M7160 Obyˇcejné diferenciální rovnice II
M7160 Ordinary Differential Equations II
Michal Veselý
Text vychází z pˇrednášek prof. Lomtatidze
Masarykova univerzita, Pˇrírodovˇedecká fakulta
Ústav matematiky a statistiky
Se sazbou v systému LATEX pomáhali Jakub Juránek a Jiˇrina Šišoláková
Tato publikace vznikla v rámci projektu Fond rozvoje Masarykovy
univerzity (projekt MUNI/FR/1133/2018) realizovaného v období
1/2019–12/2019.
Contents
1 Auxiliary results 1
1.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Carathéodory class . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Absolute continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 Existence of solutions 8
3 Extendability of solutions 13
4 Set of solutions 18
5 Upper and lower solutions 21
6 Wintner theorem 29
7 Uniqueness of solutions 31
8 Krein theorem 37
9 Kneser theorem 47
10 Fukuhara theorems 51
1. Auxiliary results
1.1 Notations
R+ non-negative real numbers
x = (xi)m
i=1 column vectors with coordinates x1,...,xm
Rm
the space of real vectors x = (xi)m
i=1 with the norm
x =
m
∑
i=1
|xi|
B[x0,r] the closed ball with the centre x0 ∈ Rm
and the radius r ≥ 0, i.e.,
B[x0,r] = {x ∈ Rm
; x−x0 ≤ r}
x·y the Euclidean inner product of vectors x,y ∈ Rm
x ≤ y the inequality between vectors x = (xi)m
i=1,y = (yi)m
i=1 ∈ Rm
such
that xi ≤ yi, i ∈ {1,...,m}
sgnx the vector (sgnxi)m
i=1
[a,b] the closed interval
(a,b) the open interval
m(A) the Lebesgue measure of a set A ⊆ Rm
int(A) the interior of a set A ⊆ Rm
∂A the boundary of a set A ⊆ Rm
I a real interval, which is not degenerated to a point
C(I,Rm
) the space of continuous and bounded vector functions u: I → Rm
with the norm
u C = sup{ u(t) ;t ∈ I}
Cloc(I,Rm
) the set of continuous vector functions u: I → Rm
˜C([a,b],Rm
) the set of absolutely continuous vector functions u: [a,b] → Rm
˜Cloc(I,Rm
) the set of vector functions u: I → Rm
, which are absolutely continuous
on any compact subinterval [a,b] ⊆ I
˜Cn
([a,b],R) the set of all functions u: [a,b] → R, whose n-th derivatives are
absolutely continuous
˜Cn
loc(I,R) the set of all functions u: I → R such that u ∈ ˜Cn
([a,b],R) for all
interval [a,b] ⊆ I
L(I,Rm
) the space of all vector functions u: I → Rm
which are strongly integrable
in the Lebesgue sense with the norm
u L =
I
u(s) ds
Lloc(I,Rm
) the set of all vector functions u: I → Rm
such that u ∈ L([a,b],Rm
)
for all interval [a,b] ⊆ I
1
Other notations are given by the range of considered values. For example, L(I,D)
is the set
{u: I → D;u ∈ L(I,Rm
)},
where D ⊆ Rm
.
1.2 Carathéodory class
Definition 1.1. Let A ⊆ Rm
and D ⊆ Rn
be given. We say that a vector function
g: I × A → D belongs to the Carathéodory class and we write g ∈ K(I × A,D) if the
following conditions:
1. the function g(t,−): A → D is continuous for almost all t ∈ I;
2. the function g(−,x): I → D is measurable for all x ∈ A;
3. for all r > 0, there exists hr ∈ L(I,R+) such that
g(t,x) ≤ hr(t), t ∈ I, x ∈ A∩B[0,r],
are satisfied.
Definition 1.2. We say that a vector function g: I ×Rm
→ Rn
is from the set
Kloc(I ×Rm
,Rn
)
if
g ∈ K([a,b]×Rm
,Rn
)
for all [a,b] ⊆ I.
Lemma 1.1. If g ∈ K(I ×Rm
,Rn
) and if x ∈ Cloc(I,Rm
), then the vector function
t → g(t,x(t)), t ∈ I,
is measurable.
Lemma 1.2. Let g ∈ K(I ×Rm
,Rn
) be given and let A ⊂ Rm
be bounded. Let
g (t) = sup{ g(t,x) ;x ∈ A}, t ∈ I.
Then, g ∈ L(I,R+).
Lemma 1.3. Let g ∈ K(I ×Rm
,Rn
). Then, there exists a function h ∈ L(I,R+) and a
non-decreasing function ϕ ∈ Cloc(R+,R+) such that
g(t,x) ≤ h(t)ϕ( x ), t ∈ I, x ∈ Rm
.
Lemma 1.4. Let g ∈ K(I ×Rm
,Rn
). For all r > 0, there exists a function hr ∈ L(I,R+)
and a non-decreasing function ϕr ∈ Cloc(R+,R+) such that ϕr(0) = 0 and
g(t,x)−g(t,y) ≤ hr(t)ϕr( x−y ), t ∈ I, x,y ∈ B[0,r].
2
Lemma 1.5. Let g ∈ K(I ×Rm
,Rn
). The operator
F(x)(−) = g(−,x(−)), x ∈ C(I,Rm
),
maps the space C(I,Rm
) into the space L(I,Rn
) continuously.
Proof. For all x ∈C(I,Rm
), the function F(x): I → Rn
is measurable (see Lemma 1.1).
According to Definition 1.1, we have that F(x) ∈ L(I,Rn
). Thus, the operator maps the
space C(I,Rm
) into the space L(I,Rn
).
Let {xk}∞
k=1 ⊆ C(I,Rm
) and x ∈ C(I,Rm
) satisfy
lim
k→∞
xk −x C = 0.
Let r > 0 be such that
xk(t) ≤ r, x(t) ≤ r, t ∈ I, k ∈ N.
Therefore, there exists a function hr ∈ L(I,R+) such that (see Lemma 1.4)
g(t,xk(t))−g(t,x(t)) ≤ hr(t), t ∈ I, k ∈ N,
i.e.,
F(xk)(t)−F(x)(t) ≤ hr(t), t ∈ I, k ∈ N.
Since the function g(t,−): Rm
→ Rn
is continuous for almost all t ∈ I, it holds
lim
k→∞
[F(xk)(t)−F(x)(t)] = 0
for almost all t ∈ I. By the Lebesgue theorem, we have
F(xk)−F(x) L =
I
F(xk)(s)−F(x)(s) ds → 0 as k → ∞.
Thus, the operator F is continuous.
Lemma 1.6. Let g ∈ K(I ×Rm
,Rn
). For arbitrary r > 0, let a function hr ∈ L(I,R+)
satisfy
g(t,x) ≤ hr(t), t ∈ I, x ∈ B[0,r].
Then, there exist functions gk : I ×Rm
→ Rn
for k ∈ N such that:
1. all functions gk have all partial derivatives with respect to the last m variables
and all functions gk and their partial derivatives belong to the class
K(I ×Rm
,Rn
);
2. for all r > 0, the inequality
gk(t,x) ≤ hr+1(t), t ∈ I, x ∈ B[0,r], k ∈ N,
holds;
3. for almost all t ∈ I and all r > 0, it holds
lim
k→∞
gk(t,x) = g(t,x)
uniformly in B[0,r].
3
Remark 1.1. If
g(t,x) ≤ h(t), t ∈ I, x ∈ Rm
,
then one can assume that hr ≡ h for all r > 0.
Remark 1.2. The functions gk from the statement of Lemma 1.6 have continuous partial
derivatives. Therefore, they are locally Lipschitz, i.e., for all r > 0 and k ∈ N, there
exists a function lr,k ∈ L(I,R+) such that
gk(t,x)−gk(t,y) ≤ lr,k(t) x−y , t ∈ I, x,y ∈ B[0,r].
Remark 1.3. The functions gk from Lemma 1.6 can be chosen in such a way that they
have the following property. For all r > 0, there exists a function ωr ∈ K(I ×R+,R+)
which is non-decreasing with respect to the second variable, ωr(−,0) ≡ 0, and
gk(t,x)−gk(t,y) ≤ ωr(t, x−y )+ωr t,
1
k
, t ∈ I, x,y ∈ B[0,r], k ∈ N.
Proof of Lemma 1.6. Let ϕk : Rm
→ R+ for k ∈ N be functions satisfying the following
conditions:
a) the functions ϕk have continuous all partial derivatives on Rm
;
b) the functions ϕk satisfy
ϕk(x) = 0, x ∈ Rm
, x ≥
1
k
, k ∈ N;
c) it holds
Rm
ϕk(x)dx = 1, k ∈ N.
Such functions can be constructed as follows. Let
ϕ(s) =
e− 1
1−s , 0 ≤ s < 1;
0, s ≥ 1.
Let ρk > 0, k ∈ N, be such that
ρk
Rm
ϕ m2
k2
x·x dx = 1.
We put
ϕk(x) = ρk ϕ m2
k2
x·x , x ∈ Rm
, k ∈ N.
Obviously, a) and c) are valid. It is known that
x 2
≤ m2
(x·x), x ∈ Rm
.
If
x ≥
1
k
,
then
m2
k2
(x·x) ≥ 1
4
and, consequently, ϕk(x) = 0. Therefore, b) is valid as well.
We define
gk(t,x) =
Rm
ϕk(y−x)g(t,y)dy, t ∈ I, x ∈ Rm
, k ∈ N.
The functions gk satisfy the condition 1. Let r > 0 and x ∈ B[0,r] be arbitrarily given.
We have
gk(t,x) =
B[0,r+1]
ϕk(y−x)g(t,y)dy, t ∈ I, k ∈ N.
We obtain
gk(t,x) ≤
B[0,r+1]
ϕk(y−x) g(t,y) dy
≤ hr+1(t)
B[0,r+1]
ϕk(y−x)dy
≤ hr+1(t)
Rm
ϕk(y)dy = hr+1(t), t ∈ I, k ∈ N,
which gives the condition 2.
It remains to prove the condition 3. Let r > 0 and x ∈ B[0,r] be arbitrarily given.
Since
Rm
ϕk(y−x)dy = 1, k ∈ N,
we have
g(t,x) = g(t,x)
Rm
ϕk(y−x)dy =
Rm
ϕk(y−x)g(t,x)dy, t ∈ I, k ∈ N.
Therefore,
gk(t,x)−g(t,x) =
Rm
ϕk(y−x)[g(t,y)−g(t,x)] dy
=
B[x, 1
k ]
ϕk(y−x)[g(t,y)−g(t,x)] dy, t ∈ I, k ∈ N.
(1.1)
Due to Lemma 1.4, there exists a function ωr+1 ∈ K(I ×R+,R+) which is non-decreasing
in the second variable, ωr+1(−,0) ≡ 0, and
g(t,y1)−g(t,y2) ≤ ωr+1(t, y1 −y2 ), t ∈ I, y1,y2 ∈ B[0,r +1].
Therefore,
g(t,x)−g(t,y) ≤ ωr+1 t,
1
k
, t ∈ I, y ∈ B x,
1
k
.
5
From (1.1), we have
gk(t,x)−g(t,x) ≤
B[x, 1
k ]
ϕk(y−x) g(t,y)−g(t,x) dy
≤ ωr+1 t,
1
k
B[x, 1
k ]
ϕk(y−x)dy
≤ ωr+1 t,
1
k
Rm
ϕk(y−x)dy = ωr+1 t,
1
k
, t ∈ I, k ∈ N.
Thus, 3. is valid.
1.3 Absolute continuity
Definition 1.3. We say that a function x: [a,b] → Rm
is absolutely continuous on [a,b]
if, for every ε > 0, there exists δ > 0 such that, for any finite system of pairwise disjoint
subintervals
(ak,bk) ⊆ [a,b], k ∈ {1,...,n},
satisfying
n
∑
k=1
(bk −ak) < δ,
it holds
n
∑
k=1
x(bk)−x(ak) < ε.
We recall that the set of all absolutely continuous functions x: [a,b] → Rm
is denoted
by
˜C([a,b],Rm).
Theorem 1.1. A function x: [a,b] → Rm
is called absolutely continuous if and only if
the following conditions:
1. the function x is differentiable almost everywhere in [a,b];
2. the derivative
x ∈ L([a,b],Rm
);
3. it holds
β
α
x (s)ds = x(β)−x(α), α,β ∈ [a,b],
are satisfied.
