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sub(super)-periodic solutions.

De¬nition 2.12 We say that x ∈ C 1 ([0, T ]; R) is a sub-periodic solution for
(3) if
x (t) + g(x(t)) = f (t), t ∈ [0, T ],
and x(0) ¤ x(T ).

Note that a necessary condition for the existence is given next.

Proposition 2.13 If equation (3) has sub-periodic solutions, then there is x0 ∈
R such that g(x0 ) ¤ f .

Proof Let x be a sub-periodic solution of (3). By integration on [0, T ] we ¬nd
T
x(T ) ’ x(0) + g(x(t))dt = T f .
0

Since g —¦ x is continuous, there is „ ∈]0, T [ such that

1
g(x(„ )) = f ’ (x(T ) ’ x(0)) ¤ f .
T
Similarly we de¬ne the notion of super-periodic solution.

De¬nition 2.14 We say that y ∈ C 1 ([0, T ]; R) is a super-periodic solution for
(3) if
y (t) + g(y(t)) = f (t), t[0, T ],
and y(0) ≥ y(T ).

The analogous necessary condition holds.
Mihai Bostan 11


Proposition 2.15 If equation (3) has super-periodic solutions, then there is
y0 ∈ R such that g(y0 ) ≥ f .
Remark 2.16 It is clear that x is periodic solution for (3) if and only if is in
the same time sub-periodic and super-periodic solution. Therefore there are
x0 , y0 ∈ R such that
g(x0 ) ¤ f ¤ g(y0 ).
Since g is continuous, we deduce that f ∈ Range(g) which is exactly the
necessary condition given by the Proposition 2.6.
As before we will prove that the necessary condition of Proposition 2.13 is
also su¬cient for the existence of sub-periodic solutions.
Theorem 2.17 Assume that g is increasing Lipschitz continuous and f is T -
periodic continuous. Then equation (3) has sub-periodic solutions if and only if
there is x0 ∈ R such that g(x0 ) ¤ f .

Proof The condition is necessary (see Proposition 2.13). Let us prove now
that it is also su¬cient. Consider z0 an arbitrary initial data and denote by
x : [0, ∞[’ R the solution for (3) with the initial condition x(0) = z0 . If there
is t0 ≥ 0 such that x(t0 ) ¤ x(t0 + T ), thus xt0 (t) := x(t0 + t), t ∈ [0, T ] is a
sub-periodic solution. Suppose now that x(t) > x(t+T ), ∀t ∈ R. By integration
on [nT, (n + 1)T ], n ≥ 0 we get
T
x((n + 1)T ) ’ x(nT ) + g(x(nT + t))dt = T f , n ≥ 0.
0

Using the hypothesis and the average formula we have
1
{x(nT ) ’ x((n + 1)T )} > g(x0 ),
g(x(nT + „n )) = f +
T
for „n ∈]0, T [ and n ≥ 0. Since g is increasing we deduce that x(nT + „n ) >
x0 , n ≥ 0. We have also x(nT + „n ) ¤ x((n ’ 1)T + „n ) ¤ · · · ¤ x(„n ) ¤
supt∈[0,T ] |x(t)| and thus we deduce that (x(nT + „n ))n≥0 is bounded:
|x(nT + „n )| ¤ K, n ≥ 0.
Consider now the functions xn : [0, T ] ’ R given by
t ∈ [0, T ].
xn (t) = x(nT + t),
By a standard computation we get
1d
|xn (t)|2 + [g(xn (t)) ’ g(0)]xn (t) = [f (t) ’ g(0)]xn (t), t ∈ [0, T ].
2 dt
Using the monotony of g we obtain
t
|xn (t)| ¤ |xn (s)| + |f (u) ’ g(0)|du, 0 ¤ s ¤ t ¤ T.
s
12 Periodic solutions for evolution equations


