The estimate of the upper bounds of eigenvalues
for a class of systems of ordinary differential equations with higher order is
considered by using the calculus theory. Several results about the upper bound
inequalities of the (n+1)th eigenvalue are obtained by the first n eigenvalues.
The estimate coefficients do not have any relation to the geometric measure of
the domain. This kind of problem is interesting and significant both in theory of
systems of differential equations and in applications to mechanics and physics.

1. Introduction

In many physical settings, such as the vibrations of the general homogeneous or nonhomogeneous string, rod and plate can yield the Sturm-Liouville eigenvalue problems or other eigenvalue problems. However, it is not easy to get the accurate values by the analytic method. Sometimes, it is necessary to consider the estimations of the eigenvalues. And since 1960s, the problems of the eigenvalue estimates had become one of the hotspots of the differential equations.

There are lots of achievements about the upper bounds of arbitrary eigenvalues for the differential equations and uniformly elliptic operators with higher orders [1–9]. However, there are few achievements associated with the estimates of the eigenvalues for systems of differential equations with higher order. In the following, we will obtain some inequalities concerning the eigenvalue λn+1 in terms of λ1,λ2,…,λn in the systems of ordinary differential equations with higher order. In fact, the eigenvalue problems have great strong practical backgrounds and important theoretical values [10, 11].

Let [a,b]⊂R1 be a bounded domain and t≥2 be an integer. The following eigenvalue problems are studied: (-1)tDt(a11Dty1+a12Dty2+⋯+a1nDtyn)=λs(x)y1,(-1)tDt(a21Dty1+a22Dty2+⋯+a2nDtyn)=λs(x)y2,⋮(-1)tDt(an1Dty1+an2Dty2+⋯+annDtyn)=λs(x)yn,Dkyi(a)=Dkyi(b)=0(i=1,2,…,n,k=0,1,2,…,t-1),
where D=d/dx,Dk=dk/dxk, aij(x)(i,j=1,2,…,n) and s(x) satisfies the following conditions:

aij(x)∈Ct[a,b],aij(x)=aji(x),i,j=1,2,…,n;

for the arbitrary ξ=(ξ1,ξ2,…,ξn)∈Rn, we have
μ1|ξ|2≤∑i,j=1naij(x)ξiξj≤μ2|ξ|2,∀x∈[a,b],
where μ2≥μ1>0, μ1,μ2 are both constants;

s(x)∈C[a,b], and there are constants ν1≤ν2, such that 0<ν1≤s(x)≤ν2.

According to the theories of the differential equations [11, 12], the eigenvalues of (1.1) are all positive real numbers, and they are discrete.

We change (1.1) to the form of matrix. Let yT=(y1y2⋮yn),DtyT=(Dty1Dty2⋮Dtyn),A(x)=(a11(x)a12(x)⋯a1n(x)a21(x)a22(x)⋯a2n(x)⋮an1(x)an2(x)⋯ann(x)).

By virtue of aij(x)=aji(x), therefore AT(x)=A(x), (1.1) can be changed into the following form: (-1)tDt(A(x)DtyT)=λs(x)yT,y(k)(a)=y(k)(b)=0,k=0,1,2,…,t-1.
Obviously, (1.4)-(1.5) is equivalent to (1.1).

Suppose that 0<λ1≤λ2≤⋯≤λn≤⋯ are eigenvalues of (1.4)-(1.5), y1,y2,…,yn,… are the corresponding eigenfunctions and satisfy the following weighted orthogonal conditions:∫abs(x)yiyjTdx=δij={1,i=j,0,i≠j,i,j=1,2,….

Multiplying yi in sides of (1.4), by using (1.5) and integration by parts, we have λi=∫abDtyiA(x)DtyiTdx,i=1,2,….
From (2°), we have ∫ab|Dtyi|2dx=∫abDtyiDtyiTdx≤λiμ1,i=1,2,….

For fixed n, let Φi=xyi-∑j=1nbijyj,i=1,2,…,n,
where bij=∫abxs(x)yiyjTdx. Obviously, bij=bji, and Φi are weighted orthogonal to y1,y2,…,yn. Furthermore, Φi(a)=Φi(b)=0, i,j=1,2,…,n.

