The Innity-Potential in the Square Karl K. Brustad Abstract. A representation formula for the solution of the 1-Laplace

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The Inﬁnity-Potential in the Square

Karl K. Brustad

Abstract. A representation formula for the solution of the ∞-Laplace

equation is constructed in a punctured square, the prescribed boundary

values being u= 0 on the sides and u= 1 at the centre. This so-called

∞-potential is obtained with a hodograph method. The heat equation is

used and one of Jacobi’s Theta functions appears. The formula disproves

a conjecture.

1. Introduction

I shall obtain a special explicit solution of the ∞-Laplace Equation

∆∞u:= u2

xuxx + 2uxuyuxy +u2

yuyy = 0

in the plane. The solution is the ∞-potential (or the ”capacitary function”)

for a square with the centre removed, at which boundary point the value

u= 1 is prescribed. On the four sides u= 0.

Figure 1. The graph of the solution over the square.

Let Ω be the square 0 <x<2,0< y < 2. I shall show that the viscosity

solution of the Dirichlet problem just described is

u(x, y) = min

θ∈[0,π/2] max

r∈[0,1] nr(xcos θ+ysin θ)−W(r, θ)o(1.1)

where

W(r, θ) = 8

πr4

6sin(2θ) + r36

210 sin(6θ) + r100

990 sin(10θ) + ···.

arXiv:2210.03447v2 [math.AP] 1 Dec 2022

2 K. K. Brustad

The formula is valid in the quadrant 0 ≤x≤1, 0 ≤y≤1, and the obvious

symmetries extends the solution to the whole Ω. So far as we know, the

only hitherto known explicit ∞-potentials are for stadium-like domains with

solutions on the form dist(x, ∂Ω).

The explicit solution (1.1) settles a conjecture, indicated in [JLM99] and

[JLM01]. A calculation shows that uis not a solution to the ∞-eigenvalue

problem

max Λ−|∇u|

u,∆∞u= 0, u|∂Ω= 0.

Se also [Yu07].

The ∞-Laplace equation ∆∞u= 0 was introduced by G. Aronsson in

1967 as the limit of the p-Laplace equation

∆pu:= div |∇u|p−2∇u= 0

when p→ ∞. See [Aro67], [Aro68]. For this 2nd order equation, the concept of

viscosity solutions was introduced by T. Bhattacharya, E. DiBenedetto, and

J. Manfredi in [BDM89], and uniqueness of such solutions with continuous

boundary values was proved by R. Jensen, cf. [Jen93]. For the p-Laplace

equation, the capacitary problem in convex rings was studied by J. Lewis,

[Lew77]. A detailed investigation of this problem for the ∞-Laplace equation

was given in [LL19], [LL21].

I use the hodograph method, according to which a linear equation is

produced for a function w=w(p, q) in the new coordinates

p=∂u(x, y)

∂x , q =∂u(x, y)

∂y .

A solution W(r, θ) = w(rcos θ, r sin θ) is constructed with judiciously ad-

justed boundary values. When transforming back we get a formula for uin

terms of its gradient, which, in turn, is shown to be a critical value of the

objective function in (1.1). The critical point – that is, the obtained minimax

(r, θ) – is the length r=|∇u|and the direction θ= arg ∇uof the gradient

at (x, y). A full account is given in Section 6.

It is interesting that the 2nd Jacobi Theta function

ϑ2(z, q) = 2

∞

k=1

q(k−1/2)2cos((2k−1)z)

appears in the formulas. For example, we shall see that

u=4

πˆarg ∇u

ϑ22ψ, |∇u|16dψ, 0≤arg ∇u≤π/2,

and that the determinant of the Hessian matrix of uis

det Hu=−π2

16 |∇u|4ϑ−2

22 arg ∇u, |∇u|16,arg ∇u6= (2k−1)π/4.

An immediate consequence of the last identity is that uis not C2on the

diagonals of the square. Indeed, the symmetries yield arg ∇u(x, x) = π/4

and ϑ2is zero for odd integer multiples of z=π/2.

The Inﬁnity-Potential in the Square 3

Write

D1:= {(r, θ)|0< r < 1,0< θ < π/2}.

Deﬁne W:D1→R, as

W(r, θ) := 8

∞

n=1

rm2

(m2

n−1) mn

sin (mnθ), mn= 4n−2,(1.2)

and let u: Ω →Rbe the extension of formula (1.1) to Ω. I prove

Theorem 1.1. The function uis the unique viscosity solution of the problem











∆∞u= 0 in Ω\ {(1,1)},

u= 0 on ∂Ω,

u= 1 at (1,1),

(1.3)

in the square Ω = {(x, y)|0< x < 2,0< y < 2}. Moreover, u∈C1(Ω \

{(1,1)})and uis real-analytic, except at the diagonals and medians.

Actually, we do not know how smooth uis across the medians.

2. Proof of the Theorem

The coeﬃcients in Wand the exponents on rare chosen so that W(1, θ) is

the Fourier series for the odd π-periodic extension of cos θ+ sin θ−1, and so

that rWr+Wθθ = 0. Thus, Wis the solution of the problem











rWr+Wθθ = 0 in D1,

W(r, 0) = W(r, π/2) = 0,0≤r≤1,

W(0, θ) = 0,

W(1, θ) = cos θ+ sin θ−1,0≤θ≤π/2.

(2.1)

I prove ﬁrst that (1.1) is the viscosity solution to the Dirichlet problem











∆∞u= 0 in Ω1,

u(0, t) = u(t, 0) = 0,

u(1, t) = u(t, 1) = tfor 0 ≤t≤1,

(2.2)

where

Ω1:= {(x, y)|0<x<1,0< y < 1}.

