Gross, O.A

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The boundary value problem on an infinite interval: Existence, uniqueness, and asymptotic behavior of bounded solutions to a class of nonlinear second order differential equations 

JOURNAL OF MATHEMATICAL ANALYSIS .ANU APPLICATIONS 7, loo-109 (1963) 
The Boundary Value Problem on an Infinite Interval: 
Existence, Uniqueness, and Asymptotic Behavior of Bounded Solutions to 
a Class of Nonlinear Second Order Differential Equations* 
0. A. GROSS 
The RAND Corporation, Santa Monica, California 
Submitted by Richard Bellman 
I. INTRODUCTION AND HYPOTHESIS 
This paper concerns itself with a study of the real continuous bounded 
solutions u, if any, of the system 
d2u - = w 4, dr2 r > 0, 
u(0) = - a, U-1) 
where cy is a given positive number. Specifically, we are interested in negative 
solutions, and we impose the following conditions (1.2), (1.3), and (1.4) on 
(b to define the hypothesis H of the theorem in the subsequent section: 
(1.2) 4 is continuous in the range u 5 0, r > 0, 
(1.3) c$(u, r) s 0 and is nondecreasing as a function of u in the range 
u 5 0, Y  > 0, where, moreover, &(u, r) exists and is continuous. 
(1.4) $(u, r) - u 2 0 and is nonincreasing as a function of u in the 
range u 5 0, Y > 0. 
One verifies that the hypothesis H implies 
s 
m  
e+ (- OL, s) ds exists and is finite. 
0 
* This research is sponsored by the United States Air Force under Project RAND- 
Contract No. AF 49(638)-7cmonitored by the Directorate of Development 
Planning, Deputy Chief of Staff, Research and Technology, Hq USAF. Views or 
conclusions contained in this Memorandum should not be interpreted as representing 
the official opinion or policy of the United States Air Force. Permission to quote from 
or reproduce portions of this Memorandum must be obtained from The RAND 
Corporation. 
100 
BOUNDARY VALUE PROBLEM ON INFINITE INTERVAL 101 
Of course, the theorem we wish to prove would be otiose, to say the least, 
if the hypothesis H were only trivially satisfied, e.g., by C$ = 0 or $(u, r) = u 
(each of which clearly satisfies N). However, a nonlinear example + which 
satisfies H is given by 
$(u, Y) = r(P - 1). 
This is precisely the example treated in the final section. We leave it as an 
exercise. The reader can easily verify that conditions (1.2), (1.3), and (1.4) 
are satisfied. 
Before proceeding to the statement of the theorem and its proof, we 
remark that some such conditions as the monotonicity requirement of (1.3) 
together with the remaining (1.4) seem to be essential for uniqueness, etc., 
granted that a solution exists. Thus, for example, the system 
d2u 
- z U(1 + U)2, 
dr2 - r2 
r > 0, 
u(0) = - 1, 
has a one-parameter family of continuous, uniformly bounded negative solu- 
tions on [0, a], namely, 
w = - & ) c 2 0. 
II. THE MAIN THEOREM 
The result we wish to establish can be phrased as follows: 
THEOREM 1. I f  q5 satisjes H, then there exists a unique, negative, bounded 
continuous solution u to (1.I) Th is solution is nondecreasing and concave and 
therefore tends to some nonpositive constant as r -+ 03. 
PROOF. We shall establish existence first. To this end we consider the 
nonempty set L of functions w satisfying 
- a 2 w(r) 5 - aecT, all y > 0, 
and such that erw(r) is monotone nonincreasing in Y. We assert that L forms a 
complete lattice under the partial ordering relation (>) given by 
102 GROSS 
To see that L is a lattice, we note first that given any two elements wi, wg EL, 
we have 
1.u.b. (wi, wa) (7) = max (wl(r), z+(r)), 
g.1.b. (wi, 20~) (Y) = min (wi(r), wz(r)). 
One can readily show that 1.u.b. (wi, ws) and g.1.b. (wr, wa) are both in L. 
Finally, the universal upper bound (I) ’ f 1s urnished by the upper bounding 
function (- ale-r), and the universal lower bound (0), by the function which 
is identically equal to - CY. 
That the lattice is complete, i.e., to say that any nonempty subset of elements 
of L has a g.1.b. and an 1.u.b. in L (in the obvious extension of these two 
functions), is easily established. 
Next, consider the nonlinear operator Tl given as follows: 
T,w(r) = - ae+ - 6 y e-W(w(4 4 - w(s)) ds, Y  r_ 0. 
