Here we study the existence of solutions of a nonlocal two-point, with parameters, boundary value problem of a first order nonlinear differential equation. The maximal and minimal solutions will be proved. The continuous dependence of the unique solution on the parameters of the nonlocal condition will be proved. The anti-periodic boundary value problem will be considered as an application.
The existence of solutions of the two-point boundary value problem of the first order differential equation \[ \frac{dx}{dt}=f(t,x(t)), ~~~~~~~~ t\in(0,T) \] have been studied in [1,2,3]. Also, some nonlocal problems of differential equations have been considered in [4,5,6,7,8,9,10,11,12,13,14,15,16,17]. Consider the nonlocal two-point boundary value problem with parameters \(\alpha\) and \(\beta\);
Lemma 1. If the solution of the Problem (1)-(2) exists, then it can be expressed by the integral equation
Proof. Let the boundary value Problem (1)-(2). Integrating Equation (1), we obtain
Theorem 1. Let the assumption (i)-(ii)-(iii) are satisfied, then there exists at least one absolutely continuous solution \(x\in AC[0,T]\) of the Problem (1)-(2).
Proof. Define the operator \(F\) by \[ Fx(t)=\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t}f(s,x(s))ds. \] Define the set \[ Q_r =\{x:||x||\leq r\}\subset C[0,T],~~~~~r=\frac{|x_0|+2||m||(\alpha+\beta)}{(\alpha+\beta)(1-2b)}. \] Let \( x\in Q_r \), then \begin{eqnarray*} |Fx(t)|&=&|\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t}f(s,x(s))ds|\\ &\leq&\frac{1}{\alpha+\beta}[|x_0|+\alpha\int_{0}^{\tau}|f(s,x(s))|ds+\beta\int_{0}^{\eta}|f(s,x(s))|ds]+\int_{0}^{t}|f(s,x(s))|ds\\ &\leq&\frac{1}{\alpha+\beta}[|x_0|+\alpha\int_{0}^{\tau}(|m(s)|+b|x|)ds+\beta\int_{0}^{\eta}(|m(s)|+b|x|)ds]+\int_{0}^{t}(|m(s)|+b|x|)ds \end{eqnarray*} \begin{eqnarray*} &\leq&\frac{1}{\alpha+\beta}[|x_0|+\alpha\int_{0}^{T}(|m(s)|+b|x|)ds+\beta\int_{0}^{T}(|m(s)|+b|x|)ds]+\int_{0}^{T}(|m(s)|+b|x|)ds\\ &\leq&\frac{1}{\alpha+\beta}[|x_0|+\frac{\alpha}{\alpha+\beta}(||m||+b||x||T)+\frac{\beta}{\alpha+\beta}(||m||+b||x||T)]+(||m||+b||x||T)\\ &\leq&\frac{1}{\alpha+\beta}|x_0|+2(||m||+b||x||T)\leq r. \end{eqnarray*} Then the class of functions \(\{F{x}\}\) is uniformly bounded on \(Q_r\), and \(F:Q_r\rightarrow Q_r\). Let \(x\in Q_r\) and \(t_1,t_2\in[0,T] \), such that \(|t_2-t_1|< \delta\), then \begin{eqnarray*} |Fx(t_2)-Fx(t_1)|&=&|\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t_2}f(s,x(s))ds \\ &&-\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t_1}f(s,x(s))ds| \\ &\leq&\int_{t_1}^{t_2}|f(s,x(s))|ds\\ &\leq&\int_{t_1}^{t_2}(|m(s)|+b|x|)ds \\ &\leq&\int_{t_1}^{t_2}|m(s)|ds+b|x|(t_2-t_1). \end{eqnarray*} So, the class of functions \(\{Fx\}\) is equi-continuous on \(Q_r\). From Arzela Theorem [18] we deduce that the class of functions \(\{Fx\}\) is compact, and \(F:Q_r\rightarrow Q_r\) is compact. Now we prove that \(F\) is continuous operator. For this let \(\{x_n\}\subset Q_r\) be convergent sequence such that \(x_n(t)\rightarrow x_0(t)\), then \begin{equation*} Fx_n(t)=\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x_n(s))ds-\beta\int_{0}^{\eta}f(s,x_n(s))ds]+\int_{0}^{t}f(s,x_n(s))ds \end{equation*} and \begin{equation*} \lim_{n\rightarrow\infty}Fx_{n}(t)=\frac{1}{\alpha+\beta}[x_0-\alpha\lim_{n\rightarrow\infty}\int_{0}^{\tau}f(s,x_n(s))ds-\beta\lim_{n\rightarrow\infty}\int_{0}^{\eta}f(s,x(s))ds] +\lim_{n\rightarrow\infty}\int_{0}^{t}f(s,x_n(s))ds. \end{equation*} From assumptions (i), (ii), we have \[ f(s,x_n(s))\rightarrow f(s,x_0(s)) \] and \[ |f(s,x(s))|\leq|m(s)|+b|x|\in L_1[0,T]. \] Applying Lebesgue dominated convergence Theorem [18], we have \begin{eqnarray*} \lim_{n\rightarrow\infty}Fx_{n}(t)&=&\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}\lim_{n\rightarrow\infty}f(s,x_n(s))ds-\beta\int_{0}^{\eta}\lim_{n\rightarrow\infty}f(s,x_n(s))ds]+\int_{0}^{t}\lim_{n\rightarrow\infty}f(s,x_n(s))ds\\ &=&\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,\lim_{n\rightarrow\infty}x_n(s))ds-\beta\int_{0}^{\eta}f(s,\lim_{n\rightarrow\infty}x_n(s))ds]+\int_{0}^{t}f(s,\lim_{n\rightarrow\infty}x_n(s))ds\\ &=&\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x_0(s))ds-\beta\int_{0}^{\eta}f(s,x_0(s))ds]+\int_{0}^{t}f(s,x_0(s))ds=Fx_0(t). \end{eqnarray*} Hence, \(F:Q_r\rightarrow Q_r\) is continuous. Now by Schauder fixed point Theorem [19] there exists at least one solution \(x\in C[0,T]\) of the Problem (1)-(2). Let \(x\in C[0,T]\) be a solution of the Problem (1)-(2). Differentiating the integral Equation (7), we obtain \begin{eqnarray*} \frac{dx}{dt}&=&\frac{d}{dt}(\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t}f(s,x(s))ds)\\ &=&\frac{d}{dt}\int_0^t f(s,x(s))ds. \end{eqnarray*} Since \(f\) is measurable in \(t\in[0,T]\) and bounded by integrable function, then \(f\in L^1[0,T]\), and \[ \frac{dx}{dt}=f(t,x(t)) ~~~~~ a.e, ~~~~~ t\in(0,T]. \] Putting \(t=\tau\) in the integral Equation (7), we get \[ x(\tau)=\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{\tau}f(s,x(s))ds. \] and \begin{eqnarray*} \alpha x(\tau)&=&\frac{\alpha}{\alpha+\beta}[x_0-\frac{\alpha_2}{\alpha+\beta}\int_{0}^{\tau}f(s,x(s))ds -\frac{\alpha\beta}{\alpha+\beta}\int_{0}^{\eta}f(s,x(s))ds]+\alpha\int_{0}^{\tau}f(s,x(s))ds\\ &=&\frac{\alpha}{\alpha+\beta}x_0+\frac{\alpha\beta}{\alpha+\beta}\int_{0}^{\tau}f(s,x(s))ds -\frac{\alpha\beta}{\alpha+\beta}\int_{0}^{\eta}f(s,x(s))ds. \end{eqnarray*} Also \[ \beta x(\eta)=\frac{\beta}{\alpha+\beta}x_0-\frac{\alpha\beta}{\alpha+\beta}\int_{0}^{\tau}f(s,x(s))ds +\frac{\alpha\beta}{\alpha+\beta}\int_{0}^{\eta}f(s,x(s))ds. \] Then \[ \alpha x(\tau)+\beta x(\eta)=x_0,~~~~~ \tau\in[0,T) , \eta\in(0,T]. \]
