Solutions with a prescribed interval of positivity for differential systems with nonlocal conditions

Abstract

Based on fixed point index, the paper develops a theory of existence, localization and multiplicity of solutions to first-order differential systems subject to linear nonlocal conditions.

The main features concern the role of the support of the nonlocal condition and the positivity of solutions which is only required on a prescribed subinterval.

Several examples of problems admitting at least one, two, or sequences of such solutions are included, and numerical solutions are obtained using the Mathematica shooting program with starting initial conditions suggested by the theoretical localization results.

Authors

Veronica Ilea
Department of Mathematics, Babeş-Bolyai University,  Cluj-Napoca, Romania
Adela Novac
Department of Mathematics, Technical University of Cluj-Napoca,  Romania
Diana Otrocol
Technical University of Cluj-Napoca,  Romania
Tiberiu Popoviciu Institute of Numerical Analysis (Romanian Academy)
Radu Precup
Babeş-Bolyai University, Romania

Keywords

Differential system; Nonlocal condition; Positive solution; Multiple solutions; Fixed point index

References

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V. Ilea, A. Novac, D. Otrocol, R. Precup, Solutions with a prescribed interval of positivity for differential systems with nonlocal conditions, Appl. Math. Comput., 375 (2020), doi: 10.1016/j.amc.2020.125092

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Solutions with a prescribed interval of positivity for differential systems with nonlocal conditions

Veronica Ilea Babeş–Bolyai University, Department of Mathematics, 1 M. Kogălniceanu Street, 400084 Cluj-Napoca, Romania ilea.veronica@gmail.com , Adela Novac Technical University of Cluj-Napoca, Department of Mathematics, 28 Memorandumului Street, 400114 Cluj-Napoca, Romania adela.chis@math.utcluj.ro , Diana Otrocol Technical University of Cluj-Napoca, Department of Mathematics, 28 Memorandumului Street, 400114 Cluj-Napoca, Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, P.O.Box. 68-1, 400110 Cluj-Napoca, Romania Diana.Otrocol@math.utcluj.ro and Radu Precup Babeş–Bolyai University, Department of Mathematics, 1 M. Kogălniceanu Street, 400084 Cluj-Napoca, Romania r.precup@math.ubbcluj.ro
Abstract.

Based on fixed point index, the paper develops a theory of existence, localization and multiplicity of solutions to first-order differential systems subject to linear nonlocal conditions. The main features concern the role of the support of the nonlocal condition and the positivity of solutions which is only required on a prescribed subinterval. Several examples of problems admitting at least one, two, or sequences of such solutions are included, and numerical solutions are obtained using the Mathematica shooting program with starting initial conditions suggested by the theoretical localization results.

Key words: differential system, nonlocal condition, positive solution, multiple solutions, fixed point index

Mathematics Subject Classification: 34B18, 47H11, 47H30

1. Introduction

Many mathematical models arising from physics, chemistry, biology, economics etc. are given by first-order differential systems whose time-dependent variables stay for specific quantities, most often subject by their nature to the condition of positivity. In this connection, there is a huge literature on positive solutions to various classes of problems. However, there are cases where the positivity is only required on a subinterval of time. For instance, in case of the Lotka-Volterra predator-prey model, we may be interested that the size of the prey population x(t)x\left(t\right) on a given time interval JJ - for example the reproduction season - be larger than a given threshold x0,x_{0}, and the size of predator population y(t)y\left(t\right) on that time interval be less than y0.y_{0}. Under the substitutions u=xx0u=x-x_{0} and v=y0y,v=y_{0}-y, the desired threshold conditions become the positivity conditions u0u\geq 0 and v0v\geq 0 on the prescribed time interval J.J. Analogous problems appear in economic models when the size of production of some goods in some periods of the year needs to be sufficiently larger to satisfy the market demand, or on the contrary, to be sufficiently smaller to avoid excessive stocks. Having in mind these examples as a main motivation, in this paper we seek solutions to differential systems on a given interval, which are necessarily positive only on a prescribed subinterval.

Additionally in this paper, we treat differential systems with nonlocal conditions. In particular, they include the initial value condition, boundary value conditions, multi-point and integral constraints. As a motivating example, we have the integral condition Ωu=1\int_{\Omega}u=1 which appears in many physical models where uu is a probability distribution. The research on differential equations with nonlocal conditions began with the pioneer works of Cioranescu [5], Whyburn [17] and Conti [6], and have gained a special attention in the last decades motivated by concrete applications in different domains (see, e.g., [1]-[4], [8]-[16] and [18]).

More exactly in this paper we study the existence, localization and multiplicity of solutions to the following problem:

(1.1) {u(t)=f(t,u(t)),t[0,1]g(u)=0u0on [a,b]\left\{\begin{array}[]{l}u^{\prime}(t)=f(t,u(t)),\ \ \ t\in[0,1]\\ g(u)=0\\ u\geq 0\,\,\,\ \ \text{on\ }\,\,[a,b]\end{array}\right.

where [a,b]\left[a,b\right] is a fixed subinterval of [0,1],\left[0,1\right], f:[0,1]×nnf:[0,1]\times\mathbb{R}^{n}\rightarrow\mathbb{R}^{n} is continuous, and g:C([0,1],n)n\ g:C([0,1],\mathbb{R}^{n})\rightarrow\mathbb{R}^{n} a continuous linear mapping. We shall seek solutions uC1([0,1],n).u\in C^{1}([0,1],\mathbb{R}^{n}).

Taking α(u):=u(a)+g(u),\alpha(u):=u(a)+g(u), the nonlocal condition g(u)=0g\left(u\right)=0 can be replaced by

u(a)=α(u).u\left(a\right)=\alpha\left(u\right).

Our approach to problem (1.1) is based on fixed point principles and fixed point index theory, and takes into consideration the support [a,c]\left[a,c\right] of α,\alpha, namely the smallest subinterval [a,c][a,b]\left[a,c\right]\subseteq\left[a,b\right] such that α(u)=α(v)\alpha(u)=\alpha(v) for every u,vC([0,1],n)u,v\in C([0,1],\mathbb{R}^{n}) having the same restriction to [a,c].\left[a,c\right].

Recall that for a closed convex subset DD of a Banach X,X, a subset UDU\subseteq D open in D,D, and a completely continuous operator N:U¯DN:\overline{U}\rightarrow D such that N(u)uN\left(u\right)\neq u on the DU=U¯U,\partial_{D}U=\overline{U}\setminus U, one can consider the fixed point index of NN over UU with respect to D,D, which is denoted by i(N,U,D)i\left(N,U,D\right) (see, e.g., [7, p. 238]) and has the following properties: i(N,U,D)0i\left(N,U,D\right)\neq 0 implies that NN has at least one fixed point in UU (existence);i(u0,U,D)=1\ i\left(u_{0},U,D\right)=1\ for the any constant operator Nu0UN\equiv u_{0}\in U (normalization); the homotopy invariance; and additivity with respect to U.U. We have the following lemma that is used in the proof of our main result.

