Notes on Differential Equations and Differential Inequalities

Adam Coffman, Yifei Pan, and Yuan Zhang^{∗}

1 Real autonomous ODE

The following Lemma gives conditions for the existence of a solution of a differential equation which is bounded on the domain ℝ.

Lemma 1.1. Given real numbers a < b, if f : (a, b) → ℝ is a continuous, nonvanishing function, and there are some constants C_{1} > 0, C_{2} > 0, δ_{1} ∈ (0, b − a), δ_{2} ∈ (0, b − a) so that |f(t)| ≤ C_{1}(t − a) for a < t < a + δ_{1} and |f(t)| ≤ C_{2}(b − t) for b − δ_{2} < t < b, then there exists a one-to-one, onto function g : ℝ → (a, b) so that y = g(t) is a solution of the equation \frac{dy}{dt} = f(y).

Proof. \frac{1}{f(x)} is continuous on (a, b), so the function

G(t) = ∫ t_{\frac{a+b}{2}} \frac{1}{f(x)}dx  (1)

is differentiable on (a, b) with a nonvanishing, nonzero derivative, \frac{d}{dt}G(t) = \frac{1}{f(t)}. Because f and \frac{1}{f} have constant sign, G(t) is monotone on (a, b). Suppose f(t) > 0, so G is increasing; the f(t) < 0 case is similar.

For t ∈ (b − δ_{2}, b),

G(t) = ∫ b−δ_{2}_{\frac{a+b}{2}} \frac{1}{f(x)}dx + ∫_{b−δ_{2}}^{t} \frac{1}{f(x)}dx ≥ ∫ b−δ_{2}_{\frac{a+b}{2}} \frac{1}{f(x)}dx + ∫_{b−δ_{2}}^{t} \frac{1}{C_{2}(b − x)}dx,

∗Supported in part by National Science Foundation DMS-1265330.

which is unbounded. Similarly, G is also unbounded at the other endpoint, so G : (a, b) → ℝ is onto and invertible. Let T be any constant, and define

g(t) = G^{−1}(t + T),  (2)

so g : ℝ → (a, b) is onto and increasing, and (by the Inverse Function Theorem, [C]), y = g(t) is differentiable with

\frac{dy}{dt} = \frac{1}{G′(G^{−1}(t + T))} = \frac{1}{\frac{1}{f(y)}} = f(y).

Such solutions with domain ℝ are unique up to translation.

Lemma 1.2. Given an open (possibly infinite) interval I, if f : I → ℝ is a continuous, nonvanishing function, and g_{1} : ℝ → I and g_{2} : ℝ → I are solutions of the equation \frac{dy}{dt} = f(y), then there exists a constant T so that g_{2}(t) = g_{1}(t + T).

Proof. Because g′_{1}(x) = f(g_{1}(x)) is continuous and nonzero, the Inverse Function Theorem applies. For t ∈ ℝ,

\frac{d}{dt} (g_{1}^{−1}(g_{2}(t)) − t) = \frac{1}{g′_{1}(g_{1}^{−1}(g_{2}(t)))}g′_{2}(t) − 1  (3)
= \frac{1}{f(g_{1}(g_{1}^{−1}(g_{2}(t))))}f(g_{2}(t)) − 1 ≡ 0.

There is also a local uniqueness theorem for solutions on an interval, with one initial condition.

Lemma 1.3. Given open (possibly infinite) intervals I_{0}, I_{1}, I_{2}, if f : I_{0} → ℝ is a continuous, nonvanishing function, and g_{1} : I_{1} → I_{0} and g_{2} : I_{2} → I_{0} are solutions of the equation \frac{dy}{dt} = f(y), and there is a point c ∈ I_{1} ∩ I_{2} such that g_{1}(c) = g_{2}(c), then there exists δ > 0 so that g_{1}(t) = g_{2}(t) for all t ∈ (c − δ, c + δ).

Proof. Because g′_{1}(x) = f(g_{1}(x)) is continuous and nonzero, the Inverse Function Theorem applies: there exists some δ_{1} > 0 so that g_{1} is one-to-one on (c − δ_{1}, c + δ_{1}). Suppose f > 0, so g_{1} is increasing; the f < 0 case is similar. Let ε = min\{g_{1}(c + \frac{1}{2}δ_{1}) − g_{1}(c), g_{1}(c) − g_{1}(c − \frac{1}{2}δ_{1})\} > 0. Because g_{2} is continuous, there is some δ_{2} > 0 corresponding to ε, so that for all t ∈ (c−δ_{2}, c+δ_{2}), |g_{2}(t)−g_{2}(c)| = |g_{2}(t)−g_{1}(c)| < ε. Let δ = min{δ_{1}, δ_{2}} > 0, then (c − δ, c + δ) ⊆ I_{1} ∩ I_{2}, where both g_{1} and g_{2} are defined. Also, for any t ∈ (c − δ, c + δ),

g_{1}(c − \frac{1}{2}δ_{1}) ≤ g_{1}(c) − ε < g_{2}(t) < g_{1}(c) + ε ≤ g_{1}(c + \frac{1}{2}δ_{1}),

and by the Intermediate Value Theorem, there is some x ∈ (c − \frac{1}{2}δ_{1}, c + \frac{1}{2}δ_{1}) so that g_{1}(x) = g_{2}(t); this shows that x = g_{1}^{−1}(g_{2}(t)), so g_{2}(t) is in the domain of g_{1}^{−1}. As in (3), for c − δ < t < c + δ,