We know in advance that the solution must be linear on the medians. It is

then straight forward to show that the gluing (2.15) of the translations and

rotations of (1.1) is the solution of the original problem (1.3).

For each (x, y)∈Ω1I denote the objective function f(x,y):D1→Rin

(1.1) by

f(x,y)(r, θ) := r(xcos θ+ysin θ)−W(r, θ).

Note that we immediately have

u(x, y)≥min

θ∈[0,π/2] f(x,y)(0, θ)=0.

4 K. K. Brustad

Figure 2. The solution to problem (2.2) over the square

Ω1= [0,1]2.

The bounds in the Proposition below show that the correct boundary values

are obtained.

Proposition 2.1. For all (x, y)∈Ω1,

1−p(1 −x)2+ (1 −y)2≤u(x, y)≤dist((x, y), ∂Ω).

Proof.

u(x, y) = min

θ∈[0,π/2] max

r∈[0,1] f(x,y)(r, θ)

≥min

θ∈[0,π/2] f(x,y)(1, θ)

= min

θ∈[0,π/2] {xcos θ+ysin θ−(cos θ+ sin θ−1)}

= 1 −max

θ∈[0,π/2] {(1 −x) cos θ+ (1 −y) sin θ}

= 1 −p(1 −x)2+ (1 −y)2max

θ∈[0,π/2] cos (θ−φ)

where the last line is due to a standard trigonometric identity. The lower

bound is conﬁrmed, since the maximum is at θ=φ:= arctan 1−y

1−x∈[0, π/2].

Since W≥0 (see Lemma 2.1),

u(x, y) = min

θ∈[0,π/2] max

r∈[0,1] f(x,y)(r, θ)

≤min

θ∈[0,π/2] max

r∈[0,1] {r(xcos θ+ysin θ)}

= min

θ∈[0,π/2] {xcos θ+ysin θ}

= min {x, y}= dist((x, y), ∂Ω).



I prove next that uis ∞-harmonic in the viscosity sense in Ω1. The

strategy is to ﬁrst show that uis a classical solution in Ω1except on the

diagonal. At the diagonal it is necessary to demonstrate that the smoothness

breaks down in such a way that no test function can touch ufrom below.

Yet, it is essential to know that uis C1, because we need the gradient of a

test function to align with the diagonal when the touching is from above. The

proof is then completed by showing that x7→ u(x, x) is convex.

The Inﬁnity-Potential in the Square 5

It turns out that the Heat Equation is helpful. Consider an insulated

rod of length π/2 with initial temperature v(0, θ) = 1, for 0 < θ < π/2.

Suppose the rod is subjected to the heat equation vt=vθθ with boundary

conditions v(t, 0) = v(t, π/2) = 0 when t > 0. The solution of this textbook

example is

v(t, θ) = 8

∞

n=1

e−m2

sin(mnθ), mn= 4n−2.

It becomes

U(r, θ) := rWr(r, θ)−W(r, θ) = 8

∞

n=1

rm2

sin(mnθ) (2.3)

under the substitution

r=e−t.

Since W∈C(D1) with W(1,0) = W(1, π/2) = 0, an immediate consequence

is that Wr(r, θ) has jumps from 0 to 1 at the two corner points r= 1, θ= 0

and r= 1, θ=π/2. Otherwise it is continuous up to the boundary ∂D1.

The temperature v(t, θ) is strictly decreasing. To see this note that h:=

vtis again caloric with initial values h(0, θ)≤0 and lateral values h(t, 0) =

h(t, π/2) = 0. The strong maximum principle then yields h < 0 at inner

points, since certainly h6≡ 0. Since Ur=rWrr it follows that

0> vt=Ur

dt=−r2Wrr

and Wrr >0 in D1. Integrating back, we see that also Wrand Ware positive

in D1. Moreover, Uθθ =vθθ =vt<0 and

θ7→ Uθ(r, θ) = rWrθ(r, θ)−Wθ(r, θ) = 8

∞

n=1

rm2

ncos(mnθ)

is therefore strictly decreasing for each 0 < r < 1. Its zero is obviously at

θ=π/4. The following Lemma is proved.

Lemma 2.1. Let W:D1→Rbe the series deﬁned in (1.2).

(I) The functions W,Wr, and Wrr are positive in D1.

(II) W, Wθ∈CD1, and Wr∈CD1\ {(1,0),(1, π/2)}.Wris discontin-

uous at the two points (r, θ) = (1,0) and (r, θ) = (1, π/2).

(III) For 0< r < 1,

Uθ(r, θ) = rWrθ(r, θ)−Wθ(r, θ)

is positive when 0≤θ < π/4, negative when π/4< θ ≤π/2, and thus

zero in D1precisely when θ=π/4.

Let

B1:= (p, q)|0< p2+q2<1, p > 0, q > 0

be the sector in the ﬁrst quadrant and deﬁne w:B1→Ras Win Cartesian

coordinates, i.e., w(rcos θ, r sin θ) := W(r, θ). I shift to vector notation

x:= [x, y]>∈Ω1,p:= [p, q]>∈B1,r:= [r, θ]>∈D1,

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TheInnity-PotentialintheSquareKarlK.BrustadAbstract.Arepresentationformulaforthesolutionofthe1-Laplaceequationisconstructedinapuncturedsquare,theprescribedboundaryvaluesbeingu=0onthesidesandu=1atthecentre.Thisso-called1-potentialisobtainedwithahodographmethod.TheheatequationisusedandoneofJacobi'sTh...

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The Innity-Potential in the Square Karl K. Brustad Abstract. A representation formula for the solution of the 1-Laplace

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