We can use the well-known Rayleigh theorem [11, 12] to obtain λn+1≤(-1)t∫abΦiDt(A(x)DtΦiT)dx∫abs(x)|Φi|2dx.
It is easy to see that (-1)tDt(A(x)DtΦiT)=(-1)ttDt(A(x)Dt-1yiT)+(-1)ttDt-1(A(x)DtyiT)+(-1)txDt(A(x)DtyiT)-(-1)t∑j=1nbijDt(A(x)DtyjT)=(-1)ttDt(A(x)Dt-1yiT)+(-1)ttDt-1(A(x)DtyiT)+λixs(x)yiT-s(x)∑j=1nλjbijyjT.

We have ∫abΦi(-1)tDt(A(x)DtΦiT)dx=λi∫abxs(x)ΦiyiTdx+(-1)tt∫abΦiDt(A(x)Dt-1yiT)dx+(-1)tt∫abΦiDt-1(A(x)DtyiT)dx-∫abs(x)Φi∑j=1nλjbijyjTdx.

In addition, using the fact that Φi are weighted orthogonal to y1,y2,…,yn and ∫abs(x)|Φi|2dx=∫abxs(x)ΦiyiTdx,
we know that the last term of (1.12) is equal to zero. Thus, we have ∫abΦi(-1)tDt(A(x)DtΦiT)dx=λi∫abs(x)|Φi|2dx+(-1)tt∫abΦiDt(A(x)Dt-1yiT)dx+(-1)tt∫abΦiDt-1(A(x)DtyiT)dx.
Set Ii=(-1)tt∫abΦiDt(A(x)Dt-1yiT)dx,I=∑i=1nIi,Ji=(-1)tt∫abΦiDt-1(A(x)DtyiT)dx,J=∑i=1nJi.
From (1.14), we have ∑i=1n∫abΦi(-1)tDt(A(x)DtΦiT)dx=∑i=1nλi∫abs(x)|Φi|2dx+I+J.
By using (1.10) and (1.16), one can give λn+1∑i=1n∫abs(x)|Φi|2dx≤∑i=1nλi∫abs(x)|Φi|2dx+I+J.
Substituting λn for λi in (1.17), we get (λn+1-λn)∑i=1n∫abs(x)|Φi|2dx≤I+J.

In order to get the estimations of the eigenvalues, we only need to show the estimates about I,J, and ∑i=1n∫abs(x)|Φi|2dx.

2. LemmasLemma 2.1.

Suppose that the eigenfunctions yi of (1.4)-(1.5) correspond to the eigenvalues λi. Then one has

(1) By induction. If p=1, using integration by parts and the Schwarz inequality, we have
∫ab|Dyi|2dx≤|∫ab|Dyi|2dx|=|∫abDyiDyiTdx|=|∫abyiD2yiTdx|≤(∫ab|yi|2dx)1/2(∫ab|D2yiT|2dx)1/2≤ν1-1/2(∫ab|D2yiT|2dx)1/2.
Therefore, when p=1, (1) is true.

If for p=k, (1) is true, that is,
∫ab|Dkyi|2dx≤ν1-1/(k+1)(∫ab|Dk+1yi|2dx)k/(k+1).
For p=k+1, using integration by parts, the Schwarz inequality and the result when p=k, one can give
∫ab|Dk+1yi|2dx≤|∫ab|Dk+1yi|2dx|=|∫abDkyi⋅Dk+2yiTdx|≤(∫ab|Dkyi|2dx)1/2(∫ab|Dk+2yiT|2dx)1/2≤ν1-1/(2(k+1))(∫ab|Dk+2yi|2dx)1/2(∫ab|Dk+1yi|2dx)k/(2(k+1)).
By further calculating, one can give
∫ab|Dk+1yi|2dx≤ν1-1/((k+1)+1)(∫ab|D(k+1)+1yi|2dx)(k+1)/((k+1)+1).
Therefore, when p=k+1, (1) is true.