6
Remark 1.4. Let t0 ∈ [a,b] be arbitrarily given. If h ∈ L([a,b],Rm
), then the function
x: [a,b] → Rm
given by
x(t) =
t
t0
h(s)ds, t ∈ [a,b],
is absolutely continuous and x (t) = h(t) for almost all t ∈ [a,b].
7
2. Existence of solutions
Let us consider the equation
x = f(t,x), (2.1)
where f ∈ Kloc(I ×Rn
,Rn
).
Definition 2.1. Let I0 ⊆ I be an interval. We say that a function x: I0 → Rn
is a solution
of Eq. (2.1) if:
1. x ∈ ˜Cloc(I0,Rn
);
2. it holds
x (t) = f(t,x(t))
for almost all t ∈ I0.
Let t0 ∈ I0, c0 ∈ Rn
. A solution x: I0 → Rn
of Eq. (2.1) satisfying the condition
x(t0) = c0 (2.2)
is called the solution of the Cauchy problem (2.1), (2.2).
Definition 2.2. Let A,B be sets of functions x: I → Rn
and let t0 ∈ I. An operator
T : A → B is called t0-Volterra if, for all x,y ∈ A and all t ∈ I such that
x(s) = y(s), min{t0,t} ≤ s ≤ max{t0,t},
it holds
T(x)(t) = T(y)(t).
Lemma 2.1. Let t0 ∈ [a,b], c0 ∈ Rn
, r > 0, and let
T : C([a,b],B[c0,r]) → C([a,b],B[c0,r])
be a continuous t0-Volterra operator such that T(c0)(t0) = c0. Let there exist a function
ω ∈ C([0,b−a],R+) such that ω(0) = 0 and that
T(x)(t)−T(x)(s) ≤ ω(|t −s|), t,s ∈ [a,b], x ∈ C([a,b],B[c0,r]).
Then, the operator T has at least one fixed point, i.e., there exists a function
x ∈ C([a,b],B[c0,r])
such that
T(x)(t) = x(t), t ∈ [a,b].
Proof. Without loss of generality, we can assume that the function ω is non-decreasing.
We denote
Ik,j = t0 −
j(t0 −a)
k
,t0 +
j(b−t0)
k
, j ∈ {1,...,k −1}, k ∈ N.
8
For k ∈ N and t ∈ [a,b], we define
sk(t) =



t +
t0 −a
k
, t < t0 −
t0 −a
k
;
t0, t ∈ Ik,1;
t −
b−t0
k
, t > t0 +
b−t0
k
.
Obviously, the functions sk : [a,b] → [a,b], k ∈ N, are continuous and
|sk(t)−sk(τ)| ≤ |t −τ|, t,τ ∈ [a,b], k ∈ N.
For t ∈ [a,b], we denote
yk,0(t) = c0
and
yk,j(t) = T(yk,j−1)(sk(t)), j ∈ {1,...,k −1}, k ∈ N.
We show that
yk,j(t) = yk,j−1(t), t ∈ Ik,j, j ∈ {1,...,k −1}, k ≥ 2, k ∈ N. (2.3)
For an arbitrarily given integer k ≥ 2, using the induction, we prove that the identity
yk,j(t) = yk,j−1(t), t ∈ Ik,j, (2.4)
is valid for all j ∈ {1,...,k −1}. Obviously,
yk,1(t) = T(yk,0)(sk(t)) = T(c0)(sk(t)) = T(c0)(t0) = c0 = yk,0(t), t ∈ Ik,1.
We assume that (2.4) is valid for some j ∈ {1,...,k −2}. We show that (2.4) is also
valid for j +1. If t ∈ Ik,j+1, then sk(t) ∈ Ik,j. Therefore,
yk,j+1(t) = T(yk,j)(sk(t)) = T(yk,j−1)(sk(t)) = yk,j(t), t ∈ Ik,j+1.
Hence, (2.4) is valid for all j ∈ {1,...,k −1}. By the induction, we have proved (2.3).
We denote
xk(t) = yk,k−1(t), t ∈ [a,b], k ∈ N.
One can see that
xk ∈ C([a,b],B[c0,r]), k ∈ N,
and that (see (2.3))
xk(t) = yk,k−2(t), t ∈ Ik,k−1, k ≥ 2, k ∈ N.
Since
sk(t) ∈ Ik,k−1, t ∈ [a,b], k ≥ 2, k ∈ N,
we obtain
xk(t) = T(yk,k−2)(sk(t)) = T(xk)(sk(t)), t ∈ [a,b], k ≥ 2, k ∈ N. (2.5)
Next, we get
xk(t)−xk(τ) = T(xk)(sk(t))−T(xk)(sk(τ))
≤ ω(|sk(t)−sk(τ)|)
≤ ω(|t −τ|), t,τ ∈ [a,b], k ≥ 2, k ∈ N.
9
Therefore, the functions xk, k ≥ 2, k ∈ N, are uniformly bounded and equicontinuous.
We can use the Arzelà–Ascoli theorem.
Without loss of generality, we can assume that
lim
k→∞
xk(t) = x(t)
uniformly for t ∈ [a,b], where
x ∈ C([a,b],B[c0,r]).
We have
T(xk)(sk(t))−T(x)(t) ≤ T(xk)(sk(t))−T(xk)(t) + T(xk)(t)−T(x)(t)
≤ ω(|sk(t)−t|)+ T(xk)(t)−T(x)(t)
for all t ∈ [a,b], k ∈ N. Since the operator T is continuous, ω(0) = 0, and since
lim
k→∞
sk(t) = t, t ∈ [a,b],
we obtain (see (2.5))
x(t) = lim
k→∞
T(xk)(sk(t)) = T(x)(t), t ∈ [a,b],
i.e., x is a fixed point of the operator T.
Theorem 2.1. Let r > 0 and [a,b] ⊆ I be such that t0 ∈ [a,b] and
t
t0
fc0
(s,r)ds ≤ r, t ∈ [a,b],
where
fc0
(t,r) = sup{ f(t,x) ;x ∈ B[c0,r]}, t ∈ [a,b].
Then, the problem (2.1), (2.2) has a solution on [a,b].
Proof. The problem (2.1), (2.2) is equivalent with the equation
x(t) = c0 +
t
t0
f(τ,x(τ))dτ, t ∈ [a,b].
Especially, if x ∈ C([a,b],Rn
) satisfies this integral equation, then x ∈ ˜C([a,b],Rn
).
Therefore, it suffices to prove that there exists a function x ∈ C([a,b],Rn
) satisfying
this integral equation.
We define
T(x)(t) = c0 +
t
t0
f(τ,x(τ))dτ, t ∈ [a,b], x ∈ C([a,b],B[c0,r]).
It is obvious that the operator
T : C([a,b],B[c0,r]) → C([a,b],Rn
)
10
is t0-Volterra. According to Lemma 1.5, the operator T is continuous. For
x ∈ C([a,b],B[c0,r]),
we have
T(x)(t)−c0 ≤
t
t0
fc0
(τ,r)dτ ≤ r, t ∈ [a,b],
i.e.,
T(x)(t) ∈ B[c0,r], t ∈ [a,b].
Thus,
T : C([a,b],B[c0,r]) → C([a,b],B[c0,r]).
Further,
T(x)(t)−T(x)(s) ≤
t
s
fc0
(τ,r)dτ , s,t ∈ [a,b], x ∈ C([a,b],B[c0,r]).
We denote
ω(δ) = max



t+δ
t
fc0
(τ,r)dτ;t ∈ [a,b−δ]



, δ ∈ [0,b−a].
The function ω : [0,b−a] → R+ is continuous, ω(0) = 0, and
T(x)(t)−T(x)(s) ≤ ω(|t −s|), s,t ∈ [a,b], x ∈ C([a,b],B[c0,r]).
Due to Lemma 2.1, there exists a function x ∈ C([a,b],B[c0,r]) with the required pro-
perty.
Corollary 2.1. For arbitrary t0 ∈ I and c0 ∈ Rn
, there exists an interval I0 ⊆ I such
that t0 ∈ I0 and the problem (2.1), (2.2) has at least one solution on the interval I0.
Moreover, if t0 is an interior point of I, then the interval I0 can be chosen so that t0 is
an interior point of I0.
Proof. The statement of the corollary follows from Theorem 2.1.
Now, we consider the equation
u(n)
= f t,u,u ,...,u(n−1)
, (2.6)
where f ∈ Kloc(I ×Rn
,R).
Definition 2.3. Let I0 ⊆ I be an interval. We say that a function u: I0 → R is a solution
of Eq. (2.6) on I0 if
1. u ∈ ˜Cn−1
loc (I0,R);
2. it holds
u(n)
(t) = f t,u(t),u (t),...,u(n−1)
(t)
for almost all t ∈ I0.
11
Let t0 ∈ I, ci ∈ R, i ∈ {0,1,...,n−1}. A solution u: I0 → R of Eq. (2.6) satisfying the
condition
u(i)
(t0) = ci, i ∈ {0,1,...,n−1}, (2.7)
is called the solution of the Cauchy problem (2.6), (2.7).
Corollary 2.2. For arbitrary t0 ∈ I and ci ∈ R, i ∈ {0,1,...,n−1}, there exists an
interval I0 ⊆ I such that t0 ∈ I0 and that the problem (2.6), (2.7) has at least one solution
on I0. Moreover, if t0 is an interior point of I, then the interval I0 can be chosen so that
t0 is an interior point of I0.
Proof. The statement of the corollary follows from Corollary 2.1.
12
3. Extendability of solutions
We consider the Cauchy problem
x = f(t,x), (3.1)
x(t0) = c0, (3.2)
where f ∈ Kloc(I ×Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
.
Definition 3.1. Let x be a solution of Eq. (3.1) on an interval (a,b) ⊆ I. We say that the
solution x is right-extendable if there exist b1 > b, b1 ∈ I, and a solution y of Eq. (3.1)
on the interval (a,b1) ⊆ I such that y(t) = x(t), t ∈ (a,b). The solution y is called a
right-extension of the solution x. If any right-extension of the solution x does not exist,
then we say that the solution x is not right-extendable.
Analogously, left-extendable solutions (and solutions which are not left-extendable)
are defined. We say that a solution x is extendable if it is right-extendable or
left-extendable. In the opposite case, we say that x is non-extendable.
Lemma 3.1. Let (α,β) ⊆ I, r > 0, δ ≥ 0, c ∈ Rn
, and let
δ +
β
α
fc (τ,r)dτ < r,
where
fc (t,r) = sup{ f(t,x) ;x ∈ B[c,r]}, t ∈ (α,β).
Let x be a solution of Eq. (3.1) on (α,β) satisfying
inf{ x(t)−c ;t ∈ (α,β)} ≤ δ.
Then,
x(t)−c < r, t ∈ (α,β),
and the limits
lim
t→α+
x(t), lim
t→β−
x(t)
exist.
Proof. We prove the lemma by contradiction. There exists a point t0 ∈ (α,β) such that
x(t0)−c +
β
α
fc (τ,r)dτ < r. (3.3)
In addition, there exists [α0,β0] ⊆ (α,β) such that t0 ∈ [α0,β0] and that
max{ x(t)−c ;t ∈ [α0,β0]} = r. (3.4)
From (3.1), we obtain
x(t)−c = x(t0)−c+
t
t0
f(s,x(s))ds, t ∈ (α,β).
13
Therefore,
x(t)−c ≤ x(t0)−c +
β0
α0
fc (s,r)ds, t ∈ [α0,β0]. (3.5)
From (3.3) and (3.5), it follows that
x(t)−c < r, t ∈ [α0,β0].
This is a contradiction with (3.4).
It remains to prove the existence of the limits. Since
f(−,x(−)) ∈ L((α,β),Rn
),
the existence of the limits comes from
x(t) = x(t0)+
t
t0
f(s,x(s))ds, t ∈ (α,β).
Theorem 3.1. Let x be a solution of Eq. (3.1) on an interval (a,b) ⊆ I. Then, x is
right-extendable if and only if b < supI and
liminf
t→b−
x(t) < ∞.
Proof. Let b < supI and let
lim
t→b−
x(t) = ∞.
There exists c ∈ Rn
such that
liminf
t→b−
x(t)−c = 0.
