We will study solutions of
uj+ fj(x,u1,u2, . . . ,up)=0 in, uj =0 on∂(j =1,2, . . . ,p) for p≥2,n ≥2, where f :Rn× Rp→Rpis a given smooth function, and the regionis assumed open, bounded andC2(or better). We assumeis con- nected during the argument, but this is not important – we may perturb each com- ponent separately. We write f(x,u) for thep×pmatrix [∂fj(x,u)/∂uk]pj,k=1. If zero is a regular value ofu→ u+ f(r,u), fromWq2∩Wq1(,Rp)0to Lq(,Rp),n <q <∞, then all solutionsuof u+ f(r,u)=0 in,u=0 on∂, are simple. This would hold for mostif zero were a regular value of (h,u)→h∗( + f)h∗−1u. We prove this by contradiction, under appropriate hypotheses on f. Effective hypotheses are far from obvious, so we start merely assuming reasonable smoothness and that, for everynear a certain0, there is a solutionu which is not simple. This will imply a sequence of additional conditions, and solvability of ever more over-determined problems. Then it will be clear how to select our hypotheses for an argument by contradiction.
Roughly speaking, we treat the cases (1) f(x,0)≡0;
(2) f(x,0)≡0 and f(x,0)≡0, depending onx;
(3) f(u) is independent ofx, f(0)=0, and f(0) has only simple eigenvalues (and pis not too large);
(4) p=2, f :R2→R2is independent ofx∈Rn, f(0)=0, f(0)=λI for some realλ, f(0)=0.
(The exact conditions are more complicated: see Theorem 8.5.1 for (1), (2);
Thm. 8.5.3 for (3); Thm. 8.5.4 for (4).) In each case, if f is fairly smooth, we conclude all solutions of u+ f(x,u)=0 in,u=0 on∂, are simple, for most choices of⊂Rn.
As in the scalar case (6.5), if there is a non-simple solutionu for every near0, then there is a solution (u, ψ) of
(∗)
u+ f(x,u)=0 in, u =0 on∂
ψ+Tf(x,u)ψ=0 in, ψ=0 on∂, ψ≡0
with∂ψ∂N r∂∂Nu ≡0 on∂. Also, by uniqueness in the Cauchy problem,∂ψ/∂N = 0 on a dense set of ∂; but since p≥2, we cannot conclude that ∂u/∂N vanishes.
8.5. Generic Simplicity of Solutions of a System 169 Applying the transversality theorem, we find: for everynear0there is a solution (u, ψ) of (∗)such thatψN ruN =0 and also
σ → ∂
∂N{BL∗ r(σ ψN)−AL∗ rT
f(r,u)ψBL(σuN)}ruN
+ψN r ∂
∂NBL(σuN)+σψN rf0
has finite rank,L = + f( r,u),L∗= +Tf(r,u), f0=( r,0). (We have used the factsψN N+HψN =0,uN N+H un+ f0=0 on∂.) Takingσ = γeiωθ,|∇∂θ| =1 in suppγ, as usual, we find
˙0 ruN +ψN rU˙0+γ ψN r f0=0, ˙k ruN +ψN rU˙k=0 fork≥1 where
Uk+LUk−1=0 (U−1=0), Uk|∂=
γuN ifk=0 0 ifk>0
k+L∗k−1= −Tf(u)ψUk−1, k|∂=
γ ψN ifk=0 0 ifk>0 and= ∇S r∇ +1/2 Sis the same forLorL∗.
It is convenient to treat a more general case, so assume: for every near 0, there is a solution (u, ψ) of (∗)such that
ψN rA=0 on∂ for alla ∈A and
ψN rMuN =0 on∂ for allM ∈M
whereA⊂C(Rn,Rp),M⊂C(Rn,Rp×p) are given sets (or linear spaces).
This certainly holds forA= {0},M= {I}, the constant functions. The same argument as above yields:
˙0 ra−γ ∂N(ψN ra)=0 ˙0 rMuN+ψN rMU˙0−σ ∂N(ψn rMuk)=0 and
˙k ra=0 ˙k rMuN +ψN rMU˙k=0, fork≥1.
Step 1. Takek=0: ifa,MareC1,
0=iγ ∂θψN ra+γ ∂N(ψN ra),
0=iγ(∂θψN rMuN +ψN rM∂θuN)+γ ∂N(ψN rMuN).
170 Chapter 8. The Method of Rapidly-Oscillating Solutions The first of these (usingψN N+HψN = Aψ|∂=0) yields
ψN r∂a/∂xj =0 on∂ (1≤ j ≤n), a ∈A∩C1. The second yields (sinceuN N +H uN+ f0=0 on∂, f0= f(r,0))
ψN r∂xjMuN−NjψN rM f0=0 on∂ (1≤ j ≤n).
This is not quite the form treated before, but a similar transversality argument yields (in the principal symbol, for a change)
0=ψN r∂xjMuN−i ej r∇∂θψN rM f0
where ej is the unit vector on the xj-axis. This shows each term vanishes separately: forM ∈M∩C1, on∂,
ψN r∂xjMuN =0 (1≤ j ≤n), ψN rM f0 =0.
Step 2. Next considerk=1, witha,MinC2. Writing f0= f(r,0), ˙1 ra= 0 yields
0=( ¨0+H˙0+A∂γ ψN+Tf0(γ ψN)ra.
