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Chapter 1: Introduction

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    214044
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    Suppose that \(\mathbb{T}\) is a time scale with forward jump operator and delta differentiation operator \(\sigma\) and \(\Delta\), respectively. Let \(t_0\in \mathbb{T}\) and \(I\subset \mathbb{T}\).

    Definition:

    A dynamic equation (DE) is a relation that contains one independent variable \(t\in \mathbb{T}\), the dependent variable \(y\) and some of its delta derivatives \(y^\Delta, y^{\Delta^2}, \ldots, y^{\Delta^n}\).

    Example

    The equation
    \[
    y^\Delta +\sin_1(t, t_0)yy^\Delta +y^2=0
    \notag\]
    is a DE.

    Example

    The equation
    \[
    y^{\Delta^2}+e_1(t, t_0)y^\sigma+\frac{1}{1+\left(\cosh_3(t, t_0)\right)^4}y^\Delta+y^3=\frac{1}{1+t^2},\quad t\in \mathbb{T},
    \notag\]
    is a DE provided that \(1-9\mu^2\ne 0\).

    Example

    The equation
    \[
    (1+\cosh_1(t, t_0))^3y^{\Delta^4}(y^\sigma)^3-y^2=\frac{1}{1+10(\sin_1(t, t_0))^8},\quad t\in \mathbb{T},
    \notag\]
    is a DE provided that \(1-\mu^2\ne 0\).

    Definition

    The order of a dynamic equation is defined as the order of the highest delta derivative in the equation.

    Example

    The equation
    \[
    yy^{\Delta^3}-3y^2+\frac{1}{1+(e_1(t, t_0))^4}y^{\Delta^2}y^\sigma= \frac{1}{1+t^8},\quad t\in \mathbb{T},
    \notag\]
    is a third order dynamic equation provided that \(1\in \mathcal{R}\).

    Example

    The equation
    \[
    y^{\Delta^4}-y^\sigma y^{\Delta^2}+y^\Delta -\sin_1(t, t_0)yy^\sigma=\frac{1}{1+t^2},\quad t\in \mathbb{T},
    \notag\]
    is a fourth order dynamic equation.

    Example

    The equation
    \[
    y^{\Delta^2}-t^3 y^\sigma +\sin_5(t, t_0)y^\Delta=y^2,\quad t\in \mathbb{T},
    \notag\]
    is a second order dynamic equation.

    Exercise

    Determine the order of the following DEs.

    1. \(\left(y^\Delta\right)^2-e_2(t, t_0)y=0,\quad 2\in\mathcal{R},\quad t\in \mathbb{T},\)

    2. \( y^{\Delta^3}-e_4(t, t_0)y^{\Delta^2}=\sin_3(t, t_0),\quad 4\in \mathcal{R},\quad t\in \mathbb{T},\)

    3. \(\left(y^{\Delta^2}\right)^3=(\sin_{-1}(t, t_0))^2,\quad t\in \mathbb{T}\),

    4. \(y^{\Delta^5}-e_3(t, t_0) y^{\Delta^2}=\cos_1(t, t_0),\quad t\in \mathbb{T},\)

    5. \(y^{\Delta^{11}}-e_2(t, t_0)y^\Delta-\cos_3(t, t_0) y=\tanh_4(t, t_0),\quad 2\in \mathcal{R},\quad 1-16\mu^2\ne 0,\quad t\in \mathbb{T}.\)

    In general, an \(n\)th order DE can be written in the form
    \begin{equation}
    \label{1.3}
    F\left(t, y, y^\Delta, \ldots, y^{\Delta^n}\right)=0,
    \end{equation}
    where \(F\) is a given function.

    Definition

    A functional relation between the real dependent variable \(y\) and the time scale independent variable \(t\) in some interval \(I\subset \mathbb{T}\) that satisfies the given DE is said to be a solution of the DE. The general solution of an \(n\)th order DE depends on \(n\) constants \(c_1, c_2, \ldots, c_n\) and the independent time scale variable \(t\). Sometimes, the solutions of a DE are said to be delta integral curves for the corresponding DE.