We put r = 1, δ = 0, and β = b. Let α ∈ (a,β) be such that the conditions of Lemma 3.1
are satisfied. Hence,
lim
t→b−
x(t) = c.
We consider the Cauchy problem
x = f(t,x), x(b) = c.
From Corollary 2.1, it follows the existence of b1 > b, b1 < supI, and the existence of
a solution ¯x of the considered Cauchy problem on the interval [b,b1]. We put
y(t) =
x(t), t ∈ (a,b);
¯x(t), t ∈ [b,b1).
Obviously,
y ∈ ˜Cloc((a,b1),Rn
)
and y is a solution of Eq. (3.1) on the interval (a,b1), i.e., y is a right-extension of the
solution x.
14
We consider the opposite implication. If x is a right-extendable solution and if
y: (a,b1) → Rn
is a right-extension of x, then
b < b1 ≤ supI
and
liminf
t→b−
x(t) = y(b) < ∞.
Theorem 3.2. Let x be a solution of Eq. (3.1) on an interval (a,b) ⊆ I. Then, x is
left-extendable if and only if a > infI and
liminf
t→a+
x(t) < ∞.
Proof. Theorem 3.2 can be proved analogously as Theorem 3.1.
Theorem 3.3. The problem (3.1), (3.2) has a non-extendable solution.
Proof. We suppose that t0 < supI. We show that the problem (3.1), (3.2) has a solution
which is not right-extendable. Similarly, one can show the second case.
We consider an increasing sequence {bk}∞
k=1 ⊂ (t0,supI) with the property that
lim
k→∞
bk = supI.
For c ∈ Rn
and r > 0, we define
fc (t,r) = sup{ f(t,x) ;x ∈ B[c,r]}, t ∈ I.
Obviously, there exists t1 ∈ (t0,b1] such that
t1
t0
fc0
(s,1)ds ≤ 1.
According to Theorem 2.1, the problem (3.1), (3.2) has a solution x0 on the interval
[t0,t1]. We define
c1 = x0(t1)
and
r1 = max{ x0(t)−c0 ;t ∈ [t0,t1]}.
If
b2
t1
fc0
(s,r1 +1)ds ≤ 1,
then we put t2 = b2. Otherwise, we choose t2 ∈ (t1,b2) such that
t2
t1
fc0
(s,r1 +1)ds = 1.
15
We have
t2
t1
fc1
(s,1)ds ≤
t2
t1
fc0
(s,r1 +1)ds ≤ 1.
According to Theorem 2.1, the problem
x = f(t,x), x(t1) = c1
has a solution x1 on the interval [t1,t2].
We continue in this process. We obtain the sequences {tk}∞
k=1, {xk}∞
k=0, {ck}∞
k=1,
{rk}∞
k=1 for which:
1. tk ∈ (tk−1,bk], k ∈ N;
2. ck = xk−1(tk), k ∈ N;
3. xk for k ∈ N∪{0} is a solution of the problem
x = f(t,x), x(tk) = ck
on the interval [tk,tk+1];
4. it holds
rk = max{ xk−1(t)−c0 ;t ∈ [tk−1,tk]}, k ∈ N;
5. if
tk+1
tk
fc0
(s,rk +1)ds < 1
for some k ∈ N, then tk+1 = bk+1.
We put
b = lim
k→∞
tk
and
x(t) = xk(t), t ∈ [tk,tk+1), k ∈ N∪{0}.
Considering 3., we see that x is a solution of the problem (3.1), (3.2) on the interval
[t0,b).
By contradiction, we show that the solution x is not right-extendable. We assume
that the finite limit
lim
t→b−
x(t)
exists and that
b < supI.
The function x is bounded, i.e.,
r = sup{ x(t)−c0 ;t ∈ [t0,b)} < ∞.
According to 4.,
rk ≤ r, k ∈ N.
16
Thus,
tk+1
tk
fc0
(s,rk +1)ds ≤
b
tk
fc0
(s,r +1)ds → 0 as k → ∞.
Therefore, there exists k0 ∈ N such that
tk+1
tk
fc0
(s,rk +1)ds < 1, k ≥ k0, k ∈ N.
Now, from 5., it follows that
tk+1 = bk+1, k ≥ k0, k ∈ N.
Hence,
b = supI.
The obtained contradiction proves that the solution x is not right-extendable.
17
4. Set of solutions
We consider the Cauchy problem
x = f(t,x), (4.1)
x(t0) = c0, (4.2)
where f ∈ Kloc(I ×Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
.
Lemma 4.1. Let r0 ≥ 0, r > 0, t0 ≤ a0 < b0, where b0 ∈ I, be such that
b0
a0
f0 (s,r0 +r)ds < r,
where
f0 (t,r) = sup{ f(t,x) ;x ∈ B[0,r]}, t ∈ I.
Then, for all solution x of Eq. (4.1) on the interval [t0,b0) which satisfies
x(t) ≤ r0, t ∈ [t0,a0],
it holds
x(t) < r0 +r, t ∈ [t0,b0).
Moreover, the limit
lim
t→b−
0
x(t)
exists.
Proof. We prove the lemma by contradiction. We suppose that there exist a solution x
of Eq. (4.1) on the interval [t0,b0) and t1 ∈ (a0,b0) such that
x(t) ≤ r0, t ∈ [t0,a0],
x(t) < r0 +r, t ∈ [a0,t1),
and that
x(t1) = r0 +r.
The contradiction comes from
x(t1) ≤ x(a0) +
t1
a0
f(s,x(s)) ds ≤ r0 +
t1
a0
f0 (s,r0 +r)ds < r0 +r.
Note that the obtained statement guarantees
f(−,x(−)) ∈ L([a0,b0),Rn
).
Therefore, the existence of the limit
lim
t→b−
0
x(t)
18
follows from
x(t) = x(a0)+
t
a0
f(s,x(s))ds, t ∈ [a0,b0).
Theorem 4.1. Let [a,b] ⊆ I, t0 ∈ [a,b], and let any non-extendable solution of the
problem (4.1), (4.2) exist on [a,b]. Let X[a,b] be the set of the restrictions of all nonextendable
solutions of the problem (4.1), (4.2) to the interval [a,b]. Then, the set X[a,b]
is bounded in the space C([a,b],Rn
).
Proof. We assume that t0 < b. We show that the set X[t0,b] is bounded in the space
C([t0,b],Rn
). For t0 > a, one can similarly show that the set X[a,t0] is bounded in the
space C([a,t0],Rn
).
We put
ρ(t) = sup x(s) ;s ∈ [t0,t],x ∈ X[t0,b] , t ∈ [t0,b].
We choose t1 ∈ (t0,b] such that
t1
t0
f0 (s, c0 +1)ds < 1,
where f0 is from Lemma 4.1. According to Lemma 4.1, we have
ρ(t) ≤ ρ(t1) ≤ c0 +1, t ∈ [t0,t1].
We show that ρ(b) < ∞. By contradiction, we assume that ρ(b) = ∞. Then, t1 < b and
there exists t ∈ (t1,b] such that
ρ(t) = ∞, t ∈ (t ,b],
and that
ρ(t) < ∞, t ∈ [t0,t ).
We assume the existence of a sequence {τk}∞
k=1 ⊂ [t0,t ) and a sequence {xk}∞
k=1 of
solutions of the problem (4.1), (4.2) on the interval [t0,t ] such that
lim
k→∞
xk(τk) = ∞.
In addition, for all β ∈ [t0,t ), we have
xk(t) ≤ ρ(β), t ∈ [t0,β], k ∈ N.
It is obvious that the functions xk, k ∈ N, are uniformly bounded on any compact subinterval
of the interval [t0,t ). At the same time, for any β ∈ [t0,t ), we have
xk(t)−xk(s) ≤
t
s
f(ξ,xk(ξ)) dξ
≤
t
s
f0 (ξ,ρ(β))dξ , s,t ∈ [t0,β], k ∈ N.
19
Therefore, the functions xk, k ∈ N, are also equicontinuous on any compact subinterval
of the interval [t0,t ). Due to the Arzelà–Ascoli theorem, without loss of generality,
we can assume that the sequence {xk}∞
k=1 is uniformly convergent on any compact
subinterval of the interval [t0,t ). We put
lim
k→∞
xk(t) = x(t), t ∈ [t0,t ).
By the Lebesgue theorem, one can verify that x is a solution of the problem (4.1),
(4.2) on the interval [t0,t ). Since t ≤ b and since any non-extendable solution of the
problem (4.1), (4.2) exists on the interval [t0,b], the finite limit
lim
t→t −
x(t)
exists. Hence,
r0 = sup{ x(t) ;t ∈ [t0,t )} < ∞. (4.3)
We choose t ∈ [t0,t ) so that
t
t
f0 (s,r0 +2)ds < 1.
From the construction of the solution x and from (4.3), it follows the existence of k0 ∈ N
with the property that
xk(t) ≤ r0 +1, t ∈ [t0,t ], k ≥ k0, k ∈ N.
Therefore, considering Lemma 4.1, we obtain
xk(t) < r0 +2, t ∈ [t0,t ), k ≥ k0, k ∈ N.
Hence,
ρ(t ) < ∞.
Now, it is enough to consider again Lemma 4.1 (for r0 +3).
Remark 4.1. From the Arzelà–Ascoli theorem, from the proof of Theorem 4.1, and
from the Lebesgue theorem, it follows that the set X[a,b] from the statement of Theorem
4.1 is even compact in C([a,b],Rn
).
20
5. Upper and lower solutions
We consider the Cauchy problem
x = f(t,x), (5.1)
x(t0) = c0, (5.2)
where f ∈ Kloc(I ×Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
.
Definition 5.1. Let f : I ×Rn
→ Rn
, where f = (fi)n
i=1. We say that f is quasi non-decreasing
in the last n variables if, for all i ∈ {1,...,n} and almost all t ∈ I, it holds
fi(t,x1,...,xn) ≤ fi(t,y1,...,yn), xk ≤ yk, k ∈ {1,...,n}, k = i, xi = yi.
Lemma 5.1. Let the map
(t,x) → f(t,x) sgn(t −t0), (t,x) ∈ I ×Rn
,
be quasi non-decreasing in the last n variables. Let [a,b] ⊆ I, t0 ∈ [a,b], and let any
non-extendable solution of the problem (5.1), (5.2) exist on the interval [a,b]. Then, for
any function y ∈ ˜C([a,b],Rn
) satisfying
y(t0) ≤ c0
and
y (t)− f(t,y(t)) sgn(t −t0) ≤ 0
for almost all t ∈ [a,b], there exists a solution x of the problem (5.1), (5.2) on the
interval [a,b] such that
y(t) ≤ x(t), t ∈ [a,b].
Proof. Let y ∈ ˜C([a,b],Rn
) be an arbitrary function from the statement of the lemma.
We suppose that t0 < b. We prove the existence of a solution x of the problem (5.1),
(5.2) on the interval [t0,b] with the property that
y(t) ≤ x(t), t ∈ [t0,b].
In the second case, we can proceed analogously.
For all i ∈ {1,...,n}, we put
χi(t,z) =
yi(t), z ≤ yi(t);
z, z > yi(t);
t ∈ [t0,b], z ∈ R.
We define
χ(t,x) = (χi(t,xi))n
i=1, t ∈ [t0,b], x = (xi)n
i=1 ∈ Rn
,
and
˜f(t,x) = f(t,χ(t,x)), t ∈ [t0,b], x ∈ Rn
.
Obviously,
˜f ∈ K([t0,b]×Rn
,Rn
).
21
We consider the Cauchy problem
x = ˜f(t,x), x(t0) = c0.
We consider that x = (xi)n
i=1 is a solution of this problem, which is not right-extendable,
and that [t0,b0) ⊆ [t0,b] is the maximal interval, where the solution x exists (see Theorem
3.3). Let i ∈ {1,...,n} be arbitrarily given. We prove that
yi(t) ≤ xi(t), t ∈ [t0,b0).
Let us consider the opposite, i.e., let there exist [α,β] ⊆ [t0,b0) such that
yi(α) = xi(α)
and that
yi(t) > xi(t), t ∈ (α,β].
We define
u(t) = yi(t)−xi(t), t ∈ [α,β].
Obviously, u(α) = 0 and
u(t) > 0, t ∈ (α,β].