We have 0= ∂(γ ψN ra) so, by previous results, 0= ∂(γ ψN)ra, and clearly 0=˙0 ra. Finally ¨0 ra = −i∂θ˙0 ra=i˙0 r∂θa=0 so
ψN r f0a=0. Similarly, 0=˙1 rMuN+ψN rMU˙1yields
0=ψN r(f0M+M f0)uN + ∂ψN rMuN+ψN rM ∂uN
−∂θ2ψN rMun−ψN rM∂θ2uN.
Ifn=2,θis arclength in the curve∂and ∂=∂2/∂θ2so 0=ψN r(fM +M f0)uN.
We will show this also holds forn≥3, by more calculation.
Step 3. From∂θ2(ψN rMuN)=0, we see
− ∂θ2ψN rMuN +ψN rM∂θ2uN
=2∂θψN rM∂θuN,
8.5. Generic Simplicity of Solutions of a System 171 so∂rψN rM∂ruN =A=constant for all directionsτ withτ⊥N,|r| =1, so (in component form)
A= p α,β=1
Mαβarg
τ r∇∂ψαNτ r∇∂uβn
= 1 n−1
α,β
Mαβ∇∂ψNα r∇∂uβN
= − 1
2(n−1)( ∂ψN rMuN+ψN rM ∂uN).
Then forτ⊥N,
2(n−2)∂τψN rM∂τuN = |τ|2ψN r(f0M+M f0)uN.
Further, previous results imply ∂τψN rMuN N+ψN N rM∂τuN =0, ψN N r
MuN N =0 so, for anyz∈Rn, with∂z=z r∇,
2(n−2)∂zψN rM∂zuN =(|z|2−(z rN)2)ψN r(f0M+M f0)uN. We apply transversality again (M ∈M∩C2) with this condition to conclude
σ →2(n−2){∂z(∂N(δψ)−N˙ r∇ψ) rM∂zuN
+∂zψN rM∂z(∂N(δu)−N˙ r∇u)}
−2z rN z rN˙ψN rM1uN
− |ztan|2((δψ)N rM1uN+ψN rM1(δu)N)
−σ∂N(2(n−2)∂zψN rM∂zuN− |ztan|2ψN rM1uN) has finite rank, whereM1 = f0M+M f0and
δu+ f(u)δu=(finite), δu|∂=σuN
δψ+Tf(u)δψ+Tf(u, ψ)δu=(finite), δψ|∂=σ ψN. Since ψ=0, on ∂, also, ˙N r∇ψ=0 on ∂ and we find ∂z( ˙N r∇ψ)= z rNN˙ r∇ψN = −z rN∇∂σ r∇∂ψN (assuming ∂N/∂N=0 on ∂); and similarly for u. We takeσ =γeiωθ as usual, and as usual, the principal part (coefficient ofω2) says nothing new. But the coefficient ofωis
0=2(n−2)γ{(−z r(N+iθ)i∂θψN+ztan r∇∂ψN)rM∂zuN
+∂zψN rM(−z r(N+iθ)i∂θuN+ztan r∇∂uN)}
+(2z rN z rθiγ− |ztan|2γ)ψN rM1uN.
172 Chapter 8. The Method of Rapidly-Oscillating Solutions
The part odd inθ must vanish, but this tells us nothing new. The even part also vanishes, so takingz=τ, θ=σ, unit tangent vectors (⊥N), we obtain 0=(n−2)γ τ rσ(∂τψN rM∂σuN+∂σψN rM∂τuN).Choosingτ =σwe see
(n−2)∂τψN rM∂τuN =0=ψN rM1uN
soψN r(M f0+ f0M)un=0 on∂forn≥2.
Thus the hypothesis.
(A,M): For allnear0, there is a solution (u, ψ) of (∗)such that on
∂, ψN ra=0 for alla ∈A, ψN rMuN =0 for allM ∈M, implies this still holds withA1,M1in place ofA,M,
A1=A∪ ∂a
∂xj
(j=1, . . . ,n) fora ∈A∩C1
∪ {Mf0,M ∈M∩C1}
∪ {f0a,a∈A∩C2} M1=M∪
∂M
∂xj
(j=1, . . . ,n) forM ∈M∩C1
∪ {f0M+M f0,M ∈M∩C2}.
We apply this, starting fromA= {0},M= {I}, to obtain firstA= {f0},M= {I,2f0}; then
A=
f0, ∂
∂xj
f0, f0f0
, M=
I,2f0,2∂xj f0,4f02
; etc.
Theorem 8.5.1. Suppose n≥2,p≥2,and let f :Rn×Rp→Rp, f:Rn× Rp →Rp×pbe at least C1and f0= f(r,0), f0= f(r,0)at least C2.Define linear spacesA⊂C(Rn,Rp),M⊂C(Rn,Rp×p)by
A=
j≥0
Aj M=
j≥0
Mj
A0= {0}, M0 = {I}, and for j ≥0,
Aj+1=span
Aj; ∂a
∂xk
(1≤k≤n,a∈Aj∩C1); f0a(a ∈Aj∩C2);
Mf0(M ∈Mj∩C1)
8.5. Generic Simplicity of Solutions of a System 173 Mj+1=span
Mj;∂M
∂xk
(1≤k≤n,M ∈Mj∩C1);
f0M+M f0(M∈Mj∩C2)
.
M contains (at least) all polynomials in f0 and A contains (polynomial in f0)× f0.
(1) Assume,for a dense set of x∈Rn, that
{a(x), all a∈A} =Rp.
Then for most C2bounded regions⊂Rn, all solutions of u+ f(x,u)=0in,u =0on∂,are simple.
(2) In case f(x,0)≡0,we haveA= {0},but assumeMis large so that: for a dense set of x∈Rn, wheneverβ=0 inRp,{M(x)β,all M∈M} = Rp.Then we have generic simplicity of solutions,as in case(1).