\frac{d}{dt}(g_{1}^{−1}(g_{2}(t))) ≡ 1 ⟹ g_{1}^{−1}(g_{2}(t)) = t + T

for some constant T. Evaluating g_{2}(t) = g_{1}(t + T) at t = c gives g_{2}(c) = g_{1}(c + T), and g_{2}(c) = g_{1}(c) by hypothesis, so c + T = c because g_{1} is oneto-one. It follows that T = 0 and g_{1}(t) = g_{2}(t) for all t ∈ (c − δ, c + δ).

Lemma 1.4. If f : ℝ → ℝ is a continuous, nonvanishing function, then there exist some open interval I and a one-to-one, onto function g : I → ℝ so that y = g(t) is a solution of the equation \frac{dy}{dt} = f(y).

Proof. \frac{1}{f(x)} is continuous on ℝ, so the function

G(t) = ∫_{0}^{t} \frac{1}{f(x)}dx

is differentiable on ℝ with a nonvanishing, nonzero derivative, \frac{d}{dt}G(t) = \frac{1}{f(t)}. Because f and \frac{1}{f} have constant sign, G(t) is monotone on ℝ. Its image is some open interval I, so G : ℝ → I is invertible.

Let T be any constant, and define g(t) = G^{−1}(t + T) on the interval I − T = {x ∈ ℝ : x + T ∈ I}, so g : I − T → ℝ is invertible, and therefore not bounded. g is a solution of the ODE as in Lemma 1.1.

Lemma 1.5. Given b ∈ ℝ, if f : (−∞, b) → ℝ is a continuous, nonvanishing function, and there are some constants C_{3} > 0, δ_{3} ∈ (0, 1) so that |f(t)| ≤ C_{3}(b − t) for b − δ_{3} < t < b, then there exist an open interval I and a oneto-one, onto function g : I → (−∞, b) so that y = g(t) is a solution of the equation \frac{dy}{dt} = f(y).

Proof. \frac{1}{f(x)} is continuous on (−∞, b), so the function

G(t) = ∫_{b−1}^{t} \frac{1}{f(x)}dx

is differentiable on (−∞, b) with a nonvanishing, nonzero derivative, \frac{d}{dt}G(t) = \frac{1}{f(t)}. Because f and \frac{1}{f} have constant sign, G(t) is monotone on (−∞, b). Suppose f(t) > 0, so G is increasing; the f(t) < 0 case is similar.

For t ∈ (b − δ_{3}, b),

G(t) = ∫_{b−1}^{b−δ_{3}} \frac{1}{f(x)}dx + ∫_{b−δ_{3}}^{t} \frac{1}{f(x)}dx ≥ ∫_{b−1}^{b−δ_{3}} \frac{1}{f(x)}dx + ∫_{b−δ_{3}}^{t} \frac{1}{C_{3}(b − x)}dx,

which is unbounded. So, the image of G is either I = (L, ∞) or I = ℝ.

Let T be any constant, and define g(t) = G^{−1}(t + T) on the interval I − T = {x ∈ ℝ : x + T ∈ I}, so g : I − T → (−∞, b) is invertible, and therefore not bounded. g is a solution of the ODE as in Lemma 1.1.

Theorem 1.6. Let f : ℝ → ℝ be a real analytic function. The following are equivalent:

1. there exists a non-constant, bounded, real analytic function g : ℝ → ℝ so that y = g(t) is a solution of the equation \frac{dy}{dt} = f(y);

2. there are at least two distinct points a_{0}, b_{0} where f(a_{0}) = f(b_{0}) = 0.

Proof. At any zero of f, say c_{0}, the real analyticity implies there is some constant C_{0} so that |f(t)| ≤ C_{0}|x − c_{0}| for x near c_{0}.

To show 2 . ⟹ 1 ., the property that f is real analytic on ℝ implies that a_{0} is an isolated zero, and because a_{0} is not the only zero of f, there is some open interval (a, b) with either a = a_{0} or b = a_{0}, satisfying the hypotheses of Lemma 1.1. The conclusion is that there exists a non-constant solution g : ℝ → (a, b), which by construction, (1) and (2), is real analytic.

To show 1 . ⟹ 2 ., suppose, toward a contradiction, that there exists a solution g_{1} as claimed, and that f has fewer than two zeros.

