Green’s Function and a Lyapunov-Type Inequality for a Second-Order Boundary Value Problem with Robin Conditions

MSC 2020: 34B05, 34B27, 34C10, 34L15.

1Introduction

Lyapunov-type inequalities provide necessary conditions for the existence of nontrivial solutions of differential equations subject to prescribed boundary conditions. They have applications in oscillation theory, disconjugacy, and eigenvalue estimation. The classical Dirichlet inequality states that if

has a nontrivial solution, then

Green’s functions offer a systematic approach to deriving related inequalities because they represent solutions as integral operators whose norms can be estimated.

The purpose of this article is to provide a rigorous Green’s-function derivation for separated Robin conditions. We emphasize that the basic norm estimate proved here is a baseline result; its presence alone does not establish novelty relative to the existing literature.

Related literature includes work on Robin Lyapunov inequalities for elliptic and quasilinear problems [1, 2], a recent paper on generalized fractional Hill equations [3], and work on Lyapunov inequalities for general second-order elliptic equations [4].

2Problem and assumptions

Let a < b and L = b − a. We consider

where q ∈ C[a, b] and α, β, γ, δ are real constants. Throughout, assume:

(A1) q ∈ C[a, b];

(A2) α, β, γ, δ ≥ 0;

(A3)

The last condition is a nonresonance condition for the auxiliary problem −u′′ = f with the same boundary conditions. The nonnegative coefficient assumption is a sufficient condition for positivity of the Green’s function in the formula below; it is not necessary for all Robin problems.

A nontrivial solution means a function x ∈ C²[a, b], not identically zero, satisfying (1) and the boundary conditions.

3Green’s function

Consider the auxiliary boundary value problem

Define

Both functions solve u′′ = 0. The function ϕ satisfies the left boundary condition, and ψ satisfies the right boundary condition. Their Wronskian is

Theorem 1 (Green’s function). Under assumptions (A1)–(A3), the Green’s function for (3) is

Equivalently,

Moreover, G is continuous and symmetric on [a, b]², and G(t, s) ≥ 0.

Proof. For fixed s, the function G(·, s) is linear on either side of t = s, hence G_tt = 0 for t ≠ s. The two branches agree at t = s. Their derivative jump is

Thus −G_tt(t, s) = δ_s(t) in the distributional sense. The endpoint conditions follow because the left branch is a multiple of ϕ(t) at t = a, and the right branch is a multiple of ψ(t) at t = b. Since D > 0, the associated homogeneous problem has only the zero solution, so the Green’s function is unique. Finally, ϕ, ψ ≥ 0 on [a, b] by (A2), proving nonnegativity. Symmetry follows directly from the minimum–maximum formula. □

Theorem 2 (Green’s representation). If f ∈ C[a, b], the unique solution u of (3) is

Consequently, every solution of (1) satisfies

Proof. Let v(t) = ∫_a^b G(t, s)f(s) ds. From (4),

Differentiation, using cancellation of the diagonal terms in the first derivative and the derivative jump in the second, gives −v′′ = f. The endpoint conditions follow from those of ϕ and ψ. Uniqueness follows from D > 0. Taking f = qx yields (6). □

4Lyapunov-type inequalities

Let

Continuity ensures that this maximum exists. Under (A2)–(A3), M_G > 0.

Theorem 3 (Green’s-function Lyapunov inequality). If (1) has a nontrivial solution, then

A more informative necessary condition is

Proof. From (6) and G ≥ 0,

Since x is nontrivial, ${\Vert x\Vert }_{\infty }$ > 0. Choose t₀ ∈ [a, b] with |x(t₀)| = ${\Vert x\Vert }_{\infty }$. Dividing the preceding inequality at t = t₀ by ${\Vert x\Vert }_{\infty }$ gives

which proves (8). Since G(t₀, s) ≤ M_G, we obtain

and hence (7). □

Corollary 1 (Explicit Robin constant). Under (A1)–(A3),

Therefore, every nontrivial solution satisfies

Proof. Put m = min{t, s} and r = max{t, s}. Since ϕ is nondecreasing and ψ is nonincreasing, ϕ(m)ψ(r) ≤ ϕ(r)ψ(r). Thus the maximum of G is attained on the diagonal, which gives (9). Substitution in (7) proves (10). □

For an explicitly computable form, put y = t − a, so 0 ≤ y ≤ L, and

Then

If αγ > 0, the maximizing point is

If αγ = 0, the expression is linear or constant, so its maximum occurs at an endpoint unless it is constant.