Remark. An important case is when f(u) is independent ofx and f(0)=0;
thenA= {0}andM= {p−1
k=0ckf(0)k(ck∈R)}, and hypothesis (2) ordinar- ily fails. (The only exception is whenp=2 andf(0) has complex eigenvalues.) We will treat this case below.
Proof. The difficult point, as usual, is density. We suppose this fails: for every near some0(connected, bounded, smooth), suppose there is a non-simple so- lution. Then there is a solution (u, ψ) of (∗)such thatψN ra =0, ψN rMuN = 0 on∂for alla∈A,M ∈M.
In case (1), we may suppose{a(x) alla∈ A} =Rpfor a dense (open) set in
∂and concludeψN =0 on∂, henceψ=0 in, a contradiction. (It suffices to consider a finite subset ofA, depending on0.)
In case (2), we may suppose the dense (open) set of the hypothesis is also dense in ∂. Since ψN =0 on a dense set of ∂, we conclude uN ≡0 on
∂. By uniqueness in the Cauchy problem,u ≡0. Thus for allnear0, there is a nontrivial solutionφofφ+ f0φ=0 in, φ=0 on∂, and it is not simple. This is a particular (linear) case of our initial hypothesis, andM is unchanged when we replace f(x,u) by f(x,0)u, so we obtain the same conclusion:φ=0, φN =0 on∂, soφ≡0 in, contrary to hypothesis.
Now suppose f(u) is independent ofx and f(0)=0. Our hypothesis for contradiction is that, for anynear0, there is a solution (u, ψ) of
(∗)
u+ f(u)=0 in, u =0 on∂
ψ+Tf(u)ψ=0 in, ψ =0 on∂, ψ≡0
174 Chapter 8. The Method of Rapidly-Oscillating Solutions
such thatψN rf(0)juN =0 on∂for all j ≥0 (or 0≤ j≤ p−1). For any ,u ≡0 solves the first equation, but the zero solution is usually simple. In fact, it fails to be simple if and only if some eigenvalueλof the matrix f(0) is also an eigenvalue of the scalar Dirichlet problem
φ+λφ=0 in, φ=0 on∂, φ≡0.
For a dense open set of, we avoid the eigenvalues of f(0) so assume hence- forth: the zero solution is simple. Then our hypothesis above includes the con- ditionu≡0.
Consider the case where all eigenvalues of f(0) are simple (distinct). Say f(0)pj =λjpj,Tf(0)qj =λjqj, withqj rpj =1,qj rpk=0 forj =k(1≤ j,k≤ p). Then the conditionψN rf(0)juN =0 for all j ≥0 becomes
k
λkjψN rpkqk ruN =0 for all j ≥0,
hence ψN rpkqk ruN =0 for k=1,2, . . . ,p. Suppose we can prove qk ruN =0 on a dense set of ∂, for every k; then ψN rpk=0 for every k, ψN =0 on∂, ψ =0 in, contrary to hypothesis.
The argument is a bit more general: if each eigenvalue of f(0) has only one eigenvector (as in the case f(0)= 0010
) andqk ruN =0 densely∂for every eigenvectorqkof Tf(0), we again concludeψ≡0.
In the following lemma we make no hypothesis about f(0).
Lemma 8.5.2. Assume f:Rp →Rpis C4with f(0)=0. Let Q⊂Cpbe any subspace= {0}and let V ⊂Rn be open and assume: for everynear0, there is a solution u of
u+ f(u)=0in, u=0on∂, u≡0,
such that q r∂u/∂N=0on∂∩V for every q∈ Q. Then, for everynear 0, there is a solution u≡0such that for all q ∈span{Tf(0)jQ: j ≥0},
q ruN =0, q rf(0)u2N =0, q r f(0)u3N =0, q r f(0)(uN,f(0)uN)=0, and q r fiv(0)u4N =0 on∂∩V.
Proof. The transversality theorem says there is a solutionu ≡0 such that, for q ∈ Q,q ruN =0 on∂∩V and also
σ →q r∂NBL(σuN)|∂∩V, L = + f(u)
8.5. Generic Simplicity of Solutions of a System 175 has finite rank. (It suffices to use a finite set of q in Q.) Taking σ = γeiθ,|∇∂θ| =1 on suppγ, we conclude
qrU˙k=0 on∂∩V fork≥0 (at least, 0≤k≤4.)
NowU0|∂=γuNandUk|∂=0 fork≥0, so alsoq rUk=0 on∂∩V, orq rUk(y+t N(y))=O(t2) ast →0 withy∈∂∩V,k≥0.
Since = ∇S r∇ +1/2S is a scalar operator, acting separately on each component, and q is constant, we have q rU0=q rU0=0 with q rU0|∂=0, and soq rU0(y+t N(y))=o(|t|N) ast →0,y∈∂, for ar- bitraryN – assuming∂, θ, γ are smooth, as we may.
Fork≥1
qrUk+ (q rUk−1)+q rf(u)Uk−1 ∼=0, q rUk=O(t2) Defineckby q r f(u)U0(y+t N(y))=c0+tc1+t2c2+t3c3+o(t3),recall- ing fisC3, and then
qrU1= −(c0+c1t+c2t2+c3t3)+o(t3) ast →0,in∂∩V.
Att =0,q rU˙1= −c0=0 orq r f(0)γuN =0 forq ∈ Q.Thus our hypothe- sis holds also with span{Q,Tf(0)Q}in place ofQ. We may, and shall, assume
Tf(0)Q⊂Q.