We establish first that T, is everywhere defined on L. Now since w E L Z- w 
is bounded and integrable, it is easy to see that 
s 
m 
e+ w(s) ds 
t 
exists and is continuous. Moreover, to see that s,” e-s+(w(s), s) ds exists and 
is integrable, we note that condition (1.3) shows that the integral is majorized 
by sp e-sC (- 01, s) ds, which in turn is dominated by Jr e-“+ (- OL, s) ds, 
which exists and is finite by (1.5). But since 4 (= 0, jf e~$(w(s), s) ds is 
monotone in N and hence tends to a finite limit as N -03. Thus Tl is 
defined on L. 
We shall show next that Tl is isotone. By the monoticity condition (1.4), 
we see that if wl(s) 2 wa(s) for all s 2 0, then 
Yw4,s) - WI(S) 5 e&), 4 - %(4 
The integral of T, is preceded by a minus sign. Hence, if wr(r) 2 We 
for all Y  2 0, then T,w,(r) 2 T,w,(r) for all Y  2 0. In other words, T, is 
isotone. 
We show next that Tl maps L into itself. First, the condition 
(p(u, r) - u 2 0 of (1.4) guarantees that the integral contribution of the map- 
ping is nonpositive, and hence that 
T,w(Y) 5 - aemT. 
BOUNDARY VALUE PROBLEM ON INFINITE INTERVAL 103 
To see that the lower bound condition is satisfied by Tlw, we consider first 
the image of 0 under T,: 
T,O(r) = - Lue-T - e-7 
s: s 
e2tdt r eP(d(- (II, s) + CX) ds 
(by virtue of the inequality of (1.3) and the negative sign preceding the inte- 
gral). The right member is readily evaluated to obtain 
T,O(r) 1 - 01, i.e. T,O > 0. = 
But w EL =+- w >= 0; hence, applying the isotone property already established 
yields 
i.e. 
Tp > T,O > 0, = zcz 
T,w(r) 2 - 01. 
It remains to show that if w EL then erTlw(r) is monotone nonincreasing 
in r. But we have 
erTlw(r) = - (Y - 
s’ s 
e2tdt 
0 
y e+(+(w(s), s) - w(s)) ds, 
whence 
(erTlw) (r) = - e2+’ lrn eP(#(w(s), s) - w(s)) ds 5 0, 
Ip 
by virtue of the inequality (1.4). Hence Tl maps L into itself. In short, we 
have established that Tl is an isotone function which maps the complete 
lattice L into itself. 
We are now in a position to apply a neat theorem due to G. Birkhoff [I], 
namely: 
THEOREM. An isotone function from a complete lattice L into itserf has a 
jixpoint in L. 
It follows that there exists a bounded function u satisfying 
u(0) = - lx 
and 
U(Y) = - me-* - e-7 
s: I 
e2 tdt O” e-8(9(U(s), 4 - 4s)) 4 r 2 0. 
t 
104 GROSS 
The usual arguments on the integral operator show that u E C, (for Y > 0). 
Multiplying this last equation through by e r, differentiating, etc. we finally 
obtain 
u”(r) = #4(r), Y) Y > 0. 
Now, since u EL, we have u(r) < 0 for all r; moreover by (1.3), u is concave. 
Finally, since u is concave and bounded, it is necessarily nondecreasing and 
therefore tends to some nonpositive constant as Y tends to infinity. Thus, 
the existence of a solution to (1.1) with the required and desired properties is 
established. 
In order to establish uniqueness, consider any bounded, negative, continu- 
ous solution u to (1 .l): 
u”(r) = q@(r), 4 Y < 0, 
u(0) = - a. 
Since u is necessarily continuously differentiable, bounded, and concave, 
u(r) --+ 0 as r -+ a. Hence we can integrate the above differential equation 
from t finite to infinity to obtain 
u’(t) = - j; #i(s), s) ds. 
Finally, integrating this last from 0 to r and making use of the initial value, 
we obtain 
u(r) = - a - 
s s 
+- dt 
0 
; c+(s), s) ds. 
Thus, any such function u is a fixpoint of the mapping T2 given by 
T2w(r) = - a - 
s s 
’ dt 
0 
; vWS), 4 ds. 
Now, by the monotonicity property of (l-3), we see that if wl(s) 1 wa(s) 
for all s 1 0, then +(wJs), s) 2 $(ws(s), s). The integral of T, is preceded by 
a minus sign. Hence if wl(r) 1 W&Y) for all r 2 0, then T,wl(r) _I T,w,(r). 
In other words, T2 is antitone. 
