Let F and G be twice differentiable functions from R 2 into R , and satis- fying the partial differential equations

9.6.10. Let F and G be twice differentiable functions from R 2 into R , and satis- fying the partial differential equations

Δ F 2 = −F(1 − F 2 −G ), Δ G 2 = −G(1 − F 2 −G ). Assume that there exist ω 1 , ω 2 ∈R 2 that are with respect to the standard basis and

such that F and G are periodic, with periods ω 1 and ω 2 .

( a ) Show that F 2 +G 2 ≤1 . ( b ) Prove that if | ω 1 | and | ω 2 | are sufficiently small, then F and G are constant.

Solution. (a) Set P

1 + µω 2 ;0 ≤ λ , µ ≤ 1} ⊂ R ; P = {x ∈ P; F (x) + G (x) ≥ 1} ⊂ R . Let ζ

∈C ∞ (R 2 ) be a nonnegative function with compact support such that ζ =0 in a neighborhood of the origin. For any n ≥ 1, set ζ n (x) = n −2 ζ (x/n). Multi-

plying the equality satisfied by F (resp., by G) by F (F 2 +G 2 − 1) + ζ n (resp., by

G (F 2 +G 2 − 1) + ζ n ) and integrating by parts, we obtain, by addition and passing to the limit as n →∞,

+ 3(|∇F| + |∇G| )(F 2 +G 2 − 1) + |∇(F 2 +G 2 )| P )|

≤− " + (F 2 +G 2 )(F 2 +G 2 − 1) 2 P .

This inequality implies (a). ⊓ ⊔ Alternative proof of (a). Define u : R 2 2 (x) + G → R by u(x) = F 2 (x) − 1.

Clearly, u is an ( ω 1 , ω 2 )-periodic function of class C 2 (R 2 ). Thus, its values are determined by the values it takes on on the fundamental cell

P ={ λω 1 + νω 2 :0 ≤ λ , ν ≤ 1}.

364 9 Riemann Integrability In particular,

sup {u(x) : x ∈ R 2 } = sup{u(x) : x ∈ P}. But the fundamental cell is clearly compact, so there exists x 0 ∈ P such that

u (x 0 ) = sup{u(x) : x ∈ P},

by continuity of u; whence

u (x 0 ) = sup{u(x) : x ∈ R 2 },

by the preceding remark. Suppose now by way of contradiction that u takes a (strictly) positive value at some point. Then u (x 0 ) > 0, so u(x) > 0 for all x ∈

B (x 0 , r), some r > 0, by continuity. Here and here in after, B(x 0 , r) denotes the open r -ball centered at x 0 , that is, the set of all points x in the plane whose distance from x 0 is (strictly) less than r. It should be clear that

u (x 0 ) = sup{u(x) : x ∈ B(x 0 , r)};

(9.28) that is, u assumes an interior maximum value on B (x 0 , r). Now a straightforward

calculation yields Δ u = 2((∇F) 2 + (∇G) 2 ) + 2u(u + 1) ≥ 2u(u + 1).

(Use was made here of the condition Δ F + F(1 − F 2 −G 2 )= Δ G + G(1 − F 2 −

G 2 ) = 0.) This shows that

(9.29) Thus, u is a subharmonic function on B (x 0 , r), assuming an interior maximum value,

Δ u > 0 on B(x 0 , r).

by (9.28). The strong maximum principle now applies to show that u must be con- stant on B (x 0 , r). But then clearly Δ u = 0 on B(x 0 , r), thus contradicting (9.29). We conclude that u

2 + (G(x)) 2 (x) ≤ 0, i.e., (F(x)) 2 ≤ 1, for all x ∈ R .

(b) Consider the Hilbert space H of doubly periodic smooth functions u : P →R of periods ω 1 and ω 2 endowed with the norm

H = (u 2 #u# 2 (x) + |∇u(x)| )dx .

Let ( ϕ n ) n ≥0

be an orthonormal basis of eigenfunctions of the Laplace operator (− Δ ) in H and denote by ( λ n ) n ≥0 the corresponding eigenvalues. We can suppose that ϕ 0 = 1, so that λ n > 0 for any n ≥ 1. If P is “small,” so if | ω 1 | and | ω 2 | are suf- ficiently small, then λ n > 2, for any n ≥ 1. If u = (F,G) : R 2 →R 2 is a doubly periodic mapping that solves the above system, then we obtain the Fourier devel- opments

n ≥0

9.6 Qualitative Results 365 and

u 2 |u| = ∑ d n ϕ n .

n ≥0

Integrating the relations Δ F + F(1 −F 2 −G 2 ) = 0 and Δ G + G(1 −F 2 −G 2 )=0 over P, we obtain c 0 =d 0 . Next, multiplying the same identities by ϕ n (for n ≥ 1), we obtain, after integration, |d n |=( λ n − 1)|c n | > |c n |, provided that c n

other hand, since F 2 +G 2 ≤ 1, it follows that

" (F " 2 (x) + G 2 (x))dx ≥ (F 2 (x) + G 2 (x)) 3 dx ,

which can be rewritten, by Parseval’s identity,

2 ∑ 2 |c

n | ≥ ∑ |d n | .