Since c0=0,q rU1= −c1t2/2+O(t3) so (q rU1)|∂= −c2 so 0= q rU2+ q rU1+q r f(u)U1|t=0=q rU˙2−c1,so c1=0. (We compute c1,c2,c3below.)
Sincec1=0,q rU1= −c2t3/3+O(t4) so
0=q rU2+(−2c2t+O(t2))+q r f(0)tU˙1+O(t2),
so q rU2=c2t2+O(|t|3) (recalling q r f(0) ˙U1 =0). Then 0=q rU3+ q rU2+q rf(u)U2|∂=q rU˙3+2c2so alsoc2=0.
Nowq rU1= −c3t4/4+o(t4), and this step is more delicate. We haveq r f(u)(tU˙1+t22Uă1+ ã ã ã)=tqr f(0) ˙U1+t2(qr f(0)uNU˙1+1/2q rf(0) ăU1)+ o(t2),and thet-coefficient vanishes, soq rU2=at3+o(t3) wherea=c3−
1/3(q rf(0)uNU˙1+1/2q r f(0) ¨U1).Then 0=q rU3+6at+tq rf(0) ˙U2+ o(t) so q rU3 = −3at2+o(t2), and 0=q rU˙4−6a, or a=0. In fact q rU1=O(t4) for anyq ∈Qso alsoq r f(0)U1=O(t4) andq rf(0) ¨U1=0 in∂∩ ∇. Thus we have
q rf(u)U0=c0+c1t+c2t2+c3t3+o(t3)
= t3
3(q r f(0) ˙U1)+o(t3)
=q r(f(u)− f(0))U0+o(t3)
176 Chapter 8. The Method of Rapidly-Oscillating Solutions
=q r f(0)
tuN+t2
2uN N+t3
6uN N N+ ã ã ã
r
γUN+tU˙0+t2
2Uă0+ ã ã ã
+1/2q rf(0)
tuN+t2
2uN N+ ã ã ã 2
r(γUN+tU˙0+ ã ã ã)
+1/6q rfiv(0)(tuN+ ã ã ã)3(γUN + ã ã ã)+o(t3).
The coefficient oftvanishes, soq rf(0)u2N =0 in∂∩V,which implies 0=
∂τ(q rf(0)u2N)=2q rf(0)uN∂τuN for anyτ⊥N, so q r f(0)uNU˙0 =0.
Thus the coefficient oft2is 0= γ
2q rf(0)uNuN N+γ
2q rf(0)u3N
since uN N+H uN =0 (= u+ f(u)|∂) we have q rf(0)u3N =0 on
∂∩V. Omitting some terms clearly zero, we find from the coefficient of t3,
γ(q r fiv(0)u4N +q rf(0)uNuN N N)
=2q r f(0)uNU˙1−3q r f(0)uNU¨0
= −5q rf(0)uNU¨0−2q rf(0)uN(HU˙0+ ∂γuN+ f(0)γuN)
=5γq r f(0)uN∂θ2uN−2γq r f(0)uN( ∂uN + f(0)uN).
Since 0=∂N( u+ f(u))|∂=uN N N −(H2+H2)uN+ ∂uN+f(0)uN, we have, in suppγ ⊂∂∩V,
5q r f(0)uN∂θ2uN =q rfiv(0)u4N+q r f(0)uN( ∂uN + f(0)uN).
Since∂θ2(q rf(0)u2N)=0, the left-side is−5q rf(0)(∂τuN)2when∇∂θ= τ⊥N,|τ| =1, and this is independent of the choice ofτ. Thus
q r f(0)(∂τuN)2= average
τ⊥N,|τ|=1
q r f(0)(τ r∇∂uN)2
= 1
n−1q r f(0)(∇∂uN)2
= − 1
n−1q r f(0)uN ∂uN
(The penultimate term is ambiguous, but clear in component form.) Then (n−6)q r f(0)(∂τuN)2=Q(uN) for|τ| =1, τ⊥N, whereQ(uN)=q r fiv(0)u4N +q rf(0)(uN, f(0)uN).
8.5. Generic Simplicity of Solutions of a System 177 A similar condition occurred above, just before Theorem 5.1, and this is treated the same way. Since q rf(0)∂τuNuN N =0=q rf(0)u2N N(uN N =
−H uN) forz∈Rn,
(n−6)q rf(0)(∂zuN)2− |z|2tanQ(uN)=0 on∂∩V.
We apply the transversality theorem, and as usual the principal part says nothing new, but the sub-principal part gives (with z=τ⊥N,∇∂θ=σ⊥N,|σ| =
|τ| =1)
2(n−6)q r f(0)(∂τuN)(∂τuN +τ rσ ∂σuN)= Q(uN)uN
=2{Q(uN)+(τ rσ)2Q(uN)}.
Choose τ =σ: then Q(uN)uN =4Q(uN), so the quadratic part vanishes, q r f(0)(uN, f(0)uN)=0. if n ≥3, we may choose τ⊥σ, so also q r
fiv(0)u4N =0.This completes the proof of the lemma.
Theorem 8.5.3. Let f :Rp →Rp be C4 with f(0)=0. Assume f(0) has only simple eigenvalues, or more generally, each eigenvalue of f(0)has only one eigenvector. Further assume, wheneverβ =0inRp, that
Rp= span
j≥o f(0)j{β,f(0)β2, f(0)β3, f(0)(β, f(0)β),fiv(0)β4}.
Then for most bounded C2regions⊂Rn, all solutions of u+ f(u)=0i n⊂Rn, u =0on∂, are simple.
