ρ
|Du|2
aw,
where ωn is the measure of the n dimensional unit ball.
Proof. Since w ≥ 0 in G(u), and Bn+1ρ (y) ⊂ Bn
ρ (y)× |xn+1 − u(y)| < ρ, we know that∫G(u)∩Bn+1
ρ (y)
w dHn ≤∫G(u)∩(Bnρ (y)×|xn+1−u(y)|<ρ)
w dHn.
By definition of Hn, we can re-write the right hand integral above as∫G(u)∩(Bnρ (y)×|xn+1−u(y)|<ρ)
w dHn =
∫Bnρ (y)∩|u−u(y)|<ρ
aw.
Now, recall that a =√
1 + |Du|2, which means we can write a = (1+|Du|2)/a, and also w = log a.Therefore,∫
Bnρ (y)∩|u−u(y)|<ρaw =
∫Bnρ (y)∩|u−u(y)|<ρ
1 + |Du|2
aw
≤∫Bnρ (y)∩|u−u(y)|<ρ
1 +
∫Bnρ (y)∩|u−u(y)|<ρ
|Du|2
aw, since
log a
a≤ 1
≤ ωnρn +
∫Bnρ (y)∩|u−u(y)|<ρ
|Du|2
aw,
which completes the proof.
Again, we want to bound the integral on the right hand side of the inequality in Proposition 5.8.
Proposition 5.9. For any y ∈ Ω, and ρ > 0 such that Bn2ρ(y) ⊂ Ω, we have the following
inequality: ∫Bnρ (y)∩|u−u(y)|<ρ
|Du|2
aw ≤ C
∫Bn2ρ(y)∩u−u(y)>−2ρ
a,
where C = C(u) is independent of the choice of y and ρ.
As before, we will split the proof up into some lemmas.
Lemma 5.10. Let y ∈ Ω, and find ρ > 0 such that Bn2ρ(y) ⊂ Ω. Let η ∈ C1
c (Ω) be a functionsatisfying
1) 0 ≤ η ≤ 1 in Ω,
2) |Dη| ≤ 2/ρ in Ω,
40
3) η = 1 in Bnρ (y),
4) η = 0 in Ω/Bn2ρ(y).
Then the following inequality holds:∫Bnρ (y)∩|u−u(y)|<ρ
|Du|2
aw ≤ 4
∫Bn2ρ(y)∩u−u(y)>−ρ
a+ 2ρ
∫Bn2ρ(y)∩u−u(y)>−ρ
η|Dw|.
Proof. Without loss of generality, we assume that y = 0 and u(y) = 0. Now, recall the result inLemma 5.3, that ∫
Ω
νiDiφ = 0, (5.6)
for φ ∈ C1c (Ω). We will prove the inequality in this lemma by choosing the test function φ carefully.
Let γ : R→ R be a function defined by
γ(t) =
0 for t < −ρ,
t+ ρ for −ρ ≤ t ≤ ρ,
2ρ for t > ρ.
Then, we let our test function φ in (5.6) be φ = ηwγ(u). Since
Di(ηwγ(u)) = wγ(u)Diη + ηγ(u)Diw + ηwγ′(u)Diu,
we can rearrange (5.6) to∫Ω
νiηwγ′(u)Diu = −
(∫Ω
νiηγ(u)Diw +
∫Ω
νiwγ(u)Diη
).
Note that the function γ′(u) is an indicator function for the set |u| < ρ. This, along with thefact that |Du|2/a = −ν ·Du, lets us write∫
Bnρ (y)∩|u−u(y)|<ρ
|Du|2
aw ≤ −
∫Ω
νiηwγ′(u)Diu
=
∫Ω
νiηγ(u)Diw +
∫Ω
νiwγ(u)Diη.
Then we use Cauchy-Schwartz to see∫Ω
νiηγ(u)Diw +
∫Ω
νiwγ(u)Diη ≤∫
Ω
|ν|ηγ(u)|Dw|+∫
Ω
|ν|wγ(u)|Dη|,
after which we can use the facts that |ν| = 1, γ < 2ρ and γ(u) = 0 when u < −ρ to see∫Ω
|ν|ηγ(u)|Dw|+∫
Ω
|ν|wγ(u)|Dη| ≤ 2ρ
∫u>−ρ
η|Dw|+ 2ρ
∫u>−ρ
w|Dη|.
