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Chapter 1

Krein-Rutman Theorem and the

Principal Eigenvalue

The Krein-Rutman theorem plays a very important role in nonlinear par-

tial differential equations, as it provides the abstract basis for the proof of
the existence of various principal eigenvalues, which in turn are crucial in
bifurcation theory, in topological degree calculations, and in stability anal-
ysis of solutions to elliptic equations as steady-state of the corresponding
parabolic equations. In this chapter, we first recall the well-known Krein-
Rutman theorem and then combine it with the classical maximum principle
of elliptic operators to prove the existence of principle eigenvalues for such
operators.
Let X be a Banach space. By a cone K ⊂ X we mean a closed convex
set such that λK ⊂ K for all λ ≥ 0 and K ∩ (−K) = {0}. A cone K in X
induces a partial ordering ≤ by the rule: u ≤ v if and only if v − u ∈ K.
A Banach space with such an ordering is usually called a partially ordered
Banach space and the cone generating the partial ordering is called the
positive cone of the space. If K − K = X, i.e., the set {u − v : u, v ∈ K}
is dense in X, then K is called a total cone. If K − K = X, K is called a
reproducing cone. If a cone has nonempty interior K 0 , then it is called a
solid cone. Any solid cone has the property that K − K = X; in particular,
it is total. Indeed, choose x0 ∈ K 0 and r > 0 such that the closed ball
Br (u0 ) := {u ∈ X : ku − u0 k ≤ r} is contained in K. Then for any
u ∈ X\{0}, v0 := u0 +ru/kuk ∈ K and hence u = (kuk/r)(v0 −u0 ) ∈ K−K.
We write u > v if u − v ∈ K \ {0}, and u  v if u − v ∈ K 0 .
Let X ∗ denote the dual space of X. The set K ∗ := {l ∈ X ∗ : l(x) ≥
0 ∀x ∈ K} is called the dual cone of K. It is easily seen that K ∗ is closed
and convex, and λK ∗ ⊂ K ∗ for any λ ≥ 0. However it is not generally true
that K ∗ ∩ (−K ∗ ) = {0}. But if K is total, this last condition is satisfied
and hence K ∗ is a cone in X ∗ . Indeed, if l ∈ K ∗ ∩ (−K ∗ ), then for every

1
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2 Maximum Principles and Applications

x ∈ K, l(x) ≥ 0, −l(x) ≥ 0, and therefore l(x) = 0 for all x ∈ K. Since

K − K = X, this implies that l(x) = 0 for all x ∈ X, i.e., l = 0.
Let Ω be a bounded domain in RN . It is easily seen that the set of
nonnegative functions K in X = Lp (Ω) is a cone satisfying K − K = X.
However, it has empty interior. Similarly the set of nonnegative func-
tions in W 1,p (Ω) gives a reproducing cone, and generally the nonnegative
functions in W k,p (Ω) (k ≥ 2, p > 1) form a total cone. On the other
hand, the nonnegative functions form a solid cone in C(Ω) but only form
a reproducing cone in C0 (Ω) := {u ∈ C(Ω) : u = 0 on ∂Ω}. If Ω has
C 1 boundary ∂Ω, then it is easy to see that the nonnegative functions in
C01 (Ω) := {u ∈ C 1 (Ω) : u = 0 on ∂Ω} form a solid cone; for example, any
function satisfying u(x) > 0 in Ω and Dν u(x) < 0 on ∂Ω is in the interior
of the cone, where ν denotes the outward unit normal of ∂Ω.

Theorem 1.1 (The Krein-Rutman Theorem, [Deimling(1985)] Theorem

19.2 and Ex.12) Let X be a Banach space, K ⊂ X a total cone and
T : X → X a compact linear operator that is positive (i.e., T (K) ⊂ K)
with positive spectral radius r(T ). Then r(T ) is an eigenvalue with an
eigenvector u ∈ K \ {0}: T u = r(T )u. Moreover, r(T ∗ ) = r(T ) is an
eigenvalue of T ∗ with an eigenvector u∗ ∈ K ∗ .

Let us now use Theorem 1.1 to derive the following useful result.

Theorem 1.2 Let X be a Banach space, K ⊂ X a solid cone, T : X → X

a compact linear operator which is strongly positive, i.e., T u  0 if u > 0.
Then

(a) r(T ) > 0, and r(T ) is a simple eigenvalue with an eigenvector

v ∈ K 0 ; there is no other eigenvalue with a positive eigenvector.
(b) |λ| < r(T ) for all eigenvalues λ 6= r(T ).

Let us recall that r is a simple eigenvalue of T if there exists v 6= 0 such

that T v = rv and (rI − T )n w = 0 for some n ≥ 1 implies w ∈ span{v}.