Lemma 2. Let the assumption of Theorem 1 are satisfied and \(x(t)\) and \(y(t)\) are two continuous functions on \([0,T]\) satisfying \begin{eqnarray*} x(t)&\leq&\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t}f(s,x(s))ds,\\ y(t)&\geq&\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,y(s))ds-\beta\int_{0}^{\eta}f(s,y(s))ds]+\int_{0}^{t}f(s,y(s))ds, \end{eqnarray*} and one of them is strict. If \(f\) is monotonic nondecreasing in \(x\), then
Proof. Let the conclusion (8) is false, then there exist \(t_1\) such that \[ x(t_1)=y(t_2)~~~~~t_1>0, \] and \[ x(t)< y(t),~~~~~0< t< t_1. \] From the monotonicity of \(f(t,x(t))\) in \(x\), we have \begin{eqnarray*} x(t_1)&\leq&\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t_1}f(s,x(s))ds,~~~t\in[0,T] \\ &<&\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}f(s,y(s))ds-\beta\int_{0}^{\eta}f(s,y(s))ds]+\int_{0}^{t_1}f(s,y(s))ds,\\ &<&y(t_1). \end{eqnarray*} This contradicts the fact that \(x(t_1)=y(t_1)\), then \[ x(t)< y(t),~~~~~~t\in[0,T]. \] For the existence of the maximal and minimal solutions we have the following theorem.
Theorem 2. Let the assumptions of Theorem 1 are satisfied. If \(f(t,x(t))\) is monotonic nondecreasing in \(x\) for each \(t\in[0,T]\), then the Equation (7) (consequently the Problem (1)-(2) has maximal and minimal solutions.
Proof. Firstly we shall prove the existence of the maximal solution of (7). Let \(\epsilon>0\) be given then consider the integral equation
Consider the problem (1)-(2) under the following assumptions
Theorem 3. Let the assumptions (i\(^*\)) be satisfied. If \( 2LT< 1, \) then the solution of the nonlocal two-point boundary value Problem (1)-(2) is unique.
Proof. From assumption (\(i^*\)) we get \[ |f(t,x)|-|f(t,0)|\leq|f(t,x)-f(t,0)|\leq b|x| \] and \[ |f(t,x)|\leq b|x|+|f(t,0)|= m(t)+ b|x|. \] Then the assumptions (ii) is satisfied, so there exists at least one solution \(x\in AC[0,T]\) of the Problem (1)-(2). Let \(x\) and \(y\) be two solutions of the Problem (1)-(2), then we have \begin{eqnarray*} |x(t)-y(t)|&=& |\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}\int_{0}^{t}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t}f(s,x(s))ds\\ && -\frac{1}{\alpha+\beta}[x_0-\alpha\int_{0}^{\tau}\int_{0}^{t}f(s,y(s))ds-\beta\int_{0}^{\eta}f(s,y(s))ds]+\int_{0}^{t}f(s,y(s))ds|\\ & \leq& \frac{\alpha}{\alpha+\beta}\int_{0}^{\tau}|f(s,x(s))-f(s,y(s))|ds+\frac{\beta}{\alpha+\beta}\int_{0}^{\eta}|f(s,x(s))-f(s,y(s))|ds\\ && +\int_{0}^{t}|f(s,x(s))-f(s,y(s))|ds\\ & \leq& \frac{\alpha}{\alpha+\beta}L\int_{0}^{\tau}|x(s)-y(s)|ds+\frac{\beta}{\alpha+\beta}L\int_{0}^{\eta}|x(s)-y(s)|ds+L\int_{0}^{t}|x(s)-y(s)|ds\\ & \leq& \frac{\alpha}{\alpha+\beta}L\int_{0}^{T}|x(s)-y(s)|ds+\frac{\beta}{\alpha+\beta}L\int_{0}^{T}|x(s)-y(s)|ds+L\int_{0}^{T}|x(s)-y(s)|ds\\ & \leq&\frac{\alpha}{\alpha+\beta}LT||x-y||+\frac{\beta}{\alpha+\beta}LT||x-y||+LT||x-y||. \end{eqnarray*} Then \[ ||x-y||\leq2LT||x-y|| \] and \[ ||x-y||(1-2LT)\leq 0 \Rightarrow ||x-y||=0\Rightarrow x=y. \] Hence, the solution of the integral Equation (7) (consequently the Problem (1)-(2) is unique solution \(x\in AC[0,T]\).
