On a certain way of proving the solvability for boundary value problems

Abstract

A certain way of replacing a given boundary value problem by another one, a solution of which solves also the original problem, is considered.

MSC:34B15.

Consider the solvability of the boundary value problem (BVP)

$${\left(\phi (t,x,x^{\prime })\right)}^{\prime }=f(t,x,x^{\prime }),t\in I=[a,b],$$

(1)

$$H_{1}x=h_1,H_{2}x=h_2,\alpha \le x\le \beta ,$$

(2)

where $\phi \in C(I\times R^2,R)$ is strictly increasing in $x^{\prime }$ for fixed t and x, $f:I\times R^2\to R$ satisfies the Caratheodory conditions, that is, $f(t,\cdot ,\cdot )$ is measurable in I for fixed $x,x^{\prime }\in R$, $f(\cdot ,x,x^{\prime })$ is continuous on $R^2$ for fixed $t\in I$, and for any compact set $P\subset R^2$ there exists function $g\in L(I,R)$ such, that for any $(t,x,x^{\prime })\in I\times P$, the estimate $|f(t,x,x^{\prime })|\le g(t)$ holds, $H_1,H_2\in C(C^1(I,R),R)$, $h_1,h_2\in R$, α is the lower function, β the upper function.

This boundary value problem is replaced by another one, which is dependent on the parameter $M\in (M_0,+\mathrm{\infty })$, $M_0>0$,

$$\begin{array}{r}{\left({\phi }_M(t,x,x^{\prime })\right)}^{\prime }=f_M(t,x,x^{\prime }),t\in I=[a,b],\\ H_{1}x=h_1,H_{2}x=h_2,\alpha \le x\le \beta ,\end{array}$$

(3)

where ${\phi }_M\in C(I\times R^2,R)$ is strictly increasing in $x^{\prime }$ for fixed t and x, and $f:I\times R^2\to R$ satisfies the Caratheodory conditions.

Definition 1 A function $x\in C^1(I,R)$ is a solution of (1), if $\phi (t,x(t),x^{\prime }(t))$ is absolutely continuous on I and (1) is satisfied almost everywhere on I.

We provide below definitions of generalized upper and lower functions and the generalized solution along with Theorem 1 from [1–3]. This is needed to prove the main result.

Definition 2 The class $B{B}^{+}(I,R)$ consists of functions $\alpha :I\to R$, which possess the property: for any $t\in (a,b]$ there exist the left derivative ${\alpha }_l^{\prime }(t)$ and the limit ${lim}_{\tau \to t-}{\alpha }_l^{\prime }(\tau )$, and ${\alpha }_l^{\prime }(t)\ge {lim}_{\tau \to t-}{\alpha }_l^{\prime }(\tau )$; for any $t\in [a,b)$ there exist the right derivative ${\alpha }_r^{\prime }(t)$ and the limit ${lim}_{\tau \to t+}{\alpha }_r^{\prime }(\tau )$, and ${\alpha }_r^{\prime }(t)\le {lim}_{\tau \to t+}{\alpha }_r^{\prime }(\tau )$, and, for any $t\in (a,b)$, ${\alpha }_l^{\prime }(t)\le {\alpha }_r^{\prime }(t)$.

The class $B{B}^{-}(I,R)$ consists of functions $\beta :I\to R$, which possess the following property: for any $t\in (a,b]$ there exist the left derivative ${\beta }_l^{\prime }(t)$ and the limit ${lim}_{\tau \to t-}{\beta }_l^{\prime }(\tau )$, and ${\beta }_l^{\prime }(t)\le {lim}_{\tau \to t-}{\beta }_l^{\prime }(\tau )$; for any $t\in [a,b)$ there exist the right derivative ${\beta }_r^{\prime }(t)$ and the limit ${lim}_{\tau \to t+}{\beta }_r^{\prime }(\tau )$, and ${\beta }_r^{\prime }(t)\ge {lim}_{\tau \to t+}{\beta }_r^{\prime }(\tau )$, and, for any $t\in (a,b)$, ${\beta }_l^{\prime }(t)\ge {\beta }_r^{\prime }(t)$.

