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Existence of Solutions for a Class of Discontinuous Differential Equations in R n
Journal of Mathematical Analysis and Applications 233, 634᎐643 Ž1999. Article ID jmaa.1999.6324, available online at http:rrwww.idealibrary.com on
Existence of Solutions for a Class of Discontinuous Differential Equations in ⺢ n Witold Rzymowski Instytut Matematyki UMCS, 20-031 Lublin, Poland Submitted by Zhi¨ ko S. Athanasso¨ Received July 10, 1997
In this paper we extend the Caratheodory existence theorem for a class of ´ discontinuous differential equations with right-hand sides in a special form but with no continuity or monotonicity assumptions. A new existence result for a scalar 䊚 1999 Academic Press discontinuous differential equation is also established.
1. INTRODUCTION It is well known that if f : w0, ⬁. = ⺢ n ª ⺢ n is bounded and satisfies the Caratheodory condition, i.e., f Ž⭈, x . is measurable for all x g ⺢ n and ´ f Ž t, ⭈ . is continuous for almost all t G 0, then the Cauchy problem x⬘ Ž t . s f Ž t , x Ž t . . , a.e. in w 0, ⬁ . , x Ž 0. s 0 has an absolutely continuous solution defined in w0, ⬁.. Existence results with f discontinuous in x involve usually semicontinuity or monotonicity assumptions Žsee, e.g., w3, 4, 1x.. In this paper we deal with a discontinuous differential equation of a special form. Namely, we shall consider the system of ordinary differential equations xX1 Ž t . s g 1 Ž t , q11 Ž x 1 Ž t . . , . . . , q1 n Ž x n Ž t . . . ,
x 1 Ž t . s 0,
xX2
x 2 Ž t . s 0,
Ž t . s g 2 Ž t , q21 Ž x 1 Ž t . . , . . . , q2 n Ž x n Ž t . . . , ⭈⭈⭈
xXn
Ž t . s g n Ž t , qn1 Ž x 1 Ž t . . , . . . , qn n Ž x n Ž t . . . , 634
0022-247Xr99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
x n Ž t . s 0,
635
DISCONTINUOUS DIFFERENTIATION EQUATIONS
where g i , i s 1, . . . , n satisfy the Caratheodory condition and qi j , i, j s ´ 1, . . . , n are measurable and finite almost everywhere. Clearly, the above function f does not need to be continuous or monotone with respect to x in any reasonable sense.
2. AUXILIARY LEMMAS Throughout this paper the symbol will stand for the Lebesgue measure in ⺢. We will write w: w0, ⬁. ª w a, b x and g: w0, ⬁. = ⺢ n ª w a, b x if w Ž t . g w a, b x and g Ž t, y . g w a, b x for almost all t G 0 and almost all x g ⺢ n. LEMMA 1. Suppose that q: w0, M x ª ⺢ is measurable. If x: w0, T x ª w0, M x satisfies the condition 0 F s F t F T « x Ž t . y x Ž s . G aŽ t y s .
Ž 1.
with a fixed a ) 0, then the composition q( x is a measurable function. Proof. Let us fix arbitrary ␣ g ⺢. Since xy1 is a Lipschitzian function and q is measurable, the set
t g w 0, T x : q Ž x Ž t . . ) ␣ 4 s xy1 Ž qy1 Ž Ž ␣ , ⬁. . . is measurable as well. LEMMA 2. If F ; w0, M x is a measurable set and x: w0, T x ª w0, M x satisfies condition Ž1. then
Ž t g w 0, T x : x Ž t . f F 4 . F
1 a
Ž w 0, M x R F . .
Proof. Since xy1 satisfies the Lipschitz condition with constant have
Ž t g w 0, T x : x Ž t . f F 4 . s Ž xy1 Ž w 0, M x R F . . F
1 a
1 a
we
Ž w 0, M x R F . .
LEMMA 3. Suppose that q: w0, M x ª ⺢ is measurable and finite almost e¨ erywhere. If x k : w0, T x ª w0, M x, k g ⺞, is a sequence of functions satisfying Ž1. and con¨ erging uniformly to an x: w0, T x ª w0, M x, then the sequence q( x k , k g ⺞ con¨ erges to q( x in measure.
636
WITOLD RZYMOWSKI
Proof. Let us fix arbitrary , ⑀ ) 0. It follows from Luzin’s theorem that there exists a compact set F ; w0, M x such that a⑀
Ž w 0, M x R F . F
2
and q < F is continuous. Since x satisfies Ž1., by Lemma 2, we obtain
Ž t g w 0, T x : x Ž t . f F 4 . F
⑀ 2
and similarly, for every k g ⺞,
Ž t g w 0, T x : x k Ž t . f F 4 . F
⑀ 2
.
