The Maximum Principle

5.4 The Maximum Principle

Nature to Newton was an open book ... In one person he combined the experimenter, the theorist, the mechanic, and, not least, the artist in exposition. Newton stands before us strong, certain, and alone.

Albert Einstein (1879–1955) The maximum principle is one of the simplest tools in nonlinear mathematical anal-

ysis. It is very efficient in the qualitative study of wide classes of ordinary differen- tial equations, as well as in the treatment of partial differential equations.

The maximum principle can be described very easily in elementary terms. We give in what follows two different statements of this basic property.

be a twice differentiable convex function. Then u attains its maximum on [a, b] either in a or in b . In particular, if u (a) ≤ 0 and u (b) ≤ 0 , then u ≤0 in [a, b] .

Weak Maximum Principle. Let u : [a, b]→R

be a twice differentiable convex function such that u (a) = u(b) = 0 . Then the following alternative holds: either

Strong Maximum Principle. Let u : [a, b]→R

(i) u ≡0 in [a, b] or (ii) u <0 in (a, b) and, moreover, u ′ (a) < 0 , and u ′ (b) > 0 .

The purpose of this section is to present some variants of the maximum prin- ciple and to illustrate the force of this method in the qualitative analysis of some ordinary differential equations or differential inequalities. We refer to the excellent monograph [94] for more details and examples on this subject.

be a continuous function that is twice differentiable on (a, b) such that for all x ∈ (a,b) we have f ′′ (x) = α f (x) , for some constant α >0 . Prove that

5.4.1. (The Maximum Principle). Let f : [a, b]→R

| f (x)| ≤ max{| f (a)|,| f (b)|}, for all x ∈ [a,b].

Solution. The continuous function f is bounded, so it achieves the minimum and the maximum on the compact interval [a, b]. The idea of the proof is to show that

f can have neither a positive maximum nor a negative minimum in (a, b). Fix x 0 ∈ [a, b] and without loss of generality suppose f (x 0 ) > 0. Let M = max x ∈[a,b] f (x) ≥

f (x 0 ) > 0. If this maximum is attained at x M ∈ (a,b), then f ′′ (x M ) ≤ 0 and hence M = f (x M ) ≤ 0, a contradiction. Thus the maximum is attained at either a or b and

f (x 0 ) ≤ M = max{ f (a), f (b)} ≤ max{| f (a)|,| f (b)|}. Since x 0 is arbitrary, the result follows. ⊓ ⊔

236 5 Differentiability

be a bounded function and let u : (a, b)→R be a function satisfying

5.4.2. (The Maximum Principle) Let f : (a, b)→R

u ′′ (x) + f (x)u ′ (x) ≥ 0,

f or all x ∈ (a,b). Assume that u (x) ≤ M , for all x ∈ (a,b) . Prove that if there exists a point c in (a, b)

such that u (c) = M , then u ≡M ˆın (a, b) . Solution. Assume by contradiction that u (c) = M and there exists d ∈ (a,b) such

that u (d) < M. Let us suppose that d > c. Define the function ϕ (x) = e α(x−c) − 1, where α > 0 is a constant to be determined. Observe that ϕ (x) < 0 for all a < x < c, ϕ (x) > 0 if c < x < b and ϕ (c) = 0. We have

Φ ϕ (x) := ϕ ′′ (x) + f (x) ϕ ′ (x) = α [ α + f (x)]e α(x−c) .

Choose α > 0 sufficiently large that Φ ϕ (x) > 0 for any a < x < d. In other words, we choose α such that α > − f (x), for all x ∈ (a,b). This is possible, since f is bounded. Define w (x) = u(x) + εϕ (x), where ε > 0 is chosen such that ε < (M − u(d))/ ϕ (d). Such a choice for ε is possible due to u (d) < M and ϕ (d) > 0. Since ϕ < 0 in (a, c) we have w (x) < M, for all a < x < c. By the choice of ε we obtain

w (d) = u(d) + εϕ (d) < u(d) + M − u(d).

Hence w (d) < M. On the other hand,

w (c) = u(c) + εϕ (c) = M.

It follows that w has a maximum value greater than or equal to M, and this maximum is attained in an interior point of (a, d). But

Φ w (x) = Φ u (x) + εΦ ϕ (x) > 0, for all x ∈ (a,d).

