Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 18, 1-9; http://www.math.u-szeged.hu/ejqtde/

A generalized Fucik type eigenvalue problem for p-Laplacian

Yuanji Cheng

School of Technology Malmö University, SE-205 06 Malmö, Sweden Email: yuanji.cheng@mah.se

Abstract

In this paper we study the generalized Fucik type eigenvalue for the boundary value problem of one dimensional p−Laplace type differential equations

\{ \begin{matrix}−(φ(u′))′ = ψ(u), −T < x < T; \\ u(−T) = 0, u(T) = 0 \end{matrix}  (∗)

where φ(s) = αs_{+}^{p−1} − βs_{−}^{p−1}, ψ(s) = λs_{+}^{p−1} − μs_{−}^{p−1}, p > 1. We obtain a explicit characterization of Fucik spectrum (α, β, λ, μ), i.e., for which the (*) has a nontrivial solution.

(1991) AMS Subject Classification: 35J65, 34B15, 49K20.

1 Introduction

In the study of nonhomogeneous semilinear boundary problem

\{ \begin{matrix}−Δu = f(u) + g(x), & in Ω \\ u = 0, & on ∂Ω \end{matrix}

it has been discovered in [3, 7] that the solvability of another boundary value problem

\{ \begin{matrix}−Δu = λu_{+} − μu_{−}, & in Ω \\ u = 0, & on ∂Ω \end{matrix}

where u_{+} = max{u, 0}, u_{−} = max{−u, 0} plays an important role. Since then there are many works devoted to this subject [5, 13, 14, references therein], and the study has also been extended to the p-Laplacian

\{ \begin{matrix}−Δ_{p}u = λu_{+}^{p−1} − μu_{−}^{p−1}, & in Ω \\ u = 0, & on ∂Ω \end{matrix}

EJQTDE, 2009 No. 18, p. 1 where Δ_{p} =: div{|∇u|^{p−2}∇u}, p > 1 [2, 4, 6, 10, 12] and even associated trigonometrical p−sine and cosine functions have been studied [9]. In this paper, we are interested in generalization of such Fucik spectrum and will consider one dimensional boundary value problem

\{ \begin{matrix}−(α(u′)_{+}^{p−1} − β(u′)_{−}^{p−1}))′ = λu_{+}^{p−1} − μu_{−}^{p−1}, −T < x < T \\ u(−T) = 0, u(T) = 0 \end{matrix}  (1.1)

where α, β, λ, μ > 0 are parameters, and call (α, β, λ, μ) the generalized Fucik spectrum, if (1.1) has a non-trivial solution. The problem is motivated by the study of two-point boundary value problem

\{ \begin{matrix}−(φ(u′)))′ = ψ(x, u), −T < x < T \\ u(−T) = 0, u(T) = 0 \end{matrix}  (1.2)

and to our knowledge it is always assumed in the literature that φ is an odd function. Thus a natural question arises: what would happen, if the function φ is merely a homeomorphism, not necessarily odd function on R ? Here we shall first investigate the autonomous eigenvalue type problem and in the forthcoming treat non-resonance problem.

By a solution of (1.2) we mean that u(x) is of C^{1} such that φ(u′(x)) is differentiable and the equation (1.2) is satisfied pointwise almost everywhere. The main results of this paper are complete characterization of Fucik type eigenvalues, their associated eigenfunctions and observations of changes of frequency, amplitude of solutions, when they pass the mini- and maximum points respective change their signs (see the figures below and (3.8) in details). Let π_{p} = \frac{2π}{p sin(π/p)}, then we have

Theorem 1 (α, β, λ, μ) belongs to the generalized Fucik spectrum of (1.1), if and only if for some integer k ≥ 0

1) (\sqrt[p]{α} + \sqrt[p]{β})((k + 1)\sqrt[p]{λ^{−1}} + k\sqrt[p]{μ^{−1}})π_{p} = 2T and corresponding eigenfunction u is initially positive and has precisely 2k nodes.