We have
u (t) = yi(t)−xi(t)
≤ fi(t,y(t))− ˜fi(t,x(t))
= fi(t,y1(t),...,yn(t))− fi(t,χ1(t,x1(t)),...,χn(t,xn(t)))
for almost all t ∈ (α,β). From the definition of the function χ, it follows that
yi(t) = χi(t,xi(t)), t ∈ [α,β],
and that
yk(t) ≤ χk(t,xk(t)), t ∈ [α,β], k = i.
Since the map
(t,x) → f(t,x)sgn(t −t0), (t,x) ∈ I ×Rn
,
is quasi non-decreasing in the last n variables, u (t) ≤ 0 for almost all t ∈ (α,β),
which gives a contradiction. The contradiction (together with the arbitrariness of
i ∈ {1,...,n}) proves that
y(t) ≤ x(t), t ∈ [t0,b0).
Thus,
˜f(t,x(t)) = f(t,x(t)), t ∈ [t0,b0),
i.e., x is a solution of the problem (5.1), (5.2) on [t0,b0).
Now, we prove that b0 = b. Let b0 < b. Since x is a solution of the problem
x = ˜f(t,x), x(t0) = c0
on the interval [t0,b0), which is not right-extendable, from Theorem 3.1, it follows
lim
t→b−
0
x(t) = ∞.
22
At the same time, from Theorem 3.1, it follows a contradiction with an assumption of
the lemma. Thus, we have proved that b0 = b, i.e.,
y(t) ≤ x(t), t ∈ [t0,b).
Due to the assumptions of the lemma, there exists the finite limit
lim
t→b−
x(t),
i.e., x is a solution of the problem (5.1), (5.2) on [t0,b] and
y(t) ≤ x(t), t ∈ [t0,b].
Lemma 5.2. Let the map
(t,x) → f(t,x) sgn(t −t0), (t,x) ∈ I ×Rn
,
be quasi non-decreasing in the last n variables. Let [a,b] ⊆ I, t0 ∈ [a,b], and let any
non-extendable solution of the problem (5.1), (5.2) exist on the interval [a,b]. Then, for
any function y ∈ ˜C([a,b],Rn
) satisfying
y(t0) ≥ c0
and
y (t)− f(t,y(t)) sgn(t −t0) ≥ 0
for almost all t ∈ [a,b], there exists a solution x of the problem (5.1), (5.2) on the
interval [a,b] with the property that
y(t) ≥ x(t), t ∈ [a,b].
Proof. The lemma is possible to prove analogously as Lemma 5.1.
Definition 5.2. Let x be a solution of the problem (5.1), (5.2) on the interval I0 ⊆ I,
where t0 ∈ I0. We say that x is the upper (lower) solution of the problem (5.1), (5.2)
on the interval I0 ⊆ I if, for all interval I1 ⊆ I0, where t0 ∈ I1, and any solution x of the
problem (5.1), (5.2) on the interval I1, it holds
x(t) ≤ x (t) (x(t) ≥ x (t)), t ∈ I1.
Theorem 5.1. Let the map
(t,x) → f(t,x) sgn(t −t0), (t,x) ∈ I ×Rn
,
be quasi non-decreasing in the last n variables. Let [a,b] ⊆ I, t0 ∈ [a,b], and let any
non-extendable solution of the problem (5.1), (5.2) exist on the interval [a,b]. Then, the
problem (5.1), (5.2) has the upper solution and the lower solution on the interval [a,b].
Proof. We prove only the existence of the upper solution on the interval [a,b]. In the
second case, it is possible to proceed analogously. As X, we denote the set of all
solutions of the problem (5.1), (5.2) on the interval [a,b]. We define
xi (t) = sup{xi(t);(xk)n
k=1 ∈ X}, t ∈ [a,b], i ∈ {1,...,n}.
23
According to Theorem 4.1, the set X is bounded in the space C([a,b],Rn
). Therefore,
xi (t) < ∞, t ∈ [a,b], i ∈ {1,...,n}.
We show that x = (xi )n
i=1 is the upper solution of the problem (5.1), (5.2) on the
interval [a,b]. It is obvious that
x(t) ≤ x (t), t ∈ [a,b], x ∈ X.
Firstly, we prove that the function x is absolutely continuous. Since the set X is
bounded in C([a,b],Rn
), there exists a function h ∈ L([a,b],R+) with the property that
f(t,x) ≤ h(t), t ∈ [a,b], x ∈ X.
Let s,t ∈ [a,b], s < t. Then,
xi(t) = xi(s)+
t
s
fi(τ,x(τ))dτ
≤ xi(s)+
t
s
h(τ)dτ
≤ xi (s)+
t
s
h(τ)dτ, x = (xi)n
i=1 ∈ X, i ∈ {1,...,n},
i.e.,
xi (t) ≤ xi (s)+
t
s
h(τ)dτ, i ∈ {1,...,n}.
Analogously, one can obtain
xi (s) ≤ xi (t)+
t
s
h(τ)dτ, i ∈ {1,...,n},
which gives
|xi (t)−xi (s)| ≤
t
s
h(τ)dτ , i ∈ {1,...,n}, s,t ∈ [a,b].
Therefore, x ∈ ˜C([a,b],Rn
).
Now, we show that x is a solution of the problem (5.1), (5.2). We know that one
can find r0 > 0 such that
x(t) ≤ r0, t ∈ [a,b], x ∈ X.
According to Lemma 1.4, for r0, there exist a function h0 ∈ L([a,b],R+) and a non-decreasing
function ω0 ∈ Cloc(R+,R+) such that ω0(0) = 0 and
f(t,x)− f(t,y) ≤ h0(t)ω0( x−y ), t ∈ [a,b], x,y ∈ B[0,r0].
As I0, we denote the set of all s ∈ [a,b] for which the following conditions:
24
1. there exists (x ) (s);
2. there exists v (s) and
v (s) = f(s,x (s)),
where
v(t) =
t
t0
f(τ,x (τ))dτ, t ∈ [a,b];
3. there exists v0(s) and
v0(s) = h0(s),
where
v0(t) =
t
t0
h0(τ)dτ, t ∈ [a,b],
are valid. Obviously, m(I0) = b − a. Let i ∈ {1,...,n} and s ∈ I0, where s > t0, be
arbitrary. Due to Remark 4.1, X is a compact set in the space C([a,b],Rn
). Therefore,
there exists ˜x = (˜xk)n
k=1 ∈ X such that ˜xi(s) = xi (s). We put
ε(t) = max{|˜xi(τ)−xi (τ)|;t ≤ τ ≤ s}, t ∈ [t0,s].
It is seen that the function ε : [t0,s] → R+ is continuous, non-increasing, and ε(s) = 0.
We have
fi(t, ˜x(t)) ≤ fi t,x1(t),...,xi−1(t), ˜xi(t),xi+1(t),...,xn(t)
= fi(t,x (t))+ fi t,x1(t),...,xi−1(t), ˜xi(t),xi+1(t),...,xn(t) − fi(t,x (t))
≤ fi(t,x (t))+h0(t)ω0(|˜xi(t)−xi (t)|)
≤ fi(t,x (t))+h0(t)ω0(ε(t)), t ∈ (t0,s].
Hence (consider ˜xi(s) = xi (s)), we obtain
xi (s) = ˜xi(s) = ˜xi(t)+
s
t
fi(τ, ˜x(τ))dτ
≤ xi (t)+
s
t
fi(τ,x (τ))dτ +
s
t
h0(τ)ω0(ε(τ))dτ
≤ xi (t)+
s
t
fi(τ,x (τ))dτ +ω0(ε(t))
s
t
h0(τ)dτ, t ∈ (t0,s].
Therefore,
xi (s)−xi (t)
s−t
≤
1
s−t
s
t
fi(τ,x (τ))dτ +
1
s−t
ω0(ε(t))
s
t
h0(τ)dτ, t ∈ (t0,s).
Since s ∈ I0 is arbitrary and ω0(ε(s)) = 0, we obtain
(xi ) (s) ≤ fi(s,x (s)), s ∈ I0, s > t0.
25
Analogously, one can show
(xi ) (s) ≥ fi(s,x (s)), s ∈ I0, s < t0.
Thus,
(x ) (t)− f(t,x (t)) sgn(t −t0) ≤ 0
for almost all t ∈ [a,b]. According to Lemma 5.1, there exists x0 ∈ X such that
x (t) ≤ x0(t), t ∈ [a,b].
At the same time,
x(t) ≤ x (t), t ∈ [a,b], x ∈ X, (5.3)
which gives
x0(t) ≤ x (t), t ∈ [a,b].
Therefore, x ≡ x0 and x is a solution of the problem (5.1), (5.2). Moreover, (5.3)
means that x is the upper solution of this problem on the interval [a,b].
Corollary 5.1. Let the map
(t,x) → f(t,x) sgn(t −t0), (t,x) ∈ I ×Rn
,
be quasi non-decreasing in the last n variables. Let [a,b] ⊆ I, t0 ∈ [a,b], and let any
non-extendable solution of the problem (5.1), (5.2) exist on the interval [a,b].
1. For any function y ∈ ˜C([a,b],Rn
) satisfying
y(t0) ≤ c0
and
y (t)− f(t,y(t)) sgn(t −t0) ≤ 0
for almost all t ∈ [a,b], it holds
y(t) ≤ x (t), t ∈ [a,b],
where x is the upper solution of the problem (5.1), (5.2) on the interval [a,b].
2. For any function y ∈ ˜C([a,b],Rn
) satisfying
y(t0) ≥ c0
and
y (t)− f(t,y(t)) sgn(t −t0) ≥ 0
for almost all t ∈ [a,b], it holds
y(t) ≥ x (t), t ∈ [a,b],
where x is the lower solution of the problem (5.1), (5.2) on the interval [a,b].
Proof. The corollary follows directly from Lemmas 5.1 and 5.2 and from Theorem 5.1.
26
Definition 5.3. Let f : I × Rn
→ Rn
be given. We say that f is non-decreasing in the
last n variables if
f(t,x) ≤ f(t,y), t ∈ I, x ≤ y.
Corollary 5.2. Let the map
(t,x) → f(t,x) sgn(t −t0), (t,x) ∈ I ×Rn
,
be non-decreasing in the last n variables. Let [a,b] ⊆ I, t0 ∈ [a,b], and let any non-extendable
solution of the problem (5.1), (5.2) exist on the interval [a,b].
1. For any function y ∈ C([a,b],Rn
) satisfying
y(t) ≤ c0 +
t
t0
f(s,y(s))ds, t ∈ [a,b],
it holds
y(t) ≤ x (t), t ∈ [a,b],
where x is the upper solution of the problem (5.1), (5.2) on the interval [a,b].
2. For any function y ∈ C([a,b],Rn
) satisfying
y(t) ≥ c0 +
t
t0
f(s,y(s))ds, t ∈ [a,b],
it holds
y(t) ≥ x (t), t ∈ [a,b],
where x is the lower solution of the problem (5.1), (5.2) on the interval [a,b].
Proof. We prove only the first part. The second part can be proved analogously. We
put
z(t) = c0 +
t
t0
f(s,y(s))ds, t ∈ [a,b].
Obviously,
z ∈ ˜C([a,b],Rn
)
and
y(t) ≤ z(t), t ∈ [a,b].
The identity
z (t) = f(t,y(t))
holds for almost all t ∈ [a,b]. We have
z (t) sgn(t −t0) = f(t,y(t)) sgn(t −t0)
≤ f(t,z(t)) sgn(t −t0)
for almost all t ∈ [a,b], i.e.,
z (t)− f(t,z(t)) sgn(t −t0) ≤ 0
27
for almost all t ∈ [a,b]. According to Corollary 5.1, we obtain
z(t) ≤ x (t), t ∈ [a,b],
where x is the upper solution of the problem (5.1), (5.2) on the interval [a,b]. Finally,
the inequality
y(t) ≤ z(t), t ∈ [a,b],
gives
y(t) ≤ x (t), t ∈ [a,b].
28
6. Wintner theorem
We consider the Cauchy problem
x = f(t,x), (6.1)
x(t0) = c0, (6.2)
where f ∈ Kloc(I ×Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
.
Theorem 6.1 (Wintner). Let there exist a function h ∈ Kloc(I ×R+,R+) such that
f(t,x)·sgn((t −t0)x) ≤ h(t, x ), (t,x) ∈ I ×Rn
,
where the problem
ρ = h(t,ρ) sgn(t −t0), (6.3)
ρ(t0) = c0 (6.4)
has the upper solution on the interval I. Then, the problem (6.1), (6.2) has a solution
on I. Moreover, all non-extendable solutions of the problem (6.1), (6.2) exist on I.