Let us take a ␦ ) 0 such that < q Ž . y q Ž .< F for all , g F with < y < F ␦ , and fix a k 0 g ⺞ such that < x k Ž t . y x Ž t .< F ␦ , for all t g w0, T x and all k G k 0 . For k G k 0 we now have < q Ž x k Ž t .. y q Ž x Ž t ..< F in the set Ek s t g w 0, T x : x Ž t . g F
xk g F 4 .
and
This completes the proof because of the inequality
Ž w 0, T x R Ek . F Ž t g w 0, T x : x Ž t . f F 4 . q Ž t g w 0, T x : x k Ž t . f F 4 . F ⑀ .
3. EXISTENCE THEOREMS For x s Ž x 1 , . . . , x n . g ⺢ n we define 5 x5 s
max < x i < .
is1, . . . , n
THEOREM 4. Suppose that for i s 1, . . . , n, g i : w0, ⬁. = ⺢ n ª w a i , bi x with 0 f w a i , bi x satisfies the Caratheodory condition, and, for i, j s 1, . . . , n, ´ qi j : ⌬ i j ª ⺢ are measurable and finite almost e¨ erywhere, where ⌬ij s
½
w 0, ⬁ . , Ž y⬁, 0 x ,
if a i ) 0, if a i - 0,
5
j s 1, . . . , n.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
637
Under the abo¨ e assumptions, the Cauchy problem Ž CP ., xX1 Ž t . s g 1 Ž t , g 11 Ž x 1 Ž t . . , . . . , q1 n Ž x n Ž t . . . ,
x 1 Ž t . s 0,
xX2
Ž t . s g 2 Ž t , q21 Ž x 1 Ž t . . , . . . , q2 n Ž x n Ž t . . . ,
x 2 Ž t . s 0,
⭈⭈⭈ xXn Ž t . s g n Ž t , qn1 Ž x 1 Ž t . . , . . . , qn n Ž x n Ž t . . . ,
x n Ž t . s 0,
has an absolutely continuous solution in the inter¨ al w0, ⬁.. Proof. Changing the coordinate system if necessary we may assume that 0 - a i F bi - ⬁,
i s 1, . . . , n.
Let us define as
min is1, . . . , n
ai ,
bs
max bi
is1, . . . , n
and fix arbitrary T ) 0. It follows from Luzin’s theorem that for every k g ⺞ there exists a compact set Fk ; w0, Tb x such that Žw0, Tb x R Fk . F 1k and qi j < F k is continuous, for i, j s 1, . . . , n. Involving Tietze’s extension theorem we may extend each of qi j < F k to a continuous qiŽ jk . : w0, Tb x ª ⺢. Let us consider, for each k g ⺞, the following Cauchy problem xXk Ž t . s g Ž t , Q Ž k . Ž x k Ž t . . . , almost everywhere in w 0, T x ,
Ž 2.
x k Ž t . s 0, where g Ž t , Q Ž k . Ž y . . s Ž g 1 Ž t , Q1Ž k . Ž y . . , . . . , g n Ž t , Q nŽ k . Ž y . . . and for i s 1, . . . , n, Q iŽ k . Ž y 1 , . . . , yn . s Ž qi1Ž k . Ž y 1 . , . . . , qiŽnk . Ž yn . . . Since each of g Ž⭈, Q Ž k . Ž⭈.., k g ⺞, satisfies the Caratheodory condition and ´ is bounded, the problem Ž2. has a solution x k . All functions x k , k g ⺞, are Lipschitzian with constant b, so that there exists a subsequence x k i , i g ⺞ convergent uniformly to an absolutely continuous x: w0, T x ª ⺢ n. In order to make the notation simpler we assume that the sequence x k , k g ⺞ is convergent to x. We claim that x is a solution of ŽCP.. It is enough to
638
WITOLD RZYMOWSKI
show that for every t g w0, T x lim
t
H g Ž s, Q
Žk.
kª⬁ 0
t
Ž x k Ž s . . . ds s H g Ž s, Q Ž x Ž s . . .
s,
0
where g Ž t , Q Ž y . . s Ž g 1 Ž t , Q1 Ž y . . , . . . , g n Ž t , Q n Ž y . . . and, for i s 1, . . . , n, Q i Ž y 1 , . . . , yn . s Ž qi1 Ž y 1 . , . . . , qi n Ž yn . . . By Lemma 1, functions Q i Ž x Ž⭈.., i s 1, . . . , n, are measurable, which implies that g Ž⭈, QŽ x Ž⭈... is measurable as well since g i , i s 1, . . . , n, satisfy Caratheodory’s condition. Since g is bounded, the integral ´ H0t g Ž s, QŽ x Ž s ... ds makes sense. For each t g w0, T x and each k g ⺞ we have t
H0 g Ž s, Q F
Žk.
t
Ž x k Ž s . . . ds y H g Ž s, Q Ž x Ž s . . .
ds
0
t H0 g Ž s, Q
q
t
H0
Žk.
Ž x k Ž s . . . y g Ž s, Q Ž x k Ž s . . .
ds
g Ž s, Q Ž x k Ž s . . . y g Ž s, Q Ž x Ž s . . . ds.