    Obviously, the general solution of an \(n\)th order DE is written in the following form
    \[
    \phi(t, y, c_1, c_2, \ldots, c_n)=0.
    \notag\]

    Example

    Consider the equation
    \[
    y^\Delta=y,\quad t\in \mathbb{T}.
    \notag\]
    We will show that
    \[
    y(t)=e_1(t, t_0),\quad t\in \mathbb{T},
    \notag\]
    is its solution. Really, we have
    \[
    \begin{aligned}
    y^\Delta(t)=& e_1(t, t_0)\\ \\
    =& y(t),\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]

    Example

    Let \(\mathbb{T}=2^{\mathbb{N}_0}\). Consider the equation
    \[
    y^{\Delta^2}+\frac{3}{1+4t}y^\Delta= \frac{3}{1+4t}\sin_1(t, t_0)+\cos_1(t, t_0),\quad t\in \mathbb{T}.
    \notag\]
    We will prove that the function
    \[
    y(t)=\frac{1}{1+t}-\cos_1(t, t_0),\quad t\in \mathbb{T},
    \notag\]
    is its solution. Here
    \[
    \sigma(t)=2t,\quad t\in \mathbb{T}.
    \notag\]
    Then
    \[
    \begin{aligned}
    y^\Delta(t)=& -\frac{1}{(1+t)(1+\sigma(t))}+\sin_1(t, t_0)\\ \\
    =& -\frac{1}{(1+t)(1+2t)}+\sin_1(t, t_0),\\ \\
    y^{\Delta^2}(t)=& \frac{2(\sigma(t)+t)+3}{(1+t)(1+2t)(1+\sigma(t))(1+2\sigma(t))}+\cos_1(t, t_0)\\ \\
    =& \frac{2(2t+t)+3}{(1+t)(1+2t)(1+2t)(1+4t)}+\cos_1(t, t_0)\\ \\
    =& \frac{3(2t+1)}{(1+t)(1+2t)^2(1+4t)}+\cos_1(t, t_0)\\ \\
    =& \frac{3}{(1+t)(1+2t)(1+4t)}+\cos_1(t, t_0),\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]
    Hence, we arrive at
    \[
    y^{\Delta^2}(t)+\frac{3}{1+4t}y^\Delta(t)
    \]
    \[
    \begin{aligned}
    =& \frac{3}{(1+t)(1+2t)(1+4t)}+\cos_1(t, t_0)+\frac{3}{1+4t}\left(-\frac{1}{(1+t)(1+2t)}+\sin_1(t, t_0)\right)\\ \\
    =& \frac{3}{(1+t)(1+2t)(1+4t)}+\cos_1(t, t_0)-\frac{3}{(1+t)(1+2t)(1+4t)}+\frac{3}{1+4t}\sin_1(t, t_0)\\ \\
    =& \frac{3}{1+4t}\sin_1(t, t_0)+\cos_1(t, t_0),\quad t\in \mathbb{T}.
    \end{aligned}
    \]

    Example

    Let \(\mathbb{T}=3^{\mathbb{N}_0}\). Consider the equation
    \[
    y^{\Delta^3}+t^2y^{\Delta^2}-4ty^\Delta -\frac{1}{t^2}y= 48t^2-21t+56-\frac{5}{t}+\frac{6}{t^2},\quad t\in \mathbb{T}.
    \notag\]
    We will prove that the function
    \[
    y(t)=t^3-4t^2+5t+6,\quad t\in \mathbb{T},
    \notag\]
    is its solution. Here
    \[
    \sigma(t)=3t,\quad t\in \mathbb{T}.
    \notag\]
    Then
    \[
    \begin{aligned}
    y^\Delta(t)=& (\sigma(t))^2+t\sigma(t)+t^2-4(\sigma(t)+t)+5\\ \\
    =& (3t)^2+t(3t)+t^2-4(3t+t)+5\\ \\
    =& 9t^2+3t^2+t^2-16t+5\\ \\
    =& 13t^2-16t+5,\\ \\
    y^{\Delta^2}(t)=& 13(\sigma(t)+t)-16\\ \\
    =& 13(3t+t)-16\\ \\
    =& 52t-16,\\ \\
    y^{\Delta^3}(t)=& 52,\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]
    Hence,
    \[
    y^{\Delta^3}(t)+t^2 y^{\Delta^2}(t)-4ty^\Delta(t)-\frac{1}{t^2}y(t)
    \notag\]
    \[
    \begin{aligned}
    =& 52+t^2(52t-16)-4t(13t^2-16t+5)-\frac{1}{t^2}(t^3-4t^2+5t+6)\\ \\
    =& 52+52t^3-16t^2-52t^3+64t^2-20t-t+4-\frac{5}{t}+\frac{6}{t^2}\\ \\
    =& 48t^2-21t+56-\frac{5}{t}+\frac{6}{t^2},\quad t\in \mathbb{T}.
    \end{aligned}
    \]

