Boundary value problems for second-order causal differential equations – Boundary Value Problems

In this part, the existence of extremal solutions to (1) is shown.

The function

$\varphi \in {}^{}$
C
2
(J,R)

is called a lower solution of (1) if

and an upper solution of (1) if the reversed inequalities hold.

To demonstrate the existence of extremal solutions to (1), the corresponding linear problem is considered, given by

where (sigma _{delta}(mathfrak{s})=(mathscr{Q}delta )(mathfrak{s})+ mathscr{Z}(mathfrak{s})delta (mathfrak{s})+(Gamma delta )( mathfrak{s})).

Theorem 3.1

Assume that

((H_{1})):

(3) holds with (mathscr{Z}in C(J,[0,+infty ])), (Phi (mathfrak{s})>0), (mathfrak{s}in (0,1)), and Γ being a positive linear operator;

((H_{2})):

$\varphi ,\psi \in {}^{}$
C
2
(J,R)

are lower and upper solutions of problem (1), respectively, and (phi leq psi );

((H_{3})):

(mathscr{Q}in C(E,E)) satisfies

$$ (mathscr{Q}bar{g}) (mathfrak{s})-(mathscr{Q}g) (mathfrak{s})geq – mathscr{Z}(mathfrak{s}) bigl(bar{g}(mathfrak{s})-g(mathfrak{s})bigr)-bigl( Gamma (bar{g}-g)bigr) (mathfrak{s}), $$

for (phi (mathfrak{s})leq g(mathfrak{s})leq bar{g}(mathfrak{s}) leq psi (mathfrak{s})), (mathfrak{s}in J);

((H_{4})):

$\delta \in {}^{}$
C
2
(J,R)

and (phi (mathfrak{s})leq delta (mathfrak{s})leq psi (mathfrak{s})), (mathfrak{s}in J).

Then problem (4) has only one solution in the sector

Proof

We need four steps to complete the proof.

Step 1. We rewrite problem (4) in the following way:

where (G(mathfrak{s},s)) is the Green’s function defined by

and (omega (mathfrak{s})=k_{1}mathfrak{s}+k_{2}) is the solution of the associated boundary value problem:

where (k_{1}= frac {alpha _{0}chi _{1}-alpha _{1}chi _{0}}{alpha _{0} (alpha _{1}+beta _{1})+alpha _{1}beta _{0}}) and (k_{2}= frac {chi _{0}(alpha _{1}+beta _{1})+chi _{1}beta _{0}}{alpha _{0}(alpha _{1}+beta _{1})+alpha _{1}beta _{0}}).

Apparently, if

$\phi (\mathfrak{s})\in {}^{}$
C
2
(J,R)

is a solution of (5), we have (alpha _{0}varphi (0)-beta _{0}varphi ‘(0)=chi _{0}), (alpha _{1}varphi (1)+beta _{1}varphi ‘(1)=chi _{1}), and (varphi ”(mathfrak{s})=mathscr{Z}(mathfrak{s})varphi ( mathfrak{s})+(Gamma varphi )(mathfrak{s})-sigma _{delta}( mathfrak{s})), so φ is a solution of problem (4).

Step 2. We show that problem (5) has a solution.

Notice that E is a Banach space and (|varphi |= max_{mathfrak{s}in J}|varphi ( mathfrak{s})|). For the purpose of using Schauder’s fixed point theorem, we consider the right-hand side of (5) and denote it using operator (mathscr{P}): (Eto E). It follows that (mathscr{Z}(mathfrak{s})varphi (mathfrak{s})+(Gamma varphi )( mathfrak{s})-sigma _{delta}(mathfrak{s})) is bounded on J, (mathscr{P}) is continuous and bounded.

Moveover, let (|mathscr{Z}(mathfrak{s})varphi (mathfrak{s})+(Gamma varphi )( mathfrak{s})-sigma _{delta}(mathfrak{s})|leq T), (T>0), and take (mathfrak{s}_{1},mathfrak{s}_{2}in J), (mathfrak{s}_{1}<mathfrak{s}_{2}). Then we have

where (Q= frac{1}{frac{beta _{0}}{alpha _{0}}+frac{beta _{1}}{alpha _{1}}+1}) and (k=|k_{1}|). As (mathfrak{s}_{2}to mathfrak{s}_{1}), the right-hand side of the above inequality tends to zero. Thus operator (mathscr{P}) is compact. It then follows from Schauder’s fixed point theorem that (mathscr{P}) has a fixed point. Apparently, this fixed point is the solution of (4).

Step 3. We show that problem (4) has at most one solution.