Proof. Let {q1, . . . ,qm}be a basic for the eigenfunctions of Tf(0) and let {V1,V2, . . .}be a countable basis for the topology ofRn– for example, all open balls of rational radius whose center has rational coordinates. By the lemma, for eachj ∈ {1,2, . . . ,m}and eachk∈ {1,2, . . .}, there is an ample (=category II) set of regionssuch that, for any solutionu≡0, we haveqj r∂u/∂N =0 at some point of∂∩Vk(unless the intersection is empty). In the countable in- tersection, which is still ample, any solution u ≡0, satisfies qj r∂u/∂N = 0 on a dense set of ∂, for all j ∈ {1, . . . ,m}. As note above (before the lemma), this implies denseness of the property “only simple solutions.”
Example. (p =2) f :R2→R2isC4with f(u)= f(0)u+
Q1(u2)+K1(u3)+L1(u4)+o(|u|4) Q2(u2)+K2(u3)+L2(u4)+o(|u|4)
178 Chapter 8. The Method of Rapidly-Oscillating Solutions
whereQ1,Q2are (real homogeneous) quadratic polynomials;K1,K2are cu- bics;L1,L2are quartics; and we consider the cases
(i) f(0) has two complex eigenvalues (ii) f(0)=(λ0
0 à) has real distinct eigenvalues (iii) f(0)=(λ1
0 λ) has one eigenvalue and only one eigenvector (iv) f(0)=(λ0
0 λ) has one real eigenvalue with two eigenvactors.
(We may always change theu-variable to put f(0) in real Jordan canonical form.)
Case (i): Theorem 8.5.1, part (2), applies. Say f(0)(p+i q)=(α+iβ) (p+i q) whereα, β ∈Randp,q ∈R2, β=0 and (p,q)=(0,0). It is easy to see p,q are independent whenβ=0, and verify{γ, f(0)γ}are independent for anyγ =0 inR2.
Case (ii): Lete1=col(1,0),e2=col(0,1): then assuming 0=Q1 e22+K1 e23+L1 e42 and
0=Q1 e12+K2 e31+L2 e41, the hypotheses of Theorem 8.5.3 hold.
Case (iii): Ifα r f(0)jβ=0 forj=0,1, andα=0, β=0 inR2, thenα1=0, β2=0, and we may suppose without loss of generalityα=e2, β=e1. The hypotheses of Theorem 8.5.3 hold if
0=Q2 e21+K2 e13+L2 e41.
Case (iv): This case is not included above, but will be permitted in the next theorem provided for someγ ∈R2
det
γ1 Q1(γ2) γ2 Q2(γ2)
≡0 and for allγ =0 inR2, the 2×4 matrix
[γ,Q(γ2),K(γ3),L(γ4)]
has rank 2.
Remark. Many cases remain open – for example Case (iv) when f is odd, f(−u)= −f(u), or f(0)=λIwithn≥3.
8.5. Generic Simplicity of Solutions of a System 179 Theorem 8.5.4. Let f :R2→R2 be C4, f(0)=0, f(0)=(λ0 1λ)for some realλ, and assume
(i) for someγ =0inR2, f(0)γ2is not a real multiple ofγ, or det[γ, f(0)γ2]≡0;
(ii) for everyγ =0inR2, the2×4matrix
[γ,f(0)γ2, f(0)γ3, fiv(0)γ4] has rank 2.
Then for most C2 bounded regions⊂R2 (n ≥2), all solutions u of u+ f(u)=0in,u=0on∂are simple.
Proof. As noted earlier, we may assume the zero solution is simple. If there is a non-simple solution for everynear some0, then there is a solution (u, ψ) of (∗)withu ≡0, ψ ≡0 andψNruN =0 on∂. We apply the transversality theorem to conclude that also (in usual notation)
˙k ruN+ψN rU˙k=0 on∂, fork≥0. The equation fork=0 says nothing new; considerk=1.
0= −˙1 ruN −ψN rU˙1
=( ¨0+H˙0+ ∂γ ψN+ f(0)γ ψN)ruN
+ψN r( ¨U0+HU˙0+ ∂γuN + f(0)γuN), which (with f(0)=λI) simplifies to
γ( ∂ψN ruN +ψN r ∂uN)=γ ∂θ2ψN ruN+ψN r∂θ2uN
.
Ifn=2, then ∂=∂θ2so we have nothing new; but assumingn ≥3, it follows (as above) that, whereγ =0,
∂θ2ψN ruN+ψN r∂θ2uN = −2∂θψN r∂θuN
= 1
n−1(ψN r ∂uN+ ∂ψN ruN) so ∂τψN r∂τuN =0 for any τ⊥N. For any tangent vector field σ,Dσ = (scalar)σ r∇∂+ (scalar) is a general first-order operator, and a second-order
180 Chapter 8. The Method of Rapidly-Oscillating Solutions (tangential) operator is a sum of termsDσ1Dσ2+Dσ3, and we have
ψN ruN =0, DσψN ruN+ψN rDσuN =0, Dσ1ψN rDσ2uN+Dσ2ψN rDσ1uN =0 and, for any second-order tangential operatorD2,
D2ψN ruN+ψN rD2uN =0.
(For example, ∂(γ ψN)ruN+ψN r ∂(γuN)=0.)
This greatly simplifies the calculation fork=2 – in fact, almost everything vanishes and 0=ψN rU˙2+˙2 ruN reduces to 0=2γ ψN rf(0)u2N.