41
We can now use the conditions on η set out in the statement of the lemma to see that
2ρ
∫u>−ρ
η|Dw|+2ρ
∫u>−ρ
w|Dη|
≤ 2ρ
∫Bn2ρ∩u>−ρ
η|Dw|+ 2ρ
∫Bn2ρ∩u>−ρ
w|Dη| by condition 4,
≤ 2ρ
∫Bn2ρ∩u>−ρ
η|Dw|+ 4
∫Bn2ρ∩u>−ρ
w by condition 2,
≤ 2ρ
∫Bn2ρ∩u>−ρ
η|Dw|+ 4
∫Bn2ρ∩u>−ρ
a since log a ≤ a,
and this completes the proof.
Lemma 5.11. Let y ∈ Ω, and find ρ > 0 such that Bn2ρ(y) ⊂ Ω, and let η ∈ C1
c (Ω) be a functionsatisfying the conditions in Lemma 5.10. Then the following inequality holds:∫
Bn2ρ(y)∩u−u(y)>−ρη|Dw| ≤ C
ρ
∫Bn2ρ(y)∩u−u(y)>−2ρ
a,
where C = C(u) is independent of the choice of y and ρ.
Proof. We will start with a relationship that we showed in (5.5) back in Lemma 5.6, namely∫G(u)
|∇w|2ϕ+
∫G(u)
∇w · ∇ϕ ≤ 0,
which holds for any ϕ ∈ C1c (Ω× R). If instead we take our test function as ϕ2 ∈ C1
c (Ω× R), wecan rewrite this as:
0 ≥∫G(u)
|∇w|2ϕ2 +
∫G(u)
∇w · ∇ϕ2
=
∫G(u)
|∇w|2ϕ2 + 2
∫G(u)
ϕ∇w · ∇ϕ.
This gives ∫G(u)
|∇w|2ϕ2 ≤ 2
(∫G(u)
ϕ2|∇w|2)1/2(∫
G(u)
|∇w|2)1/2
,
by Cauchy-Schwarz, and therefore∫G(u)
|∇w|2ϕ2 ≤ 4
∫G(u)
|∇ϕ|2. (5.7)
We will also assume, without loss of generality, that y = 0, and u(y) = 0. Now, let us define thefunction that we will be using as our ϕ. We set
ϕ(x′, xn+1) = η(x′)λ(xn+1),
where η is as in the statement of Lemma 5.10, and λ is a suitably differentiable cut off functionthat satisfies
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1) λ(x) = 1 for all x ∈ (−ρ, supB2ρu),
2) λ(x) = 0 for all x ∈ R/(−2ρ, ρ+ supB2ρu),
3) |λ′(x)| ≤ 2/ρ for all x ∈ R.
By choosing λ, and therefore ϕ, in this way, we are able to get a further bound for equation (5.7):∫G(u)
|∇w|2ϕ2 ≤ 4
∫G(u)
|∇ϕ|2 ≤ C2
ρ2Hn(B), (5.8)
whereB = x ∈ G(u) : |x′| < 2ρ and − 2ρ < xn+1 < ρ+ sup
B2ρ
u
is the set where ∇ϕ is non-zero. Now, because w is independent of xn+1, we know that
νn+1|Dw| ≤ |∇w|, (5.9)
from the definition of the tangential gradient, and recall that νn+1 = 1/a. We now have all of thefacts we need for our estimate:∫
Bn2ρ∩u>−ρη|Dw| ≤
∫Bn2ρ∩u>−ρ
aη|∇w|, by (5.9),
≤∫B
η|∇w| dHn,
≤(∫
B
η2|∇w|2)1/2
Hn(B)1/2, by Cauchy-Schwartz,
≤(∫
B
ϕ2|∇w|2)1/2
Hn(B)1/2, since λ = 1 on B,
≤(∫
G(u)
ϕ2|∇w|2)1/2
Hn(B)1/2, because B ⊂ G(u),
≤ C
ρHn(B), by (5.8),
=C
ρ
∫B2ρ∩u>−2ρ
a,
which completes the proof.
Proof of Proposition 5.9. As before, we have done most of the work now. We know that∫Bnρ (y)∩|u−u(y)|<ρ
|Du|2
aw
≤ 4
∫Bn2ρ(y)∩u−u(y)>−ρ
a+ 2ρ
∫Bn2ρ(y)∩u−u(y)>−ρ
η|Dw|, from Lemma 5.10,
≤ 4
∫Bn2ρ(y)∩u−u(y)>−ρ
a+ 2C1
∫B2ρ(y)∩u−u(y)>−2ρ
a, by Lemma 5.11,
≤ C2
∫Bn2ρ(y)∩u−u(y)>−2ρ
a,
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and we are done.