Applying Lemma 3 and taking into account the fact that g is bounded, we obtain lim
t
H
kª⬁ 0
g Ž s, Q Ž x k Ž s . . . y g Ž s, Q Ž x Ž s . . . ds s 0.
We are thus going to estimate the first integral. Let us define, for i s 1, . . . , n and k g ⺞, ⌰ Ž0k . s t g w 0, T x : x i Ž t . g Fk 4 , ⌰iŽ k . s t g w 0, T x : x k , i Ž t . g Fk 4 , where Ž x k, 1Ž t ., . . . , x k, nŽ t .. s x k Ž t .. By Lemma 2, we have
Ž w 0, T x R ⌰iŽ k . . F
1 ak
,
i s 0, 1, . . . , n.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
639
Since g Ž s, QŽ x k Ž s ... s g Ž s, Q Ž k . Ž x k Ž s ..., for s g ⌰ 0Ž k . l ⌰1Ž k . l ⭈⭈⭈ l ⌰nŽ k ., we have t H0 g Ž s, Q
Žk.
Ž x k Ž s . . . y g Ž s, Q Ž x k Ž s . . .
ds
n
F
Ý Hw
is0
F
0, t xR⌰ iŽ k .
Ž n q 1. b
g Ž s, Q Ž k . Ž x k Ž s . . . y g Ž s, Q Ž x k Ž s . . . ds
ª 0.
ak
kª⬁
We have just proved that the problem ŽCP. has a solution in w0, T x. Repeating the above procedure one can extend the solution into intervals w T, 2T x, w2T, 3T x, . . . and so on. Thus, the problem has a solution over the whole interval w0, ⬁.. COROLLARY 5. Let for i s 1, . . . , n, h i : ⺢ nq 1 ª ⺢ be continuous and let wi : w0, ⬁. ª ⺢ and qi j : ⌬ i ª ⺢, j s 1, . . . , n be measurable, where ⌬ i s Žy⬁, 0. or ⌬ i s Ž0, ⬁.. If, for e¨ ery i s 1, . . . , n, almost all t G 0 and almost all x g ⌬ i , wi Ž t . F M,
qi j Ž x . F M,
j s 1, . . . , n
and h i Ž wi Ž t . , qi1 Ž x 1 . , . . . , qi n Ž x n . . g ⌬ i , then the Cauchy problem xX1 Ž t . s h1 Ž w 1 Ž t . , q11 Ž x 1 Ž t . . , . . . , q1 n Ž x n Ž t . . . ,
x 1 Ž 0 . s 0,
xX2
x 2 Ž 0 . s 0,
Ž t . s h 2 Ž w 2 Ž t . , q21 Ž x 1 Ž t . . , . . . , q2 n Ž x n Ž t . . . , ⭈⭈⭈
xXn Ž t . s h n Ž wn Ž t . , qn1 Ž x 1 Ž t . . , . . . , qn n Ž x n Ž t . . . ,
Ž 3.
x n Ž 0. s 0
has an absolutely continuous solution in the inter¨ al w0, ⬁.. Proof. Let us define
!
n q# 1 times
"
K s w yM, M x = ⭈⭈⭈ = w yM, M x
and, for every i s 1, . . . , n, w a i , bi x s h i Ž K .. Let hUi : ⺢ nq1 ª w a i , bi x be a continuous extension of h i < K , i s 1, . . . , n. It is easy to see that g s
640
WITOLD RZYMOWSKI
Ž g 1 , . . . , g n . and Q s Ž Q1 , . . . , Q n ., where g i Ž t , x . s hUi Ž wi Ž t . , Q i Ž x . .
and
Q i Ž x . s Ž qi1 Ž x 1 . , . . . , qi n Ž x n . . , i s 1, . . . , n
satisfy the assumptions of Theorem 4. It is also easy to see that each solution of the Cauchy problem x⬘ Ž t . s g Ž t , Q Ž x Ž t . . . , almost everywhere in w 0, ⬁ . , x Ž 0. s 0 solves Ž3. as well. COROLLARY 6.
Suppose that for i s 1, . . . , n,
wi : w 0, ⬁ . ª w a i0 , bi0 x
and qi j : ⺢ ª a i j , bi j ,
j s 1, . . . , n,
are measurable and, for e¨ ery i s 1, . . . , n, n
n
Ý ai j ) 0
or
js0
Ý bi j - 0. js0
Under the abo¨ e assumptions, the Cauchy problem xX1 Ž t . s q11 Ž x 1 Ž t . . q ⭈⭈⭈ qq1 n Ž x n Ž t . . q w 1 Ž t . ,
x 1 Ž 0 . s 0,
xX2
x 2 Ž 0 . s 0,
Ž t . s q21 Ž x 1 Ž t . . q ⭈⭈⭈ qq2 n Ž x n Ž t . . q w 2 Ž t . , ⭈⭈⭈
xXn
Ž t . s qn1 Ž x 1 Ž t . . q ⭈⭈⭈ qqn n Ž x n Ž t . . q wn Ž t . ,
x n Ž 0. s 0
has an absolutely continuous solution in the inter¨ al w0, ⬁.. Proof. Let us define, for i s 1, . . . , n and s, y 1 , . . . , yn g ⺢, h i Ž s, y 1 , . . . , yn . s s q y 1 q ⭈⭈⭈ qyn and M s max
½
n
max is1, . . . , n
Ý ai j js0
n
,
max is1, . . . , n
Ý bi j js0
5
.