n ≥0

n ≥0

This last inequality implies c n =d n = 0 for any n ≥ 1, so u is a constant mapping. Alternative proof of (b). We shall prove that if the diameter δ of the open

fundamental cell Ω ={ λ 1 ω 1 + λ 2 ω 2 :0 < λ 1 , λ 2 < 1}

does not exceed 1 /2 (e.g., if | ω k |≤ 1/4, k ∈ 1,2), then F and G are both constant. To begin with, notice that Ω is a relatively compact domain with a piecewise C 1

boundary. Next, define u ,v: Ω → C by u(x) = F(x) + iG(x) and v(x) = u(x)|u(x)| 2 , respectively. Clearly, u and v are both of class C 2 ( Ω ). But by (a), |u(x)| ≤ 1 for all

x ∈ Ω , so #u# L 2 (Ω ) ≥ #v# L 2 (Ω ) .

Let us show that if #u# L 2 (Ω ) = #v# L 2 (Ω ) , then F and G are both constant on Ω , so throughout the plane. Indeed, if #u# L 2 (Ω ) = #v# L 2 (Ω ) , then

" |u| 2 (1 − |u| 2 ) = 0,

so, with reference again to (a) above, |u|(1 − |u|) = 0 on all of Ω , by continuity. Since Ω is connected and |u| is continuous, it follows that either |u| = 0 on all of Ω or |u| = 1 on all of Ω . But

Δ |u| 2 = 2((∇F) 2 + (∇G) 2 ) + 2|u| 2 (|u| 2 − 1), so ∇F = ∇G = 0 on all of Ω , and F and G are indeed both constant throughout Ω .

Assume now once and for all that F and G are not constant. Then

(5) by the preceding. Next, recall that there exists a sequence { ϕ n } n ∈Z + of functions in

#u# L 2 (Ω ) > #v# L 2 (Ω )

C ∞ ( Ω ) and an unbounded increasing sequence { λ n } n ∈Z + of positive real numbers with the following properties:

(a) − Δϕ n = λ n ϕ n on Ω , for each index n;

366 9 Riemann Integrability

(b) the ϕ n are orthonormal in the Hilbert space L 2 ( Ω ), endowed with the standard scalar product

L 2 (Ω ) =

and their C-linear span is dense in L 2 ( Ω );

√ (c) the elements ϕ n / λ n are orthonormal in the Sobolev space H 1 0 ( Ω ), endowed

with the alternative scalar product

H 0 1 (Ω ) = (∇ ϕ )(∇ ¯ ψ ), Ω

and their C-linear span is dense in H 1 0 ( Ω ).

Recall the notation δ for the diameter of Ω . The Poincar´e inequality for ϕ m ,

√ # ϕ m # L 2 (Ω ) ≤ 2 δ # ϕ m # H 0 1 (Ω ) ,

√ in conjunction with the orthonormality of the ϕ in L 2 Ω

ϕ λ √ in n ( ) and of the n / n

H 0 1 ( Ω ), yields δ 2 λ m ≥ 1 for all m. Thus, if δ ≤ 1/2, then λ n ≥ 2 for all n. Assume henceforth that δ ≤ 1/2, so λ n ≥ 2 for all n, and consider the Fourier expansions of u and v in L 2 ( Ω ) in terms of the ϕ n :

u = ∑ a n ϕ n , with a n = u¯ ϕ n ,

n ∈Z +

and

v = ∑ b n ϕ n , with b n = v¯ ϕ n .

n ∈Z +

Since u ∈C 2 ( Ω )⊂L 2 ( Ω ), it follows that Δ u ∈ C( Ω )⊂L 2 ( Ω ), so the Fourier expansion of Δ u in L 2 ( Ω ), in terms of the ϕ n , is given by

Δ u =− ∑ a n λ n ϕ n ;

n ∈Z +

of course, use was made here of condition (a) above. But Δ u + u = v in Ω , so

b n = (1 − λ n )a n for all n, by the preceding. Recalling that λ n ≥ 2 for all n, it follows that |b n | ≥ |a n | for all n, whence

2 ∑ 2 |b

n | ≥ ∑ |a n | .

n ∈Z +

n ∈Z +

Finally, rewrite (5) in terms of Parseval’s identity to get

2 ∑ 2 |a

n | > ∑ |b n | ,

n ∈Z +

n ∈Z +

9.7 Independent Study Problems 367 and thus reach a blatant contradiction. Consequently, if the diameter of the open

fundamental cell does not exceed 1 /2, then both F and G must indeed be constant throughout the plane.

Remark. This problem is related to the study of stationary periodic solutions of the coupled Allen–Cahn/Cahn–Hilliard system that describes various phase transi- tion phenomena. The framework corresponds to doubly periodic solutions in several dimensions, and the results formulated in this problem are related to those proved in the one-dimensional case.