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.