2) (\sqrt[p]{α} + \sqrt[p]{β})(k\sqrt[p]{λ^{−1}} + (k + 1)\sqrt[p]{μ^{−1}})π_{p} = 2T and the corresponding eigenfunction u is initially negative and has also 2k nodes

3) (k+1)(\sqrt[p]{α}+\sqrt[p]{β})(\sqrt[p]{λ^{−1}}+\sqrt[p]{μ^{−1}})π_{p} = 2T and the corresponding eigenfunction u_{1}, u_{2} has exact 2k + 1 nodes and u_{1} is initially positive and u_{2} is negative.

Moreover, the eigenfunctions are piecewise p−sine functions (see part 2 for definitions).

2 Review on 1D p−Laplacian

We shall review some basic results about eigenvalues and associated eigenfunctions for one dimensional p−Laplacian. Eigenfunctions are also called p−sine and -cosine functions, sin_{p}(x), cos_{p}(x), which have been discussed in details in [9, 11], but for our purpose we adopt the version in [1]. Let π_{p} = \frac{2π}{p sin \frac{π}{p}} , the p−sine, p−cosine functions sin_{p}(x), cos_{p}(x) are defined via

x = ∫_{0}^{sin_{p}(x)} \frac{dt}{\sqrt[p]{1 − t^{p}}} , 0 ≤ x ≤ π_{p}/2  (2.1)

and extended to [π_{p}/2, π_{p}] by sin_{p}(π_{p}/2+x) = sin_{p}(π_{p}/2−x) and to [−π_{p}, 0] by sin_{p}(x) = − sin_{p}(−x) then finally extended to a 2π_{p} periodic function on the whole real line; Then p−cosine function is defined as cos_{p} x = \frac{d}{dx}(sin_{p} x) and they have the properties:

sin_{p} 0 = 0, sin_{p} π_{p}/2 = 1; cos_{p} 0 = 1, cos_{p} π_{p}/2 = 0.

They share several remarkable relations as ordinary trigonometric functions, for instance

| sin_{p} x|^{p} + | cos_{p} x|^{p} = 1.

But

\frac{d}{x}(cos_{p} x) = −| tan_{p} x|^{p−2} sin_{p} x ≠ sin_{p} x, where tan_{p} x = sin_{p} x/ cos_{p} x.

The eigenvalues of one dimensional p-Laplace operator

\{ \begin{matrix}−(|u′(x)|^{p−2}u′(x))′ = λ|u(x)|^{p−2}u(x), 0 ≤ x ≤ π_{p} \\ u(0) = 0, u(π_{p}) = 0 \end{matrix}  (2.2)

are 1^{p}, 2^{p}, 3^{p}, ⋯ and the corresponding eigenfunctions are precisely

sin_{p}(x), sin_{p}(2x), sin_{p}(3x), ⋯

and therefore we have another relation between p−cosine and sine functions

−(| cos_{p}(kx)|^{p−2} cos_{p}(kx))′ = k| sin_{p}(kx)|^{p−2} sin_{p}(kx), k = 1, 2, ⋯

EJQTDE, 2009 No. 18, p. 3 Here we note that function sin_{p}(x) is a solution to the following problem

\{ \begin{matrix}−(|u′(x)|^{p−2}u′(x))′ = |u(x)|^{p−2}u(x), 0 < x < π_{p}/2 \\ u(0) = 0, u′(π_{p}/2) = 0 \end{matrix}  (2.3)

and any solution is of form C sin_{p}(x) for some constant C.

If the p−Laplacian is associated to another interval [a, b], different from [0, π_{p}], then by change of variables we see that eigenvalues and the associated eigenfunctions are

(\frac{kπ_{p}}{b − a})^{p}, sin_{p}(k \frac{x − a}{b − a}π_{p}), k = 1, 2, 3, · · · .  (2.4)

3 Proof of Theorem 1

To understand the generalized Fucik spectrum of (1.1), we need to examine the following Dirichlet-Neumann boundary value problem

\{ \begin{matrix}−(α(u′)_{+}^{p−1} − β(u′)_{−}^{p−1}))′ = λu_{+}^{p−1} − μu_{−}^{p−1}, a < x < b \\ u(a) = 0, u′(b) = 0 \end{matrix}  (3.1)

We shall focus only on the constant sign solutions of (3.1) and note that there exist essentially only ’two’ solutions, one positive and another negative, due to the positive homogeneity of (3.1).