    Exercise

    Let \(\mathbb{T}=2\mathbb{Z}\). Prove that
    \[
    y(t)=t^4-3t^2,\quad t\in \mathbb{T},
    \notag\]
    is a solution to the equation
    \[
    y^{\Delta^4}+ty^{\Delta^3}-2y^{\Delta^2}+ty^\Delta -4y= 12t^3+22t^2-6t-76,\quad t\in \mathbb{T}.
    \notag\]

    Definition

    If a solution can be obtained for some values of the constants \(c_1, c_2, \ldots, c_n\), then it is called a particular solution. Otherwise, it is said to be a singular solution.

    Example

    Consider the equation
    \[
    y^{\Delta^2}-3y^\Delta+2y=0,\quad t\in \mathbb{T}.
    \notag\]
    Note that
    \[
    y(t)=c_1 e_1(t, t_0)+c_2e_2(t, t_0),\quad t\in \mathbb{T},\quad c_1, c_2\in \mathbb{R},
    \notag\]
    is a general solution of the considered equation. Really, we have
    \[
    \begin{aligned}
    y^\Delta(t)=& c_1 e_1(t, t_2)+2c_2 e_2(t, t_0),\\ \\
    y^{\Delta^2}(t)=& c_1 e_1(t, t_0)+4c_2 e_2(t, t_0),\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]
    Hence,
    \[
    y^{\Delta^2}(t)+3y^{\Delta}(t)+2y(t)
    \notag\]
    \[
    \begin{aligned}
    =& c_1e_1(t, t_0)+4c_2 e_2(t, t_0)-3(c_1e_1(t, t_0)+2c_2 e_2(t, t_0))+2(c_1e_1(t, t_0)+c_2e_2(t, t_0))\\ \\
    =& c_1e_1(t, t_0)+4c_2 e_2(t, t_0)-3c_1e_1(t, t_0)-6c_2 e_2(t, t_0)+2c_1e_1(t, t_0)+2c_2 e_2(t, t_0)\\ \\
    =& 0,\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]
    The functions
    \[
    y_1(t)=7e_1(t, t_0)\quad \mbox{and}\quad y_2(t)=-3e_2(t, t_0),\quad t\in \mathbb{T},
    \notag\]
    are particular solutions of the considered equation.

    Example

    Let \(\mathbb{T}=\mathbb{R}\). Consider the equation
    \[
    \left(y^\Delta\right)^2-ty^\Delta+y=0,\quad t\in \mathbb{T}.
    \notag\]
    Here
    \[
    \sigma(t)=2t,\quad t\in \mathbb{T}.
    \notag\]
    The general solution of the considered equation is given by
    \begin{equation}
    \label{1.2} y(t)=ct^2-c^2,\quad t\in \mathbb{T},\quad c\in \mathbb{R}.
    \end{equation}
    Really, we have
    \[
    y^\Delta(t)=c,\quad t\in \mathbb{T}.
    \notag\]
    Then
    \[
    \begin{aligned}
    \left(y^\Delta(t)\right)^2-ty^\Delta(t)+y(t)=& c^2-ct+ct-c^2\\ \\
    =& 0,\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]
    On the other hand, we will show that the function
    \begin{equation}
    \label{1.1} y(t)=\frac{2}{9}t^2,\quad t\in \mathbb{T},
    \end{equation}
    is a solution of the given equation. Indeed, we have
    \[
    \begin{aligned}
    y^\Delta(t)=& \frac{2}{9}(\sigma(t)+t)\\ \\
    =& \frac{2}{9}(2t+t)\\ \\
    =& \frac{2}{3}t,\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]
    Hence,
    \[
    \begin{aligned}
    \left(y^\Delta(t)\right)^2-ty^\Delta(t)+y(t)=& \left(\frac{2}{3}t\right)^2-t\left(\frac{2}{3}t\right)+\frac{2}{9}t^2\\ \\
    =& \frac{4}{9}t^2-\frac{2}{3}t^2+\frac{2}{9}t^2\\ \\
    =& \frac{4-6+2}{9}t^2\\ \\
    =& 0,\quad t\in \mathbb{T}.
    \end{aligned}
    \notag\]
    We can not get \eqref{1.1} by \eqref{1.2} for any constant \(c\in \mathbb{R}\). Therefore \eqref{1.1} is a singular solution of the considered equation.