Definition 1. The solution of the nonlocal two-point boundary value Problem (1)-(2) depends continuously on \(x_{0}\), if \(\forall~\epsilon~>0,~~~\exists ~\delta~>0\), we have \[ |x-x^*_{0}|\leq\delta\Rightarrow ~\|x-x^{*}\| \leq \epsilon. \] where \(x^{*}\in AC[0,T]\) is the unique solution of the nonlocal two-point boundary value Problem (1)-(2).
Theorem 4. Let the assumption of Theorem 3 are satisfied, then the solution of a nonlocal two-points boundary value Problem (1)-(2) is dependence continuously on \(x_0\).
Proof. let \(x , x^*\) be the solutions of a nonlocal two-points boundary value Problem (1)-(2), then \begin{eqnarray*} |x(t)-x^*(t)|&= & |\frac{1}{\alpha+\beta}[x_0-\beta\int_{0}^{\tau}f(s,x(s))ds-\beta\int_{0}^{\eta}f(s,x(s))ds]+\int_{0}^{t}f(s,x(s))ds\\ && -\frac{1}{\alpha+\beta}[x^*_0-\beta\int_{0}^{\tau}f(s,x^*(s))ds-\beta\int_{0}^{\eta}f(s,x^*(s))ds]+\int_{0}^{t}f(s,x^*(s))ds|\\ & \leq& \frac{1}{\alpha+\beta}|x_0-x^*_0|+\frac{\alpha}{\alpha+\beta}\int_{0}^{\tau}|f(s,x(s))-f(s,x^*(s))|ds\\ && +\frac{\beta}{\alpha+\beta}\int_{0}^{\eta}|f(s,x(s))-f(s,x^*(s))|ds+\int_{0}^{t}|f(s,x(s))-f(s,x^*(s))|ds\\ & \leq& \frac{1}{\alpha+\beta}|x_0-x^*_0|+\frac{\alpha}{\alpha+\beta}L\int_{0}^{T}|x(s)-x^*(s)|ds\\ && +\frac{\beta}{\alpha+\beta}L\int_{0}^{T}|x(s)-x^*(s)|ds+L\int_{0}^{T}|x(s)-x^*(s)|ds\\ & \leq& \frac{\delta}{\alpha+\beta}+2LT||x-x^*||. \end{eqnarray*} Then \begin{eqnarray*} ||x-x^*||(1-2LT)\leq\frac{\delta}{\alpha+\beta}\;\;\;\text{implies}\;\;\;||x-x^*||\leq\frac{\delta}{(1-2LT)(\alpha+\beta}=\epsilon. \end{eqnarray*} This prove the continuous dependence of solution of the nonlocal two-point boundary value Problem (1)-(2) on \(x_0\).
Definition 2. The solution of the nonlocal two-point boundary value Problem (1)-(2) depends continuously on \(\alpha\) and \(\beta\), if \(\forall~\epsilon~>0 ,~ \exists ~\delta~>0 \), we have \[ |\alpha-\alpha^*|\leq\delta_1,~~~|\beta-\beta^*|\leq\delta_2 \Rightarrow ~\|x-x^{*}\| \leq \epsilon, \] where \(x^{*}\) is the unique solution of the nonlocal two-points boundary value Problem (1)-(2).