If u(x) is a positive solution of (3.1), then u must be increasing on [a, b), because the equation (3.1) says that the function g(x) =: φ(u′(x)) is decreasing on (a, b] and satisfies g(b) = 0, thus g(x) is positive for all x ∈ [a, b) and thus u′(x) has to be positive, due to strict monotonicity of function φ(s) and φ(0) = 0. It follows that u satisfies

\{ \begin{matrix}−(|u′(x)|^{p−2}u′(x))′ = \frac{λ}{α}|u(x)|^{p−2}u(x), a < x < b \\ u(a) = 0, u′(b) = 0 \end{matrix}  (3.2)

It follows from (2.3) that u(x) = C sin_{p}(\frac{x−a}{2} \sqrt[p]{\frac{λ}{α}}) and α, λ satisfy π_{p}\sqrt[p]{α/λ} = b − a. Likely if u is negative solution to (3.1), then

\{ \begin{matrix}−(|u′(x)|^{p−2}u′(x))′ = \frac{μ}{β}|u(x)|^{p−2}u(x), a < x < b \\ u(a) = 0, u′(b) = 0 \end{matrix}  (3.3)

u(x) = −C sin_{p}(\frac{x−a}{2} \sqrt[p]{\frac{μ}{β}}) and β, μ satisfy π_{p}\sqrt[p]{β/μ} = b − a.

In analogy we see that for the following boundary value problem

\{ \begin{matrix}−(α(u′)_{+}^{p−1} − β(u′)_{−}^{p−1}))′ = λu_{+}^{p−1} − μu_{−}^{p−1}, a < x < b \\ u′(a) = 0, u(b) = 0 \end{matrix}  (3.4)

the positive and negative solutions are u(x) = D sin_{p}(\frac{b−x}{b−a} \frac{π_{p}}{2} ) respectively u(x) = −D sin_{p}(\frac{b−x}{b−a} \frac{π_{p}}{2} ) and α, β, λ, μ satisfy π_{p}\sqrt[p]{β/λ} = b − a or π_{p}\sqrt[p]{α/μ} = b − a.

It follows from the above analysis that if u is a positive solution to (1.1) and u(x_{0}) = max u(x) := C, then x_{0} is determined by

\sqrt[p]{α/λ}π_{p} = x_{0} + T

and α, β, λ, μ satisfy

L^{+} := (\sqrt[p]{α/λ} + \sqrt[p]{β/λ})π_{p} = 2T.

Furthermore the solution u(x) is given by

u(x) = \{ \begin{matrix}C sin_{p}(\frac{x+T}{2} \sqrt[p]{\frac{λ}{α}}), & −T ≤ x ≤ −T + π_{p}\sqrt[p]{\frac{α}{λ}} \\ C sin_{p}(\frac{T −x}{2} \sqrt[p]{\frac{λ}{β}}), & T − π_{p}\sqrt[p]{\frac{β}{λ}} ≤ x ≤ T \end{matrix}  (3.5)

and for the negative solution u of (1.1), then it holds

L^{−} := (\sqrt[p]{α/μ} + \sqrt[p]{β/μ})π_{p} = 2T

u(x) = \{ \begin{matrix}−D sin_{p}(\frac{x+T}{2} \sqrt[p]{\frac{μ}{β}}), & −T ≤ x ≤ −T + π_{p}\sqrt[p]{\frac{β}{μ}} \\ −D sin_{p}(\frac{T −x}{2} \sqrt[p]{\frac{μ}{α}}), & T − π_{p}\sqrt[p]{\frac{α}{μ}} ≤ x ≤ T \end{matrix}  (3.5′)

It is clear from (3.5) that u(x) changes its frequency, when it passes its maximum point and is not symmetric anymore, which is in contrast to the symmetry principle of Gidas, Ni and Nirenberg [8].