Proof. Step 1: There exists v0 > 0 such that T v0 = r(T )v0 .

Fix u ∈ K 0 . Then αT u ≥ u for some α > 0, and we can find σ > 0
such that Bσ (u) ⊂ K. It follows that w ≤ (σ)−1 kwku for any w ∈ X. Let
S = αT . Then

u ≤ S n u ≤ σ −1 kS n uku ≤ σ −1 kS n kkuku, ∀n ≥ 1.
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Krein-Rutman Theorem and the Principal Eigenvalue 3

Hence

kS n k ≥ σ/kuk and r(S) = lim kS n k1/n > 0.

n→∞

By Theorem 1.1, r(S) is an eigenvalue of S corresponding to a positive

eigenvector v0 ∈ K \ {0}. Clearly r(T ) = r(S)/α > 0 and T v0 = r(T )v0 .
Step 2: To prove that r(T ) is simple, we show a more general conclu-
sion: If r > 0 and T v = rv for some v > 0, then r is a simple eigenvalue
of T .
Let us first show that (rI − T )w = 0 implies w ∈ span{v}. Suppose
T w = rw with w 6= 0. Then T (v ± tw) = r(v ± tw) for all t > 0. Since T
is strongly positive, v ∈ K 0 and the above identity implies v ± tw 6∈ ∂K
unless v ± tw = 0. But v ± tw ∈ K 0 for small t and this cannot hold for all
large t for otherwise w ∈ K ∩ (−K) = {0}. Therefore there exists t0 6= 0
such that v + t0 w ∈ ∂K and hence v + t0 w = 0. This proves w ∈ span{v}.
Let (rI − T )2 w = 0. By what has just been proved, rw − T w = t0 v for
some t0 ∈ R1 . If t0 6= 0, then we may assume t0 > 0 (otherwise change w
to −w). Since

T (v + sw) = r(v + sw) − st0 v  r(v + sw) for all s > 0,

and v + sw ∈ K 0 for all small s ≥ 0, we easily deduce v + sw ∈ K 0 for all

s ≥ 0. This implies that w ∈ K, and hence w = r −1 (t0 v + T w) ∈ K 0 . We
now have

w − tv ∈ K 0 for all small t > 0,

but not for all large t > 0 as this would imply v = 0. Therefore there exists
t1 > 0 such that w − t1 v ∈ ∂K. But then

rw − t0 v − t1 rv = T (w − t1 v) ≥ 0, w − t1 v ≥ r−1 t0 v  0,

contradicting w − t1 v ∈ ∂K. Therefore we must have t0 = 0 and hence

rw − T w = 0, w ∈ span{v}. This proves that r is a simple eigenvalue.
Step 3: Next we show that T cannot have two positive eigenvalues
r1 > r2 corresponding to positive eigenvectors:

T v1 = r 1 v1 , T v 2 = r 2 v2 .

Let v(t) = v2 − tv1 , t ≥ 0. Since T is strongly positive, we have v1 , v2 ∈

K 0 . As before we have v(t) ∈ K 0 for small t but not for all large t.
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4 Maximum Principles and Applications

Therefore there exists t0 > 0 such that v(t0 ) ∈ K but v(t) 6∈ K for t > t0 .
We now have

v2 − t0 (r1 /r2 )v1 = r2−1 T (v2 − t0 v1 ) ∈ K,

which implies r1 ≤ r2 due to the maximality of t0 . This contradiction

proves step 3.
Step 4: If T w = λw with w 6= 0 and λ 6= r(T ), then |λ| < r(T ).
If λ > 0, then by Step 3, w 6∈ K. It follows that v0 + tw ∈ K for
all small t > 0 but not for all large t. Therefore there exists t0 > 0 such
that v0 + t0 w ∈ K and v0 + tw 6∈ K for t > t0 . It then follows that
v0 + t0 (λ/r(T ))w = r(T )−1 T (v0 + t0 w) ∈ K. The maximality of t0 implies
that λ ≤ r(T ) and hence λ < r(T ).
If λ < 0, then from T 2 w = λ2 w and T 2 v0 = r(T )2 v0 and the above
argument (applied to T 2 ) we deduce λ2 < r(T )2 and hence |λ| < r(T ).
Consider now the case that λ = σ + iτ with τ 6= 0. Then necessarily
w = u + iv and

T u = σu − τ v, T v = τ u + σv. (1.1)

