Georgian Mathematical Journal 1(1994), No. 1, 1-8

ON CONJUGACY OF HIGH-ORDER LINEAR ORDINARY DIFFERENTIAL EQUATIONS

T. CHANTURIA †

Abstract. It is shown that the differential equation

u^{(n)} = p(t)u, where n ≥ 2 and p : [a, b] → ℝ is a summable function, is not conjugate in the segment [a, b], if for some l ∈ {1, …, n − 1} , α ∈]a, b[ and β ∈]α, b[ the inequalities

n ≥ 2 + \frac{1}{2} (1 + (−1)^{n−l}), (−1)^{n−l}p(t) ≥ 0 for t ∈ [a, b],

∫_{α}^{β}(t − a)^{n−2}(b − t)^{n−2}|p(t)|dt ≥ l!(n − l)! \frac{(b − a)^{n−1}}{(b − β)(α − a)} ,

hold.

Consider the differential equation

u^{(n)} = p(t)u,  (1)

where n ≥ 2, p ∈ L_{loc}(I), and I ⊂ ℝ is an interval.

The following definitions will be used below.

Equation (1) is said to be conjugate in I if there exists a nontrivial solution of this equation with at least n zeroes (each zero counted accordingly to its multiplicity) in I.

Let l ∈ {1, …, n − 1}. Equation (1) is said to be (l, n − l) conjugate in I if there exists a nontrivial solution u of this equation satisfying

u^{(i)}(t_{1}) = 0 (i = 0, …, l − 1),
u^{(i)}(t_{2}) = 0 (i = 0, …, n − l − 1),

with t_{1}, t_{2} ∈ I and t_{1} < t_{2}.

Suppose first that −∞ < a < b < +∞ and p ∈ L([a, b]).

1991 Mathematics Subject Classification. 34C10.

2 T. CHANTURIA

Lemma. Let a < α < β < b. Then the Green’s function G of the problem

u^{(n)}(t) = 0 for t ∈ [a, b],
u^{(j)}(a) = 0 (j = 0, …, l − 1),
u^{(j)}(b) = 0 (j = 0, …, n − l − 1),

satisfies the inequality (−1)^{n−l}G(t, s) >

> \frac{(b − β)(α − a)(s − a)^{n−l−1}(b − s)^{l−1}(t − a)^{l−1}(b − t)^{n−l−1}}{(b − a)^{n−1}} ×

× ∑_{i=1}^{n−l} \frac{(−1)^{n−l−i}}{(i − 1)!(n − i)!} for α ≤ t < s ≤ β.  (2)

Proof. The function G can be written in the form

G(t, s) = \{ \begin{matrix}∑_{i=n−l+1}^{n}(−1)^{i−1}x_{i}(t)x_{n−i+1}(s) & for a ≤ s < t ≤ b, \\ − ∑_{i=1}^{n−l}(−1)^{i−1}x_{i}(t)x_{n−i+1}(s) & for a ≤ t ≤ s ≤ b, \end{matrix}

where

x_{i}(t) = \frac{(t − a)^{n−i}(b − t)^{i−1}}{(i − 1)!(b − a)^{n−i}} .

It is easy to verify that for any fixed s ∈]a, b[ the function \frac{(−1)^{n−l}G(·,s)}{x_{n−l}(·)x_{l+1}(s)} decreases on ]a, b[ and the function \frac{(−1)^{n−l}G(·,s)}{x_{n−l+1}(·)x_{l}(s)} increases on ]a, b[. Thus

(−1)^{n−l}G(t, s) ≥ (−1)^{n−l}G(s, s)\frac{x_{n−l}(t)}{x_{n−l}(s)} for t ≤ s.  (3)

Taking into account that

(−1)^{n−l}G(s, s) = (−1)^{n−l−1} ∑_{i=1}^{n−l}(−1)^{i−1}x_{i}(s)x_{n−i+1}(s) =

= \frac{(s − a)^{n−1}(b − s)^{n−1}}{(b − a)^{n−1}} ∑_{i=1}^{n} \frac{(−1)^{n−l−i}}{(i − 1)!(n − i)!}

and

\frac{x_{n−l}(t)}{x_{n−l}(s)} = \frac{(t − a)^{l}(b − t)^{n−l−1}}{(s − a)^{l}(b − s)^{n−l−1}} ,

from the inequality (3) we deduce

(−1)^{n−l}G(t, s) ≥

≥ \frac{(s − a)^{n−l−1}(b − s)^{l}(t − a)^{l}(b − t)^{n−l−1}}{(b − a)^{n−1}} ∑_{i=1}^{n−l} \frac{(−1)^{n−l−i}}{(i − 1)!(n − i)!} >

> (b − β)(α − a)\frac{(s − a)^{n−l−1}(b − s)^{l−1}(t − a)^{l−1}(b − t)^{n−l−1}}{(b − a)^{n−1}} ×

× ∑_{i=1}^{n−l} \frac{(−1)^{n−l−i}}{(i − 1)!(n − i)!} for α ≤ t < s ≤ β.