Case 1. If f is nonvanishing on ℝ then Lemma 1.4 applies and there is some nonempty open interval I and some onto function g_{2} : I → ℝ which is a solution of \frac{dy}{dt} = f(y). By the construction of Lemma 1.4, g_{2} is real analytic. On the interval I, g_{2} − g_{1} is continuous, and because g_{1} is bounded and g_{2} is onto, g_{2} − g_{1} attains some negative value and some positive value, so by the Intermediate Value Theorem, there is some c so that g_{1}(c) = g_{2}(c). By Lemma 1.3, g_{1}(t) ≡ g_{2}(t) on some interval (c − δ, c + δ), and because both functions are real analytic, g_{1}(t) ≡ g_{2}(t) on I. The contradiction is that g_{2} is unbounded on I while g_{1} is bounded.

Case 2. If f has exactly one zero, b ∈ ℝ, then Lemma 1.5 applies to f on (−∞, b): there is some interval I_{2} and some one-to-one, onto solution g_{2} : I_{2} → (−∞, b). By an analogous existence result for f on (b, ∞), there is some interval I_{3} and some one-to-one, onto solution g_{3} : I_{3} → (b, ∞). Because g_{1} is non-constant, there is some x_{0} where g(x_{0}) ≠ b.

If g_{1}(x_{0}) < b, then there is some T ∈ I_{2} with g_{2}(T) = g_{1}(x_{0}), and the function g_{4}(t) = g_{2}(t + T − x_{0}) is, by construction, a real analytic solution of \frac{dy}{dt} = f(y) on an open interval I_{2} − (T − x_{0}) satisfying g_{4}(x_{0}) = g_{1}(x_{0}). As in Case 1., the uniqueness from Lemma 1.3 shows that g_{4} = g_{1} on I − (T − x_{0}), contradicting the boundedness of g_{1}.

The g_{1}(x_{0}) > b case is similar, using g_{3}.

2 Ordinary differential inequalities

2.1 Linear differential inequalities

Lemma 2.1. If u(t) is continuous on [a, b] and u(a) > 0, then either u(t) > 0 for all t ∈ [a, b], or there is some t_{1} ∈ (a, b] so that u(t) > 0 on [a, t_{1}) and u(t_{1}) = 0.

Proof. Consider the set S = {x ∈ [a, b] : u(t) > 0 for all t ∈ [a, x)}; it is non-empty (by continuity of u and u(a) > 0), and bounded above, so it has a least upper bound t_{1} ∈ (a, b]. If there were some t_{0} with a < t_{0} < t_{1} and u(t_{0}) ≤ 0, then there would be no x ∈ S with x > t_{0}, and t_{0} would be an upper bound, contradicting the least property of t_{1}. So, u(t) > 0 for all t ∈ [a, t_{1}). Case 1: If u(t_{1}) = 0, then u > 0 on [a, t_{1}) as claimed. Case 2: If u(t_{1}) < 0 then by continuity, there is some t_{2} with a < t_{2} < t_{1} and u(t_{2}) < 0, contradicting the above property that u(t) > 0 for all t ∈ [a, t_{1}). Case 3: If t_{1} = b and u(t_{1}) > 0, then u > 0 on [a, b] as claimed. Case 4: If t_{1} < b and u(t_{1}) > 0, then by continuity, u(t) would be positive on some interval [a, t_{1} + δ), contradicting the property that t_{1} is an upper bound for S. Only Cases 1 and 3 do not lead to a contradiction.

Lemma 2.2. Suppose a(t) is a real function on [0, 1) such that a(t) ≥ 0 and a is bounded on every subinterval [0, x] ⊆ [0, 1). If y is continuous on [0, 1] with y(0) ≥ 0, lim_{t→0^{+}} y′(t) ≥ 0, and y′′(t) ≥ a(t)y(t) for 0 < t < 1, then

y(t) ≥ 0 for all 0 ≤ t ≤ 1 and y′(t) ≥ 0 for all 0 < t < 1.

Proof. Case 1: y(0) > 0 and lim_{t→0^{+}} y′(t) > 0. In this case, we can show that

y > 0 on [0, 1] and y′ > 0 on (0, 1). By Lemma 2.1 applied to y on [0, 1], either y > 0 on [0, 1], or there is some t_{1} ∈ (0, 1] so that y(t) > 0 for all t ∈ [0, t_{1}) and y(t_{1}) = 0. In the latter case, y attains some positive maximum value on [0, t_{1}]. If y(0) is the maximum, then by the Mean Value Theorem, for any 0 < δ < t_{1}, there is some t_{2} with 0 < t_{2} < δ and y′(t_{2}) = \frac{y(δ)−y(0)}{δ} ≤ 0, which contradicts lim_{t→0^{+}} y′(t) > 0. If the maximum is at an interior point t_{3}

with 0 < t_{3} < t_{1}, then y′(t_{3}) = 0. From lim_{t→0^{+}} y′(t) > 0, there is some t_{4} with

0 < t_{4} < t_{3} and y′(t_{4}) > 0. Applying the Mean Value Theorem to y′ on [t_{4}, t_{3}], there is some t_{5} with t_{4} < t_{5} < t_{3} and y′′(t_{5}) = \frac{y′(t_{3})−y′(t_{4})}{t_{3}−t_{4}} < 0. This contradicts y′′(t_{5}) ≥ a(t_{5})y(t_{5}) ≥ 0. We can conclude that y(t_{1}) must be the maximum value, and y(t_{1}) > 0, which contradicts y(t_{1}) = 0. This shows y(t) > 0 for all t ∈ [0, 1].