Proof. We suppose that t0 ∈ int(I). If t0 is an endpoint of the interval I, then one can
proceed analogously.
From Theorem 3.3, it follows that there exists at least one non-extendable solution
of the problem (6.1), (6.2) on a subinterval J of I. Let x be an arbitrary non-extendable
solution of the problem (6.1), (6.2) on int(J) = (a,b) ⊆ I. We show that a = infI,
b = supI. We denote
u(t) = x(t) , t ∈ (a,b).
Obviously,
u ∈ ˜Cloc((a,b),R+),
u(t0) = c0 ,
and
u (t) = x (t)·sgnx(t)
for almost all t ∈ (a,b). Therefore,
u (t) sgn(t −t0) ≤ h(t,u(t))
for almost all t ∈ (a,b). Thus,
u (t)−h(t,u(t)) sgn(t −t0) sgn(t −t0) ≤ 0 (6.5)
for almost all t ∈ (a,b). Since the problem (6.3), (6.4) has the upper solution on I, all
non-extendable solutions of the problem (6.3), (6.4) exist on I. Therefore, considering
u(t0) = c0 together with (6.5), from Corollary 5.1, it follows that
u(t) ≤ ρ (t), t ∈ (a,b),
29
where ρ is the upper solution of the problem (6.3), (6.4) on the interval I. Using
Theorems 3.1 and 3.2, we obtain
a = infI, b = supI.
If a /∈ I, b /∈ I, then the proof is done. Let us consider that
a ∈ I or b ∈ I.
We show that the finite limit
lim
t→a+
x(t) or lim
t→b−
x(t)
exists. We know that
f ∈ K([a,t0]×Rn
,Rn
) or f ∈ K([t0,b]×Rn
,Rn
).
Therefore, from
u(t) ≤ ρ (t), t ∈ (a,b),
and from the fact that ρ is the upper solution of the problem (6.3), (6.4) on I, it follows
f(−,x(−)) ∈ L([a,t0],Rn
) or f(−,x(−)) ∈ L([t0,b],Rn
).
The existence of the limit
lim
t→a+
x(t) or lim
t→b−
x(t)
comes from
x(t) = c0 +
t
t0
f(s,x(s))ds, t ∈ (a,b).
Corollary 6.1. Let h0 ∈ Lloc(I,R+) and let ω ∈ Cloc(R+,(0,∞)) satisfy
∞
0
ds
ω(s)
= ∞.
If the inequality
f(t,x)·sgn((t −t0)x) ≤ h0(t)ω( x )
holds on the set I ×Rn
, then the problem (6.1), (6.2) has a solution on I. Moreover, all
non-extendable solutions of the problem (6.1), (6.2) exist on I.
Proof. Since the problem
ρ = h0(t)ω(ρ) sgn(t −t0),
ρ(t0) = c0
has the upper solution on the interval I, the statement of the corollary comes directly
from Theorem 6.1.
30
7. Uniqueness of solutions
We consider the Cauchy problem
x = f(t,x), (7.1)
x(t0) = c0, (7.2)
where f ∈ Kloc(I ×Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
.
Definition 7.1. We say that the problem (7.1), (7.2) is uniquely solvable if, for arbitrary
solutions x1 and x2 on intervals I1 and I2, respectively, it holds
x1(t) = x2(t), t ∈ I1 ∩I2.
Definition 7.2. We say that a function g: I {t0} × R+ → R is an element of the set
Kloc(I {t0}×R+,R) if g ∈ K(I0 ×R+,R) for any compact interval I0 ⊆ I {t0}.
Definition 7.3. Let g ∈ Kloc(I {t0} × R+,R) and let I0 ⊆ I be such that t0 ∈ I0. We
say that a function x: I0 {t0} → R+ is a solution of the equation x = g(t,x) if the
following conditions:
1. x ∈ ˜C(J,R) for any compact interval J ⊆ I0 {t0};
2. x (t) = g(t,x(t)) for almost all t ∈ I0
are fulfilled.
Lemma 7.1. Let ∆ > 0, λ ∈ [0,1), h ∈ L([t0,t0 + ∆],(0,∞)), and let a function ϕ ∈
Cloc(R+,R+) be such that
lim
s→0+
ϕ(s)
sλ
= 0.
If a function u ∈ C([t0,t0 +∆],R+) satisfies
u(t) ≤
t
t0
h(τ)ϕ(u(τ))dτ, t ∈ [t0,t0 +∆],
then
lim
t→t+
0
[u(t)]1−λ
t
t0
h(s)ds
= 0.
Proof. It is seen that u(t0) = 0. Let ε > 0 be arbitrary. There exists tε ∈ (t0,t0 +∆] with
the property that
ϕ(u(τ)) ≤ ε [u(τ)]λ
, τ ∈ [t0,tε ].
Next, we obtain
u(t) ≤ ε
t
t0
h(τ)[u(τ)]λ
dτ, t ∈ [t0,tε ]. (7.3)
31
We denote
v(t) =

ε(1−λ)
t
t0
h(τ)dτ


1
1−λ
, t ∈ [t0,tε ].
One can directly verify that v is “the unique positive solution” of the Cauchy problem
v = εh(t)vλ
, v(t0) = 0
for t ∈ [t0,tε ]. Therefore, Corollary 5.2 and (7.3) give
u(t) ≤ v(t), t ∈ [t0,tε ].
Thus,
limsup
t→t+
0
[u(t)]1−λ
t
t0
h(s)ds
≤ limsup
t→t+
0
ε(1−λ)
t
t0
h(s)ds
t
t0
h(s)ds
≤ ε.
Now, it suffices to consider the arbitrariness of ε > 0.
Lemma 7.2. Let ∆ > 0, λ ∈ [0,1), h ∈ L([t0 − ∆,t0],(0,∞)), and let a function ϕ ∈
Cloc(R+,R+) be such that
lim
s→0+
ϕ(s)
sλ
= 0.
If a function u ∈ C([t0 −∆,t0],R+) satisfies
u(t) ≤
t0
t
h(τ)ϕ(u(τ))dτ, t ∈ [t0 −∆,t0],
then
lim
t→t−
0
[u(t)]1−λ
t0
t
h(s)ds
= 0.
Proof. The lemma can be proved analogously as Lemma 7.1.
Theorem 7.1. Let δ > 0 and ε > 0 be such that
[f(t,x)− f(t,y)]·sgn[(t −t0)(x−y)] ≤ h(t)ϕ( x−y ), t ∈ J, x,y ∈ B[c0,δ],
where J = [t0 −ε,t0 +ε]∩I, h ∈ Lloc(J,(0,∞)), and ϕ ∈C([0,2δ],R+) has the property
that
lim
s→0+
ϕ(s)
sλ
= 0
for some λ ∈ [0,1). If, for any r > 0, there exists a function
ωr ∈ Kloc(I {t0}×R+,R)
such that
ωr(−,0) ≡ 0
32
and that
[f(t,x)− f(t,y)]·sgn[(t −t0)(x−y)] ≤ ωr(t, x−y ), t ∈ I {t0}, x,y ∈ B[c0,r],
and if the problem
ρ = ωr(t,ρ) sgn(t −t0), lim
t→t0
[ρ(t)]1−λ
t
t0
h(s)ds
= 0
has only the zero solution, then the problem (7.1), (7.2) is uniquely solvable.
Proof. Let x1 and x2 be non-extendable solutions of the problem (7.1), (7.2) on intervals
I1 and I2, respectively. Our aim is to prove
x1(t) = x2(t), t ∈ I1 ∩I2.
We denote
u(t) = x1(t)−x2(t) , t ∈ I1 ∩I2.
We consider that there exists t1 ∈ I1 ∩I2 such that u(t1) = 0. Without loss of generality,
we can assume that t1 > t0. It is obvious that
u ∈ ˜C([t0,t1],R+)
and that
u (t) = x1(t)−x2(t) ·sgn(x1(t)−x2(t)) (7.4)
for almost all t ∈ [t0,t1]. Thus, we obtain
u(t) =
t
t0
u (τ)dτ ≤
t
t0
h(τ)ϕ(u(τ))dτ, t ∈ [t0,t0 +∆],
where ∆ > 0 is sufficiently small. According to Lemma 7.1, we have
lim
t→t+
0
[u(t)]1−λ
t
t0
h(s)ds
= 0.
Next, we denote
r = max{ x1(t)−c0 + x2(t)−c0 ;t ∈ [t0,t1]}.
For this number r, there exists a function ωr ∈ Kloc(I {t0} × R+,R) from the statement
of the theorem. From (7.4), it follows
u (t) ≤ ωr(t,u(t))
for almost all t ∈ (t0,t1]. We put
¯ωr(t,y) =



ωr(t,u(t)), y > u(t);
ωr(t,y), 0 ≤ y ≤ u(t);
0, y < 0;
t ∈ (t0,t1], y ∈ R.
33
Evidently,
¯ωr ∈ Kloc((t0,t1]×R,R).
From Theorem 3.3, it follows that there exist a ∈ [t0,t1) and a non-extendable solution ρ
of the problem
ρ = ¯ωr(t,ρ), ρ(t1) =
1
2
u(t1)
on the interval (a,t1].
We show that
ρ(t) ≤ u(t), t ∈ (a,t1].
By contradiction, we assume that there exist t2,t3 ∈ (a,t1), where t2 < t3, such that
ρ(t) > u(t), t ∈ [t2,t3),
and that
ρ(t3) = u(t3).
From the definition of ¯ωr, it follows
ρ (t) = ωr(t,u(t))
for almost all t ∈ [t2,t3]. From the inequality
u (t) ≤ ωr(t,u(t))
which is valid for almost all t ∈ (t0,t1], we obtain
ρ(t) ≤ u(t), t ∈ [t2,t3].
We have a contradiction.
Now, we show that
ρ(t) > 0, t ∈ (a,t1].
By contradiction, we assume that there exists t4 ∈ (a,t1) for which
ρ(t) > 0, t ∈ (t4,t1],
and for which
ρ(t4) = 0.
We recall that
ρ(t) ≤ u(t), t ∈ (a,t1].
Due to the definition of ¯ωr, the function
¯ρ(t) =
0, t ∈ (t0,t4];
ρ(t), t ∈ (t4,t1]
is a non-zero solution of the problem
ρ (t) = ωr(t,ρ) sgn(t −t0), lim
t→t0
[ρ(t)]1−λ
t
t0
h(s)ds
= 0 (7.5)
34
on the interval (t0,t1], which is a contradiction.
We have proved that
0 < ρ(t) ≤ u(t), t ∈ (a,t1].
From Theorem 3.2, it follows that a = t0, i.e.,
0 < ρ(t) ≤ u(t), t ∈ (t0,t1].
Due to the definition of ¯ωr, we have a non-zero solution of the problem (7.5) on the
interval (t0,t1], which is a contradiction with the assumption of the theorem.
Corollary 7.1 (Osgood). If there exist functions lr ∈ L(I,R+) and ηr ∈ Cloc(R+,R+)
for any r > 0 such that
ηr(0) = 0, ηr(s) > 0, s > 0,
2r
0
ds
ηr(s)
= ∞,
and that
[f(t,x)− f(t,y)]·sgn[(t −t0)(x−y)] ≤ lr(t)ηr( x−y ), t ∈ I, x,y ∈ B[c0,r],
then the problem (7.1), (7.2) is uniquely solvable.
Proof. We put
δ = 1, J = I, λ = 0,
h(t) = l1(t)+1, t ∈ J,
ϕ(x) = η1(x), x ∈ [0,2],
and
ωr(t,x) = lr(t)ηr(x), t ∈ I, x ∈ R+, r > 0.
One can directly verify that all conditions of Theorem 7.1 are fulfilled. Note that, for
any r > 0, the problem
ρ = lr(t)ηr(ρ) sgn(t −t0), ρ(t0) = 0
has only the zero solution. Thus, for any r > 0, the problem
ρ = lr(t)ηr(ρ) sgn(t −t0), lim
t→t0
ρ(t)
t
t0
h(s)ds
= 0
has only the zero solution as well. Therefore, the corollary follows from Theorem 7.1.