Theorem 1. Let l ∈ {1, …, n − 1},

n ≥ 2 + \frac{1 + (−1)^{n−l}}{2} and (−1)^{n−l}p(t) ≥ 0 for t ∈ [a, b].  (4)

If, in addition, there exist α, β ∈]a, b[ such that a < α < β < b and

∫_{α}^{β}(t − a)^{n−2}(b − t)^{n−2}|p(t)|dt ≥ l!(n − l)!\frac{(b − a)^{n−1}}{(b − β)(α − a)},  (5)

then Eq. (1) is (l, n − l) conjugate in [a, b].

Note that analogous results are given in [3,5] for the case where n = 2.

Proof. Put p(t) = 0 for t > b and consider Eq. (1) in the interval [a, +∞[. For any γ > a, let u_{γ} be the solution of (1) satisfying

u_{γ}^{(i)}(a) = 0 (i = 0, …, l − 1),

u_{γ}^{(i)}(γ) = 0 (i = 0, …, n − l − 2),

∑_{i=0}^{n−1} |u_{γ}^{(i)}(a)| = 1, max{u_{γ}(t) : a ≤ t ≤ γ} > 0.

Suppose now that in spite of the statement of the theorem Eq. (1) is not (l, n − l) conjugate in [a, b].

Note that if γ ∈]a, b], then u_{γ}(t) > 0 for t ∈]a, γ[ and (−1)^{n−l−1}× ×u_{γ}^{(n−l−1)}(γ) > 0. Indeed, if it is not so, there exists t_{0} ∈]a, γ[ such that u_{γ}(t_{0}) = 0. Let γ_{0} = inf{γ > a : u_{γ}(t) = 0 for a certain t ∈]a, γ[}. Then u_{γ_{0}}(t) > 0 for t ∈]a, γ_{0}[ and

u_{γ_{0}}^{(i)}(a) = 0 (i = 0, …, l − 1),
u_{γ_{0}}^{(i)}(γ_{0}) = 0 (i = 0, …, n − l − 1),

which contradicts our assumption.

Let γ^{0} = sup{γ > b : u_{γ}(t) > 0 for t ∈]a, γ[}. Consider first the case where γ^{0} = +∞. There exists the sequence {γ_{k}}_{k=1}^{+∞} such that

lim_{k→+∞} γ_{k} = +∞, lim_{k→+∞} u_{γ_{k}}(t) = u_{0}(t)

where u_{0} is the solution of Eq. (1). Show that

u_{0}(t) > 0 for t > a.  (6)

It is clear that u_{0}(t) ≥ 0 for t > a. If now u_{0}(t_{∗}) = 0 for some t_{∗} > a, then for any k large enough the function u′_{γ_{k}} will have at least one zero in ]a, γ_{k}[. Taking into account the multiplicities of zeroes of u_{γ_{k}} in a and γ_{k}, it is easy to show that u_{γ_{k}}^{(n−1)} has at least two zeroes in ]a, γ_{k}[. Hence u_{γ_{k}}^{(n)} changes sign in this interval and this is impossible.

Thus inequality (6) is proved. This inequality and the results of [1] imply that there exist l_{0} ∈ {1, …, n} (l − l_{0} is even) and t_{1} > b such that

\begin{matrix}u_{0}^{(i)}(t) > 0 & for t ≥ t_{1} (i = 0, …, l_{0} − 1), \\ (−1)^{i+l_{0}}u_{0}^{(i)}(t) ≥ 0 & for t ≥ t_{1} (i = l_{0}, …, n). \end{matrix}  (7)

Clearly,

\begin{matrix}(−1)^{i+l_{0}}u_{0}^{(i)}(t) ≥ 0 for t ≥ a (i = l_{0}, …, n), \\ (−1)^{i+l_{0}}u_{0}^{(i)}(a) > 0  (i = l_{0}, …, n − 1). \end{matrix}  (8)

Hence l ∈ {1, …, l_{0}}.

Suppose that l < l_{0}. Then for any k large enough we have γ_{k} > t_{1}, u_{γ_{k}}^{(i)}(t_{1}) > 0(i = 0, …, l_{0} − 1). This means that the function u_{γ_{k}}^{(i)} has at least one zero in ]t_{1}, γ_{k}[. Taking into account the multiplicity of zero in γ_{k}, it is easy to see that u_{γ_{k}}^{(n−1)} has at least two zeroes in ]t_{1}, γ_{k}[, and u_{γ_{k}}^{(n)} changes sign in this interval. But this is impossible. Thus l = l_{0}.