For any t_{7} ∈ (0, 1), there is some t_{8} with 0 < t_{8} < t_{7} and y′(t_{8}) > 0. By the Mean Value Theorem, there is some t_{9} with \frac{y′(t_{7})−y′(t_{8})}{t_{7}−t_{8}} = y′′(t_{9}) ≥ a(t_{9})y(t_{9}) ≥ 0. It follows that y′(t_{7}) ≥ y′(t_{8}) > 0.

Case 1 did not use the boundedness of a, just a ≥ 0.

Case 2: y(0) ≥ 0 and lim_{t→0^{+}} y′(t) ≥ 0. Suppose, toward a contradiction,

that there is some t_{0} with 0 < t_{0} < 1 and y(t_{0}) < 0. For 0 ≤ t ≤ t_{0}, there is a bound A > 0 with 0 ≤ a(t) ≤ A. For t ∈ [0, t_{0}], define u(t) = y(t) − \frac{1}{2}y(t_{0})e^{\sqrt{A}(t−t_{0})}. Then, by construction, u(0) > 0 and u(t_{0}) < 0. For 0 < t < t_{0},

u′(t) = y′(t) − \frac{1}{2}y(t_{0})\sqrt{A}e^{\sqrt{A}(t−t_{0})}
⟹ lim_{t→0^{+}} u′(t) = ( lim_{t→0^{+}} y′(t)) − \frac{1}{2}y(t_{0})\sqrt{A} > 0,
u′′(t) = y′′(t) − \frac{1}{2}y(t_{0})Ae^{\sqrt{A}(t−t_{0})}
≥ a(t)y(t) − \frac{1}{2}y(t_{0})a(t)e^{\sqrt{A}(t−t_{0})}

+\frac{1}{2}y(t_{0})a(t)e^{\sqrt{A}(t−t_{0})} − \frac{1}{2}y(t_{0})Ae^{\sqrt{A}(t−t_{0})}

= a(t)u(t) + \frac{1}{2}y(t_{0})(a(t) − A)e^{\sqrt{A}(t−t_{0})}
≥ a(t)u(t).

Let w(t) = u(t_{0}t), just horizontally re-scaling u to the domain [0, 1] so that w(0) > 0, w(1) < 0, lim_{t→0^{+}} w′(t) > 0, and w′′(t) ≥ t_{0}^{2}a(t_{0}t)w(t), so Case 1

applies to w, contradicting w(1) < 0. The conclusion is that y(t) ≥ 0 on [0, 1), and on [0, 1] by continuity.

To establish the inequality y′ ≥ 0, for any t_{1} with 0 < t_{1} < 1 and any ϵ > 0, from lim_{t→0^{+}} y′(t) ≥ 0, there is some t_{2} with 0 < t_{2} < t_{1} and y′(t_{2}) > −ϵ. By the Mean Value Theorem, there is some t_{3} with t_{2} < t_{3} < t_{1} and \frac{y′(t_{1})−y′(t_{2})}{t_{1}−t_{2}} = y′′(t_{3}) ≥ a(t_{3})y(t_{3}) ≥ 0. It follows that y′(t_{1}) ≥ y′(t_{2}) > −ϵ.

Here’s a higher order generalization, using only the Mean Value Theorem, not the maximum value.

Lemma 2.3. Let k ≥ 2 be an integer. Suppose a(t) is a real function on [0, 1) such that a(t) ≥ 0 and a is bounded on every subinterval [0, x] ⊆ [0, 1). If y is continuous on [0, 1] with y(0) ≥ 0, and lim_{t→0^{+}} y^{(j)} ≥ 0 for j = 1, …, k − 1,

and y^{(k)}(t) ≥ a(t)y(t) for 0 < t < 1, then y(t) ≥ 0 for all 0 ≤ t ≤ 1 and y^{(j)}(t) ≥ 0 for all 0 < t < 1, j = 1, …, k.

Proof. Case 1: y(0) > 0 and lim_{t→0^{+}} y^{(j)}(t) > 0. In this case, we can show that

y > 0 on [0, 1] and y^{(j)} > 0 on (0, 1) for j = 1, …, k − 1.

By Lemma 2.1 applied to y on [0, 1], either y > 0 on [0, 1], or there is some t_{1} ∈ (0, 1] so that y(t) > 0 for all t ∈ [0, t_{1}) and y(t_{1}) = 0.