Lemma 1.1.

If UDU\subseteq D is bounded, convex, open in D,D, and N(U¯)U,N\left(\overline{U}\right)\subseteq U, then i(N,U,D)=1.i\left(N,U,D\right)=1.

Proof.

Since N(U¯)U,N\left(\overline{U}\right)\subseteq U, one has N(u)uN\left(u\right)\neq u on the DU,\partial_{D}U, so the index i(N,U,D)i\left(N,U,D\right) is defined. Let u0Uu_{0}\in U be arbitrarily fixed. Then the homotopy Hλ(u)=λN(u)+(1λ)u0H_{\lambda}\left(u\right)=\lambda N\left(u\right)+\left(1-\lambda\right)u_{0} (λ[0,1])\left(\lambda\in\left[0,1\right]\right) is admissible since Hλ(U¯)UH_{\lambda}\left(\overline{U}\right)\subseteq U for all λ[0,1],\lambda\in\left[0,1\right], due to the convexity of U.U. Finally, by the homotopy invariance and the normalization properties of the index, we have

i(N,U,D)=i(H1,U,D)=i(H0,U,D)=i(u0,U,D)=1.i\left(N,U,D\right)=i\left(H_{1},U,D\right)=i\left(H_{0},U,D\right)=i\left(u_{0},U,D\right)=1.

Basic notations. We conclude the introduction by some nonstandard notations for vectors and vector-valued functions that will be used throughout the paper. First we make the convection that all vectors in n\mathbb{R}^{n} are identified to column matrices, and for two vectors x,yn,x=(x1,,xn),y=(y1,,yn),x,y\in\mathbb{R}^{n},\ x=(x_{1},\cdot\cdot,x_{n}),\ y=(y_{1},\cdot\cdot,y_{n}), we let xyx\leq y (x<y)\left(x<y\right) if xiyix_{i}\leq y_{i} (xi<yi,x_{i}<y_{i}, respectively) for i=1,,n.i=1,\cdot\cdot,n. Also by max{x,y}\max\left\{x,y\right\} we mean the vector of components max{xi,yi},\max\left\{x_{i},y_{i}\right\}, i=1,,n.i=1,\cdot\cdot,n., and by the notation x>iyx>_{i}y we mean that the ii-th components xi,x_{i}, yiy_{i} of the two vectors satisfy the inequality xi>yi.x_{i}>y_{i}.

We shall use the notation α[1],\alpha\left[1\right], for the matrix whose columns are

α((1,0,,0)),α((0,1,,0)),,α((0,0,,1)).\alpha\left((1,0,\cdot\cdot,0)\right),\ \alpha\left((0,1,\cdot\cdot,0)\right),\ \cdot\cdot,\ \alpha\left((0,0,\cdot\cdot,1)\right).

Thus, in virtue of the linearity of α,\alpha, for any constant vector-valued function c=(c1,,cn),c=(c_{1},\cdot\cdot,c_{n}), using the decomposition

c=c1(1,0,,0)+c2(0,1,,0)++cn(0,0,,1)c=c_{1}(1,0,\cdot\cdot,0)+c_{2}(0,1,\cdot\cdot,0)+\cdot\cdot+c_{n}(0,0,\cdot\cdot,1)

one has that α(c)\alpha\left(c\right) is the product of matrix α[1]\alpha[1] with the column vector c,c,\ i.e.,

(1.2) α(c)=α[1]c.\alpha\left(c\right)=\alpha\left[1\right]c.

Also the notation II will stay for the identity matrix of size n.n.

Finally, for a vector x=(x1,,xn)n,x=\left(x_{1},\cdot\cdot,x_{n}\right)\in\mathbb{R}^{n}, by |x|\left|x\right| we shall mean the column vector whose elements are |x1|,,|xn|;\left|x_{1}\right|,\cdot\cdot,\left|x_{n}\right|; for a scalar function vC[0,1],v\in C[0,1], we shall denote by |v|\left|v\right|_{\infty} its supremum norm, and for a vector-valued function uC([0,1],n)u\in C([0,1],\mathbb{R}^{n}) of scalar components u1,,un,u_{1},\cdot\cdot,u_{n}, we shall denote by |u|\left|u\right|_{\infty} the column vector of elements |u1|,,|un|.\left|u_{1}\right|_{\infty},\cdot\cdot,\left|u_{n}\right|_{\infty}. Also, for any subinterval [a,b]\left[a,b\right] of [0,1],\left[0,1\right], by abu(t)𝑑t\int_{a}^{b}u\left(t\right)dt we shall mean the column vector of the elements abu1(t)dt,,abun(t)dt.\int_{a}^{b}u_{1}\left(t\right)dt,\cdot\cdot,\int_{a}^{b}u_{n}\left(t\right)dt. Thus, when referred to vectors or vector-valued functions, the equalities and inequalities from below have to be seen as equalities and inequalities between column vectors.

2. Main result

First, if the matrix Iα[1]I-\alpha\left[1\right] is non-singular, then our differential system subject to the condition u(a)=α(u)u\left(a\right)=\alpha\left(u\right) is equivalent to the integral type equation in C([0,1],n),C\left(\left[0,1\right],\mathbb{R}^{n}\right), namely

(2.1) u(t)=(Iα[1])1α(af(s,u(s))𝑑s)+atf(s,u(s))𝑑s,t[0,1].u(t)=\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}f(s,u(s))ds\right)+\int_{a}^{t}f(s,u(s))ds,\ \ \ t\in[0,1].

To prove this it suffices to integrate the differential equation from aa to an arbitrary t,t, to obtain

(2.2) u(t)=u(a)+atf(s,u(s))𝑑s,u\left(t\right)=u\left(a\right)+\int_{a}^{t}f\left(s,u\left(s\right)\right)ds,

then to apply the linear mapping α\alpha and use the condition u(a)=α(u),u\left(a\right)=\alpha\left(u\right), to obtain

α(u)=α(u(a))+α(af(s,u(s))𝑑s)=α[1]u(a)+α(af(s,u(s))𝑑s),\alpha\left(u\right)=\alpha\left(u\left(a\right)\right)+\alpha\left(\int_{a}^{\cdot}f\left(s,u\left(s\right)\right)ds\right)=\alpha\left[1\right]u\left(a\right)+\alpha\left(\int_{a}^{\cdot}f\left(s,u\left(s\right)\right)ds\right),

whence

u(a)=(Iα[1])1α(af(s,u(s))𝑑s).u\left(a\right)=\left(I-\alpha\left[1\right]\right)^{-1}\alpha\left(\int_{a}^{\cdot}f\left(s,u\left(s\right)\right)ds\right).