It is enough to check that functions h i and qi j satisfy the assumptions of Corollary 5.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
641
As an intermediate consequence we obtain COROLLARY 7. If w: w0, ⬁. ª w0, b 0 x and q: w0, ⬁. ª w a, b x, with a ) 0, are measurable then the Cauchy problem x⬘ Ž t . s q Ž x Ž t . . q w Ž t . ,
almost e¨ erywhere in w 0, ⬁ . ,
x Ž 0. s 0 has an absolutely continuous solution in the inter¨ al w0, ⬁.. It is noteworthy that the above result does not follow from w2, 5x.
4. FINAL REMARKS Remark 1. We have dealt with differential equations of right-hand sides bounded from the origin. This is rather an essential assumption. For x⬘ Ž t . s q Ž x Ž t . . q w Ž t . , a.e. in w 0, ⬁ . , x Ž 0. s 0 with qŽ x. s
½
1, y1,
x-0 xG0
w Ž t . s 0, t G 0,
5
we arrive at the standard example of the Cauchy problem without solutions. Remark 2. The assumptions of the main theorem and corollaries are not invariant for changes of coordinates. This allows us to formulate a more general result. Let f : w0, ⬁. = ⺢ n ª ⺢ n be as in Theorem 4 and ⌽: ⺢ n ª ⺢ n be a C 1-diffeomorphism such that ⌽ Ž⺢ n . s ⺢ n and ⌽ Ž0. s 0. Let us define g Ž t , y . s ⌽⬘ Ž ⌽y1 Ž y . . f Ž t , ⌽y1 Ž y . . . It is easy to see that the Cauchy problem y⬘ s g Ž t , y . ,
y Ž 0. s 0
has a solution y s ⌽ Ž x ., where x is a solution of the problem x⬘ s f Ž t , x . ,
x Ž 0 . s 0.
Clearly, g does not need to satisfy the assumptions of Theorem 4.
642
WITOLD RZYMOWSKI
EXAMPLE 8. For any measurable q1 , q2 : ⺢ ª w a, b x, with a ) 0, the Cauchy problem yX1 s q1 Ž y 1 y y 2 . q q2 Ž y 1 q y 2 . ,
y 1 Ž 0 . s 0,
yX2
y 2 Ž 0. s 0
s q1 Ž y 1 y y 2 . y q 2 Ž y 1 q y 2 . ,
has a solution
Ž y1 , y 2 . s
ž
x1 q x 2 2
,
x1 y x 2 2
/
,
where Ž x 1 , x 2 . solves the problem xX1 s 2 q1 Ž x 2 . ,
x 1 Ž 0 . s 0,
xX2
x 2 Ž 0. s 0
s 2 q2 Ž x 1 . ,
Žsee Corollary 6.. Remark 3. The assumptions of Theorem 4 do not guarantee uniqueness even with smooth g i ’s. Let us look at the following example suggested by the referee Žsee also Example 2 of w4x.. Define
qŽ x. s
¡
1,
~
1,
if x F 0, 1 1 if 2 n - x F 2 ny1 , n g ⺞, 2 2 1 1 if 2 nq1 - x F 2 n , n g ⺞. 2 2
¢y1,
Let g: ⺢ 2 ª w1, 2x be a smooth function such that g Ž u, ¨ . s
½
2, 1,
if u¨ s y1, if u¨ s 1.
The Cauchy problem xX1 s 1,
x 1 Ž 0 . s 0,
xX2 s g Ž q Ž x 1 . , q Ž x 2 . . ,
x 2 Ž 0. s 0
has at least two different solutions, Ž t, t . and Ž t, 2 t ..
ACKNOWLEDGMENT The author thanks the referee for helpful suggestions.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
643
REFERENCES 1. D. C. Biles, Existence of solutions for discontinuous differential equations, Differential Integral Equations 8 Ž1995., 1525᎐1532. 2. P. Binding, The differential equation ˙ x s f ( x, J. Differential Equations 31 Ž1979., 183᎐199. 3. A. Bressan, Unique solution for a class of discontinuous differential equations, Proc. Amer. Math. Soc. 104 Ž1988., 772᎐778. 4. A. Bressan and Wen Shen, Uniqueness for discontinuous ODE and conservation laws, Nonlinear Anal. 34 Ž1998., 637᎐652. 5. E. R. Hassan and W. Rzymowski, Extremal solutions of a discontinuous scalar differential equation, Nonlinear Anal., to appear.
