Assume that there exists a smooth positive function f on (0, 1) satisfying the differential equation

5.5.9. Assume that there exists a smooth positive function f on (0, 1) satisfying the differential equation

−f − +

= f (1 − f 2 ) in (0, 1) together with with boundary conditions f (0) = 0 and f (1) = 1 . Prove that f is

unique. Solution. We give an argument of I. Shafrir [6] that is based on a method

introduced by H. Brezis and L. Oswald[12]. Let f 1 and f 2 be two positive functions satisfying the hypotheses. Dividing the differential equation by f and subtracting the corresponding equations, we obtain

Multiplying the above equality by r (f 2 −f 1 2 2 ) and integrating over (0, 1) yields " 1 2 " 1 2 " 1

Therefore f 1 =f 2 on (0, 1). ⊓ ⊔ The next property is very useful for proving uniqueness results.

be functions such that f is continuous, g is differen- tiable, and g vanishes at least once. Assume that there exists a real constant λ such that for all x ∈ [a,b] , | f (x)g(x) + λ g ′ (x)| ≤ |g(x)| . Prove that g =0 in [a, b] .

5.5.10. Let f , g : [a, b]→R

Solution. Arguing by contradiction, there exists x 0 ∈ (a,b) such that g(x 0 ) > 0. Let V = (c, d) ⊂ [a,b] be a neighborhood of x 0 such that g > 0 on V . Since g vanishes at least once on [a, b], we can assume that g(c) = 0. For any x ∈ V we have, by hypothesis,

1 + sup | f (x)| .

g (x) | λ |

g (x)

x ∈[a,b]

It follows that the mapping V as x ց c, we obtain the desired contradiction.

The above property implies the following classical result, which is very useful to establish the uniqueness of the solution in the theory of differential equations. ⊓ ⊔

Gronwall’s lemma. Let I ⊂R

be an interval. Assume that f :I →R is a differ- entiable function that does not have a constant sign on I and satisfying, for all x ∈I , |f ′ (x)| ≤ C | f (x)| , where C is a positive constant. Then f =0 on I .

250 5 Differentiability The following second-order differential equation with cubic nonlinearity has

only bounded solutions.

5.5.11. Consider the differential equation x ′′ (t) + a(t)x 3 (t) = 0 on 0 ≤t<∞ , where a (t) is continuously differentiable and a (t) ≥ κ >0 . If a ′ (t) has only finitely many changes of sign, prove that any solution x (t) is bounded.

Ph. Korman, Math. Magazine, Problem 1577

Solution. For any t ≥ 0, define the function

2 x 4 (t)

E (t) =

(t) + a(t)

Using the differential equation, we obtain

Assuming that a ′ (t) ≤ 0 for all t ∈ [t 1 ,t 2 ], then E ′ (t) ≤ 0 on [t 1 ,t 2 ]. Thus,

E (t) ≤ E(t 1 ) for all t ∈ [t 1 ,t 2 ]. If a ′ (t) ≥ 0 on [t 1 ,t 2 ], then E ′ (t)/E(t) ≤ a ′ (t)/a(t), for all t 1 ≤t≤t 2 . Hence

E (t )

a (t

E 1 a 2 (t) ≤ ) (t) ≤ E (t 1 ), for all t 1 ≤t≤t 2 . (5.37)

a (t 1 )

a (t 1 )

This gives an idea about the growth rate of E. More precisely, E (t) can increase by at most a factor of a (t 2 )/a(t 1 ) on [t 1 ,t 2 ]. Relation (5.37) combined with the definition of E (t) yields

x 4 (t)

E (t 1 )

, for all t 1 ≤t≤t 2 . (5.38)

4 a (t 1 )

Let us now assume that a ′ (t) changes sign at points c 1 ,c 2 ,...,c n . Since E (t) is nonnegative and nonincreasing on any interval on which a ′ (t) ≤ 0, and increases by at most a factor a (c k +1 )/a(c k ) on any interval [c k ,c k +1 ] on which a ′ (t) ≥ 0, it follows that E (t) remains bounded on [0, c n ]. We distinguish two cases:

(i) a ′ (t) ≤ 0 on [c n , ∞). Then E ′ (t) ≤ 0 for all t ∈ [c n , ∞). Thus, E(t) is nonincreas- ing on [c n , ∞), so remains bounded. This implies that x(t) is bounded on [0, ∞).

(ii) a ′ (t) ≥ 0 on (c n , ∞). Thus, by (5.38), it follows that x 4 (t) ≤ 4E(c n )/a(c n ), for all t ≥c n . Therefore x (t) is bounded.

The next problem concerns the differential equation

(5.39) where h is a positive number (either fixed or variable) and x ∈ R. This equa-

f (x + h) − f (x − h) = 2h f ′ (x),

tion appears in the study of central forces and it is called the gravity equation.

5.5 Differential Equations and Inequalities 251 Archimedes observed that any quadratic polynomial satisfies (5.39), and he gave

the following geometric interpretation of this fact. Select any two points A and

B on a parabola and let C be the point of the parabola at which the tangent to the parabola is parallel to the chord AB. Then the line through C parallel to the axis of the parabola is midway between the lines through A and B parallel to the axis. ⊓ ⊔

We also observe that if V (x) is a potential function associated with a radially symmetric central force and if a sphere of radius h attracts exterior particles as if all the mass of the sphere were at its center, then the function f (x) := (xV (x)) ′ satisfies the differential equation (5.39). The next result gives a sufficient condition to ensure that a solution of (5.39) has at most exponential growth.

5.5.12. Let f (x) satisfy (5.39) for all x and for two values of h , say for h =a and h =b with 0 <a<b . Then there exist positive constants A and c such that | f (x)| < Ae c |x| , for all real x .

Sherman Stein Solution. We assume for the sake of notational simplicity that x ≥ 0 (a similar

argument can be applied for x < 0). Since (5.39) is fulfilled for h = a and h = b, we obtain

a [ f (x + a) − f (x − a)] . Replacing x with y − a in the above relation, we obtain

f (x + b) = f (x − b) +

f (y + b − a) = f (y − b − a) + [ f (y) − f (y − 2a)] .

This identity relates f (y + b − a) to values f (x) for x no larger than y. Set M (x) := max{ f (t); 0 ≤ t ≤ x}. It follows that for all y ≥ a + b,

2b

M (y + b − a) ≤ M(y) + M (y) .

Define M : = max{M(x); 0 ≤ x ≤ a + b}. Then, considering M(x) over successive intervals of length b − a, we obtain

2b x /(b−a)

M (x) ≤ A 1 +

which concludes the proof. ⊓ ⊔ If f is twice differentiable on an interval I and f ′′ = 0 on I, then f is linear. Thus,

f ≡ 0 in I, provided f has at least two zeros in I. This simple property is extended in the next exercise to a larger class of linear differential equations.