Now replacing in (2.2) gives (2.1).

The solutions uC([0,1],n)u\in C\left(\left[0,1\right],\mathbb{R}^{n}\right) of (2.1) are the fixed points of the operator N:C([0,1],n)C([0,1],n)N:C\left(\left[0,1\right],\mathbb{R}^{n}\right)\rightarrow C\left(\left[0,1\right],\mathbb{R}^{n}\right) given by

N(u)(t)=(Iα[1])1α(af(s,u(s))𝑑s)+atf(s,u(s))𝑑s(t[0,1]),N\left(u\right)\left(t\right)=\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}f(s,u(s))ds\right)+\int_{a}^{t}f(s,u(s))ds\ \ \ \left(t\in[0,1]\right),

which is completely continuous due to the continuity of ff and α.\alpha.

Looking for solutions that are positive on the prescribed subinterval [a,b]\left[a,b\right] of [0,1],\left[0,1\right], we are let to consider the set

K={uC([0,1],n):u(t)0 for t[a,b]}.K=\left\{u\in C([0,1],\mathbb{R}^{n}):\ u(t)\geq 0\text{ \ for\ \ }t\in[a,b]\right\}.

Obviously, KK is a wedge (i.e., closed, convex and with λKK\lambda K\subset K for every λ+\lambda\in\mathbb{R}_{+}). In order to use fixed point principles and index, we need that the condition N(K)KN\left(K\right)\subset K is satisfied. It is easily seen that this happens if additionally the following positivity properties hold:

(2.3) f(t,u)0for all t[a,b] and u+n,f(t,u)\geq 0\ \ \text{for all \ }t\in[a,b]\text{ and }u\in\mathbb{R}_{+}^{n},
(2.4) α(u)0for every uC([0,1],n) with u|[a,c]0,\alpha\left(u\right)\geq 0\ \ \text{for every \ }u\in C([0,1],\mathbb{R}^{n})\ \text{ with \ }u|_{[a,c]}\geq 0,
(2.5) (Iα[1])1n×n(+),\left(I-\alpha[1]\right)^{-1}\in\mathcal{M}_{n\times n}(\mathbb{R}_{+}),

where [a,c]\left[a,c\right] is the support of α.\alpha. Thus f(t,)f\left(t,\cdot\right) is not necessarily positive on +n\mathbb{R}_{+}^{n} for t[0,1][a,b],t\in\left[0,1\right]\setminus\left[a,b\right], but α(u)\alpha\left(u\right) is positive even for functions uu which are not positive on the whole interval [0,1].\left[0,1\right]. Also note that under condition (2.4), one has α[1]n×n(+)\alpha\left[1\right]\in\mathcal{M}_{n\times n}(\mathbb{R}_{+}) and then the hypothesis (2.5) is equivalent to the fact that the eigenvalues of the matrix α[1]\alpha\left[1\right] are located inside the unit circle.

Our main result is the following theorem on the existence, localization and multiplicity of solutions to problem (1.1).

For two vectors x,y+nx,y\in\mathbb{R}_{+}^{n} with yx,y\leq x, and i=1,,n,i=1,\cdot\cdot,n, we denote:

f¯ix(s)\displaystyle\underline{f}_{i}^{x}\left(s\right) :\displaystyle: =maxu[0,x1]××[0,xn]fi(s,u)(s[a,b]),\displaystyle=\underset{u\in[0,x_{1}]\times\cdot\cdot\times[0,x_{n}]}{\max}f_{i}(s,u)\ \ \ \left(s\in\left[a,b\right]\right),
f¯ix(s)\displaystyle\overline{f}_{i}^{x}\left(s\right) :\displaystyle: =maxu[x1,x1]××[xn,xn]|fi(s,u)|(s[0,1](a,b)),\displaystyle=\underset{u\in[-x_{1},x_{1}]\times\cdot\cdot\times[-x_{n},x_{n}]}{\max}\left|f_{i}(s,u)\right|\ \ \ \left(s\in\left[0,1\right]\setminus\left(a,b\right)\right),
f¯iy,x(s)\displaystyle\underline{f}_{i}^{y,x}\left(s\right) :\displaystyle: =minu[y1,x1]××[yn,xn]fi(s,u)(s[a,b]).\displaystyle=\underset{u\in[y_{1},x_{1}]\times\cdot\cdot\times[y_{n},x_{n}]}{\min}f_{i}(s,u)\ \ \ \left(s\in\left[a,b\right]\right).

Also f¯x(s),f¯x(s),μx\underline{f}^{x}\left(s\right),\ \ \overline{f}^{x}\left(s\right),\ \mu^{x} and f¯y,x(s)\underline{f}^{y,x}\left(s\right) are the vector-valued functions of the corresponding scalar functions from the above table. For instance,

f¯x(s)=(f¯1x(s),,f¯nx(s)),\underline{f}^{x}\left(s\right)=\left(\underline{f}_{1}^{x}\left(s\right),\ \cdot\cdot,\ \underline{f}_{n}^{x}\left(s\right)\right),

which is looked as a column vector.

Theorem 2.1.

Let the conditions (2.3), (2.4) and (2.5) hold.

(10)(1^{0}):

If for a vector R+n,R\in\mathbb{R}_{+}^{n}, the following condition

(Iα[1])1α(af¯R(s)𝑑s)\displaystyle\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}\underline{f}^{R}\left(s\right)ds\right)
+max{0af¯R(s)𝑑s,abf¯R(s)𝑑s+b1f¯R(s)𝑑s}\displaystyle+\max\left\{\int_{0}^{a}\overline{f}^{R}\left(s\right)ds,\ \int_{a}^{b}\underline{f}^{R}\left(s\right)ds+\int_{b}^{1}\overline{f}^{R}\left(s\right)ds\right\}
\displaystyle\leq R\displaystyle R

holds, then problem (1.1) has a solution uKu\in K such that

|u|R.\left|u\right|_{\infty}\leq R.
(20)(2^{0}):

If in addition to ((10)(1^{0}): ) there is a vector r+n,r\in\mathbb{R}_{+}^{n}, such that 0r<R0\neq r<R and

(2.7) (Iα[1])1α(af¯r,R(s)𝑑s)>irfor all i with ri>0,\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}\underline{f}^{r,R}\left(s\right)ds\right)>_{i}r\ \ \ \text{for all }i\text{ with }r_{i}>0,

then problem (1.1) has a solution uKu\in K such that

(2.8) |u|Rand mint[a,b]ui(t)rifor i=1,,n.\left|u\right|_{\infty}\leq R\ \ \ \text{and\ \ }\underset{t\in[a,b]}{\min}u_{i}(t)\geq r_{i}\ \ \text{for }i=1,\cdot\cdot,n.
(30)(3^{0}):