Existence of Solutions for a Class of Discontinuous Differential Equations in ⺢ n Witold Rzymowski Instytut Matematyki UMCS, 20-031 Lublin, Poland Submitted by Zhi¨ ko S. Athanasso¨ Received July 10, 1997
In this paper we extend the Caratheodory existence theorem for a class of ´ discontinuous differential equations with right-hand sides in a special form but with no continuity or monotonicity assumptions. A new existence result for a scalar 䊚 1999 Academic Press discontinuous differential equation is also established.
1. INTRODUCTION It is well known that if f : w0, ⬁. = ⺢ n ª ⺢ n is bounded and satisfies the Caratheodory condition, i.e., f Ž⭈, x . is measurable for all x g ⺢ n and ´ f Ž t, ⭈ . is continuous for almost all t G 0, then the Cauchy problem x⬘ Ž t . s f Ž t , x Ž t . . , a.e. in w 0, ⬁ . , x Ž 0. s 0 has an absolutely continuous solution defined in w0, ⬁.. Existence results with f discontinuous in x involve usually semicontinuity or monotonicity assumptions Žsee, e.g., w3, 4, 1x.. In this paper we deal with a discontinuous differential equation of a special form. Namely, we shall consider the system of ordinary differential equations xX1 Ž t . s g 1 Ž t , q11 Ž x 1 Ž t . . , . . . , q1 n Ž x n Ž t . . . ,
x 1 Ž t . s 0,
xX2
x 2 Ž t . s 0,
Ž t . s g 2 Ž t , q21 Ž x 1 Ž t . . , . . . , q2 n Ž x n Ž t . . . , ⭈⭈⭈
xXn
Ž t . s g n Ž t , qn1 Ž x 1 Ž t . . , . . . , qn n Ž x n Ž t . . . , 634
0022-247Xr99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
x n Ž t . s 0,
635
DISCONTINUOUS DIFFERENTIATION EQUATIONS
where g i , i s 1, . . . , n satisfy the Caratheodory condition and qi j , i, j s ´ 1, . . . , n are measurable and finite almost everywhere. Clearly, the above function f does not need to be continuous or monotone with respect to x in any reasonable sense.
2. AUXILIARY LEMMAS Throughout this paper the symbol will stand for the Lebesgue measure in ⺢. We will write w: w0, ⬁. ª w a, b x and g: w0, ⬁. = ⺢ n ª w a, b x if w Ž t . g w a, b x and g Ž t, y . g w a, b x for almost all t G 0 and almost all x g ⺢ n. LEMMA 1. Suppose that q: w0, M x ª ⺢ is measurable. If x: w0, T x ª w0, M x satisfies the condition 0 F s F t F T « x Ž t . y x Ž s . G aŽ t y s .
Ž 1.
with a fixed a ) 0, then the composition q( x is a measurable function. Proof. Let us fix arbitrary ␣ g ⺢. Since xy1 is a Lipschitzian function and q is measurable, the set
t g w 0, T x : q Ž x Ž t . . ) ␣ 4 s xy1 Ž qy1 Ž Ž ␣ , ⬁. . . is measurable as well. LEMMA 2. If F ; w0, M x is a measurable set and x: w0, T x ª w0, M x satisfies condition Ž1. then
Ž t g w 0, T x : x Ž t . f F 4 . F
1 a
Ž w 0, M x R F . .
Proof. Since xy1 satisfies the Lipschitz condition with constant have
Ž t g w 0, T x : x Ž t . f F 4 . s Ž xy1 Ž w 0, M x R F . . F
1 a
1 a
we
Ž w 0, M x R F . .
LEMMA 3. Suppose that q: w0, M x ª ⺢ is measurable and finite almost e¨ erywhere. If x k : w0, T x ª w0, M x, k g ⺞, is a sequence of functions satisfying Ž1. and con¨ erging uniformly to an x: w0, T x ª w0, M x, then the sequence q( x k , k g ⺞ con¨ erges to q( x in measure.
636
WITOLD RZYMOWSKI
Proof. Let us fix arbitrary , ⑀ ) 0. It follows from Luzin’s theorem that there exists a compact set F ; w0, M x such that a⑀
Ž w 0, M x R F . F
2
and q < F is continuous. Since x satisfies Ž1., by Lemma 2, we obtain
Ž t g w 0, T x : x Ž t . f F 4 . F
⑀ 2
and similarly, for every k g ⺞,
Ž t g w 0, T x : x k Ž t . f F 4 . F
⑀ 2
.
Let us take a ␦ ) 0 such that < q Ž . y q Ž .< F for all , g F with < y < F ␦ , and fix a k 0 g ⺞ such that < x k Ž t . y x Ž t .< F ␦ , for all t g w0, T x and all k G k 0 . For k G k 0 we now have < q Ž x k Ž t .. y q Ž x Ž t ..< F in the set Ek s t g w 0, T x : x Ž t . g F
xk g F 4 .
and
This completes the proof because of the inequality
Ž w 0, T x R Ek . F Ž t g w 0, T x : x Ž t . f F 4 . q Ž t g w 0, T x : x k Ž t . f F 4 . F ⑀ .