We observe that u and v are linearly independent for otherwise we neces-

sarily have τ = 0. Let X1 := span{u, v}. Then (1.1) implies that X1 is an
invariant subspace of T . We claim that K1 := X1 ∩K = {0}. Otherwise K1
is a positive cone in X1 with nonempty interior, as for any w ∈ K1 \ {0},
T w ∈ X1 ∩ K 0 = K10 . We can now apply Step 1 above to T on X1 to
conclude that there exists r > 0 and w0 ∈ K10 such that T w0 = rw0 . By
Steps 2 and 3, we necessarily have r = r(T ) and w0 ∈ span{v0 }. In other
words, v0 ∈ K1 and v0 = αu + βv for some real numbers α and β. But
then one can use (1.1) and T v0 = r(T )v0 to easily derive α = β = 0, a
contradiction. Therefore K1 = {0}.
From span{u, v} ∩ K = {0} we find that the set

Σ := {(ξ, η) ∈ R2 : v0 + ξu + ηv ∈ K}

is bounded and closed. Since v0 ∈ K 0 , M := sup{ξ 2 + η 2 : (ξ, η) ∈ Σ} > 0

and is achieved at some (ξ0 , η0 ) ∈ Σ. Let z0 = v0 + ξ0 u + η0 v. Then
z0 ∈ K \ {0} and T z0 ∈ K 0 . Therefore we can find α ∈ (0, r(T )) such that
T z0 ≥ αv0 , i.e.,

(r(T ) − α)v0 + (ξ1 u + η1 v) ≥ 0, (1.2)

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Krein-Rutman Theorem and the Principal Eigenvalue 5

where

ξ1 = ξ0 σ + η0 τ, η1 = η0 σ − ξ0 τ.

Clearly

ξ12 + η12 = (σ 2 + τ 2 )(ξ02 + η02 ) = M |λ|2 .

By (1.2), we find that (ξ1 , η1 )/(r(T ) − α) ∈ Σ and hence

ξ12 + η12 ≤ M (r(T ) − α)2 ,

that is,

|λ|2 ≤ (r(T ) − α)2 ,

and hence |λ| < r(T ). The proof of Step 4 and hence the theorem is now
complete. 
Suppose now L is the elliptic operator and Ω the bounded domain as
given in Theorem A.4, namely

Lu = aij (x)Dij u + bi (x)Di u + c(x)u

has C α (Ω) coefficients and is strictly uniformly elliptic in the bounded

domain Ω which has C 2,α boundary. Choose ξ > 0 large enough so that
c − ξ < 0 in Ω, and denote Lξ u = Lu − ξu. Let K be the positive cone in
X := C01,α (Ω) consisting of nonnegative functions. For any v ∈ X, Theorem
A.1 guarantees that the problem

−Lξ u = v in Ω, u = 0 on ∂Ω

has a unique solution u satisfying

kuk2,α ≤ Ckvkα ≤ C1 kvk1,α

for some constant C1 > 0 independent of u and v. It follows that T : X → X

defined by T v = u is a compact linear operator. Moreover, by the weak
maximum principle Theorem A.34, T v ≥ 0 if v ∈ K. The strong maximum
principle Theorem A.36 then implies that u = T v > 0 in Ω if v ∈ K \ {0},
and the Hopf boundary lemma (Lemma A.35) gives further Dν u < 0 on
∂Ω. This implies that T v ∈ K 0 . Therefore T is strongly positive. It now
follows from Theorem 1.2 that r(T ) > 0 is a simple eigenvalue of T with
an eigenfunction v ∈ K 0 : T v = r(T )v. Thus u = T v satisfies

−Lu + ξu = r(T )−1 u in Ω, u = 0 on ∂Ω,

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6 Maximum Principles and Applications

i.e.,

Lu + λ1 u = 0 in Ω, u = 0 on ∂Ω,

with λ1 = r(T )−1 − ξ.

Generally, it is easily checked that µ is an eigenvalue of T if and only if
λ = µ−1 − ξ is an eigenvalue of

Lu + λu = 0 in Ω, u = 0 on ∂Ω. (1.3)

Theorem 1.2 now implies the following result.

Theorem 1.3 Under the conditions of Theorem A.4 for L and Ω, the
eigenvalue problem (1.3) has a simple eigenvalue λ1 ∈ R1 which corresponds
to a positive eigenfunction; none of the other eigenvalues corresponds to a
positive eigenfunction.