If in addition to ((10)(1^{0}): ) and (2.7) there is a nonempty set of indices Λ{1,,n}\Lambda\subseteq\left\{1,\cdot\cdot,n\right\} and a vector τ(0,+)n\tau\in\left(0,+\infty\right)^{n} such that τi<ri\tau_{i}<r_{i} for all iΛ,i\in\Lambda, τi=Ri\tau_{i}=R_{i} for iΛ,i\notin\Lambda, and

(Iα[1])1α(af¯τ(s)𝑑s)\displaystyle\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}\underline{f}^{\tau}\left(s\right)ds\right)
+max{0af¯τ(s)𝑑s,abf¯τ(s)𝑑s+b1f¯τ(s)𝑑s}\displaystyle+\max\left\{\int_{0}^{a}\overline{f}^{\tau}\left(s\right)ds,\ \int_{a}^{b}\underline{f}^{\tau}\left(s\right)ds+\int_{b}^{1}\overline{f}^{\tau}\left(s\right)ds\right\}
<i\displaystyle<_{i} τ for iΛ,\displaystyle\tau\text{ \ \ for }i\in\Lambda,

then problem (1.1) has at least three solutions u,v,wKu,v,w\in K such that

|u|\displaystyle\left|u\right|_{\infty} \displaystyle\leq R,|v|τ,|w|R,\displaystyle R,\ \ \left|v\right|_{\infty}\leq\tau,\ \ \left|w\right|_{\infty}\leq R,\
mint[a,b]ui(t)\displaystyle\min_{t\in\left[a,b\right]}u_{i}(t) \displaystyle\geq rifor i=1,,n,\displaystyle r_{i}\ \ \ \text{for }i=1,\cdot\cdot,n,
mint[a,b]wi(t)\displaystyle\underset{t\in[a,b]}{\min}w_{i}(t) <\displaystyle< ri for at least one i{1,,n},\displaystyle r_{i}\text{ \ for at least one }i\in\{1,\cdot\cdot,n\},
|wi|\displaystyle\left|w_{i}\right|_{\infty} >\displaystyle> τi for at least one iΛ.\displaystyle\tau_{i}\text{ \ for at least one }i\in\Lambda.
Proof.

Consider the bounded closed convex subset of C([0,1],n),C([0,1],\mathbb{R}^{n}),

D:={uK:|u|R}.D:=\left\{u\in K:\ \left|u\right|_{\infty}\leq R\right\}.

We use the fixed point index of the operator NN over a number of open sets in DD and with respect to D.D.

(101^{0}) First we prove that under condition ((10)(1^{0}): ), one has N(D)D,N\left(D\right)\subseteq D, which, according to Lemma 1.1 guarantees that the fixed point index of NN over DD with respect to DD is equal to one, i.e., i(N,D,D)=1.i\left(N,D,D\right)=1. Hence NN has at least one fixed point in D,D, that is a solution as wished. Notice that the existence of a fixed point also follows from Schauder’s fixed point theorem. For doing estimations on N,N, it is convenient to represent NN as sum of two operators, N=N1+N2,N=N^{1}+N^{2}, where

N1(u)\displaystyle N^{1}\left(u\right) :\displaystyle: =(Iα[1])1α(af(s,u(s))𝑑s),\displaystyle=\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}f(s,u(s))ds\right),
N2(u)(t)\displaystyle N^{2}\left(u\right)\left(t\right) :\displaystyle: =atf(s,u(s))𝑑s(t[0,1]),\displaystyle=\int_{a}^{t}f(s,u(s))ds\ \ \ \left(t\in\left[0,1\right]\right),

and to recall the notations introduced before the statement of the theorem.

First we estimate from above N1(u).N^{1}\left(u\right). For uDu\in D and t[a,c]t\in[a,c] we have that 0u(s)R0\leq u\left(s\right)\leq R for every s[a,t],s\in\left[a,t\right], and so

atf(s,u(s))𝑑satf¯R(s)𝑑s.\int_{a}^{t}f(s,u(s))ds\leq\int_{a}^{t}\underline{f}^{R}\left(s\right)ds.

Then, using (2.4) and the linearity of α,\alpha, we deduce that

α(af(s,u(s))𝑑s)α(af¯R(s)𝑑s),\alpha\left(\int_{a}^{\cdot}f(s,u(s))ds\right)\leq\alpha\left(\int_{a}^{\cdot}\underline{f}^{R}\left(s\right)ds\right),

which together with (2.5) yields

(2.10) |N1(u)|=N1(u)(Iα[1])1α(af¯R(s)𝑑s).\left|N^{1}(u)\right|=N^{1}(u)\leq\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}\underline{f}^{R}\left(s\right)ds\right).

Next we estimate from above N2(u)N^{2}(u) for uD.u\in D. For t[a,b],t\in[a,b], we still have 0u(s)R0\leq u\left(s\right)\leq R for every s[a,t],s\in\left[a,t\right], and so

(2.11) |atf(s,u(s))𝑑s|=atf(s,u(s))𝑑sabf¯R(s)𝑑s(t[a,b]).\left|\int_{a}^{t}f(s,u(s))ds\right|=\int_{a}^{t}f(s,u(s))ds\leq\int_{a}^{b}\underline{f}^{R}\left(s\right)ds\ \ \ \left(t\in\left[a,b\right]\right).

For t[0,1][a,b],t\in\left[0,1\right]\setminus[a,b], since u(t)u\left(t\right) is not necessarily positive, we only have Ru(s)R.-R\leq u\left(s\right)\leq R. Hence

(2.12) |atf(s,u(s))𝑑s|0af¯R(s)𝑑s(t[0,a]),\left|\int_{a}^{t}f(s,u(s))ds\right|\leq\int_{0}^{a}\overline{f}^{R}\left(s\right)ds\ \ \ \left(t\in\left[0,a\right]\right),
|atf(s,u(s))𝑑s|\displaystyle\left|\int_{a}^{t}f(s,u(s))ds\right| \displaystyle\leq |abf(s,u(s))𝑑s|+|b1f(s,u(s))𝑑s|\displaystyle\left|\int_{a}^{b}f(s,u(s))ds\right|+\left|\int_{b}^{1}f(s,u(s))ds\right|
\displaystyle\leq abf¯R(s)𝑑s+b1f¯R(s)𝑑s(t[b,1]).\displaystyle\int_{a}^{b}\underline{f}^{R}\left(s\right)ds+\int_{b}^{1}\overline{f}^{R}\left(s\right)ds\ \ \ \ \left(t\in[b,1]\right).