3. EXISTENCE THEOREMS For x s Ž x 1 , . . . , x n . g ⺢ n we define 5 x5 s
max < x i < .
is1, . . . , n
THEOREM 4. Suppose that for i s 1, . . . , n, g i : w0, ⬁. = ⺢ n ª w a i , bi x with 0 f w a i , bi x satisfies the Caratheodory condition, and, for i, j s 1, . . . , n, ´ qi j : ⌬ i j ª ⺢ are measurable and finite almost e¨ erywhere, where ⌬ij s
½
w 0, ⬁ . , Ž y⬁, 0 x ,
if a i ) 0, if a i - 0,
5
j s 1, . . . , n.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
637
Under the abo¨ e assumptions, the Cauchy problem Ž CP ., xX1 Ž t . s g 1 Ž t , g 11 Ž x 1 Ž t . . , . . . , q1 n Ž x n Ž t . . . ,
x 1 Ž t . s 0,
xX2
Ž t . s g 2 Ž t , q21 Ž x 1 Ž t . . , . . . , q2 n Ž x n Ž t . . . ,
x 2 Ž t . s 0,
⭈⭈⭈ xXn Ž t . s g n Ž t , qn1 Ž x 1 Ž t . . , . . . , qn n Ž x n Ž t . . . ,
x n Ž t . s 0,
has an absolutely continuous solution in the inter¨ al w0, ⬁.. Proof. Changing the coordinate system if necessary we may assume that 0 - a i F bi - ⬁,
i s 1, . . . , n.
Let us define as
min is1, . . . , n
ai ,
bs
max bi
is1, . . . , n
and fix arbitrary T ) 0. It follows from Luzin’s theorem that for every k g ⺞ there exists a compact set Fk ; w0, Tb x such that Žw0, Tb x R Fk . F 1k and qi j < F k is continuous, for i, j s 1, . . . , n. Involving Tietze’s extension theorem we may extend each of qi j < F k to a continuous qiŽ jk . : w0, Tb x ª ⺢. Let us consider, for each k g ⺞, the following Cauchy problem xXk Ž t . s g Ž t , Q Ž k . Ž x k Ž t . . . , almost everywhere in w 0, T x ,
Ž 2.
x k Ž t . s 0, where g Ž t , Q Ž k . Ž y . . s Ž g 1 Ž t , Q1Ž k . Ž y . . , . . . , g n Ž t , Q nŽ k . Ž y . . . and for i s 1, . . . , n, Q iŽ k . Ž y 1 , . . . , yn . s Ž qi1Ž k . Ž y 1 . , . . . , qiŽnk . Ž yn . . . Since each of g Ž⭈, Q Ž k . Ž⭈.., k g ⺞, satisfies the Caratheodory condition and ´ is bounded, the problem Ž2. has a solution x k . All functions x k , k g ⺞, are Lipschitzian with constant b, so that there exists a subsequence x k i , i g ⺞ convergent uniformly to an absolutely continuous x: w0, T x ª ⺢ n. In order to make the notation simpler we assume that the sequence x k , k g ⺞ is convergent to x. We claim that x is a solution of ŽCP.. It is enough to
638
WITOLD RZYMOWSKI
show that for every t g w0, T x lim
t
H g Ž s, Q
Žk.
kª⬁ 0
t
Ž x k Ž s . . . ds s H g Ž s, Q Ž x Ž s . . .
s,
0
where g Ž t , Q Ž y . . s Ž g 1 Ž t , Q1 Ž y . . , . . . , g n Ž t , Q n Ž y . . . and, for i s 1, . . . , n, Q i Ž y 1 , . . . , yn . s Ž qi1 Ž y 1 . , . . . , qi n Ž yn . . . By Lemma 1, functions Q i Ž x Ž⭈.., i s 1, . . . , n, are measurable, which implies that g Ž⭈, QŽ x Ž⭈... is measurable as well since g i , i s 1, . . . , n, satisfy Caratheodory’s condition. Since g is bounded, the integral ´ H0t g Ž s, QŽ x Ž s ... ds makes sense. For each t g w0, T x and each k g ⺞ we have t
H0 g Ž s, Q F
Žk.
t
Ž x k Ž s . . . ds y H g Ž s, Q Ž x Ž s . . .
ds
0
t H0 g Ž s, Q
q
t
H0
Žk.
Ž x k Ž s . . . y g Ž s, Q Ž x k Ž s . . .
ds
g Ž s, Q Ž x k Ž s . . . y g Ž s, Q Ž x Ž s . . . ds.