For an initially positive nodal solution u to (1.1) with only one node at T_{1}, we get that T_{1}, α, β, λ, ν satisfy

(\sqrt[p]{α/λ} + \sqrt[p]{β/λ})π_{p} = T_{1} + T.
(\sqrt[p]{α} + \sqrt[p]{β})(\sqrt[p]{λ^{−1}} + \sqrt[p]{μ^{−1}})π_{p} = 2T  (3.6)

If C = max_{x∈[−T,T_{1}]} u(x), −D = max_{x∈[T_{,}T]} |u(x)|, then in view of identity

\frac{α}{p′} (u′(x))_{+}^{p} + \frac{β}{p′} (u′(x))_{+}^{p} + \frac{λ}{p}(u(x))_{+}^{p} + \frac{μ}{p}(u′(x))_{−}^{p} = constant, ∀x ∈ [−T, T] (3.7)

we deduce

λC^{p} = μD^{p}, C = \sqrt[p]{μ}t, D = \sqrt[p]{λ}t, for some t > 0.

Using the positive homogeneity of (1.1) we derive that the solution u(x) is given in by

u(x) = \{ \begin{matrix}t\sqrt[p]{μ} sin_{p}(\frac{x+T}{2} \sqrt[p]{\frac{λ}{α}}), & −T ≤ x ≤ −T + π_{p}\sqrt[p]{\frac{α}{λ}} \\ t\sqrt[p]{μ} sin_{p}(\frac{T_{1}−x}{2} \sqrt[p]{\frac{λ}{β}}), & T_{1} − π_{p}\sqrt[p]{\frac{β}{λ}} ≤ x ≤ T_{1} \\ −t\sqrt[p]{λ} sin_{p}(\frac{x−T_{1}}{2} \sqrt[p]{\frac{μ}{β}}), & T_{1} ≤ x ≤ T_{1} + π_{p}\sqrt[p]{\frac{β}{μ}} \\ −t\sqrt[p]{λ} sin_{p}(\frac{T −x}{2} \sqrt[p]{\frac{μ}{α}}), & T − π_{p}\sqrt[p]{\frac{α}{μ}} ≤ x ≤ T \end{matrix}  (3.8)

It follows from (3.8) that the differences between α and β are reflected by change of amplitudes between positive and negative waves at the switch between plus and minus.

For the switch from negative wave to positive wave, it holds

u(x) = \{ \begin{matrix}−\sqrt[p]{λ} sin_{p}(\frac{x+T}{2} \sqrt[p]{\frac{μ}{β}}), & −T ≤ x ≤ −T + π_{p}\sqrt[p]{\frac{β}{μ}} \\ −\sqrt[p]{λ} sin_{p}(\frac{T_{1}−x}{2} \sqrt[p]{\frac{μ}{α}}), & T_{1} − π_{p}\sqrt[p]{\frac{α}{μ}} ≤ x ≤ T_{1} \\ \sqrt[p]{μ} sin_{p}(\frac{x−T_{1}}{2} \sqrt[p]{\frac{λ}{α}}), & T_{1} ≤ x ≤ T_{1} + π_{p}\sqrt[p]{\frac{α}{λ}} \\ \sqrt[p]{μ} sin_{p}(\frac{T_{1}−x}{2} \sqrt[p]{\frac{λ}{β}}), & T − π_{p}\sqrt[p]{\frac{β}{λ}} ≤ x ≤ T \end{matrix}  (3.8′)

where T_{1} = −T + (\sqrt[p]{α/μ} + \sqrt[p]{β/μ})π_{p}. In view of (3.8) and (3.8’) we see that u′(−T) = u′(T) and therefore can extend u(x) to a 2T -periodic function on the whole real line R.

For any given integer k ≥ 1. If u is a solution of (1.1) with (2k + 1) nodes, then it must have equal number of positive and negative (k+1) waves.

Let T_{1} < T_{2} < ⋯ < T_{2k+1} be the nodal points of u, it follows from (3.7) that all positive waves of u have same height and so are the same for negative waves. Moreover similarly as deriving (3.6) we get

(\sqrt[p]{α} + \sqrt[p]{β})(\sqrt[p]{λ^{−1}} + \sqrt[p]{μ^{−1}})π_{p} = T_{i+2} − T_{i}, i = 1, ⋯, 2k − 1.