We are now in a position to establish uniqueness. Therefore let ‘or and ws 
be a pair of solutions with the required properties. 
Case 1. One of the pair (zlr, oa) strictly dominates the other on the range 
r > 0. 
Without loss of generality, we can assume or > z1.J~) for all r > 0. Then, 
by the fixpoint and antitone properties, we have 
q(y) = T,v,(r) 5 T&r) = q(r). 
BOUNDARY VALUE PROBLEM ON INFINITE INTERVAL 105 
But this last contradicts the assumption, and therefore this case is vacuous, 
i.e., neither function can strictly dominate the other. Since, moreover, o1 
and va are continuous, the only remaining case is 
Case 2. There exists an r, > 0 such that vr(r,,) = va(rJ. By hypo- 
thesis, ~~(0) = v,(O) = - OL, and vl(yO) = va(r,,) = - j3 (say). We are now 
in a position to apply, with minor modification, a uniqueness theorem (see 
[2]) f or t e oun ar va ue ro h 6 d y  I p bl em on the finite interval to obtain 
vi(r) = v~(Y) if y E [O, r,]. 
But since both functions are in C, and agree on the initial interval, which 
implies that v;(rJ = vi(rJ, we can apply a standard’ uniqueness theorem 
for the initial value problem to show that 
Vl(4 = w for r > Yg. 
Thus vr = v2 and uniqueness is established. 
Remark 1. An alternative approach would be to assume another “case 1” 
for which T+(Y) > v2(r) if r > y. and proceed as before, but perhaps the argu- 
ment would be a bit more complicated. 
Remark 2. We have shown that the unique required solution to (1.1) 
tends to some nonpositive constant as Y + 00. If the constant happens to be 
negative, its evaluation solves the “first term” asymptotic problem; if zero, 
one would ordinarily require more. This latter situation is exemplified by the 
equation in the subsequent and final section. 
Remark 3. In a certain restricted sense, the conclusions of Theorem 1 
and the properties of the negative bounded solutions to (1.1) as obtained 
in the course of its proof, form a complete presentation; i.e., corresponding 
to the hypothesis H imposed on 4 and the given number (Y, we have derived 
all relevant properties of the desired bounded solution. To see this, let f be 
an arbitrary concave function in C, n L and such that f” -f 2 0 for all 
Y > 0. (That this last condition is implied by H for u =f to be a negative 
solution to (1.1) is an immediate consequence of the inequality (1.4)). Next, 
set +(u, r) = u [j”(r),@)]. Th en one readily verifies that this 4 satisjes H 
and, correspondingly, that u =f is a solution to (1.1). 
We, of course, do not wish to infer that the theorem cannot be generalized. 
Thus,, 4 given by r#(u, r) = - l/(1 + r)” does not satisfy H, and yet the 
conclusion of the theorem holds for this 4, provided CY > &. 
1 For example, see E. A. CODDINGTON and N. LEVINSON, “Theory of Ordinary 
Differential Equations,” Chapter 1, Theorem 7.5. McGraw-Hill, New York, 1955. 
106 GROSS 
III. AN APPLICATION 
Physicists have long considered the differential equation 
d”u 
- = r(e”” - l), 
dr2 
Y > 0, 
as representing the motion of a particle in an ionized field under the influence 
of the Ukawa potential. One of their assumptions is that the solution u is 
initially negative and that for some constant A, u is asymptotic to Ae-r as 
r --f 03. Since it is not even a priori evident that these assumptions are mathe- 
matically consistent (i.e., nonvacuously satisfied), it seems a worthwhile 
project to validate them. Moreover, since the assumed asymptotic behavior 
implies that the assumed solution is bounded, it would appear that the 
physicist could have gotten by with much less, as the following theorem 
indicates. 
THEOREM 2. Let a be a given positive number. Then there exists a unique 
continuous bounded solution u to the da$rential equation d%/dr2 = r(e”lr - 1) 
for r > 0, and hating u(O) = - CL. This solution is negative increasing and 
concave; moreover, for some negative constant A, u is asymptotic to Ae-r as 
Y  - 00. (Remark. The uniqueness of u automatically implies the uniqueness of A.) 
PROOF. First of all, consider a required bounded solution u, if any exists. 
Then so long as u(r) I 0, by the trivial inequality on the exponential function, 
u is concave on that set and is convex on the set where u > 0. Moreover, by 
the well-known inequality (e5 2 1 + x for all x real) we have 
hence 
,, a4 - uzo for all r >O; 
e@u” - u) 2 0. 