Taking s = „n ∈]0, T [ we can write
t T
|xn (t)| ¤ |xn („n )| + |f (u) ’ g(0)|du ¤ K + |f (u) ’ g(0)|du, t ∈ [„n , T ].
„n 0

For t ∈ [0, „n ], n ≥ 1 we have
nT +t
|xn (t)| = |x(nT + t)| ¤ |x((n ’ 1)T + „n’1 )| + |f (u) ’ g(0)|du
(n’1)T +„n’1
(n+1)T
¤ K+ |f (u) ’ g(0)|du
(n’1)T
T
¤ K +2 |f (u) ’ g(0)|du.
0

Therefore, the sequence (xn )n≥0 is uniformly bounded in L∞ (R) and
T
¤K +2 |f (t) ’ g(0)|dt := M.
xn L∞ (R)
0

Moreover, (xn )n≥0 is also uniformly bounded in L∞ (R). Indeed we have

|xn (t)| = |f (t) ’ g(xn (t))| ¤ f + max{g(M ), ’g(’M )},
L∞ (R)

and hence, by Arzela-Ascoli™s theorem we deduce that (xn )n≥0 converges in
C 0 ([0, T ], R):
lim xn (t) = u(t), uniformly for t ∈ [0, T ].
n’∞
As usual, by passing to the limit for n ’ ∞ we ¬nd that u is also solution for
(3). Moreover since (x(nT ))n≥0 is decreasing and bounded, it is convergent and
we can prove that u is periodic:

u(0) = lim xn (0) = lim x(nT ) = lim x((n + 1)T ) = lim xn (T ) = u(T ).
n’∞ n’∞ n’∞ n’∞

Therefore, u is a sub-periodic solution for (3). An analogous result holds for
super-periodic solutions.

Proposition 2.18 Under the same assumptions as in Theorem 2.17 the equa-
tion (3) has super-periodic solutions if and only if there is y0 ∈ R such that
g(y0 ) ≥ f .

We state now a comparison result between sub-periodic and super-periodic
solutions.

Proposition 2.19 If g is increasing, x is a sub-periodic solution and y is a
super-periodic solution we have

x(t) ¤ y(t), ∀t ∈ [0, T ],

provided that x and y are not both periodic.
Mihai Bostan 13


Proof Both x and y verify (3), thus
(x ’ y) (t) + g(x(t)) ’ g(y(t)) = 0, t ∈ [0, T ].
With the notation
g(x(t))’g(y(t))
t ∈ [0, T ], x(t) = y(t)
x(t)’y(t)
r(t) = (24)
t ∈ [0, T ], x(t) = y(t),
0

we can write g(x(t)) ’ g(y(t)) = r(t)(x(t) ’ y(t)), t ∈ [0, T ] and therefore,
(x ’ y) (t) + r(t)(x(t) ’ y(t)) = 0, t ∈ [0, T ]
which implies
t
x(t) ’ y(t) = (x(0) ’ y(0))e’ r(s)ds
. (25)
0


Now it is clear that if x(0) ¤ y(0) we also have x(t) ¤ y(t), t ∈ [0, T ]. Suppose
now that x(0) > y(0). Taking t = T in (25) we obtain
T
x(T ) ’ y(T ) = (x(0) ’ y(0))e’ r(t)dt
. (26)
0