In casen=2, the equations fork=0,k=1, tell us nothing. We must go directly toψN rU˙2+˙2 ruN =0, knowing onlyψN ruN =0 andn=2. For- tunately, the calculation simplifies forn =2 sinceq =H=κ(the scalar cur- vature of∂),θis arc-length along∂, ∂=∂2/∂θ2, andσ0= −q+H+ i ∂θ =0. Again everything cancels but one term, though the calculations are more difficult and we conclude, forn≥2, ψN rf(0)u2N =0 on∂.
Now let {β1. . . , βm} be the finite set of β ∈R2 with |β| =1 and det[β,f(0)β2]=0. Chooseγk =βk⊥∈R2, βkrotated 90◦. By Lemma 8.5.2 and hypothesis (ii), we may suppose any non-zero solution u satisfies γk ruN =0 on a dense set of∂for each k=1, . . . ,m, so{uN, f(0)u2N} is a basis forRn. ButψNruN =0 andψN r f(0)u2N =0 on∂, soψN ≡0 on
∂, ψ ≡0 in, contrary to hypothesis.
Summary of 8.2 (1)|∇∂θ| ≡1,∇S r∇S =0,S|∂=iθ,Re∂∂Ns >0
S(y+t N(y))=
k≥0
tkSk(y) k!
=iθ(y)+t−qt2 2! + t3
3!(−q2+3r+i∂θq) + t4
4!(−i∂θS3+3qS3−12s−6i∇∂q rK∇∂θ)+O(t5) whereq=θ r Kθ,r=θ r K2θ,S=θ r K3θ, θ= ∇∂θ, ∂θ =θ r∇∂.
(2)= ∇S r∇ +1/2σ, σ = S+b r∇S,b(y+t N(y))=b(y)+tb(y)˙ +
t2
2!b(y)ă + ã ã ã
σ(y+t N(y))=σ0(y)+tσ1(y)+t2
2!σ2(y)+ ã ã ã
= −q+H+i ∂θ+b r(N+iθ)
+t{S3−q H−H2+i∂θH−2idiv∂(Kθ)
8.5. Generic Simplicity of Solutions of a System 181 +b˙ r(N+iθ)+b r(−q N−i Kθ)}
+t2
2!{S4+q H S3+2q H2+2H3−i∂θH2
−4i(Kθ)rH+b¨ r(N+iθ)+2 ˙b r(−q N−i Kθ) +6idiv∂(Kθ)− ∂q+b r(S3N−q+2i K2θ)}
+O(t3)
Hm= traceKm,H =H1,K =D N =curvature.
∇S r∇ =∂t+i∂θ+t(−q∂t−2i Kθ r∂y) + t2
2(S3∂t+6i K2θ r∂y−q r∂y)+ ã ã ã
=L
(3)
( + b r∇ +c)eωS(U0 + 2ω1 U1 + (2ω)12U2 + ã ã ã)=2ωeωS(F0 +
2ω1 F1 + ã ã ã)
Uj|∂=Gj, Uj+LUj−1=Fj (U−1=0)
U0=F0, U0(y+t N(y))=G0(y)+tU˙0(y)+t2
2!Uă0(y)+ ã ã ã Att =0:
F0 =U˙0+i∂θG0+1/2σ0G0 F˙0 =U¨0+ i∂θ+1/2σ0−q
U˙0+ −2i Kθ r∂y+1/2σ1
G0
F¨0 = U˙¨0+ i∂θ+1/2σ0−2q
U¨0+(−4i Kθ r∂y+σ1+S3
U˙0
+ 6i K2θ r∂y−q r∂y+1/2σ2
G0
U1+LU0=F1, U1|∂=G1, U1=G1+tU˙1+t2
2Uă1+ ã ã ã Att =0:
F1=U˙1 + i∂θ+1/2σ0
G1+U¨0+(H+b rN) ˙U0 + ∂G0+b r∇∂G0+cG0
F˙1=U¨1 + i∂θ+1/2σ0−q
U˙1+ −2i Kθ r∂y+1/2σ1
G1
+U˙¨0+(H+b rN) ¨U0+ ∂U˙0+b r∇∂U˙0+cU˙0
182 Chapter 8. The Method of Rapidly-Oscillating Solutions +( ˙b rN −H2) ˙U0−2 div∂(K G0)
+( ˙b+H) rG0−b rK G0+cG˙ 0 U2+LU1 =F1 U2|∂=G2 Att =0:
F2=U˙2+ i∂θ+1/2σ0
G2+U¨1+(H+b rN) ˙U1+ ∂G1
+b r∇∂G1+cG1
Appendix 1
Eigenvalues of the Laplacian in the Presence of Symmetry
A. L. Pereira, in his doctoral thesis investigated more general symmetry condi- tions than the one discussed in 6.2. We describe some of his results (and sketch some of the arguments) that appear with more details in [28], which is the general reference for this section. We proved (Example 6.1) that, for mostC2- regular bounded regions⊂Rn, all eigenvalues of
u+λu =0 in, u=0 in
are simple. Given this result, it may be surprising to find higher multiplicity in many specific examples. This is well-known in the case of the disc or the rectangle inR2. Another example is the equilateral triangle, where eigenvalues with arbitrarily high multiplicity exist (see Pinsky [29]). It seems that higher multiplicity appears whenever a detailed analysis can be done. Now, a common characteristic in these cases is the symmetry of the regions. In fact, it is the symmetry that makes a detailed analysis possible. To be more precise, we start with some definitions. LetGbe a subgroup of the orthogonal groupO(n) onRn. We say thatisG-symmetric(orG-invariant) ifg()=for everyg∈G.