Definition 3 We call a bounded function $\alpha \in B{B}^{+}(I,R)$ a generalized lower function and write $\alpha \in AG(I,R)$, if in any interval $[c,d]\in I$, where this function satisfies the Lipschitz condition, for any $t_1\in (c,d)$ and $t_2\in (t_1,d)$ where the derivative exists, the inequality

$$\phi (t_2,\alpha (t_2),{\alpha }^{\prime }(t_2))-\phi (t_1,\alpha (t_1),{\alpha }^{\prime }(t_1))\ge {\int }_{t_1}^{t_2}f(s,\alpha (s),{\alpha }^{\prime }(s))ds$$

holds. We will call a bounded function $\beta \in B{B}^{-}(I,R)$ a generalized upper function and write $\beta \in BG(I,R)$, if in any interval $[c,d]\in I$, where this function satisfies the Lipschitz condition, for any $t_1\in (c,d)$ and $t_2\in (t_1,d)$ where the derivative exists, the inequality

$$\phi (t_2,\beta (t_2),{\beta }^{\prime }(t_2))-\phi (t_1,\beta (t_1),{\beta }^{\prime }(t_1))\le {\int }_{t_1}^{t_2}f(s,\beta (s),{\beta }^{\prime }(s))ds$$

holds.

A function $x:I\to R$ will be called a generalized solution, if $x\in AG(I,R)\cap BG(I,R)$.

A generalized solution has a derivative at any point, possibly infinite, either −∞ or +∞, and $x^{\prime }$ is continuous on $[-\mathrm{\infty },+\mathrm{\infty }]$; if in some interval the derivative $x^{\prime }$ does not attain the values −∞ or +∞, then x is a solution of (1) in this interval.

Theorem 1 Let $\alpha \in AG(I,R)$, $\beta \in BG(I,R)$ and $\alpha \le \beta$. Then for any $A\in [\alpha (a),\beta (a)]$ and $B\in [\alpha (b),\beta (b)]$ there exists a generalized solution of the Dirichlet problem

$${\left(\phi (t,x,x^{\prime })\right)}^{\prime }=f(t,x,x^{\prime }),x(a)=A,x(b)=B,\alpha \le x\le \beta .$$

(4)

In addition to conditions on α and β the compactness conditions are needed for solvability of the boundary value problem (1)-(2). The Nagumo condition [4] for φ-Laplacian and the Schrader condition [5] are sufficient conditions for compactness of a set of solutions. We accept the following compactness conditions.

Definition 4 We say that the compactness condition is fulfilled, if for all $A\in [\alpha (a),\beta (a)]$ and $B\in [\alpha (b),\beta (b)]$ any generalized solution of the Dirichlet problem (4) is a solution.

It is clear that this condition is weaker than the Schrader condition.

A set of solutions of the Dirichlet problem (4) will be denoted by S.

Remark 1 If $\alpha \in AG(I,R)$, $\beta \in BG(I,R)$, $\alpha \le \beta$ and the compactness condition is fulfilled, then the Dirichlet problem (4) has a solution.

Theorem 2 Let $\alpha \in AG(I,R)$, $\beta \in BG(I,R)$ and the compactness condition be fulfilled. If the boundary value problem (3) has a solution $u_M$ for all $M\in (M_0,+\mathrm{\infty })$ and for $t\in I$

$${\phi }_M(t,x,x^{\prime })=\phi (t,x,x^{\prime }),f_M(t,x,x^{\prime })=f(t,x,x^{\prime }),\alpha \le x\le \beta ,\left|x^{\prime }\right|\le M,$$

then there exists $M_1\in (M_0,+\mathrm{\infty })$ such that $u_{M_1}$ solves the boundary value problem (1)-(2).

Proof Notice that the results in [6] imply that $sup\{{\parallel x^{\prime }\parallel }_C:x\in S\}=M_0<+\mathrm{\infty }$. Suppose the contrary. Let the sequence $\{M_i\}$, where $M_i\in (M_0,+\mathrm{\infty })$, $i=1,2,\dots$ tend to infinity. Consider the sequence $\{u_i\}$, where $u_i=u_{M_i}$, $i=1,2,\dots$ . We can assume, without loss of generality, that it converges in any rational points of the interval $(a,b)$ to the function u, located between α and β. Notice that without loss of generality for any interval $(a_1,b_1)\subset (a,b)$ it follows from the boundedness of u and the Mean Value Formula that there exists an interval $[c,d]\subset (a_1,b_1)$ such that

$$sup\left\{\right|u_i^{\prime }(t)|:i\in \{1,2,\dots \},t\in [c,d]\}=L<+\mathrm{\infty }.$$