Theorem 4 (Weighted eigenvalue estimate). Let w ∈ C[a, b] be nonnegative and not identically zero. If

with the same Robin boundary conditions has a nontrivial solution, then

Proof. Apply Theorem 3 with q(t) = λw(t). Since w ≥ 0,

The stated estimate follows. □

Theorem 5 (L^p-Lyapunov inequality). Let 1 < p < ∞ and p′ = p/(p − 1). If (1) has a nontrivial solution, then

Proof. Choose t₀ with |x(t₀)| = ${\Vert x\Vert }_{\infty }$. As in the proof of Theorem 3, using Hölder’s inequality,

Bounding the first factor by its maximum over t ∈ [a, b] proves (12). □

5Special cases

5.1Dirichlet boundary conditions

Taking α = γ = 1 and β = δ = 0 gives x(a) = x(b) = 0, D = L, and

The diagonal maximum is M_G = L/4. Therefore Theorem 3 gives

The classical Dirichlet Lyapunov inequality is usually stated with strict inequality under its standard hypotheses; the norm argument above alone yields the non-strict version.

5.2Dirichlet–Neumann conditions

For x(a) = 0 and x′(b) = 0, choose α = δ = 1 and β = γ = 0. Then D = 1 and G(t, s) = min{t, s} − a, so M_G = L. Thus

5.3Neumann–Dirichlet conditions

For x′(a) = 0 and x(b) = 0, choose β = γ = 1 and α = δ = 0. Then G(t, s) = b − max{t, s}, M_G = L, and

Remark 1. For pure Neumann conditions, the auxiliary operator −u′′ is resonant because constants belong to its kernel. In that case D = 0, so the Green’s function above does not apply without a compatibility condition or a modified inverse.

6Worked example

Consider

The Green’s function for the auxiliary problem is G(t, s) = min{t, s}, and M_G = 2. Hence any nontrivial solution must satisfy

If q(t) = λ is constant, this implies |λ| ≥ 1/4.

For positive λ = k², the equation has solutions x(t) = A sin(kt). The boundary condition x′(2) = 0 requires cos(2k) = 0, so

Thus the first positive eigenvalue is π²/16, which is consistent with, but stronger than, the necessary bound 1/4.

7Discussion and research directions

The results above establish a reusable baseline for a broader study. Several directions require further investigation:

1. 1. Sharp constants. Determine whether the Green’s-function constant can be improved under symmetry, monotonicity, or sign assumptions on q, and establish whether equality is possible.
2. 2. General Robin parameters. When some boundary coefficients are negative, positivity of G may fail. One can instead seek estimates based on |G|, while proving invertibility separately.
3. 3. Variable coefficients. Study −(p(t)x′(t))′ = λw(t)x(t) with p(t) > 0 and w(t) ≥ 0.
4. 4. Nonlinear problems. For x′′ + f(t, x) = 0, additional growth or sign assumptions are needed; the linear integral representation cannot be used unchanged.
5. 5. Comparison with the literature. Any proposed refinement should be compared with the existing literature, including the Robin inequalities and recent generalized-operator papers cited here.

8Conclusion

We constructed the Green’s function for a second-order equation with separated Robin boundary conditions under explicit nonresonance and positivity assumptions. The integral representation yields an unweighted Lyapunov-type inequality, a weighted eigenvalue estimate, and an L^p-version. Dirichlet and mixed conditions are recovered as special cases, while pure Neumann conditions require separate treatment because of resonance.

The main necessary condition is

This provides a rigorous starting point for research into sharper constants, optimality, and extensions to more general operators.

References

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2. [2]Kantar ÜD, Özden T. Lyapunov-type inequalities for quasilinear elliptic equations with Robin boundary condition. Journal of Inequalities and Applications. 2017;2017:45. https://doi.org/10.1186/s13660-017-1320-4
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