This shows that there is no x 0 ∈ (a,d) such that w ′ (x 0 ) = 0 and w ′′ (x 0 ) ≤ 0. Thus, w cannot achieve its maximum in (a, d), contradiction. If d < c, then we use the auxiliary function ϕ (x) = e −α(x−c) − 1, with α > f (x) for all x ∈ (a,c). ⊓ ⊔

be a function that is bounded on every closed interval contained in (a, b) and let u

5.4.3. Let f : (a, b)→R

be a nonconstant function satisfying the differential inequality u ′′ (x) + f (x)u ′ (x) ≥ 0 , for all x ∈ (a,b) . Moreover, we suppose that u has

one-sided derivatives in a and in b .

(i) Assume that the maximum of u is achieved at x =a and that f is bounded from below near x =a . Prove that u ′ (a+) < 0 . (ii) Assume that the maximum of u is achieved at x =b and that f is bounded from above near x =b . Prove that u ′ (b−) > 0 .

5.4 The Maximum Principle 237 Solution. (i) Assume that u (a) = M and u(x) ≤ M for all a ≤ x ≤ b. Let d ∈ (a,b)

be such that u (d) < M. Consider the auxiliary function ϕ (x) = e α(x−a) − 1, with α > 0. Choose α >−f (x), for all a ≤ x ≤ d. Hence Φ ϕ (x) := ϕ ′′ (x) + f (x) ϕ ′ (x) > 0. Let w (x) = u(x) + εϕ (x), where 0 < ε < (M − u(d))/ ϕ (d). Since Φ w (x) > 0, it follows that the maximum of w in [a, d] is achieved at one of the endpoints. Since w (a) = M > w(d), the maximum is achieved for x = a. So, the right derivative in

a cannot be positive, so w ′ (a) = u ′ (a) + εϕ ′ (a) ≤ 0. But ϕ ′ (a) = α > 0. Therefore u ′ (a) < 0.

(ii) Apply the same arguments as above. ⊓ ⊔ We give in what follows a variant of the maximum principle for nonlinear

differential inequalities. More precisely, the nonlinearity is u −u 3 , and this goes back to the Ginzburg–Landau theory arising in superconductivity.

5.4.4. (a) Fix R >0 and let u i =u i (r) : [R, ∞)→[3 −1/2 , ∞) ( i = 1, 2 ) be twice differentiable functions satisfying

1 −u ′′ 1 − u 1 ′ 1 1 + 1 2 u 2 1 −u 1 (1 − u 1 ) ≥ 0 ≥ −u ′′ 2 − u ′

r 2 r + r 2 r u 2 −u 2 (1 − u 2 ). We also assume that u 1 (R) ≥ u 2 (R) and lim sup r → ∞ (u 1 −u 2 ) (r) ≥ 0 . Prove that

u 1 ≥u 2 on [R, ∞) . (b) Let f = f (r)

be a twice differentiable function such that lim inf r → ∞ f (r) = 1 and

2 f = f (1 − f ) in r (0, ∞). r Prove that

Solution. (a) Set u : =u 1 −u 2 . Then

1 −u ′′ (r) − u ′ r (r) + a(r)u(r) ≥ 0 for all r ∈ (R,∞),

where a (r) := u 2 1 (r) + u 2 2 (r) + u 1 (r)u 2 (r) + r −2 − 1. Our hypotheses imply a(r) > 0 on (R, ∞), u(R) ≥ 0, and limsup r → ∞ u (r) ≥ 0. So, by the maximum principle, u ≥ 0 on [R, ∞), which concludes the proof.

(b) Define, for all r > 0, ϕ (r) = 1 − (2r 2 ) −1 −2 −3 9r −4 . For a positive constant C, set u 1 (r) = ϕ (r)+Cr −5 and u 2 (r) = ϕ (r)−Cr −5 . Then u 1 (R) ≥ u 2 (R) ≥ 3 −1/2 , provided that R > 0 is sufficiently large. We fix R with this property and choose

C > 0 such that u 1 (R) ≥ f (R) ≥ f 2 (R). Thus, the pairs (u 1 , f ) and (u 2 , f ) satisfy the hypotheses imposed at (a). Hence u 1 ≥f≥u 2 in (R, ∞). In particular, this implies that for all r > R, | ϕ (r) − f (r)| ≤ Cr −5 , and the conclusion follows. ⊓ ⊔

238 5 Differentiability