It is clear that $u_i$, $i\in \{1,2,\dots \}$, and u satisfy the Lipschitz condition with constant L in $[c,d]$. The u can be extended by continuity to the entire interval $[c,d]$, and thus we obtain a function u that satisfies the Lipschitz condition. It follows from the Lipschitz condition that $\{u_i(t)\}$ converges to $u(t)$ for any $t\in [c,d]$. It is clear that the derivatives $\{u_i^{\prime }(t)\}$ converge to the derivative $u^{\prime }(t)$ for any $t\in [c,d]$. Therefore, $u(t)$ is a solution of (1) in the interval $[c,d]$. Continuing the construction of $u(t)$ on both sides, one gets a solution of (1) on the maximal interval $(c_1,d_1)$. If $c_1>a$, then ${lim}_{t\to c_1+}{u}^{\prime }(t)$ is either −∞ or +∞. Similarly, if $d_1<b$, then ${lim}_{t\to d_1-}{u}^{\prime }(t)$ is either −∞ or +∞. If $c_1=a$ and ${lim}_{t\to a+}{u}^{\prime }(t)$ is not −∞ or +∞, then $u(t)$ can be continued to a. Similarly, if $d_1=b$ and ${lim}_{t\to b-}{u}^{\prime }(t)$ is not −∞ nor +∞, then $u(t)$ can be continued to b. By repeating this construction, find an open set $I_1$ in I, where the function $u(t)$ is defined and $u(t)$ is a solution of (1) on intervals from $I_1$. A set $I_2=I\mathrm{\setminus }{I}_1$ is closed and nowhere dense. For $t\in I_2$ the limit ${lim}_{i\to \mathrm{\infty }}{u}_i^{\prime }(t)$ is equal to −∞ or +∞. Indeed, assuming the contrary and acting as above, we get $t\in I_1$. Extend $u(t)$ to irrational points of $I_2$. If $a\in I_2$, then $u(a)={lim}_{t\to a+}u(t)$, and in the remaining cases $u(\tau )={lim}_{t\to \tau -}u(t)$. The above limits exist since $u(t)$ is monotone in neighborhood of any point from $I_2$. Similarly we get for $t\in I_2$, $u^{\prime }(t)={lim}_{i\to \mathrm{\infty }}{u}_i^{\prime }(t)$ and ${lim}_{\tau \to t}{u}^{\prime }(\tau )=u^{\prime }(t)$. Therefore $u(t)$ is a generalized solution of (1). It follows from the compactness condition that $u(t)$ is a solution of (1). Let us show that the sequence $\{u_i^{\prime }(t)\}$ uniformly converges to $u^{\prime }(t)$. Suppose the contrary is true. We assume, without loss of generality, that there exist $\epsilon >0$ and a sequence $\{t_i\}$, where $t_i\in I$, $i=1,2,\dots$ such that $|u^{\prime }(t_i)-u_i^{\prime }(t_i)|>\epsilon$, $i=1,2,\dots$ and ${lim}_{i\to \mathrm{\infty }}{t}_i=t_0$. Consider the case $u_i^{\prime }(t_i)>u^{\prime }(t_i)+\epsilon$, $i=1,2,\dots$ . We can assume, without loss of generality, that $u_i^{\prime }(t_0)>u^{\prime }(t_0)+\epsilon /2$, $i=1,2,\dots$ , and this contradicts the equality ${lim}_{i\to \mathrm{\infty }}{u}_i^{\prime }(t_0)=u^{\prime }(t_0)$. The uniform convergence is proved. We can conclude now that all $u_i(t)$ are the solutions of the boundary value problem (1)-(2). □

Remark 2 Theorem 2 gives the possibility to prove the solvability of boundary value problems if the solvability of more simple boundary value problems is known.

Remark 3 If ${\alpha }^{\prime }(a)\ge {\beta }^{\prime }(a)$ and the inequalities ${\alpha }^{\prime }(a)\ge x^{\prime }(a)\ge {\beta }^{\prime }(a)$ hold for a solution x of the boundary value problem (1)-(2), then the compactness condition (Definition 4) can be weakened.

Definition 5 We will say that the compactness condition holds if for any $A_1\in [{\beta }^{\prime }(a),{\alpha }^{\prime }(a)]$ and $B\in [\alpha (b),\beta (b)]$ all generalized solutions of the problem

$${\left(\phi (t,x,x^{\prime })\right)}^{\prime }=f(t,x,x^{\prime }),x^{\prime }(a)=A_1,x(b)=B,\alpha \le x\le \beta ,$$

are classical solutions.

Example One way to use Theorem 2 is to verify that for all $t\in I$, $x,x^{\prime }\in R$ and $M\in (M_0,+\mathrm{\infty })$, $M_0>0$, the following conditions are satisfied:

$$\begin{array}{c}{\phi }_M(t,x,x^{\prime })=\phi (t,x,x^{\prime }),\hfill \\ f_M(t,x,x^{\prime })=f(t,x,\delta (-M,x^{\prime },M)),\hfill \end{array}$$

where $\delta (u,v,w)=u$ if $v<u$, $\delta (u,v,w)=v$ if $u\le v\le w$, $\delta (u,v,w)=w$ if $w<v$.