A maph:→Rnis (G-)equivariantifh◦g=g◦hfor allg∈G, soh() isG-symmetric whenisG-symmetric.
A functionφonRn isevenorG-invariantifφ◦g =φfor allg∈G. If isG-symmetric anduis an eigenfunction thenu◦g−1is also an eigenfunction for every g∈G, so the multiplicity of λ is at least the dimension of span {u◦g−1|g∈G}.
In fact, suppose there exists a pointx∈such that{g ∈G:gx =x} = {Id}
(such a point is called a free point under the natural action ofGin; there is always a free point inifGis, for example, finite or commutative). Then, if some irreducible representation ofGhas real dimensionm, there are infinitely many eigenvalues of the Laplacian with multiplicity≥min anyG-symmetric bounded open set.
183
184 Appendix 1. Eigenvalues of the Laplacian in the Presence
If we want only simple eigenvalues, the irreducible representations of G must have real dimension one, i.e., G is isomorphic to Z2⊕Z1⊕ ã ã ã ⊕Z2
(mtimes) orGis generated bymcommuting reflections. In any other case the symmetry forces higher multiplicity.
SupposeisG-symmetric,λis an eigenvalue of the Laplacian inwith an eigenfunctionu; if span{u◦g−1:g∈G} =N(λ+); i.e., the orbit of some eigenfunction generates all eigenfunctions corresponding toλ, we say thatλis aG-simple eigenvalue. [If we use complex-valued functions, we should also include theG-orbit of the conjugate ¯u.] Equivalently,λisG-simple if the natural action ofG(u→u◦g−1) onN(+λ) is irreducible.
Supposeis aG-symmetric bounded open set inRn which isC2regular, andλis aG-simple eigenvalue of multiplicitym. Then, for every,C2,G- equivariant region ˜nearthere is a unique eigenvalueλ( ˜) for the Laplacian in ˜which is close toλ=λ(), λ( ˜) is alsoG-simple of multiplicitym, and ˜ →λ( ˜) is analytic.
Thus aG-simple eigenvalue in aG-symmetric region behaves like a simple eigenvalue when no symmetry is imposed. The most we could hope for then is: in generic G-symmetricC2-domains, all eigenvalues of the Laplacian are G-simple. This is proved in certain cases – for finite commutative groups (argument sketched below) and for finite subgroups of O(2)⊂O(n) (such as the non-commutative dihedral groups) along with partial results for some other cases (see [28]).
The commutative case is comparatively simple, since the irreducible representations all have real or complex dimension one it suffices to use characters. AcharacterofGis a mapχ:G→Cwith|χ(g)| =1, χ(Id)=1, χ(g1 •g2)=χ(g1)•χ(g2) forg,g1,g2∈G. The character is real if it only has values in{−1,1}. Ifχis a character, define
Mχ= {u∈L2(,C)|u◦g−1=χ(g)u,g∈G}
Ifχ1,χ2are distinct characters, thenMχ1⊥Mχ2. In fact, ifu∈ Mχ1, v∈ Mχ2, then
uv¯ =
u◦g−1v¯◦g−1=χ1(g)χ2(g)
uv,¯ for allg∈G, so if
uv¯ =0, we must haveχ1(g)=χ2(g) for allg∈G, or χ1=χ2. SinceGis commutativeL2(,C)= ⊕χ Mχ (see [20] p. 51). Further (H2∩H01(,C)∩Mχ)⊂Mχ for every characterχandG-symmetric. It suffices then to prove the following:
Appendix 1. Eigenvalues of the Laplacian in the Presence 185 (A) For every characterχand integerk, there is a dense open set of
G-invariantC3regionssuch that the restriction |H2∩H01(,C)∩Mχ →Mχ has only simple eigenvaluesλwith|λ| ≤k.
(B) For an integerkand every pair of characters withχ1 =χ2andχ1 =χ¯2, there is a dense open set ofG-invariantC3regionssuch that there is no eigenvalueλwith|λ| ≤kwith both an eigenfunction inMχ1and one in Mχ2.
For real characters the argument is like in Example 6.1. In any case, openness is easy to obtain; to obtain density we use transversality arguments.
For (A) with a complexχ, given aC3,G-invariant0 ⊂Rn consider the map
(u, λ,h)−→F h∗(+λ)h∗−1u∈ Mχ
withu∈ H2∩H01(,C)∩Mχ\{0}, λ∈C,|λ| ≤k,h∈ E3whereE3= {C3 G-equivariant imbeddings of0 ⊂Rn}. It is enough to show that 0 is a regular value of F. If this is not true, we find that there exists λ∈C,|λ| ≤k and u, ψ∈ H2∩H01(,C)∩Mχ\ {0}such that
u+λu =0 in, ψ+λψ=0 in
ψu¯ =0, Re ∂ψ¯
∂N ∂u
∂N
=0 on∂.
(∗)
(Note thatψ=i ku,k∈R\{0}satisfies all conditions except that
ψu¯ =0).