Theorem 5. Let the assumption of Theorem 3 is satisfied, then the solution of the nonlocal two-point boundary value Problem (1)-(2) is depends continuously on \(\alpha\), \(\beta\).
Proof. let \(x , x^*\) be the solutions of the nonlocal two-points boundary value Problem (1)-(2), then \begin{eqnarray*} |x(t)-x^*(t)|&=&|\frac{x_0}{\alpha+\beta}-\frac{\alpha}{\alpha+\beta}\int_{0}^{\tau}f(s,x(s))ds-\frac{\beta}{\alpha+\beta}\int_{0}^{\eta}f(s,x(s))ds\\ &&+\int_{0}^{t}f(s,x(s))ds-\frac{1}{\alpha^*+\beta^*}x_0+\frac{\alpha^*}{\alpha^*+\beta^*}\int_{0}^{\tau}f(s,x^*(s))ds\\ &&+\frac{\beta^*}{\alpha^*+\beta^*}\int_{0}^{\eta}f(s,x^*(s))ds-\int_{0}^{t}f(s,x^*(s))ds|\\ &=&|\frac{(\alpha^*-\alpha)(\beta^*-\beta)}{(\alpha+\beta)(\alpha^*+\beta)}x_0-\frac{\alpha}{\alpha+\beta}\int_{0}^{\tau}f(s,x(s))ds+\frac{\alpha}{\alpha+\beta}\int_{0}^{\tau}f(s,x^*(s))ds\\ &&-\frac{\alpha}{\alpha+\beta}\int_{0}^{\tau}f(s,x^*(s))ds-\frac{\beta}{\alpha+\beta}\int_{0}^{\eta}f(s,x(s))ds-\frac{\beta}{\alpha+\beta}\int_{0}^{\eta}f(s,x^*(s))ds\\ &&-\frac{\beta}{\alpha^*+\beta}\int_{0}^{\eta}f(s,x^*(s))ds+\frac{\beta}{\alpha^*+\beta}\int_{0}^{\tau}f(s,x^*(s))ds\\ &&+\frac{\beta^*}{\alpha^*+\beta}\int_{0}^{\eta}f(s,x^*(s))ds+\int_{0}^{t}(f(s,x(s))-f(s,x^*(s)))ds|\\ &\leq&\frac{(\alpha^*-\alpha)(\beta^*-\beta)}{(\alpha+\beta)(\alpha^*+\beta)}|x_0|+\frac{\alpha}{\alpha+\beta}\int_{0}^{\tau}|f(s,x^*(s))-f(s,x(s))|ds\\ &&+\frac{\beta}{\alpha+\beta}\int_{0}^{\eta}|f(s,x(s))-f(s,x^*(s))|ds+\frac{\alpha^*(\alpha+\beta)-\alpha(\alpha^*+\beta^*)}{(\alpha+\beta)(\alpha^*+\beta)}\int_{0}^{\tau}|f(s,x^*(s))|ds\\ &&+\frac{\beta^*(\alpha+\beta)-\beta(\alpha^*+\beta^*)}{(\alpha+\beta)(\alpha^*+\beta)}\int_{0}^{\eta}|f(s,x^*(s))|ds+\int_{0}^{t}(f(s,x(s))-f(s,x^*(s)))ds|\\ &\leq&\frac{\delta_1+\delta_2}{(\alpha+\beta)(\alpha^*+\beta)}|x_0|+\frac{\alpha}{\alpha+\beta}L\int_{0}^{T}|x(s)-x^*(s)|ds\\ &&+\frac{\beta}{\alpha+\beta}L\int_{0}^{T}|x(s)-x^*(s)|ds+\frac{\alpha^*(\alpha+\beta)-\alpha(\alpha^*+\beta^*)}{(\alpha+\beta)(\alpha^*+\beta)}\int_{0}^{T}|f(s,x^*(s))|ds\\ &&+\frac{\beta^*(\alpha+\beta)-\beta(\alpha^*+\beta^*)}{(\alpha+\beta)(\alpha^*+\beta)}\int_{0}^{T}|f(s,x^*(s))|ds+L\int_{0}^{T}|x(s)-x^*(s)|ds. \end{eqnarray*} Then \begin{eqnarray*} ||x-x^*||&\leq&\frac{(\delta_1+\delta_2}{(\alpha+\beta)(\alpha^*+\beta^*)}|x_0|+\frac{\alpha}{\alpha+\beta}LT||x-x^*||+\frac{\beta}{\alpha+\beta}LT||x-x^*||+\frac{\alpha^*\beta-\alpha\beta^*}{(\alpha+\beta)(\alpha^*+\beta^*)}M\\ &&+\frac{\beta^*\alpha-\beta\alpha^*}{(\alpha+\beta)(\alpha^*+\beta^*)}M+LT||x-x^*||\\ &\leq&\frac{\delta_1+\delta_2}{(\alpha+\beta)(\alpha^*+\beta^*)}|x_0|+2LT||x-x^*|| \end{eqnarray*} and \begin{eqnarray*} (1-2LT)||x-x^*||\leq\frac{\delta_1+\delta_2}{(\alpha+\beta)(\alpha^*+\beta^*)}|x_0|\;\;\;\text{implies}\;\;\; ||x-x^*||\leq\frac{|\delta_1+\delta_2}{(\alpha+\beta)(\alpha^*+\beta^*)(1-2LT)}|x_0|=\epsilon. \end{eqnarray*} This prove that \(\forall\epsilon>0\), \(\exists \delta(\epsilon)>0\) such that \[ |\alpha-\alpha^*|\leq\delta_1, |\beta-\beta^*|\leq\delta_2\Rightarrow||x-x^*||\leq\epsilon. \]
Corollary 6. If \(\alpha=1\), \(\beta=1\) and \(\eta=1-\tau\) and \(x_0=0~\) in Theorem 1, then the anti-periodic boundary value problem \begin{eqnarray*} \frac{dx}{dt}&=&f(t,x(t)),~~~~~t\in (0,T)\\ x(\tau)&=&-x(1-\tau),~~~~~~~\tau \in (0,T) \end{eqnarray*} has the at lease one solution \(x\in AC[0,T]\) \[ x(t)=-\int_{0}^{\tau}f(s,x(s))ds-\int_{0}^{1-\tau}f(s,x(s))ds+\int_{0}^{t}f(s,x(s))ds. \] Now, let \(\tau=\frac{1}{2},\) then \begin{eqnarray*} x(t)&=&-\int_{0}^{\frac{1}{2}}f(s,x(s))ds-\int_{0}^{\frac{1}{2}}f(s,x(s))ds+\int_{0}^{t}f(s,x(s))ds\\ &=&\int_{0}^{t}f(s,x(s))ds-2\int_{0}^{\frac{1}{2}}f(s,x(s))ds. \end{eqnarray*}
Example 1. Consider the differential Equation (1) with the backward condition \(x(T)=x_0, \) we have the following corollary;
Corollary 7. Let \(\alpha=1\), \(\beta=0\) and \(\eta=T\) in Theorem 1, then the backward problem \[ \frac{dx}{dt}=f(t,x(t)) \] \[ x(T)=x_0 \] has the solution \(x\in AC[0,T]\) \begin{eqnarray*} x(t)&=&x_0-\int_{0}^{T}f(s,x(s))ds+\int_{0}^{t}f(s,x(s))ds\\ &=&x_0-\int_{t}^{T}f(s,x(s))ds. \end{eqnarray*}
Example 2. Consider the differential Equation (1) with the forward condition \( x(0)=x_0,\) we have the following corollary;
Corollary 8. Let \(\alpha=0\), \(\beta=1\) and \(\tau=0\) in Theorem 1, then the initial value problem \[ \frac{dx}{dt}=f(t,x(t)) \] \[ x(0)=x_0 \] has the solution \(x\in AC[0,T]\) \[ x(t)=x_0+\int_{0}^{t}f(s,x(s))ds \]