Estimates for Eigenvalues of Poly-Harmonic Operators

Conjecture 1.1 (Pólya)

If Ω is a bounded domain in ℝn, then the k-th eigenvalue λk of the fixed membrane problem satisfies

λ k ≥ 4 ⁢ π 2 ( B n ⁢ V ⁢ ( Ω ) ) 2 n ⁢ k 2 n , k = 1 , 2 , … .

Let Ω be a bounded domain in an n-dimensional Euclidean space ℝn. Assume that l≥2 and λi is the i-th eigenvalue of the eigenvalue problem (1.1). Then, the eigenvalues satisfy

When l=2, Theorem 1.2 reduces to the result of Cheng and Wei [5].

When l≥2, we give an important improvement of the result in (1.9) due to Cheng, Qi and Wei [3], since the inequality (1.10) is sharper than (1.9). We will give a proof of this fact in Section 2.

Let b≥2 be a positive real number and μ>0. If ψ:[0,+∞)→[0,+∞) is a decreasing function such that

- μ ≤ ψ ′ ⁢ ( s ) ≤ 0

and

A := ∫ 0 ∞ s b - 1 ⁢ ψ ⁢ ( s ) ⁢ 𝑑 s > 0 ,

then, for any positive integer l≥2, we have

Proof of Theorem 1.2.

In this proof, we will use the same notation as in [3]. Let φ^j⁢(z) be the Fourier transform of the trial function φj⁢(x) defined as

φ j ⁢ ( x ) = { u j ⁢ ( x ) , x ∈ Ω , 0 , x ∈ ℝ n ∖ Ω ,

where uj⁢(x) is an orthonormal eigenfunction corresponding to the eigenvalue λj, f⁢(z):=∑j=1k|φ^j⁢(z)|2, and f∗ is the symmetric decreasing rearrangement of f. We assume that A=kn⁢Bn, b=n, and ψ⁢(s)=ϕ⁢(s), where ϕ:[0,+∞)→[0,(2⁢π)-n⁢V⁢(Ω)] is a non-increasing function of |x| and ϕ⁢(x) is defined by ϕ⁢(|x|):=f∗⁢(x). Then, from Lemma 2.1 we can obtain that

Now, we define a function ξ⁢(t) as

Here, we assume that l≤n+1. The other cases (i.e., n+1<l<2⁢(n+1), l≥2⁢(n+1)) can be discussed by using a similar method. After differentiating (2.2) with respect to the variable t, we derive

Putting

ζ ⁢ ( t ) = ξ ′ ⁢ ( t ) ⁢ n + 2 ⁢ l B n ⁢ ( k B n ) - n + 2 ⁢ l n ⁢ t 2 ⁢ l n + 1

and noticing that μ≥(2⁢π)-n⁢Bn-1n⁢V⁢(Ω)n+1n, from (2.3) we have

Since the right-hand side of (2.4) is an increasing function of t, if the right-hand side of (2.4) is not larger than 0 at t=(2⁢π)-n⁢V⁢(Ω), that is,

we can claim from (2.5) that ξ′⁢(t)≤0 on (0,(2⁢π)-n⁢V⁢(Ω)]. If ξ′⁢(t)≤0, then ξ⁢(t) is a decreasing function on (0,(2⁢π)-n⁢V⁢(Ω)]. In fact, by a direct calculation, we can obtain

ζ ⁢ ( t ) ≤ - 2 ⁢ l + l ⁢ ( 2 ⁢ n - 2 ⁢ l + 2 ) 6 ⁢ n + l ⁢ ( 4 ⁢ n - 2 ⁢ l + 4 ) ⁢ ( n + 2 ⁢ ( l - 1 ) ) 2 144 ⁢ n 2 ⁢ ( n + 2 ⁢ l ) ≤ 0 ,

since

B n 4 n ( 2 ⁢ π ) 2 < 1 .

On the other hand, since 0<ϕ⁢(0)≤(2⁢π)-n⁢V⁢(Ω) and the right-hand side of (2.1) is ξ⁢(ϕ⁢(0)), which is a decreasing function of ϕ⁢(0) on (0,(2⁢π)-n⁢V⁢(Ω)], we can replace ϕ⁢(0) by (2⁢π)-n⁢V⁢(Ω) in (2.1), which gives the inequality

This completes the proof of the theorem. ∎

Proof of Remark 1.4.

Under the same assumptions as in Lemma 2.1, letting b=n and A=kn⁢Bn, from

μ ≥ ( 2 ⁢ π ) - n ⁢ B n - 1 n ⁢ V ⁢ ( Ω ) n + 1 n

we obtain that

( b ⁢ A ) - 2 b ⁢ ψ ⁢ ( 0 ) 2 + 2 b μ 2 ≤ ( 2 ⁢ π ) - 2 ⁢ n - 2 ⁢ V ⁢ ( Ω ) 2 + 2 n ( 2 ⁢ π ) - 2 ⁢ n ⁢ B n - 2 n ⁢ V ⁢ ( Ω ) 2 ⁢ ( n + 1 ) n ⁢ ( k B n ) - 2 n = ( 2 ⁢ π ) - 2 ( B n ) - 4 n ⁢ k - 2 n < ( 2 ⁢ π ) - 2 ( B n ) - 4 n < 1

and then we have

By a direct calculation, we derive

l ⁢ ( 6 ⁢ ( b + 2 ) - 1 ) ⁢ ( b + 2 ⁢ ( l - 1 ) ) 2 > 24 ⁢ b ⁢ ( b + 2 ⁢ l ) ⁢ ( l - 1 ) > 0 .

In fact,

that is,

Therefore, from (2.6) and (2.7) we get

Taking

and substituting (2.9) into (2.8), we have

This completes the proof of the remark. ∎