To see this is not possible for most(i.e., except in a closed set with empty interior), we consider the map
(u, ψ, λ,h)−→G
h∗(+λ)h∗−1u,h∗(+λ)h∗−1ψ,
ψu,¯ h∗Bh∗−1(u, ψ)|∂
where B(u, v)=Reu•v,¯ u, ψ∈ H2∩H01(,C)∩Mχ\{0}, λ∈C,|λ| ≤ k,h∈ E3 and the values are inMχ ×Mχ ×C×L1(,C). Again, it is easy to see that G−1 (0, 0, 0, 0) is closed. If it is (0, 0, 0, 0) for some choice of (u, ψ, λ) for everyh in an open set, we may supposei0 is in its interior and then, condition (2.β) of Theorem 5.4 must fail, so the quotient
R(DG(u, ψ, λ,h)) R
∂G
∂(u,ψ,λ)
186 Appendix 1. Eigenvalues of the Laplacian in the Presence
must be finite dimensional. This implies that for every,C3-close to0, there is a solution (u, ψ, λ) of (∗) such that
σ−→ Re{uN∂NB+λ(σψ¯N)+ψN∂NB+λ(σuN)}
withσeven, has finite rank. HereB+λis given byv=B+λg, whereB+λis defined by
(+λ)v∈N(+λ) v|∂=g, v⊥N(+λ) The method of rapidly oscillating solutions shows, asω→ ∞
(γcosωθ)=ω−1γcosωθRe(∇∂uN•∇∂ψ¯N−∂θuN∂θψ¯N)+O(ω−2) whereγ, θare even|∇∂θ| =1 in suppγandγ is supported in a small neigh- borhood of{G x}for somex∈∂– recallGis finite. Then Re(∂τuN∂τψN)= Re(∇∂uN •∇∂ψ¯N) for allτ⊥N,|τ| =1 which implies
Re(∂τuN∂τψ¯N)= 1
n−1 Re(∇∂uN •∇∂ψ¯N).
Ifn ≥3 we conclude Re(∂τuN∂τu¯N)=0 for allτ⊥N. Ifn =2, more calcu- lation is needed – we only quote the result:
(γcosωθ)= −ω−3λγRe(∂θψ¯N∂θuN)+O(ω−4).
Thus in every case, on∂
Re(uN,ψ¯N)=0 and Re(unψ¯N)=0
where indicates any tangential derivative. Consider a point where uN =0, ψN =0
uN =q1+i q2 ψN =r1+ir2 (qj,rjreal),
then q1r1+q2r2=0,q1r1+q2r2 =0 so, for some α,r1=αq2,r2 = −αq1
and 0=α(q2q1 −q1q2). If∇∂α =0, almost all tangential directions have α1 =0 soq1/q2(orq2/q1) is locally constant. It follows, for somez∈C− {0}, that u/zis real valued in. But this is impossible in Mχ\{0}, whenχ is a complex character. Ifαis locally constant, for someα∈R, ψN =iαuN on a nonempty open set of∂, soψ= −iαuin, 0=
ψ¯ •u=iα
|u|2. Thus α=0 soψ=0 which is false.
This contradiction shows, for an open dense set of E3(∗) is not solvable;
restrictingFto this set, we obtain (A).
Appendix 1. Eigenvalues of the Laplacian in the Presence 187 (B) Suppose, for everyG,C3-near0, there existsu, v, λwith
u∈Mχ1, v∈Mχ2, u, vinH2∩H01(,C)\{0} |λ| ≤k u+λu =0 in, v+λv=0 in
(∗∗)
We may assume λ is a simple eigenvalue for the restriction of− in Mχ1 and in Mχ2. For ˜G-symmetric nearthere is a unique eigenvalueλj( ˜) nearλfor the restriction oftoMχj andλ1( ˜)=λ2( ˜). Thus the derivative ofλ1( ˜)−λ2( ˜) vanishes at ˜=so (assuming
|u|2=
|v|2=1) we also have|∂∂Nu|2= |∂∂vN|2 on∂. Thus, for everyh∈ E3neari0, there exists u, v∈ H2∩H01(,C)\{0}withu∈ Mχ1, v∈ Mχ2, andλ∈C, with|λ| ≤k and
(h∗(+λ)h∗−1u,h∗(+λ)h∗−1v,h∗C h∗−1(u, v)|∂)=(0,0,0) where C(u, v)= |∇u|2− |∇v|2; the values are in Mχ1×Mχ2×L1(∂).
Theorem 5.4 (2.β) then implies for everyG-symmetric region,C3-near0, there is a solution (u, v, λ) of (∗∗) which also satisfies
|u|2=1=
|v|2,
|uN|2= |vN|2on∂and
σ −→Re( ¯uN∂N(B+λ(σuN)+v¯N∂NB+λ(σ vN)) has finite rank. This implies (withγ, θeven and|∇∂θ| =1 is suppγ)
(γcosωθ)=2ω−1γcosωθ(|∇∂uN|2− |∇∂vN|2
− |∂θuN|2+ |∂θvN|2)+O(ω−2) Ifn=2, we have
(γcosωθ)= −ω−3λ
2cosωθ(|∂θuN|2− |∂θvN|2)+O(ω−4) Then – sinceGis finite – we find|∂τuN| = |∂τvN|on∂for allτ⊥N.
LetuN =Reiρ, vN =Reiσ, then
|∂τuN| = |Rτ+i Rρτ| = |Rτ+i Rστ|
soρτ2 =στ2, ∂τ(ρ−σ)∂τ(ρ+σ)=0 whenτ⊥N. Eitherρ−σ (anduN/vN) is locally constant orρ+σ(anduN/¯vN) is locally constant, thus eitheru, vare linearly dependent (soχ1=χ2) oru,v¯are linearly dependent (so χ1=χ¯2).
Therefore we have a contradiction and (B) follows.
With (A) and (B) we conclude: for some G-symmetric C2 regions⊂ Rn(n≥2), all eigenvalues areG-simple: simple eigenvalues when the eigen- function corresponds to a real character, double when it corresponds to a com- plex character with independent eigenfunctionsu,u¯(or Re,u, Imu).