    Exercise

    Let \(\mathbb{T}=\mathbb{R}\). Consider the equation
    \[
    ty+(t+1)y^\Delta=0,\quad t\in \mathbb{T}.
    \notag\]

    1. Prove that the general solution of the considered equation is given by
    \[
    y(t)=c(t+1) e_{-1}(t, 0),\quad t\in \mathbb{T},\quad c\in \mathbb{R}.
    \notag\]

    2. Prove that \(t=-1\) is a singular solution of the considered equation.

    Definition

    The function
    \[
    y=\phi(t),\quad t\in I,
    \notag\]
    is said to be an explicit solution of the equation \eqref{1.3} provided
    \begin{equation}
    \label{1.4}
    F\left(t, \phi(t), \phi^\Delta(t), \phi^{\Delta^2}(t), \ldots, \phi^{\Delta^n}(t)\right)=0,\quad t\in I.
    \end{equation}

    Definition

    A relation of the form
    \begin{equation}
    \label{1.5} \psi(t, y)=0,\quad t\in I,
    \end{equation}
    is said to be an implicit solution of the equation \eqref{1.3} provided that it determines one or more functions \(y=\phi(t)\), \(t\in I\), which satisfy \eqref{1.4}.

    Sometimes it is difficult to solve the equation \eqref{1.5}. In these cases, if \(y(\mathbb{T})=\widetilde{\mathbb{T}}\) is a time scale with forward jump operator and delta differentiation operator \(\widetilde{\sigma}\) and \(\widetilde{\Delta}\), respectively, and the function \(\psi\) is enough times completely delta differentiable, then
    \[
    \psi_{t_1}^\Delta(t, y(t))+\psi_{t_2}^{\widetilde{\Delta}}(\sigma(t), y(t))y^\Delta(t)=0,\quad t\in I.
    \notag\]
    From here, we find
    \[
    y^\Delta(t)=-\frac{\psi_{t_1}^\Delta(t, y(t))}{\psi_{t_2}^{\widetilde{\Delta}}(\sigma(t), y(t))},\quad t\in I,
    \notag\]
    provided that
    \[
    \psi_{t_2}^{\widetilde{\Delta}}(\sigma(t), y(t))\ne 0,\quad t\in I.
    \notag\]
    From here, we express \(y^{\Delta^2}(t)\), \(y^{\Delta^3}(t)\), \(\ldots\), \(y^{\Delta^n}(t)\), \(t\in I\). Then, we check if
    \[
    F\left(t, y(t), y^\Delta(t), \ldots, y^{\Delta^n}(t)\right)=0,\quad t\in I.
    \notag\]

    Definition

    A DE is said to be linear if it is linear in \(y\) and its delta derivatives. Otherwise, it is said to be nonlinear.

    Example

    The DE
    \[
    y^{\Delta^2}+2y^\Delta+\sin_3(t, t_0)y=1,\quad t\in \mathbb{T},
    \notag\]
    is a linear DE.

    Example

    The DE
    \[
    \left(y^\Delta\right)^3+y^{\Delta^3}+\cos_{-5}(t, t_0)=0,\quad t\in \mathbb{T},
    \notag\]
    is a nonlinear DE.

    Example

    The DE
    \[
    y^\Delta y^{\Delta^2}+\sinh_{-3}(t, t_0)y=0,\quad t\in I,
    \notag\]
    is a nonlinear DE provided that \(1-9\mu^2\ne 0\) on \(\mathbb{T}\).

    Exercise

    Classify each of the following DEs as linear or nonlinear.

    1. \( y^\Delta+\cos_1(t, t_0)y=\sin _3(t, t_0),\quad t\in \mathbb{T}\).

    2. \(\left(y^\Delta\right)^4-(e_4(t, t_0))^2=0,\quad t\in \mathbb{T}\).

    3. \( y^\Delta y^{\Delta^2}=t,\quad t\in \mathbb{T}\).

    The general form of \(n\)th order linear DEs is as follows.
    \begin{equation}
    \label{1.7} p_0(t)y^{\Delta^n}+p_1(t)y^{\Delta^{n-1}}+\cdots+ p_n(t)y=f(t),\quad t\in \mathbb{T},
    \end{equation}
    where \(p_j, f: \mathbb{T}\mathbb{T}o \mathbb{R}, j\in \{1, 2, \ldots, n\}\), are given functions.

    Definition

    The equation \eqref{1.7} is said to be homogeneous if \(f\equiv 0\) on \(\mathbb{T}\). Otherwise, the equation \eqref{1.7} is said to be nonhomogeneous.