If the boundary operator is of Neumann or Robin type,

Bu = Dν u + σ(x)u, σ ≥ 0, σ ∈ C 1,α (∂Ω),

then we let X = C 1,α (Ω) and let K be the cone of nonnegative functions
in this space. We define the operator T analogously as in the Dirichlet
case and again find that it is compact on X and maps K to itself, due
to the weak maximum principle. Suppose now v ∈ K \ {0}. Then by
the strong maximum principle, u = T v > 0 in Ω. Moreover, by the Hopf
boundary lemma, if u(x0 ) = 0 for some x0 ∈ ∂Ω, then Dν u(x0 ) < 0 and
hence Bu(x0 ) < 0, contradicting the boundary condition. Therefore u > 0
on ∂Ω. Therefore T v > 0 on Ω, which implies that T v ∈ K 0 , i.e., T is
strongly positive. Therefore we can apply Theorem 1.2 to conclude that
Theorem 1.3 holds also for the Neumann and Robin boundary conditions.
The eigenvalue λ1 in Theorem 1.3 is usually called the principle eigenvalue.

Theorem 1.4 If λ 6= λ1 is an eigenvalue of (1.3) but the boundary con-

dition is either Dirichlet, or Neumann, or Robin type, then Re(λ) ≥ λ1 .

Proof. Suppose w > 0 is an eigenvector corresponding to λ1 and u is an

eigenvector corresponding to λ. Set v := u/w. Then

−λv = w−1 L(vw) = Lv − cv + 2w −1 aij Dj wDi v − λ1 v.

Writing

Kv := aij Dij v + b̃i Di v, b̃i := bi + 2w−1 aij Dj w,

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Krein-Rutman Theorem and the Principal Eigenvalue 7

we obtain

Kv + (λ − λ1 )v = 0.

Take complex conjugates to yield

Kv + (λ − λ1 )v = 0.

Next we compute

K(|v|2 ) = K(vv) = vKv + vKv + 2aij Di vDj v ≥ vKv + vKv,

since

aij ξi ξ j = aij (Re(ξi )Re(ξj ) + Im(ξi )Im(ξj )) ≥ 0

for any complex vector ξ ∈ C N . We now easily obtain

K(|v|2 ) ≥ 2(Re(λ) − λ1 )|v|2 in Ω.

Suppose now the boundary operator B is either Neumann or Robin

type. Then w > 0 over Ω and a direct computation shows Dν v = 0 and
Dν |v|2 = 0. If Re(λ) ≤ λ1 , then φ := |v|2 ≥ 0 satisfies

Kφ ≥ 0 in Ω, Dν φ = 0 on ∂Ω.

We now apply the strong maximum principle and Hopf boundary lemma
and conclude that φ ≡ constant, that is u = cw and hence λ = λ1 , a
contradiction. Therefore we must have Re(λ) > λ1 .
To prove the Dirichlet case, we replace w by w := w1− , 0 <  < 1, in
the above discussion and obtain
Lw
K(|v|2 ) ≥ −2(Re(λ) + )|v|2 in Ω.
w
Since

Lw = (1 − )w− Lw − (1 − )w −1− aij Di wDj w + cw1−

≤ (1 − )w− Lw + cw1− ≤ (C − (1 − )λ1 )w ,

where C = maxΩ c, we deduce

K(|v|2 ) ≥ 2((1 − )λ1 − C − Re(λ))|v|2 in Ω.

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8 Maximum Principles and Applications

By the Hopf boundary lemma, we know Dν w < 0 on ∂Ω. Therefore,

since u|∂Ω = 0,
u(x) Dν u(x0 )
lim = .
x→x0 ∈∂Ω w(x) Dν w(x0 )
It follows that
u(x)
lim v = lim w(x) = 0.
x→∂Ω x→∂Ω w(x)
If Re(λ) ≤ (1 − )λ1 − C, then,
K(|v|2 ) ≥ 0 in Ω.
Let Ωn be a sequence of smooth domains enlarging to Ω, e.g., Ωn := {x ∈
Ω : d(x, ∂Ω) > δ0 /n} with δ0 > 0 small. We apply the maximum principle
on Ωn and deduce maxΩn |v|2 ≤ max∂Ωn |v|2 . Letting n → ∞, it results
|v|2 = 0 and hence u = 0, a contradiction. Therefore Re(λ) > (1−)λ1 −C.
Letting  → 0 we obtain Re(λ) ≥ λ1 . 
Remark 1.5 The above proof shows that in the Neumann and Robin
boundary conditions case, Re(λ) > λ1 . This is also true in the Dirichlet
case; see Theorem 2.7 in the next chapter.
Remark 1.6 Instead of (1.3), sometimes one also needs to consider the
weighted eigenvalue problem
Lu + λh(x)u = 0 in Ω, Bu = 0 on ∂Ω,

where h(x) is a weight function, B is the Dirichlet, Neumann or Robin

boundary operator. If h(x) is positive and suitably smooth, similar results
to those in Theorems 1.3 and 1.4 can be analogously proved. If h(x) changes
sign, similar results can still be proved by considerably different techniques,
see [Hess-Kato(1980)].