Proposition 5.12. For any y ∈ Ω, and ρ > 0 such that Bn3ρ(y) ⊂ Ω, we have the following
inequality: ∫Bn
2ρ(y)×u−u(y)>−2ρ
a ≤ ωn(2ρ)n + Cρn
(1 +
1
ρsup
x∈B3ρ(y)
[u(x)− u(y)]
).
Proof. As before, we assume without loss of generality that y = 0 and u(y) = 0. Now, by writinga = (1 + |Du|2)/a, we can see that∫
B2ρ∩u>−2ρa ≤ ωn(2ρ)n +
∫B2ρ∩u>−2ρ
|Du|2
a.
As usual, our task is to estimate the integral on the right hand and we do this by a careful choiceof test function in the weak form of the minimal surface equation, from Lemma 5.3. Let µ be afunction defined by
µ(t) =
t+ 2ρ, for t ≥ −2ρ,
0, otherwise,
and let φ be a suitably differentiable function which satisfies
1. 0 ≤ φ ≤ 1 everywhere,
2. φ = 1 on B2ρ,
3. φ = 0 on R/B3ρ,
4. |Dφ| ≤ 2/ρ everywhere.
We then take µ(u)φ as our test function, so∫Ω
νiDi(µ(u)φ) = 0,
which, when we expand the derivative, gives∫Ω
νiµ′(u)Diuφ = −
∫Ω
νiµ(u)Diφ.
So, using the fact that |Du|2/a = −ν ·Du and our choice of test function, we know that∫B2ρ∩u>−2ρ
|Du|2
a≤ −
∫Ω
νiµ′(u)Diuφ,
=
∫Ω
νiµ(u)Diφ,
≤ 2
ρ
∫B3ρ∩u≥−2ρ
u+ 2ρ, by Cauchy Schwartz and the definition of µ,
≤ Cρn
[1 +
1
ρsupB3ρ
u
],
which completes the proof.
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We have now done most of the work we need in order to complete the proof of Theorem 5.1.
Proof of Theorem 5.1. We first collect the results we have proven in the propositions in this sectionto get a bound on w = log
√1 + |Du|2. Let y ∈ Ω, and choose ρ > 0 such that Bn
3ρ ⊂ Ω. Then
w(y) ≤ 1
ωnρn
∫G(u)∩Bn+1
ρ (y)
w dHn by Proposition 5.5
≤ 1 +1
ωnρn
∫Bρ(y)n×|u−u(y)|<ρ
|Du|2
aw by Proposition 5.8
≤ 1 +C1
ωnρn
∫Bn
2ρ(y)×u−u(y)>−2ρ
a by Proposition 5.9
≤ 1 +C1
ωnρn
(ωn(2ρ)n + C2ρ
n
(1 +
1
ρsup
x∈B3ρ(y)
[u(x)− u(y)]
))by Proposition 5.12
≤ C3
(1 + sup
x∈B3ρ(y)
[u(x)− u(y)
ρ
]).
Now notice that we can take ρ = d/3, where d = dist(y, ∂Ω), to see that
w(y) ≤ C
(1 + sup
x∈Bd(y)
[u(x)− u(y)
d
]).
By exponentiating both sides, we finally come to the conclusion that
|Du(y)| ≤ expw(y) ≤ exp
(C
(1 + sup
x∈Bd(y)
[u(x)− u(y)
d
])),
which completes the proof.
References
[1] L.C. Evans. Partial differential equations. graduate studies in mathematics 19. AmericanMathematical Society, 1998.
[2] D. Gilbarg and N.S. Trudinger. Elliptic partial differential equations of second order, volume224. Springer Verlag, 2001.
[3] E. Giusti. Minimal surfaces and functions of bounded variation. 1984.
[4] Q. Han and F. Lin. Elliptic Partial Differential Equations. Courant Institute of MathematicalSciences, 2000.
[5] S.T Hughes. The dirichlet problem for the minimal surface equation. 2011.
[6] S.T Hughes. Topics in geometric analysis. 2011.
[7] J. Jost. Partial Differential Equations. Springer, 2002.
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