    Example

    The DE
    \[
    y^{\Delta^2}+y^\Delta=\cos_1(t, t_0),\quad t\in \mathbb{T},
    \notag\]
    is nonhomogeneous.

    Example

    The DE
    \[
    y^{\Delta^3}+\sin_2(t, t_0)y^{\Delta^2}+y^\Delta=0,\quad t\in \mathbb{T},
    \notag\]
    is a homogeneous DE.

    Example

    The DE
    \[
    y^{\Delta^4}+y^\Delta=t^2,\quad t\in \mathbb{T},
    \notag\]
    is a nonhomogeneous DE.

    Exercise

    Classify each of the following DEs as homogeneous or nonhomogeneous.

    1. \( y^{\Delta}+t^2y=t,\quad t\in \mathbb{T}.\)

    2. \(y^{\Delta^4}+y=0,\quad t\in \mathbb{T}.\)

    3. \(y^{\Delta^2}-y^\Delta=1\), \(t\in \mathbb{T}\).

    Now, we assume that the equation \eqref{1.3} can be solved explicitly. More precisely, suppose
    \begin{equation}
    \label{1.8} y^{\Delta^n}=f\left(t, y, y^\Delta, \ldots, y^{\Delta^{n-1}}\right),\quad t\in \mathbb{T},
    \end{equation}
    where \(f: \mathbb{T}\mathbb{T}o \mathbb{R}\) is a given function.

    Definition

    By initial conditions for the DE \eqref{1.8} we mean conditions of the form
    \begin{equation}
    \label{1.9}
    \begin{array}{lll}
    y(t_0)=& y_0,\\ \\
    y^\Delta(t_0)=& y_1,\\ \\
    \vdots&\\ \\
    y^{\Delta^{n-1}}(t_0)=& y_{n-1},
    \end{array}
    \end{equation}
    where \(y_j\in \mathbb{R}, j\in \{0, 1, \ldots, n-1\}\), are given constants.

    Definition

    A problem consisting of the DE \eqref{1.8} together with the initial conditions \eqref{1.9} is said to be initial value problem (IVP).

    Definition

    Observe that the initial conditions \eqref{1.9} are prescribed at a point \(t_0\). In many problems, these conditions are prescribed into two or more points. These conditions are called boundary conditions A problem consisting of the DE \eqref{1.8} together with boundary conditions is said to be boundary value problem (BVP).

    Definition

    A problem consiting of the DE \eqref{1.8} together with the initial conditions \eqref{1.9} and boundary conditions is said to be initial boundary value problem (IBVP).

    Example

    Let \(\mathbb{T}=\mathbb{Z}\). The problem
    \[
    \begin{aligned}
    y^{\Delta^3}-3y=& 0,\quad t\in \mathbb{T},\\ \\
    y(0)=& 0,\\ \\
    y^\Delta(0)=& -1,\\ \\
    y^{\Delta^2}(0)=& 2,
    \end{aligned}
    \notag\]
    is an IVP.

    Example

    Let \(\mathbb{T}=2^{\mathbb{N}_0}\). The problem
    \[
    \begin{aligned}
    y^{\Delta^2}-4y^\Delta+ty=& t^2,\quad t\in \mathbb{T},\\ \\
    y(1)-3y^\Delta(32)=&0
    \end{aligned}
    \notag\]
    is a BVP.

    Example

    Let \(\mathbb{T}=4^{\mathbb{N}_0}\). The problem
    \[
    \begin{aligned}
    y^{\Delta^2}+ty^\Delta-t^2y=& \sin_1(t, 1),\quad t\in \mathbb{T},\\ \\
    y(1)=& 1,\\ \\
    y^\Delta(1)=& -3,\\ \\
    3y(1)-5y(256)=& 0,
    \end{aligned}
    \notag\]
    is an IBVP.

    The main questions for IVPs, BVPs and IBVPs are as follows.

    1. Existence of solutions.

    2. Uniqueness of solutions.

    3. Continuous dependence on initial conditions and parameters.

    4. The interval under which the solutions are defined.

    5. Extension of solutions.

    6. Upper and lower bounds of the unknown solutions and existence of minimal and maximal solutions.

    7. Periodicity of the solutions.

    8. Behaviour of the solutions.

    9. Stability of the solutions.

    10. Oscillations of solutions.


    This page titled Chapter 1: Introduction is shared under a GNU General Public License 3.0 license and was authored, remixed, and/or curated by Svetlin G. Georgiev.