As l = l_{0}, inequalities (7) and (8) imply

(−1)^{i+l}u_{0}^{(i)}(t) ≥ 0 for t ≥ a (i = l, …, n),
u_{0}^{(i)} > 0 for t > a (i = 0, …, l − 1).

Let

v(t) = u_{0}^{(l−1)}(t) − ∑_{j=l}^{n−1} \frac{(−1)^{j−l}}{(j − l + 1)!}(t − a)^{j−l+1}u_{0}^{(j)}(t);

then

v′(t) = \frac{(−1)^{n−l}}{(n − l)!} (t − a)^{n−l}u_{0}^{(n)}(t) ≥ 0 for t ≥ a.

Hence

\begin{matrix}u_{0}^{(l−1)}(t) ≥ ∑_{j=l}^{n−1} \frac{(−1)^{j−l}(t − a)^{j−l+1}u_{0}^{(j)}(t)}{(j − l + 1)!} ≥ \\ ≥ \frac{(t − a)(−1)^{n−l}}{(n − l)!} ∫_{t}^{+∞}(s − a)^{n−l−1}p(s)u_{0}(s)ds for t ≥ a. \end{matrix} (9)

Denote ρ_{i}(t) = iu_{0}^{(l−i)}(t) − (t − a)u_{0}^{(l−i+1)}(t) for t ≥ a (i = 0, 1, …, l). Then ρ′_{i}(t) = ρ_{i−1}(t) for t ≥ a (i = 1, …, l). Since ρ_{0}(t) = −(t − a)u_{0}^{(l+1)}(t) ≥ 0 for t ≥ a and ρ_{i}(a) = 0 (i = 1, …, l), we have ρ_{i}(t) ≥ 0 for t ≥ a (i = 0, 1, …, l). This implies

u_{0}(t) ≥ \frac{(t − a)^{l−1}}{l!} u_{0}^{(l−1)}(t).  (10)

From (9) and (10) we obtain

1 ≥ \frac{(t − a)}{l!(n − l)!} ∫_{t}^{β}(s − a)^{n−2}|p(s)|ds for t ≥ a,

which contradicts (5). The case γ^{0} = +∞ is thus eliminated.

Now consider the case where γ^{0} < +∞. As we have already noted, γ^{0} > b, u_{γ^{0}}(t)>0 for t ∈]a, γ^{0}[ and

\begin{aligned}u_{γ^{0}}^{(i)}(a) & = 0 (i = 0, …, l − 1), \\ u_{γ^{0}}^{(i)}(γ^{0}) & = 0 (i = 0, …, n − l − 1). \end{aligned}  (11)

Hence

u_{γ^{0}}(t) = ∫_{a}^{γ^{0}}G(t, s)p(s)u_{γ^{0}}(s)ds,

where G is the Green’s function of the boundary value problem (11) for the equation u^{(n)} = 0.

Let t_{0} ∈]α, β[ be such that

\frac{u_{γ^{0}}(t)}{(t − a)^{l−1}(γ^{0} − t)^{n−l−1}} ≥ \frac{u_{γ^{0}}(t_{0})}{(t_{0} − a)^{l−1}(γ^{0} − t_{0})^{n−l−1}}

for t ∈ [α, β].  (12)

Then from the lemma and the inequality (12) it follows that

\begin{aligned}u_{γ^{0}}(t_{0}) & ≥ (γ^{0} − β)(α − a) ∑_{i=1}^{n−l} \frac{(−1)^{n−l−i}}{(i − 1)!(n − i)!} × \\ & × ∫_{α}^{β} \frac{(s − a)^{n−2}(γ^{0} − s)^{n−2}}{(γ^{0} − a)^{n−1}} |p(s)|dsu_{γ^{0}}(t_{0}) > \\ & > \frac{(b − β)(α − a)}{(b − a)^{n−1}} ∑_{i=1}^{n−l} \frac{(−1)^{n−l−i}}{(i − 1)!(n − i)!} × \\ & × ∫_{α}^{β}(s − a)^{n−2}(b − s)^{n−2}|p(s)|dsu_{γ^{0}}(t_{0}). \end{aligned}  (13)

Since ∑_{i=1}^{n−l} \frac{(−1)^{n−l−i}}{(i−1)!(n−i)!} ≥ \frac{1}{l!(n−l)!} the inequality (13) contradicts (5).

Denote

μ_{n}^{−} = min{l!(n − l)! : l ∈ {1, …, n − 1}, n − l odd},
μ_{n}^{+} = min{l!(n − l)! : l ∈ {1, …, n − 1}, n − l even}.