Applying Lemma 3 and taking into account the fact that g is bounded, we obtain lim
t
H
kª⬁ 0
g Ž s, Q Ž x k Ž s . . . y g Ž s, Q Ž x Ž s . . . ds s 0.
We are thus going to estimate the first integral. Let us define, for i s 1, . . . , n and k g ⺞, ⌰ Ž0k . s t g w 0, T x : x i Ž t . g Fk 4 , ⌰iŽ k . s t g w 0, T x : x k , i Ž t . g Fk 4 , where Ž x k, 1Ž t ., . . . , x k, nŽ t .. s x k Ž t .. By Lemma 2, we have
Ž w 0, T x R ⌰iŽ k . . F
1 ak
,
i s 0, 1, . . . , n.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
639
Since g Ž s, QŽ x k Ž s ... s g Ž s, Q Ž k . Ž x k Ž s ..., for s g ⌰ 0Ž k . l ⌰1Ž k . l ⭈⭈⭈ l ⌰nŽ k ., we have t H0 g Ž s, Q
Žk.
Ž x k Ž s . . . y g Ž s, Q Ž x k Ž s . . .
ds
n
F
Ý Hw
is0
F
0, t xR⌰ iŽ k .
Ž n q 1. b
g Ž s, Q Ž k . Ž x k Ž s . . . y g Ž s, Q Ž x k Ž s . . . ds
ª 0.
ak
kª⬁
We have just proved that the problem ŽCP. has a solution in w0, T x. Repeating the above procedure one can extend the solution into intervals w T, 2T x, w2T, 3T x, . . . and so on. Thus, the problem has a solution over the whole interval w0, ⬁.. COROLLARY 5. Let for i s 1, . . . , n, h i : ⺢ nq 1 ª ⺢ be continuous and let wi : w0, ⬁. ª ⺢ and qi j : ⌬ i ª ⺢, j s 1, . . . , n be measurable, where ⌬ i s Žy⬁, 0. or ⌬ i s Ž0, ⬁.. If, for e¨ ery i s 1, . . . , n, almost all t G 0 and almost all x g ⌬ i , wi Ž t . F M,
qi j Ž x . F M,
j s 1, . . . , n
and h i Ž wi Ž t . , qi1 Ž x 1 . , . . . , qi n Ž x n . . g ⌬ i , then the Cauchy problem xX1 Ž t . s h1 Ž w 1 Ž t . , q11 Ž x 1 Ž t . . , . . . , q1 n Ž x n Ž t . . . ,
x 1 Ž 0 . s 0,
xX2
x 2 Ž 0 . s 0,
Ž t . s h 2 Ž w 2 Ž t . , q21 Ž x 1 Ž t . . , . . . , q2 n Ž x n Ž t . . . , ⭈⭈⭈
xXn Ž t . s h n Ž wn Ž t . , qn1 Ž x 1 Ž t . . , . . . , qn n Ž x n Ž t . . . ,
Ž 3.
x n Ž 0. s 0
has an absolutely continuous solution in the inter¨ al w0, ⬁.. Proof. Let us define
!
n q# 1 times
"
K s w yM, M x = ⭈⭈⭈ = w yM, M x
and, for every i s 1, . . . , n, w a i , bi x s h i Ž K .. Let hUi : ⺢ nq1 ª w a i , bi x be a continuous extension of h i < K , i s 1, . . . , n. It is easy to see that g s
640
WITOLD RZYMOWSKI
Ž g 1 , . . . , g n . and Q s Ž Q1 , . . . , Q n ., where g i Ž t , x . s hUi Ž wi Ž t . , Q i Ž x . .
and
Q i Ž x . s Ž qi1 Ž x 1 . , . . . , qi n Ž x n . . , i s 1, . . . , n
satisfy the assumptions of Theorem 4. It is also easy to see that each solution of the Cauchy problem x⬘ Ž t . s g Ž t , Q Ž x Ž t . . . , almost everywhere in w 0, ⬁ . , x Ž 0. s 0 solves Ž3. as well. COROLLARY 6.
Suppose that for i s 1, . . . , n,
wi : w 0, ⬁ . ª w a i0 , bi0 x
and qi j : ⺢ ª a i j , bi j ,
j s 1, . . . , n,
are measurable and, for e¨ ery i s 1, . . . , n, n
n
Ý ai j ) 0
or
js0
Ý bi j - 0. js0
Under the abo¨ e assumptions, the Cauchy problem xX1 Ž t . s q11 Ž x 1 Ž t . . q ⭈⭈⭈ qq1 n Ž x n Ž t . . q w 1 Ž t . ,
x 1 Ž 0 . s 0,
xX2
x 2 Ž 0 . s 0,
Ž t . s q21 Ž x 1 Ž t . . q ⭈⭈⭈ qq2 n Ž x n Ž t . . q w 2 Ž t . , ⭈⭈⭈
xXn
Ž t . s qn1 Ž x 1 Ž t . . q ⭈⭈⭈ qqn n Ž x n Ž t . . q wn Ž t . ,
x n Ž 0. s 0
has an absolutely continuous solution in the inter¨ al w0, ⬁.. Proof. Let us define, for i s 1, . . . , n and s, y 1 , . . . , yn g ⺢, h i Ž s, y 1 , . . . , yn . s s q y 1 q ⭈⭈⭈ qyn and M s max
½
n
max is1, . . . , n
Ý ai j js0
n
,
max is1, . . . , n
Ý bi j js0
5
.