Thereby α, β, λ, μ satisfy

(k + 1)(\sqrt[p]{α} + \sqrt[p]{β})(\sqrt[p]{λ^{−1}} + \sqrt[p]{μ^{−1}})π_{p} = 2T  (3.6′)

and the nodes are T_{1+2i} = −T +(1+i)L^{+}+iL^{−}, T_{2i} = −T +i(L^{+}+L^{−}), i = 0, 1, 2, ⋯, k. Furthermore let T_{0} = −T, T_{2k+2} = T then the initially positive solution u(x) on [T_{2i}, T_{2i+2}], i = 0, 1, 2, ⋯ k, is given by

u(x) = \{ \begin{matrix}t\sqrt[p]{μ} sin_{p}(\frac{x−T_{i}}{2} \sqrt[p]{\frac{λ}{α}}), & T_{i} ≤ x ≤ T_{i} + π_{p}\sqrt[p]{\frac{α}{λ}} \\ t\sqrt[p]{μ} sin_{p}(\frac{T_{i+1}−x}{2} \sqrt[p]{\frac{λ}{β}}), & T_{i+1} − π_{p}\sqrt[p]{\frac{β}{λ}} ≤ x ≤ T_{i+1} \\ −t\sqrt[p]{λ} sin_{p}(\frac{x−T_{1}}{2} \sqrt[p]{\frac{μ}{β}}), & T_{i+1} ≤ x ≤ T_{i+1} + π_{p}\sqrt[p]{\frac{β}{μ}} \\ −t\sqrt[p]{λ} sin_{p}(\frac{T_{i+2}−x}{2} \sqrt[p]{\frac{μ}{α}}), & T_{i+2} − π_{p}\sqrt[p]{\frac{α}{μ}} ≤ x ≤ T_{i+2} \end{matrix}  (3.9)

and the initially negative solution u by

u(x) = \{ \begin{matrix}−t\sqrt[p]{λ} sin_{p}(\frac{x−T_{i}}{2} \sqrt[p]{\frac{μ}{β}}), & T_{i} ≤ x ≤ T_{i} + π_{p}\sqrt[p]{\frac{β}{μ}} \\ −t\sqrt[p]{λ} sin_{p}(\frac{T_{i+1}−x}{2} \sqrt[p]{\frac{μ}{α}}), & T_{i+1} − π_{p}\sqrt[p]{\frac{α}{μ}} ≤ x ≤ T_{i+1} \\ t\sqrt[p]{μ} sin_{p}(\frac{x−T_{i+1}}{2} \sqrt[p]{\frac{λ}{α}}), & T_{i+1} ≤ x ≤ T_{i+1} + π_{p}\sqrt[p]{\frac{α}{λ}} \\ t\sqrt[p]{μ} sin_{p}(\frac{T_{i+2}−x}{2} \sqrt[p]{\frac{λ}{β}}), & T_{i+2} − π_{p}\sqrt[p]{\frac{β}{λ}} ≤ x ≤ T_{i+2} \end{matrix}  (3.9′)

where T_{1+2i} = −T + (1 + i)L^{−} + iL^{+}, T_{2i} = −T + i(L^{+} + L^{−}), i = 0, 1, 2, ⋯, k.

EJQTDE, 2009 No. 18, p. 6

If u has 2k nodes, then there are two possibilities 1) (k + 1) positive waves and k negative waves, 2) k positive waves and (k + 1) negative waves. In 1) the solution should be initially positive and be initially negative in 2). It follows then

1)

(\sqrt[p]{α} + \sqrt[p]{β})((k + 1)\sqrt[p]{λ^{−1}} + k\sqrt[p]{μ^{−1}})π_{p} = 2T  (3.10)

and the solution u on [−T, T − L^{−}] is (2T − L_{−})/k-periodic and is given by (3.9) on [T_{i}, T_{i+2}], i = 0, 1, ⋯, 2k − 2, and on [T_{2k}, T], u(x) is given by

u(x) = \{ \begin{matrix}t\sqrt[p]{μ} sin_{p}(\frac{x−T_{2k}}{2} \sqrt[p]{\frac{λ}{α}}), & T_{2k} ≤ x ≤ T_{2k} + π_{p}\sqrt[p]{\frac{α}{λ}} \\ t\sqrt[p]{μ} sin_{p}(\frac{T −x}{2} \sqrt[p]{\frac{λ}{β}}), & T − π_{p}\sqrt[p]{\frac{β}{λ}} ≤ x ≤ T \end{matrix}  (3.11)