Now (2.11), (2.12) and (2) give

(2.14) |N2(u)(t)|max{0af¯R(s)𝑑s,abf¯R(s)𝑑s+b1f¯R(s)𝑑s} for all t[0,1].\left|N^{2}(u)(t)\right|_{\infty}\leq\max\left\{\int_{0}^{a}\overline{f}^{R}\left(s\right)ds,\ \int_{a}^{b}\underline{f}^{R}\left(s\right)ds+\int_{b}^{1}\overline{f}^{R}\left(s\right)ds\right\}\text{ \ \ for all }t\in[0,1].

From (2.10) and (2.14) we deduce the final estimate from above

|N(u)|\displaystyle\left|N\left(u\right)\right|_{\infty} \displaystyle\leq (Iα[1])1α(af¯R(s)𝑑s)\displaystyle\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}\underline{f}^{R}\left(s\right)ds\right)
+max{0af¯R(s)𝑑s,abf¯R(s)𝑑s+b1f¯R(s)𝑑s},\displaystyle+\max\left\{\int_{0}^{a}\overline{f}^{R}\left(s\right)ds,\ \int_{a}^{b}\underline{f}^{R}\left(s\right)ds+\int_{b}^{1}\overline{f}^{R}\left(s\right)ds\right\},

which together with condition ((10)(1^{0}): ) shows that N(D)D.N(D)\subset D. Thus the proof of (101^{0}) is finished.

(202^{0}) Let

U:={uD:mint[a,b]ui(t)>ri for all i with ri>0}.U:=\left\{u\in D:\ \min_{t\in\left[a,b\right]}u_{i}\left(t\right)>r_{i}\text{ for all }i\text{ with }r_{i}>0\right\}.

The set UU is open in D.D. We prove that N(U¯)U,N(\overline{U})\subseteq U, which implies i(N,U,D)=1,i\left(N,U,D\right)=1, whence the conclusion. To this aim we need to estimate N(u)N\left(u\right) from below. Thus, for uU¯,u\in\overline{U},\

N1(u)(Iα[1])1α(af¯r,R(s)𝑑s),N^{1}(u)\geq\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}\underline{f}^{r,R}(s)ds\right),

and for t[a,b],t\in\left[a,b\right], N2(u)(t)0.N^{2}\left(u\right)\left(t\right)\geq 0. Hence, from (2.7) we have that N(U¯)UN(\overline{U})\subseteq U as desired.

(303^{0}) Now consider the set

V:={uD:|ui|<τi for iΛ}.V:=\left\{u\in D:\ \left|u_{i}\right|_{\infty}<\tau_{i}\text{ \ \ for }\ i\in\Lambda\right\}.

Clearly, the set VV is open in DD and the sets U¯\overline{U} and VV are disjoint. Since τR,\tau\leq R, we have f¯τ(s)f¯R(s)\underline{f}^{\tau}\left(s\right)\leq\underline{f}^{R}\left(s\right) for s[a,b]s\in\left[a,b\right], and f¯τ(s)f¯R(s)\overline{f}^{\tau}\left(s\right)\leq\overline{f}^{R}\left(s\right) for s[0,1](a,b).s\in\left[0,1\right]\setminus\left(a,b\right). Then from ((30)(3^{0}): ) and ((10)(1^{0}): ), we see that

(Iα[1])1α(af¯τ(s)𝑑s)+max{0af¯τ(s)𝑑s,abf¯τ(s)𝑑s+b1f¯τ(s)𝑑s}τ.\left(I-\alpha[1]\right)^{-1}\alpha\left(\int_{a}^{\cdot}\underline{f}^{\tau}\left(s\right)ds\right)+\max\left\{\int_{0}^{a}\overline{f}^{\tau}\left(s\right)ds,\ \int_{a}^{b}\underline{f}^{\tau}\left(s\right)ds+\int_{b}^{1}\overline{f}^{\tau}\left(s\right)ds\right\}\leq\tau.

Then, as at case (101^{0}), we may infer that N(V¯)VN(\overline{V})\subset V and so i(N,V,D)=1.i(N,V,D)=1.\ Hence a second fixed point vv exists in V.V.

In the end, from

i(N,D(U¯V¯),D)=i(N,D,D)i(N,U,D)i(N,V,D)=111=10,i(N,D\setminus(\overline{U}\cup\overline{V}),D)=i(N,D,D)-i(N,U,D)-i(N,V,D)=1-1-1=-1\neq 0,

we find a third fixed point ww in D(U¯V¯).D\setminus(\overline{U}\cup\overline{V}). The localization properties of the three solutions come from the definitions of the sets U,U, VV and D(U¯V¯).D\setminus(\overline{U}\cup\overline{V}).\

Remark 2.1.

Under the assumptions of Theorem 2.1 (303^{0}), the solutions uu and ww are different from zero. If in addition f(,0)0,f\left(\cdot,0\right)\neq 0, then the solution vv is also different from zero.

Remark 2.2.

For only one autonomous equation (n=1n=1), conditions ((10)(1^{0}): ), (2.7) and ((30)(3^{0}): ) read as:

α(ta)1α(1)max[0,R]f+max{amax[R,R]|f|,(ba)max[0,R]f+(1b)max[R,R]|f|}R,\frac{\alpha\left(t-a\right)}{1-\alpha\left(1\right)}\max_{\left[0,R\right]}f+\max\left\{a\max_{\left[-R,R\right]}\left|f\right|,\ \left(b-a\right)\max_{\left[0,R\right]}f+\left(1-b\right)\max_{\left[-R,R\right]}\left|f\right|\right\}\leq R,
α(ta)1α(1)min[r,R]f>r,\frac{\alpha\left(t-a\right)}{1-\alpha\left(1\right)}\min_{\left[r,R\right]}f>r,
α(ta)1α(1)max[0,τ]f+max{amax[τ,τ]|f|,(ba)max[0,τ]f+(1b)max[τ,τ]|f|}<τ.\frac{\alpha\left(t-a\right)}{1-\alpha\left(1\right)}\max_{\left[0,\tau\right]}f+\max\left\{a\max_{\left[-\tau,\tau\right]}\left|f\right|,\ \left(b-a\right)\max_{\left[0,\tau\right]}f+\left(1-b\right)\max_{\left[-\tau,\tau\right]}\left|f\right|\right\}<\tau.

In particular, if

(2.15) max[R,R]|f|=max[0,R]f=f(R),max[τ,τ]|f|=max[0,τ]f=f(τ)and min[r,R]f=f(r),\max_{\left[-R,R\right]}\left|f\right|=\max_{\left[0,R\right]}f=f\left(R\right),\ \ \ \max_{\left[-\tau,\tau\right]}\left|f\right|=\max_{\left[0,\tau\right]}f=f\left(\tau\right)\ \ \text{\emph{and}\ }\ \min_{\left[r,R\right]}f=f\left(r\right),

the above inequalities reduce to

(2.16) f(R)R1α(1)α(ta)+(1α(1))max{a,1a}=:A,\frac{f\left(R\right)}{R}\leq\frac{1-\alpha\left(1\right)}{\alpha\left(t-a\right)+\left(1-\alpha\left(1\right)\right)\max\left\{a,1-a\right\}}=:A,
(2.17) f(r)r>1α(1)α(ta)=:B,\frac{f\left(r\right)}{r}>\frac{1-\alpha\left(1\right)}{\alpha\left(t-a\right)}=:B,
(2.18) f(τ)τ<1α(1)α(ta)+(1α(1))max{a,1a}=A\frac{f\left(\tau\right)}{\tau}<\frac{1-\alpha\left(1\right)}{\alpha\left(t-a\right)+\left(1-\alpha\left(1\right)\right)\max\left\{a,1-a\right\}}=A

showing the oscillation of the function f(u)/u.f\left(u\right)/u.