Corollary 7.2 (Nagumo–Perron). Let t0 ∈ int(I). Let h ∈ Lloc(I,(0,∞)) be such that
the function
f0(t,x) =
f(t,x)
h(t)
, (t,x) ∈ I ×Rn
,
is continuous in a neighbourhood of [t0,c0]. If the inequality
[f(t,x)− f(t,y)]·sgn[(t −t0)(x−y)] ≤
h(t)
t
t0
h(τ)dτ
x−y , t = t0,
is valid on the set I ×Rn
, then the problem (7.1), (7.2) is uniquely solvable.
35
Proof. We put λ = 0 and
ωr(t,x) =
h(t)x
t
t0
h(τ)dτ
, t = t0, t ∈ I, x ∈ R+, r > 0.
Since the function f0 is continuous in some neighbourhood of the point [t0,c0], there
exist
ε,δ ∈ 0,
1
2
such that the function f0 is uniformly continuous on [t0 − ε,t0 + ε] × B[c0,δ]. Let
J = [t0 −ε,t0 +ε]. For s ∈ [0,2δ] ⊆ [0,1], we put
ϕ(s) = max{ f0(t,x)− f0(t,y) ;t ∈ J,x,y ∈ B[c0,δ], x−y ≤ s}.
Obviously, ϕ ∈ C([0,2δ],R+) and ϕ(0) = 0. One can easily verify that all conditions
of Theorem 7.1 are fulfilled. For example, by a direct computation, one can verify that
the problem
ρ =
h(t)
t
t0
h(τ)dτ
ρ, lim
t→t0
ρ(t)
t
t0
h(τ)dτ
= 0
has only the zero solution. Therefore, the corollary also follows from Theorem 7.1.
36
8. Krein theorem
Let I be a compact interval. Now, we consider the Cauchy problem
x = f0(t,x), (8.1.0)
x(t0) = c0, (8.2.0)
where f0 ∈ K(I ×Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
.
For m ∈ N, together with the problem (8.1.0), (8.2.0), we consider the perturbed
problem
x = fm(t,x), (8.1.m)
x(tm) = cm, (8.2.m)
where fm ∈ K(I ×Rn
,Rn
), tm ∈ I, cm ∈ Rn
.
For m ∈ N∪{0}, by the symbol X(fm,tm,cm), we denote the set of all non-extendable
solutions of the problem (8.1.m), (8.2.m).
Definition 8.1. Let Y ⊆ C(I,Rn
). Then, the ε-neighbourhood of the set Y is the set
Yε =
y∈Y
B(y,ε),
where
B(y,ε) = {x ∈ C(I,Rn
); x−y C < ε}.
Definition 8.2. Let Y ⊆ C(I,Rn
) and let Ym
⊆ C(I,Rn
) for all sufficiently large m ∈ N.
If, for any ε > 0, there exists m0 ∈ N such that
Ym
⊆ Yε , m ≥ m0,
then we write
lim
m→∞
Ym
⊆ Y.
Lemma 8.1. Let the following conditions:
¯1. for all x ∈ Rn
, it holds
lim
m→∞
t
t0
fm(τ,x)dτ =
t
t0
f0(τ,x)dτ
uniformly on I;
¯2. for any r > 0, there exists ωr ∈ K(I ×R+,R+) such that
ωr(−,0) ≡ 0
and that
fm(t,x)− fm(t,y) ≤ ωr(t, x−y ), t ∈ I, x,y ∈ B[0,r], m ∈ N,
37
be fulfilled. If {xm}∞
m=0 is a sequence of functions from the space C(I,Rn
) such that
lim
m→∞
xm(t) = x0(t)
uniformly on I, then
lim
m→∞
t
t0
fm(τ,xm(τ))dτ =
t
t0
f0(τ,x0(τ))dτ
uniformly on I.
Proof. Since the sequence {xm}∞
m=1 is uniformly convergent, { xm C}∞
m=1 is bounded.
Let r ∈ R satisfy
1+ xm C ≤ r, m ∈ N∪{0}.
Let ωr be from the condition ¯2. Without loss of generality, we can assume that the
function ωr is non-decreasing in the second variable. Using Lemma 1.4, we can also
assume that
f0(t,x)− f0(t,y) ≤ ωr(t, x−y ), t ∈ I, x,y ∈ B[0,r].
We denote
ym(t) =
t
t0
fm(τ,xm(τ))dτ −
t
t0
f0(τ,x0(τ))dτ, t ∈ I, m ∈ N.
Let ε > 0 be arbitrarily given. We choose η ∈ (0,1] so that (see also Lemma 4.1)
I
ωr(τ,η)dτ <
ε
2
.
Since the function x0 is continuous on the compact interval, there exists a system
{ti}k
i=1 (k ≥ 2, k ∈ N) of points of the interval I such that
minI = t1 < t2 < ··· < tk = maxI
and that
x0(t)−x0(ti) <
η
2
, t ∈ [ti,ti+1], i ∈ {1,...,k −1}.
We put
˜x(t) = x0(ti), t ∈ [ti,ti+1), i ∈ {1,...,k −1},
and
˜x(tk) = x0(tk−1).
Obviously,
˜x(t)−x0(t) <
η
2
, t ∈ I.
There exists m0 ∈ N with the property that
xm −x0 C <
η
2
, m ≥ m0, m ∈ N.
38
Therefore,
xm(t)− ˜x(t) < η, t ∈ I, m ≥ m0, m ∈ N.
Next, we obtain
ym(t) ≤
t
t0
fm(τ,xm(τ))− fm(τ, ˜x(τ)) dτ
+
t
t0
fm(τ, ˜x(τ))− f0(τ, ˜x(τ))dτ
+
t
t0
f0(τ, ˜x(τ))− f0(τ,x0(τ)) dτ
≤
t
t0
fm(τ, ˜x(τ))− f0(τ, ˜x(τ))dτ +2
I
ωr(τ,η)dτ
for all t ∈ I, m ≥ m0, m ∈ N. We define
γi,m = max



t
t0
fm(τ,x0(ti))− f0(τ,x0(ti))dτ ;t ∈ I



for all i ∈ {1,...,k −1}, m ∈ N.
Using the condition ¯1, one can easily verify that
lim
m→∞
γi,m = 0, i ∈ {1,...,k −1}.
Since
I
ωr(τ,η)dτ <
ε
2
,
the definitions of ˜x and γi,m give
ym(t) ≤ ε +
k−1
∑
i=1
γi,m, t ∈ I, m ≥ m0, m ∈ N.
From this inequality, we obtain the statement of the lemma. It is enough to consider
the arbitrariness of ε > 0 and
lim
m→∞
γi,m = 0, i ∈ {1,...,k −1}.
Theorem 8.1 (Krein). If the following conditions:
1. it holds
lim
m→∞
tm = t0, lim
m→∞
cm = c0;
39
2. for all x ∈ Rn
, it holds
lim
m→∞
t
t0
fm(τ,x)dτ =
t
t0
f0(τ,x)dτ
uniformly on I;
3. for any r > 0, there exists a function ωr ∈ K(I ×R+,R+) such that
ωr(−,0) ≡ 0
and that
fm(t,x)− fm(t,y) ≤ ωr(t, x−y ), t ∈ I, x,y ∈ B[0,r], m ∈ N;
4. every non-extendable solution of the problem (8.1.0), (8.2.0) exists on I, i.e.,
X(f0,t0,c0) ⊂ C(I,Rn
),
are fulfilled, then there exists m0 ∈ N such that, for all m > m0, m ∈ N, all non-extendable
solution of the problem (8.1.m), (8.2.m) exists on I and
lim
m→∞
X (fm,tm,cm) ⊆ X (f0,t0,c0).
Proof. According to Theorem 4.1, the set X (f0,t0,c0) is bounded in the spaceC(I,Rn
).
Thus, there exists r0 > 0 such that
x C ≤ r0, x ∈ X (f0,t0,c0).
We define the function
χ(x) =



x, x ≤ r0 +1, x ∈ Rn
;
(r0 +1)
x
x
, x > r0 +1, x ∈ Rn
,
and we put
˜fm(t,x) = fm(t,χ(x)), (t,x) ∈ I ×Rn
, m ∈ N∪{0}.
It is obvious that
˜fm ∈ K(I ×Rn
,Rn
), m ∈ N∪{0},
and that
˜fm(t,x) ≤ fm(t), t ∈ I, x ∈ Rn
, m ∈ N∪{0},
where
fm(t) = sup{ fm(t,x) ;x ∈ B[0,r0 +1]}, t ∈ I.
Due to Lemma 1.2, fm(t) ∈ L(I,R+) for all considered m. For all m ∈ N ∪ {0}, we
consider the equation
x = ˜fm(t,x). (8.3.m)
As X ˜fm,tm,cm , we denote the set of all non-extendable solutions of the problem
(8.3.m), (8.2.m).
40
Now, we show that
X ˜f0,t0,c0 = X (f0,t0,c0).
Obviously,
X(f0,t0,c0) ⊆ X ˜f0,t0,c0 .
We assume the existence of ˜x ∈ X ˜f0,t0,c0 such that ˜x /∈ X(f0,t0,c0). We have
˜x C > r0 +1, ˜x(t0) = c0 < r0 +1.
Therefore, there exists an interval I0 ⊂ I such that t0 ∈ I0 and that
sup{ ˜x(t) ;t ∈ I0} = r0 +1.
Now, it is seen that the function ˜x is a solution of (8.1.0), (8.2.0) on the interval I0. Due
to 4. from the statement of the theorem, we know that there exists an extension y of the
solution ˜x to I. Then,
y ∈ X(f0,t0,c0)
and
y C ≥ r0 +1.
This is a contradiction which proves
X ˜f0,t0,c0 ⊆ X(f0,t0,c0).
Since
˜fm(t,x) ≤ fm(t), t ∈ I, x ∈ Rn
, m ∈ N,
according to Corollary 6.1, for arbitrary m ∈ N, any element of the set X ˜fm,tm,cm
exists on I, i.e.,
X ˜fm,tm,cm ⊂ C(I,Rn
), m ∈ N.
Next, we prove that
lim
m→∞
X ˜fm,tm,cm ⊆ X(f0,t0,c0).
We assume the opposite. Then, there exist ε0 > 0, an increasing sequence {mk}∞
k=1 of
positive integers, and a sequence {xk}∞
k=1 of functions from the space C(I,Rn
) such
that
xk ∈ X ˜fmk
,tmk
,cmk
, k ∈ N,
and that
xk /∈ Xε0 (f0,t0,c0), k ∈ N.
Evidently,
xk(t) = cmk
+yk(t)+zk(t), t ∈ I, k ∈ N,
where
yk(t) =
t
tmk
˜fmk
(τ,xk(τ))− fmk
(τ,0)dτ, t ∈ I, k ∈ N,
and
zk(t) =
t
tmk
fmk
(τ,0)dτ, t ∈ I, k ∈ N.
41
From the condition 3., we obtain
˜fm(t,x)− fm(t,0) = fm(t,χ(x))− fm(t,0)
≤ ωr0+1(t, χ(x) )
≤ ωr0+1(t,r0 +1), t ∈ I, x ∈ Rn
, m ∈ N,
because, without loss of generality, we can assume that the function ωr0+1 is non-decreasing
in the second variable. Thus,
yk(t) ≤ ωr0+1(t,r0 +1)
for almost all t ∈ I, k ∈ N. Hence, the functions yk, k ∈ N, are equicontinuous. Moreover,
we have
yk(t) ≤
t
tmk
yk(τ) dτ
≤
I
ωr0+1(τ,r0 +1)dτ, t ∈ I, k ∈ N,
i.e., the functions yk, k ∈ N, are uniformly bounded. According to the Arzelà–Ascoli
theorem, without loss of generality, we can assume that the sequence {yk}∞
k=1 is
uniformly convergent.
At the same time, for t ∈ I, k ∈ N, we have
zk(t)−
t
t0
f0(τ,0)dτ ≤
tmk
t0
fmk
(τ,0)dτ +
t
t0
fmk
(τ,0)dτ −
t
t0
f0(τ,0)dτ
≤
tmk
t0
fmk
(τ,0)dτ −
tmk
t0
f0(τ,0)dτ +
tmk
t0
f0(τ,0)dτ
+
t
t0
fmk
(τ,0)dτ −
t
t0
f0(τ,0)dτ
≤
tmk
t0
f0(τ,0) dτ
+2max



s
t0
fmk
(τ,0)dτ −
s
t0
f0(τ,0)dτ ;s ∈ I



.