In the latter case, y(t_{1}) = 0 < y(0), so by the Mean Value Theorem for y on [0, t_{1}], there is some t_{2} with 0 < t_{2} < t_{1} and y′(t_{2}) < 0. Then, the MVT applies to y′ on [t_{3}, t_{2}] for some t_{3} > 0 where y′(t_{3}) > 0, using lim_{t→0^{+}} y′(t) > 0,

so there is some t_{4} > 0 where y′′(t_{4}) = \frac{y′(t_{2})−y′(t_{3})}{t_{2}−t_{3}} < 0. Repeatedly applying this MVT argument to y^{(j)} until j = k, gives some t_{N} with 0 < t_{N} < t_{1}, y^{(k)}(t_{N}) < 0, contradicting y^{(k)}(t_{N}) ≥ a(t_{N})y(t_{N}) ≥ 0.

So, the only case not leading to a contradiction is that y(t) > 0 on [0, 1]. The above MVT argument also shows that all y^{(j)} are positive on (0, 1)

for j = 1, …, k − 1, since any point t_{n} with 0 < t_{n} < t_{1} = 1 and y^{(j)}(t_{n}) ≤ 0 leads to another point t_{m} with 0 < t_{m} < t_{n} and y^{(j+1)}(t_{m}) < 0, eventually contradicting y^{(k)}(t_{N}) ≥ a(t_{N})y(t_{N}) ≥ 0.

Case 1 did not use the boundedness of a, just a ≥ 0.

Case 2: y(0) ≥ 0 and lim_{t→0^{+}} y^{(j)} ≥ 0, j = 1, …, k − 1. Suppose, toward

a contradiction, that there is some t_{0} with 0 < t_{0} < 1 and y(t_{0}) < 0. For 0 ≤ t ≤ t_{0}, there is a bound A > 0 with 0 ≤ a(t) ≤ A. For t ∈ [0, t_{0}], define u(t) = y(t) − \frac{1}{2}y(t_{0})e^{A^{1/k}(t−t_{0})}. Then, by construction, u(0) > 0 and u(t_{0}) < 0. For 0 < t < t_{0}, 1 ≤ j ≤ k − 1,

u^{(j)}(t) = y^{(j)}(t) − \frac{1}{2}y(t_{0})A^{j/k}e^{A^{1/k}(t−t_{0})}
⟹ lim_{t→0^{+}} u^{(j)}(t) = ( lim_{t→0^{+}} y^{(j)}(t)) − \frac{1}{2}y(t_{0})A^{j/k} > 0,
u^{(k)}(t) = y^{(k)}(t) − \frac{1}{2}y(t_{0})Ae^{A^{1/k}(t−t_{0})}
≥ a(t)y(t) − \frac{1}{2}y(t_{0})a(t)e^{A^{1/k}(t−t_{0})}

+\frac{1}{2}y(t_{0})a(t)e^{A^{1/k}(t−t_{0})} − \frac{1}{2}y(t_{0})Ae^{A^{1/k}(t−t_{0})}

= a(t)u(t) + \frac{1}{2}y(t_{0})(a(t) − A)e^{A^{1/k}(t−t_{0})}
≥ a(t)u(t).

Let w(t) = u(t_{0}t), just horizontally re-scaling u to the domain [0, 1] so that w(0) > 0, w(1) < 0, lim_{t→0^{+}} w^{(j)}(t) > 0, and w^{(k)}(t) ≥ t_{0}^{k}a(t_{0}t)w(t), so Case 1

applies to w, contradicting w(1) < 0. The conclusion is that y(t) ≥ 0 on [0, 1), and on [0, 1] by continuity.

To establish the inequalities y^{(j)} ≥ 0, start with j = k − 1. Then, for any t_{1} with 0 < t_{1} < 1 and any ϵ > 0, from lim_{t→0^{+}} y^{(k−1)}(t) ≥ 0, there is some

t_{2} with 0 < t_{2} < t_{1} and y′(t_{2}) > −ϵ. By the MVT, there is some t_{3} with t_{2} < t_{3} < t_{1} and \frac{y^{(k−1)}(t_{1})−y^{(k−1)}(t_{2})}{t_{1}−t_{2}} = y^{(k)}(t_{3}) ≥ a(t_{3})y(t_{3}) ≥ 0. It follows that y^{(k−1)}(t_{1}) ≥ y^{(k−1)}(t_{2}) > −ϵ. A similar argument applies for j decreasing from k − 1 to 1.

Lemma 2.4. If the left-side limit lim_{t→b^{−}} f(t) = −∞, then there is no interval

(b − δ, b) on which f′(t) is bounded below.

Proof. (See [C].)

Lemma 2.5. Suppose a(t) is a real function on [0, 1) such that a is bounded above on every subinterval [0, x] ⊆ [0, 1) and bounded below on every subinterval [x_{1}, x_{2}] ⊆ (0, 1). If y is continuous on [0, 1] with y(0) ≥ 0 and y′(t) ≥ a(t)y(t) for 0 < t < 1, then y(t) ≥ 0 for all 0 ≤ t ≤ 1.