Since g is increasing, by the de¬nition of the function r we have r ≥ 0. Two
T T
cases are possible: (i) either 0 r(t)dt > 0, (ii) either 0 r(t)dt = 0 in which
case r(t) = 0, t ∈ [0, T ] (r vanishes in all points of continuity t such that
x(t) = y(t) and also in all points t with x(t) = y(t) by the de¬nition). Let us
analyse the ¬rst case (i). By (26) we deduce that x(T ) ’ y(T ) < x(0) ’ y(0) or
x(T ) ’ x(0) < y(T ) ’ y(0). Since x is sub-periodic we have x(0) ¤ x(T ) which
implies that y(T ) > y(0) which is in contradiction with the super-periodicity of
y ( y(T ) ¤ y(0)).
In the second case (ii) we have g(x(t)) = g(y(t)), t ∈ [0, T ] so (x ’ y) = 0 and
therefore there is a constant C ∈ R such that x(t) = y(t) + C, t ∈ [0, T ]. Taking
t = 0 and t = T we obtain
0 ≥ x(0) ’ x(T ) = y(0) ’ y(T ) ≥ 0,
and thus x and y are both periodic which is in contradiction with the hypothesis.
In the following we will see how it is possible to retrieve the existence result for
periodic solutions by using the method of sub(super)-periodic solutions. Sup-
pose that f ∈ Range(g). Obviously both su¬cient conditions for existence of
sub(super)-periodic solutions are satis¬ed and thus there are x0 (y0 ) sub(super)-
periodic solutions. If y0 is even periodic the proof is complete. Assume that y0
is not periodic (y0 (0) > y0 (T )). Denote by M the set of sub-periodic solutions
for (3):
M = {x : [0, T ] ’ R : x sub-periodic solution , x0 (t) ¤ x(t), t ∈ [0, T ]}.
Since x0 ∈ M we have M = …. Moreover, from the comparison result since y0
is super-periodic but not periodic we have x ¤ y0 , ∀x ∈ M. We prove that M
contains a maximal element in respect to the order:
x2 (if and only if) x1 (t) ¤ x2 (t), t ∈ [0, T ].
x1
14 Periodic solutions for evolution equations


Finally we show that this maximal element is even a periodic solution for (3)
since otherwise it would be possible to construct a sub-periodic solution greater
than the maximal element. We state now the following generalization.

Theorem 2.20 Assume that g : R — R ’ R is increasing Lipschitz continuous
function in x, T -periodic and continuous in t and f : R ’ R is T -periodic and
continuous in t. Then the equation

t ∈ R,
x (t) + g(t, x(t)) = f (t), (27)

has periodic solutions if and only if there is x0 ∈ R such that
T T
1 1
f := f (t)dt = g(t, x0 )dt = G(x0 ). (28)
T T
0 0

Moreover, in this case we have the estimate
T
∀ x0 ∈ G’1 f .
¤ |x0 | + |f (t) ’ g(t, x0 )|dt,
x L∞ (R)
0


Proof Consider the average function G : R ’ R given by
T
1
g(t, x)dt, x ∈ R.
G(x) =
T 0

It is easy to check that G is also increasing and Lipschitz continuous with the
same constant. Let us prove that the condition (28) is necessary. Suppose that
x is a periodic solution for (27). By integration on [0, T ] we get
T
1
g(t, x(t))dt = f . (29)
T 0

We can write
m ¤ x(t) ¤ M, t ∈ [0, T ],
and thus
g(t, m) ¤ g(t, x(t)) ¤ g(t, M ), t ∈ [0, T ],
which implies
T T T
1 1 1
g(t, m)dt ¤ g(t, x(t))dt ¤
G(m) = g(t, M )dt = G(M ).
T T T
0 0 0

Since G is continuous it follows that there is x0 ∈ [m, M ] such that G(x0 ) =
1T
T 0 g(t, x(t))dt and from (29) we deduce that f = G(x0 ).
Let us show that the condition (28) is also su¬cient. As before let us consider
the unique periodic solution for

t ∈ [0, T ], ± > 0,
±x± (t) + x± (t) + g(t, x± (t)) = f (t),
Mihai Bostan 15


(existence and uniqueness follow by the Banach™s ¬xed point theorem exactly as
before). All we need to prove is that (x± )±>0 is uniformly bounded in L∞ (R)
(then (x± )±>0 is also uniformly bounded in L∞ (R) and by Arzela-Ascoli™s the-
orem we deduce that x± converges to a periodic solution for (27)). Taking the
average on [0, T ] we get
T
1
{±x± (t) + g(t, x± (t))}dt = f = G(x0 ), ± > 0.
T 0

As before we can write

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