2) For an initially negative solution with 2k nodes, then

(\sqrt[p]{α} + \sqrt[p]{β})(k\sqrt[p]{λ^{−1}} + (k + 1)\sqrt[p]{μ^{−1}})π_{p} = 2T  (3.12)

and on [T_{i}, T_{i+2}] the solution u is given by (3.9’) for i = 0, 1, ⋯, 2k − 2, and on [T_{2k}, T], u(x) is given by

u(x) = \{ \begin{matrix}−t\sqrt[p]{λ} sin_{p}(\frac{x−T_{2k}}{2} \sqrt[p]{\frac{μ}{β}}), & T_{2k} ≤ x ≤ T_{2k} + π_{p}\sqrt[p]{\frac{β}{μ}} \\ −t\sqrt[p]{λ} sin_{p}(\frac{T −x}{2} \sqrt[p]{\frac{μ}{α}}), & T − π_{p}\sqrt[p]{\frac{α}{μ}} ≤ x ≤ T. \end{matrix}  (3.11′)

So the proof is complete.

4 A final remark

In the study of nontrivial solutions to one dimensional nonlinear differential equation

−(φ(x, u′))′ = f(x, u)  (4.1)

one usually adopt the notation of solution by (4.1) being satisfied pointwise, which in turn ensures per definition only C^{1} smoothness of solution. Of course, one expects higher order smoothness of the solutions. Here we shall examine this question for a very special case, namely φ(x, s) = α(x)s_{+}^{p−1} − β(x)s_{−}^{q−1}, p, q > 1 e.g.,

−(α(x)(u′)_{+}^{p−1} − β(x)(u′)_{−}^{q−1})′ = f(x, u)  (4.2)

If we assume that the equation (4.2) is satisfied pointwise and moreover α, β > 0 are also C^{1}, then any solution u is obviously C^{2} for any point x where u′(x) ≠ 0. So in order to get differentiability of u we need a closer examination at those points where u′(x) = 0.

Let x_{0} be a critical point of u(x), u_{0} = u(x_{0}), if p = q = 2, then we easily deduce from the equation (4.2) that for small δ > 0

u′(x_{0} − δ) = f(x_{0}, u_{0})/α(x_{0})(δ + o(δ)); −u′(x_{0} + δ) = f(x_{0}, u_{0})/β(x_{0})(δ + o(δ))

EJQTDE, 2009 No. 18, p. 7 thereafter u′′(x) has a jump at x_{0} since α ≠ β and thus u ∈ C^{1,1}, but not C^{2}.

In general, for any critical point x = x_{0} of u(x), we have the following asymptotic as δ → 0

\{ \begin{matrix}u′(x_{0} − δ) & = & C_{1}δ^{1/(p−1)}(1 + o(1)) \\ u′(x_{0} + δ) & = & −C_{2}δ^{1/(q−1)}(1 + o(1)) \end{matrix}

where C_{1} = \sqrt[p−1]{f(x_{0}, u_{0})/α(x_{0})}, C_{2} = \sqrt[q−1]{f(x_{0}, u_{0})/β(x_{0})}. In view of the above estimates, we deduce that

1. If 1 < p, q < 2 then u ∈ C^{2}

2. If max{p, q} = 2 then u ∈ C^{1,1}

3. If 2 < p, q then u ∈ C^{1,ε}, ε = min\{\frac{1}{p−1}, \frac{1}{q−1}\}.

Acknowledgement

This work is partially supported by the project Nr 20-06/120 for promotion of research at Malmö University and also the G S Magnusons fond, the Swedish Royal Academy of Science.

The author thanks the referee for careful reading and suggestions which lead to improvement of the presentation.

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(Received January 15, 2008)