t

’1

u=W γ (s) ds

t0

is the unique solution of (4.8) and depends continuously on u 0 .

4.2 Estimating solutions

Theorem 4.4. Assume, in addition to the hypotheses of Theorem 4.1, that

g(t, x) ¤ γ (t)ρ( x ), x ∈ B0 , t ∈ I0 ,

(4.9)

where γ (t) and ρ(t) satisfy the hypotheses of Lemma 4.2 with T = t0 + T1 . Let u(t)

be the positive solution of

u (t) = γ (t)ρ(u(t)), t ∈ [t0 , T ), u(t0 ) = u 0 ≥ x 0

(4.10)

provided by that lemma. Then the unique solution of (4.4) satis¬es

x(t) ¤ u(t), t ∈ [t0 , T ).

(4.11)

Proof. Assume that there is a t1 ∈ [t0 , T ) such that

u(t1 ) < x(t1 ) .

For µ > 0, let u µ (t) be the solution of

u (t) = [γ (t) + µ]ρ(u(t)), t ∈ [t0 , T ), u(t0 ) = u 0 .

(4.12)

By Lemma 4.2, a solution exists for µ > 0 suf¬ciently small. Moreover, u µ (t) ’ u(t)

uniformly on any compact subset of [t0 , T ). Let

w(t) = x(t) ’ u µ (t).

Then we may take µ suf¬ciently small so that

w(t0 ) ¤ 0, w(t1 ) > 0.

Let t2 be the largest number in [t0 , t1 ) such that w(t2 ) = 0 and

w(t) > 0, t ∈ (t2 , t1 ].

For h > 0 suf¬ciently small, we have

w(t2 + h) ’ w(t2 )

> 0.

h

34 4. Ordinary Differential Equations

Consequently,

D + w(t2 ) ≥ 0.

But

D + w(t2 ) = D + x(t2 ) ’ u µ (t2 )

(4.13)

¤ x (t2 ) ’ u µ (t2 )

= g(t2 , x(t2 )) ’ [γ (t2 ) + µ]ρ(u µ (t2 ))

¤ γ (t2 )ρ( x(t2 ) ) ’ [γ (t2 ) + µ]ρ(u µ (t2 ))

= ’ µρ(u µ (t2 ))

< 0.

This contradiction proves the theorem.

4.3 Extending solutions

Theorem 4.5. Let g(t, x) be a continuous map from R— H to H , where H is a Banach

space. Assume that for each point (t0 , x 0 ) ∈ R — H, there are constants K , b > 0 such

that

(4.14) g(t, x)’ g(t, y) ¤ K x ’ y , |t ’t0 | < b, x ’ x 0 < b, y ’ x 0 < b.

Assume also that

g(t, x) ¤ γ (t)ρ( x ), x ∈ H, t ∈ [t0 , TM ),

(4.15)

where TM ¤ ∞, and γ (t), ρ(t) satisfy the hypotheses of Lemma 4.2. Then, for each

x 0 ∈ H and t0 > 0, there is a unique solution x(t) of the equation

d x(t)

= g(t, x(t)), t ∈ [t0 , TM ), x(t0 ) = x 0 .

(4.16)

dt

Moreover, x(t) depends continuously on x 0 and satis¬es

x(t) ¤ u(t), t ∈ [t0 , TM ),

(4.17)

where u(t) is the solution of (4.6) in that interval satisfying u(t0 ) = u 0 ≥ x 0 .

Before proving Theorem 4.5, we note that the following is an immediate conse-

quence.

Corollary 4.6. Let V (y) be a locally Lipschitz continuous map from H to itself satis-

fying

V (y) ¤ C(1 + y ), y ∈ H.

(4.18)

Then, for each y0 ∈ H, there is a unique solution of

t ∈ R+ , y(0) = y0 .

y (t) = V (y(t)),

(4.19)

4.4. The proofs 35

4.4 The proofs

We now give the proof of Theorem 4.5.

Proof. By Theorems 4.1 and 4.4, there is an interval [t0 , t0 + m], m > 0, in which a

unique solution of

d x(t)

= g(t, x(t)), t ∈ [t0 , t0 + m], x(t0 ) = x 0

(4.20)

dt

exists and satis¬es

x(t) ¤ u(t), t ∈ [t0 , t0 + m],

(4.21)

where u(t) is the unique solution of

u (t) = γ (t)ρ(u(t)), t ∈ [t0 , TM ), u(t0 ) = u 0 = x 0 .

(4.22)

Let T ¤ TM be the supremum of all numbers t0 + m for which this holds. If t1 < t2 <

T, then the solution in [t0 , t2 ] coincides with that in [t0 , t1 ], since such solutions are

unique. Thus, a unique solution of (4.20) satisfying (4.21) exists for each t0 < t < T.

Moreover, we have

t2

x(t2 ) ’ x(t1 ) = g(t, x(t)) dt.

t1

Consequently,

t2

x(t2 ) ’ x(t1 ) ¤ g(t, x(t)) dt

t1

t2

¤ γ (t)ρ( x(t) ) dt

t1

t2

¤ γ (t)ρ(u(t)) dt

t1

= u(t2 ) ’ u(t1 ).

Assume that T < TM . Let tk be a sequence such that t0 < tk < T and tk ’ T . Then,