Notice the dependence of the numbers AA and BB on the support [a,c]\left[a,c\right] of the mapping α\alpha through the values α(1)\alpha\left(1\right) and α(ta).\alpha\left(t-a\right). Also note a sufficient condition for (2.15) to hold, namely that ff is nonnegative and increasing on [0,R]\left[0,R\right] and fλf-\lambda is odd or even on [R,R]\left[-R,R\right] for some λ0.\lambda\geq 0.

Example 2.1.

The problem

{u=51+e8(u+1),t[0,1]u(23)=14u(1)u0on [23,1]\left\{\begin{array}[]{l}u^{\prime}=\frac{5}{1+e^{-8\left(u+1\right)}},\ \ \ t\in[0,1]\\ u\left(\frac{2}{3}\right)=\frac{1}{4}u\left(1\right)\\ u\geq 0\ \ \ \text{\emph{on}\ \ }\left[\frac{2}{3},1\right]\end{array}\right.

has a solution (see Figure 1).

Refer to caption
Figure 1. Solution to the problem in Example 2.1.

It is easy to show that the function f(t,u)=f(u)=5/(1+e8(u+1))f\left(t,u\right)=f\left(u\right)=5/\left(1+e^{-8\left(u+1\right)}\right) is positive and increasing on .\mathbb{R}. Hence (2.15) holds whatever would be rr and R.R.~Such numbers r,Rr,R with 0<r<R0<r<R and satisfying (2.16) and (2.17) exist since

limx0f(x)x=+and limx+f(x)x=0.\lim_{x\rightarrow 0}\frac{f\left(x\right)}{x}=+\infty\ \ \ \text{\emph{and}\ \ \ }\lim_{x\rightarrow+\infty}\frac{f\left(x\right)}{x}=0.

The result then follows from Theorem 2.1 (202^{0}). We underline the fact that the solution is not positive on the whole interval [0,1],\left[0,1\right], but, as predicted by our theory, it is positive on the subinterval [2/3,1],\left[2/3,1\right], the support of the nonlocal condition.

Example 2.2.

Consider the problem

(2.19) {u=f(u),t[0,1]u(12)=25u(1)u0on [12,1].\left\{\begin{array}[]{l}u^{\prime}=f\left(u\right),\ \ \ t\in[0,1]\\ u\left(\frac{1}{2}\right)=\frac{2}{5}u\left(1\right)\\ u\geq 0\,\,\,\ \ \text{\emph{on}\ }\,\,[\frac{1}{2},1].\end{array}\right.

For

f(u)=|u|+λ1|u|+λ2and 0<λ1<λ2,f\left(u\right)=\frac{\left|u\right|+\lambda_{1}}{\left|u\right|+\lambda_{2}}\ \ \ \text{\emph{and}\ \ \ }0<\lambda_{1}<\lambda_{2},

the problem has at least one solution uu with u>0u>0 on [1/2,1].\left[1/2,1\right].

Indeed, the function f(t,u)=f(u)f\left(t,u\right)=f\left(u\right) is even on \mathbb{R} and nonnegative, increasing on +\mathbb{R}_{+} since 0<λ1<λ2,0<\lambda_{1}<\lambda_{2}, hence (2.15) holds whatever will be rr and R.R.~We have α(u)=(2/5)u(1)\alpha\left(u\right)=\left(2/5\right)u\left(1\right) and then α(1)=2/5\alpha\left(1\right)=2/5 and α(t1/2)=(2/5)(11/2)=1/5.\alpha\left(t-1/2\right)=\left(2/5\right)\left(1-1/2\right)=1/5. Also A=6/5A=6/5 and B=3.B=3. Looking for a number r>0r>0 with f(r)/r>B,f\left(r\right)/r>B, we are led to the inequality 3r2+(3λ21)rλ1<0,3r^{2}+\left(3\lambda_{2}-1\right)r-\lambda_{1}<0, which has positive solutions for λ1>0.\lambda_{1}>0. Finally, a number R>rR>r can be found since f(x)/xf\left(x\right)/x 0\rightarrow 0 as x+.x\rightarrow+\infty. The conclusion follows from Theorem 2.1 (202^{0}).

Example 2.3.

For

f(u)=5u2+λ1u2+λ2,f\left(u\right)=\frac{5u^{2}+\lambda_{1}}{u^{2}+\lambda_{2}}\ ,

λ2(0,25/36)\lambda_{2}\in\left(0,25/36\right) and sufficiently small λ1(0,λ2),\lambda_{1}\in\left(0,\lambda_{2}\right), problem (2.19) has at least three nonzero solutions. As in the previous example, the function ff is even on \mathbb{R} and nonnegative, increasing on +,\mathbb{R}_{+}, hence (2.15) holds whatever would be τ,r\tau,r and R.R.~Also A=6/5A=6/5 and B=3.B=3. First we prove that there are two numbers rr and τ\tau with 0<τ<r0<\tau<r such that f(τ)/τ<Af\left(\tau\right)/\tau<A and f(r)/r>B,f\left(r\right)/r>B, that is

(2.20) 6τ325τ2+6λ2τ5λ1\displaystyle 6\tau^{3}-25\tau^{2}+6\lambda_{2}\tau-5\lambda_{1} >\displaystyle> 0,\displaystyle 0,
3r35r2+3λ2rλ1\displaystyle 3r^{3}-5r^{2}+3\lambda_{2}r-\lambda_{1} <\displaystyle< 0.\displaystyle 0.