It is enough to check that functions h i and qi j satisfy the assumptions of Corollary 5.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
641
As an intermediate consequence we obtain COROLLARY 7. If w: w0, ⬁. ª w0, b 0 x and q: w0, ⬁. ª w a, b x, with a ) 0, are measurable then the Cauchy problem x⬘ Ž t . s q Ž x Ž t . . q w Ž t . ,
almost e¨ erywhere in w 0, ⬁ . ,
x Ž 0. s 0 has an absolutely continuous solution in the inter¨ al w0, ⬁.. It is noteworthy that the above result does not follow from w2, 5x.
4. FINAL REMARKS Remark 1. We have dealt with differential equations of right-hand sides bounded from the origin. This is rather an essential assumption. For x⬘ Ž t . s q Ž x Ž t . . q w Ž t . , a.e. in w 0, ⬁ . , x Ž 0. s 0 with qŽ x. s
½
1, y1,
x-0 xG0
w Ž t . s 0, t G 0,
5
we arrive at the standard example of the Cauchy problem without solutions. Remark 2. The assumptions of the main theorem and corollaries are not invariant for changes of coordinates. This allows us to formulate a more general result. Let f : w0, ⬁. = ⺢ n ª ⺢ n be as in Theorem 4 and ⌽: ⺢ n ª ⺢ n be a C 1-diffeomorphism such that ⌽ Ž⺢ n . s ⺢ n and ⌽ Ž0. s 0. Let us define g Ž t , y . s ⌽⬘ Ž ⌽y1 Ž y . . f Ž t , ⌽y1 Ž y . . . It is easy to see that the Cauchy problem y⬘ s g Ž t , y . ,
y Ž 0. s 0
has a solution y s ⌽ Ž x ., where x is a solution of the problem x⬘ s f Ž t , x . ,
x Ž 0 . s 0.
Clearly, g does not need to satisfy the assumptions of Theorem 4.
642
WITOLD RZYMOWSKI
EXAMPLE 8. For any measurable q1 , q2 : ⺢ ª w a, b x, with a ) 0, the Cauchy problem yX1 s q1 Ž y 1 y y 2 . q q2 Ž y 1 q y 2 . ,
y 1 Ž 0 . s 0,
yX2
y 2 Ž 0. s 0
s q1 Ž y 1 y y 2 . y q 2 Ž y 1 q y 2 . ,
has a solution
Ž y1 , y 2 . s
ž
x1 q x 2 2
,
x1 y x 2 2
/
,
where Ž x 1 , x 2 . solves the problem xX1 s 2 q1 Ž x 2 . ,
x 1 Ž 0 . s 0,
xX2
x 2 Ž 0. s 0
s 2 q2 Ž x 1 . ,
Žsee Corollary 6.. Remark 3. The assumptions of Theorem 4 do not guarantee uniqueness even with smooth g i ’s. Let us look at the following example suggested by the referee Žsee also Example 2 of w4x.. Define
qŽ x. s
¡
1,
~
1,
if x F 0, 1 1 if 2 n - x F 2 ny1 , n g ⺞, 2 2 1 1 if 2 nq1 - x F 2 n , n g ⺞. 2 2
¢y1,
Let g: ⺢ 2 ª w1, 2x be a smooth function such that g Ž u, ¨ . s
½
2, 1,
if u¨ s y1, if u¨ s 1.
The Cauchy problem xX1 s 1,
x 1 Ž 0 . s 0,
xX2 s g Ž q Ž x 1 . , q Ž x 2 . . ,
x 2 Ž 0. s 0
has at least two different solutions, Ž t, t . and Ž t, 2 t ..
ACKNOWLEDGMENT The author thanks the referee for helpful suggestions.
DISCONTINUOUS DIFFERENTIATION EQUATIONS
643
REFERENCES 1. D. C. Biles, Existence of solutions for discontinuous differential equations, Differential Integral Equations 8 Ž1995., 1525᎐1532. 2. P. Binding, The differential equation ˙ x s f ( x, J. Differential Equations 31 Ž1979., 183᎐199. 3. A. Bressan, Unique solution for a class of discontinuous differential equations, Proc. Amer. Math. Soc. 104 Ž1988., 772᎐778. 4. A. Bressan and Wen Shen, Uniqueness for discontinuous ODE and conservation laws, Nonlinear Anal. 34 Ž1998., 637᎐652. 5. E. R. Hassan and W. Rzymowski, Extremal solutions of a discontinuous scalar differential equation, Nonlinear Anal., to appear.