Thus, from the conditions 1. and 2., we obtain
lim
k→∞
zk(t) =
t
t0
f0(τ,0)dτ
uniformly on I. From
xk(t) = cmk
+yk(t)+zk(t), t ∈ I, k ∈ N,
42
from the condition 1., and from the uniform convergences of the sequences {yk}∞
k=1
and {zk}∞
k=1 obtained above, it follows that there exists ˜x ∈ C(I,Rn
) such that
lim
k→∞
xk(t) = ˜x(t)
uniformly on I.
Now, we show that
˜x ∈ X ˜f0,t0,c0 .
From
xk ∈ X ˜fmk
,tmk
,cmk
, k ∈ N,
it follows
xk(t) = xk(t0)+
t
t0
fmk
(τ,χ(xk(τ)))dτ, t ∈ I, k ∈ N. (8.1)
Since
xk(t0)−c0 ≤ xk(t0)− ˜x(t0) + ˜x(t0)− ˜x(tmk
)
+ ˜x(tmk
)−xk(tmk
) + xk(tmk
)−c0
≤ 2 xk − ˜x C + ˜x(t0)− ˜x(tmk
) + cmk
−c0
for all k ∈ N, from the condition 1., from the uniform limit
lim
k→∞
xk(t) = ˜x(t)
on I, and from the continuity of ˜x, we obtain
lim
k→∞
xk(t0) = c0.
Next,
lim
k→∞
χ(xk(t)) = χ(˜x(t))
uniformly on I. Thus, (8.1) and Lemma 8.1 give
˜x(t) = c0 +
t
t0
˜f0(τ, ˜x(τ))dτ, t ∈ I,
i.e.,
˜x ∈ X ˜f0,t0,c0 .
Therefore,
˜x ∈ X (f0,t0,c0).
Since ˜x is the uniform limit of the sequence {xk}∞
k=1, there exists k0 ∈ N such that
˜x−xk C < ε0, k ≥ k0, k ∈ N,
and, consequently,
xk ∈ Xε0 (f0,t0,c0), k ≥ k0, k ∈ N,
which is a contradiction with
xk /∈ Xε0 (f0,t0,c0), k ∈ N.
43
The contradiction proves
lim
m→∞
X ˜fm,tm,cm ⊆ X (f0,t0,c0).
From Definition 8.2, it follows the existence of m0 ∈ N for which
X ˜fm,tm,cm ⊆ X1 (f0,t0,c0), m ≥ m0, m ∈ N.
Let m ≥ m0 (m ∈ N) and y ∈ X ˜fm,tm,cm be arbitrary. Then, y ∈ X1 (f0,t0,c0), i.e.,
there exists x ∈ X (f0,t0,c0) such that y−x C < 1. Thus,
y C < x C +1 ≤ r0 +1,
i.e.,
y ∈ X(fm,tm,cm).
We have obtained that
X ˜fm,tm,cm ⊆ X (fm,tm,cm), m ≥ m0, m ∈ N.
The opposite inclusion is also valid, which gives
X ˜fm,tm,cm = X (fm,tm,cm), m ≥ m0, m ∈ N.
We have proved that
X ˜fm,tm,cm ⊂ C(I,Rn
), m ∈ N.
Hence,
X (fm,tm,cm) ⊂ C(I,Rn
), m ≥ m0, m ∈ N.
Now, it is enough to consider
lim
m→∞
X ˜fm,tm,cm ⊆ X (f0,t0,c0),
i.e.,
lim
m→∞
X (fm,tm,cm) ⊆ X (f0,t0,c0).
Corollary 8.1. Let the assumptions 1.–4. from Theorem 8.1 be valid and let
xm ∈ X(fm,tm,cm), m ∈ N.
Then, the sequence {xm}∞
m=1 has a subsequence xmk
∞
k=1
such that
lim
k→∞
xmk
(t) = x(t)
uniformly on I, where
x ∈ X(f0,t0,c0).
44
Proof. From Theorem 8.1, it follows the existence of m0 ∈ N such that
xm ∈ C(I,Rn
), m ≥ m0, m ∈ N.
In addition,
lim
m→∞
X (fm,tm,cm) ⊆ X (f0,t0,c0).
Thus, there exists r0 > 0 for which
xm(t) ≤ r0, t ∈ I, m ≥ m0, m ∈ N.
We recall that the set X(f0,t0,c0) is bounded in the space C(I,Rn
) (due to Theorem
4.1). We have obtained that the functions of the sequence {xm}∞
m=m0
are uniformly
bounded.
Obviously,
xm(t) = xm(t0)+ym(t)+zm(t), t ∈ I, m ≥ m0, m ∈ N,
where
ym(t) =
t
t0
fm(τ,xm(τ))− fm(τ,0)dτ, t ∈ I, m ≥ m0, m ∈ N,
and
zm(t) =
t
t0
fm(τ,0)dτ, t ∈ I, m ≥ m0, m ∈ N.
From the condition 3., we obtain
fm(t,x)− fm(t,0) ≤ ωr0 (t, x )
≤ ωr0 (t,r0), t ∈ I, x ∈ B[0,r0], m ≥ m0, m ∈ N,
because we can assume (without loss of generality) that the function ωr0 is non-decreasing
in the second variable. At the same time, we have
ym(t) ≤ ωr0 (t,r0)
for almost all t ∈ I and all m ≥ m0, m ∈ N. Hence, the functions of the sequence
{ym}∞
m=m0
are equicontinuous. From the condition 2., we get
lim
m→∞
zm(t) =
t
t0
f0(τ,0)dτ
uniformly on I. Thus, the functions of the sequence {zm}∞
m=m0
are equicontinuous.
Therefore, the functions of the sequence {xm}∞
m=m0
are equicontinuous as well.
Using the Arzelà–Ascoli theorem, from the sequence {xm}∞
m=m0
, one can extract a
subsequence xmk
∞
k=1
satisfying
lim
k→∞
xmk
(t) = x(t)
uniformly on I, where x ∈ C(I,Rn
).
45
It remains to prove that x ∈ X(f0,t0,c0). Let ε > 0 be arbitrary. Then, there exists
k0 ∈ N such that
xmk0
∈ Xε
2
(f0,t0,c0)
and that
xmk0
−x
C
<
ε
2
.
Therefore, one can choose y ∈ X(f0,t0,c0) satisfying
xmk0
∈ B y,
ε
2
,
i.e.,
y−xmk0 C
<
ε
2
.
Thus, x ∈ B(y,ε). From the arbitrariness of ε > 0, it follows that x is an element of the
closure of the set X(f0,t0,c0). This set is closed (see Remark 4.1). Therefore,
x ∈ X(f0,t0,c0).
Remark 8.1. From the proof of Theorem 8.1 and from Corollary 8.1, it follows that the
condition 3. can be replaced by the following one:
¯3. for any r > 0, there exists a function ωr ∈ K(I ×R+,R+) such that
ωr(−,0) ≡ 0
and that
fm(t,x)− fm(t,y) ≤ ωr(t, x−y )+ωr t,
1
m
for all t ∈ I, x,y ∈ B[0,r], m ∈ N.
46
9. Kneser theorem
Let I be a compact interval. We consider the Cauchy problem
x = f(t,x), (9.1)
x(t0) = c0, (9.2)
where f ∈ K(I × Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
. As X(f,t0,c0), we denote the set of all
non-extendable solutions of the problem (9.1), (9.2).
Theorem 9.1 (Kneser). If any non-extendable solution of the problem (9.1), (9.2) exists
on I, then the set X(f,t0,c0) is compact and connected in the space C(I,Rn
).
Proof. According to Remark 4.1, the set X(f,t0,c0) is compact in the space C(I,Rn
).
We show that it is connected. We suppose the opposite, i.e., let X(f,t0,c0) be disconnected.
Thus, there exist non-empty closed sets X1,X2 ⊂ X(f,t0,c0) such that
X1 ∩X2 = /0, X1 ∪X2 = X(f,t0,c0).
We denote
δ = inf{ y−x C ;y ∈ X1,x ∈ X2}.
We know that δ > 0. The sets X1,X2 are compact. Hence, there exist x1 ∈ X1, x2 ∈ X2
such that
x1 −x2 C = δ.
From the compactness of X(f,t0,c0) in C(I,Rn
), it follows the existence of r > 0 such
that
x C ≤ r −1, x ∈ X(f,t0,c0).
Now, we use Lemma 1.6 and Remarks 1.2 and 1.3. There exists a sequence
{ fm}∞
m=1 of functions satisfying the following conditions:
a. it holds
fm ∈ K(I ×Rn
,Rn
), m ∈ N;
b. for any ρ > 0 and m ∈ N, there exists a function lρ,m ∈ L(I,R+) such that
fm(t,x)− fm(t,y) ≤ lρ,m(t) x−y , t ∈ I, x,y ∈ B[0,ρ];
c. for almost all t ∈ I and any ρ > 0, it holds
lim
m→∞
fm(t,x) = f(t,x)
uniformly on B[0,ρ];
d. for any ρ > 0, there exists a function hρ ∈ L(I,R+) such that
fm(t,x) ≤ hρ (t), t ∈ I, x ∈ B[0,ρ], m ∈ N;
47
e. for any ρ > 0, there exists a function ωρ ∈ K(I ×R+,R+) such that
ωρ (−,0) ≡ 0
and that
fm(t,x)− fm(t,y) ≤ ωρ (t, x−y )+ωρ t,
1
m
for all t ∈ I, x,y ∈ B[0,ρ], m ∈ N.
For all λ ∈ [0,1] and m ∈ N, we put
fλ,m(t,x) = fm(t,x)+(1−λ)[f(t,x1(t))− fm(t,x1(t))]
+λ [f(t,x2(t))− fm(t,x2(t))], t ∈ I, x ∈ Rn
.
One can easily verify that the functions fλ,m satisfy the following conditions:
1. it holds
fλ,m ∈ K(I ×Rn
,Rn
), λ ∈ [0,1], m ∈ N;
2. for any ρ > 0 and m ∈ N, there exists a function lρ,m ∈ L(I,R+) such that
fλ,m(t,x)− fλ,m(t,y) ≤ lρ,m(t) x−y , t ∈ I, x,y ∈ B[0,ρ], λ ∈ [0,1];
3. for almost all t ∈ I and any ρ > 0, it holds
lim
m→∞
fλ,m(t,x) = f(t,x)
uniformly with respect to x ∈ B[0,ρ], λ ∈ [0,1];
4. for any ρ > 0, it holds
fλ,m(t,x) ≤ ˜hρ (t), t ∈ I, x ∈ B[0,ρ], λ ∈ [0,1], m ∈ N,
where
˜hρ (t) = hρ (t)+hr−1(t)+ f(t,x1(t)) + f(t,x2(t)) , t ∈ I;
5. for any ρ > 0, there exists a function ωρ ∈ K(I ×R+,R+) with the property that
ωρ (−,0) ≡ 0
and that
fλ,m(t,x)− fλ,m(t,y) ≤ ωρ (t, x−y )+ωρ t,
1
m
for all t ∈ I, x,y ∈ B[0,ρ], λ ∈ [0,1], m ∈ N.
For all λ ∈ [0,1] and m ∈ N, we consider the problem
x = fλ,m(t,x), (9.3)
x(t0) = c0. (9.2)
48
Due to the condition 2., from Corollary 7.1, it follows that the problem (9.3), (9.2) is
uniquely solvable for any λ ∈ [0,1] and m ∈ N. As xλ,m, we denote the non-extendable
solution of this problem. Since the convergence is uniform with respect to λ in the
condition 3. and since the majorants are also independent on λ in the conditions 4. and
5., the assumptions of Theorem 8.1 (Remark 8.1) are satisfied (uniformly with respect
to λ). Thus, there exists m0 ∈ N such that, for all m ≥ m0, m ∈ N, and λ ∈ [0,1], the
solution xλ,m exists on the interval I. Moreover, due to the inequality
x C ≤ r −1, x ∈ X(f,t0,c0),
the proof of Theorem 8.1 guarantees that, without loss of generality, we can assume
the inequality
xλ,m C
≤ r, m ≥ m0, m ∈ N, λ ∈ [0,1].