Since λ2<1/36,\lambda_{2}<1/36, there is r>0r>0 with 3r35r2+3λ2r<0,3r^{3}-5r^{2}+3\lambda_{2}r<0, which makes true the second inequality in (2.20) for every λ1>0.\lambda_{1}>0. Next we choose τ(0,r)\tau\in\left(0,r\right) such that 6τ225τ+6λ2>0,6\tau^{2}-25\tau+6\lambda_{2}>0, and then λ1(0,λ2)\lambda_{1}\in\left(0,\lambda_{2}\right) such that 5λ1<(6τ225τ+6λ2)τ.5\lambda_{1}<\left(6\tau^{2}-25\tau+6\lambda_{2}\right)\tau. Thus conditions (2.20) are fulfilled. Finally, since f(x)/x0f\left(x\right)/x\rightarrow 0 as x+,x\rightarrow+\infty, there exists R>rR>r such that f(R)/RA.f\left(R\right)/R\leq A. From Theorem 2.1 (303^{0}), the problem has three solutions u,v,wu,v,w with

(2.21) min[1/2,1]ur,|u|R;|v|τ;|w|>τ,min[1/2,1]w<r.\min_{\left[1/2,1\right]}u\geq r,\ \ \left|u\right|_{\infty}\leq R;\ \ \ \left|v\right|_{\infty}\leq\tau;\ \ \ \left|w\right|_{\infty}>\tau,\ \ \min_{\left[1/2,1\right]}w<r.

These solutions are nonzero since f(0)0.f\left(0\right)\neq 0. To test our theoretical result, we consider λ1=0.01\lambda_{1}=0.01 and λ2=0.5.\lambda_{2}=0.5. Then the conditions on the three radii R,rR,r and τ\tau are fulfilled by the values R=4.2,r=1,R=4.2,\ r=1, τ=0.1,\tau=0.1, and using the Shooting program of Mathematica, we obtain three numerical solutions u,v,wu,v,w satisfying (2.21), as shown in Figure 2.

Refer to caption
Figure 2. Three solutions to the problem in Example 2.3, where λ1=0.01\lambda_{1}=0.01 and λ2=0.5.\lambda_{2}=0.5.
Example 2.4.

For

f(u)=λ+2110u1920usin(35lnu2)for u0and  f(0)=λ,f\left(u\right)=\lambda+\frac{21}{10}u-\frac{19}{20}u\sin\left(\frac{3}{5}\ln u^{2}\right)\ \ \text{\emph{for}\ }u\neq 0\ \text{\emph{and\ }\ }f\left(0\right)=\lambda,

and any λ+,\lambda\in\mathbb{R}_{+}, problem (2.19) has a sequence of solutions uku_{k} with |uk|+.\left|u_{k}\right|_{\infty}\rightarrow+\infty. If λ=0,\lambda=0, then in addition there exists a sequence of nonzero solutions vkv_{k} with vk0.v_{k}\rightarrow 0.

The function ff is nonnegative and increasing on +,\mathbb{R}_{+}, and fλf-\lambda is odd on .\mathbb{R}. Hence (2.15) holds. As in the previous two examples, one have A=6/5A=6/5 and B=3.B=3. Since

lim infx+f(x)x=2320<Aand lim supx+f(x)x=6120>B,\liminf_{x\rightarrow+\infty}\frac{f\left(x\right)}{x}=\frac{23}{20}<A\ \ \ \text{\emph{and}\ \ \ }\limsup_{x\rightarrow+\infty}\frac{f\left(x\right)}{x}=\frac{61}{20}>B,

there exist sequences rkr_{k} and RkR_{k} tending to ++\infty and such that

 rk<Rk<rk+1<Rk+1, f(Rk)Rk<Aand  f(rk)rk>B\emph{\ }r_{k}<R_{k}<r_{k+1}<R_{k+1},\emph{\ \ \ \ }\frac{f\left(R_{k}\right)}{R_{k}}<A\ \ \ \text{\emph{and}\ \ }\emph{\ }\frac{f\left(r_{k}\right)}{r_{k}}>B

for all k. \emph{for\ all\ }k.\emph{\ }Hence, from Theorem 2.1 (202^{0}), for each k,k, the problem has a solution uku_{k} with |uk|Rkand min[1/2,1]ukrk.\left|u_{k}\right|_{\infty}\leq R_{k}\ \emph{and\ }\min_{\left[1/2,1\right]}u_{k}\geq r_{k}. From

|uk|min[1/2,1]ukrk>Rk1\left|u_{k}\right|_{\infty}\geq\min_{\left[1/2,1\right]}u_{k}\geq r_{k}>R_{k-1}

and Rk+,R_{k}\rightarrow+\infty, we find that |uk|+.\left|u_{k}\right|_{\infty}\rightarrow+\infty. For λ=0.1,\lambda=0.1, two such solutions are numerically obtained with the Shooting program of Mathematica, and are represented in Figure 3.

Refer to caption
Figure 3. Two functions from the sequence of solutions to the problem in Example 2.4, where λ=0.1.\lambda=0.1.

If λ=0,\lambda=0, then we also have

lim infx0f(x)x=2320<Aand lim supx0f(x)x=6120>B.\liminf_{x\rightarrow 0}\frac{f\left(x\right)}{x}=\frac{23}{20}<A\ \ \ \text{\emph{and}\ \ \ }\limsup_{x\rightarrow 0}\frac{f\left(x\right)}{x}=\frac{61}{20}>B.

Consequently, there exist sequences rkr_{k} and RkR_{k} tending to 0 and such that

rk+1<Rk+1<rk<Rk, f(Rk)Rk<Aand  f(rk)rk>Br_{k+1}<R_{k+1}<r_{k}<R_{k},\emph{\ \ \ }\frac{f\left(R_{k}\right)}{R_{k}}<A\ \ \ \text{\emph{and}\ \ }\emph{\ }\frac{f\left(r_{k}\right)}{r_{k}}>B

for all k.k. Then for each k,k, there is a solution vkv_{k} satisfying |vk|Rk\left|v_{k}\right|_{\infty}\leq R_{k} and min[1/2,1]vkrk.\min_{\left[1/2,1\right]}v_{k}\geq r_{k}. Clearly from |vk|Rk\left|v_{k}\right|_{\infty}\leq R_{k} and Rk0,R_{k}\rightarrow 0, we have that vk0.v_{k}\rightarrow 0.

Example 2.5.

Consider the following problem related to a bidimensional system

(2.22) {u=f(u),t[0,1]u(12)=25u(1)u0on [12,1].\left\{\begin{array}[]{l}u^{\prime}=f\left(u\right),\ \ \ t\in\left[0,1\right]\\ u\left(\frac{1}{2}\right)=\frac{2}{5}u\left(1\right)\\ u\geq 0\,\,\,\ \ \text{\emph{on}\ }\,\,[\frac{1}{2},1].\end{array}\right.