It is clear that

μ_{n}^{−} = \{ \begin{matrix}(\frac{n}{2} − 1)!(\frac{n}{2} + 1)! & for n ≡ 0 (mod 4), \\ [(\frac{n}{2})!]^{2} & for n ≡ 2 (mod 4), \\ (\frac{n − 1}{2})!(\frac{n + 1}{2})! & for n ≡ 1 (mod 2), \end{matrix}

μ_{n}^{+} = \{ \begin{matrix}[(\frac{n}{2})!]^{2} & for n ≡ 0 (mod 4), \\ (\frac{n}{2} − 1)!(\frac{n}{2} + 1)! & for n ≡ 2 (mod 4), \\ (\frac{n − 1}{2})!(\frac{n + 1}{2})! & for n ≡ 1 (mod 2). \end{matrix}

Corollary 1. Let either n ≥ 2, μ_{n} = μ_{n}^{−}, and p(t) ≤ 0 for t ∈ [a, b] or n≥3, μ_{n} = μ_{n}^{+}, and p(t) ≥ 0 for t ∈ [a, b]. Let, moreover, α, β ∈]a, b[ exist such that a < α < β < b and

∫_{α}^{β}|p(t)|dt ≥ μ_{n}(\frac{b − a}{(b − β)(α − a)})^{n−1}.

Then Eq. (1) is conjugate in [a, b].

Note that max{l!(n − l)! : l ∈ {1, …, n − 1}} = (n − 1)!. Thus from Theorem 1 easily follows

Corollary 2. Let l ∈ {1, …, n − 1}, let the conditions (4) be fulfilled, and let α, β ∈]a, b[ exist such that a < α < β < b and

∫_{α}^{β}|p(t)|dt ≥ (n − 1)!(\frac{b − a}{(b − β)(α − a)})^{n−1}.  (14)

Then Eq. (1) is (l, n − l) conjugate in [a, b].

Note that in the inequality (14) the factor (n − 1)! cannot be replaced by (n − 1)! − ε with ε ∈]0, 1[. This is shown by the following

Example. Let ε ∈]0, 1[ be given beforehand and choose α ∈]a, b[ and β ∈]α, b[ such that

(n − 1)!(\frac{α − a}{β − a})^{n−1} (\frac{b − β}{b − a} )^{n−1} > (n − 1)! − ε.  (15)

Put

v(t) = \begin{cases}t − a & for t ∈ [a, α], \\ \frac{α + β}{2} − a − \frac{1}{2(β − α)}(t − β)^{2} & for t ∈]α, β[, \\ \frac{α + β}{2} − a & for t ∈ [β, b], \end{cases}

u_{0}(t) = \frac{1}{(n − 3)!} ∫_{a}^{t}(t − s)^{n−3}v(s)ds for t ∈ [a, b], n ≥ 3,

u_{0}(t) = v(t) for t ∈ [a, b], n = 2,

and

p(t) = \frac{v′′(t)}{u_{0}(t)} for a < t < b.

Then the function u_{0} is non-decreasing for t ∈ [a, β], and the inequality

u_{0}(t) ≤ u_{0}(β) ≤ \frac{1}{(n − 3)!} ∫_{a}^{β}(β − s)^{n−3}(s − a)ds = \frac{(β − a)^{n−1}}{(n − 1)!}

is valid. Taking into account the inequality (15), we obtain

∫_{α}^{β} p(t)dt = \frac{1}{β − α} ∫_{α}^{β} \frac{dt}{u_{0}(t)} > \frac{(n − 1)!}{(β − a)^{n−1}} >

> ((n − 1)! − ε)(\frac{b − a}{(b − β)(α − a)})^{n−1}.

On the other hand, in the case considered, Eq. (1) is not conjugate in [a, b] because it has a solution u_{0} satisfying the following conditions:

u_{0}^{(i)}(a) = 0 (i = 0, …, n − 2), u_{0}^{(n−1)}(a) = 1, u_{0}(t) > 0 for a < t ≤ b.

This example shows that in Corollary 2 inequality (14) cannot be replaced by the inequality

∫_{α}^{β} |p(t)|dt ≥ ((n − 1)! − ε)(\frac{b − a}{(b − β)(α − a)})^{n−1}

no matter how small ε > 0 is.

Now consider Eq. (1) on the whole axis ℝ with p ∈ L_{loc}(ℝ). From Corollary 2 easily follows

Corollary 3. Let l ∈ {1, …, n − 1}, p is not zero on the set of the positive measure and

n ≥ 2 + \frac{1 + (−1)^{n−l}}{2} , (−1)^{n−l}p(t) ≥ 0 for t ∈ ℝ.

Then Eq. (1) is (l, n − l) conjugate in ℝ.

References

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(Received 1.07.1992)^{1}

1The original manuscript was prepared for publication by D. Paatashvili