Proof. Case 1: y(0) > 0.

By Lemma 2.1 applied to y on [0, 1], either y > 0 on [0, 1], or there is some t_{1} ∈ (0, 1] so that y(t) > 0 for all t ∈ [0, t_{1}) and y(t_{1}) = 0. If t_{1} = 1, then y ≥ 0 as claimed.

So, suppose toward a contradiction that t_{1} < 1. On the interval [\frac{1}{2}t_{1}, t_{1}], a is bounded below: there is some K < 0 so that K ≤ a(t). Define the function f(t) = ln(y(t)) for t in the interval (\frac{1}{2}t_{1}, t_{1}). f has left-side limit lim_{t→t_{1}^{−}} f(t) = −∞. For all t in (\frac{1}{2}t_{1}, t_{1}), the derivative is bounded below: f′(t) =

\frac{1}{y(t)}y′(t) ≥ \frac{1}{y(t)}a(t)y(t) = a(t) ≥ K, but this contradicts Lemma 2.4. Case 2: y(0) = 0. Suppose toward a contradiction that there is some p ∈ [0, 1] with y(p) < 0.

p ≠ 0 by assumption, and if p = 1, then by continuity of y, there is some nearby point p − δ_{1}/2 with y(p − δ_{1}/2) < 0. So by re-labeling if necessary, we can assume 0 < p < 1. On the interval [0, p], a is bounded above: there is some A > 0 so that a(t) ≤ A.

Define g(t) = −y(p−pt) on the domain 0 ≤ t ≤ 1, so that g(0) = −y(p) > 0, g(1) = −y(0) = 0, and g is continuous on [0, 1]. The derivative satisfies

g′(t) = py′(p − pt) ≥ pa(p − pt)y(p − pt) = −pa(p − pt)g(t),

and the coefficient −pa(p − pt) is bounded below by −pA. Case 1 applies to g, so g(t) > 0 on [0, 1) and g(1) = 0. Define the function f(t) = ln(g(t)) for t in the interval (0, 1). f has left-side limit lim_{t→1^{−}} f(t) = −∞. For all t in (0, 1),

the derivative is bounded below: f′(t) = \frac{1}{g(t)}g′(t) ≥ \frac{1}{g(t)}(−pa(p − pt))g(t) = −pa(p − pt) ≥ −pA, but this contradicts Lemma 2.4.

Lemma 2.6. Suppose a(t) is a bounded real function on [0, X]. Then there is some δ with 0 < δ ≤ X, with the property that if y is continuous on [0, X] with y(0) ≥ 0, lim_{t→0^{+}} y′(t) ≥ 0, and y′′(t) ≥ a(t)y(t) for 0 < t < X, then

y(t) ≥ 0 for all 0 ≤ t ≤ δ.

Proof. Step 1. Pick any x in (0, X], so that by hypothesis, there are some A > 0 and K < 0 so that K ≤ a(t) ≤ A for t ∈ [0, x]. Let δ = min\{x, \frac{1}{\sqrt{A}}, \frac{1}{\sqrt{−K}} \} > 0. (Remark: depending on a, it may be possible to choose x that optimizes δ.) To show that this is a δ as claimed by the Lemma, suppose toward a contradiction that there is some c with 0 ≤ c ≤ δ and y(c) < 0. c > 0 by hypothesis.

Step 2. Let y(b) be the minimum value of y on [0, c], so y(b) ≤ y(c) < 0 and 0 < b ≤ c ≤ δ. The MVT applies to y on [0, b]: there is some t_{0} with 0 < t_{0} < b and y′(t_{0}) = \frac{y(b)−y(0)}{b−0}. The MVT applies to y′ (extended to y′(0) = lim_{t→0^{+}} y(t) ≥ 0) on [0, t_{0}]: there is some t_{1} with 0 < t_{1} < t_{0} and

y′′(t_{1}) = \frac{y′(t_{0}) − y′(0)}{t_{0} − 0} = \frac{\frac{y(b)−y(0)}{b} − y′(0)}{t_{0}}.

By hypothesis,

a(t_{1})y(t_{1}) ≤ y′′(t_{1}) = \frac{y(b) − y(0) − by′(0)}{bt_{0}} ≤ \frac{y(b)}{bt_{0}} < 0.

If a(t_{1}) > 0 and y(t_{1}) < 0, then y(t_{1}) ≤ \frac{y(b)}{bt_{0}a(t_{1})} < \frac{y(b)}{δ^{2}A} ≤ y(b), contradicting the minimum property of y(b). So, a(t_{1}) < 0 and y(t_{1}) > 0.

Step 3. Lemma 2.1 applies to y on the interval [t_{1}, b], so there is some t_{3} with t_{1} < t_{3} < b, y(t_{3}) = 0, and y(t) > 0 for all t ∈ [t_{1}, t_{3}). A left-side version of Lemma 2.1 applies to y on the interval [0, t_{1}]; there are two cases:

Case 1. There is some t_{2} with 0 ≤ t_{2} < t_{1}, y(t_{2}) = 0, and y(t) > 0 for all t ∈ (t_{2}, t_{1}].