If

f1(u)=u12+u22u12+u22+λ1,f2(u)=|u1|+|u2||u1|+|u2|+λ2,f_{1}\left(u\right)=\frac{u_{1}^{2}+u_{2}^{2}}{u_{1}^{2}+u_{2}^{2}+\lambda_{1}},\ \ \ \ f_{2}\left(u\right)=\frac{\left|u_{1}\right|+\left|u_{2}\right|}{\left|u_{1}\right|+\left|u_{2}\right|+\lambda_{2}},

λ1(0,1/18)\lambda_{1}\in\left(0,1/18\right) and λ2(0,1/3],\lambda_{2}\in(0,1/3], the problem has at least one solution satisfying u>0u>0 on [1/2,1]\left[1/2,1\right]. In this case we have

α[1]=[α1(1)00α2(1)],(Iα[1])1=[11α1(1)0011α2(1)],\alpha\left[1\right]=\left[\begin{array}[]{cc}\alpha_{1}\left(1\right)&0\\ 0&\alpha_{2}\left(1\right)\end{array}\right],\ \ \left(I-\alpha\left[1\right]\right)^{-1}=\left[\begin{array}[]{cc}\frac{1}{1-\alpha_{1}\left(1\right)}&0\\ 0&\frac{1}{1-\alpha_{2}\left(1\right)}\end{array}\right],

α1=α2\alpha_{1}=\alpha_{2} and α1(1)=α2(1)=2/5.\alpha_{1}\left(1\right)=\alpha_{2}\left(1\right)=2/5. Also, taking R1=R2=:R0R_{1}=R_{2}=:R_{0} and r1=r2=:r0,r_{1}=r_{2}=:r_{0}, the vector conditions in Theorem 2.1 (202^{0}) are

(Iα[1])1α((t12)f(R))+12f(R)R,\left(I-\alpha\left[1\right]\right)^{-1}\alpha\left(\left(t-\frac{1}{2}\right)f\left(R\right)\right)+\frac{1}{2}f\left(R\right)\leq R,
(Iα[1])1α((t12)f(r))>r.\left(I-\alpha\left[1\right]\right)^{-1}\alpha\left(\left(t-\frac{1}{2}\right)f\left(r\right)\right)>r.

Explicitly, the first condition gives

(2.23) γf1(R0,R0)R0,γf2(R0,R0)R0,\gamma f_{1}\left(R_{0},R_{0}\right)\leq R_{0},\ \ \ \gamma f_{2}\left(R_{0},R_{0}\right)\leq R_{0},

where γ=α1(t1/2)/(1α1(1))+1/2=1/3+1/2=5/6.\gamma=\alpha_{1}\left(t-1/2\right)/\left(1-\alpha_{1}\left(1\right)\right)+1/2=1/3+1/2=5/6. Since for both i=1,2,i=1,2, one has fi(x,x)/x0f_{i}\left(x,x\right)/x\rightarrow 0 as x+,x\rightarrow+\infty, such a number R0R_{0} exists sufficiently large. The second vector condition gives

(2.24) ηf1(r0,r0)>r0,ηf2(r0,r0)>r0,\eta f_{1}\left(r_{0},r_{0}\right)>r_{0},\ \ \ \eta f_{2}\left(r_{0},r_{0}\right)>r_{0},

where η=α1(t1/2)/(1α1(1))=1/3.\eta=\alpha_{1}\left(t-1/2\right)/\left(1-\alpha_{1}\left(1\right)\right)=1/3. These inequalities yield to the system

6r022r0+3λ1<0, 6r0<23λ26r_{0}^{2}-2r_{0}+3\lambda_{1}<0,\ \ \ 6r_{0}<2-3\lambda_{2}

which has a positive solution if λ1(0,1/18)\lambda_{1}\in\left(0,1/18\right) and λ2(0,1/3].\lambda_{2}\in(0,1/3]. Thus Theorem 2.1 (202^{0}) applies and gives the conclusion.

Example 2.6.

Consider problem (2.22) where

f1(u)=λ1u12(|u2|+1)(u12+1)(|u2|+2),f2(u)=λ2|u1|+|u2||u1|+|u2|+1.f_{1}\left(u\right)=\lambda_{1}\frac{u_{1}^{2}\left(\left|u_{2}\right|+1\right)}{\left(u_{1}^{2}+1\right)\left(\left|u_{2}\right|+2\right)},\ \ \ \ f_{2}\left(u\right)=\lambda_{2}\frac{\left|u_{1}\right|+\left|u_{2}\right|}{\left|u_{1}\right|+\left|u_{2}\right|+1}.

For each number r0>0r_{0}>0 and any large enough λ1,λ2>0,\lambda_{1},\lambda_{2}>0, the problem has at least three nonzero solutions. Indeed, with the notations from the previous example, the conditions (2.24) are satisfied for sufficiently large λ1\lambda_{1} and λ2.\lambda_{2}. Next conditions (2.23) allow us to find R0>r0R_{0}>r_{0} large enough since for both i=1,2,i=1,2, one has fi(x,x)/x0f_{i}\left(x,x\right)/x\rightarrow 0 as x+.x\rightarrow+\infty. Now we put Λ={1},\Lambda=\left\{1\right\}, so τ2=R2(=R0),\tau_{2}=R_{2}\left(=R_{0}\right), and we look for τ1(0,r0)\tau_{1}\in\left(0,r_{0}\right) such that the additional condition in Theorem 2.1 (303^{0}) is satisfied. Thus, we need that

γf1(τ1,R0)<τ1.\gamma f_{1}\left(\tau_{1},R_{0}\right)<\tau_{1}.

Such a number τ1\tau_{1} exists since f1(x,R0)/x0f_{1}\left(x,R_{0}\right)/x\rightarrow 0 as x0.x\rightarrow 0. Now the conclusion follows from Theorem 2.1 (303^{0}).

In the above examples we have considered autonomous equations. We emphasize the more general applicability of Theorem 2.1 to nonautonomous equations, as the following example shows.

Example 2.7.

In problem (2.19), consider the tt-depending function

f(t,u)=5u2+λ1(t+1)u2+λ2.f\left(t,u\right)=\frac{5u^{2}+\lambda_{1}\left(t+1\right)}{u^{2}+\lambda_{2}}\ .

As in Example 2.3, if λ2(0,25/36)\lambda_{2}\in\left(0,25/36\right) and λ1(0,λ2)\lambda_{1}\in\left(0,\lambda_{2}\right) is sufficiently small, there exists at least three nonzero solutions. Indeed, by direct computation involving maximization and minimization of f(t,u)f\left(t,u\right) also with respect to t,t, one can see that conditions ((30)(3^{0}): ) and (2.7) of Theorem 2.1 are satisfied providing that

(2.25) 6τ325τ2+6λ2τ10λ1\displaystyle 6\tau^{3}-25\tau^{2}+6\lambda_{2}\tau-10\lambda_{1} >\displaystyle> 0,\displaystyle 0,
3r35r2+3λ2rλ1\displaystyle 3r^{3}-5r^{2}+3\lambda_{2}r-\lambda_{1} <\displaystyle< 0.\displaystyle 0.

Similarly to Example 2.3, these inequalities make possible the choice of rr and τ.\tau. In addition, condition ((10)(1^{0}): ) is fulfilled by a sufficiently large R.R.

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