(2) Using (1) and the inductive method, we have
∫ab|Dpyi|2dx≤ν1-1/(p+1)(∫ab|Dp+1yi|2dx)p/(p+1)≤ν1-2/(p+2)(∫ab|Dp+2yi|2dx)p/(p+2)≤⋯≤ν1-(1-(p/t))(∫ab|Dtyi|2dx)p/t.
From (1.8) and (2.5), we get
∫ab|Dpyi|2dx≤ν1-(1-(p/t))(∫ab|Dtyi|2dx)p/t≤ν1-(1-(p/t))(λiμ1)p/t,p=1,2,…,t.
Taking p=1, we have
∫ab|Dyi|2dx≤ν1-(1-(1/t))(λiμ1)1/t.
So Lemma 2.1 is true.

Lemma 2.2.

Let λ1,λ2,…,λn be the eigenvalues of (1.4)-(1.5). Then one has
I+J≤t(2t-1)μ1-(1-(1/t))ν1-1/tμ2∑i=1nλi1-(1/t).

Proof.

Since
Ii=(-1)tt∫abΦiDt(A(x)Dt-1yiT)dx=(-1)tt∫ab(xyi-∑j=1nbijyj)Dt(A(x)Dt-1yiT)dx=(-1)tt∫abxyiDt(A(x)Dt-1yiT)dx-(-1)tt∑j=1nbij∫abyjDt(A(x)Dt-1yiT)dx=t2∫abDt-1yiA(x)Dt-1yiTdx+t∫abxDtyiA(x)Dt-1yiTdx-t∑j=1nbij∫abDtyj(A(x)Dt-1yiT)dx,Ji=(-1)tt∫abΦiDt-1(A(x)DtyiT)dx=-t(t-1)∫abDt-2yiA(x)DtyiTdx-t∫abxDt-1yiA(x)DtyiTdx+t∑j=1nbij∫abDt-1yjA(x)DtyiTdx,
we have
I+J=∑i=1n(Ii+Ji)=∑i=1nt∫ab(tDt-1yiA(x)Dt-1yiT-(t-1)Dt-2yiA(x)DtyiT)dx-t∑i,j=1nbij∫ab(DtyjA(x)Dt-1yiT-Dt-1yjA(x)DtyiT)dx.
By aij(x)=aji(x), the last term of (2.11) is zero. Then we can get
I+J=∑i=1nt∫ab(tDt-1yiA(x)Dt-1yiT-(t-1)Dt-2yiA(x)DtyiT)dx.
Using (2°), Lemma 2.1, (1) and (2.6), we have
∫abDt-1yiA(x)Dt-1yiTdx≤μ2∫ab|Dt-1yi|2dx≤μ2ν1-1/t(λiμ1)1-(1/t).
Using (2°), the Schwarz inequality, Lemma 2.1 (1), and (2.6), one can give
|-∫abDt-2yiA(x)DtyiTdx|≤μ2(∫ab|Dt-2yi|2dx)1/2(∫ab|DtyiT|2dx)1/2≤μ2ν1-1/t(λiμ1)1-(1/t).
Therefore, we obtain
I+J≤t(2t-1)μ1-(1-(1/t))ν1-1/tμ2∑i=1nλi1-(1/t).

Lemma 2.3.

If Φi and λi(i=1,2,…,n) as above, then one has
∑i=1n∫abs(x)|Φi|2dx≥μ11/tν12-(1/t)n24ν22(∑i=1nλi1/t)-1.

Proof.

By the definition of Φi, one has
∑i=1n∫abΦiDyiTdx=∑i=1n∫abxyiDyiTdx-∑i,j=1nbij∫abyjDyiTdx.
Using bij=bji and ∫abyjDyiTdx=-∫abyiDyjTdx, it is easy to see that the last term of (2.17) is zero. Then we have
∑i=1n∫abΦiDyiTdx=∑i=1n∫abxyiDyiTdx.
Using integration by parts, one can give
∫abxyiDyiTdx=-∫ab|yi|2dx-∫abxyiDyiTdx,∫abxyiDyiTdx=-12∫ab|yi|2dx.
By 1/ν2≤∫ab|yi|2dx≤1/ν1, we have
|∫abxyiDyiTdx|=12∫ab|yi|2dx≥12ν2.
From (2.18) and (2.21), we can get
∑i=1n|∫abΦiDyiTdx|≥n2ν2.
Using the Schwarz inequality, Lemma 2.1 (2), and (3°), we have
n24ν22≤(∑i=1n∫abs(x)|Φi|2dx)(∑i=1n∫ab|Dyi|2s(x)dx)≤(∑i=1n∫abs(x)|Φi|2dx)ν1-(2-(1/t))μ1-1/t∑i=1nλi1/t.
By further calculating, we can easily get Lemma 2.3.