We suppose that problem (4) has two different solutions (varphi _{1},varphi _{2}in C^{2}(J,R)). Set (lambda =varphi _{1}-varphi _{2}), then (lambda ”(mathfrak{s})=mathscr{Z}(mathfrak{s})lambda ( mathfrak{s})+(Gamma lambda )(mathfrak{s})) and (mathcal{P}lambda (theta )=0), (theta =0,1) on J. From ((H_{1})) and Lemma 2.1, we obtain (varphi _{1}(mathfrak{s})geq varphi _{2}(mathfrak{s})), (mathfrak{s}in J). Now letting (lambda =varphi _{2}-varphi _{1}), we get (varphi _{2}(mathfrak{s})geq varphi _{1}(mathfrak{s})), (mathfrak{s}in J), based on Lemma 2.1. Hence (varphi _{1}(mathfrak{s})=varphi _{2}(mathfrak{s})), (mathfrak{s}in J).

By the above steps, we obtain that problem (4) has a unique solution. Denote it as (varphi =varphi (mathfrak{s})).

Step 4. We can prove (varphi in [phi ,psi ]).

Setting (lambda (mathfrak{s})=phi (mathfrak{s})-varphi (mathfrak{s})), due to ((H_{2})), ((H_{3})), and ((H_{4})), we acquire

Noticing (mathcal{P}lambda (theta )leq 0), (theta =0,1), it then follows from Lemma 2.1 that (phi leq varphi ). Analogously, we can prove that (varphi leq psi ), and we have (varphi in [phi ,psi ]). The proof is then completed. □

Theorem 3.2

Let the assumptions ((H_{1}))–((H_{3})) be satisfied. Then problem (1) has extremal solutions in the sector ([phi ,psi ]).

Proof

For each (delta in [phi ,psi ]), consider the boundary value problem (4). By Theorem 3.1, problem (4) possesses a unique solution

$\phi \in {}^{}$
C
2
(J,R)

, and we define the mapping (mathcal{F}) by (mathcal{F}delta =varphi ). We shall use this mapping to construct two sequences ({phi _{n}}), ({psi _{n}}). For this purpose, we shall prove that

1. (1)

   (phi leq mathcal{F}phi ), (psi geq mathcal{F}psi );

2. (2)

   (mathcal{F}) is a monotone mapping in ([phi ,psi ]).

In order to prove (1), set (lambda (mathfrak{s})=phi (mathfrak{s})-phi _{1}(mathfrak{s})), where (phi _{1}=mathcal{F}phi ). Then we have

and

By virtue of Lemma 2.1, one arrives at (lambda leq 0), so (phi leq phi _{1}). Analogously, one attains (mathcal{F}psi leq psi ).

In order to prove (2), let (delta _{1}leq delta _{2}) be such that (delta _{1}, delta _{2}in [phi ,psi ]). Assume that (varphi _{1}=mathcal{F}delta _{1}), (varphi _{2}=mathcal{F}delta _{2}), and (lambda (mathfrak{s})=varphi _{1}(mathfrak{s})-varphi _{2}( mathfrak{s})). We obtain

and

In view of Lemma 2.1, we derive (lambda leq 0), which implies (mathcal{F}delta _{1}leq mathcal{F}delta _{2}).

Now, define the sequences ({phi _{n}(mathfrak{s})}), ({varphi _{n}(mathfrak{s})}) by (phi _{n}=mathcal{F}phi _{n-1}), (psi _{n}=mathcal{F}psi _{n-1}), and (phi _{0}=phi ), (psi _{0}=psi ), and conclude from previous arguments that

By means of standard arguments, we derive that (lim_{nto infty}phi _{n}(mathfrak{s})=xi ( mathfrak{s})) and (lim_{nto infty}psi _{n}(mathfrak{s})=zeta ( mathfrak{s})) uniformly and monotonically on J. It is easy to see that ξ, ζ are solutions of problem (1).

To demonstrate ξ and ζ are extremal solutions to (1), we set u be an arbitrarily solution to (1) with (phi leq uleq psi ). Suppose that for some

$n\in \mathbb{N}$

, (phi _{n}leq uleq psi _{n}). Employing the monotonic nondecreasingness property of (mathcal{F}), we acquire (phi _{n+1}=mathcal{F}phi _{n}leq mathcal{F}u=u), hence (phi _{n+1}leq u) on J. Analogously, we have (uleq psi _{n+1}) on J. Note that (phi _{0}leq uleq psi _{0}), so by induction we see that (phi _{n}leq uleq psi _{n}) for every n. Taking the limit as (nto infty ), one concludes (xi leq uleq zeta ), and the proof is complete. □