We denote
ηm(λ) = inf xλ,m −y C
;y ∈ X1 , λ ∈ [0,1], m ≥ m0, m ∈ N.
From the definition of the function fλ,m and the uniqueness of the solution xλ,m, it
follows that
x0,m ≡ x1, x1,m ≡ x2, m ∈ N.
Therefore, ηm(0) = 0, ηm(1) = δ, m ≥ m0, m ∈ N. Let m ≥ m0 (m ∈ N) be arbitrarily
given. Let λ0 ∈ [0,1] be also arbitrarily given and let {λk}∞
k=1 ⊂ [0,1] be a sequence
for which
lim
k→∞
λk = λ0.
Using Theorem 8.1, one can verify that
lim
k→∞
xλk,m(t) = xλ0,m(t)
uniformly on I. Since the set X1 is compact, from the definition of the function ηm, it
follows that
lim
k→∞
ηm(λk) = ηm(λ0),
i.e., the function ηm is continuous on [0,1]. Therefore,
ηm(0) = 0, ηm(1) = δ, m ≥ m0, m ∈ N,
implies that, for any m ≥ m0, m ∈ N, there exists λm ∈ (0,1) with the property that
ηm(λm) =
δ
2
,
i.e.,
inf xλm,m −y C
;y ∈ X1 =
δ
2
, m ≥ m0, m ∈ N.
Due to the condition 4. and
xλ,m C
≤ r, m ≥ m0, m ∈ N, λ0 ∈ [0,1],
the functions of the sequence xλm,m
∞
m=m0
have to be uniformly bounded and equicontinuous.
According to the Arzelà–Ascoli theorem, without loss of generality, we can
assume that this sequence is convergent, i.e., there exists ˜x ∈ C(I,Rn
) such that
lim
m→∞
xλm,m(t) = ˜x(t)
49
uniformly on I.
Next, Theorem 8.1 gives
˜x ∈ X(f,t0,c0).
Since the set X1 is compact, we have
inf{ ˜x−y C ;y ∈ X1} =
δ
2
.
Thus, we know that ˜x /∈ X1 and, consequently, ˜x ∈ X2, which is a contradiction. The
contradiction proves that the set X(f,t0,c0) is connected.
50
10. Fukuhara theorems
Let I be a compact interval. We consider the Cauchy problem
x = f(t,x), (10.1)
x(t0) = c0, (10.2)
where f ∈ K(I × Rn
,Rn
), t0 ∈ I, c0 ∈ Rn
. As X(f,t0,c0), we denote the set of all
non-extendable solutions of the problem (10.1), (10.2). For M ⊆ Rm
, we denote
X(f,t0,M) =
c∈M
X(f,t0,c).
Theorem 10.1 (1. Fukuhara). Let M be a closed and connected subset of Rn
. If, for
any c0 ∈ M, all non-extendable solution of the problem (10.1), (10.2) exists on I, then
the set X(f,t0,M) is closed and connected in the space C(I,Rn
). Moreover, if the set
M is bounded, then the set X(f,t0,M) is compact.
Proof. The set X(f,t0,M) is closed in the space C(I,Rn
). Indeed, it suffices to consider
Theorem 8.1 and the fact that the set M is closed.
We show that the set X(f,t0,M) is connected. We suppose the opposite. Then,
there exist non-empty closed sets X1,X2 ⊂ X(f,t0,M) such that
X1 ∩X2 = /0, X1 ∪X2 = X(f,t0,M).
Let c0 ∈ M be arbitrary. We denote
Yi = X (f,t0,c0)∩Xi, i ∈ {1,2}.
If Y1 = /0 and Y2 = /0, then Y1 and Y2 are non-empty closed subsets of X(f,t0,c0) such
that
Y1 ∩Y2 = /0, Y1 ∪Y2 = X(f,t0,c0).
Next, from Theorem 9.1, it follows that the set X(f,t0,c0) is a subset of Xi for some
i ∈ {1,2}. We denote
Mi = {c0 ∈ M;X(f,t0,c0) ⊆ Xi}, i ∈ {1,2}.
Obviously, M1 = /0, M2 = /0, M1 ∩M2 = /0, and M1 ∪M2 = M.
We prove that the set M1 is closed. Let {cm}∞
m=1 ⊆ M1 be such that
lim
m→∞
cm = c0.
Then, c0 ∈ M (M is closed) and there exists a sequence {xm}∞
m=1 ⊆ X1 such that
xm ∈ X(f,t0,cm), m ∈ N.
Without loss of generality (see Corollary 8.1), we can assume that
lim
m→∞
xm(t) = ˜x(t)
51
uniformly on I, where
˜x ∈ X (f,t0,c0).
Since the set X1 is closed, we see that ˜x ∈ X1. Thus, c0 ∈ M1, which means that M1 is
closed.
Analogously, one can prove that M2 is closed as well. Hence, we have proved that
M1,M2 are non-empty closed subsets of the set M satisfying
M1 ∩M2 = /0, M1 ∪M2 = M,
which is a contradiction with the assumption that M is connected. The contradiction
means that the set X(f,t0,M) is connected.
Now, we consider that the set M is bounded. Let {xm}∞
m=1 be a sequence of elements
of X(f,t0,M). Then, there exists a sequence {cm}∞
m=1 ⊆ M such that
xm ∈ X(f,t0,cm), m ∈ N.
Due to the compactness of the set M, without loss of generality, we can assume that
lim
m→∞
cm = c0,
where c0 ∈ M. From Corollary 8.1, it follows the existence of a subsequence xmk
∞
k=1
of the sequence {xm}∞
m=1 satisfying
lim
k→∞
xmk
(t) = x(t)
uniformly on I, where
x ∈ X(f,t0,c0).
Since c0 ∈ M, we see that
x ∈ X(f,t0,M),
i.e., the set X(f,t0,M) is compact in the space C(I,Rn
).
Let
X(f,t0,c0) ⊆ C(I,Rn
).
We denote
W(f,t0,c0) = {(t,x(t));t ∈ I,x ∈ X(f,t0,c0)}
and, for t ∈ I, we put
Wt(f,t0,c0) = {x ∈ Rn
;(t,x) ∈ W(f,t0,c0)}.
Theorem 10.2 (2. Fukuhara). If any non-extendable solution of the problem (10.1),
(10.2) exists on I, then, for all ¯t ∈ I and ¯c ∈ ∂W¯t(f,t0,c0), there exists ¯x ∈ X(f,t0,c0)
such that ¯x(¯t) = ¯c and the graph of the function ¯x between t0 and ¯t is on the boundary
of the set W(f,t0,c0), i.e.,
(t, ¯x(t)) ∈ ∂W(f,t0,c0), min{t0, ¯t} ≤ t ≤ max{t0, ¯t}.
52
Proof. If we use the boundedness of the set X(f,t0,c0) (see Theorem 4.1), then, without
loss of generality, we can assume the existence of a function h ∈ L(I,R+) for which
f(t,x) ≤ h(t), t ∈ I, x ∈ Rn
.
We prove the theorem in the case when ¯t > t0 (in the second case, the proof is
analogical). According to Theorem 9.1, W¯t(f,t0,c0) is a connected and compact subset
of Rn
. Therefore, there exists a sequence {ck}∞
k=1 ⊂ Rn
such that
ck /∈ W¯t(f,t0,c0), k ∈ N,
and that
lim
k→∞
ck = ¯c.
For k ∈ N, we consider the problem
x = f(t,x), (10.1)
x(¯t) = ck. (10.3)
We assume that
f(t,x) ≤ h(t), t ∈ I, x ∈ Rn
.
Therefore, for all k ∈ N, there exists a solution xk of the problem (10.1), (10.3) on the
interval I, where (see also Corollary 6.1)
xk ∈ X(f, ¯t,ck), k ∈ N.
By contradiction, we show that
(t,xk(t)) /∈ W(f,t0,c0), t ∈ [t0, ¯t], k ∈ N.
If, for some k ∈ N, there exists t ∈ [t0, ¯t) such that
(t ,xk(t )) ∈ W(f,t0,c0),
then one can find ˜x ∈ X(f,t0,c0) satisfying ˜x(t ) = xk(t ). We denote
y(t) =
˜x(t), t ∈ [infI,t ];
xk(t), t ∈ (t ,supI].
Obviously,
y ∈ X(f,t0,c0).
Therefore,
y(¯t) ∈ W¯t(f,t0,c0).
At the same time,
y(¯t) = ck /∈ W¯t(f,t0,c0),
which is a contradiction.
Corollary 6.1 gives that all non-extendable solution of the problem
x = f(t,x), (10.1)
x(¯t) = ¯c (10.4)
53
exists on the interval I. Therefore, with regard to Corollary 8.1, without loss of generality,
we can assume that
lim
k→∞
xk(t) = x0(t)
uniformly on I, where
x0 ∈ X(f, ¯t, ¯c).
Let s ∈ (t0, ¯t) be arbitrarily given. We consider the sets Ws(f,t0,c0) and Ws(f, ¯t, ¯c). We
have proved that x0(s) is not an inner point of the set Ws(f,t0,c0), i.e.,
x0(s) /∈ Ws(f,t0,c0) or x0(s) ∈ ∂Ws(f,t0,c0).
But, it is valid that
Ws(f,t0,c0)∩Ws(f, ¯t, ¯c) = /0.
Indeed, since (¯t, ¯c) ∈ W(f,t0,c0), there exists ˜x ∈ X(f,t0,c0) such that ˜x(¯t) = ¯c. This
fact means that
˜x ∈ X(f, ¯t, ¯c).
Thus,
˜x(s) ∈ Ws(f,t0,c0)∩Ws(f, ¯t, ¯c).
The sets Ws(f,t0,c0) and Ws(f, ¯t, ¯c) are connected and closed and they are not disjoint.
There exists a point belonging to the set Ws(f, ¯t, ¯c) which is not in the interior of
the set Ws(f,t0,c0). Therefore, there exists a point cs ∈ Rn
such that
cs ∈ ∂Ws(f,t0,c0)∩Ws(f, ¯t, ¯c).
Hence, there exists a solution ˜x1 of the problem (10.1), (10.2) passing through the point
[s,cs] and, at the same time, there exists a solution ˜x2 of the problem (10.1), (10.4)
passing through the point [s,cs]. If we put
y(t) =
˜x1(t), t ∈ [infI,s];
˜x2(t), t ∈ (s,supI],
then
y ∈ X(f,t0,c0), y(s) = cs, y(¯t) = ¯c.
We have proved that, for any s ∈ (t0, ¯t), there exists y ∈ X(f,t0,c0) such that
y(¯t) = ¯c, (s,y(s)) ∈ ∂W(f,t0,c0).
Thus, for an arbitrary set
{s1,...,sm} ⊂ (t0, ¯t),
there exists ym ∈ X(f,t0,c0) with the property that
ym(¯t) = ¯c, (si,ym(si)) ∈ ∂W(f,t0,c0), i ∈ {1,...,m}. (10.5)
Let {sk}∞
k=1 be dense in (t0, ¯t). It is obvious that, for any m ∈ N, there exists ym ∈
X(f,t0,c0) satisfying (10.5). Since the set X(f,t0,c0) is compact in the space C(I,Rn
),
without loss of generality, we can assume that the sequence {ym}∞
m=1 is convergent,
i.e.,
lim
m→∞
ym(t) = ¯x(t)
54
uniformly on I, where
¯x ∈ X(f,t0,c0).
Using (10.5), one can easily verify that
(t, ¯x(t)) ∈ ∂W(f,t0,c0), t ∈ [t0, ¯t].
The proof is complete.
55
REFERENCES
[1] Y. Deng: Lectures, problems and solutions for ordinary differential equations.
Hackensack: World Scientific, 2018.
[2] S. P. Hastings: Classical methods in ordinary differential equations: with applications
to boundary value problems. Providence: American Mathematical
Society, 2012.
[3] D. W. Jordan: Nonlinear ordinary differential equations—problems and solutions:
a sourcebook for scientists and engineers. Oxford: Oxford University
Press, 2007.
[4] A. K. Nandakumaran, P. S. Datti, R. K. George: Ordinary differential equations:
principles and applications. New York: Cambridge University Press, 2017.
[5] C. E. Roberts: Ordinary differential equations: applications, models, and computing.
Boca Raton: Chapman and Hall/CRC, 2010.
[6] D. G. Schaeffer: Ordinary differential equations: basics and beyond. New York:
Springer, 2016.