3. Main ResultsTheorem 3.1.

If λi(i=1,2,…,n+1) are the eigenvalues of (1.4)-(1.5), then
(1)λn+1≤λn+4t(2t-1)μ2ν22μ1ν12n2∑i=1nλi1-(1/t)∑i=1nλi1/t;(2)λn+1≤(1+4t(2t-1)μ2ν22μ1ν12)λn.

Proof.

From (1.18), we can get
(λn+1-λn)≤(I+J)(∑i=1n∫abs(x)|Φi|2dx)-1.
Using Lemmas 2.2 and 2.3, we can easily get (3.1). In (3.1), Replacing λi with λn, by further calculating, we can get (3.2).

Theorem 3.2.

For n≥1, one has
∑i=1nλi1/tλn+1-λi≥μ1ν12n24t(2t-1)μ2ν22(∑i=1nλi1-(1/t))-1.

Proof.

Choosing the parameter σ>λn, using (1.17), one can give
λn+1∑i=1n∫abs(x)|Φi|2dx≤σ∑i=1n∫abs(x)|Φi|2dx+∑i=1n∫ab(λi-σ)s(x)|Φi|2dx+I+J.
By (2.22) and the Young inequality, we obtain
n2ν2≤δ2∑i=1n(σ-λi)∫abs(x)|Φi|2dx+12δ∑i=1n(σ-λi)-1∫ab|Dyi|2s(x)dx,
where δ>0 is a constant to be determined. Set
V=∑i=1n∫abs(x)|Φi|2dx,T=∑i=1n(σ-λi)∫abs(x)|Φi|2dx.
Using Lemma 2.1, (3.5), and (3.6), we can get the following results, respectively,
(λn+1-σ)V+T≤I+J,nν2≤δT+1δμ1-1/tν1-(2-(1/t))∑i=1n(σ-λi)-1λi1/t.
In order to get the minimum of the right of (3.9), we can take
δ=T-1/2(μ1-1/tν1-(2-(1/t))∑i=1n(σ-λi)-1λi1/t)1/2.
By (3.9), and (3.10), we can easily get
T≥μ11/tν12-(1/t)n24ν22(∑i=1nλi1/tσ-λi)-1.
Using Lemma 2.2, (3.8), and (3.11), we have
(λn+1-σ)V+μ11/tν12-(1/t)n24ν22(∑i=1nλi1/tσ-λi)-1≤t(2t-1)μ1-(1-(1/t))ν1-1/tμ2∑i=1nλi1-(1/t),
that is,
(λn+1-σ)V≤t(2t-1)μ1-(1-(1/t))ν1-1/tμ2∑i=1nλi1-(1/t)-μ11/tν12-(1/t)n24ν22(∑i=1nλi1/tσ-λi)-1.
Let the right term of (3.13) be f(σ). It is easy to see that
limσ→+∞f(σ)=-∞,limσ→λn+f(σ)=t(2t-1)μ1-(1-(1/t))ν1-1/tμ2∑i=1nλi1-(1/t)>0.
Hence, there is σ0∈(λn,+∞), such that
∑i=1nλi1/tσ0-λi=μ1ν12n24t(2t-1)μ2ν22(∑i=1nλi1-(1/t))-1.
On the other hand, letting
g(σ)=∑i=1nλi1/tσ-λi,
we have
g′(σ)=-∑i=1nλi1/t(σ-λi)2≤0.
It implies that g(σ) is the monotone decreasing and continuous function, and its value range is (0,+∞). Therefore, there exits exactly one σ0 to satisfy (3.15). From (3.13), we know that σ0>λn+1. Replacing σ0 with λn+1 in (3.15), we can get the result.

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