Case 2. y(t) > 0 for all t ∈ [0, t_{1}]. In this case denote t_{2} = 0.

In either case, there is some interval [t_{2}, t_{3}] where 0 ≤ t_{2} < t_{1} < t_{3} < b, y(t_{3}) = 0, y(t_{2}) ≥ 0, and y(t) > 0 for all t ∈ (t_{2}, t_{3}). Let y(t_{4}) be the maximum value of y on [t_{2}, t_{3}]. In Case 1, t_{2} < t_{4} < t_{3}, so the maximum occurs at an interior point and y′(t_{4}) = 0. In Case 2, t_{4} is either an interior point of [t_{2}, t_{3}], or the maximum occurs at the endpoint t_{4} = t_{2} = 0, where there is a right-side derivative y′(0) ≥ 0 as in Step 2. In either case, y′(t_{4}) ≥ 0.

Step 4. The MVT applies to y on [t_{4}, t_{3}]: there is some t_{5} with t_{4} < t_{5} < t_{3} and y′(t_{5}) = \frac{y(t_{3})−y(t_{4})}{t_{3}−t_{4}}. The MVT applies to y′ on [t_{4}, t_{5}]: there is some t_{6} with t_{4} < t_{6} < t_{5} and

y′′(t_{6}) = \frac{y′(t_{5}) − y′(t_{4})}{t_{5} − t_{4}} = \frac{\frac{y(t_{3})−y(t_{4})}{t_{3}−t_{4}} − y′(t_{4})}{t_{5} − t_{4}} < \frac{−y(t_{4})}{b^{2}}.

Using the lower bound for a and the property y(t_{6}) > 0,

Ky(t_{6}) ≤ a(t_{6})y(t_{6}) ≤ y′′(t_{6}) < \frac{−y(t_{4})}{b^{2}}
⟹ y(t_{6}) > \frac{−y(t_{4})}{b^{2}K} ≥ \frac{−y(t_{4})}{δ^{2}K} ≥ y(t_{4}),

contradicting the maximum property of y(t_{4}).

Theorem 2.7. Suppose a(t) and b(t) are real functions on [0, 1), and there is a point X such that 0 < X < 1 and a, b, and b′ are bounded on (0, X]. Then there is some δ with 0 < δ ≤ X, with the property that if y is continuous on [0, 1) with y(0) ≥ 0, lim_{t→0^{+}} y′(t) ≥ 0, and y′′(t) ≥ a(t)y(t) + b(t)y′(t) for

0 < t < X, then y(t) ≥ 0 for all 0 ≤ t ≤ δ.

Proof. By hypothesis, there are some A > 0 and K < 0 so that K ≤ a(t) ≤ A for t ∈ [0, X], and there are some B > 0 and L < 0 so that L ≤ b(t) ≤ B for t ∈ [0, X]. Let x = min\{X, \frac{1}{\sqrt{2A}}, \frac{1}{4B} \} > 0.

Recall the elementary calculus fact that if b is continuous and bounded on (0, x), then b is (Riemann) integrable on [0, x]. Let p(t) = ∫_{0}^{t} −\frac{1}{2}b(x)dx, so p is continuous on [0, x] and for 0 < t < x, p′(t) = −\frac{1}{2}b(t).

Let

f(t) = e^{p(t)} [y(t) − Ky(0)t^{2} − y(0)] ,

so f is continuous on [0, x] and f(0) = 0. For 0 < t < x,

f′(t) = e^{p(t)} [y′(t) − 2Ky(0)t]

+e^{p(t)}(−\frac{1}{2}b(t)) [y(t) − Ky(0)t^{2} − y(0)] ,

and using lim_{t→0^{+}} (y(t) − y(0)) = 0 and the boundedness of b, the limit exists:

lim_{t→0^{+}} f′(t) = lim_{t→0^{+}} y′(t) ≥ 0.

For 0 < t < x,

f′′(t) = e^{p(t)} [y′′(t) − 2Ky(0)] + e^{p(t)}(−\frac{1}{2}b(t)) [y′(t) − 2Ky(0)t]

+e^{p(t)}(−\frac{1}{2}b(t))^{2} [y(t) − Ky(0)t^{2} − y(0)]

+e^{p(t)}(−\frac{1}{2}b′(t)) [y(t) − Ky(0)t^{2} − y(0)]

+e^{p(t)}(−\frac{1}{2}b(t)) [y′(t) − 2Ky(0)t]

≥ e^{p(t)} [a(t)y(t) + b(t)y′(t) − 2Ky(0)] − e^{p(t)}b(t) [y′(t) − 2Ky(0)t]

+e^{p(t)}(−\frac{1}{2}b(t))^{2} [y(t) − Ky(0)t^{2} − y(0)]

+e^{p(t)}(−\frac{1}{2}b′(t)) [y(t) − Ky(0)t^{2} − y(0)]