Now, the boundedness of u and the previous remarks imply e-r(u’ + u) -+ 0 
as r --+ 00. Thus, integrating the last inequality from r to 00 gives 
hence, 
e-‘(u’ + u) 2 0, 
er(u’ + u) 5 0. 
Integrating this last from 0 to r, and making use of continuity and the initial 
value, we obtain 
u(r) 5 - ale-r. 
Thus, any bounded continuous solution is necessarily negative. The conclusions 
BOUNDARY VALUE PROBLEM ON INFINITE INTERVAL 107 
of the theorem, except fbr the asymptotic behavior, are now trivial consequences 
of Theorem 1. 
It remains to determine the asymptotic behavior of the solution u. Now 
since u < 0, we can write the differential equation in the form 
21” eU/T - 1 -=-* 
U uir 
Moreover, since u is bounded, u/r + 0 as r -+ 00. Using L’Hospital’s rule in 
the right member of the above equation then gives 
U’I 
u 
-+l as r-+00, 
It follows that given E > 0, there exists an rO > 0 such that 
$1-C for all r 2 r,. 
But since u < 0, this implies 
U” -((1-6)Ugo for all r 1 r,. 
Moreover, we can choose E < 1. Proceeding in the same manner as before 
(but with the inequalities reversed this time), using the boundedness and 
appropriate integrating factors, we obtain 
u(r) 2 Ce-“l-’ r, (3.1) 
for some negative constant C and all r 2 r,,. Next, since u is a fixpoint of TI 
for this special case of +, we obtain 
e%(r) = - 01 - j: e2tdt 1: sem8 (eu(S)/g - 1 - F) ds. (3.2) 
Then, if r > rO, by splitting up the integral into 2 parts, we have 
e%(r) = C, - 11, e2Vt 1: se+ (e”(a)/s - 1 - $) ds (3.3) 
for some constant C,. 
Next, using the isotone property of the integral operator and (3.1) gives 
e%(r) 2 C, - ]I0 e2$dt 1: se+ [exp ( 
ce-dCc 8 
s ) - 1 - 
ce4cz 8 
s ] ds, 
for all r 2 rO. 
(3.4) 
108 GROSS 
Now, it is a trivial exercise in integration to verify that if x < 0, then 
Applying this inequality to the right member of (3.4) with 
gives for Y 2 r0 
e’u(r) 2 Cl - T 
C2 j-IO e2tdt j-r e(-l-yir;)s ds. 
(3.5) 
But, since s 2 t 2 r,,, the above yields 
Upon performing the integrations, we observe that if we choose any positive 
E < 3, the right hand member of the above will be bounded below by some 
constant for all r 2 Y,,. Consequently e%(r) is bounded below. Finally, we 
recall that u is necessarily in L and hence e%(r) is monotone nonincreasing 
in r. Applying the principle of bounded monotone convergence then gives 
the asymptotic result. This completes the proof of Theorem 2. 
A few remarks are in order regarding the asymptotic coefficient A. Since 
the convergence is appropriately monotone, it follows that u(r) > Ae-+ for 
all Y 2 0. Using this fact and a further refinement gives in addition, 
U(T) < Ae-7 + $ e+, r > 0. 
We shall omit the details, which are based on the power series expansion 
of the exponential function. Thus, if / A j is known and not unduly large, 
one can estimate the tail of u very accurately. We assert also that the second 
term in the right member of the above inequality is not the second term in an 
asymptotic expansion of u but is an estimate of a certain function involving 
the exponential integral. Without going into the details, which are predomi- 
nantly computational, on letting r + 0 and using certain properties of the 
exponential integral, we obtain the inequality 
As-asA++log3. 
BOUNDARY VALUE PROBLEM ON INFINITE INTERVAL 109 
Thus, we have some information about the behavior of the mapping: 
cy -+ - A as defined by Theorem 2. It is no doubt the case that this mapping 
is a homeomorphism taking R+ onto itself. We have not, however, taken 
the pains to investigate this point. Perhaps this will be the subject of another 
paper. 
REFERENCES 
1. BIRKHOFF, GARRETT, “Lattice Theory,” p. 54. Am. Math. Sm. Colloq. Publ., 
Vol. 25 (1948). 
2. HENRICI, P. “Discrete Variable Methods in Ordinary Differential Equations,” 
Theorem 7.1, p. 347, et seq.; also, p. 15, et seq. Wiley, New York, 1962. 


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The boundary value problem on an infinite interval: Existence, uniqueness, and asymptotic behavior of bounded solutions to a class of nonlinear second order differential equations

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