= e^{p(t)} [y(t) − Ky(0)t^{2} − y(0)] (a(t) + \frac{1}{4}(b(t))^{2} − \frac{1}{2}b′(t))  (4)
+e^{p(t)}y(0) (K(a(t)t^{2} + 2b(t)t − 2) + a(t)) .  (5)

In the last step, the y′ terms cancel by construction. Term (4) is equal to ã(t)f(t), where ã(t) = (a(t) + \frac{1}{4}(b(t))^{2} − \frac{1}{2}b′(t)) is bounded by hypothesis. The upper bounds a(t) ≤ A and b(t) ≤ B and the initial choice of x imply, for 0 < t < x,

a(t)t^{2} + 2b(t)t − 2 ≤ At^{2} + 2Bt − 2
≤ A (\frac{1}{2A}) + 2B (\frac{1}{4B} ) − 2 = −1
⟹ K(a(t)t^{2} + 2b(t)t − 2) + a(t) ≥ −K + a(t) ≥ 0,

so the entire term (5) is non-negative, and for 0 < t < x, f′′(t) ≥ ã(t)f(t). Lemma 2.6 applies to f, so there is some δ_{1} depending on a, b, b′, X, but not on y, with f ≥ 0 on [0, δ_{1}]. The factor [y(t) − Ky(0)t^{2} − y(0)] is non-negative on the same interval, where

y(t) − Ky(0)t^{2} − y(0) ≥ 0 ⟹ y(t) ≥ y(0)(1 + Kt^{2}),

so y(t) ≥ 0 for 0 ≤ t ≤ δ = min\{δ_{1}, \frac{1}{\sqrt{−K}} \}.

2.2 A nonlinear differential inequality

Theorem 2.8. Given a function f that satisfies f′′f − (f′)^{2} ≥ 0 on (a, b), at every critical point c with f(c) ≠ 0, there is either a positive local min. or a negative local max.

Proof. Suppose c is a critical point, meaning f′(c) = 0. Suppose also that f(c) ≠ 0, so that the function g(x) = f′(x)/f(x) is defined on a neighborhood N = (c − δ, c + δ) ⊆ (a, b). By the quotient rule,

g′(x) = \frac{f(x)f′′(x) − (f′(x))^{2}}{(f(x))^{2}}

which is ≥ 0 on N by hypothesis. It follows that g(x) is weakly increasing on N. For c < x < c + δ, f′(x)/f(x) = g(x) ≥ g(c) = 0, and for c − δ < x < c, f′(x)/f(x) = g(x) ≤ g(c) = 0.

If f(c) > 0, then f(x) > 0 on N so f′(x) ≥ 0 on the right and f′(x) ≤ 0 on the left. f(c) is a local min. by the first derivative test.

If f(c) < 0, then f(x) < 0 on N so f′(x) ≤ 0 on the right and f′(x) ≥ 0 on the left. f(c) is a local max.

Note that C^{2} is not used, just the existence of f′′. Constant functions trivially satisfy both the hypotheses and conclusions.

Lemma 2.9. If p(x) satisfies p′′(x) ≥ 0 on (a, b) then for any c ∈ (a, b), p satisfies p(x) ≥ p(c) + p′(c)(x − c) for all x ∈ (a, b).

Proof. (See [C].)

Theorem 2.10. Given a function f that satisfies f′′f − (f′)^{2} ≥ 0 on (a, b), if there is a point c in (a, b) with f(c) > 0, then f satisfies

f(x) ≥ f(c) · exp(\frac{f′(c)}{f(c)}(x − c))

for all x ∈ (a, b).

Proof. By continuity, there is some neighborhood (s, t) ⊆ (a, b) so that s < c < t and f(x) > 0 on (s, t). Suppose f(z) = 0 for some z ∈ (c, b). Then, the set {t : f(x) > 0 on (c, t)} is non-empty and has sup = T ≤ z < b. By construction and using continuity again, f(T) = 0 and f(x) > 0 on (s, T).

Consider h(x) = ln(f(x)), which is well-defined on (s, T). h′ = f′/f = g, from Theorem 2.8, so h′′ = g′ ≥ 0 on (s, T). By Lemma 2.9, h(x) ≥ h(c) + h′(c) · (x − c) on (s, T):

ln(f(x)) ≥ ln(f(c)) + \frac{f′(c)}{f(c)} · (x − c)
f(x) ≥ f(c) · exp(\frac{f′(c)}{f(c)} · (x − c))

for all x in (s, T). This implies lim_{x→T^{−}} f(x) = f(T) > 0, which contradicts the

construction of T. We can conclude that f is never zero on (c, b), and always positive there, so the inequality holds on (s, b). The inequality on the other side of c follows from an analogous inf argument.

It follows that if c is a critical point with f(c) > 0, then f(c) is a global minimum. It further follows that either f is constant or there is at most one point c where f′(c) = 0 and f(c) > 0.

References

[C] A. Coffman, Notes on first semester calculus

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